Student Study Guide/Solutions Manual
to accompany Organic Chemistry, Fourth Edition
University of Pretoria
http://create.mcgraw-hill.com
Copyright 2014 by McGraw-Hill Education. All rights
reserved. Printed in the United States of America. Except as
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of this publication may be reproduced or distributed in any form
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ISBN-10: 1308115040
ISBN-13: 9781308115047
Contents
1. Structure and Bonding 1
2. Acids and Bases 35
3. Introduction to Organic Molecules and Functional Groups 61
4. Alkanes 81
5. Stereochemistry 117
6. Understanding Organic Reactions 145
7. Alkyl Halides and Nucleophilic Substitution 167
8. Alkyl Halides and Elimination Reactions 201
9. Alcohols, Ethers, and Epoxides 233
10. Alkenes 267
11. Alkynes 299
12. Oxidation and Reduction 323
13. Mass Spectrometry and Infrared Spectroscopy 353
14. Nuclear Magnetic Resonance Spectroscopy 371
15. Radical Reactions 397
16. Conjugation, Resonance, and Dienes 425
17. Benzene and Aromatic Compounds 453
18. Reactions of Aromatic Compounds 477
19. Carboxylic Acids and the Acidity of the O-H Bond 517
20. Introduction to Carbonyl Chemistry 539
21. Aldehydes and Ketones—Nucleophilic Addition 573
22. Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 609
23. Substitution Reactions of Carbonyl Compounds at the a Carbon 645
24. Carbonyl Condensation Reactions 675
25. Amines 707
26. Carbon-Carbon Bond-Forming Reactions in Organic Synthesis 741
27. Pericyclic Reactions 765
28. Carbohydrates 789
29. Amino Acids and Proteins 825
30. Lipids 861
31. Synthetic Polymers 877
iii
Credits
1. Structure and Bonding: Chapter 1 from Student Study Guide/Solutions Manual to accompany Organic
Chemistry, Fourth Edition by Smith, Berk, 2014 1
2. Acids and Bases: Chapter 2 from Student Study Guide/Solutions Manual to accompany Organic Chemistry,
Fourth Edition by Smith, Berk, 2014 35
3. Introduction to Organic Molecules and Functional Groups: Chapter 3 from Student Study Guide/Solutions
Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 61
4. Alkanes: Chapter 4 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth
Edition by Smith, Berk, 2014 81
5. Stereochemistry: Chapter 5 from Student Study Guide/Solutions Manual to accompany Organic Chemistry,
Fourth Edition by Smith, Berk, 2014 117
6. Understanding Organic Reactions: Chapter 6 from Student Study Guide/Solutions Manual to accompany
Organic Chemistry, Fourth Edition by Smith, Berk, 2014 145
7. Alkyl Halides and Nucleophilic Substitution: Chapter 7 from Student Study Guide/Solutions Manual to
accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 167
8. Alkyl Halides and Elimination Reactions: Chapter 8 from Student Study Guide/Solutions Manual to
accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 201
9. Alcohols, Ethers, and Epoxides: Chapter 9 from Student Study Guide/Solutions Manual to accompany
Organic Chemistry, Fourth Edition by Smith, Berk, 2014 233
10. Alkenes: Chapter 10 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth
Edition by Smith, Berk, 2014 267
11. Alkynes: Chapter 11 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth
Edition by Smith, Berk, 2014 299
12. Oxidation and Reduction: Chapter 12 from Student Study Guide/Solutions Manual to accompany Organic
Chemistry, Fourth Edition by Smith, Berk, 2014 323
13. Mass Spectrometry and Infrared Spectroscopy: Chapter 13 from Student Study Guide/Solutions Manual to
accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 353
14. Nuclear Magnetic Resonance Spectroscopy: Chapter 14 from Student Study Guide/Solutions Manual to
accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 371
15. Radical Reactions: Chapter 15 from Student Study Guide/Solutions Manual to accompany Organic
Chemistry, Fourth Edition by Smith, Berk, 2014 397
16. Conjugation, Resonance, and Dienes: Chapter 16 from Student Study Guide/Solutions Manual to
accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 425
17. Benzene and Aromatic Compounds: Chapter 17 from Student Study Guide/Solutions Manual to accompany
Organic Chemistry, Fourth Edition by Smith, Berk, 2014 453
18. Reactions of Aromatic Compounds: Chapter 18 from Student Study Guide/Solutions Manual to accompany
Organic Chemistry, Fourth Edition by Smith, Berk, 2014 477
iv
19. Carboxylic Acids and the Acidity of the O-H Bond: Chapter 19 from Student Study Guide/Solutions Manual
to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 517
20. Introduction to Carbonyl Chemistry: Chapter 20 from Student Study Guide/Solutions Manual to accompany
Organic Chemistry, Fourth Edition by Smith, Berk, 2014 539
21. Aldehydes and Ketones—Nucleophilic Addition: Chapter 21 from Student Study Guide/Solutions Manual to
accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 573
22. Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution: Chapter 22 from Student Study
Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 609
23. Substitution Reactions of Carbonyl Compounds at the a Carbon: Chapter 23 from Student Study
Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 645
24. Carbonyl Condensation Reactions: Chapter 24 from Student Study Guide/Solutions Manual to accompany
Organic Chemistry, Fourth Edition by Smith, Berk, 2014 675
25. Amines: Chapter 25 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth
Edition by Smith, Berk, 2014 707
26. Carbon-Carbon Bond-Forming Reactions in Organic Synthesis: Chapter 26 from Student Study
Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 741
27. Pericyclic Reactions: Chapter 27 from Student Study Guide/Solutions Manual to accompany Organic
Chemistry, Fourth Edition by Smith, Berk, 2014 765
28. Carbohydrates: Chapter 28 from Student Study Guide/Solutions Manual to accompany Organic Chemistry,
Fourth Edition by Smith, Berk, 2014 789
29. Amino Acids and Proteins: Chapter 29 from Student Study Guide/Solutions Manual to accompany Organic
Chemistry, Fourth Edition by Smith, Berk, 2014 825
30. Lipids: Chapter 30 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth
Edition by Smith, Berk, 2014 861
31. Synthetic Polymers: Chapter 31 from Student Study Guide/Solutions Manual to accompany Organic
Chemistry, Fourth Edition by Smith, Berk, 2014 877
v
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
1
Structure and Bonding 1–1
Chapter 1 Structure and Bonding
Chapter Review
Important facts
•
The general rule of bonding: Atoms strive to attain a complete outer shell of valence electrons
(Section 1.2). H “wants” 2 electrons. Second-row elements “want” 8 electrons.
nonbonded electron pair
H
C
N
O
X
Usual number of bonds
in neutral atoms
1
4
3
2
1
Number of nonbonded
electron pairs
0
0
1
2
3
X = F, Cl, Br, I
The sum (# of bonds + # of lone pairs) = 4 for all elements except H.
•
Formal charge (FC) is the difference between the number of valence electrons on an atom and
the number of electrons it “owns” (Section 1.3C). See Sample Problem 1.3 for a stepwise
example.
Definition:
Examples:
formal charge
C
• C shares 8 electrons.
• C "owns" 4 electrons.
• FC = 0
•
=
number of
valence electrons
–
number of electrons
an atom "owns"
C
• C shares 6 electrons.
• C "owns" 3 electrons.
• FC = +1
C
• C shares 6 electrons.
• C has 2 unshared electrons.
• C "owns" 5 electrons.
• FC = −1
Curved arrow notation shows the movement of an electron pair. The tail of the arrow always
begins at an electron pair, either in a bond or a lone pair. The head points to where the electron
pair “moves” (Section 1.6).
Move an electron pair to O.
O
H C N H
A
O
H C N H
B
Use this electron pair to form a double bond.
•
Electrostatic potential plots are color-coded maps of electron density, indicating electron rich
(red) and electron deficient (blue) regions (Section 1.12).
2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–2
The importance of Lewis structures (Sections 1.3–1.5)
A properly drawn Lewis structure shows the number of bonds and lone pairs present around each
atom in a molecule. In a valid Lewis structure, each H has two electrons, and each second-row
element has no more than eight. This is the first step needed to determine many properties of a
molecule.
Geometry
[linear, trigonal planar, or tetrahedral] (Section 1.7)
Hybridization
Lewis structure
Types of bonds
[sp, sp2, or sp3] (Section 1.9)
[single, double, or triple] (Sections 1.3, 1.10)
Resonance (Section 1.6)
The basic principles:
• Resonance occurs when a compound cannot be represented by a single Lewis structure.
• Two resonance structures differ only in the position of nonbonded electrons and π bonds.
• The resonance hybrid is the only accurate representation for a resonance-stabilized compound.
A hybrid is more stable than any single resonance structure because electron density is
delocalized.
O
O
O
CH3CH2 C
CH3CH2 C
O
delocalized charges
CH3CH2 C
O
δ−
O
δ−
delocalized π bonds
resonance structures
hybrid
The difference between resonance structures and isomers:
• Two isomers differ in the arrangement of both atoms and electrons.
• Resonance structures differ only in the arrangement of electrons.
O
O
CH3 C
O
CH3CH2 C
O CH3
isomers
CH3CH2 C
O H
O H
resonance structures
Geometry and hybridization
The number of groups around an atom determines both its geometry (Section 1.7) and hybridization
(Section 1.9).
Number of
groups
2
3
4
Geometry
linear
trigonal planar
tetrahedral
Bond angle (o)
180
120
109.5
Hybridization
sp
sp2
sp3
Examples
BeH2, HC≡CH
BF3, CH2=CH2
CH4, NH3, H2O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
3
Structure and Bonding 1–3
Drawing organic molecules (Section 1.8)
•
Shorthand methods are used to abbreviate the structure of organic molecules.
CH3 H
=
C
C
H
H
CH3
skeletal structure
•
CH3
CH3 C
CH3
(CH3)2CHCH2C(CH3)3
=
isooctane
condensed structure
A carbon bonded to four atoms is tetrahedral in shape. The best way to represent a tetrahedron is
to draw two bonds in the plane, one in front, and one behind.
Four equivalent drawings for CH4
H
H
H
HH
C
C
C
H
H
H
H
H
H
H
H
C
H
HH
Each drawing has two solid lines, one wedge, and one dashed line.
Bond length
•
Bond length decreases across a row and increases down a column of the periodic table
(Section 1.7A).
C H
>
N H
>
O H
H F
Increasing bond length
•
H Cl
<
<
H Br
Increasing bond length
Bond length decreases as the number of electrons between two nuclei increases (Section 1.11A).
CH3 CH3
<
CH2 CH2 < H C C H
Increasing bond length
•
Bond length increases as the percent s-character decreases (Section 1.11B).
Csp H
Csp2 H
Csp3 H
Increasing bond length
•
Bond length and bond strength are inversely related. Shorter bonds are stronger bonds (Section 1.11).
longest C–C bond
weakest bond
C C
C C
Increasing bond strength
C C
shortest C–C bond
strongest bond
4
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–4
•
Sigma (σ) bonds are generally stronger than π bonds (Section 1.10).
C C
1 strong σ bond
C C
1 stronger σ bond
1 weaker π bond
C C
1 stronger σ bond
2 weaker π bonds
Electronegativity and polarity (Sections 1.12, 1.13)
•
•
•
Electronegativity increases from left to right across a row and decreases down a column of the
periodic table.
A polar bond results when two atoms of different electronegativity are bonded together.
Whenever C or H is bonded to N, O, or any halogen, the bond is polar.
A polar molecule has either one polar bond, or two or more bond dipoles that reinforce.
Drawing Lewis structures: A shortcut
Chapter 1 devotes a great deal of time to drawing valid Lewis structures. For molecules with many
bonds, it may take quite awhile to find acceptable Lewis structures by using trial-and-error to place
electrons. Fortunately, a shortcut can be used to figure out how many bonds are present in a
molecule.
Shortcut on drawing Lewis structures—Determining the number of bonds:
[1] Count up the number of valence electrons.
[2] Calculate how many electrons are needed if there are no bonds between atoms and every
atom has a filled shell of valence electrons; that is, hydrogen gets two electrons, and secondrow elements get eight.
[3] Subtract the number obtained in Step [1] from the sum obtained in Step [2]. This difference
tells how many electrons must be shared to give every H two electrons and every secondrow element eight. Since there are two electrons per bond, dividing this difference by two
tells how many bonds are needed.
To draw the Lewis structure:
[1] Arrange the atoms as usual.
[2] Count up the number of valence electrons.
[3] Use the shortcut to determine how many bonds are present.
[4] Draw in the two-electron bonds to all the H’s first. Then, draw the remaining bonds between
other atoms making sure that no second-row element gets more than eight electrons and that
you use the total number of bonds determined previously.
[5] Finally, place unshared electron pairs on all atoms that do not have an octet of electrons, and
calculate formal charge. You should have now used all the valence electrons determined in
the first step.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
5
Structure and Bonding 1–5
Example: Draw all valid Lewis structures for CH3NCO using the shortcut procedure.
[1] Arrange the atoms.
•
H
H C N C O
In this case the arrangement of atoms is implied by the way the
structure is drawn.
H
[2] Count up the number of valence electrons.
3H's
2C's
1N
1O
x
x
x
x
1 electron per H
4 electrons per C
5 electrons per N
6 electrons per O
=
=
=
=
3 electrons
8 electrons
5 electrons
+ 6 electrons
22 electrons total
[3] Use the shortcut to figure out how many bonds are needed.
• Number of electrons needed if there were no bonds:
3 H's
x
4 second-row elements x
•
2 electrons per H
=
8 electrons per element =
6 electrons
+ 32 electrons
38 electrons needed if
there were no bonds
Number of electrons that must be shared:
38 electrons
– 22 electrons
16 electrons must be shared
•
Since every bond takes two electrons, 16/2 = 8 bonds are needed.
[4] Draw all possible Lewis structures.
• Draw the bonds to the H’s first (three bonds). Then add five more bonds. Arrange them
between the C’s, N, and O, making sure that no atom gets more than eight electrons. There
are three possible arrangements of bonds; that is, there are three resonance structures.
• Add additional electron pairs to give each atom an octet and check that all 22 electrons are used.
H
H C N C O
H
H
H C N C O
H
Bonds to H's added.
H
H C N C O
H
H
H C N C O
H
All bonds drawn in.
Three arrangements possible.
H
H C N C O
H
H
H C N C O
H
H
H C N C O
H
Electron pairs drawn in.
Every atom has an octet.
6
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–6
•
Calculate the formal charge on each atom.
H
H
H C N C O
H
H C N C O
H
+1
•
H
H C N C O
H
–1
–1
+1
You can evaluate the Lewis structures you have drawn. The middle structure is the best
resonance structure, since it has no charged atoms.
Note: This method works for compounds that contain second-row elements in which every element
gets an octet of electrons. It does NOT necessarily work for compounds with an atom that does not
have an octet (such as BF3), or compounds that have elements located in the third row and later in
the periodic table.
Practice Test on Chapter Review
1. a. Which compound(s) contain a labeled atom with a +1 formal charge? All lone pairs of electrons
have been drawn in.
1.
2.
N(CH3)2
O
CH3
3. CH3 C
CH3
4. Both (1) and (2) have labeled atoms with a +1 charge.
5. Compounds (1), (2), and (3) all contain labeled atoms with a +1 formal charge.
b. Which of the following compounds is a valid resonance structure for A?
OCH3
OCH3
1.
OCH3
2.
OCH3
3.
A
4. Both (1) and (2) are valid resonance structures for A.
5. Cations (1), (2), and (3) are all valid resonance structures for A.
c. Which species contains a labeled carbon atom that is sp2 hybridized?
1.
2.
C CH2
3.
4. Both (1) and (2) contain labeled sp2 hybridized atoms.
5. Species (1), (2), and (3) all contain labeled sp2 hybridized carbon atoms.
7
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Structure and Bonding 1–7
d. Which of the following compounds has a net dipole?
1. CH3CH2NHCH2CH3
2. CH3CH2CH2OH
3. FCH2CH2CH2F
2.
4. Compounds (1) and (2) both have net dipoles.
5. Compounds (1), (2), and (3) all have net dipoles.
Rank the labeled bonds in order of increasing bond length. Label the shortest bond as 1, the
longest bond as 4, and the bonds of intermediate length as 2 and 3.
C
B
A
D
HC
3.
C
Answer the following questions about compounds A–D.
[1]
NCH2CH3
C O
A
a.
b.
c.
d.
e.
B
[2]
D
C
What is the hybridization of the labeled atom in A?
What is the molecular shape around the labeled atom in B?
In what type of orbital does the lone pair in C reside?
What orbitals are used to form bond [1] in D?
Which orbitals are used to form the carbon–oxygen double bond [2] in D?
4. Draw an acceptable Lewis structure for CH3NO3. Assume that the atoms are arranged as drawn.
H
O
H C O N O
H
5. Follow the curved arrows and draw the product with all the needed charges and lone pairs.
O
H
H
O
O
H
Answers to Practice Test
1. a. 4
b. 1
c. 1
d. 5
2. A – 1
B–4
C–2
D–3
3. a. sp3
b. trigonal planar
c. sp2
d. Csp3–Csp2
e. Csp–Osp2,
Cp–Op
4. One possibility:
5.
H
O
H
O
H C O N O
H
CH3
H
C
O
CH3
O
H
8
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–8
Answers to Problems
1.1
The mass number is the number of protons and neutrons. The atomic number is the number
of protons and is the same for all isotopes.
Nitrogen-14
Nitrogen-13
a. number of protons = atomic number for N = 7
7
7
b. number of neutrons = mass number – atomic number
7
6
c. number of electrons = number of protons
7
7
d. The group number is the same for all isotopes.
5A
5A
1.2
Ionic bonds form when an element on the far left side of the periodic table transfers an
electron to an element on the far right side of the periodic table. Covalent bonds result when
two atoms share electrons.
a.
F
F
covalent
1.3
1.4
b. Li+ Br
ionic
H H
c. H C C H
d. Na+ N H
H
H H
All C–H and C–C
bonds are covalent.
ionic
Both N–H bonds
are covalent.
Atoms with one, two, three, or four valence electrons form one, two, three, or four bonds,
respectively. Atoms with five or more valence electrons form [8 – (number of valence electrons)]
bonds.
a. O 8 − 6 valence e− = 2 bonds
c. Br 8 − 7 valence e− = 1 bond
b. Al 3 valence e− = 3 bonds
d. Si 4 valence e− = 4 bonds
[1] Arrange the atoms with the H’s on the periphery.
[2] Count the valence electrons.
[3] Arrange the electrons around the atoms. Give the H’s 2 electrons first, and then fill the
octets of the other atoms.
[4] Assign formal charges (Section 1.3C).
a.
[1]
H H
H C C H
H H
b.
[1]
H
H C N H
H H
[2] Count valence e−.
2C x 4 e− = 8
6H x 1 e− = 6
total e− = 14
[2] Count valence e−.
1C x 4 e− = 4
5H x 1 e− = 5
1N x 5 e− = 5
total e− = 14
[3]
H H
H C C H
H H
[3]
All 14 e− used.
All second-row elements
have an octet.
H
H
H C N H
H C N H
H H
H H
e−
12 used.
N needs 2 more
electrons for an octet.
9
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Structure and Bonding 1–9
c.
[1]
[2] Count valence e−.
=4
1C x 4 e −
3H x 1 e −
=3
negative charge = 1
total e−
=8
H
H C
H
d.
[1]
[3]
H
H
H C
H C
H
H
e−
6 used.
C needs 2 more
electrons for an octet.
[The –1 charge on C is
explained in Section 1.3C.]
H
H
[2] Count valence e−.
[3]
1C x 4 e −
= 4
H C Cl
H C Cl
3H x 1 e −
= 3
H
H
1Cl x 7 e–
= 7
−
8 e used.
Complete octet.
total e−
= 14
Cl needs 6 more
electrons for an octet.
H
H C Cl
H
1.5 Follow the directions from Answer 1.4.
a. HCN
Count valence e−.
1C x 4 e− = 4
1H x 1 e− = 1
1N x 5 e− = 5
total e− = 10
H C N
b. H2CO
Count valence e−.
1C x 4 e− = 4
2H x 1 e− = 2
1O x 6 e− = 6
total e−
= 12
H C O
H
H O
Count valence e−.
H O C C O H 2C x 4 e− = 8
H
4H x 1 e− = 4
c. HOCH2CO2H
H C N
4 e− used.
H C O
H
H
6 e− used.
Complete O and C octets.
H O
H O C C O H
H
H O
H O C C O H
H
16 e− used.
Complete octets.
Formal charge (FC) = number of valence electrons – [number of unshared electrons +
(½)(number of shared electrons)]
H
a.
Complete N and C octets.
H C O
3O x 6 e− = 18
total e− = 30
1.6
H C N
6 − [2 + 1/2(6)] = +1
5 − [0 + 1/2(8)] = +1
+
b.
H N H
c.
CH3 N C
O O O
H
5 − [0 + (1/2)(8)] = +1
4 − [0 + (1/2)(8)] = 0
4 − [2 + (1/2)(6)] = −1
6 − [4 + (1/2)(4)] = 0
6 − [6 + (1/2)(2)] = −1
1.7
H
a. CH3O
[1] H C O
H
H
[2] Count valence e−.
[3] H C O
1C x 4 e− = 4
H
3H x 1 e− = 3
− used.
8
e
1O x 6 e− = 6
total e− = 13
Add 1 for (−) charge = 14
H
H C O
H
H
[4] H C O
H
Assign charge.
10
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–10
[1] H C C
b. HC2
c. (CH3NH3)
[1] H H
H C N H
H H
d. (CH3NH)–
[1] H
H C N H
H
[2] Count valence e−.
[3] H C C
2C x 4 e− = 8
1H x 1 e− = 1
4 e− used.
total e− = 9
Add 1 for (−) charge = 10
H C C
[4] H C C
Assign charge.
[2] Count valence e−.
[3]
H H
1C x 4 e− = 4
H C N H
6H x 1 e− = 6
H H
1N x 5 e− = 5
14 e− used.
−
total e
= 15
Subtract 1 for (+) charge = 14
[4]
[2] Count valence e−.
[3]
H
1C x 4 e− = 4
H C N H
4H x 1 e− = 4
H
1N x 5 e− = 5
10 e− used.
−
total e
= 13
Add 1 for (−) charge = 14
[4]
H H
H C N H
H H
Assign charge.
H
H C N H
H
Complete octet and
assign charge.
1.8
a. C2H4Cl2 (two isomers)
Count valence e−.
2C x 4 e− = 8
4H x 1 e− = 4
2Cl x 7 e− = 14
total e− = 26
H
H C
H
H
H
C Cl
H C
Cl
Cl
H
C Cl
H
H
b. C3H8O (three isomers)
Count valence e−.
3C x 4 e− = 12
8H x 1 e− = 8
1O x 6 e− = 6
total e−
= 26
c. C3H6 (two isomers)
Count valence e−.
3C x 4 e− =12
6H x 1 e− = 6
total e− = 18
1.9
H H H
H O H
H C C C O H
H H H
H
H H
H C C C H
H H H
H
H C C O C H
H H
H
H H
C
H C C H
H
H
H
H C C C
H H H
Two different definitions:
• Isomers have the same molecular formula and a different arrangement of atoms.
• Resonance structures have the same molecular formula and the same arrangement of atoms.
2 lone pairs
N in the middle
N at the end
O
a.
N C O
and
C N O
b.
different arrangement of atoms = isomers
HO C O
3 lone pairs
O
and
HO C O
same arrangement of atoms =
resonance structures
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
11
Structure and Bonding 1–11
1.10 Isomers have the same molecular formula and a different arrangement of atoms.
Resonance structures have the same molecular formula and the same arrangement of atoms.
2 lone pairs 3 lone pairs
CH3 bonded to C=O H bonded to C=O
a.
O
O
CH3 C OH
CH3 C OH
A
B
O
c.
b.
O
O
d. CH3 C OH
CH3 C OH
H
A
D
different arrangement of atoms = isomers
O
CH3 C OH
O
H C CH2OH
A
same arrangement of atoms =
resonance structures
C2H4O2
(C2H5O2)–
O
CH3 C OH
H C CH2OH
B
D
C
different arrangement of atoms = isomers
different molecular formulas = neither
1.11 Curved arrow notation shows the movement of an electron pair. The tail begins at an electron
pair (a bond or a lone pair) and the head points to where the electron pair moves.
a.
H C O
H C O
H
H
b.
CH3 C C CH2
CH3 C C CH2
H H
H H
The net charge is the same
in both resonance structures.
The net charge is the same
in both resonance structures.
1.12 Compare the resonance structures to see what electrons have “moved.” Use one curved arrow
to show the movement of each electron pair.
a.
CH2 C
H
C CH3
CH2 C
H
H
C CH3
b.
H
One electron pair moves:
one curved arrow.
O C O
O C O
O
O
Two electron pairs move:
two curved arrows.
1.13 To draw another resonance structure, move electrons only in multiple bonds and lone pairs
and keep the number of unpaired electrons constant.
a.
CH3 C C
H H
b.
C CH3
CH3 C C
H
H H
CH3 C CH3
CH3 C CH3
Cl
Cl
C CH3
H
c.
H C C Cl
H H
H C C Cl
H H
12
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–12
1.14 A “better” resonance structure is one that has more bonds and fewer charges. The better
structure is the major contributor and all others are minor contributors. To draw the resonance
hybrid, use dashed lines for bonds that are in only one resonance structure, and use partial
charges when the charge is on different atoms in the resonance structures.
a.
CH3 C N CH3
CH3 C N CH3
H H
H H
hybrid:
All atoms have octets.
one more bond
major contributor
δ+ δ+
CH3 C N CH3
b.
CH2
C
CH2
C
CH2
H
These two resonance structures are equivalent.
They both have one charge and the same number
of bonds. They are equal contributors to the hybrid.
hybrid:
−
−
δ
H H
CH2
H
CH2
C
δ
CH2
H
1.15 Draw a second resonance structure for nitrous acid.
δ+
H O N O
H O N O
major contributor
fewer charges
minor contributor
H O
δ–
N
O
hybrid
1.16 All representations have a carbon with two bonds in the plane of the page, one in front of the
page (solid wedge) and one behind the page (dashed line). Four possibilities:
H
C
H
H
Cl
Cl
Cl
Cl
Cl Cl
C
C
C
H
H
Cl
Cl
H
HH
1.17 To predict the geometry around an atom, count the number of groups (atoms + lone pairs),
making sure to draw in any needed lone pairs or hydrogens: 2 groups = linear, 3 groups =
trigonal planar, 4 groups = tetrahedral.
N has 2 atoms + 2 lone pairs
4 groups = tetrahedral (or bent
molecular shape)
3 groups = trigonal planar
4 groups = tetrahedral
4 groups = tetrahedral
O
CH3 C CH3
a.
c.
NH2
3 groups = trigonal planar
4 groups = tetrahedral
b.
4 groups = tetrahedral
4 groups = tetrahedral
CH3 O CH3
4 groups = tetrahedral (or bent molecular shape)
d.
2 groups = linear
CH3 C N
2 groups = linear
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
13
Structure and Bonding 1–13
1.18 To predict the bond angle around an atom, count the number of groups (atoms + lone pairs),
making sure to draw in any needed lone pairs or hydrogens: 2 groups = 180°, 3 groups = 120°,
4 groups = 109.5°.
This C has 3 groups, so
both angles are 120°.
2 groups = 180°
H
a. CH3 C C Cl
H
b. CH2 C Cl
c. CH3 C Cl
H
This C has 4 groups, so
both angles are 109.5°.
2 groups = 180°
1.19 To predict the geometry around an atom, use the rules in Answer 1.17.
3 groups
H
H trigonal planar H
C
HO C C
C C
C C C C C C
H
H
H
H
4 groups
tetrahedral (or bent molecular shape)
H
4 groups
H H O H H tetrahedral
H
C C C C C CH3
H H H H H
3 groups
trigonal planar
4 groups
2 groups
tetrahedral
linear
(or bent molecular shape)
enanthotoxin
1.20 Reading from left to right, draw the molecule as a Lewis structure. Always check that carbon
has four bonds and all heteroatoms have an octet by adding any needed lone pairs.
(CH3)2CHCH(CH2CH3)2
(CH3)3CCH(OH)CH2CH3
H
H C H
H H H H H
a. H C C C C C C H
H H H H H
H C H
(CH3)2CHCHO
H C H
H
H
H
H C H
HH C H
H C H
b. H C
C H
c. H C
H
C
H C H
O H H
C C C H
HH C H
H C H
H
CH3(CH2)4CH(CH3)2
H
H H H
H
H H
d. H C C
O
C H
H
H C H
H
H C H
double bond
needed to
give C an octet
H
1.21 Simplify each condensed structure using parentheses.
CH2CH3
a.
CH3CH2CH2CH2CH2Cl
b. CH3CH2CH2 C CH2CH3
H
CH3(CH2)4Cl
CH3(CH2)2CH(CH2CH3)2
CH2OH
CH3
CH3
c. HOCH2 C CH2CH2CH2 C CH2 C CH3
H
CH3
CH3
(HOCH2)2CH(CH2)3C(CH3)2CH2C(CH3)3
14
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–14
1.22 Draw the Lewis structure of lactic acid.
H
H O
H C C
CH3CH(OH)CO2H
O
C O H
H H
1.23 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms.
The carbons are all tetravalent.
1H
1H
1H
O
O
O
3 H's
0 H's
O
a.
b.
0 H's
O
0 H's
O
octinoxate
(2-ethylhexyl 4-methoxycinnamate)
C18H26O3
3 H's
avobenzone
C20H22O3
1.24 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms.
Convert by writing in all carbons, and then adding hydrogen atoms to make the carbons
tetravalent.
H
CH3
H C C
b.
CH3
H C C H
a.
H H
H
H
O
H C
C H
H
H C
C
C
H
H
H H
CH3 H
c.
CH3
C
C
C
H
H
HH
H H
C
Cl
C
C
H H H
d.
CH3
N
C
C
N
CH3
CH3 H H
1.25 A charge on a carbon atom takes the place of one hydrogen atom. A negatively charged C
has one lone pair, and a positively charged C has none.
H
positive charge
no lone pairs
one H needed
negative charge
one lone pair
one H needed
positive charge
no lone pairs
no H's needed
d.
c.
b.
a.
O
H
H
negative charge
one lone pair
one H needed
1.26 Draw each indicated structure. Recall that in the skeletal drawings, a carbon atom is located at
the intersection of any two lines and at the end of any line.
O
a. (CH3)2C CH(CH2)4CH3 =
c.
N
= (CH3)2CH(CH2)2CONHCH3
H
H
b.
H
CH3 C
C CH2CH2Cl
H2N C
H
C H
H
O
Cl
=
H2N
d.
HO
O
= HO(CH2)2CH=CHCO2CH(CH3)2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
15
Structure and Bonding 1–15
1.27 To determine the orbitals used in bonding, count the number of groups (atoms + lone pairs):
4 groups = sp3, 3 groups = sp2, 2 groups = sp, H atom = 1s (no hybridization).
All covalent single bonds are σ, and all double bonds contain one σ and one π bond.
Each H uses a
1s orbital.
H H H
H C C C H
All single bonds are
σ bonds.
Each C–C bond is Csp3–Csp3.
Each C–H bond is Csp3–H1s.
H H H
Total of 10 σ bonds.
Each C has 4 groups and is
sp3 hybridized.
1.28 [1] Draw a valid Lewis structure for each molecule.
[2] Count the number of groups around each atom: 4 groups = sp3, 3 groups = sp2, 2
groups = sp, H atom = 1s (no hybridization).
Note: Be and B (Groups 2A and 3A) do not have enough valence e– to form an octet, and do
not form an octet in neutral molecules.
H
a. [1] H C Be
H
[2] Count groups around each atom:
H
H
H C Be
Be has 2 bonds.
H
[3] All C–H bonds: Csp3–H1s
C–Be bond: Csp3–Besp
Be–H bond: Besp–H1s
H
4 groups
sp3
2 groups
sp
CH3
b. [1] CH3 B CH3
[2] Count groups around each atom:
CH3
CH3 B CH3
B forms 3 bonds.
4 groups
sp3
H
3 groups
sp2
H
c. [1] H C O C H
H
[3] All C–H bonds: Csp3–H1s
C–B bonds: Csp3–Bsp2
[2] Count groups around each atom:
H
H
H
[3] All C–H bonds: Csp3–H1s
C–O bonds: Csp3–Osp3
H C O C H
4 groups
sp3
H
H
4 groups
sp3
1.29 To determine the hybridization, count the number of groups around each atom: 4 groups = sp3,
3 groups = sp2, 2 groups = sp, H atom = 1s (no hybridization).
a.
CH3 C CH
4 groups
sp3
2 groups
sp
b.
N CH3
3 groups 3 groups
sp2
sp2
c. CH2 C CH2
3 groups
sp2
2 groups
sp
16
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–16
1.30 All single bonds are σ. Multiple bonds contain one σ bond, and all others are π bonds.
All C−H bonds are σ bonds.
O
a.
CH3
C
H
σ bond
one σ bond,
one π bond
σ bond
b. CH3 C N
c.
C
σ bond
one σ + two π bonds
σ bond σ bond
H
O
one σ bond, one π bond
O CH3
σ bond
1.31 Single bonds are weaker and longer than double bonds, which are weaker and longer than
triple bonds. Increasing percent s-character increases bond strength and decreases bond length.
a.
C C
H
H
double bond
c.
H C OH
or
C O
H
Csp3–H1s
25% s-character
H
Csp2–H1s
33% s-character
shorter bond
triple bond
The triple bond is shorter than the
double bond.
single bond
b.
CH3
double bond
H
N
d.
N
CH2 N H
or
H
CH3 N H
Nsp2–H1s
33% s-character
shorter bond
C N
The C=N bond is shorter than the
C–N bond.
Nsp3–H1s
25% s-character
1.32 Electronegativity increases from left to right across a row of the periodic table and
decreases down a column. Look at the relative position of the atoms to determine their
relative electronegativity.
most electropositive
most electronegative
a.
Se < S < O
most electropositive
most electronegative
Na < P < Cl
b.
increasing
electronegativity
most electropositive
most electronegative
c.
increasing
electronegativity
S < Cl < F
increasing
electronegativity
most electropositive
most electronegative
d.
P<N<O
increasing
electronegativity
1.33 Dipoles result from unequal sharing of electrons in covalent bonds. More electronegative
atoms “pull” electron density towards them, making a dipole. Dipole arrows point towards
the atom of higher electron density.
δ+ δ−
a. H F
b.
δ+
δ−
B C
c.
δ−
δ+
C Li
d.
δ+
δ−
C Cl
1.34 Polar molecules result from a net dipole. To determine polarity, draw the molecule in three
dimensions around any polar bonds, draw in the dipoles, and look to see whether the dipoles
cancel or reinforce.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
17
Structure and Bonding 1–17
Electronegative
atom pulls e− density.
δ−
δ−
c.
Br
δ+C
a.
H
net dipole
C
δ− F
H
H
δ+
F
no resulting dipole =
nonpolar molecule
F δ−
F−
Four polar bonds cancel.
δ
All C–H bonds have no dipole.
one polar bond
net dipole = polar molecule
δ+
H
b.
re-draw
Br C Br
H
δ−
C
resulting dipole =
polar molecule
Br
δ− Br
resulting dipole =
polar molecule
d.
H H
δ−
Note: You must draw the molecule in three
dimensions to observe the net dipole. In the
Lewis structure, it appears the dipoles would
cancel out, when in fact they add to make a
polar molecule.
−
Cl δ
+
Cδ
H
Cl
δ+ C
H
δ− Cl
H
δ+
δ+
C C
e.
H
no resulting dipole =
nonpolar molecule
Cl δ− Two polar bonds are
equal and opposite
and cancel.
1.35
a. The two circled C’s are sp3 hybridized.
b. All the C–H bonds are nonpolar. All H’s bonded to O and N bear a partial positive charge (δ+).
δ+ H O
c.
H
H H
C C
δ+ H O C
C C C
C C
H
H
H
H H
C C
C
H O C
+
H N H+ O H δ
H
H O
O
δ
δ+
C C C
H N H O H
C C
H
O
C
H
H
one possibility
1.36
a.
O
OH
HO
H2NHN
OH
skeletal structure
b. Circled carbons are sp3 hybridized. All others are sp2 hybridized.
c. Each N is surrounded by three atoms and a lone pair, making it sp3 hybridized and trigonal
pyramidal in molecular shape.
d.
O
O H
H
O
H N N
H
11 polar bonds shown in bold
H
O H
18
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–18
1.37
O
a, b, c.
OH
O
O HO
H
O
C 6H 8O7
O
H
14 lone pairs, 2 lone pairs on each O
d. Each C that is part of a C=O is sp2 hybridized, so there are three sp2 C’s.
e. Orbitals:
[1] C=O, Csp2–Osp2 and Cp–Op
[2] C–C, Csp3–Csp2
[3] O–H, Osp3–H1s
[4] C–O, Csp3–Osp3
1.38
OH
a, b, c.
C11H14O3
OCH3
6 lone pairs, 2 lone pairs on each O
O
d. The sp2 hybridized C’s (seven) are labeled with circles.
e. Orbitals:
[1] C–C, Csp3–Csp2
[2] C–C, Csp3–Csp3
[3] C–H, Csp3–H1s
[4] C–H, Csp2– H1s
1.39 Use the definitions in Answer 1.1.
Iodine-123
53
70
53
7A
a. number of protons = atomic number for I = 53
b. number of neutrons = mass number – atomic number
c. number of electrons = number of protons
d. The group number is the same for all isotopes.
Iodine-131
53
78
53
7A
1.40 Use bonding rules in Answer 1.2.
H
a. Na+ I–
b. Br Cl
c.
H
e. Na+
d. H C N H
H Cl
O C H
H H
ionic
covalent
covalent
all covalent bonds
ionic
H
All other bonds are covalent.
1.41 Formal charge (FC) = number of valence electrons – [number of unshared electrons +
(½)(number of shared electrons)]. C is in group 4A.
H H
a.
CH2 CH
b.
H C H
c.
H C H
H
d. H C C
H H
4 − [0 + (1/2)(8)] = 0 4 − [2 + (1/2)(6)] = −1 4 − [2 + (1/2)(4)] = 0 4 − [1 + (1/2)(6)] = 0 4 − [0 + (1/2)(8)] = 0 4 − [0 + (1/2)(6)] = +1
19
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Structure and Bonding 1–19
1.42 Formal charge (FC) = number of valence electrons – [number of unshared electrons +
(½)(number of shared electrons)]. N is in group 5A and O is in group 6A.
a.
e. CH3 O
CH3 N N
c.
CH3 N CH3
6 − [5 + (1/2)(2)] = 0
5 − [0 + (1/2)(8)] = +1 5 − [2 + (1/2)(6)] = 0
5 − [4 + (1/2)(4)] = −1
6 − [2 + (1/2)(6)] = +1
5 − [0 + (1/2)(8)] = +1
5 − [2 + (1/2)(6)] = 0
OH
b.
d.
N N N
f.
CH3 C CH3
CH3 N O
6 − [4 + (1/2)(4)] = 0
5 − [4 + (1/2)(4)] = −1 5 − [4 + (1/2)(4)] = −1
1.43 Follow the steps in Answer 1.4 to draw Lewis structures.
a. CH2N2
valence e−
1C x 4 e− = 4
2H x 1 e− = 2
2N x 5 e− = 10
total e− = 16
H
H
valence e−
2C x 4 e− =
3H x 1 e− =
1N x 5 e− =
1O x 6 e− =
total e− =
H
H
valence e−
1C x 4 e− = 4
3H x 1 e− = 3
1N x 5 e− = 5
2O x 6 e− = 12
total e− = 24
or
H C N O
H O
H O
H
H
H C C N O
H C O
O
O
e. HCO3−
valence e−
or H O C O
H O C O
1C x 4 e−
= 4
O
= 1
1H x 1 e−
= 18
3O x 6 e−
or H O C O
1 for (−) charge = 1
= 24
total e−
f. −CH2CN
or H C C N O
H
H
8
3
5
6
22
or
valence
H C O
1C x 4 e−
= 4
= 1
1H x 1 e−
= 12
2O x 6 e−
1 for (−) charge = 1
total e−
= 18
or
H C N O
O
O
e−
H C N N
b. CH3NO2
c. CH3CNO
d. HCO2−
H C N N
H 2–
or H C C N O
H
H C C N
valence e−
2C x 4 e−
= 8
= 2
2H x 1 e−
= 5
1N x 5 e−
1 for (−) charge = 1
= 16
total e−
H
or H C C N
H
1.44 Follow the steps in Answer 1.4 to draw Lewis structures.
a. N2
[1] N N
[2] Count valence e−.
2N x 5 e− = 10
total e− = 10
[3]
N N
2
e−
used.
N N
Complete N octets.
20
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–20
b. (CH3OH2)+ [1]
H
H C O H
H H
[1]
c. (CH3CH2)
H
H C C H
H H
[1] H N N H
d. HNNH
[1]
e. H6BN
H H
H B N H
H H
[2] Count valence e−.
1C x 4 e− = 4
5H x 1 e− = 5
1O x 6 e− = 6
total e− = 15
Subtract 1 for
(+) charge = 14
[3]
[2] Count valence e−.
2C x 4 e − = 8
5H x 1 e − = 5
total e− = 13
Add 1 for
(−) charge = 14
[3]
[2] Count valence e−.
2H x 1 e− = 2
2N x 5 e− = 10
total e− = 12
[3] H N N H
[2] Count valence e−.
1B x 3 e− = 3
6H x 1 e− = 6
1N x 5 e− = 5
total e− = 14
[3]
H
H
[4]
H C O H
H C O H
H H
12
e−
H H
Add charge
and lone pair.
used.
[4]
H
H C C H
H
H C C H
H H
H H
12 e− used.
e−
6
Add charge
and lone pair.
H N N H
Complete N octets.
used.
[4] H H
H H
H B N H
H B N H
H H
H H
14 e− used.
Add charges.
1.45 Follow the steps in Answer 1.4 to draw Lewis structures.
a. (CH3CH2)2O
[1]
H H
H H
H C C O C C H
H H
b. CH2CHCN
[1]
H H
H H
C C C N
H
[2] Count valence e−.
1O x 6 e− = 6
10H x 1 e− = 10
4C x 4 e− = 16
total e−
= 32
[3] H H
[2] Count valence e−.
1N x 5 e− = 5
3H x 1 e− = 3
3C x 4 e− = 12
total e− = 20
[3] H H
[2] Count valence e−.
3O x 6 e− = 18
H O C C C O H
6H
x 1 e− = 6
H
H
3C x 4 e− = 12
total e− = 36
c. (HOCH2)2CO [1]
d. (CH3CO)2O
H O H
[2] Count valence e−.
3O x 6 e− = 18
H C C O C C H
6H
x 1 e− = 6
H
H
4C x 4 e− = 16
total e− = 40
[1] H O
O H
H H
[4]
H C C O C C H
H H
H H
H H
H H
28 e− used.
H H
Add lone pairs.
[4]
C C C N
H H
C C C N
H
H
12 e− used.
[3]
H H
H C C O C C H
Add lone pairs
and π bonds.
[4]
H O H
H O C C C O H
H
H
H
22 e− used.
[3] H O
H O H
H O C C C O H
O H
H C C O C C H
H
24 e− used.
H
Add lone pairs
and π bonds.
H
[4]
H O
O H
H C C O C C H
H
H
Add lone pairs
and π bonds.
21
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Structure and Bonding 1–21
1.46
a.
b. Two of the possible resonance structures:
NH2
H2N
NH2
O
N
H2N
NH2
O
N
O
H2N
O
N
O
O
1.47 Isomers must have a different arrangement of atoms.
a. Two isomers of molecular formula C3H7Cl
Cl H H
H Cl H
H C C C H
H C C C H
H H H
H H H
c. Four isomers of molecular formula C3H9N
H H H
O H
H C C H
H C C H
H
H
H
H
O
C C
H
H C C N C H
H H H H
H H H H
H
b. Three isomers of molecular formula C2H4O
O H
H H
H C C C N H
H
H H H
H C N C H
H C C C H
H
H
H
H C H
H
H
H N
H
H
H
1.48
Nine isomers of C3H6O:
H H O
H O H
H C C C H
H C C C H
H H
H
H H
H H
H C C C OH
H
H
H
HO
H
H C O
H O H
C
H
C C H
H
H
HO H
H C C C H
H C C H
H
H
H
H C C C H
H
H
H
H C C O C H
H
H
H
H
O
H C C C H
H H H
1.49 Use the definition of isomers and resonance structures in Answer 1.9.
O
O
O
O
O
CH2
a.
b.
d.
5 membered
ring
C6H8O
A
c.
C6H8O
isomers
C6H10O
different molecular formula
neither isomers nor
resonance structures
C6H8O
C6H8O
same arrangement
different arrangement
of atoms
of atoms
resonance structures
isomers
22
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–22
1.50 Use the definitions of isomers and resonance structures in Answer 1.9.
a.
(C8H11)+
B
b.
d.
c.
(C8H11)+
(C8H11)+
(C8H11)+
(C8H9)+
different arrangement
same arrangement
different arrangement different molecular formula
neither
of atoms
of atoms
of atoms
isomers
resonance structures
isomers
1.51 Use the definitions of isomers and resonance structures in Answer 1.9.
CH3
a.
CH3 O CH2CH3
c.
and CH3 C OH
and
H
one O–H bond
two C–O bonds
Different arrangement of atoms.
Both have molecular formula C3H8O = isomers
and
b.
Same arrangement of atoms.
Both have molecular formula (C4H7)−.
Different arrangement of electrons = resonance structures
CH3 C C CH3
d.
H H
CH3CH2CH3
and
CH3CH2CH2
molecular formula C3H8 molecular formula (C3H7)−
ring
double bond
Different arrangement of atoms.
Both have molecular formula C4H8 = isomers
different molecular formulas = neither
1.52 Compare the resonance structures to see what electrons have “moved.” Use one curved arrow
to show the movement of each electron pair.
H
a.
H
CH3 C N CH3
CH3 C N CH3
H
H
One electron pair moves = one arrow
O
b.
O
H C NH2
H C NH2
CH2
CH2
c.
Four electron pairs move = four arrows
Two electron pairs move = two arrows
1.53 Curved arrow notation shows the movement of an electron pair. The tail begins at an electron
pair (a bond or a lone pair) and the head points to where the electron pair moves.
a.
CH3 N N
O
b.
CH3 C CH CH2
CH3 N N
c.
CH3
NH2
O
CH3 C CH CH2
CH3
d.
NH2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
23
Structure and Bonding 1–23
1.54 Use the rules in Answer 1.13.
O
O
O
a. CH3 C O
O
c.
CH3 C O
Two electron pairs move = two arrows
Two electron pairs move = two arrows
Charge is on both atoms.
δ−
Charge is on both O's.
Double bond can
be in 2 locations.
δ− O
CH3 C
O δ−
O
δ− hybrid
hybrid
Double bond can be in 2 locations.
OH
b. CH2 NH2
CH2 NH2
d.
One electron pair moves = one arrow
OH
H C H
H C H
One electron pair moves = one arrow
δ+
partial double bond character
δ+
Charge is on both atoms.
OH
δ+
hybrid
CH2 NH2 hybrid
H C H
δ+
C–O bond has
partial double bond character.
Charge is on both atoms.
1.55 For the compounds where the arrangement of atoms is not given, first draw a Lewis structure.
Then use the rules in Answer 1.13.
a. O3
Count valence e−.
3O x 6 e− = 18
total e− = 18
b. NO3
(a central N atom)
Count valence e−.
1N x 5 e− = 5
3O x 6 e− = 18
(−) charge = 1
total e−
= 24
Count valence e−.
3N x 5 e− = 15
(−) charge = 1
= 16
total e−
c. N3
d.
O O O
O
O N
O
O
O N
O
O N
O
O
2–
2–
N N N
N N N
N N N
H O H O H
H O H O H
H O H O H
H C C C C C H
H C C C C C H
H C C C C C H
H
e.
O O O
H
H
H
H
H
24
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–24
f.
CH2 CH CH
CH CH2
O
CH2 CH CH CH CH2
O
CH2 CH CH CH CH2
O
g.
1.56 To draw the resonance hybrid, use the rules in Answer 1.14.
H
H
H
H
CH2
H
H
H
H
H
CH2
H
H
H
H
H
CH2
H
H
H
H
H
CH2
H
H
H
H
CH2
H
H
resonance hybrid
H
H
δ+
δ+
CH2
δ+
δ+
H
H
H
1.57 A “better” resonance structure is one that has more bonds and fewer charges. The better
structure is the major contributor and all others are minor contributors.
O
a.
O
CH3 C O CH3
3 C–O bonds
no charges
contributes the most = 3
b.
CH3
CH3 C N NH2
3 bonds for this N
no charges
contributes the most = 3
O
CH3 C O CH3
CH3 C O CH3
2 C–O bonds
2 charges
contributes the least = 1
3 C–O bonds
2 charges
2
CH3
CH3 C N NH2
3 bonds for this N
2 charges
2
CH3
CH3 C N NH2
2 bonds for this N
2 charges
contributes the least = 1
1.58
This C would have 5 bonds.
H H
H
H
a.
invalid
b.
c. CH3CH2 C N
CH3CH2 C N d.
O
H H
O
H H
This C would have 5 bonds.
invalid
[Note: The pentavalent C's in (a) and (d) bear a (–1) formal charge.]
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
25
Structure and Bonding 1–25
1.59 Use the rules in Answer 1.18.
3 groups = 120°
H
Cl
a. CH3Cl
C
H
c. 120°
H
H
H
4 groups = 109.5°
C N
H
CH3
H
H
H 120°
All C atoms have
3 groups = 120°.
4 groups = 109.5°
4 groups = ~109.5°
Cl
e.
3 groups = 120°
4 groups = 109.5°
H
H
HC C C OH
d.
b. H N O H
H
H
4 groups = ~109.5°
both C's surrounded by
2 groups = 180°
4 groups = ~109.5°
1.60 To predict the geometry around an atom, use the rules in Answer 1.17.
O
a.
CH3CH2CH2CH3
c.
3 groups
(3 atoms)
trigonal planar
4 groups
(4 atoms)
tetrahedral
b.
d.
(CH3)2N
BF4
4 groups
(4 atoms)
tetrahedral
4 groups
(2 atoms, 2 lone pairs)
tetrahedral
(bent molecular shape)
e.
CH3 C OH
3 groups
(3 atoms)
trigonal planar
f.
(CH3)3N
4 groups
(3 atoms, 1 lone pair)
tetrahedral
(trigonal pyramidal molecular shape)
1.61 Each C has two bonds in the plane of the page, one in front of the page (solid wedge) and one
behind the page (dashed line).
F F
CF3CHClBr
F
C
C
H
Br Cl
1.62 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms.
The C’s are all tetravalent. All H’s bonded to C’s are drawn in the following structures. C’s
labeled with (*) have no H’s bonded to them.
26
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–26
H
CH3
O
O
H H
*
*
H H H H
H
H
HO *
H
H H H H
H
H CH3
H
H
OH
CH3
N *
a.
H
b.
H HH
*
H
N
* H
H
H
H
H
H
*
H
H
H
H
H
*
*
H H H H
H
H H H OH
H
H
CH3
CH3
*
CO2H
*
H
H
H
1.63 In shorthand or skeletal drawings, all line junctions or ends of lines represent carbon atoms.
Convert by writing in all C’s, and then adding H’s to make the C’s tetravalent.
a.
H H
H C
CH3 C
C
H
H
HO H
C H CH3
C C H
CH3
C
H H
menthol
(isolated from peppermint oil)
CH3 H H
b.
CH3
C
C
C
H
H
C
C
C
H H
H
C
H
H H O
C H H
H
N
CH3
C
C
C
N
C
H H
HH
H H
H
c.
H
C
H H
C
C
CH3
H
C
OH
H
ethambutol
(drug used to treat tuberculosis)
H
H
H
H
d.
H
O
C
C
C
C
H
H H CH3 O H
H
C
H
C
C C
H
C
C C
H
C
C
C
H
H
C
CH H
C
H
H
H H
H
H
estradiol
(a female sex hormone)
myrcene
(isolated from bayberry)
1.64 In skeletal formulas, leave out all C’s and H’s, except H’s bonded to heteroatoms.
H
a. (CH3)2CHCH2CH2CH(CH3)2
H
d.
H
C
C
C
C
Cl
H
c. (CH3)3C(CH2)5CH3
C
N
f.
CH3(CH2)2C(CH3)2CH(CH3)CH(CH3)CH(Br)CH3
H
H
b. CH3CH(Cl)CH(OH)CH3
OH
N
e.
CH3
H H
H
H CH3
C
C C
CH2
C C
H
HH
C C
limonene
(oil of lemon)
Br
27
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Structure and Bonding 1–27
1.65 For Lewis structures, all atoms including H’s and all lone pairs must be drawn in.
a. CH3CH2COOH
H H
c. CH3COCH2Br
O
H C C
C O H
H H
b. CH3CONHCH3
H
O
O
H
C
C Br
H C
C N
H
H
CH3 C
H
CH3 C O H
H
H
h. HO2CCH(OH)CO2H
O H O
O
H C
CH3
H
H C C C C O H
f. CH3COCl
CH3
C H
H O H O
C H
CH3
H
d. (CH3)3COH
H
g. CH3COCH2CO2H
CH3 O
H
H C
H
e. (CH3)3CCHO
H O C C C O H
C Cl
O H
H
1.66 A charge on a C atom takes the place of one H atom. A negatively charged C has one lone
pair, and a positively charged C has none.
H
a.
H
H
O
H
b.
H
H
H
H
H
c.
H
H
H
H
H
d.
H
H
H
H
H
CH3
H
H
H
H
1.67
H
CH3
a.
CH
NCH3
b.
CH3 C NH
CH3
H
H
H H H H
H
c. O
O
d. CH3
H
H
O
H
H
H
N
CH3
CH3
CH3
1.68 Examine each structure to determine the error.
This C has 3 bonds.
a. CH3CH=CH=CHCH3
CH2CH3
OH
c.
d.
e. CH2CH3
This C has 5 bonds.
b. (CH3)3CHCH2CH2CH3
This C has 5 bonds.
This H has 2 bonds.
It should be HO–.
This C has 4 bonds. The
negative charge means a lone
pair, which gives C 10 electrons.
This C has 5 bonds.
It should be CH3CH2–.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–28
1.69 To determine the hybridization around the labeled atoms, use the procedure in Answer 1.29.
c. (CH3)3O+
a. CH3CH2
4 groups
(3 atoms, 1 lone pair)
sp3, tetrahedral
(trigonal pyramidal
molecular shape)
e. CH3 C C H
2 groups
(2 atoms)
sp, linear
4 groups
(3 atoms, 1 lone pair)
sp3, tetrahedral
(trigonal pyramidal
molecular shape)
d.
b.
4 groups
(4 atoms)
sp3, tetrahedral
g. CH3CH=C=CH2
3 groups
(3 atoms)
sp2
trigonal planar
2 groups
(2 atoms)
sp, linear
f. CH2=NOCH3
CH2Cl
3 groups
(2 atoms, 1 lone pair)
sp2, trigonal planar
4 groups
(4 atoms)
sp3, tetrahedral
(Each C has 2 H's.)
1.70 To determine what orbitals are involved in bonding, use the procedure in Answer 1.27.
Csp3–H1s
a.
H
Csp–Csp2
σ: Csp2–Csp2
π: Cp–Cp
b.
σ: Csp2–Osp2
π: Cp–Op
c.
Csp3–Csp3
Csp2–H1s
O
Csp2–Csp3
H H
H C C C N CH3
d.
H
Csp2–Csp3
Csp–H1s σ: Csp–Csp
π: Cp–Cp
π: Cp–Cp
σ: Csp3–Nsp2
1.71
σ: Csp–Csp2
π: Cp–Cp
sp sp2
sp2
π bonds
π bond
σ bonds
H
CH2 C O
ketene
sp2
C
1s
σ: Csp–Osp2
π: Cp–Op
C
sp2
O
H
sp
H
sp2
π bond
H
C
C
sp2
[For clarity, only the large bonding lobes of the hybrid orbitals are drawn.]
1.72
CH2 CH
sp2
sp
σ: Csp2–Csp
π: Cp–Cp
CH2 CH
sp2
sp2
σ: Csp2–Csp2
π: Cp–Cp
C
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
29
Structure and Bonding 1–29
1.73 To determine relative bond length, use the rules in Answer 1.31.
H
H
a.
C C H
C C H
C CH3
H H
H
H C C CH2 C C C H
H H H
Csp3–H1s
Csp–H1s
single bond
Csp2–H1s
lowest
highest
longest
middle
% s-character
% s-character
% s-character
longest
shortest
middle
double bond
middle
triple bond
shortest
b.
1.74
b. and d. bond (1)
longest,
weakest C–C
bond (2)
single bond
c. shortest, strongest C–C bond
H
a. shortest C–C
single bond
e. strongest C–H bond
f. Bond (1) is a Csp3–Csp3 bond, and bond (2) is a Csp3–Csp2 bond. Bond
(2) is shorter due to the increased percent s-character in the sp2
hybridized carbon.
1.75 Percent s-character determines the strength of a bond. The higher percent s-character of an
orbital used to form a bond, the stronger the bond.
vinyl chloride
CH2 CH
Csp2
chloroethane (ethyl chloride)
CH3 CH2 Cl
Cl
33% s-character
higher percent s-character
stronger bond
25% s-character
Csp3
1.76 Dipoles result from unequal sharing of electrons in covalent bonds. More electronegative
atoms “pull” electron density towards them, making a dipole.
δ+
δ−
b. NH2 OH
a. δ+ Br Cl δ−
c.
δ+
δ−
δ−
d.
CH3 NH2
Li δ+
1.77 Use the directions from Answer 1.34.
Br
c. CBr4
a. CHBr3
Br
H
C
Br
Br
C
net dipole
Br
b. CH3CH2OCH2CH3
O
net dipole
Br
d.
e.
Br
Br no net dipole
Br
net dipole
Cl
CH3
C O
CH3
net dipole
f.
no net dipole
Cl
30
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–30
1.78
1 σ and 2 π bonds
polar bond
σ
(essentially) nonpolar
tetrahedral
sp3 hybridized
CH3 C N
linear
sp hybridized
The lone pair is in an
sp hybrid orbital.
linear
sp hybridized
All C–H bonds are nonpolar σ bonds.
All H's use a 1s orbital in bonding.
1.79
a. sp2
b. Each C is trigonal planar; the ring is flat, drawn as a hexagon.
c.
H
H
H
H
sp2 hybridized C
H
H
H
H
H
H
H
[Only the larger bonding lobe
of each orbital is drawn.]
H
p orbitals on C's overlap
sp2 hybrid orbitals on C
π bonds
σ bonds
d.
e. Benzene is stable because of its two resonance structures that contribute
equally to the hybrid. [This is only part of the story. We'll learn more about
benzene's unusual stability in Chapter 17.]
1.80
3 groups
sp2
a. trigonal planar
3 groups
sp2
trigonal
planar
H
H
4 groups
sp3
tetrahedral
H O H H H
HO
S
C C N
NH2
H
N
O
H
CH3
H C O
HO
4 groups
sp3
tetrahedral
(trigonal pyramidal molecular shape)
H
CH3
3 groups
sp2
trigonal planar
H
H O H H H
b.
HO
S
C C N
NH2
H
H
N
O
CH3
CH3
H C O
H O
All C–O, C–N, C–S, N–H, and O–H bonds are
polar and labeled with arrows.
All partial positive charges lie on the C.
All partial negative charges lie on the O, N, or S.
In OH and NH bonds, H bears a δ+.
31
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Structure and Bonding 1–31
c.
skeletal structure:
e. 33% s-character = sp2 hybridized
NH2 H
H
N
S
O
HO
H
N
O
O
These C–H bonds have 33% s-character.
HO
H
H
OH
NH2 H
N
d.
S
O
HO
N
O
6 π bonds
O
OH
1.81
a, b, c.
d.
4 groups
CH3
sp3
tetrahedral
(trigonal pyramidal molecular shape)
lone pair in sp3 orbital
N
N
3 groups
sp2
trigonal planar
lone pair in sp2 orbital
CH3
N
e.
N
H
N
N
constitutional isomer
CH3
resonance structure
1.82 a.
longest C–N bond
H H
H
H O
H O
C
C
C
C
H
C
C
C
C
O
C C H
C N
H H
H H
*
H
C N
H H
N O
O
C C H
shortest longest C–C bond
C–N bond
b. The C–C bonds in the CH2CH3 groups are the longest because they are formed from sp3
hybridized C’s.
c. The shortest C–C bond is labeled with a (*) because it is formed from orbitals with the
highest percent s-character (Csp–Csp2).
d. The longest C–N bond is formed from the sp3 hybridized C atom bonded to a N atom
[labeled in part (a)].
e. The shortest C–N bond is the triple bond (C≡N); increasing the number of electrons between
atoms decreases bond length.
H
f.
H
O
H O
H
N CH2CH3
H O
H
C N
H
O
H O
N CH2CH3
CH2CH3
H O
H
N O
N O
O
O
C N
CH2CH3
32
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–32
H
g.
H
O
H
H O
N CH2CH3
H O
H
H
N CH2CH3
CH2CH3
C N
O
H O
H O
H
N O
N O
O
O
C N
CH2CH3
1.83
4 groups
CH3
sp3 tetrahedral
(The molecular shape is trigonal pyramidal.)
plot B
3 groups
sp2 trigonal planar
plot A
CH3
The blue region is evidence of
the electron-poor cation.
The red region is evidence of
the electron-rich anion.
1.84 If the N atom is sp2 hybridized, the lone pair occupies a p orbital, which can overlap with the π
bond of the adjacent C=O. This allows electron density to delocalize, which is a stabilizing
feature.
σ bonds in –CONH2
O
C, O, and N use sp2 hybrid orbitals to form σ bonds.
C
H
N
H
π bond formed by overlap of Cp–Op
O
π bonds in –CONH2
C
N
The lone pair occupies the unhybridized p orbital.
1.85
[1]
a.
O
CH3
OH
CH3
sp3
25% s-character
The lower percent s-character
makes this bond longer.
[3]
O
b.
CH3
C
O
C
[2]
OH
sp2
33% s-character
O δ−
O
CH3
C
[4]
two resonance structures
O
CH3
C
O δ−
hybrid
Bonds [3] and [4] are both equivalent in length, because the anion is resonance-stabilized, and
the C–O bond of the hybrid is a composite of one single bond and one double bond. Both
resonance structures contribute equally to the hybrid. Since each C–O bond in the hybrid has
partial double bond character, it is shorter than the C–O single bond labeled [2].
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
33
Structure and Bonding 1–33
1.86 Ten additional resonance structures are drawn. (There are more possibilities.)
H
HO
H
N
HO
H
N
O
HO
N
O
O
H
HO
H
N
HO
H
N
O
HO
O
O
H
HO
N
H
N
HO
H
N
O
HO
O
N
O
H
HO
H
N
HO
O
N
O
1.87 Polar bonds result from unequal sharing of electrons in covalent bonds. Normally we
think of more electronegative atoms “pulling” more of the electron density towards them,
making a dipole. In looking at a Csp2–Csp3 bond, the atom with a higher percent s-character
will “pull” more of the electron density towards it, creating a small dipole.
δ−
33% s-character
higher percent s-character
pulls more electron density
more electronegative
Csp2
δ+
Csp3
25% s-character
1.88
Isomers of C4H8:
H
C
H
H
H
C
C
CH2CH3
CH3
H
C
H
CH3
CH3
C
H
C
H
CH3
These two compounds are different
because of restricted rotation around
the C=C (Section 8.2B).
H
C
C
CH3
CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 1–34
1.89 Carbocation A is more stable than carbocation B because resonance distributes the positive
charge over two carbons. Delocalizing electron density is stabilizing. B has no possibility of
resonance delocalization.
No resonance structures
A
B
1.90
H
HO
O
+
a.
HO
b.
H
[1]
H
H
H Cl
+
OH
H
H
OH
H
[2]
Cl
Cl
H
+
[3]
X
phenol
H Cl
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
35
Acids and Bases 2–1
Chapter 2 Acids and Bases
Chapter Review
A comparison of Brønsted–Lowry and Lewis acids and bases
Type
Brønsted–Lowry acid
(2.1)
Definition
proton donor
Structural feature
a proton
Brønsted–Lowry base
(2.1)
proton acceptor
a lone pair or a π bond
Lewis acid (2.8)
electron pair
acceptor
a proton, or an unfilled
valence shell, or a partial
(+) charge
Lewis base (2.8)
electron pair
donor
a lone pair or a π bond
Examples
HCl, H2SO4, H2O,
CH3COOH, TsOH
–
OH, –OCH3, H–, –NH2,
CH2=CH2
BF3, AlCl3, HCl,
CH3COOH, H2O
–
OH, –OCH3, H–, –NH2,
CH2=CH2
Acid–base reactions
[1] A Brønsted–Lowry acid donates a proton to a Brønsted–Lowry base (2.2).
H
Example
H
O
H
acid
proton donor
+
H N H
H
base
proton acceptor
O
+
conjugate base
H N H
conjugate acid
[2] A Lewis base donates an electron pair to a Lewis acid (2.8).
CH3
Example
CH3 C
CH3
+
Br
CH3 C Br
CH3
Lewis acid
electrophile
•
•
CH3
Lewis base
nucleophile
Electron-rich species react with electron-poor ones.
Nucleophiles react with electrophiles.
Important facts
• Definition: pKa = −log Ka. The lower the pKa, the stronger the acid (2.3).
NH3
pKa = 38
versus
H2O
pKa = 15.7
lower pKa = stronger acid
36
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–2
•
The stronger the acid, the weaker the conjugate base (2.3).
Increasing pKa
Increasing pKa of the conjugate acid
CH2 CH2
CH3COOH
HCl
pKa = 44
pKa = 4.8
pKa = –7
Cl
CH3COO
Increasing basicity
Increasing acidity
•
CH2 CH
In proton transfer reactions, equilibrium favors the weaker acid and the weaker base (2.4).
H C C H
+
NH2
H C C
+
NH3
pKa = 38
weaker acid
pKa = 25
stronger acid
Equilibrium favors
the products.
unequal equilibrium arrows
•
An acid can be deprotonated by the conjugate base of any acid having a higher pKa (2.4).
Acid
pKa
Conjugate base
CH3COO–H
4.8
CH3COO—
CH3CH2O–H
16
CH3CH2O—
HC≡CH
H–H
25
HC≡C
35
H
—
These bases
can deprotonate
—
CH3COO–H.
higher pKa
than
CH3COO–H
Factors that determine acidity (2.5)
[1] Element effects (2.5A)
The acidity of H−A increases both left to right across a row and
down a column of the periodic table.
C H
N H
O H
H F
Increasing electronegativity
Increasing acidity
H F
H Cl
H Br
Increasing size
Increasing acidity
H
I
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
37
Acids and Bases 2–3
[2] Inductive effects (2.5B)
The acidity of H−A increases with the presence of electronwithdrawing groups in A.
CH3CH2OH
weaker acid
CF3CH2OH
CH3CH2O
No additional electronegative
atoms stabilize the conjugate base.
δ–
F
stronger acid
F
δ–
H
C
δ+
C
O
Fδ– H
CF3 withdraws electron density,
stabilizing the conjugate base.
[3] Resonance effects (2.5C)
The acidity of H−A increases when the conjugate base A:– is
resonance stabilized.
CH3CH2O H
ethanol
CH3CH2O
ethoxide
conjugate base
only one Lewis structure
O
CH3 C
OH
acetic acid
more acidic
[4] Hybridization effects
(2.5D)
O
O
CH3 C
O
CH3 C
acetate
conjugate base
O
two resonance structures
The acidity of H−A increases as the percent s-character of
the A:– increases.
H C C H
CH3CH3
CH2=CH2
ethane
ethylene
acetylene
pKa = 50
pKa = 44
pKa = 25
Increasing acidity
38
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–4
Practice Test on Chapter Review
1. a Given the pKa data, which of the following bases is strong enough to deprotonate C6H5OH
(pKa = 10) so that the equilibrium lies to the right?
Compound
H3O+
NH4+
H2O
NH3
pKa
–1.7
9.4
15.7
38
1. NaOH
2. NaNH2
3. NH3
4. Compounds (1) and (2) are strong enough to deprotonate
C6H5OH.
5. Compounds (1), (2), and (3) are all strong enough to
deprotonate C6H5OH.
b. Which of the following statements is true about pKa, acidity, and basicity?
1. A higher pKa means the acid is less acidic.
2. In an acid–base reaction, the equilibrium lies on the side of the acid with the higher pKa.
3. A lower pKa value for the acid means the conjugate base is more basic.
4. Statements (1) and (2) are both true.
5. Statements (1), (2), and (3) are all true.
c. Which of the following species can be Lewis acids?
1. BCl3
2. CH3OH
3. (CH3)3C+
4. Both (1) and (2) can be Lewis acids.
5. Species (1), (2), and (3) can all be Lewis acids.
2. Answer the following questions about compounds A–D.
O
CH2CH3
A
CH2OH
CH2NH2
B
C
C
OH
D
a. Which compound is the strongest acid?
b. Which compound forms the strongest conjugate base?
c. The conjugate base of C is strong enough to remove a proton on which compound(s)
such that the equilibrium favors the products?
3. (a) Which compound is the strongest Brønsted–Lowry acid? (b) Which compound is the weakest
Brønsted–Lowry acid?
Br
Cl
NH2
A
O
Cl
OH
Cl
B
C
OH
D
39
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Acids and Bases 2–5
4. Draw all the products formed in the following reactions.
Cl
NHCH3
Na+ –NH2
b.
HBr
a.
HO
NO2
5. Draw the product(s) formed in the following Lewis acid–base reaction.
(CH 3)2CH
CH3OH
Answers to Practice Test
1. a. 4
b. 4
c. 5
2. a. D
b. A
c. B, D
3. a. C
b. B
4. a.
5.
NH2CH3
(CH3)2CH
Br–
O
CH3
H
b.
Cl
NH3
O
Na
NO2
Answers to Problems
2.1 Brønsted–Lowry acids are proton donors and must contain a hydrogen atom.
Brønsted–Lowry bases are proton acceptors and must have an available electron pair
(either a lone pair or a π bond).
a.
b.
HBr
NH3
CCl4
acid
acid
not an acid—no H
CH3CH3
(CH3)3CO
H C C H
lone pairs
on O
base
base—π bonds
no lone pairs
or π bonds
not a base
O
c.
CH3CH2CH2CH3
CH3CH2OH
CH3 C
OCH3
not a base—no lone pairs
or π bonds
acid—contains H atoms
base—lone pairs on O
acid—contains H atoms
base—lone pairs on O's, π bond
acid—contains H atoms
2.2 A Brønsted–Lowry base accepts a proton to form the conjugate acid. A Brønsted–Lowry
acid loses a proton to form the conjugate base.
a.
NH3
NH4+
Cl
HCl
(CH3)2C=O
b.
HBr
HSO4
(CH3)2C=OH
CH3OH
Br−
SO42–
CH3O−
40
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–6
2.3
The Brønsted–Lowry base accepts a proton to form the conjugate acid. The Brønsted–Lowry
acid loses a proton to form the conjugate base. Use curved arrows to show the movement of
electrons (NOT protons). Re-draw the starting materials if necessary to clarify the electron
movement.
a.
H Cl
+
conjugate base conjugate acid
base
acid
O
b.
CH3
O
C
+
CH2
OCH3
CH3
H
acid
2.4
C
CH2
CH3OH
conjugate base conjugate acid
base
O
O
+
Cl3C C
O CH3
H C C H
CH3 NH2
H
+
+
H C C
(−)1 charge on each side
H2
base
+
CH3 NH3
H Cl
base
d.
(−)1 charge on each side
base
acid
c.
HO CH3
O
acid
b.
+
Cl3C C
O H
+
net neutral on each side
Cl
acid
+
CH3CH2 O H
base
H OSO 3H
CH 3CH2 O H +
OSO 3H
net neutral on each side
H
acid
Draw the products in each reaction as in Answer 2.4.
a. CH3OH
HCl
b. (CH3CH2)2O
2.6
+
To draw the products:
[1] Find the acid and base.
[2] Transfer a proton from the acid to the base.
[3] Check that the charges on each side of the arrows are balanced.
a.
2.5
H3 O
+
Cl
H2 O
CH3OH2 + Cl–
HCl
(CH3CH2)2OH + Cl–
c.
(CH3)3N
d.
NH
HCl
HCl
(CH3)3NH + Cl–
–
NH2 + Cl
The smaller the pKa, the stronger the acid. The larger the Ka, the stronger the acid.
OH
a.
CH3CH2CH3
pKa = 50
or
CH3CH2OH
CH3
or
b.
pKa = 16
Ka = 10–10
smaller pKa
stronger acid
larger Ka
stronger acid
Ka = 10–41
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
41
Acids and Bases 2–7
2.7
To convert from Ka to pKa, take (–) the log of the Ka; pKa = –log Ka.
To convert pKa to Ka, take the antilog of (–) the pKa.
a. Ka = 10–10
pKa = 10
2.8
Ka = 10–21
Ka = 5.2 x 10–5
pKa = 21
pKa = 4.3
b. pKa = 7
Ka = 10–7 Ka = 10–11 Ka = 6.3 x 10–4
Increasing acidity
−OH, −NH
conjugate bases:
2,
−CH
Increasing acidity
b.
H2O, NH3, CH4
pKa = 15.7 38 50
HC≡CH, CH2=CH2, CH4
pKa =
3
conjugate bases:
44
−C≡CH, −CH=CH
50
2,
−CH
3
Use the definitions in Answer 2.8 to compare the acids. The smaller the pKa, the larger the
Ka and the stronger the acid. When a stronger acid dissolves in water, the equilibrium lies
further to the right.
HCO2H
formic acid
pKa = 3.8
a. smaller pKa = larger Ka
b. smaller pKa = stronger acid
d. stronger acid = equilibrium further to the right
(CH3)3CCO2H
pivalic acid
pKa = 5.0
c. weaker acid = stronger conjugate base
To estimate the pKa of the indicated bond, find a similar bond in the pKa table (H bonded to
the same atom with the same hybridization).
H
N H
a.
b.
c. BrCH2COO H
O H
For CH3CH2OH,
pKa is 16.
estimated pKa = 16
For NH3, pKa is 38.
estimated pKa = 38
2.11
25
Increasing basicity
Increasing basicity
2.10
pKa = 3.2
Since strong acids form weak conjugate bases, the basicity of conjugate bases increases
with increasing pKa of their acids. Find the pKa of each acid from Table 2.1 and then rank
the acids in order of increasing pKa. This will also be the order of increasing basicity of their
conjugate bases.
a.
2.9
pKa = 11
For CH3COOH, pKa is 4.8.
estimated pKa = 5
Label the acid and the base and then transfer a proton from the acid to the base. To
determine if the reaction will proceed as written, compare the pKa of the acid on the left with
the conjugate acid on the right. The equilibrium always favors the formation of the
weaker acid and the weaker base.
a.
CH2 CH2
acid
pKa = 44
weaker acid
+
H
CH2 CH
base
conjugate base
+
H2
conjugate acid
pKa = 35
Equilibrium favors
the starting materials.
42
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–8
b.
CH4
+
acid
pKa = 50
weaker acid
c.
CH3COOH
OH
CH3
base
conjugate base
+
CH3CH2O
d.
Cl
+
base
2.12
acid
pKa = 16
weaker acid
Base
NaH
Na2CO3
NaOH
NaNH2
NaHCO3
Equilibrium favors
the starting materials.
conjugate base
pK a
35
10.2
15.7
38
6.4
Conjugate acid
H2
HCO3–
H2O
NH3
H2CO3
Only NaH and NaNH2
are strong enough to
deprotonate acetonitrile.
The acidity of H–Z increases left to right across a row and down a column of the periodic
table.
b. H2S, HBr
Br is farther across and down
the periodic table.
stronger acid
O is farther to the right in
the periodic table.
stronger acid
Compare the most acidic protons in each compound to determine the stronger acid.
a.
CH3CH2CH2NH2
(CH3)3N
or
N–H bond
C–H bond
b.
CH3CH2OCH3
C–H bond
N is farther to the right in
the periodic table.
stronger acid
2.15
Equilibrium favors
the products.
An acid can be deprotonated by the conjugate base of any acid with a higher pKa.
a. NH3, H2O
2.14
conjugate acid
pKa = 16
weaker acid
CH3CH2O
conjugate acid
pKa = −7
CH3CN
pKa = 25
Any base having a conjugate
acid with a pKa higher than
25 can deprotonate this acid.
2.13
+
HCl
CH3CH2OH
+
conjugate base
CH3CH2OH
Equilibrium favors
the starting materials.
conjugate acid
pKa = 15.7
CH3COO
base
acid
pKa = 4.8
H2 O
+
or
CH3CH2CH2OH
O–H bond
O is farther to the right in
the periodic table.
stronger acid
Look at the element bonded to the acidic H and decide its acidity based on the periodic
trends. Farther to the right and down the periodic table is more acidic.
most acidic
a. CH3CH2CH2CH2OH
Molecule contains C–H
and O–H bonds.
O is farther right; therefore,
O–H hydrogen is the most acidic.
most acidic
b. HOCH2CH2CH2NH2
Molecule contains C–H,
N–H, and O–H bonds.
O is farthest right; therefore,
O–H hydrogen is the most acidic.
most acidic
c. (CH3)2NCH2CH2CH2NH2
Molecule contains C–H and N–H
bonds.
N is farther right; therefore,
N–H hydrogen is the most acidic.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
43
Acids and Bases 2–9
2.16
The acidity of HA increases left to right across the periodic table. Pseudoephedrine contains
C–H, N–H, and O–H bonds. The O–H bond is most acidic.
H O H
H
N
H
pseudoephedrine
skeletal structure
2.17
More electronegative atoms stabilize the conjugate base, making the acid stronger.
Compare the electron-withdrawing groups on the acids below to decide which is a stronger
acid (more electronegative groups = more acidic).
or
ClCH2COOH
a.
FCH2COOH
c.
more acidic
F is more electronegative than Cl, making the
O–H bond in the acid on the right more acidic.
b.
Cl2CHCH2OH
or
Cl is closer
to the acidic O–H bond.
more acidic
2.18
CH3COOH
or
O2NCH2COOH
more acidic
NO2 is electron withdrawing, making the
O–H bond in the acid on the
right more acidic.
Cl2CHCH2CH2OH
Cl is farther from the
O–H bond.
More electronegative groups stabilize the conjugate base, making the acid stronger.
HOCH2CO2H
an α-hydroxy acid
The extra OH group contains an electronegative
O, which stabilizes the conjugate base.
stronger acid
CH3CO2H
acetic acid
2.19
HBr is a stronger acid than HCl because Br is farther down a column of the periodic table,
and the larger Br– anion is more stable than the smaller Cl– anion. In these acids the H is
bonded directly to the halogen. In HOCl and HOBr, the H is bonded to O, and the halogens
Cl and Br exert an inductive effect. In this case, the more electronegative Cl stabilizes –OCl
more than the less electronegative Br stabilizes –OBr. Thus, HOCl forms the more stable
conjugate base, making it the stronger acid.
2.20
The acidity of an acid increases when the conjugate base is resonance stabilized. Compare
the conjugate bases of acetone and propane to explain why acetone is more acidic.
O
CH3
C
O
CH3
acetone
pKa = 19.2
O
2 resonance structures
more stable conjugate base
Acetone is more acidic.
One resonance structure places the (–) charge on the
more electronegative O atom. This is especially good.
base
CH3
C
CH2
CH3
C
CH2
44
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–10
base
CH3CH2CH3
only one Lewis structure
less stable conjugate base
(Any C–H bond in the starting
material can be removed.)
CH3CH2CH2
propane
pKa = 50
2.21
The acidity of an acid increases when the conjugate base is resonance stabilized. Acetonitrile
has a resonance-stabilized conjugate base, which accounts for its acidity.
H
H
H
base
H C C N
H C C N
H C C N
H
acetonitrile
(one Lewis structure)
2.22
The negative charge is stabilized by
delocalization on the C and N atoms.
Having the (–) charge on the electronegative N atom adds stability.
Increasing percent s-character makes an acid more acidic. Compare the percent
s-character of the carbon atoms in each of the C–H bonds in question. A stronger acid has a
weaker conjugate base.
H
a.
or
CH3CH2 C C H
base
CH3CH2CH2CH2–H
sp hybridized C
50% s-character
base
more acidic
sp3 hybridized C
25% s-character
b.
base
H
or
H
sp3
sp2 hybridized C
33% s-character
more acidic
hybridized C
25% s-character
base
CH3CH2 C C
or
H
CH3CH2CH2CH2
or
stronger
conjugate base
2.23
stronger
conjugate base
To compare the acids, first look for element effects. Then identify electron-withdrawing
groups, resonance, or hybridization differences.
a. CH3CH2CH3
C is farthest left in
the periodic table.
CH bond is
least acidic.
CH3CH2NH2
intermediate
acidity
CH3CH2OH
O is farthest right in
the periodic table.
OH bond is
most acidic.
b. CH3CH2CH2OH
CH3CH2COOH
OH group
least acidic
intermediate
acidity
c. (CH3)3N
CH3CH2NH2
C is farthest left in
the periodic table.
CH bond is
least acidic.
BrCH2COOH
Br is electron withdrawing
and the conjugate base
is resonance stabilized.
most acidic
intermediate
acidity
CH3CH2OH
O is farthest right in
the periodic table.
OH bond is
most acidic.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
45
Acids and Bases 2–11
2.24
Look at the element bonded to the acidic H and decide its acidity based on the periodic
trends. Farther to the right and down the periodic table is more acidic.
most acidic
O
most acidic
OH
b.
a.
COOH
O
THC
tetrahydrocannabinol
The molecule contains C–H
and O–H bonds.
O is farther right; therefore,
O–H hydrogen is the most acidic.
2.25
ketoprofen
The molecule contains C–H
and O–H bonds.
O is farther right; therefore,
O–H hydrogen is the most acidic.
Draw the products of proton transfer from the acid to the base.
(CH3)2CHO Na+
a. (CH3)2CHO−H + Na+ H
acid
base
b. (CH3)2CHO−H + H–OSO3H
base
acid
(CH3)2CHOH2
2.26
HSO4–
+
conjugate base
(CH3)2CHO Li+
+
HN[CH(CH3)2]2
conjugate base
(CH3)2CHOH2
conjugate acid
+
conjugate acid
–OCOCH
3
conjugate base
To cross a cell membrane, amphetamine must be in its neutral (not ionic) form.
CH3
CH2
CH3
protonation
CH2 C H
C H
NH2
by HCl in
the stomach
NH2
CH3
deprotonation
CH2 C H
in the intestines
H
amphetamine
2.27
H2
conjugate acid
conjugate acid
c. (CH3)2CHO−H + Li+ –N[CH(CH3)2]2
acid
base
d. (CH3)2CHO−H + H–OCOCH3
base
acid
+
conjugate base
NH2
absorption here
in the neutral form
Lewis bases are electron pair donors: they contain a lone pair or a π bond.
a. NH3
yes - has
lone pair
b. CH3CH2CH3
no - no lone pair
or π bond
c. H
yes - has
lone pair
d. H C C H
yes - has
2 π bonds
46
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–12
2.28
Lewis acids are electron pair acceptors. Most Lewis acids contain a proton or an unfilled
valence shell of electrons.
yes
unfilled valence shell
on B
2.29
c. (CH3)3C+
b. CH3CH2OH
a. BBr3
yes
contains a proton
d. Br
yes
no
unfilled valence shell
no proton
on C
no unfilled valence shell
Label the Lewis acid and Lewis base and then draw the curved arrows.
new bond
F
a.
BF3
+ CH3 O CH3
b.
CH3
F
Lewis base
Lewis acid
unfilled valence shell lone pairs
on B
on O
2.30
CH3
F B O
(CH3)2CH
+
(CH3)2CHOH
OH
Lewis acid Lewis base
unfilled valence lone pairs
shell on C
on O
A Lewis acid is also called an electrophile. When a Lewis base reacts with an electrophile
other than a proton, it is called a nucleophile. Label the electrophile and nucleophile in the
starting materials and then draw the products.
Cl
Br
O
Br B Br
a. CH3CH2 O CH2CH3
Lewis base
nucleophile
lone pairs
on O
2.31
+
BBr3
CH3CH2 O CH2CH3
Cl Al Cl
b.
CH3
C
CH3
+
AlCl3
Lewis acid
CH3
Lewis base
electrophile
nucleophile unfilled valence shell
lone pairs
on Al
on O
Lewis acid
electrophile
unfilled valence shell
on B
O
C
CH3
Draw the product of each reaction by using an electron pair of the Lewis base to form a new
bond to the Lewis acid.
CH3
CH3 B CH3
a. CH3CH2 N CH2CH3
+
B(CH3)3
CH2CH3
Lewis base
nucleophile
lone pair
on N
CH3CH2 N CH2CH3
CH2CH3
Lewis acid
electrophile
unfilled valence shell
on B
CH3
CH3 C CH3
b. CH3CH2 N CH2CH3
+
+C(CH )
3 3
CH2CH3
Lewis base
nucleophile
lone pair
on N
CH3CH2 N CH2CH3
CH2CH3
Lewis acid
electrophile
unfilled valence shell
on C
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
47
Acids and Bases 2–13
Cl
Cl Al Cl
+
c. CH3CH2 N CH2CH3
AlCl3
CH3CH2 N CH2CH3
CH2CH3
Lewis base
nucleophile
lone pair
on N
2.32
CH2CH3
Lewis acid
electrophile
unfilled valence shell
on Al
Curved arrows begin at the Lewis base and point towards the Lewis acid.
CH2 C
new C–H bond
H
CH3
H C C
H
H
H
CH3
H2O
+
H
+
H2O
Lewis acid
Lewis base
contains a π bond contains a proton
2.33
a, b. Since acidity increases from left to right across a row of the periodic table and
propranolol has C–H, N–H, and O–H bonds, the O–H bond is most acidic. NaH is a
base and removes the most acidic OH proton.
O
O
H
most acidic
N
H
O
O
H
propranolol
Na
N
H
H2
H
Na H
skeletal structure
c, d. Of the atoms with lone pairs (N and O), N is to the left in the periodic table, making it
the most basic site. HCl is an acid, which protonates the most basic site.
O
HO
2.34
N
H
H
Cl
Cl
O
HO
H
H
N
H2
a, b. Using periodic trends, the N–H bond of amphetamine is most acidic. NaH is a base that
removes a proton on N.
H
N
H
Na H
H CH3
amphetamine
skeletal structure
N
H CH3
most acidic
H
Na
H2
48
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–14
c. HCl protonates the lone pair on N.
H
NH2
Cl
NH3
H CH3
2.35
Cl
H CH3
To draw the conjugate acid of a Brønsted–Lowry base, add a proton to the base.
H
a. H2O
H
d. CH3CH2NHCH3
H3O
CH3CH2NH2CH3
H
b. NH2
H
H
c. HCO3
2.36
H
f. CH3COO
CH3 O CH3
H
CH3COOH
CO32
H
c. (CH3)2NH2
H
d. HC CH
CN
H
b. HCO3
2.38
H2CO3
H
To draw the conjugate base of a Brønsted–Lowry acid, remove a proton from the acid.
a. HCN
2.37
NH3
e. CH3OCH3
HC C
H
e. CH3CH2COOH
(CH3)2NH
H
f. CH3SO3H
CH3CH2COO
CH3SO3
Use the definitions from Answer 2.2.
CH2 CH2
CH2 CH3
ethylene
accepts a proton
conjugate acid
CH2 CH2
CH2 CH
loses a proton
conjugate base
To draw the products of an acid–base reaction, transfer a proton from the acid (H2SO4 in this
case) to the base.
a.
+
OH + H OSO3H
+
O H
HSO4
H
b.
NH2
c.
OCH3
+
+
NH3 + HSO4
H OSO3H
+ H OSO3H
+
+ HSO4
OCH3
H
d.
N CH3
+
H OSO3H
+ H
N
CH3
+
HSO4
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
49
Acids and Bases 2–15
2.39
To draw the products of an acid–base reaction, transfer a proton from the acid to the base
(–OH in this case).
a.
O H + K+ –OH
O
+ H2O
O
+ K+ –OH
b.
+ H 2O
O K+
O H
+ K+ –OH
C C H
c.
d. CH3
2.40
K+
O
O H
+
C C K+
+ K+ –OH
H2O
O K+
CH3
+
H 2O
Label the Brønsted–Lowry acid and Brønsted–Lowry base in the starting materials and
transfer a proton from the acid to the base for the products.
a.
CH3O H
NH2
CH3O
acid
base
conjugate base
+
NH3
conjugate acid
O
b.
O
CH3CH2 C
+
O H
O CH3
CH3CH2 C
+
O
base
acid
HO CH3
conjugate acid
conjugate base
c.
CH3CH2 O H
+
base
d.
CH3C C
H Br
acid
+
H OH
base
CH3CH2 O H
+
Br
H
conjugate acid
CH3C CH
+
conjugate base
HO
conjugate acid conjugate base
acid
2.41 Label the acid and base in the starting materials and then draw the products of proton transfer
from acid to base.
O
a.
F 3C C
acid
b.
O
Na+ HCO3
+
O H
CH3 C C H
acid
base
+
Na+ NH2
base
F3 C C
O
+
H2CO3
Na
+ NH3
Na
CH3 C C
50
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–16
O
+
CH3 N H
c.
CH3
H
S O H
CH3
CH3 N H
O
acid
base
O
d.
CH3
C
CH3
CH3
C
+
CH3
acid
Draw the products of proton transfer from acid to base.
H CH
3
COO– Na+
COO–H
a.
Na+
+
CH3O
acid
OH
base
H CH2CH2NH2CH3
CF3 + H–Cl
O
b.
O
CF3
+ Cl
acid
base
Draw the products of proton transfer from acid to base.
CF3
CF3
CH2CH(CH3)NHCH2CH3
CH2C(CH3)2NH2
CH2CH(CH3)NHCH2CH3
+ H OCOCH3
base
H
acid
+ H OCOCH3
CH2C(CH3)2NH3 +
acid
base
To convert pKa to Ka, take the antilog of (–) the pKa.
a. H2S
pKa = 7.0
Ka = 10–7
2.45
+ H2 O
CH3O
H CH2CH2NHCH3
2.44
SO3
HSO4
H CH
3
2.43
CH3
O H
H OSO3H
base
2.42
+
CH3
b. ClCH2COOH
pKa = 2.8
Ka = 1.6 x 10–3
c. HCN
pKa = 9.1
Ka = 7.9 x 10–10
To convert from Ka to pKa, take (–) the log of the Ka; pKa = –log Ka.
+
a.
CH2NH3
Ka = 4.7 x 10–10
pKa = 9.3
+
b.
NH3
Ka = 2.3 x 10–5
pKa = 4.6
c. CF3COOH
Ka = 5.9 x 10–1
pKa = 0.23
OCOCH3
+
OCOCH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
51
Acids and Bases 2–17
2.46
An acid can be deprotonated by the conjugate base of any acid with a higher pKa.
CH3CH2CH2C≡CH
pKa = 25
Any base having a conjugate
acid with a pKa higher than
25 can deprotonate this acid.
2.47
–
Conjugate acid
H3O+
H 2O
NH3
NH4+
H2
CH4
Base
H 2O
NaOH
NaNH2
NH3
NaH
CH3Li
p Ka
–1.7
15.7
38
9.4
35
50
Only NaNH2, NaH, and
CH3Li are strong enough
to deprotonate the acid.
OH can deprotonate any acid with a pKa < 15.7.
b. H2S
a. HCOOH
pKa = 3.8
stronger acid
deprotonated
pKa = 7.0
stronger acid
deprotonated
c.
d. CH3NH2
CH3
pKa = 41
weaker acid
pKa = 40
weaker acid
These acids are too weak to be
deprotonated by −OH.
2.48
Draw the products and then compare the pKa of the acid on the left and the conjugate acid on
the right. The equilibrium lies towards the side having the acid with a higher pKa
(weaker acid).
O
a.
O
+
CF3 C
OCH2CH3
+ HOCH2CH3
pKa = 16
CF3 C
O
O H
products favored
pKa = 0.2
O
b.
O
CH3CH2 C
+
O H
Na+ Cl
+
CH3CH2 C
Na+
O
HCl
pKa = –7
starting material favored
pKa = ~5
c.
(CH3)3COH +
H−OSO3H
+
(CH3)3COH2
pKa = –9
pKa = ~ –3
+ Na+ HCO
3
+
H2CO3
pKa = 6.4
pKa = 10
e.
H C C H
+
Li+ CH2CH3
H C C
products favored
starting material favored
Li+ + CH3CH3
pKa = 50
pKa = 25
f.
HSO4
O Na+
O H
d.
+
CH3NH2 + H−OSO3H
pKa = –9
+
CH3NH3 + HSO4
pKa = 10.7
products favored
products favored
52
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–18
2.49
Compare element effects first and then resonance, hybridization, and electron-withdrawing
groups to determine the relative strengths of the acids.
a. Acidity increases across a row:
NH3 < H2O < HF
f. increasing acidity: H2O < H2S < HCl
Compare HCl and SH bonds first:
acidity increases across a row.
H–Cl is more acidic.
b. Acidity increases down a column:
HF < HCl < HBr
Compare OH and SH bonds:
acidity increases down a column.
SH is more acidic.
c. increasing acidity: −OH < H2O < H3O +
g. CH3CH2CH3, ClCH2CH2OH, CH3CH2OH
d. increasing acidity: NH3 < H2O < H2S
O–H bond and
O–H bond
electron-withdrawing Cl
strongest acid
increasing acidity: CH3CH2CH3 < CH3CH2OH < ClCH2CH2OH
only C–H bonds
weakest acid
Compare NH and OH bonds first:
acidity increases across a row.
OH is more acidic.
Then compare OH and SH bonds:
acidity increases down a column.
SH is more acidic.
e. Acidity increases across a row:
CH3CH3 < CH3NH2 < CH3OH
2.50
h. HC CCH2CH3
CH3CH2CH2CH3
CH3C CCH3
H H
sp C–H
strongest acid
sp3 C–H
sp2 C–H
all
weakest acid
increasing acidity: CH3CH2CH2CH3 < CH3CH=CHCH3 < HC CCH2CH3
The strongest acid has the weakest conjugate base.
a. Draw the conjugate acid.
Increasing acidity of conjugate acids:
CH3CH3 < CH3NH2 < CH3OH
d. Draw the conjugate acid.
Increasing acidity of conjugate acids:
increasing basicity: CH3O < CH3NH < CH3CH2
b.
CH2CH3 <
Draw the conjugate acid.
Increasing acidity of conjugate acids:
CH4 < H2O < HBr
CH CH2 <
C CH
increasing basicity:
C C
increasing basicity: Br < HO < CH3
<
CH CH <
CH2CH2
c. Draw the conjugate acid.
Increasing acidity of conjugate acids:
CH3CH2OH < CH3COOH < ClCH2COOH
increasing basicity: ClCH2COO < CH3COO < CH3CH2O
2.51
More electronegative atoms stabilize the conjugate base by an electron-withdrawing
inductive effect, making the acid stronger. Thus, an O atom increases the acidity of an acid.
NH2
pKa = 11.1
2.52
O
NH2
The O atom makes this cation the stronger acid.
pKa = 8.33
In both molecules the OH proton is the most acidic H. In addition, compare the percent
s-character of the carbon atoms in each molecule. Nearby C’s with a higher percent
s-character can help to stabilize the conjugate base.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
53
Acids and Bases 2–19
HC CCO2H
CH3CH2CO2H
pKa = 1.8
The sp hybridized C's of the triple bond have a higher percent
s-character than an sp3 hybridized C, so they pull electron
density towards them, stabilizing the conjugate base.
stronger acid
pKa = 4.9
2.53
CH3
CH3CH2CH2 H
O
CH2 C
CH3
CH2 H
pKa = 50
pKa = 43
C
The negative charge on O
is good. This makes this
resonance structure
especially good.
CH2 H
pKa = 19.2
conjugate base:
CH3
CH2 C
CH3CH2CH2
CH3
CH2
one Lewis structure
weakest acid
2.54
CH3
CH2
two resonance structures
negative charge delocalized
on two carbons
C
O
CH2
CH3
C
CH2
two resonance structures
negative charge delocalized
on one O and one C
strongest acid
To draw the conjugate acid, look for the most basic site and protonate it. To draw the
conjugate base, look for the most acidic site and remove a proton.
NH3
OH
conjugate acid
2.55
O
CH2 C
NH2
OH
A
most basic site
NH2
O
conjugate base
most acidic proton
Estimate the pKa of B as 16. A difference of 105 in acidity is a difference of 5 pKa units.
O
a.
HO
b.
O
c.
O
O
most acidic H
pKa ~ 5
most acidic
pKa ~ 25
OH
most acidic
pKa ~ 16
54
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–20
Remove the most acidic proton to form the conjugate base. Protonate the most basic electron
pair to form the conjugate acid.
2.56
only O–H bond
most acidic proton
a.
N
most basic electron pair
b.
O
OCH3
Increasing basicity:
COOH
O
O
N
ibuprofen
O
cocaine
conjugate base:
conjugate acid:
H N
O
OCH3
COO
O
O
Compare the isomers.
2.57
dimethyl ether
CH3
O
CH3CH2OH
CH3
All H's are on C.
H
B
N
H
O
O
H
N
OH
less acidic proton
N
O
N
H
O
N
OH
B
O
N
OH
H
H
H
N
O
N
OH
N
O
O
H
more acidic proton
O
H
O
N
OH
H
One O–H bond
O–H bonds are more acidic
than C–H bonds.
more acidic
Compare the Lewis structures of the conjugate bases when each H is removed. The more
stable base makes the proton more acidic.
2.58
H
ethanol
no resonance stabilization
formed from the weaker acid
H
The negative charge on N is
stabilized by resonance.
This conjugate base is more
stable, so it is formed by
removal of the more acidic H.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
55
Acids and Bases 2–21
2.59
Draw the conjugate base to determine the most acidic hydrogen.
O
O
O
O
H H
O
H
H
resonance-stabilized conjugate base
Resonance stabilization of the conjugate base
makes this the most acidic proton.
In Appendix A, the closest compound for
comparison is CH3CO2CH2CH3 with a pKa of
24.5; therefore, the estimated pKa of ethyl
butanoate is 25.
2.60
O
Look at the element bonded to the acidic H and decide its acidity based on the periodic
trends. Farther to the right across a row and down a column of the periodic table is
more acidic.
F
a.
CO2H
most acidic
O
c.
O
O
The molecule contains C–H and O–H
bonds. O is farther right in the
periodic table; therefore, the O–H
hydrogen is the most acidic.
2.61
OH
b.
NH
O
H
N
most acidic
CH3O
most acidic
The molecule contains C–H and N–H
bonds. N is farther right in the
periodic table; therefore, the N–H
hydrogen is the most acidic.
The molecule contains C–H, N–H,
and O–H bonds. O is farthest right in
the periodic table; therefore, the O–H
hydrogen is the most acidic.
Use element effects, inductive effects, and resonance to determine which protons are the
most acidic. The H’s of the CH3 group are least acidic since they are bonded to an sp3
hybridized C and the conjugate base formed by their removal is not resonance stabilized.
O
lactic acid
OH
H OH
c
a
b
O
conjugate base
by loss of (c):
c>b>a
Both O–H protons [(b) and (c)] are more acidic than the C–H proton
(a) by the element effect. The most acidic proton has added
resonance stabilization when it is removed, making its conjugate
base the most stable.
O
O
H OH
O
H OH
resonance stabilization
negative charge on O in both resonance structures
This makes (c) most acidic.
O
conjugate base
by loss of (b): H O
OH
O
conjugate base
by loss of (a):
no resonance stabilization, but
negative charge on O, an electronegative atom
O
OH
OH
OH
OH
This conjugate base has two
resonance structures, but one
places a negative charge on C.
56
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–22
Lewis bases are electron pair donors: they contain a lone pair or a π bond. Brønsted–
Lowry bases are proton acceptors: to accept a proton they need a lone pair or a π bond.
This means Lewis bases are also Brønsted–Lowry bases.
2.62
lone pairs on O
both
O
a.
H
C
CH3 Cl
b.
d.
c.
H
Lewis acid is an electron pair acceptor and usually contains a proton or an unfilled
valence shell of electrons. A Brønsted–Lowry acid is a proton donor and must contain a
hydrogen atom. All Brønsted–Lowry acids are Lewis acids, though the reverse may not be
true.
2.63
a. H3O+
b.
both
contains a H
Cl3C+
Cl
+ BCl3
Cl
Lewis base
CH3
Cl
Lewis acid
CH3
CH3
C
CH3
CH3
+
C
CH3
H OSO3H
Lewis acid
C
O
Cl
+ OH
CH3 C Cl
Lewis base
OH
new bond
Lewis acid
CH3
C C H
CH3
CH3
Lewis base
2.65
neither
no H or unfilled
valence shell
O
c.
Cl B Cl
new bond
b.
d. BF4
c. BCl3
Lewis acid
unfilled valence
shell on B
Lewis acid
unfilled valence
shell on C
Label the Lewis acid and Lewis base and then draw the products.
2.64
a.
π bonds
both
neither = no lone pairs
or π bond
lone pairs on Cl
both
+
HSO4
new bond
A Lewis acid is also called an electrophile. When a Lewis base reacts with an electrophile
other than a proton, it is called a nucleophile. Label the electrophile and nucleophile in the
starting materials and then draw the products.
a.
CH3CH2OH
+
BF3
nucleophile electrophile
BF3
CH3CH2 O H
OH2
+ H2O
d.
electrophile
b.
CH3SCH3 + AlCl3
nucleophile
c.
CH3
C O
CH3
nucleophile
AlCl3
CH3 S CH3
electrophile
C O BF3
CH3
electrophile
Br
e. Br Br
CH3
+ BF3
nucleophile
nucleophile
+
FeBr3
electrophile
Br Br Fe Br
Br
57
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Acids and Bases 2–23
2.66
Draw the product of each reaction.
CH2CH3
CH2CH3
+ H2O
a. CH3CH2 C
CH2CH3
CH2CH3
CH2CH3
CH3CH2 C
CH2CH3
+ NH3
CH2CH3
CH2CH3
+ (CH3)2NH
CH3CH2 CH3
e. CH3CH2 C
OH
CH2CH3
CH2CH3
CH3CH2 C
CH2CH3
NHCH3
CH2CH3
CH3CH2 CH3
+ (CH3)2O
c. CH3CH2 C
CH2CH3
CH3CH2 C NH3
CH2CH3
CH3CH2 CH3
+ CH3OH
b. CH3CH2 C
CH2CH3
d. CH3CH2 C
CH3CH2 C OH2
CH3CH2 C
CH2CH3
O CH3
CH2CH3
2.67
nucleophile
OH
proton transfer
H Br
a.
OH2
Br
+ Br
+ H2 O
electrophile
Br
b.
H Br
proton transfer
+ Br
electrophile
nucleophile
H
H
Br
+ Br
c.
H
Br
+ Br
proton transfer
+ HBr
H
H
nucleophile electrophile
2.68
Draw the products of each reaction. In part (a), –OH pulls off a proton and thus acts as a
Brønsted–Lowry base. In part (b), –OH attacks a carbon and thus acts as a Lewis base.
a.
CH2
OH
2.69
+
C(CH3)2
H2O
b. OH + (CH3)3C+
+ CH2=C(CH3)2
(CH3)3COH
H
Answer each question about esmolol.
most acidic
OH
a.
O
Esmolol contains C–H, N–H, and O–H bonds.
Since acidity increases across a row of the
periodic table, the OH bond is most acidic.
H
N
CH3O
O
esmolol
O H
b.
O
CH3O
H
N
Na+ O
O
Na+ H
+ H2
CH3O
O
H
N
O
58
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–24
O H
c.
O
OH H H
H
N
O
H Cl
CH3O
N
+ Cl–
CH3O
O
O
OH
d, e, f.
O
*
*
CH3O
*
*
H
N
All sp2 C's are indicated with an arrow.
The N is the only trigonal pyramidal atom.
The δ+ C's are indicated with a (*).
*
*
*
O
2.70
H
H
N
H
N
H H
N
H
N
H H
P
Q
pKa = 1.2
much lower pKa
much stronger acid
weaker base
pKa = 11.1
The four CH3 groups near the electron pair make it difficult
to donate that electron pair to a proton. This makes the
conjugate base of Q much weaker, and Q a stronger acid.
2.71
N
[1]
N
N
H
H
or
N
[2]
H
N
H
H
N
N
N
more basic N
2.72
no additional stabilization
N
N
resonance-stabilized cation
Path [2] is favored because a resonance-stabilized conjugate acid is formed. The N that is
part of the C=N is therefore more basic.
Draw the product of protonation of either O or N and compare the conjugate acids. When
acetamide reacts with an acid, the O atom is protonated because it results in a resonancestabilized conjugate acid.
O
protonate O
O
CH3
C
NH2
acetamide
CH3
C
H
NH2
O
CH3
C
H
NH2
resonance stabilization of the + charge
O is more readily protonated
because the product is
resonance stabilized.
O
protonate N
CH3
C
no other resonance structure
NH3
59
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Acids and Bases 2–25
2.73
O
O
HO
O
OH
pKa = 2.86
HO
O
O
O
δ+
O
O
pKa = 5.70
O
O
O
O
O
δ+
+
δ stabilizes the (−) charge
of the conjugate base.
COO− now acts as an electron-donor group
which destabilizes the conjugate base,
making removal of the second proton more
difficult and thus it is less acidic than CH3COOH.
The nearby COOH group serves
as an electron-withdrawing group to
stabilize the negative charge. This
makes the first proton more acidic
than CH3COOH.
The COOH group of glycine gives up a proton to the basic NH2 group to form the zwitterion.
O
a.
acts as a base
proton transfer
NH2CH2 C
O
O
zwitterion form
H Cl
O
NH3CH2 C
b.
O
NH3CH2 C
acts as an acid
OH
glycine
+ Cl −
NH3CH2 C
O
OH
most basic site
O
c.
H
NH2
Na+
CH2 C
OH
O
NH2CH2 C
+ Na+ + H2O
O
O
most acidic site
2.75
O
This group destabilizes the
second negative charge.
O
HO
2.74
O
Use curved arrows to show how the reaction occurs.
O
[1]
O
H
O
H
H
OH
O
O
[2]
+
H O H
OH
H
Protonate the negative charge on this carbon to form the product.
O
60
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 2–26
2.76
Compare the OH bonds in vitamin C and decide which one is the most acidic.
OH
O
HO
O
vitamin C
ascorbic acid
This is the most acidic proton
since the conjugate base is
most resonance stabilized.
HO
OH
loss of H+
OH
OH
O
HO
O
O
OH
O
O
HO
O
OH
O
O
OH
The most delocalized anion
with 3 resonance structures.
Removal of either
of these H's does not
give a resonancestabilized anion.
OH
OH
O
HO
OH
O
HO
O
HO
OH
loss of H+ HO
OH
O
HO
O
O
HO
O
HO
O
O
only 2 resonance structures
This proton is less acidic since its conjugate base is less
resonance stabilized.
2.77
OH
– H+
O
M
OH
O
O
– H+
OCH3
N
OCH3
OCH3
only a minor contributor to
the hybrid because of
charges
less resonance stabilized than N
OH
OCH3
N has two resonance structures with the same number of bonds and charges, so both contribute
approximately equally to the hybrid. This makes N more resonance stabilized than its conjugate
base, and less willing to give up a proton than M, which has no similar resonance stabilization.
Thus M is a stronger acid than N. (Resonance structures that break the C=O bond are not drawn
in this solution, since they are possible for each compound.)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
61
Introduction to Organic Molecules and Functional Groups 3–1
Chapter 3 Introduction to Organic Molecules and Functional Groups
Chapter Review
Increasing strength
Types of intermolecular forces (3.3)
Type of
force
van der
Waals
(VDW)
dipole–dipole
(DD)
hydrogen
bonding (HB
or H-bonding)
ion–ion
Cause
Examples
Due to the interaction of temporary dipoles
• Larger surface area, stronger forces
• Larger, more polarizable atoms, stronger
forces
Due to the interaction of permanent dipoles
All organic
compounds
Due to the electrostatic interaction of a H atom
in an O–H, N–H, or H–F bond with another N,
O, or F atom.
H2O
Due to the interaction of two ions
NaCl, LiF
(CH3)2C=O, H2O
Physical properties
Property
Boiling
point
(3.4A)
•
Observation
For compounds of comparable molecular weight, the stronger the forces the
higher the bp.
CH3CH2CH2CH2CH3
VDW
MW = 72
bp = 36 oC
CH3CH2CH2CHO
VDW, DD
MW = 72
bp = 76 oC
CH3CH2CH2CH2OH
VDW, DD, HB
MW = 74
bp = 118 oC
Increasing strength of intermolecular forces
Increasing boiling point
•
For compounds with similar functional groups, the larger the surface area, the
higher the bp.
CH3CH2CH2CH3
bp = 0 oC
CH3CH2CH2CH2CH3
bp = 36 oC
Increasing surface area
Increasing boiling point
•
For compounds with similar functional groups, the more polarizable the atoms,
the higher the bp.
CH3F
bp = −78 oC
CH3I
bp = 42 oC
Increasing polarizability
Increasing boiling point
62
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–2
Property
Melting
point
(3.4B)
•
Observation
For compounds of comparable molecular weight, the stronger the forces the
higher the mp.
CH3CH2CH2CH2CH3
CH3CH2CH2CHO
VDW, DD
MW = 72
mp = −96 oC
VDW
MW = 72
mp = −130 oC
CH3CH2CH2CH2OH
VDW, DD, HB
MW = 74
mp = −90 oC
Increasing strength of intermolecular forces
Increasing melting point
•
For compounds with similar functional groups, the more symmetrical the
compound, the higher the mp.
CH3CH2CH(CH3)2
mp = −160
(CH3)4C
mp = −17 oC
oC
Increasing symmetry
Increasing melting point
Solubility
(3.4C)
Types of water-soluble compounds:
• Ionic compounds
• Organic compounds having ≤ 5 C’s, and an O or N atom for hydrogen bonding
(for a compound with one functional group).
Types of compounds soluble in organic solvents:
• Organic compounds regardless of size or functional group.
• Examples:
CCl4
soluble
CCl4
soluble
H2O
soluble
O
CH3CH2CH2CH3
butane
CH3
H2O
insoluble
C
CH3
acetone
Key: VDW = van der Waals, DD = dipole–dipole, HB = hydrogen bonding
MW = molecular weight
Reactivity (3.8)
•
•
•
Nucleophiles react with electrophiles.
Electronegative heteroatoms create electrophilic carbon atoms, which tend to react with nucleophiles.
Lone pairs and π bonds are nucleophilic sites that tend to react with electrophiles.
CH3CH2 Cl
δ+
electrophilic site
OH
δ+
CH3
CH3 O CH3
CH3 N CH3
basic and nucleophilic site
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
63
Introduction to Organic Molecules and Functional Groups 3–3
Practice Test on Chapter Review
1. a. Which of the following compounds exhibits dipole–dipole interactions?
H
1.
Cl
C
4. Compounds (1) and (2) both exhibit dipoledipole interactions.
Cl
H
5. Compounds (1), (2), and (3) all exhibit dipoledipole interactions.
2.
CH3
3.
CH3CH2CH2
O
CH3
OH
b. Which of the following compounds can hydrogen bond to both another molecule like itself and
to water?
O
1.
4. Compounds (1) and (2) can each hydrogen bond to
another molecule like itself and water.
CH2CH2OCH3
O
2.
3.
NH
5. Each of the three compounds, (1), (2), and (3), can
hydrogen bond to another molecule like itself and to water.
CH3C N
c. Which of the following compounds has the highest boiling point?
1. CH3CH2CH2OCH2CH3
2. CH3(CH2)4CH3
3. CH3(CH2)4NH2
4. (CH3)2CHCH(CH3)NH2
5. (CH3)2CHCH(CH3)2
d. Which statement(s) is (are) true about the following compounds?
A
1. The boiling point of A is higher than the
boiling point of B.
2. The melting point of B is higher than the
melting point of C.
3. The boiling point of A is higher than the
boiling point of D.
4. Statements (1) and (2) are both true.
5. Statements (1), (2), and (3) are all true.
C
D
B
2.
Consider the anti-hypertensive agent atenolol drawn below.
E
H2N
A
Ha
C
O
F
N
O
B
D
O
Hb
Hc
64
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–4
a.
b.
c.
d.
e.
f.
What is the hybridization of the N atom labeled with A?
What is the shape around the C atom labeled with F?
What orbitals are used to form the bond labeled with E?
Which bond (B, C, or D) is the longest bond?
Would you predict atenolol to be soluble in water?
Which of the labeled H atoms (Ha, Hb, or Hc) is most acidic?
Answers to Practice Test
2. a. sp3
b. trigonal planar
c. Csp2–Csp3
d. D
e. yes
f. Hb
1. a. 5
b. 2
c. 3
d. 5
Answers to Problems
3.1
CH3CH2 OH
CH3CH2 O H
3.2
H OSO3H
Na+ H
CH3CH2 OH2
+ HSO4
H2SO4
CH3CH3
CH3CH2 O Na+ + H2
NaH
CH3CH3
no reaction
Identify the functional groups based on Tables 3.1, 3.2, and 3.3.
alkene
(double bond)
HO
alkene
(double bond)
CO2H
ether
O
CO2CH2CH3
O
HO
carboxylic acid
N
H
OH
alcohols
(hydroxy groups)
3.3
no reaction
shikimic acid
ester
NH2
oseltamivir
amide
amine
One possible structure for each functional group:
a. aldehyde = R
b. ketone =
R
O
O
O
C
C
C
H
CH3CH2CH2
H
c. carboxylic acid =
O
O
O
C
C
C
R
CH3
CH2CH3
d. ester =
R
R
O
CH3CH2CH2
OH
O
O
R
CH3CH2
C
O
CH3
C
OH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
65
Introduction to Organic Molecules and Functional Groups 3–5
3.4
One possible structure for each description:
O
a. C5H10O CH3CH2CH2CH2
C
O
H
CH3CH2CH2
aldehyde
ketone
b. C6H10O
CH3CH2
O
H
C
C
C
H
3.5
C
CH3
ketone
ketone
CH3
CH3CH2
O
C
H
C
C
H H
alkene
H
C
H
alkene
Summary of forces:
• All compounds exhibit van der Waals forces (VDW).
• Polar molecules have dipole–dipole forces (DD).
• Hydrogen bonding (H-bonding) can occur only when a H is bonded to an O, N, or F.
a.
c.
• only nonpolar C–C
and C–H bonds
• VDW only
e. CH3CH2CH2COOH
(CH3CH2)3N
• VDW forces
• polar C–N bonds – DD
• no H on N so
no H-bonding
• VDW forces
• polar C–O bonds
and a net dipole – DD
• H bonded to O –
H-bonding
O
b.
d.
• VDW forces
• 2 polar C–O bonds
and a net dipole – DD
• no H on O so
no H-bonding
3.6
f.
CH2 CHCl
• VDW forces
• polar C–Cl bond – DD
CH3 C C CH3
• only nonpolar C–H and
C–C bonds
• VDW only
One principle governs boiling point:
• Stronger intermolecular forces = higher bp.
Increasing intermolecular forces: van der Waals < dipole–dipole < hydrogen bonding
Two factors affect the strength of van der Waals forces, and thus affect bp:
• Increasing surface area = increasing bp.
Longer molecules have a larger surface area. Any branching decreases the surface area
of a molecule.
• Increasing polarizability = increasing bp.
a. (CH3)2C=CH2 or (CH3)2C=O
only VDW
VDW and DD
polar, stronger intermolecular forces
higher boiling point
b. CH3CH2COOH or CH3COOCH3
no H-bonding
VDW, DD, and H-bonding
stronger intermolecular forces
higher boiling point
c. CH3(CH2)4CH3 or CH3(CH2)5CH3
longer molecule, more surface area
higher boiling point
d.
CH2 CHCl
or
CH2 CHI
I is more polarizable.
higher boiling point
66
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–6
3.7
Increasing intermolecular forces: van der Waals < dipole–dipole < hydrogen bonding
O
C
CH3CH2
O
NH2
H
C
N
CH3
CH3
N–H bonds allow for hydrogen bonding.
stronger intermolecular forces
higher boiling point
no hydrogen bonding
weaker intermolecular forces
3.8
a.
or
b.
NH2
more polar
stronger intermolecular forces
(H-bonding)
higher mp
3.9
or
more spherical
packs better
higher mp
Compare the intermolecular forces to explain why sodium acetate has a higher melting point
than acetic acid.
O
CH3
C
O
OH
CH3
acetic acid
C
O– Na+
sodium acetate
a. VDW, DD, and H-bonding
b. not ionic, lower melting point
a. VDW, DD, ionic bonds
b. Ionic bonds are the strongest: higher melting point.
3.10 A compound is water soluble if it is ionic or if it has an O or N atom and ≤ 5 C’s.
a. CH3CH2OCH2CH3
b. CH3CH2CH2CH2CH3
an O atom that
can H-bond with water
≤ 5 C's
water soluble
c. (CH3CH2CH2CH2)3N
nonpolar
not water soluble
an N atom that can
H-bond to H2O, but
> 5 C's
not water soluble
3.11 Hydrophobic portions will primarily be hydrocarbon chains. Hydrophilic portions will be polar.
Circled regions are hydrophilic because they are polar.
All other regions are hydrophobic since they have only C and H.
OH
C C H
COOH
a.
b.
O
norethindrone
arachidonic acid
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
67
Introduction to Organic Molecules and Functional Groups 3–7
3.12 Like dissolves like.
• To be soluble in water, a molecule must be ionic, or have a polar functional group capable
of H-bonding for every 5 C’s.
• Organic compounds are generally soluble in organic solvents regardless of size or
functional group.
nonpolar
long hydrocarbon chain
polar
O
O
OH
a.
N
b.
polar
nonpolar
vitamin B3
(niacin)
soluble in water due to
two polar functional groups
and only 6 C's in the molecule
O
vitamin K1
(phylloquinone)
soluble in organic solvents
two polar C–O bonds but the
compound has > 10 C's
water insoluble
3.13 a.
OH
alcohol
HO
H
N
O amide
OH
O
carboxylic acid
b. The amide, carboxylic acid, and both alcohols can all hydrogen bond with water.
c. Since pantothenic acid has only nine carbons with four functional groups that can hydrogen
bond, pantothenic acid is a water-soluble vitamin.
3.14 A soap contains both a long hydrocarbon chain and a carboxylic acid salt.
ionic salt
a. CH3CO2–Na+
short chain
carboxylic acid salt
b. CH3(CH2)14CO2–Na+
long chain
This is a soap because it contains
both a long chain and a carboxylic
acid salt.
ionic salt
c. CH3(CH2)12COOH
no salt
d. CH3(CH2)9CO2–Na+
long chain
This is a soap because it contains
both a long chain and a carboxylic
acid salt.
3.15 Detergents have a polar head consisting of oppositely charged ions, and a nonpolar tail
consisting of C–C and C–H bonds, just like soaps do. Detergents clean by having the
hydrophobic ends of molecules surround grease, while the hydrophilic portion of the
molecule interacts with the polar solvent (usually water).
68
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–8
a detergent
+
SO3 Na
nonpolar tail
hydrophobic
This end interacts with
the grease to dissolve it.
polar head
ionic – hydrophilic
This end interacts with the water solvent
to maintain the micelle's solubility in water.
3.16
amide
ester
ester
ester
ether
ether
ester
H
O
ester
O
O
O
O
O
O
O
HN
O
ether
O
ether
H
N
O
O
O
O
O
O
N
O
O
O
O
ester amide
H
N
amide
O
O
O
O
O
O
O
N
H
O
N
O
H
O
ester
ester
ester
ester
amide
amide
valinomycin
nonactin
3.17 Because the interior of a cell membrane is nonpolar, aspirin crosses a cell membrane as a
neutral carboxylic acid, by the general rule “Like dissolves like.”
3.18 Electronegative heteroatoms like N, O, or X make a carbon atom an electrophile.
A lone pair on a heteroatom makes it basic and nucleophilic.
Pi (π) bonds create nucleophilic sites and are more easily broken than σ bonds.
nucleophilic
a.
Br
C bonded to Br
electrophilic
electrophilic
b. H O H
nucleophilic
nucleophilic
nucleophilic
c.
d.
N
CH3
electrophilic
amide
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
69
Introduction to Organic Molecules and Functional Groups 3–9
3.19 Electrophiles and nucleophiles react with each other.
O
a.
CH3CH2 Br
+
electrophile
b.
OH
YES
c.
CH3
+
nucleophile
Br
NO
d.
YES
nucleophile
CH3 C C CH3
nucleophile
OCH3
+
Cl
electrophile
nucleophile
CH3 C C CH3
C
nucleophile
+
Br
YES
electrophile
3.20
amide
a.
b, c.
O
H2N
OCH3
N
H
amine
H
aromatic ring
O
COOH
*
O
N
H *
*
N
H
O*
ester
*O
carboxylic acid
OCH3
*
O*
H
H's that are boxed in can hydrogen bond to O of H2O.
Atoms labeled with (*) can hydrogen bond to H of H2O.
3.21
*
a, c. CH3
*
CH
3
O
b.
*
O *
*
O
*
alkene
aromatic ring
HOCH2
HOCH2
CH3
CH2OH
Three OH's allow for H-bonding.
Stronger intermolecular forces mean
a higher bp and mp.
three ethers
Electrophilic carbons are labeled with (*).
3.22
a, c.
aldehyde
alkene
H
*C
O
The most electrophilic C is labeled with *.
b.
OH
This isomer has an OH, more opportunities
for H-bonding (through both O and H
atoms), and so probably more H2O soluble.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–10
3.23
Identify the functional groups based on Tables 3.1, 3.2, and 3.3.
amide
aromatic rings
aromatic ring
aromatic ring
a.
CH2
HO
c.
CH3CH2CO2 C
CH3 C H
e.
Darvon
CH3
N
CH3
O
COOH
penicillin G
ibuprofen
carboxylic acid
amine
amine
NH2
alkene
amine
N
CO2H
b.
sulfide
S
O
carboxylic acid
CH2N(CH3)2
ester
amide
H
N
O
d.
alkene
carboxylic acid
pregabalin
f.
OH
H
alkene
alkenes
O
O
alcohol
alkyne
O
ester
ketone
pyrethrin I
histrionicotoxin
3.24
CH3
OH
CH2OH
CH3 CH CH2 CH3
CH3CH2CH2CH2 OH
alcohol
alcohol
CH3 C OH
CH3 CH CH3
CH3
alcohol
alcohol
CH3
3.25
CH3CH2 O CH2CH3
CH3 O CH CH3
ether
ether
CH3 O CH2CH2CH3
ether
A cyclic ester is called a lactone. A cyclic amide is called a lactam.
O
a.
N CH3
amine
3.26
O
b.
c.
ether
d.
O
NH
O
ester
lactone
amide
lactam
Draw the constitutional isomers and identify the functional groups.
O
ketone
OH
carboxylic acid
H
O
OH
ester
O
aldehyde
O
ester
O
H
alcohol
O
OH
alcohol
O
aldehyde
O
H
O
aldehyde
OH
alcohol
H
O
ether
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
71
Introduction to Organic Molecules and Functional Groups 3–11
3.27
Use the rules from Answer 3.5.
O
O
OH
a.
OCH3
b.
c.
d.
N
VDW
VDW
VDW
no dipole–dipole
dipole–dipole
dipole–dipole
no H-bonding (no O–H bond) no H-bonding (no N–H bond) (nonpolar C–C, C–H bonds)
no H-bonding (no O, N, F)
VDW
dipole–dipole
H-bonding (O–H bond)
3.28
Increasing intermolecular forces: van der Waals < dipole–dipole < H-bonding
a. increasing intermolecular forces:
CH3CH3 < CH3Cl < CH3NH2
c. increasing intermolecular forces:
(CH3)2C=C(CH3)2 < (CH3)2CHCOCH3 < (CH3)2CHCOOH
VDW
VDW
dipole–dipole dipole–dipole
H-bonding
b. increasing intermolecular forces:
VDW
VDW
dipole–dipole
VDW
VDW
dipole–dipole
H-bonding
d. increasing intermolecular forces:
CH3Cl < CH3Br < CH3I
CH3Cl < CH3OH < NaCl
Increasing polarizability
stronger intermolecular forces
VDW
VDW
ionic
dipole–dipole dipole–dipole
H-bonding
3.29
O
CH3
H O
hydrogen bonding between
two acetic acid molecules
C CH3
C
O H
O
3.30 A = VDW forces; B = H-bonding; C = ion–ion interactions; D = H-bonding; E = H-bonding;
F = VDW forces.
3.31
O
H
a.
OH
O
O
aldehyde
ketone
ether
alcohol
b. The alcohol is the highest boiling since it is the only isomer that can hydrogen bond with
another molecule like itself.
3.32
Use the principles from Answer 3.6.
a.
CH3(CH2)4
I
CH3(CH2)5
I
CH3(CH2)6
I
Increasing size, increasing surface area, increasing boiling point
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–12
b.
CH3CH2CH2CH3 < (CH3)3N < CH3CH2CH2NH2
VDW
VDW
dipole–dipole
VDW
dipole–dipole
H-bonding
Increasing boiling point
c.
(CH3)3COC(CH3)3 < CH3(CH2)3O(CH2)3CH3 < CH3(CH2)7OH
VDW
VDW
VDW
dipole–dipole
dipole–dipole
dipole–dipole
smaller surface area
larger surface area
H-bonding
highest bp
Increasing boiling point
d.
<
Br <
VDW
VDW
dipole–dipole
OH
VDW
dipole–dipole
H-bonding
OH
<
VDW
dipole–dipole
H-bonding
larger surface area
Increasing boiling point
<
e.
<
smallest surface area
most branching
largest surface area
Increasing boiling point
OH
O
f.
<
VDW
<
VDW
dipole–dipole
H-bonding
VDW
dipole–dipole
Increasing boiling point
3.33
In CH3CH2NHCH3, there is a N–H bond so the molecules exhibit intermolecular hydrogen
bonding, whereas in (CH3)3N the N is bonded only to C, so there is no hydrogen bonding.
The hydrogen bonding in CH3CH2NHCH3 makes it have much stronger intermolecular
forces than (CH3)3N. As intermolecular forces increase, the boiling point of a molecule of
the same molecular weight increases.
3.34
Stronger forces, higher mp.
CH3
O
CH(CH3)2
menthone
VDW
dipole–dipole
lower melting point
CH3
OH
CH(CH3)2
menthol
VDW
dipole–dipole
H-bonding
stronger forces
higher melting point
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
73
Introduction to Organic Molecules and Functional Groups 3–13
3.35
Stronger forces, higher mp. More symmetrical compounds, higher mp.
CH3
a. (CH3)3CH < (CH3)2C=O < (CH3)2CHOH
VDW
VDW
DD
c.
VDW
DD
H-bonding
VDW
NH2
<
VDW
DD
Increasing intermolecular forces
Increasing melting point
Cl
<
VDW
DD
H-bonding
Increasing intermolecular forces
b. CH3F < CH3Cl < CH3I
Increasing melting point
Increasing polarizability
Increasing melting point
3.36
−119 oC
not symmetrical
−118 oC
not symmetrical
In both compounds the CH3 group dangling
from the chain makes packing in the solid
difficult, so the mp is low.
3.37
−25 oC
most spherical
highest mp
−91 oC
symmetrical
higher mp
This molecule can pack somewhat better
since it has no CH3 group dangling
from the chain, so the mp is somewhat
higher. It also has the most surface area
and this increases VDW forces compared
to the first two compounds.
This compound
packs the best
since it is the
most spherical in
shape, increasing
its mp.
Boiling point is determined solely by the strength of the intermolecular forces. Since
benzene has a smaller size, it has less surface area and weaker VDW interactions and
therefore a lower boiling point than toluene. The increased melting point for benzene can be
explained by symmetry: benzene is much more symmetrical than toluene. More symmetrical
molecules can pack more tightly together, increasing their melting point. Symmetry has no
effect on boiling point.
benzene
bp = 80 oC
mp = 5 oC
very symmetrical
closer packing in solid form
higher mp
CH3
and
toluene
bp = 111 oC
mp = −93 oC
less symmetrical
lower mp
74
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–14
3.38
Increasing polarity = increasing water solubility.
Neither compound is very H2O soluble.
a.
CH3CH2CH2CH3 <
(CH3)3CH
<
CH3OCH2CH3 < CH3CH2CH2OH
VDW
VDW
DD
more spherical
(This nonpolar, hydrophobic
molecule is more compact,
making it more water soluble than its
straight-chain isomer, drawn to the left.)
VDW
Br
b.
polar
no H-bonding
3.39
OH
O
polar
H-bonding to H2O,
not itself
polar and
H-bonding
More opportunities
for H-bonding with its
O atom and its H on O.
Look for two things:
• To H-bond to another molecule like itself, the molecule must contain a H bonded to O, N, or F.
• To H-bond with water, a molecule needs only to contain an O, N, or F.
These molecules can H-bond with water. All of
these molecules have an O or N atom.
b. CH3NH2, c. CH3OCH3, d. (CH3CH2)3N,
e. CH3CH2CH2CONH2, g. CH3SOCH3,
h. CH3CH2COOCH3
Each of these molecules can H-bond to another
molecule like itself. Both compounds have
N–H bonds.
b. CH3NH2, e. CH3CH2CH2CONH2
3.40
VDW
DD
H-bonding
Draw the molecules in question and look at the intermolecular forces involved.
no H bonded to O
O
diethyl ether
OH
H bonded to O:
hydrogen bonding
1-butanol
VDW forces
VDW forces
dipole–dipole forces
dipole–dipole forces
H-bonding
• Both have ≤ 5 C's and an electronegative O atom, so they can H-bond to water,
making them soluble in water.
• Only 1-butanol can H-bond to another molecule life itself, and this increases its boiling point.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
75
Introduction to Organic Molecules and Functional Groups 3–15
3.41
Use the solubility rule from Answer 3.12.
OH
O
CH3
a.
O
N
N
c.
N
O
CH3
HO
O
OH
OH
sucrose
many polar bonds with O
11 O's and 12 C's
many opportunities for H-bonding with H2O
water soluble
many polar bonds with N and O atoms
many opportunities for H-bonding
water soluble
OH
OH
d.
b.
carotatoxin
1 polar functional group
but > 10 C's
not water soluble
mestranol
CH3O
2 polar functional groups
but > 10 C's
not water soluble
Water solubility is determined by polarity. Polar molecules are soluble in water, while
nonpolar molecules are soluble in organic solvents.
Arrows indicate polar functional groups.
CH3
CH2NH2
b.
HO
HOCH2
CH3
CH3
O
CH3
CH3
CH3
CH3
CH3
vitamin E
only 2 polar functional groups
many nonpolar C–C and C–H bonds (29 C's)
soluble in organic solvents
insoluble in H2O
3.43
OH
O
HO
CH3
a.
HO
N
caffeine
3.42
HO
OH
N
CH3
pyridoxine
vitamin B6
many polar bonds and few nonpolar bonds
soluble in H2O
It is also soluble in organic solvents since it
is organic, but is probably more soluble in H2O.
Compare the functional groups in the two components of sunscreen. Dioxybenzone will most
likely be washed off in water because it contains two hydroxy groups and is more water
soluble.
O
OH
O
CH3O
avobenzone
two ketones
one ether
C(CH3)3
O
OH
OCH3
dioxybenzone
two hydroxy groups
one ketone
one ether
more water soluble
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–16
3.44
Because of the O atoms, PEG is capable of hydrogen bonding with water, which makes PEG
water soluble and suitable for a product like shampoo. PVC cannot hydrogen bond to water,
so PVC is water insoluble, even though it has many polar bonds. Since PVC is water
insoluble, it can be used to transport and hold water.
H
O
H-bond
no H-bonding
H
O
O
O
O
Cl
poly(ethylene glycol)
PEG
water soluble
3.45
Cl
Cl
Cl
poly(vinyl chloride)
PVC
water insoluble
Molecules that dissolve in water are readily excreted from the body in urine, whereas less
polar molecules that dissolve in organic solvents are soluble in fatty tissue and are retained
for longer periods. Compare the solubility properties of THC and ethanol to determine why
drug screenings can detect THC and not ethanol weeks after introduction to the body.
CH3
OH
CH3CH2
CH3
O
CH3
(CH2)4CH3
OH
ethanol
tetrahydrocannabinol
THC
THC has relatively few polar
bonds compared to the number
of nonpolar bonds, making it
soluble in organic solvents
and therefore soluble in fatty tissue.
Ethanol has 1 O atom and
only 2 C's, making it
soluble in water.
Due to their solubilities, THC is retained much longer in the fatty tissue of the body, being
slowly excreted over many weeks, while ethanol is excreted rapidly in urine after ingestion.
3.46
Compare the intermolecular forces of crack and cocaine hydrochloride. Stronger
intermolecular forces increase both the boiling point and the water solubility.
CH3
ionic bond
N
COOCH3
H
CH3
N
Cl
COOCH3
O
O
O C
H
cocaine (crack)
neutral organic molecule
O C
H
cocaine hydrochloride
a salt
77
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Organic Molecules and Functional Groups 3–17
The molecules are identical except for the ionic bond in cocaine hydrochloride. Ionic forces
are extremely strong forces, and therefore the cocaine hydrochloride salt has a much higher
boiling point and is more water soluble. Since the salt is highly water soluble, it can be
injected directly into the bloodstream, where it dissolves. Crack is smoked because it can
dissolve in the organic tissues of the nasal passages and lungs.
3.47
A laundry detergent must have both a highly polar end of the molecule and a nonpolar end of
the molecule. The polar end will interact with water, while the nonpolar end surrounds the
grease or other organic material.
H O H
H O H
a.
OCH2CH2O
CH2CH2O H
polar
interacts with water
by H-bonding at all O atoms,
as well as H's bonded to O's.
nonpolar
interacts with organic material
H
H O H
H O H
O
CH2CH2O H
N CH2CH2O
b.
O
H
H
H O H
H
O
H
H O H
nonpolar
interacts with
organic material
polar
interacts with water
by H-bonding
at O and H atoms
3.48
O
a.
OH
O
O
O
H
N
HCl
O
N
H
OH
O
O
N
Cl–
N
H
O
b.
OH H H
O
O
H
N
N
H
five functional groups that have
many opportunities for H-bonding
water soluble
OH HH
O
O
N
H
ionic salt
more water soluble
c. Since the hydrochloride salt is ionic and therefore more water soluble, it is more readily
transported in the bloodstream.
N
Cl–
78
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–18
3.49
Use the rules from Answer 3.18.
nucleophilic
δ+
−
δ+ I δ
a.
nucleophilic
nucleophilic
c.
δ+
δ+
δ+
CH
OH
e.
3
O
electrophilic
electrophilic
electrophilic
O
b.
d.
CH2
f.
nucleophilic
All the C=C's are
nucleophilic.
C
CH3 δ+ Cl
nucleophilic
electrophilic
(All lone pairs on O and
Cl are nucleophilic.)
3.50
NO
+ Br
a.
d.
nucleophilic
nucleophilic
+
CH2Cl
b.
CN
YES
+
+
e.
nucleophilic
nucleophilic
OH
NO
nucleophilic
nucleophilic
H3O
YES
electrophilic
electrophilic
O
c.
CH3
C
CH3
+
CH3
YES
nucleophilic
electrophilic
3.51
More rigid cell membranes have phospholipids with fewer C=C’s. Each C=C introduces a
bend in the molecule, making the phospholipids pack less tightly. Phospholipids without
C=C’s can pack very tightly, making the membrane less fluid, and more rigid.
The double bonds introduce kinks in the
chain, making packing of the hydrocarbon
chains less efficient. This makes the cell
membrane formed from them more fluid.
O
CH2O
O
(CH3)3NCH2CH2
H C O
O P O CH2
O
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
79
Introduction to Organic Molecules and Functional Groups 3–19
3.52
amine
can H-bond
hydroxy group
can H-bond
NH2
HO
OH
HO
O
OH
O
O
O
Cl
O
HO
Cl
H
N
O
O
N
H
HN
HO
O
O
O
O
OH
OH
OH
O
H
N
N
H
H2N
C
O
most
acidic
proton HO
3.53
O
N
H
NHCH3
amide
can H-bond
a. Seven amide groups [regular
(unbolded) arrows]
b. OH groups bonded to sp3 C’s are
circled. OH groups bonded to sp2
C’s have a square.
c. Despite its size, vancomycin is
water soluble because it contains
many polar groups and many N and
O atoms that can H-bond to H2O.
d. The most acidic proton is labeled
(COOH group).
e. Four functional groups capable of
H-bonding are ROH, RCOOH,
amides, and amines.
vancomycin
Because the O atom in tetrahydrofuran is in a ring, the C atoms bonded to it are kept away
from the lone pairs on O. This allows the O atom to more readily hydrogen bond with water,
thus increasing its solubility in water.
O
Electron pairs are more exposed.
This facilitates H-bonding.
3.54
amide
O
a.
amine
N
N
e. An isomer that can hydrogen bond should
have a higher boiling point.
hydrogen bonding possible
O
N
aromatic ring
NH
aromatic ring
b, c, f, g.
* N
H
*
most basic atom
*
O
N
*
*
*
The N atom of the amide is less basic because
the lone pair is part of resonance with the C=O.
Atoms that can hydrogen bond to H2O are boxed in.
Electrophilic carbons are labeled with (*).
most acidic H, pKa ~ 25
All H's are bonded to C's. Removal of the H on the C adjacent to the C=O results in a
resonance-stabilized anion.
d. Fentanyl exhibits van der Waals and dipole–dipole interactions but no hydrogen bonding,
since there is no H bonded to O or N.
80
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 3–20
3.55
O
OH
=
C
CHO
O
OH
H
H
CHO
A
B
The OH and CHO groups are close enough that they
can intramolecularly H-bond to each other. Since the
two polar functional groups are involved in
intramolecular H-bonding, they are less available for
H-bonding to H2O. This makes A less H2O soluble than
B, whose two functional groups are both available for
H-bonding to the H2O solvent.
H
C
O
The OH and the CHO are too far
apart to intramolecularly H-bond
to each other, leaving more
opportunity to H-bond with
solvent.
3.56
a. melting point
HOOC
H
Fumaric acid has its two larger COOH groups on opposite
ends of the molecule, and in this way it can pack better
in a lattice than maleic acid, giving it a higher mp.
H
COOH
fumaric acid
C C
b. solubility
H
H
C C
HOOC δ+ δ+COOH
Maleic acid is more polar, giving it
greater H2O solubility. The bond
dipoles in fumaric acid cancel.
maleic acid
c. removal of the first proton (pKa1)
H
H
HOOC
C C
HOOC
H
C C
COOH
H
COOH
loss of 1 proton
loss of 1 proton
H
H
C C
O C
HOOC
C O
O
H
C C
O
H
COO
H
In maleic acid, intramolecular H-bonding
stabilizes the conjugate base after one H is
removed, making maleic acid more acidic
than fumaric acid.
d. removal of the second proton (pKa2)
H
Intramolecular H-bonding
is not possible here.
H
C C
O C
H
=
O
C O
O O
Now the dianion is held in close proximity
in maleic acid, and this destabilizes the conjugate
base. Thus, removing the second H in maleic
acid is harder, making it a weaker acid than
fumaric acid for removal of the second proton.
OOC
H
C C
H
COO
The two negative charges are
much farther apart. This makes the
dianion from fumaric acid more
stable and thus pKa2 is lower for
fumaric acid than maleic acid.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
81
Alkanes 4–1
Chapter 4 Alkanes
Chapter Review
General facts about alkanes (4.1–4.3)
•
•
•
•
Alkanes are composed of tetrahedral, sp3 hybridized C’s.
There are two types of alkanes: acyclic alkanes having molecular formula CnH2n + 2, and
cycloalkanes having molecular formula CnH2n.
Alkanes have only nonpolar C−C and C−H bonds and no functional group so they undergo
few reactions.
Alkanes are named with the suffix -ane.
Classifying C’s and H’s (4.1A)
•
Carbon atoms are classified by the number of C’s bonded to them; a 1o C is bonded to one
other C, and so forth.
C C
C
C
C C
C C
C
C
1o C
•
2o C
CH3 CH3
C C C
CH3CH2 C
C
3o C
H
1o C
4o C
C CH3
CH3
4o C
3o C
o
2 C
Hydrogen atoms are classified by the type of carbon atom to which they are bonded; a 1o H is
bonded to a 1o C, and so forth.
H C C
C
C
H C C
H C C
1o H
C
1o H
o
2 H
H
CH3
o
3 H
2o H
Names of alkyl groups (4.4A)
CH3
=
methyl
CH3CH2
ethyl
CH3CH2CH2
isopropyl
=
CH3CH2CHCH3
=
sec-butyl
=
propyl
(CH3)2CH
CH3CH2CH2CH2
butyl
=
(CH3)2CHCH2
=
isobutyl
=
3o H
CH3CH2 C CH3
(CH3)3C
tert-butyl
=
82
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–2
Conformations in acyclic alkanes (4.9, 4.10)
•
Alkane conformations can be classified as staggered, eclipsed, anti, or gauche depending on the
relative orientation of the groups on adjacent carbons.
eclipsed
staggered
anti
HH
H
CH3
H
H
H
H
H
H
H
H
•
•
•
H
H
H
CH3
H
CH3
H
•
H
H
H
H
gauche
CH3
•
Dihedral angle of • Dihedral angle of
2 CH3’s = 180o
2 CH3’s = 60o
A staggered conformation is lower in energy than an eclipsed conformation.
An anti conformation is lower in energy than a gauche conformation.
Dihedral angle = 0o
Dihedral angle = 60o
Types of strain
•
•
•
Torsional strain—an increase in energy due to eclipsing interactions (4.9).
Steric strain—an increase in energy when atoms are forced too close to each other (4.10).
Angle strain—an increase in energy when tetrahedral bond angles deviate from 109.5o (4.11).
Two types of isomers
[1] Constitutional isomers—isomers that differ in the way the atoms are connected to each other (4.1A).
[2] Stereoisomers—isomers that differ only in the way atoms are oriented in space (4.13B).
cis
trans
CH3
CH3
CH3
constitutional
isomers
CH3
CH3
CH3
stereoisomers
Conformations in cyclohexane (4.12, 4.13)
•
•
Cyclohexane exists as two chair conformations in rapid equilibrium at room temperature.
Each carbon atom on a cyclohexane ring has one axial and one equatorial hydrogen. Ringflipping converts axial H’s to equatorial H’s, and vice versa.
An axial H flips equatorial.
Hax
Heq
Ring-flip.
Heq
Hax
An equatorial H flips axial.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
83
Alkanes 4–3
•
In substituted cyclohexanes, groups larger than hydrogen are more stable in the more roomy
equatorial position.
The larger CH3 group is equatorial.
H
CH3
H
CH3
Conformation 1
Conformation 2
more stable
95%
•
axial
5%
Disubstituted cyclohexanes with substituents on different atoms exist as two possible
stereoisomers.
• The cis isomer has two groups on the same side of the ring, either both up or both down.
• The trans isomer has two groups on opposite sides of the ring, one up and one down.
CH3
CH3
H
H
trans isomer
CH3
H
CH3
H
cis isomer
Oxidation–reduction reactions (4.14)
•
Oxidation results in an increase in the number of C−Z bonds or a decrease in the number of
C−H bonds.
O
CH3CH2 OH
ethanol
•
CH3
C
OH
Increase in C–O bonds = oxidation
acetic acid
Reduction results in a decrease in the number of C−Z bonds or an increase in the number of
C−H bonds.
H
H
C C
H
H H
H C C H
H
ethylene
H H
ethane
Increase in C–H bonds = reduction
84
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–4
Practice Test on Chapter Review
1. a. Which statement is true about compounds A–D below?
OH
HO
OH
OH
HO
A
OH
HO
B
1. A and C are stereoisomers.
2. B and D are identical.
3. A and B are stereoisomers.
HO
C
D
4. Statements (1) and (2) are both true.
5. Statements (1), (2), and (3) are all true.
b. Which of the following statements is true about cis-1-isopropyl-2-methylcyclohexane?
1. The more stable conformation has the isopropyl group in the equatorial position and the
methyl group in the axial position.
2. The more stable conformation has both the methyl and isopropyl groups in the equatorial
position.
3. cis-1-Isopropyl-2-methylcyclohexane is a meso compound.
4. Statements (1) and (2) are true.
5. Statements (1), (2), and (3) are all true.
c. Rank the following conformations in order of increasing energy.
CH3
H
CH3CH2
CH2CH3
H
CH3
CH2CH3
CH3
H
H CH3
CH3
CH2CH3
A
H
B
1. C < B < A
2. C < A < B
3. A < C < B
4. B < A < C
H
CH3CH2
CH3
CH2CH3
C
5. A < B < C
2. Give the IUPAC name for each of the following compounds.
a.
b.
3. How are the molecules in each pair related? Are they constitutional isomers, stereoisomers,
identical, or not isomers?
Cl
a.
and
CH3
Cl
CH3
85
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkanes 4–5
Cl
and
b.
Cl
Cl
Cl
CH3
c.
CH3
and
CH3
CH3
4. Rank the following conformations in order of increasing energy. Label the conformation of lowest
energy as 1, the highest energy as 4, and the conformations of intermediate energy as 2 and 3.
H
CH3
CH3CH2
CH3
A
CH2CH3
H
CH2CH3
CH3
CH3CH2
CH3
H
H
H
H
H
CH2CH3
CH3
B
CH3
CH2CH3
CH2CH3
CH3CH2
CH3
CH3
H
D
C
5. Consider the following disubstituted cyclohexane drawn below:
CH3
CH(CH3)2
a. Draw the more stable chair conformation for the cis isomer.
b. Draw the more stable chair conformation for the trans isomer.
Answers to Practice Test
1. a. 4
b. 1
c. 2
2. a. 5-isobutyl-2,6dimethyl-6propyldecane
b. 1-sec-butyl-4propylcyclooctane
3. a. identical
b. constitutional
isomers
c. stereoisomers
4. A–3
B–4
C–2
D–1
5. a.
CH3
CH(CH3)2
b.
CH3
H
CH(CH3)2
86
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–6
Answers to Problems
4.1 The general molecular formula for an acyclic alkane is CnH2n + 2.
Number of C atoms = n
23
25
27
2n + 2
2(23) + 2 =
2(25) + 2 =
2(27) + 2 =
Number of H atoms
48
52
56
4.2 2-Methylbutane has 4 C’s in a row with a 1 C branch.
H
H
a. CH3CH2 C CH3
b.
CH3
H
H C CH3
c. CH3CH2CH(CH3)2
CH3 C CH3
re-draw
H
2-methylbutane
d.
2-methylbutane
H C
H
C H
e.
f. CH3CH(CH3)CH2CH3
CH3 C H CH3
H
re-draw
2-methylbutane
5 C's in a row
pentane
2-methylbutane
2-methylbutane
4.3 To classify a carbon atom as 1°, 2°, 3°, or 4° determine how many carbon atoms it is bonded
to (1° C = bonded to one other C, 2° C = bonded to two other C’s, 3° C = bonded to three other
C’s, 4° C = bonded to four other C’s). Re-draw if necessary to see each carbon clearly.
To classify a hydrogen atom as 1°, 2°, or 3°, determine if it is bonded to a 1°, 2°, or 3° C
(a 1° H is bonded to a 1° C; a 2° H is bonded to a 2° C; a 3° H is bonded to a 3° C).
Re-draw if necessary.
1° C
a.
1° C's
1° C
[1] CH3CH2CH2CH3
[2] (CH3)3CH
[3]
[4]
4° C
CH3
1° C's
2° C's
1° C
CH3 C H
CH3 3° C
4° C's
All other C's are 1° C's.
All other C's are 2° C's.
3° C
re-draw
re-draw
1° H's
1° H's
HH
1° H's
CH3
b.
[1] CH3CH2CH2CH3
[2]
[3] CH C
3
CH3 C H
CH3
2° H's
CH3 CH3
1° H's
1° H's
3° H
C CH3
CH3 CH3
all 1° H's
[4]
H
H
CH3
CH3
H
H
CH3
H H
1° H's
H
3° H
All others are 2° H's.
87
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkanes 4–7
4.4 Use the definition of 1°, 2°, 3°, or 4° carbon atoms from Answer 4.3.
1°
4°
O
O
O
O
2°
OH
OH
C(CH3)3
O
4°
O
2° 2°
2°
3°
4° 1°
4.5 Constitutional isomers differ in the way the atoms are connected to each other. To draw all
the constitutional isomers:
[1] Draw all of the C’s in a long chain.
[2] Take off one C and use it as a substituent. (Don’t add it to the end carbon: this re-makes the
long chain.)
[3] Take off two C’s and use these as substituents, etc.
Five constitutional isomers of molecular formula C6H14:
[1] long chain
[2] with one C as a substituent
CH3
CH3CH2CH2CH2CH2CH3
[3] using two C's as substituents
CH3
CH3
CH3CH2CH2 C CH3 CH3CH2 C CH2CH3
H
H
CH3CH2 C CH3
H
H
CH3 C
CH3
C CH3
CH3 CH3
4.6 Draw each alkane to satisfy the requirements.
CH3
a.
b.
4° C
CH3
c. CH C CH CCH
3
2
3
1° C
1° C
All other C's are 2° C's.
H
1° H
H
3° H
2° H
4.7 Draw each compound as a skeletal structure to compare the compounds.
C3
C2
C3
CH3(CH2)3CH(CH3)2 =
CH3CH2CH(CH3)CH2CH2CH3 =
A
B
C
CH3 bonded to C3
identical to compound C
CH3 bonded to C2
CH3 bonded to C3
identical to compound A
4.8 Use the steps from Answer 4.5 to draw the constitutional isomers.
Five constitutional isomers of molecular formula C5H10 having one ring:
[1]
[2]
[3]
CH3
CH3
CH3
CH3
CH2CH3
CH3
88
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–8
4.9 Follow these steps to name an alkane:
[1] Name the parent chain by finding the longest C chain.
[2] Number the chain so that the first substituent gets the lower number. Then name and
number all substituents, giving like substituents a prefix (di, tri, etc.).
[3] Combine all parts, alphabetizing the substituents, ignoring all prefixes except iso.
4-tert-butyl
[1]
[2]
CH3
CH3 C CH3
a. CH3CH2CH2 C CH2CH2CH2CH3
CH3
4
8 carbons = octane
[1]
b.
[3] 4-tert-butyl-4-methyloctane
CH3
CH3 C CH3
CH3CH2CH2 C CH2CH2CH2CH3
CH3 6 7 8
1 2 3
H
H C CH3
H C CH2 CHCH3
CH3
CH3
4-methyl
5 H 6
[2]
[3] 2,4-dimethylhexane
H C CH3
1
H C CH2 CHCH3
4 CH3 2 CH3
6 carbons = hexane
4-methyl 2-methyl
6-isopropyl
[1]
H
[2]
CH3 C CH3
CH2CH3
CH3
CH3CH2CH2 C CH2 CH2 C CH3 CH3CH2CH2
H
H
9 8 7 6
c.
H
C CH3
C CH2 CH2
H 5
4
d.
[2]
CH3 CHCH2
CH3
5
CH2CH2CH3
1
C CH3
3
CH3 CHCH2
7
3-methyl
[3] 2,4-dimethylheptane
CH2CH2CH3
2 3
H
6
[3] 6-isopropyl-3-methylnonane
CH2CH3
C CH3
H
9 carbons = nonane
[1]
1
2
C CH3
H 4
CH3
4-methyl
7 carbons = heptane
2-methyl
4.10 Use the steps in Answer 4.9 to name each alkane.
a. (CH3)3CCH2CH(CH2CH3)2
re-draw
[1]
CH3
CH3 C CH2 CH CH2CH3
CH3
[2] 2
CH3
1
[3] 4-ethyl-2,2-dimethylhexane
4
5
6
CH3 C CH2 CH CH2CH3
CH2CH3
CH3
6 carbons = hexane
CH2CH3
4-ethyl
2,2-dimethyl
b. CH3(CH2)3CH(CH2CH2CH3)CH(CH3)2
re-draw
[1]
4-isopropyl [3] 4-isopropyloctane
[2]
CH3
CH3CH2CH2CH2 CH CH CH3
CH2CH2CH3
8 carbons = octane
CH3
CH3CH2CH2CH2 CH CH CH3
8
7
6
5
4
CH2CH2CH3
3
2
1
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
89
Alkanes 4–9
2-methyl
[2]
[1]
1
[3] 3-ethyl-2,5-dimethylheptane
34 5 6 7
c.
2
or
3-ethyl 5-methyl
longest chain = 7 carbons = heptane
Number so there are more substituents.
Pick the upper option.
2-methyl
[2] 1
[1]
d.
3
[3] 5-sec-butyl-3-ethyl-2,7-dimethyldecane
5
5-sec-butyl
2
6
3-ethyl
10 carbons = decane
8 9 10
7-methyl
4.11 To work backwards from a name to a structure:
[1] Find the parent name and draw that number of C’s. Use the suffix to identify the functional
group (-ane = alkane).
[2] Arbitrarily number the C’s in the chain. Add the substituents to the appropriate C’s.
[3] Re-draw with H’s to make C’s have four bonds.
a. 3-methylhexane
[1] 6 carbon alkane
[2]
[3]
CH3
C
C
C C C C C
methyl on C3
C C C C C
CH3
CH3CH2 CH CH2CH2CH3
b. 3,3-dimethylpentane
[1]
5 carbon alkane
[2]
methyl groups on C3
[3]
CH3
CH3
C C C C C
C
C
CH3CH2 C CH2CH3
C C C
CH3
CH3
c. 3,5,5-trimethyloctane
[1]
8 carbon alkane
[2]
C C C C C C C C
C C C C C C C C
methyl groups on C3 and C5
CH3
[3]
CH3
CH3
CH3
CH3CH2 CH CH2 C
CH3
CH2CH2CH3
CH3
d. 3-ethyl-4-methylhexane
[1]
[2]
6 carbon alkane
ethyl group on C3
CH2CH3
C
C C C C C
C C C C C C
CH3
[3]
CH2CH3
CH3CH2 CH CH CH2CH3
CH3
methyl group on C4
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–10
e. 3-ethyl-5-isobutylnonane
[1]
[2]
C C C C C C C C C
[3]
isobutyl group on C5
9 carbon alkane
CH3
CH3
CH2 CH CH3
CH2 CH CH3
C C C C C C C C C
CH3CH2 CH CH2 CH CH2CH2CH2CH3
CH2CH3
CH2CH3
ethyl group on C3
4.12 Use the steps in Answer 4.9 to name each alkane.
[1]
[3] hexane
[2]
H H H H H H
H C C C C C C H
no substituents, skip [2]
H H H H H H
6 carbons = hexane
2-methyl
[1]
[2]
H H H CH3 H
H C C C C
C H
H H H H
H
H H H CH3 H
H C C C C
5
1
[3] 2-methylpentane
1
[3] 3-methylpentane
C H
H H H H
H
5 carbons = pentane
3-methyl
[1]
H H CH3 H H
H C C C
H H H
[2] H H CH3 H H
C C H
H H
H C C C
5
C C H
H H H
H H
5 carbons = pentane
2,2-dimethyl
[1]
[2]
H H CH3 H
H C C C
C
H
H H CH3 H
4
1
H H CH3 H
H C C C
C
[3] 2,2-dimethylbutane
H
H H CH3 H
4 carbons = butane
[1]
H H
H
H
H C C
C
C H
H CH3 CH3 H
[2]
4
H H
H
H
H C C
C
C H
1
[3] 2,3-dimethylbutane
H CH3 CH3 H
4 carbons = butane
2,3-dimethyl
4.13 Follow these steps to name a cycloalkane:
[1] Name the parent cycloalkane by counting the C’s in the ring and adding cyclo[2] Numbering:
[2a] Number around the ring beginning at a substituent and giving the second
substituent the lower number.
[2b] Number to assign the lower number to the substituents alphabetically.
[2c] Name and number all substituents, giving like substituents a prefix (di, tri, etc.).
[3] Combine all parts, alphabetizing the substituents, ignoring all prefixes except iso.
(Remember: If a carbon chain has more C’s than the ring, the chain is the parent,
and the ring is a substituent.)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
91
Alkanes 4–11
1
2
CH3
[2] 3 C C C CH
3
[1]
a.
C
4
1,1-dimethyl
[3] 1,1-dimethylcyclohexane
C 6
C
5
6 carbons in ring =
cyclohexane
Number so the
substituents are at C1.
1,2,3-trimethyl
[1]
[2]
3
4C
b.
CH3
[3] 1,2,3-trimethylcyclopentane
C 2
C CH3
5C C
CH3
5 carbons in ring =
cyclopentane
[1]
1
Number so the first substituent
is at C1, second at C2.
1
2
3 C
C
[2] C
c.
C
C
CH3 4 C
6 carbons in ring =
cyclohexane
5
1-butyl
4-methyl
Number so the earlier alphabetical
substituent is at C1, butyl before methyl.
1
[1]
5C
d.
4C
C
C
[3] 1-sec-butyl-2-isopropylcyclohexane
C
C
3
[1]
1-sec-butyl
6
[2]
6 carbons in ring =
cyclohexane
2
2-isopropyl
Number so the earlier alphabetical
substituent is at C1, butyl before isopropyl.
[2]
e.
1
3
5
C
C
C
C
[3] 1-cyclopropylpentane
Number so the
cyclopropyl is at C1.
5
[2]
4
3
f.
6 carbons in ring =
cyclohexane
C
4
2
1-cyclopropyl
longest chain =
5 carbons =
pentane
[1]
[3] 1-butyl-4-methylcyclohexane
6
3-butyl
6
1
2
1,1-dimethyl
[3] 3-butyl-1,1-dimethylcyclohexane
Number so the two
methyls are at C1.
92
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–12
4.14 To draw the structures, use the steps in Answer 4.11.
a. 1,2-dimethylcyclobutane
[1] 4 carbon cycloalkane
[2]
CH3 1
C C 4
methyl groups
on C1 and C2
C C
C C
[3]
C C 3
CH3 2
CH3
CH3
b. 1,1,2-trimethylcyclopropane
[1] 3 carbon cycloalkane
CH3
[2]
C
C
3
C
C
[3] CH CH
3
2
CH3
C
C
ethyl 5 C 1 CH3
6
on C4
2 CH3's
C
[1] 5 carbon cycloalkane
CH3
CH3CH2 4 C 2 CH3
C
C
d. 1-sec-butyl-3-isopropylcyclopentane
C
CH3
3
c. 4-ethyl-1,2-dimethylcyclohexane
[1] 6 carbon cycloalkane
[2]
C
CH3
[3]
C C CH3
1 CH
3
C
C C
C
3 CH3's
2C
CH3
CH3
CH
[2]
4C
CH3
isopropyl
C3
[3]
C 2
5 C C1
CH CH3
C
C C
sec-butyl
CH3 CH2
e. 1,1,2,3,4-pentamethylcycloheptane
[1] 7 carbon cycloalkane
[2]
5 CH3's
CH3
CH3
C
C
C
C
C
C C
CH3
3 C4
C
2C
CH3
CH3
[3]
CH3
C5
CH3
1 C6
C C7
CH3
CH3
CH3
4.15 Compare the number of C’s and surface area to determine relative boiling points. Rules:
[1] Increasing number of C’s = increasing boiling point.
[2] Increasing surface area = increasing boiling point (branching decreases surface area).
8 C's
linear
largest number of C's
no branching
highest bp
7 C's
linear
7 C's
one branch
7 C's
three branches
increasing branching
decreasing surface area
decreasing bp
Increasing boiling point: (CH3)3CCH(CH3)2 < CH3CH2CH2CH2CH(CH3)2 < CH3(CH2)5CH3 < CH3(CH2)6CH3
4.16 To draw a Newman projection, visualize the carbons as one in front and one in back of each
other. The C−C bond is not drawn. There is only one staggered and one eclipsed
conformation.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkanes 4–13
Br
rotation here
H H
H C C Br
C in front
H H
H
H
H
H
60o
H
H
H Br
H
H
H
1
staggered
2
eclipsed
C behind
4.17
4.0 kJ/mol
H
H
H CH3
To calculate H,CH3 destabilization:
H
H,H eclipsing
4.0 kJ/mol of destabilization
H
4.0 kJ/mol
14 kJ/mol (total) −
8.0 kJ/mol for 2 H,H eclipsing interactions
= 6 kJ/mol for one H,CH3 eclipsing interaction
4.18 To determine the energy of conformations keep two things in mind:
[1] Staggered conformations are more stable than eclipsed conformations.
[2] Minimize steric interactions: keep large groups away from each other.
The highest energy conformation is the eclipsed conformation in which the two largest
groups are eclipsed. The lowest energy conformation is the staggered conformation in
which the two largest groups are anti.
rotation here
H
H
CH3
CH3
CH3 C CH2CH3
CH3
60o
CH3
H
CH3
H
CH3
H
H
H
1
staggered
most stable
H
60o
CH3
CH3
H
2
eclipsed
3
staggered
most stable
60o
H
H
6
eclipsed
least stable
CH3
60o
CH3
CH3
H
H
CH3
CH3
60o
H
CH3
CH3
CH3
60o
CH3
CH3
CH3
H
H
H
H
H
5
staggered
4
eclipsed
least stable
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–14
4.19 To determine the most and least stable conformations, use the rules from Answer 4.18.
Cl
a. 1,2-dichloroethane
H
H
H
60o
ClCH2 CH2Cl
H
H
H Cl
H
H
Cl
H
Cl
1
staggered, anti
2
eclipsed
rotation here
60o
H
H
Cl
H
Cl
3
staggered, gauche
60o
60o
H
Cl
H
H H
H
H
60o
H
Cl
60o
H
H
H
Cl
6
eclipsed
H H
Cl
Cl
Cl
5
staggered, gauche
4
eclipsed
highest energy
Cl groups eclipsed
least stable
b.
4
2
3
5
1
most stable
180o
Eclipsed forms are
higher in energy.
6
Energy
94
Staggered forms
are lower in energy.
1
most stable
Cl groups anti
60o 120o 180o
0o
120o 60o
Dihedral angle between 2 Cl's
4.20 Add the energy increase for each eclipsing interaction to determine the destabilization.
HH
CH3
H
H CH
3
H CH
3
b.
a.
HCH3
H
1 H,H interaction =
2 H,CH3 interactions
(2 x 6.0 kJ/mol) =
12.0 kJ/mol
Total destabilization =
16 kJ/mol
4.0 kJ/mol
CH3
3 H,CH3 interactions
(3 x 6.0 kJ/mol) = 18 kJ/mol
Total destabilization
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
95
Alkanes 4–15
4.21 Two points:
• Axial bonds point up or down, while equatorial bonds point out.
• An up carbon has an axial up bond, and a down carbon has an axial down bond.
equatorial
axial up
CH3
CH3 Br
H H
HO
H
H
axial up
equatorial
HO
H
Cl H
H
OH
Br
axial down
Cl
OH
Up carbons are dark circles.
Down carbons are clear circles.
equatorial
4.22 Draw the second chair conformation by flipping the ring.
• The up carbons become down carbons, and the axial bonds become equatorial
bonds.
• Axial bonds become equatorial, but up bonds stay up that is, an axial up bond
becomes an equatorial up bond.
• The conformation with larger groups equatorial is the more stable conformation and is
present in higher concentration at equilibrium.
axial
H eq
Br
a.
more stable
Br is equatorial.
Draw in the H
Draw second conformation.
Br
and label the C
as up or down.
Axial bond is up =
up carbon
eq
H
Up carbons switch
to down carbons.
Br
axial
Axial bond is down =
down carbon
axial
eq
Cl
b.
Cl
Draw in the H
and label the C
as up or down.
H
Draw second conformation.
Up carbons switch
to down carbons.
eq
Axial bond is up =
up carbon
Cl
axial
H
Axial bond is down =
down carbon
more stable
Cl is equatorial.
Axial bond is up =
up carbon
eq
H
c.
CH2CH3
more stable
CH2CH3 is equatorial.
Draw second conformation.
Draw in the H
CH2CH3
and label the C
as up or down.
eq
Up carbons switch
to down carbons.
H
CH2CH3
Axial bond is down =
down carbon
4.23 Larger axial substituents create unfavorable diaxial interactions, whereas equatorial groups
have more room and are favored.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–16
H
H H
H
C C H
CH2CH3
H H
larger substituent
more important to be equatorial
equatorial CH2CH3
H
H
H
C CH3
The H's and CH3 of the sp3 hybridized
C have severe 1,3-diaxial interactions
with the two other axial H's.
H
H
C C H
H
C CH
more compact substituent
less important to be equatorial
equatorial C CH
H
C
H
The sp hybridized C's are linear and
point down. The 1,3-diaxial
interactions with the two other axial
H's are less severe.
C
H
The axial conformation containing the C CH group is not as
unstable as the axial conformation containing the CH2CH3, so it
is present in higher concentration at equilibrium.
4.24 Wedges represent “up” groups in front of the page, and dashes are “down” groups in back of
the page. Cis groups are on the same side of the ring, and trans groups are on opposite sides of
the ring.
a.
cis-1,2-dimethylcyclopropane
CH3
CH3
or
CH3
b.
trans-1-ethyl-2-methylcyclopentane
or
CH3
CH3CH2
cis = same side of the ring
both groups on wedges or
both on dashes
CH3
CH3CH2
CH3
trans = opposite sides of the ring
one group on a wedge,
one group on a dash
4.25 Cis and trans isomers are stereoisomers.
cis-1,3-diethylcyclobutane
a. trans-1,3-diethylcyclobutane
cis = same side of the ring
both groups on wedges or
both on dashes
b. cis-1,2-diethylcyclobutane
trans = opposite sides of the ring
one group on a wedge,
one group on a dash
constitutional isomer
different arrangement of atoms
4.26 To classify a compound as a cis or trans isomer, classify each non-hydrogen group as up or
down. Groups on the same side = cis isomer; groups on opposite sides = trans isomer.
down bond (up)
(equatorial) HH(up)
HO
a.
HO
down bond
(equatorial)
both groups down =
cis isomer
down bond
(up) (equatorial)
H
up bond
(axial)
H
OH
OH
b.
Cl
Cl
up bond
H
(down) (equatorial)
one group up, one down =
trans isomer
H
H
(up)
Br
Cl
H
Cl
c. H
(down)
H
H
Br
Br
down bond
H
(equatorial)
one group up, one down =
trans isomer
Br
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
97
Alkanes 4–17
4.27
CH3
a.
trans: CH3
CH3
c.
CH3
CH3
groups on same side
cis isomer
CH3
H
H
CH3
H
CH3
both groups equatorial
more stable
two chair conformations for the trans isomer
CH3
H
CH3
b.
H
H
CH3
groups on opposite sides
trans isomer
(one possibility)
CH3
cis:
H
d. The trans isomer is more stable because
it can have both methyl groups in the more roomy
equatorial position.
H
two chair conformations for the cis isomer
Same stability since they both have
one equatorial, one axial CH3 group.
4.28
CH2CH3
up
axial CH2CH3
CH3CH2
H
CH3
a.
H
down
(equatorial)
CH3
c.
1,1-disubstituted
trans-1,3-disubstituted
CH3
up (axial)
H
H
CH2CH3
b.
up
CH3
d.
CH3CH2
up
(equatorial)
H
cis-1,2-disubstituted
down
H
trans-1,4-disubstituted
4.29 Oxidation results in an increase in the number of C−Z bonds, or a decrease in the number of C−H
bonds.
Reduction results in a decrease in the number of C−Z bonds, or an increase in the number of C−H
bonds.
a.
CH3
O
O
C
C
H
CH3
O
c.
OH
Decrease in the number of C–H bonds.
Increase in the number of C–O bonds.
Oxidation
CH3
C
CH3
HO OH
C
CH3
CH3
No change in the number of C–O
or C–H bonds. Neither
O
b.
CH3
C
CH3CH2CH3
CH3
d.
Decrease in the number of C–O bonds.
Increase in the number of C–H bonds.
Reduction
O
OH
Decrease in the number of C–O bonds.
Increase in the number of C–H bonds.
Reduction
4.30 The products of a combustion reaction of a hydrocarbon are always the same: CO2 and H2O.
a.
flame
CH3CH2CH3
+
5 O2
3 CO2 +
4 H2O + heat
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–18
b.
+
9 O2
flame
6 CO2 +
6 H2O + heat
4.31 “Like dissolves like.” Beeswax is a lipid, so it will be more soluble in nonpolar solvents.
H2O is very polar, ethanol is slightly less polar, and chloroform is least polar. Beeswax is
most soluble in the least polar solvent.
Increasing polarity
H2O
CH3CH2OH
CHCl3
Increasing solubility of beeswax
4.32 Re-draw the model as a skeletal structure. The longest chain has 15 C’s, making it a
derivative of pentadecane. Numbering from either direction gives the same numbers.
1
2
6
10
14
15 C's
pentadecane
4 methyl groups at C2, C6, C10, and C14
2,6,10,14-tetramethylpentadecane
4.33 Re-draw each alkane as a skeletal structure, and use the steps in Answer 4.9 to name each
compound. Use the definitions in Answer 4.3 to classify C’s.
3
4
a.
7 C's in longest chain = heptane
Number from left to right to put two
substituents on C3.
= 2° C
= 3° C
5
Four CH3 groups
tetramethyl
Answer: 3,3,4,5-tetramethylheptane
3
5
8 C's in longest chain = octane
Number from right to left to put first
substituent at C3.
ethyl at C3; 2 methyls (C5)
dimethyl
Answer: 3-ethyl-5,5-dimethyloctane
4.34 Re-draw the ball-and-stick model as a chair form.
equatorial
H
H
1
= 4° C
1
b.
a.
1° C's unlabeled
2
Br
CH3
axial
H
4
CH(CH3)2
equatorial
= 2° C
= 3° C
= 4° C
1° C's unlabeled
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkanes 4–19
b. CH3 on Cl and Br on C2 are both down, making them cis.
c. Br on C2 and CH(CH3)2 on C4 are both down, making them cis.
d.
Second chair form:
This chair form is less stable because two
groups [Br and CH(CH3)2] are in the more
crowded axial position.
H
CH3
1
H
4
2
Br
H
CH(CH3)2
4.35
b.
a.
c.
Br
Cl
H
Cl
Cl
H
H
H
H
H
H
H
CH3
H
H
Cl
H
CH3
4.36 Use the rules from Answer 4.3.
1°
4°
3°
CH3 CH3
2°
a.
[1]
2°
[2]
4°
C
b. [1] CH3
CH
CH3
CH3
3°
1°
CH2
CH2
2°
1°
All CH3's have 1° H's.
All CH2's have 2° H's.
All CH's have 3° H's.
[2]
CH2
CH3
C
CH2
CH
CH3
CH2
CH2
CH
CH3
CH3
4.37
One possibility:
CH3
a. CH3 C CH3
b.
d. (CH3)3CH
c. CH3CH2CH3
CH3
4.38
a. Five constitutional isomers of molecular formula C4H8:
CH2
CH3CH CHCH3
CH3
CH2 CHCH2CH3
CH3
C
CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 4–20
b. Nine constitutional isomers of molecular formula C7H16:
H
H
CH3 C CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH3
CH3CH2 C CH2CH2CH3
CH3
CH3
CH3
CH3
CH3 C CH2CH2CH3
H
CH3CH2 C CH2CH3
CH3
CH3
H
H
CH3CH2 C CH2CH3
CH3 C
CH2CH3
H
CH3 C
H
C CH2CH3
CH3 C
CH3 CH3
H
CH2 C CH3
CH3
CH3
CH3
C CH3
CH3 CH3
c. Twelve constitutional isomers of molecular formula C6H12 containing one ring:
4.39 Use the steps in Answers 4.9 and 4.13 to name the alkanes.
a. CH3CH2CHCH2CHCH2CH2CH3
[1]
CH3
CH2CH3
1 2 3 4 5 6 7 8
[2] CH3CH2CHCH2CHCH2CH2CH3
CH3
8 carbons = octane
3-methyl
CH2CH3
[1]
CH2CH3
5-ethyl
CH3
b. CH3CH2CCH2CH2CHCHCH2CH2CH3
7
[2] 1
CH2CH3 CH2CH3
[3] 5-ethyl-3-methyloctane
2 3 CH2CH3
7-methyl
CH3
[3] 3,3,6-triethyl-7-methyldecane
CH3CH2CCH2CH2CHCHCH2CH2CH3
CH2CH3 CH2CH3
10 carbons = decane
3,3,6-triethyl
c. CH3CH2CH2C(CH3)2C(CH3)2CH2CH3
[1]
re-draw
CH3 CH3
CH3CH2CH2 C
C CH2CH3
CH3 CH3
7 carbons = heptane
[2]
[3] 3,3,4,4-tetramethylheptane
CH3 CH3 3
CH3CH2CH2 C
4
C CH2CH3
CH3 CH32
1
3,3,4,4-tetramethyl
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
101
Alkanes 4–21
d. CH3CH2C(CH2CH3)2CH(CH3)CH(CH2CH2CH3)2
3,3-diethyl
re-draw
[1]
CH3CH2 C
4
[2]
CH3CH2 H
CH3CH2 H
H
C
CH3CH2 C
C CH2CH2CH3
1
CH3CH2 CH3 CH2CH2CH3
C CH2CH2CH3
5-propyl
3,3-diethyl
e. (CH3CH2)3CCH(CH3)CH2CH2CH3
4
CH3CH2 H
[2]
CH3CH2 C
CH3CH2 H
CH3CH2 C
C
4-methyl
re-draw
[3] 3,3-diethyl-4-methyl-5-propyloctane
CH3CH2 CH3 CH2CH2CH3
8 carbons = octane
[1]
5
H
1
CH3CH2 CH3
C CH2CH2CH3
[3] 3,3-diethyl-4-methylheptane
C CH2CH2CH3
7
6
4-methyl
CH3CH2 CH3
7 carbons = heptane
f. CH3CH2CH(CH3)CH(CH3)CH(CH2CH2CH3)(CH2)3CH3
3 4 5
re-draw
H H H
[1]
[2]
CH3CH2 C
H
H
H
CH3CH2 C
C
C CH2CH2CH2CH3
1 2
C
CH3 CH3 CH2CH2CH3
5-propyl
3,4-dimethyl
CH3 CH3 CH2CH2CH3
[3] 3,4-dimethyl-5-propylnonane
C CH2CH2CH2CH3
9 carbons = nonane
g. (CH3CH2CH2)4C
re-draw
[1]
4
4,4-dipropyl
[2]
CH2CH2CH3
CH3CH2CH2 C CH2CH2CH3
CH2CH2CH3
[3] 4,4-dipropylheptane
CH3CH2CH2 C CH2CH2CH3
CH2CH2CH3
1
2
3
CH2CH2CH3
7 carbons = heptane
1
h.
[1]
[2]
5
3
10 carbons = decane
i.
[1]
[1]
7
6-isopropyl
3-methyl
[2]
10 carbons = decane
j.
[3] 6-isopropyl-3-methyldecane
6
8
6
4-isopropyl
4
[3] 8-ethyl-4-isopropyl-2,6-dimethyldecane
1
8-ethyl
[2]
2,6-dimethyl
4-isopropyl
4
[3]
1
8 carbons = octane
4-isopropyloctane
102
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Chapter 4–22
2,2,5-trimethyl
k.
2,2,5-trimethylheptane
1 2
1
2
CH(CH2CH3)2
l.
3
=
3-cyclobutylpentane
4
5
3-cyclobutyl
5
m. 4
1
1-sec-butyl
2
3
1-sec-butyl-2-isopropylcyclopentane
2-isopropyl
5
6
1
n.
4
1-isobutyl-3-isopropylcyclohexane
3
2
1-isobutyl
3-isopropyl
4.40
2,2-dimethyl
CH3
3,3-dimethyl
CH3
4,4-dimethyl
CH3
CH3 C CH2CH2CH2CH2CH3
CH3CH2 C CH2CH2CH2CH3
CH3CH2CH2 C CH2CH2CH3
1
1
1
CH3
H
CH3
3
2
2,2-dimethylheptane
3,3-dimethylheptane
H
CH3 C
1
1
CH3
2
4
CH3
CH2CH2 C CH2CH3
CH3
2
1
H
H
H
CH3 C CH2CH2CH2 C CH3
CH3CH2 C
1
1
6 CH3
3
2,6-dimethyl
2,6-dimethylheptane
3
H
C CH2CH2CH2CH3
CH3CH3
2,3-dimethyl
2,3-dimethylheptane
2,5-dimethylheptane
2,4-dimethylheptane
2
5
H
CH3 C
CH3
2,5-dimethyl
2,4-dimethyl
CH3
2
H
H
CH3 C CH2 C CH2CH2CH3
CH3
4
4,4-dimethylheptane
H
H
C CH2CH2CH3
CH3 CH3
1
3
4
3,4-dimethyl
3,4-dimethylheptane
H
CH3CH2 C CH2 C CH2CH3
CH3
CH3
5
3,5-dimethyl
3,5-dimethylheptane
4.41 Use the steps in Answer 4.11 to draw the structures.
a. 3-ethyl-2-methylhexane
[1]
6 C chain
C C C C C C
C C
[2]
C C C C
CH3 CH2CH3
methyl
on C2
ethyl on C3
H
[3] CH3 C
H
C CH2CH2CH3
CH3 CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
103
Alkanes 4–23
b. sec-butylcyclopentane
5 C ring
[1]
[2]
isopropyl on C4
c. 4-isopropyl-2,4,5-trimethylheptane
[1]
CH3
[2]
7 C chain
CH3
CH
C C C C
CH3
C C C C C C C
C C C
CH3
CH3
[3]
CH
CH3 CH CH2 C
CH3 CH3
CH3
CHCH2CH3
CH3 CH3
methyls on C2, C4, and C5
d. cyclobutylcycloheptane
[1] 7 C cycloalkane
[2]
e. 3-ethyl-1,1-dimethylcyclohexane
[1]
6 C cycloalkane
[2]
[3]
CH3CH2
C
C
C
C
C
ethyl on C3 C
2 ethyl groups
f. 4-butyl-1,1-diethylcyclooctane
[1]
8 C cycloalkane
C
C
g. 6-isopropyl-2,3-dimethylnonane
[1]
9 C alkane
C
[2]
C
C
C C
C
C
C
C
C
C
C
C
C
C
[1]
C
C
C
C
C
C
C
C
[1]
C
C
C
CH
C
[3]
CH3
C
C
C
methyl
5 methyl groups
[2] CH3 CH3 CH3 CH3
C
C
C
C
C
C
C
C
[3]
CH3
i. cis-1-ethyl-3-methylcyclopentane
5 C ring
C
CH3
CH3
h. 2,2,6,6,7-pentamethyloctane
8 C alkane
isopropyl
methyl
CH3
C
[3]
C
C
C
CH3
C C
C
C C
2 methyl
groups on
C1
C
C C
C
C
CH3
C
C C
[2]
C
C
C
C
C
C
CH2CH3
ethyl on C1
[2]
CH2CH3
or
CH3
methyl on C3
CH3
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Chapter 4–24
j. trans-1-tert-butyl-4-ethylcyclohexane
[1]
C(CH3)3
[2]
6 C ring
CH3CH2
4.42 Undecane has 11 C’s in an unbranched long chain, with molecular formula C11H24. A
compound that is not an isomer must have a different molecular formula. Only (c) has a
different molecular formula.
a.
c.
e.
2-methyldecane
C11H24
isomer
3-ethyloctane
C10H22
not an isomer
2,2,3,3,4,4-hexamethylpentane
C11H24
isomer
b.
d.
3,3-dimethylnonane
C11H24
isomer
3,4-diethylheptane
C11H24
isomer
4.43 Draw the compounds.
a. 2,2-dimethyl-4-ethylheptane
CH3
CH2CH3
CH3 C CH2 C CH2CH2CH3
alphabetized incorrectly
ethyl before methyl
CH3
H
H
c. 2-methyl-2-isopropylheptane
CH3C CH3
Longest chain was not
chosen = octane
CH3 C CH2CH2CH2CH2CH3
CH3
2,3,3-trimethyloctane
4-ethyl-2,2-dimethylheptane
d. 1,5-dimethylcyclohexane
CH3
1
b. 5-ethyl-2-methylhexane
Numbered incorrectly.
Re-number so methyls
are at C1 and C3.
Longest chain was not
chosen = heptane
2,5-dimethylheptane
3
CH3
1,3-dimethylcyclohexane
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Alkanes 4–25
e. 1-ethyl-2,6-dimethylcycloheptane
CH2CH3
Numbered incorrectly.
Re-number so methyls
are at C1 and C4.
g. 3-butyl-2,2-dimethylhexane
2
CH3
1
4
f. 5,5,6-trimethyloctane
3
1
4-tert-butyloctane
2-ethyl-1,4-dimethylcycloheptane
Numbered incorrectly.
Re-number so methyls
are at C3 and C4.
4
Longest chain not
chosen = octane
CH3
h. 1,3-dimethylbutane
Longest chain not
chosen = pentane
4
H
CH3 CCH2
CH3
3,4,4-trimethyloctane
CH2
CH3
2-methylpentane
4.44
CH3
a.
H
CH3
CH2CH2CH3
CH3
b.
CH3
H
H
CH2CH2CH3
CH3CH2
c.
re-draw
3
4
3-ethyl-3-methylpentane
CH3CH2CH3
CH3CH2CH2CH3 CH3CH2CH2CH2CH3
4 C's
3C's
5 C's
lowest boiling point
highest boiling point
b. (CH3)2CHCH(CH3)2
CH3CH2CH2CH(CH3)2
most branching
lowest boiling point
CH3(CH2)4CH3
least branching
highest boiling point
4.46 a.
CH3(CH2)6CH3
no branching = higher surface area
higher boiling point
5
4,4-diethyl-5-methyloctane
4.45 Use the rules from Answer 4.15.
a.
H
CH2CH3
re-draw
4
CH2CH2CH3
CH3CH2CH2
H
CH2CH3
re-draw
4-isopropylheptane
CH3
CH2CH3
(CH3)3CC(CH3)3
branching = lower surface area
lower boiling point
more spherical, better packing =
higher melting point
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Chapter 4–26
b. There is a 159° difference in the melting points, but only a 20° difference in the boiling
points because the symmetry in (CH3)3CC(CH3)3 allows it to pack more tightly in the solid,
thus requiring more energy to melt. In contrast, once the compounds are in the liquid state,
symmetry is no longer a factor, the compounds are isomeric alkanes, and the boiling points
are closer together.
4.47
CH3
a.
H
H
CH3
H
CH3
H
or
H
H
b.
CH3
H
CH3
or
H CH3
CH3
CH3
H
CH3
CH3
CH3
1 gauche CH3,CH3
= 3.8 kJ/mol
of destabilization
CH3
H
H
H
higher energy
2 gauche CH3,CH3
3.8 kJ/mol x 2 = 7.6 kJ/mol
of destabilization
higher energy
3 eclipsed H,CH3
6 kJ/mol x 3 = 18 kJ/mol
of destabilization
2 gauche CH3,CH3
3.8 kJ/mol x 2 =
7.6 kJ/mol
of destabilization
Energy difference =
Energy difference =
7.6 kJ/mol – 3.8 kJ/mol = 3.8 kJ/mol
18 kJ/mol – 7.6 kJ/mol = 10.4 kJ/mol
4.48 Use the rules from Answer 4.18 to determine the most and least stable conformations.
a. CH3 CH2CH2CH2CH3
b. CH3CH2CH2 CH2CH2CH3
H CH CH CH
2
2
3
H
H
H
CH2CH2CH3
H
H
HH
H
H
H
H
eclipsed
least stable
staggered
most stable
CH3CH2
CH2CH3
CH2CH3
H
H
CH2CH3
HH
staggered
ethyl groups anti
most stable
All staggered conformations are equal in energy.
All eclipsed conformations are equal in energy.
eclipsed
ethyl groups eclipsed
least stable
4.49
CH3
a.
C
H
C
H
Br
H
H
CH3
H
C
Br
Cl
H
Br
H
Cl
CH3
H
Br
H
Br
c.
Cl H
H
H
CH3
Cl H
Cl
C
b.
Br
Cl
H
H
H
CH2CH3
Cl
Cl
CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
107
Alkanes 4–27
4.50
CH3CH2
H
H
60°
H
CH3CH2 CH2CH2CH3
H
CH3
H
60°
H
H
CH3
H
1
CH2CH3
CH3
60°
H
H
H
CH3CH2
H
3
2
60°
H
60°
H
H
CH3
6
H
H
H
H
60°
H
CH3CH2
CH3
H
CH3
CH2CH3
5
4
least stable
4
2
Energy
[1]
H
CH2CH3
H
H
Eclipsed forms are
higher in energy.
6
Staggered forms
are lower in energy.
5
3
1
most stable
180o
120o
1
most stable
60o
60o
0o
120o
180o
Dihedral angle between two alkyl groups
most stable
CH3CH2
H
[2] CH3CH2 CHCH2CH3
H
H
CH3
60°
H
CH2CH3
H
H
H
H
H
60°
CH3
CH3
CH3
CH3
CH3
1
CH2CH3
CH3
3
2
60°
CH3CH2
H
CH3
H
CH3
H
6
60°
CH3
60°
H
H
CH3CH2
H
CH3
5
H
CH3
60°
H
H
CH3
CH2CH3
least stable
4
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Chapter 4–28
least stable
4
Energy
2
3
Staggered forms
are lower in energy.
5
1
most stable
180o
120o
Eclipsed forms are
higher in energy.
6
1
most stable
60o
60o
0o
120o
180o
Dihedral angle
(between CH3CH2 in back and CH3 in front)
4.51 Two types of strain:
• Torsional strain is due to eclipsed groups on adjacent carbon atoms.
• Steric strain is due to overlapping electron clouds of large groups (e.g. gauche interactions).
H
CH3
H
CH3
H
a.
H
H
H CH3
b.
CH3
H
H
CH3
CH3
two sites
three bulky methyl groups close =
steric strain
HH
c.
CH2CH3
CH3CH2
two bulky ethyl groups close =
steric strain
eclipsed conformation =
torsional strain
eclipsed conformation =
torsional strain
4.52 The barrier to rotation is equal to the difference in energy between the highest energy eclipsed
and lowest energy staggered conformations of the molecule.
a. CH3 CH(CH3)2
CH3
H
H
CH3
H
H
most stable
b. CH3 C(CH3)3
CH3
CH3
H
H CH3
H
H
least stable
Destabilization energy =
2 H,CH3 eclipsing interactions
2(6.0 kJ/mol) = 12.0 kJ/mol
1 H,H eclipsing interaction = 4.0 kJ/mol
Total destabilization = 16.0 kJ/mol
16.0 kJ/mol = rotation barrier
H
H
CH3
CH3
CH3
H
H CH3
H
CH3
H
most stable
least stable
Destabilization energy =
3 H,CH3 eclipsing interactions
3(6.0 kJ/mol) = 18.0 kJ/mol
Total destabilization = 18.0 kJ/mol
18.0 kJ/mol = rotation barrier
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Alkanes 4–29
4.53
Cl
HH
H
H
H
H
H Cl
2 H,H eclipsing interactions = 2(4.0 kJ/mol) = 8.0 kJ/mol
Since the barrier to rotation is 15 kJ/mol,
the difference between this value and the
destabilization due to H,H eclipsing is the
destabilization due to H,Cl eclipsing.
H
H
H
most stable
least stable
15.0 kJ/mol – 8.0 kJ/mol = 7.0 kJ/mol
destabilization due to H,Cl eclipsing
4.54 The gauche conformation can intramolecularly hydrogen bond, making it the more stable
conformation.
H
OH
hydrogen bonding
O
HOCH2 CH2OH
H
H
H
O
rotation here
H
H
H
H
OH
H
anti
gauche
H
Hydrogen bonding can occur only
in the gauche conformation,
making it more stable.
4.55
axial
H
OH axial
a.
[1]
c. HO
eq
Br
H eq
CH3 eq
[2]
b.
H
CH3 up
H
H
H
d.
H eq
H
H
ax
ax
OH
c.
down
up
ax
CH3
eq Br
H eq
CH3eq
d.
OH
HO
ax
Br
CH3
H
b.
OH eq
axial
Br
both up =
cis
axial
OH ax
H
eq
ax
c.
H
axial
HO
eq HO
OH
Br
a.
H eq
H
one up, one down =
trans
up
axial
eq
eq HO
H
H
a.
H ax
OH
d.
down HO
HO
eq
[3]
ax
H up
OH
b.
HO
ax
ax
H
OH
OH eq
eq HO
eq H
H eq
H
one up, one down =
trans
H
ax
OH
ax
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Chapter 4–30
4.56
ax
ax
H
H
CH3
eq H
eq
CH3
eq
H eq
CH3
CH3
ax
ax
both groups equatorial
more stable
4.57 A cis isomer has two groups on the same side of the ring. The two groups can be drawn both
up or both down. Only one possibility is drawn. A trans isomer has one group on one side of
the ring and one group on the other side. Either group can be drawn on either side. Only one
possibility is drawn.
[1]
a.
cis
[2]
[3]
a.
a.
trans
cis
trans
cis
b. cis isomer
b. cis isomer
trans
b. cis isomer
ax
ax
ax
ax
ax
ax
eq
eq
eq
both groups equatorial
more stable
c. trans isomer
ax
eq
eq
eq
larger group equatorial
more stable
c. trans isomer
larger group equatorial
more stable
c. trans isomer
ax
ax
eq
eq
eq
eq
eq
ax
eq
ax
larger group equatorial
more stable
d.
ax
both groups equatorial
more stable
d.
The cis isomer is more
stable than the trans
since one conformation has
both groups equatorial.
both groups equatorial
more stable
d.
The trans isomer is more
stable than the cis
since one conformation has
both groups equatorial.
The trans isomer is more
stable than the cis
since one conformation has
both groups equatorial.
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Alkanes 4–31
4.58
re-draw to see
axial and equatorial
OH
OH
CH3
CH3
HO
HO
all equatorial
menthol
4.59
a. HO
or
OH
O
OH
HO
HO
HO
HO
HO
OH
O
b. HO
HO
or
HO
O
O
OH
OH
HO
most stable
All groups are equatorial.
HO
OH
HO
OH
OH
4.60
OH
OH OH
a.
O
HO
OH
OH
OH
OH
O
HO
c.
O
OH
OH
OH
more stable
More groups are equatorial.
O
HO
HO
OH
HO
constitutional isomer
OH OH
OH
O
OH
OH OH
b.
OH
HO
HO
O
d.
HO
OH
galactose
OH
glucose
All groups are equatorial.
more stable
OH
OH
stereoisomer
4.61
a.
CH3
and
c.
CH3
H
H
and
H
same molecular formula C4H8
different connectivity
constitutional isomers
CH3
CH3
and
different arrangement in three dimensions
stereoisomers
H
H
=
H
CH3
1 down, 1 up =
1 down, 1 up =
trans
trans
same arrangement in three dimensions
identical
CH2CH3
b.
CH3
d.
CH2CH3
and
CH2CH3
CH2CH3
same molecular formula C10H20
different connectivity
constitutional isomers
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Chapter 4–32
and
e.
CH3
molecular formula: C6H12
molecular formula: C6H10
different molecular formulas
not isomers
ax
ax
g.
eq
eq
H
CH3CH2
and
H
up
CH2CH3
up
CH3
H
CH3
up
both up = cis
up
H
both up = cis
same arrangement in three dimensions
identical
CH2CH3
CH2CH3
and
f.
CH3CH2
h.
and
CH3CH2
1 down, 1 up =
both down =
cis
trans
different arrangement in three dimensions
stereoisomers
3,4-dimethylhexane
2,4-dimethylhexane
same molecular formula C8H18
different IUPAC names
constitutional isomers
4.62
CH3
a.
CH
CH3
CH3
H
CH3CH2
H
H
CH2CH3
CH3
CH3
H
CH3
CH3
b.
and
H
H
H
and
CH3
CH2CH3
CH(CH3)2
re-draw
re-draw
CH3 CH2CH3
CH3
CH3 CH CH
CH3
CH2
CH3 CH
CH
CH2CH3
CH2CH3
3-ethyl-2-methylpentane
3-ethyl-2-methylpentane
same molecular formula
same name
identical molecules
same molecular formula
different arrangement of atoms
constitutional isomers
4.63
constitutional isomer
One possibility:
stereoisomer
a.
trans
cis
H
b.
H
H
H
OH
OH
HO
cis
OH
H
OH
c.
OH
trans
H
Cl
cis
Cl
Cl
Cl
Cl
Cl
trans
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113
Alkanes 4–33
4.64
Three constitutional isomers of C7H14:
1,1-dimethylcyclopentane
1,2-dimethylcyclopentane
1,3-dimethylcyclopentane
or
or
trans
cis
cis
trans
4.65 Use the definitions from Answer 4.29 to classify the reactions.
O
a.
CH3CHO
=
CH3
C
CH3
CH3CH2OH
H
Decrease in the number of C–O
bonds. Reduction
b.
CH2 CH2
H C C H
CH2Br
c.
Increase in the number of C–Z
bonds. Oxidation
d.
CH3CH2OH
Decrease in the number of C–H
bonds. Oxidation
CH2 CH2
Loss of one C–O
bond and one C–H
bond. Neither
4.66 Use the rule from Answer 4.30.
flame
a. CH3CH2CH2CH2CH(CH3)2
flame
7 CO2 + 8 H2O + heat
4 CO2 + 5 H2O + heat
b.
11 O2
(13/2) O2
4.67
1 C–O bond
2 C–O bonds
[1]
H
O
a.
[2]
2 C–H bonds
H
benzene
an arene oxide
[1] increase in C–O bonds
oxidation reaction
OH
H
1 C–H bond
phenol
[2] loss of 1 C–O bond,
loss of 1 C–H bond
neither
b. Phenol is more water soluble than benzene because it is polar (contains an O–H group)
and can hydrogen bond with water, whereas benzene is nonpolar and cannot hydrogen
bond.
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Chapter 4–34
4.68 Lipids contain many nonpolar C–C and C–H bonds and few polar functional groups.
OH
O
HO
HO
OH
a. HO
OH
mevalonic acid
c.
O HO
O
d. HO
estradiol
O
HO
many polar functional groups
not a lipid
OH
HO
few polar functional groups
a lipid
OH
OH
sucrose
many polar functional groups
not a lipid
b.
squalene
no polar functional groups
a lipid
4.69
O
OH
O
COH
OH
CNHCH2CH2SO3−Na+
This polar part of the molecule
interacts with water.
HO
OH
HO
cholic acid
a bile acid
OH
a bile salt
This nonpolar part of the molecule
can interact with lipids to create
micelles that allow for transport
of lipids through aqueous environments.
4.70 The mineral oil can prevent the body’s absorption of important fat-soluble vitamins. The
vitamins dissolve in the mineral oil, and are thus not absorbed. Instead, they are expelled with
the mineral oil.
4.71 Cyclopropane has larger angle strain than cyclobutane because the internal angles in the threemembered ring (60°) are smaller than they are in cyclobutane. Although cyclobutane is not
flat, as shown in Figure 4.11, there are more C–H bonds than there are in cyclopropane, so
there are more sites of torsional strain. Thus cyclopropane has more angle strain but less
torsional strain. The result is that both cyclopropane and cyclobutane have roughly similar
strain energies.
4.72 The amide in the four-membered ring has 90° bond angles giving it angle strain, which makes
it more reactive.
amide
H
N
penicillin G
S
O
N
CH3
CH3
O
COOH
strained amide
more reactive
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
115
Alkanes 4–35
4.73
Cl
H
Example:
HH
I
HH
HH
H
Cl
C
C
H
H
H
I
C
H
Although I is a much bigger atom than Cl, the C–I bond
is also much longer than the C–Cl bond. As a result the
eclipsing interaction of the H and I atoms is not very much
different in magnitude from the H,Cl eclipsing interaction.
HH
H
C
H
H
longer bond
H
H
H
4.74
H
H
H
decalin
H
trans-decalin
cis-decalin
H
H
H
H
H
H
1,3-diaxial interaction
cis
trans
The trans isomer is more stable since
the carbon groups at the ring junction are
both in the favorable equatorial position.
This bond is axial, creating
unfavorable 1,3-diaxial interactions.
4.75 Re-draw the ball-and-stick model using chair forms.
axial
(above)
a.
CH3
b. All bonds above the ring are on wedges and
all bonds below the ring are on dashed lines.
H
HO
H
H
equatorial
(above)
H
OH
CH3
axial
(below)
A
H
H
HO
H
OH
4.76
CH3
Cl
H
a, b.
axial
axial
Cl
H
H
B
equatorial
c. The circled H's at one ring fusion are cis. The
boxed in CH3 and H at the second ring fusion are trans.
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Chapter 4–36
4.77
4
2
1
1
1
5
3
pentylcyclopentane
4
(1-ethylpropyl)cyclopentane
2
2
3
1
1
(1,2-dimethylpropyl)cyclopentane
8
7
a.
5
4
2
5
1
6
c.
6
7
1-methyl-7-propylbicyclo[3.2.1]octane
7
4
b.
6
3
1
2
7
d.
1
2
3
6
5
2-ethyl-7,7-dimethylbicyclo[2.2.1]heptane
2
3
2,3-dimethylbicyclo[3.1.1]heptane
5
4
1
3
4
(3-methylbutyl)cyclopentane
4.78
3
3
(2-methylbutyl)cyclopentane (2,2-dimethylpropyl)cyclopentane
3
2
1
(1-methylbutyl)cyclopentane
2
3
(1,1-dimethylpropyl)cyclopentane
4
3
2
1
1
2
2
3
4
6-ethyl-3,3-dimethylbicyclo[3.2.0]heptane
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
117
Stereochemistry 5–1
Chapter 5 Stereochemistry
Chapter Review
Isomers are different compounds with the same molecular formula (5.2, 5.11).
[1] Constitutional isomers—isomers that differ in the way the atoms are connected to each other.
They have:
• different IUPAC names
• the same or different functional groups
• different physical and chemical properties.
[2] Stereoisomers—isomers that differ only in the way atoms are oriented in space. They have the
same functional group and the same IUPAC name except for prefixes such as cis, trans, R, and S.
• Enantiomers—stereoisomers that are nonsuperimposable mirror images of each other (5.4).
• Diastereomers—stereoisomers that are not mirror images of each other (5.7).
CH3
C
CH2CH3
CH3CH2
C
H
Br
CH3
C
H
Br
C
H
Br
A
CH3
C
H
Br
Br
CH2CH3
CH3CH2
C
C
H
Br
H
C
H
Br
C
B
CH3
H
Br
D
enantiomers
enantiomers
A and B are diastereomers of C and D.
Assigning priority (5.6)
•
•
•
•
Assign priorities (1, 2, 3, or 4) to the atoms directly bonded to the stereogenic center in order of
decreasing atomic number. The atom of highest atomic number gets the highest priority (1).
If two atoms on a stereogenic center are the same assign priority based on the atomic number of
the atoms bonded to these atoms. One atom of higher atomic number determines a higher priority.
If two isotopes are bonded to the stereogenic center, assign priorities in order of decreasing mass
number.
To assign a priority to an atom that is part of a multiple bond, treat a multiply bonded atom as an
equivalent number of singly bonded atoms.
highest atomic number =
highest priority
1
3
1
Br
CH2CH2CH3
OH
4 H C CH2I
*
3
Cl
2
I is NOT bonded directly
to the stereogenic center.
4 CH3 C CH2CH2CH2CH2CH3
*
2
CH(CH3)2
1
This is the highest priority C since
it is bonded to 2 other C's.
* = stereogenic center
4 H C CH2OH 3
*
COOH
2
This C is considered
to be bonded to 3 O's.
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Chapter 5–2
Some basic principles
•
•
•
•
When a compound and its mirror image are superimposable, they are identical achiral
compounds. A plane of symmetry in one conformation makes a compound achiral (5.3).
When a compound and its mirror image are not superimposable, they are different chiral
compounds called enantiomers. A chiral compound has no plane of symmetry in any
conformation (5.3).
A tetrahedral stereogenic center is a carbon atom bonded to four different groups (5.4, 5.5).
For n stereogenic centers, the maximum number of stereoisomers is 2n (5.7).
plane of
symmetry
CH3
C
H
H
CH3
CH3
*C
C
H
plane of
symmetry
[* = stereogenic center]
CH3CH2
H
no stereogenic centers
H
Cl
1 stereogenic center
CH3
*C
Cl
H
CH3
CH3
C*
H
Cl
2 stereogenic centers
*C
Cl
H
CH3
C*
H
Cl
2 stereogenic centers
Chiral compounds generally contain stereogenic centers.
A plane of symmetry makes these compounds achiral.
Optical activity is the ability of a compound to rotate plane-polarized light (5.12)
•
•
An optically active solution contains a chiral compound.
An optically inactive solution contains one of the following:
• an achiral compound with no stereogenic centers.
• a meso compound—an achiral compound with two or more stereogenic centers.
• a racemic mixture—an equal amount of two enantiomers.
The prefixes R and S compared with d and l
The prefixes R and S are labels used in nomenclature. Rules on assigning R,S are found in Section 5.6.
• An enantiomer has every stereogenic center opposite in configuration. If a compound with
two stereogenic centers has the R,R configuration, then its enantiomer has the S,S
configuration.
• A diastereomer of this same compound has either the R,S or S,R configuration; one
stereogenic center has the same configuration and one is opposite.
The prefixes d (or +) and l (or –) tell the direction a compound rotates plane-polarized light (5.12).
• d (or +) stands for dextrorotatory, rotating polarized light clockwise.
• l (or –) stands for levorotatory, rotating polarized light counterclockwise.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
119
Stereochemistry 5–3
The physical properties of isomers compared (5.12)
Type of isomer
Constitutional isomers
Enantiomers
Diastereomers
Racemic mixture
Physical properties
Different
Identical except the direction of rotation of polarized light
Different
Possibly different from either enantiomer
Equations
•
Specific rotation (5.12C):
specific =
rotation
•
[α]
α = observed rotation (o)
l = length of sample tube (dm)
c = concentration (g/mL)
α
=
l x c
Enantiomeric excess (5.12D):
ee
=
% of one enantiomer – % of other enantiomer
=
[α] mixture
[α] pure enantiomer
x 100%
Practice Test on Chapter Review
1. a. Which of the following statements is true for compounds A–D below?
1.
2.
3.
4.
5.
O
O
O
O
A
B
C
D
A and B are separable by physical methods such as distillation.
A and C are separable by physical methods such as distillation.
A and D are separable by physical methods such as distillation.
Statements (1) and (2) are both true.
Statements (1), (2), and (3) are all true.
b. Which of the following statements is true about compounds A–C below?
CH2CH3
C
CH3
A
Cl
Br
Cl
CH3
CH3
CH2CH3
C
Br
B
Br
Cl
C
CH2CH3
C
dm = decimeter
1 dm = 10 cm
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 5–4
1.
2.
3.
4.
5.
A and B are enantiomers.
A and C are enantiomers.
An equal mixture of B and C is optically active.
Statements (1) and (2) are true.
Statements (1), (2), and (3) are all true.
c. Which compound is a diastereomer of A?
CH3
CH3
A
1.
2.
3.
4.
5.
B
C
D
B only
C only
D only
Both B and C
Compounds B, C, and D
2. Rank the following four groups around a stereogenic center in order of decreasing priority. Rank
the highest priority group as 1, the lowest priority group as 4, and the two groups of intermediate
priority as 2 and 3.
CH2Cl
CH2CH2Br
A
COOH
B
C
CH2OH
D
3. Label each stereogenic center in the following compound as R or S.
a.
BrCH2 H
H
Cl2CH
b.
H2N
CH=CH2
4. State how the compounds in each pair are related to each other. Choose from constitutional
isomers, enantiomers, diastereomers, or identical compounds.
CHO
H
H
OH
a.
H
and
HO
H
OHC
CH2OH
b.
CH2OH
Cl
OH
OH
and
Cl
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
121
Stereochemistry 5–5
c.
and
5. The enantiomeric excess of a mixture of A and B is 62% with A in excess. How much of A and
B are present in the mixture?
Answers to Practice Test
1. a. 1
b. 2
c. 3
3. a. S
b. S
2. A–1
B–4
C–2
D–3
4. a. diastereomers
b. enantiomers
c. identical
5. 81% A
19% B
Answers to Problems
5.1 Cellulose consists of long chains held together by intermolecular hydrogen bonds forming
sheets that stack in extensive three-dimensional arrays. Most of the OH groups in cellulose are
in the interior of this three-dimensional network, unavailable for hydrogen bonding to water.
Thus, even though cellulose has many OH groups, its three-dimensional structure prevents
many of the OH groups from hydrogen bonding with the solvent and this makes it water
insoluble.
5.2 Constitutional isomers have atoms bonded to different atoms.
Stereoisomers differ only in the three-dimensional arrangement of atoms.
and
a.
2,3-dimethylpentane
2,4-dimethylpentane
different connectivity of atoms
different names
constitutional isomers
b.
O
and
and
c.
OH
four-membered ring three-membered ring
different connectivity of atoms
constitutional isomers
different connectivity of atoms
constitutional isomers
d.
and
trans isomer
cis isomer
Both are 1,2-dimethylcyclobutane,
but the CH3 groups are oriented differently.
stereoisomers
5.3 Draw the mirror image of each molecule by drawing a mirror plane and then drawing the
molecule’s reflection. A chiral molecule is one that is not superimposable on its mirror
image. A molecule with one stereogenic center is always chiral. A molecule with zero
stereogenic centers is not chiral (in general).
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Chapter 5–6
CH3
a.
C
CH3
CH3 C
Br
CH3
Br
Cl
c.
Cl
O
CH3
CH3
identical
Br
H
CH3
achiral molecule
CH3
CH3
C
C
Cl
O
identical
achiral molecule
b.
CH3
Cl
H Br
d.
Br
H
Br H
C
C
F
CH2CH3
CH3CH2
F
stereogenic center
stereogenic center
nonsuperimposable mirror images
nonsuperimposable mirror images
chiral molecules
chiral molecules
5.4 A plane of symmetry cuts the molecule into two identical halves.
2 H's are behind
one another.
H
H
a.
CH3
C
CH3
b.
CH3
H
CH3
H
H
c.
H
CH3
C
H
CH3
d.
C
H
Cl
H
CH3
C
H
Cl
one possible
plane of symmetry
plane of symmetry
plane of symmetry
plane of symmetry
5.5 Rotate around the middle C−C bond so that the Br atoms are eclipsed.
rotate
CH3 here H
C
Br
Br
Br
C
H
C
CH3
CH3
H
H
Br
C
CH3
C2 C3
plane of symmetry
5.6 To locate a stereogenic center, omit all C’s with two or more H’s, all sp and sp2 hybridized atoms,
and all heteroatoms. (In Chapter 25, we will learn that the N atoms of ammonium salts [R4N+X–]
can sometimes be stereogenic centers.) Then evaluate any remaining atoms. A tetrahedral
stereogenic center has a carbon bonded to four different groups.
H
H
a.
CH3CH2 C CH2CH3
Cl
bonded to 2 identical
ethyl groups
0 stereogenic centers
b.
CH3 C CH=CH2
OH
This C is bonded to
4 different groups.
1 stereogenic center
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
123
Stereochemistry 5–7
H
CH3CH2CH2OH
c.
e.
(CH3)2CHCH2
C COOH
0 stereogenic centers
NH2
stereogenic center
CH3
d.
(CH3)2CHCH2CH2
C CH2CH3
f.
H
This C is bonded to
4 different groups.
1 stereogenic center
3 C's bonded to 4
different groups
3 stereogenic centers
5.7 Use the directions from Answer 5.6 to locate the stereogenic centers.
O
OH
H2N
CH3O
O
NH2
H
N
O
4 C's bonded to 4
different groups
4 stereogenic centers
CH3O
aliskiren
5.8 Find the C bonded to four different groups in each molecule. At the stereogenic center, draw
two bonds in the plane of the page, one in front (on a wedge), and one behind (on a dash). Then
draw the mirror image (enantiomer).
stereogenic center
stereogenic center
c. CH3SCH2CH2CH(NH2)COOH
a. CH3CH(Cl)CH2CH3
H
CH3
C
H
Cl
Cl
CH2CH3
CH3CH2
C
CH3
CH3SCH2CH2
mirror images
nonsuperimposable
enantiomers
CH3CH2CH(OH)CH2OH
H
CH3CH2
OH
C
CH2OH
C
COOH
H
HO
HOCH2
mirror images
nonsuperimposable
enantiomers
C
CH2CH3
H
H2N
HOOC
mirror images
nonsuperimposable
enantiomers
stereogenic center
b.
NH2
H
C
CH2CH2SCH3
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Chapter 5–8
5.9 Use the directions from Answer 5.6 to locate the stereogenic centers.
O
C bonded to
H and 3 different C's:
1 stereogenic center
a.
O
c.
4 C's bonded to 4
different groups:
4 stereogenic centers
O
O
e.
O
CO2
H
N
H
C bonded to
H, 2 different O's and 1 C:
1 stereogenic center
Cl
b.
Each labeled C
is bonded to:
d.
H, Cl, CH2, CHCl:
Cl 2 stereogenic centers
NH2
no stereogenic centers
CO2H
5.10
O
a.
b.
O
O
O
cholesterol
HO
simvastatin
All stereogenic C's are circled. Each C is sp3
hybridized and bonded to 4 different groups.
5.11 Assign priority based on atomic number: atoms with a higher atomic number get a higher
priority. If two atoms are the same, look at what they are bonded to and assign priority based
on the atomic number of these atoms.
a. –CH3, –CH2CH3
higher priority
e. –CH2CH2Cl, –CH2CH(CH3)2
c. –H, –D
higher mass
higher priority
higher priority
H
b. –I, –Br
d. –CH2Br, –CH2CH2Br
f. –CH2OH, –CHO
=
C O
H
=
C O
O C
higher priority
higher priority
2 H's, 1 O
2 O's, 1 H
2 C–O bonds
C bonded to 2 O's has
higher priority.
5.12 Rank by decreasing priority. Lower atomic number = lower priority.
a. –COOH
–H
–NH2
–OH
Highest priority = 1, Lowest priority = 4
priority
b. –H
H = lowest
C = second lowest
3
atomic number
atomic number
–CH3
C bonded to 3 H's
4
H = lowest atomic number
N = second highest
atomic number
O = highest atomic number
2
–Cl
1
–CH2Cl
decreasing priority: –OH, –NH2, –COOH, –H
Cl = highest
atomic number
C bonded to 2 H's + 1 Cl
priority
4
3
1
2
decreasing priority: –Cl, –CH2Cl, –CH3, –H
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Stereochemistry 5–9
c. –CH2CH3
priority
2
C bonded to 2 H's + 1 C
d. –CH=CH2
priority
2
C bonded to 1 H + 2 C's
–CH3
C bonded to 3 H's
3
–CH3
C bonded to 3 H's
3
–H
H = lowest
atomic number
C bonded to 1 H + 2 C's
4
–C≡CH
C bonded to 3 C's
1
1
–H
H = lowest
atomic number
4
–CH(CH3)2
decreasing priority: –CH(CH3)2, –CH2CH3, –CH3, –H
decreasing priority: –C≡CH, –CH=CH2, –CH3, –H
5.13 To assign R or S to the molecule, first rank the groups. The lowest priority group must be
oriented behind the page. If tracing a circle from (1) → (2) → (3) proceeds in the clockwise
direction, then the stereogenic center is labeled R if the circle is counterclockwise, then it is
labeled S.
2
2
Cl
a.
C
CH3
3
2
COOH
b.
H
Br
C
CH3
3
1
counterclockwise
S isomer
CH2Br
c.
H
OH
ClCH2
CH3
1
counterclockwise
S isomer
CH2Br
rotate
C
CH3
HO
OH
lowest priority group
now back
H 3
3
C
CH2Cl
d.
1
2
1
counterclockwise
S isomer
clockwise
R isomer
5.14
CH3O
O
2
Cl
CH3OOC
N
1
S
H Cl
N
3
Cl H
3
2
COOCH3
N
1
S
S
clopidogrel
counterclockwise
S isomer
Plavix
clockwise
R isomer
5.15 a, b. Re-draw lisinopril as a skeletal structure, locate the stereogenic centers, and assign R,S.
NH2
HO2C H
H
N
N
H
S
O
S
three stereogenic centers
All have the S configuration.
H CO2H
S
5.16 The maximum number of stereoisomers = 2n where n = the number of stereogenic centers.
a. 3 stereogenic centers
23 = 8 stereoisomers
b. 8 stereogenic centers
28 = 256 stereoisomers
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Chapter 5–10
5.17
a. CH3CH2CH(Cl)CH(OH)CH2CH3
b. CH3CH(Br)CH2CH(Cl)CH3
2 stereogenic centers = 4 possible stereoisomers
CH3CH2
CH2CH3
C
H
Cl
CH3CH2
C
A
CH3CH2
H
HO
CH2CH3
C
H
Cl
CH2CH3
C
H
OH
HO
H
Br H Cl H
A
B
H Cl Br H
H Br Cl H
C
D
CH2CH3
C
H
H Cl H Br
H
Cl
B
C
C
C
CH3CH2
OH
2 stereogenic centers = 4 possible stereoisomers
C
H
Cl
D
5.18
a. CH3CH(OH)CH(OH)CH3
b. CH3CH(OH)CH(Cl)CH3
2 stereogenic centers = 4 possible stereoisomers
CH3
CH3
C
HO
H
CH3
C
H
OH
A
CH3
H
HO
CH3
C
H
HO
B
C
H
OH
H
HO
CH3
C
H
CH3
C
C
CH3
C
2 stereogenic centers = 4 possible stereoisomers
OH
C
HO
CH3
H
OH
H
HO
C
CH3
H
Cl
H
Cl
C
A
CH3
C
C
identical
C is a meso compound.
A and B are enantiomers.
Pairs of diastereomers: A and C, B and C.
5.19
H
CH3
C
C
B
CH3
CH3
H
Cl
H
Cl
C
C
CH3
H
OH
CH3
C
C
D
H
OH
Pairs of enantiomers: A and B, C and D.
Pairs of diastereomers: A and C, A and D,
B and C, B and D.
A meso compound must have at least two stereogenic centers. Usually a meso
compound has a plane of symmetry. You may have to rotate around a C–C bond to see the
plane of symmetry clearly.
CH3CH2
a.
C
HO
H
CH3
CH2CH3
b.
C
H
OH
2 stereogenic centers
plane of symmetry
meso compound
C
HO
H
OH
H Br
H
C
rotate
H Br Br H
c.
CH3
2 stereogenic centers
no plane of symmetry
not a meso compound
Br H
2 stereogenic centers
plane of symmetry
meso compound
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Stereochemistry 5–11
5.20 Use the definition in Answer 5.19 to draw the meso compounds.
a. BrCH2CH2CH(Cl)CH(Cl)CH2CH2Br
Cl
Br
H H
Cl
b.
HO
OH
HO
Br
c.
OH
H H
plane of symmetry
NH2
H2N
H H
H2N
plane of symmetry
NH2
plane of symmetry
5.21 The enantiomer must have the exact opposite R,S designations. Diastereomers with two
stereogenic centers have one center the same and one different.
If a compound is R,S:
Exact opposite: R and S interchanged.
Its enantiomer is: S,R
Its diastereomers are: R,R and S,S
One designation remains the same,
the other changes.
5.22 The enantiomer must have the exact opposite R,S designations. For diastereomers, at least
one of the R,S designations is the same, but not all of them.
a. (2R,3S)-2,3-hexanediol and (2R,3R)-2,3-hexanediol
One changes; one remains the same:
diastereomers
b. (2R,3R)-2,3-hexanediol and (2S,3S)-2,3-hexanediol
Both R's change to S's:
enantiomers
c. (2R,3S,4R)-2,3,4-hexanetriol and (2S,3R,4R)-2,3,4-hexanetriol
Two change; one remains the same:
diastereomers
5.23 The enantiomer must have the exact opposite R,S designations. For diastereomers, at least
one of the R,S designations is the same, but not all of them.
HO H HO H
a.
R
HO
R
R
S
H OH HO H
sorbitol
HO H HO H
b.
OH
HO
R
R
H OH
S
S
H OH H OH
c.
OH
H OH
HO
S
HO H
S
S
R
OH
H OH
A
B
One changes; three remain the same.
diastereomer
All stereogenic centers change.
enantiomers
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Chapter 5–12
5.24 Meso compounds generally have a plane of symmetry. They cannot have just one
stereogenic center.
Cl
a.
b.
c.
OH
no plane of symmetry
not a meso compound
plane of symmetry
meso compound
no plane of symmetry
not a meso compound
5.25
Cl
2 stereogenic centers =
4 stereoisomers maximum
a.
c.
Cl
Draw the cis and trans isomers:
CH3
CH3
CH3
CH3
Draw the cis and trans isomers:
Cl
Cl
cis
Cl
Cl
A
identical
A
CH3
CH3
CH3
CH3
identical
Cl
Cl
trans
B
C
Cl
Cl
B
identical
Pair of enantiomers: B and C.
Pairs of diastereomers: A and B, A and C.
Pair of diastereomers: A and B.
Only 3 stereoisomers exist.
Only 2 stereoisomers exist.
b.
2 stereogenic centers =
4 stereoisomers maximum
HO
Draw the cis and trans isomers:
cis
trans
OH
CH3
A
HO
CH3
B
Pairs of enantiomers: A and B, C and D.
Pairs of diastereomers: A and C, A and D,
B and C, B and D.
OH
CH3
HO
C
All 4 stereoisomers exist.
CH3
D
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129
Stereochemistry 5–13
5.26
Four facts:
• Enantiomers are mirror image isomers.
• Diastereomers are stereoisomers that are not mirror images.
• Constitutional isomers have the same molecular formula but the atoms are bonded to
different atoms.
• Cis and trans isomers are always diastereomers.
CH3
a.
C
Br
S
Br
and
H
CH2OH
C
HOCH2
S
same molecular formula
same R,S designation:
identical
b.
H
CH3
and
c.
HO
OH
and
OH
HO
1,3-isomer
1,4- isomer
constitutional isomers
d.
OH
HO
same molecular formula,
opposite configuration at one
stereogenic center
enantiomers
and
OH
HO
trans
cis
Both 1,3 isomers,
cis and trans:
diastereomers
5.27
a. Mp = same as the S isomer.
b. The mp of a racemic mixture is often different from the melting
point of the enantiomers.
c. –8.5, same as S but opposite sign
d. Zero. A racemic mixture is optically inactive.
e. Solution of pure (S)-alanine: optically active
Equal mixture of (R)- and (S)-alanine: optically inactive
75% (S)- and 25% (R)-alanine: optically active
COOH
C
CH3
H
NH2
(S)-alanine
[α] = +8.5
mp = 297 oC
5.28
[α] =
α
l xc
α = observed rotation
l = length of tube (dm)
c = concentration (g/mL)
[α] =
10°
= +100 = specific rotation
1 dm x (1 g/10 mL)
5.29 Enantiomeric excess = ee = % of one enantiomer − % of other enantiomer.
a. 95% − 5% = 90% ee
b. 85% − 15% = 70% ee
5.30
a. 90% ee means 90% excess of A and 10% racemic mixture of A and B (5% each); therefore,
95% A and 5% B.
b. 99% ee means 99% excess of A and 1% racemic mixture of A and B (0.5% each); therefore,
99.5% A and 0.5% B.
c. 60% ee means 60% excess of A and 40% racemic mixture of A and B (20% each); therefore,
80% A and 20% B.
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Chapter 5–14
5.31
a.
x 100% = 42% ee
+24
[α] mixture
ee =
+10
x 100%
[α] pure enantiomer
b.
[α] solution
x 100% = 80% ee
+24
[α] solution = +19.2
5.32
[α] mixture
a.
x 100% = 60% ee
+3.8
b. % one enantiomer – % other enantiomer = ee
80% – 20% = 60% ee
[α] mixture = +2.3
80% dextrorotatory (+) enantiomer
20% levorotatory (–) enantiomer
5.33 • Enantiomers have the same physical properties (mp, bp, solubility), and rotate the
plane of polarized light to an equal extent, but in opposite directions.
• Diastereomers have different physical properties.
• A racemic mixture is optically inactive.
cis isomer
trans isomers
CH3
CH3
A
CH3
CH3
enantiomers
CH3
B
CH3
C
A and B are diastereomers of C.
a. The bp’s of A and B are the same. The bp’s of A and C are different.
b. Pure A: optically active
Pure B: optically active
Pure C: optically inactive
Equal mixture of A and B: optically inactive
Equal mixture of A and C: optically active
c. There would be two fractions: one containing A and B (optically inactive), and one containing C
(optically inactive).
5.34
OH
HO
a, b.
F
H
S
R
H H
N
S
O
three stereogenic centers
F
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Stereochemistry 5–15
5.35
S
S
H
CH3
C
H
Cl
C
R
B
R
Cl
R
H
C
R
CH2OH
identical
CH3
C
H
Cl
Cl
A
Cl
H
CH3
CH2OH
C
C
H
H
CH3
Cl
C
S
Cl
CH2OH
H
C
CH2OH
C
a. A and B, same R,S assignment, identical
b. A and C, opposite R,S assignment, enantiomers
c. A and D, one stereogenic center different,
diastereomers
d. C and D, one stereogenic center different,
diastereomers
R
Cl
D
5.36 Use the definitions from Answer 5.2.
CH3
O
a.
and
c.
H
CH3
and
O
one up, one down
trans
same molecular formula C4H8O
different connectivity
constitutional isomers
Both compounds are
1,2-dimethylcyclohexane.
one cis, one trans = stereoisomers
O
and
b.
H
both up
cis
and
d.
O
same molecular formula C7H14
different connectivity
constitutional isomers
C5H8O
C5H10O
different molecular formulas
not isomers
5.37 Use the definitions from Answer 5.3.
CH3
a.
C
CH3
OH
CH3
CH2OH
H
HOCH2
H
O
c.
C
COOH
C
HSCH2
H
NH2
e. OHC
CH3
identical
achiral
chiral
d.
C
CH2SH
OH
HO
H
Br
Br
identical
achiral
H
CHO
OH
chiral
COOH
H
H2N
OH
OH
identical
achiral
b.
O
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Chapter 5–16
5.38
CHO
CH3
a.
C
OH
H
CH3
C
HO
A
R isomer
CH3
b.
H
CHO
C
OHC
S
enantiomer
HO H
c.
H
OH
CH3
R
identical
C
CHO
S
enantiomer
5.39 A plane of symmetry cuts the molecule into two identical halves.
H
CH3CH2
a.
C
Cl
b.
C
H
C
HOOC
CH2CH3
d.
C
C
COOH
H
HO
CH3CH2
C
H
H
CH3
H and OH
are aligned.
The plane of symmetry
bisects the molecule.
plane of symmetry
CH2CH3
C
Cl
CH3
HO H HO H
Cl
CH3CH2
Cl
c.
C
H
Cl
H
Cl
e.
C
CH2CH3
no plane of symmetry
no plane of symmetry
plane of symmetry
5.40 Use the directions from Answer 5.6 to locate the stereogenic centers.
OH
a. CH3CH2CH2CH2CH2CH3
All C's have 2 or more H's.
0 stereogenic centers
OH
f.
OH
OH
OH
Each indicated C bonded to 4 different groups =
6 stereogenic centers
H
b.
OH
CH3CH2O C CH2CH3
CH3
g.
1 stereogenic center
O
c. (CH3)2CHCH(OH)CH(CH3)2
0 stereogenic centers
H
bonded to 4 different groups
1 stereogenic center
H
H
d. (CH3)2CHCH2 C CH2 C
C
CH3
h.
CH2CH3
All C's have 2 or more H's or
are sp2 hybridized.
0 stereogenic centers
CH3 CH3
3 stereogenic centers
Cl
H
e.
CH3 C CH2CH3
i.
D
bonded to 4 different groups
1 stereogenic center
Each indicated C bonded to
4 different groups =
2 stereogenic centers
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
133
Stereochemistry 5–17
OH
HO
O
j.
HO
OH
OH
Each indicated C bonded to 4 different groups =
5 stereogenic centers
5.41 Stereogenic centers are circled.
Eight constitutional isomers:
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
5.42
a.
H2N H
NH2
H NH2
amphetamine
O
CH3
O
COOH
b.
H
COOH
ketoprofen
5.43 Draw a molecule to fit each description.
a.
c.
b.
OH
O
O
alcohol
ketone
cyclic ether
5.44 Assign priority based on the rules in Answer 5.11.
a. −CD3, −CH3
c. –CH2Cl, –CH2CH2CH2Br
D higher mass than H
higher priority
b. −CH(CH3)2, −CH2OH
C bonded to O
higher priority
C bonded to Cl
higher priority
d. −CH2NH2, −NHCH3
higher atomic number
higher priority
H CH3
HOOC
O
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Chapter 5–18
5.45 Assign priority based on the rules in Answer 5.11.
a. −F > −OH > −NH2 > −CH3
d. –COOH > –CHO > –CH2OH > –H
b. −(CH2)3CH3 > −CH2CH2CH3 > −CH2CH3 > −CH3
e. –Cl > –SH > –OH > –CH3
c. −NH2 > −CH2NHCH3 > −CH2NH2 > −CH3
f. –C≡CH > –CH=CH2 > –CH(CH3)2 > –CH2CH3
5.46 Use the rules in Answer 5.13 to assign R or S to each stereogenic center.
1
1
I
a.
H
c.
C
C
H
CH3
CH3CH2
2
T
CH3
switch H and CH3
C
CH3
D
H
D
T
2
3
3
counterclockwise
It looks like an S isomer, but we
must reverse the answer, S to R.
counterclockwise
S isomer
R isomer
1
2
NH2
b.
CH3
Cl
C
d.
CH2CH3
3 H
Br
ICH2
2
Cl
switch H and Br
C
H
ICH2
H
C
1
Br
3
clockwise, but H in front
S isomer
counterclockwise
It looks like an S isomer, but we
must reverse the answer, S to R.
R isomer
CH3
e.
CH(CH3)2
C
C
CH3
SH
H
HO
HOOC
f.
C
CH3
C
NH2
H
H
g.
H
CH3
HO
H
S, R
R, R
S
5.47
a. (3R)-3-methylhexane
c. (3R,5S,6R)-5-ethyl-3,6-dimethylnonane
CH3 H
R
CH3 H
CH3 H
b. (4R,5S)-4,5-diethyloctane
4R
R
H CH2CH3
d. (3S,6S)-6-isopropyl-3-methyldecane
H CH(CH3)2
H CH2CH3
CH3CH2 H
S
S
5S
H CH3
S
Cl
h.
Cl
S
S
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Stereochemistry 5–19
5.48
4
9
1
3
a.
6
6R
H
5
b.
S
2
4R
5
7
6R
3
3R
c.
10
9
(4R,6R)-4-ethyl-6-methyldecane
(3S)-3-methylhexane
5S
1
7
1
(3R,5S,6R)-5-isobutyl-3,6-dimethylnonane
5.49 Two enantiomers of the amino acid leucine.
COOH
C
(CH3)2CHCH2
COOH
H
NH2
H
H2N
S isomer
naturally occurring
C
CH2CH(CH3)2
R isomer
5.50
S
S CH3
COOH
a.
OH
L-dopa
a. 1R,2S C1
OH
NH Cl
b.
H NH2
HO
CH3CH2O2C
N
N
H
c.
ketamine
enalapril
O
S
O
CO2H
S
S
5.51
OH
d.
NHCH3
S
R NHCH3
e.
enantiomer of (–)-ephedrine
C2
ephedrine
b. 1S,2S C1
OH
OH
NHCH3
R
R NHCH3
diastereomer of (–)-ephedrine
C2
pseudoephedrine
c. Ephedrine and pseudoephedrine
are diastereomers (one stereogenic
center is the same; one is different).
5.52
OH
C C H
NH2 H
N
a.
HO
O
S
b.
O
c.
O
O
N
O
amoxicillin
COOH
O
N
O
norethindrone
CH3
O
heroin
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Chapter 5–20
5.53
OH
O
a. CH3CH(OH)CH(OH)CH2CH3
c.
b. CH3CH2CH2CH(CH3)2
OH
HO
OH
HO
0 stereogenic centers
2 stereogenic centers
22 = 4 possible stereoisomers
4 stereogenic centers
24 = 16 possible stereoisomers
5.54
a. CH3CH(OH)CH(OH)CH2CH3
CH3
CH2CH3
C
CH3CH2
C
H
HO
CH3
C
H
OH
C
H
HO
A
CH3
CH2CH3
C
H
OH
HO
C
CH3
C
H
OH
H
B
CH3CH2
C
H
HO
C
H
OH
D
Pairs of enantiomers: A and B, C and D.
Pairs of diastereomers: A and C, A and D, B and C, B and D.
b. CH3CH(OH)CH2CH2CH(OH)CH3
OH
OH
OH
OH
OH
OH
A
C
B
OH
OH
identical
meso compound
Pair of enantiomers: A and B.
Pairs of diastereomers: A and C, B and C.
c. CH3CH(Cl)CH2CH(Br)CH3
Cl
Br
Br
A
Cl
Cl
B
Br
Br
C
Cl
D
Pairs of enantiomers: A and B, C and D.
Pairs of diastereomers: A and C, A and D, B and C, B and D.
d. CH3CH(Br)CH(Br)CH(Br)CH3
Br
Br
Br
Br
Br
Br
A
identical
Br
Br
Br
Br
Br
Br
Br
Br
Br
B
C
D
meso compound
Pair of enantiomers: B and C.
Pairs of diastereomers: A and B, A and C, A and D, B and D, C and D.
Br
Br
Br
identical
meso compound
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
137
Stereochemistry 5–21
5.55
HOCH2
a.
C
CH3
CH3
H
OH
H
HO
C
CH2OH
C
H
HO
CH3
C
C
H
OH
HO
H I
I H
H I
I H
c.
OH
HO
CH3
d.
C
H
HO
OH
H
diastereomer
I H
diastereomer
H 2N
H 2N
or
HO
enantiomer
CH2OH
C
H
OH
I H
H 2N
or
C
H
enantiomer
NH2
CH3
diastereomer
enantiomer
b.
CH2OH
HO
diastereomer
CH3
diastereomer
CH3
CH3
or
CH2CH3
CH3CH2
CH3CH2
CH3CH2
diastereomer
enantiomer
diastereomer
5.56
CH3
CH3
CH3
CH3
a.
CH3
CH3
A
identical
CH3
CH3
B
C
meso compound
Pair of enantiomers: B and C.
Pairs of diastereomers: A and B, A and C.
b.
CH3
CH3
CH3
CH3
CH3
A
CH3
CH3
CH3
B
identical
identical
Pair of diastereomers: A and B.
Meso compounds: A and B.
c.
Cl
Cl
Br
Cl
Br
Br
B
A
C
Cl
Br
D
Pairs of enantiomers: A and B, C and D.
Pairs of diastereomers: A and C, A and D, B and C, B and D.
5.57
achiral
achiral
chiral
chiral
achiral
achiral
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Chapter 5–22
5.58 Explain each statement.
All molecules have a mirror image, but only chiral molecules have enantiomers. A is not chiral, and
therefore, does not have an enantiomer.
a.
OH
B has one stereogenic center, and therefore, has an enantiomer. Only compounds with two or
more stereogenic centers have diastereomers.
b.
Cl
C is chiral and has two stereogenic centers, and therefore, has both an enantiomer and a
diastereomer.
c.
Cl
OH
HO
rotate
OH
D has two stereogenic centers, but is a meso compound.
Therefore, it has a diastereomer, but no enantiomer since
it is achiral.
plane of symmetry
d.
OH
e. HO
E has two stereogenic centers, but is a meso compound. Therefore, it has a diastereomer,
but no enantiomer since it is achiral.
OH
plane of symmetry
5.59
C2 C3
OHC
OH
H
C
R
H
HO
C
R
OHC
a.
CH2OH
D-erythrose
HO
H
HOCH2
CH2OH
C
S
C
R
b.
HO
H
OH
2S,3R
diastereomer
2R,3R
H
C
R
CHO
C
R
OHC
c.
H
C
S
HO
H
OH
2R,3R
identical
OHC
H
OH
C
S
d.
H
HO
CH2OH
2S,3S
enantiomer
H
OH
C
R
C
S
CH2OH
2R,3S
diastereomer
5.60 Re-draw each Newman projection and determine the R,S configuration. Then determine how
the molecules are related.
CHO
CHO
H
CH2OH
H
OH
HO
H
OH
A
OHC
re-draw
OH
OH
H
H
HO
CH2OH
R,R
a. A and B are identical.
b. A and C are enantiomers.
CHO
OH
HO
H
H
CH2OH
HO
H
CH2OH
B re-draw
OHC
CH2OH
R,R
OH
H
CH2OH
H
H
HO
CHO
H
OH
C re-draw
OHC
H
OH
HO
H
D re-draw
OHC
H
CH2OH
HO
CH2OH
S,S
c. A and D are diastereomers.
d. C and D are diastereomers.
H
OH
S,R
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Stereochemistry 5–23
5.61
R
trans
R
NH2
A
R NH2
NH2
S
cis
trans
trans
B
C
D
NH2
NH2
E
NH2 group is in a
different location.
A (trans, R) and B (cis, R) are diastereomers.
A (trans, R) and C (trans, R) are identical molecules.
A (trans, R) and D (trans, S) are enantiomers.
A (trans, R) and E are constitutional isomers.
5.62
Cl
and
a.
f.
C
I
enantiomers
Br
H
C
Br
I
H
enantiomers
CH3
CH3
b.
Cl
and
g.
and
CH3
and
C
H
Br
same molecular formula
different connectivity
constitutional isomers
CH3
CH3
C
H
2S,3S
H
Br
C
C
H
Br
Br
identical
CH3
2S,3S
OH
H
CH3
c.
CHO
C
and
C
H
HO
OHC
H
OH
CH3
C
h.
and
H
HO
OH
HO
C
H
H
H
OH
one different configuration
diastereomers
1,4-cis
1,4-trans
2R,3R
2R,3S
diastereomers
i.
and
H
CH3
CH3
d.
diastereomers
different molecular formulas
not isomers
Br on end
HO CH3
Cl
H
1,3-trans
1,3-cis
and
e.
H
HO
Cl
j.
and
mirror images
not superimposable
enantiomers
5.63
a. A and B are constitutional isomers.
A and C are constitutional isomers.
B and C are diastereomers (cis and trans).
C and D are enantiomers.
and
C
H
CH2Br
CH3
CH2OH
C
Br H
Br in middle
different connectivity
constitutional isomers
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Chapter 5–24
plane of symmetry
b.
A
A has two
planes of symmetry.
achiral
B
C
D
achiral
chiral
chiral
mirror images and not
superimposable
enantiomers
c. Alone, C and D would be optically active.
d. A and B have a plane of symmetry.
e. A and B have different boiling points.
B and C have different boiling points.
C and D have the same boiling point.
f. B is a meso compound.
g. An equal mixture of C and D is optically inactive because it is a racemic mixture.
An equal mixture of B and C would be optically active.
5.64
H
HO
H
N
ee =
CH3O
[α] mixture
x 100%
[α] pure enantiomer
quinine = A
quinine's enantiomer = B
N
quinine
a.
−50
x 100% = 30% ee
−165
b. 30% ee = 30% excess one compound (A)
remaining 70% = mixture of 2 compounds (35% each A and B)
Amount of A = 30 + 35 = 65%
Amount of B = 35%
−83
x 100% = 50% ee
−165
50% ee = 50% excess one compound (A)
remaining 50% = mixture of 2 compounds (25% each A and B)
Amount of A = 50 + 25 = 75%
Amount of B = 25%
73% ee = 73% excess of one compound (A)
remaining 27% = mixture of 2 compounds (13.5% each A and B)
Amount of A = 73 + 13.5 = 86.5%
Amount of B = 13.5%
[α] mixture
x 100%
e. 60% =
–165
−120
x 100% = 73% ee
−165
c. [α] = +165
d. 80% − 20% = 60% ee
[α] mixture = −99
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141
Stereochemistry 5–25
5.65
OH
HO
OH
O
HO
O
O
OH
O
HO
HCl, H2O
OH
amygdalin
COOH
only one of the
products formed
CN
mandelic acid
OH
a. The 11 stereogenic centers are circled. Maximum number of stereoisomers = 2 11 = 2048
b. Enantiomers of mandelic acid:
HO H
H OH
HOOC
COOH
R
c. 60% − 40% = 20% ee
20% = [α] mixture/−154 x 100%
[α] mixture = −31
+50
d. ee =
+154
S
x 100% = 32% ee
[α] for (S)-mandelic acid = +154
32% excess of the S enantiomer
68% of racemic R and S = 34% S and 34% R
S enantiomer: 32% + 34% = 66%
R enantiomer = 34%
5.66
sp2
a. Each stereogenic center is circled.
b. The stereogenic centers in mefloquine are
labeled.
c. Artemisinin has seven stereogenic centers.
2n = 27 = 128 possible stereoisomers
d. One N atom in mefloquine is sp2 and one is
sp3.
e. Two molecules of artemisinin cannot
intermolecularly H-bond because there are
no O–H or N–H bonds.
CF3
H
N
CF3
O
O
O
H
H
O
R
HO
O
H H
N
sp3
H S
H
artemisinin
mefloquine
CF3
CF3
N
f.
CF3
N
CF3
H Cl
H H
N
HO
H
H
HO
H
HH
N
+
Cl
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Chapter 5–26
5.67
c. diastereomer
a. Each stereogenic center is circled.
H S
N
N
O
N S
H
O
HS
R
OH
CONH2 O
saquinavir
Trade name: Invirase
S
N
O
H
N
N
S
H
N
H
O
N
OH
CONH2 O
H
NH
H
NH
(CH3)3C
(CH3)3C
b. enantiomer
d. constitutional isomer
H
N
N
O
O
H
N
H
N
N
OH
CONH2 O
H
N
H
O
H
O
H
NH
(CH3)3C
new amine
NH
O
H
N
N
OH
O
H
NH
new aldehyde
(CH3)3C
5.68 Allenes contain an sp hybridized carbon atom doubly bonded to two other carbons. This makes
the double bonds of an allene perpendicular to each other. When each end of the allene has two
like substituents, the allene contains two planes of symmetry and it is achiral. When each end
of the allene has two different groups, the allene has no plane of symmetry and it becomes
chiral.
CH3
H
C C C
CH3
H
CH3
H
CH3
C C C
B
H
HC C C C CH C CH CH CH CH CHCH2CO2H
allene
no plane of symmetry
A
mycomycin
re-draw
chiral
These two substituents are at 90o to these two substituents.
Allene A contains two planes of symmetry,
making it achiral.
CH CH CH CHCH2CO2H
HC C C C
C C C
H
H
The substituents on each end of the allene in mycomycin
are different. Therefore, mycomycin is chiral.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
143
Stereochemistry 5–27
5.69
NH2
N
S
N
S
CH3
N
O
R
H
HOOC
OH
S
N
+
H NH2 S
OH
R
S
six stereogenic centers
5.70
HO
O
O
OH
O
OH
NH2
O
discodermolide
OH
a. The 13 tetrahedral stereogenic centers are circled.
b. Because there is restricted rotation around a C–C double bond, groups on the end of the
double bond cannot interconvert. Whenever the substituents on each end of the double bond
are different from each other, the double bond is a stereogenic site. Thus, the following two
double bonds are isomers:
R
R
R
C C
H
H
C C
H
H
R
These compounds are isomers.
There are three stereogenic double bonds in discodermolide, labeled with arrows.
c. The maximum number of stereoisomers for discodermolide must include the 13 tetrahedral
stereogenic centers and the three double bonds. Maximum number of stereoisomers
= 216 = 65,536.
5.71 When the spiro compound has a plane of symmetry, it is achiral.
a.
O
achiral
b.
c.
chiral
d.
achiral
chiral
144
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Chapter 5–28
5.72
racemic mixture of
2-phenylpropanoic acid
salts formed by proton transfer
COO
COOH
C
S
H NH2
H
CH3
C
+
S
(R)-sec-butylamine
COO
COOH
H NH2
C
+
R
R
diastereomers
enantiomers
H
CH3
H
CH3
+
H NH3
(R)-sec-butylamine
H
CH3
C
+
H NH3
R
R
These salts are diastereomers, and they are now
separable by physical methods since they have
different physical properties.
145
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Understanding Organic Reactions 6–1
Chapter 6 Understanding Organic Reactions
Chapter Review
Writing organic reactions (6.1)
•
Use curved arrows to show the movement of electrons. Full-headed arrows are used for electron
pairs and half-headed arrows are used for single electrons.
+
C
C Z
C Z
Z
+
Z
full-headed arrow
half-headed arrows
•
C
Reagents can be drawn either on the left side of an equation or over an arrow. Catalysts are
drawn over or under an arrow.
Types of reactions (6.2)
[1] Substitution
C Z
+ Y
C Y
+ Z
Z = H or a heteroatom
Y replaces Z
[2] Elimination
C
C
X
Y
+ reagent
Two σ bonds are broken.
[3] Addition
C C
+
+
C C
X Y
π bond
X Y
C C
X Y
This π bond is broken.
Two σ bonds are formed.
Important trends
Values compared
Bond dissociation
energy and bond
strength
Trend
The higher the bond dissociation energy, the stronger the bond (6.4).
Increasing size of the halogen
CH3 F
ΔHo = 456 kJ/mol
CH3 Cl
CH3 Br
351 kJ/mol
293 kJ/mol
Increasing bond strength
CH3
I
234 kJ/mol
146
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 6–2
Ea and reaction rate
The larger the energy of activation, the slower the reaction (6.9A).
Energy
Ea
larger Ea → slower reaction
slower reaction
Ea
faster reaction
Reaction coordinate
Ea and rate constant
The higher the energy of activation, the smaller the rate constant (6.9B).
Go products
ΔGo > 0
Go reactants
Keq < 1
more stable reactants
Free energy
Free energy
Equilibrium always favors the species lower in energy.
Go reactants
ΔGo < 0
Go products
Equilibrium favors the starting materials.
Keq > 1
more stable products
Equilibrium favors the products.
Reactive intermediates (6.3)
•
•
•
Breaking bonds generates reactive intermediates.
Homolysis generates radicals with unpaired electrons.
Heterolysis generates ions.
Reactive
intermediate
General structure
Reactive feature
Reactivity
radical
C
unpaired electron
electrophilic
carbocation
C
carbanion
C
positive charge;
only six electrons around C
net negative charge;
lone electron pair on C
electrophilic
nucleophilic
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
147
Understanding Organic Reactions 6–3
Energy diagrams (6.7, 6.8)
Energy
transition state
Ea
reactants
Ea determines the rate.
ΔHo is the difference in bonding energy
between the reactants and products.
ΔHo
products
Reaction coordinate
Conditions favoring product formation (6.5, 6.6)
Variable
Keq
Value
Keq > 1
Meaning
More product than starting material is present at equilibrium.
ΔGo
ΔGo < 0
The energy of the products is lower than the energy of the reactants.
ΔHo
ΔHo < 0
Bonds in the products are stronger than bonds in the reactants.
ΔS
ΔS > 0
The product is more disordered than the reactant.
o
o
Equations (6.5, 6.6)
ΔGo = −2.303RT log Keq
ΔGo
Keq depends on the energy difference
between reactants and products.
R = 8.314 J/(K•mol), the gas constant
T = Kelvin temperature (K)
free energy
change
=
ΔHo
TΔSo
change in
bonding energy
T = Kelvin temperature (K)
Factors affecting reaction rate (6.9)
Factor
energy of activation
concentration
temperature
−
Effect
higher Ea → slower reaction
higher concentration → faster reaction
higher temperature → faster reaction
change in
disorder
148
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Chapter 6–4
Practice Test on Chapter Review
1. Label each statement as TRUE (T) or FALSE (F) for a reaction with Keq = 0.5 and Ea = 18 kJ/mol.
Ignore entropy considerations.
a.
b.
c.
d.
e.
The reaction is faster than a reaction with Keq = 8 and Ea = 18 kJ/mol.
The reaction is faster than a reaction with Keq = 0.5 and Ea = 12 kJ/mol.
ΔG° for the reaction is a positive value.
The starting materials are lower in energy than the products of the reaction.
The reaction is exothermic.
2. a. Which of the following statements is true about an endothermic reaction, ignoring entropy
considerations?
1.
2.
3.
4.
5.
The bonds in the products are stronger than the bonds in the starting materials.
Keq < 1.
A catalyst speeds up the rate of the reaction and gives a larger amount of product.
Statements (1) and (2) are both true.
Statements (1), (2), and (3) are all true.
b. Which of the following statements is true about a reaction with Keq = 103 and Ea = 2.5 kJ/mol?
Ignore entropy considerations.
1.
2.
3.
4.
5.
The reaction is faster than a reaction with Ea = 4 kJ/mol.
The starting materials are higher in energy than the products of the reaction.
ΔG° is positive.
Statements (1) and (2) are both true.
Statements (1), (2), and (3) are all true.
3. a. Draw the transition state for the following reaction.
CH3
C
CH3
CH3
H 2O
CH3
C
CH2
H 3O +
CH3
b. Draw the transition state for the following one-step elimination reaction.
H H
CH3CH2 C C
H H
Br
OCH3
CH3CH2CH CH2
CH3OH
Br
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
149
Understanding Organic Reactions 6–5
Answers to Practice Test
1. a. F
b. F
c. T
d. T
e. F
2. a. 2
b. 4
3. a.
b.
CH3 +
δ
C CH2
CH3
H
δ−
H
δ+
O
H
OCH3
H
CH3CH2 C
C
H
H
H
Br
δ−
Answers to Problems
6.1
[1] In a substitution reaction, one group replaces another.
[2] In an elimination reaction, elements of the starting material are lost and a π bond is formed.
[3] In an addition reaction, elements are added to the starting material.
OH
O
Br
c.
a.
C
CH3
Br replaces OH =
substitution reaction
b.
CH3
CH3
C
CH2Cl
Cl replaces H =
substitution reaction
H
O
O
d.
CH3CH2CH(OH)CH3
CH3CH CHCH3
OH
π bond formed
elements lost
(H + OH)
elimination reaction
addition of 2 H's
addition reaction
6.2 Heterolysis means one atom gets both of the electrons when a bond is broken. A carbocation
is a C with a positive charge, and a carbanion is a C with a negative charge.
heterolysis
heterolysis
CH3
a.
CH3 C OH
b.
heterolysis
c. CH3CH2 Li
Br
CH3
Electrons go to the more
electronegative atom, O.
Electrons go to the more
electronegative atom, Br.
Electrons go to the more
electronegative atom, C.
CH3
CH3 C
OH
Br
CH3
carbocation
carbocation
CH3CH2
carbanion
Li
150
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Chapter 6–6
6.3 Use full-headed arrows to show the movement of electron pairs, and half-headed arrows to
show the movement of single electrons.
CH3
a. (CH3)3C
N N
(CH3)3C
+
c.
N N
CH3
+
CH3 C
Br
CH3 C Br
CH3
b.
CH3
+
CH3
CH3 CH3
d.
CH3
2 HO
HO OH
6.4 Increasing number of electrons between atoms = increasing bond strength = increasing bond
dissociation energy = decreasing bond length.
Increasing size of an atom = increasing bond length = decreasing bond strength.
a. H Cl
or
H Br
Br is larger than Cl.
longer,
weaker bond
higher bond
dissociation energy
b. CH3 OH
or
CH3 SH
c. (CH3)2C O
S is larger than O.
longer,
higher bond weaker bond
dissociation energy
or
higher bond
dissociation energy
CH3 OCH3
single bond
fewer electrons
6.5 To determine ΔHo for a reaction:
[1] Add the bond dissociation energies for all bonds broken in the equation (+ values).
[2] Add the bond dissociation energies for all of the bonds formed in the equation (− values).
[3] Add the energies together to get the ΔHo for the reaction.
A positive ΔHo means the reaction is endothermic. A negative ΔHo means the reaction is
exothermic.
a. CH3CH2 Br + H2O
[1] Bonds broken
CH3CH2 OH
+ HBr
[2] Bonds formed
ΔHo (kJ/mol)
ΔHo (kJ/mol)
CH3CH2 Br
+ 285
CH3CH2 OH
− 393
H OH
+ 498
H Br
− 368
+ 783 kJ/mol
Total
− 761 kJ/mol
Total
[3] Overall ΔHo =
sum in Step [1]
+
sum in Step [2]
+ 783 kJ/mol
− 761
kJ/mol
ANSWER: + 22 kJ/mol
endothermic
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
151
Understanding Organic Reactions 6–7
b. CH4 + Cl2
CH3Cl + HCl
[1] Bonds broken
[3] Overall ΔHo =
[2] Bonds formed
ΔHo (kJ/mol)
ΔHo (kJ/mol)
CH3 H
+ 435
CH3 Cl
− 351
Cl Cl
+ 242
H Cl
− 431
Total
+ 677 kJ/mol
sum in Step [1]
+
sum in Step [2]
+ 677 kJ/mol
− 782 kJ/mol
Total
− 782 kJ/mol
ANSWER: − 105 kJ/mol
exothermic
6.6 Use the directions from Answer 6.5. In determining the number of bonds broken or formed,
you must take into account the coefficients needed to balance an equation.
a. CH4 + 2 O2
CO2 + 2 H2O
ΔHo (kJ/mol)
CH3 H
O O
ΔHo (kJ/mol)
sum in Step [1]
+
sum in Step [2]
OC O − 535 x 2 = − 1070
+ 435 x 4 = + 1740
+ 497 x 2 = + 994
HO H − 498 x 4 = − 1992
+ 2734 kJ/mol
Total
[3] Overall ΔHo =
[2] Bonds formed
[1] Bonds broken
+ 2734 kJ/mol
− 3062 kJ/mol
Total
− 3062 kJ/mol
ANSWER:
b. 2 CH3CH3 + 7 O2
[3] Overall ΔHo =
[2] Bonds formed
[1] Bonds broken
ΔHo (kJ/mol)
CH3CH2 H + 410 x 12 = + 4920
O O
+ 497 x 7 = + 3479
C C
+ 368 x 2 =
Total
− 328 kJ/mol
4 CO2 + 6 H2O
+736
ΔHo (kJ/mol)
OC O
sum in Step [1]
+
sum in Step [2]
− 535 x 8 = − 4280
HO H − 498 x 12 = − 5976
Total
+ 9135 kJ/mol
− 10256 kJ/mol
− 10256 kJ/mol
+ 9135 kJ/mol
ANSWER: − 1121 kJ/mol
6.7 Use the following relationships to answer the questions:
If Keq = 1, then G° = 0; if Keq > 1, then G° < 0; if Keq < 1, then G° > 0.
a. A negative value of G° means the equilibrium favors the product and Keq is > 1.
Therefore, Keq = 1000 is the answer.
b. A lower value of G° means a larger value of Keq, and the products are more favored.
Keq = 10–2 is larger than Keq = 10–5, so G° is lower.
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Chapter 6–8
6.8 Use the relationships from Answer 6.7.
a. Keq = 5.5. Keq > 1 means that the equilibrium favors the product.
b. G° = 40 kJ/mol. A positive G° means the equilibrium favors the starting material.
6.9 When the product is lower in energy than the starting material, the equilibrium favors the product.
When the starting material is lower in energy than the product, the equilibrium favors the
starting material.
a. G° is positive, so the equilibrium favors the starting material. Therefore the starting
material is lower in energy than the product.
b. Keq is > 1, so the equilibrium favors the product. Therefore the product is lower in energy
than the starting material.
c. G° is negative, so the equilibrium favors the product. Therefore the product is lower in
energy than the starting material.
d. Keq is < 1, so the equilibrium favors the starting material. Therefore the starting material is
lower in energy than the product.
6.10
H
H
OCH3
Keq = 2.7
OCH3
a. The Keq is > 1, so the product (the conformation on the right) is favored at equilibrium.
b. The G° for this process must be negative, because the product is favored.
c. G° is somewhere between 0 and −6 kJ/mol.
6.11 A positive H° favors the starting material. A negative H° favors the product.
a. H° is positive (80 kJ/mol). The starting material is favored.
b. H° is negative (–40 kJ/mol). The product is favored.
6.12
a.
b.
c.
d.
e.
False.
True.
False.
True.
False.
The reaction is endothermic.
This assumes that G° is approximately equal to H°.
Keq < 1.
a.
b.
c.
d.
e.
True.
False. G° for the reaction is negative.
True.
False. The bonds in the product are stronger than the bonds in the starting material.
True.
The starting material is favored at equilibrium.
6.13
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Understanding Organic Reactions 6–9
6.14
transition
state
Energy
Ea
product
² H°
starting
material
Reaction coordinate
6.15 A transition state is drawn with dashed lines to indicate the partially broken and partially formed
bonds. Any atom that gains or loses a charge contains a partial charge in the transition state.
CH3
CH3
a. CH3 C OH2
H2O
CH3 C
CH3
b. CH3O H
CH3
CH3
+
CH3O
δ−
transition state:
δ+
transition state: CH3 δ C
OH
CH3O
H2O
δ−
H
OH
OH2
CH3
6.16
transition
state 2: E
Energy
transition
state 1: D
Ea(A–C)
Ea(A–B)
B
H°A–B = endothermic
H°A–C = exothermic
A
a. Reaction A–C is exothermic.
Reaction A–B is endothermic.
b. Reaction A–C is faster.
c. Reaction A–C generates a lowerenergy product.
d. See labels.
e. See labels.
f. See labels.
C
Reaction coordinate
6.17
reactive
intermediate
Energy
transition
state 1
Ea1
a.
b.
c.
d.
e.
transition
state 2
Ea2
H°1
Reaction coordinate
H°2
H°overall
Two steps, because there are two energy barriers.
See labels.
See labels.
One reactive intermediate is formed (see label).
The first step is rate-determining, because its
transition state is at higher energy.
f. The overall reaction is endothermic, because the
energy of the products is higher than the energy
of the reactants.
154
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Chapter 6–10
6.18
Energy
Ea(B–C)
B
Ea(A–B)
relative energies: C < A < B
B
C is rate-determining.
A
C
Reaction coordinate
6.19 Ea, concentration, and temperature affect reaction rate. H°, G°, and Keq do not affect
reaction rate.
a. Ea = 4 kJ/mol corresponds to a faster reaction rate.
b. A temperature of 25 °C will have a faster reaction rate, because a higher temperature
corresponds to a faster reaction.
c. No change: Keq does not affect reaction rate.
d. No change: H° does not affect reaction rate.
6.20
a.
b.
c.
d.
e.
False. The reaction occurs at the same rate as a reaction with Keq = 8 and Ea = 80 kJ/mol.
False. The reaction is slower than a reaction with Keq = 0.8 and Ea = 40 kJ/mol.
True.
True.
False. The reaction is endothermic.
6.21 All reactants in the rate equation determine the rate of the reaction.
[1] rate = k[CH3CH2Br][–OH]
[2] rate = k[(CH3)3COH]
a. Tripling the concentration of CH3CH2Br
only → The rate is tripled.
b. Tripling the concentration of –OH only →
The rate is tripled.
c. Tripling the concentration of both
CH3CH2CH2Br and –OH → The rate
increases by a factor of 9 (3 × 3 = 9).
a. Doubling the concentration of (CH3)3COH →
The rate is doubled.
b. Increasing the concentration of (CH3)3COH by
a factor of 10 → The rate increases by a
factor of 10.
6.22 The rate equation is determined by the rate-determining step.
a.
CH3CH2 Br
b.
(CH3)3C
+
CH2 CH2
OH
(CH3)3C+
Br
slow
+
Br
OH
fast
+ H2O + Br
(CH3)2C
one step
rate = k[CH3CH2Br][–OH]
CH2 + H2O
two steps
The slow step determines the rate equation.
rate = k[(CH3)3CBr]
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Understanding Organic Reactions 6–11
6.23 A catalyst is not used up or changed in the reaction. It only speeds up the reaction rate.
OH and H are added to
the starting material.
H2O
CH2 CH2
a.
H2 adds to the starting material.
I– not used up = catalyst.
b.
CH3CH2OH
I–
CH3Cl
CH3OH
–
H2SO4
O
Pt
OH
−OH
H2SO4 is not used up = catalyst.
substitutes for Cl−.
OH
H2
c.
Pt not used up = catalyst.
6.24
H
H
H
a.
H
radical
H
b.
H
CH3 C OH
OH
CH3 C
H
H
carbocation
6.25
propane
propene
CH3 CH2CH3
CH3
ΔHo = 356 kJ/mol
This bond is formed from two sp3
hybridized C's.
CH2CH3
CH3 CH CH2
CH3
CH CH2
ΔHo = 385 kJ/mol
This bond is formed from one sp2 and one
sp3 hybridized C. The higher percent scharacter in one C makes a stronger bond;
thus, the bond dissociation energy is higher.
6.26 Use the directions from Answer 6.1.
O
HO OH
O
H OH
c.
a.
π bond formed
elements lost
(H + OH)
elimination reaction
b.
H
H
Cl replaces H =
substitution reaction
H
Cl
addition of 2 H's
addition reaction
O
d.
C
O
Cl
H replaces Cl =
substitution reaction
C
H
156
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Chapter 6–12
6.27 Use the rules in Answer 6.3 to draw the arrows.
Br
a.
+
Br
O
b.
CH3
Cl
+
CH3
d.
+ Cl
e.
+
+ Br
Br Br
O
CH3 C CH3
c.
Br
Cl
C
CH3
f.
CH3 Cl
CH3CH2 Br
+
CH3CH2OH +
OH
H
CH3
C
CH3
CH3
C H +
H
+ H2O
C C
OH
CH3
H
Br
H
6.28
a.
+
I
OH +
OH
O
I
OH
O
b. CH3 C CH2CH2CH3
CH3
OCH2CH3
C
CH2CH2CH3
H
H C
c.
d.
H
H
C H
H
Br
H
C
H
H
+ H2O + Br
C C
H
+
H
H
C
Cl
H
H + HCl
OCH2CH3
6.29 Draw the curved arrows to identify the product X.
O
+
H Br
[1]
O H
+
Br
Br
[2]
OH
A
B
X
6.30 Follow the curved arrows to identify the intermediate Y.
CO2H
CO2H [1]
[2]
O
COOH
O
O
O
O
O
D
C
Y
6.31 Use the rules from Answer 6.4.
a.
I
CCl3
Br CCl3
largest halogen intermediate
weakest bond bond strength
Cl CCl3
smallest halogen
strongest bond
b. H2N NH2
single bond
weakest bond
HN NH
double bond
intermediate
bond strength
N N
triple bond
strongest bond
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Understanding Organic Reactions 6–13
6.32 Use the directions from Answer 6.5.
a. CH3CH2 H
+ Br2
CH3CH2 Br
[1] Bonds broken
+ HBr
[3] Overall ΔHo =
[2] Bonds formed
ΔHo (kJ/mol)
ΔHo (kJ/mol)
CH3CH2 H
+ 410
CH3CH2 Br
− 285
+ 602 kJ/mol
Br Br
+ 192
H Br
− 368
− 653 kJ/mol
+ 602 kJ/mol
Total
− 653 kJ/mol
Total
b. OH + CH4
CH3 + H2O
[1] Bonds broken
CH3 H
[3] Overall ΔHo =
[2] Bonds formed
ΔHo (kJ/mol)
c.
ANSWER: − 51 kJ/mol
− 498 kJ/mol
H OH
+ 435 kJ/mol
CH3 OH + HBr
+ 435 kJ/mol
ΔHo (kJ/mol)
CH3 Br
ANSWER: − 63 kJ/mol
+ H2O
[3] Overall ΔHo =
[2] Bonds formed
[1] Bonds broken
− 498 kJ/mol
ΔHo (kJ/mol)
ΔHo (kJ/mol)
CH3 OH
+ 389
CH3 Br
− 293
+ 757 kJ/mol
H Br
+ 368
H OH
− 498
− 791 kJ/mol
Total
+ 757 kJ/mol
d. Br + CH4
ANSWER: − 34 kJ/mol
H + CH3Br
[1] Bonds broken
[3] Overall ΔHo =
[2] Bonds formed
ΔHo (kJ/mol)
CH3 H
− 791 kJ/mol
Total
+ 435 kJ/mol
ΔHo (kJ/mol)
+ 435 kJ/mol
CH3 Br
−293 kJ/mol
− 293 kJ/mol
ANSWER: + 142 kJ/mol
6.33
H
CH2 CH C H
H
H
H
CH2 CH C H
CH2 CH
C H
hybrid:
δ
H
CH2 CH C H
δ
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Chapter 6–14
6.34 The more stable radical is formed by a reaction with a smaller ΔH°.
H
CH3 CH2
C H
CH3 CH2
H
C H
ΔHo = 410 kJ/mol = less stable radical
H
This C–H bond is stronger.
A
H
CH3 C
CH3
CH3 C
H
CH3
ΔHo = 397 kJ/mol = more stable radical
H
This C–H bond is weaker.
B
Because the bond dissociation for cleavage of the C–H bond to form radical A is higher, more
energy must be added to form it. This makes A higher in energy and therefore less stable than B.
6.35 Use the rules from Answer 6.9.
a. Keq = 0.5. Keq is less than one, so the starting material is favored.
b. ΔGo = −100 kJ/mol. ΔGo is less than 0, so the product is favored.
c. ΔHo = 8.0 kJ/mol. ΔHo is positive, so the starting material is favored.
d. Keq = 16. Keq is greater than one, so the product is favored.
e. ΔGo = 2.0 kJ/mol. ΔGo is greater than zero, so the starting material is favored.
f. ΔHo = 200 kJ/mol. ΔHo is positive, so the starting material is favored.
g. ΔSo = 8 J/(K•mol). ΔSo is greater than zero, so the product is more disordered and favored.
h. ΔSo = −8 J/(K•mol). ΔSo is less than zero, so the starting material is more disordered and
favored.
6.36
a. A negative G° must have Keq > 1. Keq = 102.
b. Keq = [products]/[reactants] = [1]/[5] = 0.2 = Keq. G° is positive.
c. A negative G° has Keq > 1, and a positive G° has Keq < 1. G° = −8 kJ/mol will have a
larger Keq.
6.37
R
axial
equatorial
H
R
H
R
–CH3
–CH2CH3
–CH(CH3)2
–C(CH3)3
Keq
18
23
38
4000
a. The equatorial conformation is always present in the larger amount at equilibrium, because
the Keq for all R groups is greater than 1.
b. The cyclohexane with the –C(CH3)3 group will have the greatest amount of equatorial
conformation at equilibrium, because this group has the highest Keq.
c. The cyclohexane with the –CH3 group will have the greatest amount of axial conformation
at equilibrium, because this group has the lowest Keq.
d. The cyclohexane with the –C(CH3)3 group will have the most negative G°, because it has
the largest Keq.
e. The larger the R group, the more favored the equatorial conformation.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
159
Understanding Organic Reactions 6–15
f. The Keq for tert-butylcyclohexane is much higher because the tert-butyl group is bulkier than the
other groups. With a tert-butyl group, a CH3 group is always oriented over the ring when the
group is axial, creating severe 1,3-diaxial interactions. With all other substituents, the larger CH3
groups can be oriented away from the ring, placing a H over the ring, making the 1,3-diaxial
interactions less severe. Compare:
tert-butylcyclohexane
isopropylcyclohexane
CH3
CH3
H
C CH3
H
H
H
C
H
severe 1,3-diaxial interactions with
the CH3 group and the axial H's
CH3
CH3
less severe 1,3-diaxial interactions
6.38 Calculate Keq, and then find the percentage of axial and equatorial conformations present at
equilibrium.
F (axial)
a.
H
F (equatorial)
H
fluorocyclohexane
1 part
1.5 parts
G° = –5.9log Keq
G° = –1.0 kJ/mol
–1.0 kJ/mol = –5.9log Keq
Keq = 1.5
b. Keq = [products]/[reactants]
1.5 = [products]/[reactants]
1.5[reactants] = [products]
[reactants] = 0.4 = 40% axial
[products] = 0.6 = 60% equatorial
6.39 Reactions resulting in an increase in entropy are favored. When a single molecule forms two
molecules, there is an increase in entropy.
increased number of molecules
ΔS° is positive.
products favored
decreased number of molecules
ΔS° is negative.
starting material favored
increased number of molecules
H2O
ΔS° is positive.
products favored
+
a.
b.
CH3
c.
(CH3)2C(OH)2
d.
+
CH3COOCH3
CH3CH3
CH3
(CH3)2C=O
+
H2 O
+
CH3COOH
+
CH3OH
no change in the number of molecules
neither favored
6.40 Use the directions in Answer 6.15 to draw the transition state. Nonbonded electron pairs are
drawn in at reacting sites.
Br
+
a.
transition
state:
δ−
δ+ Br
F
+
Br
b.
BF3
+
transition
state:
Cl
F B Cl
F
F
F B
F
Cl
δ− δ−
160
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 6–16
OH
c.
O
+
NH2
δ−
+
δ−
NH3
C
C H
CH3
+ H2O
transition
state:
H
CH3
H
CH3 δ+
C
CH3
+
C C
H
O H NH2
transition
state:
CH3
H
CH3
d.
H3O
H
δ+
C H OH2
H
6.41
A
H°
B
B
EaA–B
A
Reaction coordinate
• one step A B
• endothermic since
B higher than A
• high Ea
(large energy barrier)
Energy
Ea
A
B
Reaction coordinate
• one step A B
• exothermic since
B lower than A
• low Ea
(small energy barrier)
Energy
Ea
H°
d.
c.
b.
Energy
Energy
a.
EaB–C
Ea = 16 kJ/mol
A
C
Reaction coordinate H°overall
• two steps
• A lowest energy
• B highest energy
• Ea(A-B) is ratedetermining,
since the transition
state for Step [1] is
higher in energy.
H° = −80 kJ/mol
B
Reaction coordinate
• one step A
B
• exothermic since
B lower than A
6.42
a.
CH3 H
b.
Cl + CH4
CH3 + HCl
Cl
CH3 + HCl
[1] Bonds broken
ΔHo (kJ/mol)
+ 435 kJ/mol
Energy
c.
Ea = 16 kJ/mol
H°
= 4 kJ/mol
Reaction coordinate
+ 435 kJ/mol
ΔHo (kJ/mol)
H Cl
− 431 kJ/mol
− 431 kJ/mol
ANSWER:
+ 4 kJ/mol
d. The Ea for the reverse reaction is the difference in energy
between the products and the transition state, 12 kJ/mol.
16 kJ/mol
Energy
CH3 H
[3] Overall ΔHo =
[2] Bonds formed
12 kJ/mol
4 kJ/mol
Reaction coordinate
+
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
161
Understanding Organic Reactions 6–17
6.43
D
Energy
B
a.
b.
c.
d.
B, D, and F are transition states.
C and E are reactive intermediates.
The overall reaction has three steps.
A–C is endothermic.
C–E is exothermic.
E–G is exothermic.
e. The overall reaction is exothermic.
F
C
E
A
G
Reaction coordinate
6.44
O
CH3
C
O
CH3
CH3 C O
OH
CH3
CH3
C
CH3
CH3 C OH
O
CH3
Since pKa (CH3CO2H) = 4.8 and pKa [(CH3)3COH] = 18, the weaker acid is formed as product,
and equilibrium favors the products. Thus, ΔH° is negative, and the products are lower in
energy than the starting materials.
transition
state
O
CH3
−
C δ
O
Energy
transition state:
δ− CH3
H
O C CH3
CH3
Ea
starting
materials
² H°
products
Reaction coordinate
6.45
H
H
H
H
H Cl
[1]
H
+H
H
+ Cl
[2]
Cl
H
a. Step [1] breaks one π bond and the H−Cl bond, and one C−H bond is formed. The H° for
this step should be positive, because more bonds are broken than formed.
b. Step [2] forms one bond. The H° for this step should be negative, because one bond is
formed and none is broken.
c. Step [1] is rate-determining, because it is more difficult.
d. Transition state for Step [1]:
Transition state for Step [2]:
δ−
H
H
δ+ H
Cl
δ+
H
H
H
−
Cl δ
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 6–18
e.
Energy
Ea2
H°1 is positive.
Ea1
H°2 is negative.
H°overall is negative.
Reaction coordinate
6.46
B
C
Ea
Energy
162
Ea
(CH3)3C +
A
+ H2O
+ I
Ea
(CH3)3C
+ HI
OH (CH3)3C
+
H°3
H°2
H°1 = 0
OH2
H°overall
+ I
(CH3)3C
I
Reaction coordinate
a. The reaction has three steps, because there are three energy barriers.
b. See above.
c. Transition state A (see graph for location): Transition state B:
Transition state C:
δ+
(CH3)3C
OH
H
δ−
I
(CH3)3C
δ+
OH2
δ+
−
δ+ δ
I
(CH3)3C
d. Step [2] is rate-determining, because this step has the highest energy transition state.
6.47 Ea, concentration, catalysts, rate constant, and temperature affect reaction rate so (c), (d), (e), (g),
and (h) affect rate.
6.48
a.
b.
c.
d.
rate = k[CH3Br][NaCN]
Double [CH3Br] = rate doubles.
Halve [NaCN] = rate halved.
Increase both [CH3Br] and [NaCN] by factor of 5 = [5][5] = rate increases by a factor of 25.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
163
Understanding Organic Reactions 6–19
6.49
O
acetyl
chloride
CH3
C
[1]
Cl
slow
O
CH3O
O
[2]
CH3 C Cl
fast
CH3
C
+ Cl–
OCH3
methyl acetate
CH3O
a. Only the slow step is included in the rate equation: Rate = k[CH3O–][CH3COCl]
b. CH3O– is in the rate equation. Increasing its concentration by 10 times would increase the rate
by 10 times.
c. When both reactant concentrations are increased by 10 times, the rate increases by 100 times
(10 × 10 = 100).
d. This is a substitution reaction (OCH3 substitutes for Cl).
6.50
a.
b.
c.
d.
e.
f.
True: Increasing temperature increases reaction rate.
True: If a reaction is fast, then it has a large rate constant.
False: Corrected—There is no relationship between G° and reaction rate.
False: Corrected—When the Ea is large, the rate constant is small.
False: Corrected—There is no relationship between Keq and reaction rate.
False: Corrected—Increasing the concentration of a reactant increases the rate of a reaction
only if the reactant appears in the rate equation.
6.51
a. The first mechanism has one step: Rate = k[(CH3)3CI][–OH]
b. The second mechanism has two steps, but only the first step would be in the rate equation,
because it is slow and therefore rate-determining: Rate = k[(CH3)3CI]
c. Possibility [1] is second order; possibility [2] is first order.
d. These rate equations can be used to show which mechanism is plausible by changing the
concentration of –OH. If this affects the rate, then possibility [1] is reasonable. If it does not
affect the rate, then possibility [2] is reasonable.
CH3
e.
CH3 C
CH2
Energy
δ– I
H
δ–
OH
Ea
H°
B
A
Reaction coordinate
A = (CH3)3CI + –OH
B = (CH3)2C=CH2 + I– + H2O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 6–20
transition state
[1]
f.
transition state
[2]
B
Energy
164
Ea[1]
H°[1]
(CH3)3C+
+ I–
+ –OH
CH3
[1] CH3 C +CH3
δ
I δ–
Ea[2]
H°[2]
H°overall
(CH3)3CI
(CH3)2C=CH2
+ H2O
CH3
[2] CH3 C CH2
δ+
H
OH
δ–
Reaction coordinate
6.52 The difference in both the acidity and the bond dissociation energy of CH3CH3 versus HC≡CH
is due to the same factor: percent s-character. The difference results because one process is
based on homolysis and one is based on heterolysis.
Bond dissociation energy:
CH3CH2 H
sp3 hybridized
25% s-character
HC C H
sp hybridized
50% s-character
Higher percent s-character makes
this bond shorter and stronger.
Acidity: To compare acidity, we must compare the stability of the conjugate bases:
CH3CH2
sp3 hybridized
25% s-character
HC C
sp hybridized
50% s-character
Now a higher percent s-character
stabilizes the conjugate base making
the starting acid more acidic.
165
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Understanding Organic Reactions 6–21
6.53 a. Re-draw A to see more clearly how cyclization occurs.
σ bonds formed
+
re-draw
+
+
σ bond formed
A
O
B
O
b.
6.54
Ha Hb
a.
Hb
Hb
C C CH3
C C CH3
C C CH3
H H
H H
H H
H H
Hb
b.
Hb
C C CH3
Hb
C C CH3
C C CH3
H H
H H
Ha Hb
Ha
Ha
C C CH3
C C CH3
C C CH3
H H
H H
H H
c. C–Ha is weaker than C–Hb since the carbon radical formed when the C–Ha bond is broken is
highly resonance stabilized. This means the bond dissociation energy for C–Ha is lower.
6.55 In Reaction [1], the number of molecules of reactants and products stays the same, so entropy is
not a factor. In Reaction [2], a single molecule of starting material forms two molecules of
products, so entropy increases. This makes G° more favorable, thus increasing Keq.
6.56
O
CH3
C
O
OH
+ CH3CH2OH
CH3
C
OCH2CH3
+ H 2O
Keq = 4
ethyl acetate
To increase the yield of ethyl acetate, H2O can be removed from the reaction
mixture, or there can be a large excess of one of the starting materials.
166
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Chapter 6–22
6.57
a.
O H
O
O
CH3CH2 O H
ethanol
phenol
O
resonance stabilized
less energy for homolysis
O
O H
Csp2–O
higher % s-character
shorter bond
no resonance stabilization
Less energy is required for cleavage
of C6H5O–H because homolysis
forms the more stable radical.
O
b.
CH3CH2 O
CH3CH2 O H
Csp3–O
lower % s-character
longer bond
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
167
Alkyl Halides and Nucleophilic Substitution 7–1
Chapter 7 Alkyl Halides and Nucleophilic Substitution
Chapter Review
General facts about alkyl halides
•
•
•
•
Alkyl halides contain a halogen atom X bonded to an sp3 hybridized carbon (7.1).
Alkyl halides are named as halo alkanes, with the halogen as a substituent (7.2).
Alkyl halides have a polar C–X bond, so they exhibit dipole–dipole interactions but are
incapable of intermolecular hydrogen bonding (7.3).
The polar C–X bond containing an electrophilic carbon makes alkyl halides reactive towards
nucleophiles and bases (7.5).
The central theme (7.6)
•
Nucleophilic substitution is one of the two main reactions of alkyl halides. A nucleophile
replaces a leaving group on an sp3 hybridized carbon.
R X
+
Nu
R Nu
nucleophile
+
X
leaving group
The electron pair in the C−Nu bond
comes from the nucleophile.
•
•
One σ bond is broken and one σ bond is formed.
There are two possible mechanisms: SN1 and SN2.
SN1 and SN2 mechanisms compared
SN2 mechanism
One step (7.11B)
•
[2] Alkyl halide
•
Order of reactivity: CH3X > RCH2X •
> R2CHX > R3CX (7.11D)
[3] Rate equation
•
•
rate = k[RX][:Nu–]
second-order kinetics (7.11A)
•
•
rate = k[RX]
first-order kinetics (7.13A)
[4] Stereochemistry
•
backside attack of the nucleophile
(7.11C)
inversion of configuration at a
stereogenic center
favored by stronger nucleophiles
(7.17B)
better leaving group → faster
reaction (7.17C)
•
trigonal planar carbocation
intermediate (7.13C)
racemization at a stereogenic
center
favored by weaker nucleophiles
(7.17B)
better leaving group → faster
reaction (7.17C)
favored by polar aprotic solvents
(7.17D)
•
•
[5] Nucleophile
•
[6] Leaving group
•
[7] Solvent
•
•
SN1 mechanism
Two steps (7.13B)
[1] Mechanism
•
•
•
Order of reactivity: R3CX >
R2CHX > RCH2X > CH3X
(7.13D)
favored by polar protic
solvents (7.17D)
168
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Chapter 7–2
Increasing rate of an SN1 reaction
H
H
R
H C Br
H
R C Br
R C Br
R C Br
H
H
R
R
methyl
1
o
3o
both
SN1 and SN2
SN1
o
2
SN2
Increasing rate of an SN2 reaction
Important trends
•
The best leaving group is the weakest base. Leaving group ability increases left to right across a
row and down a column of the periodic table (7.7).
Increasing basicity
NH3
Increasing basicity
H2O
F
Increasing leaving group ability
•
Cl
Br
I
Increasing leaving group ability
Nucleophilicity decreases across a row of the periodic table (7.8A).
For 2nd row elements
with the same charge:
CH3
NH2
OH
F
Increasing basicity
Increasing nucleophilicity
•
Nucleophilicity decreases down a column of the periodic table in polar aprotic solvents (7.8C).
Down a column
of the periodic table
F
Cl
Br
I
Increasing nucleophilicity
in polar aprotic solvents
•
Nucleophilicity increases down a column of the periodic table in polar protic solvents (7.8C).
Down a column
of the periodic table
F
Cl
Br
I
Increasing nucleophilicity
in polar protic solvents
•
The stability of a carbocation increases as the number of R groups bonded to the positively
charged carbon increases (7.14).
+
CH3
+
RCH2
+
R2CH
+
R3C
methyl
1o
2o
3o
Increasing carbocation stability
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
169
Alkyl Halides and Nucleophilic Substitution 7–3
Important principles
•
Principle
Electron-donating groups (such as R
groups) stabilize a positive charge
(7.14A).
•
Example
3 arbocations (R3C+) are more stable than
2o arbocations (R2CH+), which are more
stable than 1o carbocations (RCH2+).
o
•
Steric hindrance decreases nucleophilicity
but not basicity (7.8B).
•
(CH3)3CO– is a stronger base but a weaker
nucleophile than CH3CH2O–.
•
Hammond postulate: In an endothermic
reaction, the more stable product is
formed faster. In an exothermic reaction,
this fact is not necessarily true (7.15).
•
SN1 reactions are faster when more stable
(more substituted) carbocations are formed,
because the rate-determining step is
endothermic.
•
Planar, sp2 hybridized atoms react with
reagents from both sides of the plane
(7.13C).
•
A trigonal planar carbocation reacts with
nucleophiles from both sides of the plane.
Practice Test on Chapter Review
1. Give the IUPAC name for the following compound, including the appropriate R,S prefix.
Cl
2. a. Which of the following carbocations is the most stable?
1.
2.
3.
4.
5.
b. Which of the following anions is the best leaving group?
1. CH3–
2. –OH
3. H–
4. –NH2
5. Cl–
c. Which species is the strongest nucleophile in polar protic solvents?
1. F–
2. –OH
3. Cl–
4. H2O
5. –SH
d. Which of the following statements is true about the given reaction?
Br
OCH2CH3
Na+ –OCH2CH3
Br
170
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 7–4
1.
2.
3.
4.
5.
The reaction follows second-order kinetics.
The rate of the reaction increases when the solvent is changed from CH3CH2OH to DMSO.
The rate of the reaction increases when the leaving group is changes from Br to F.
Statements (1) and (2) are both true.
Statements (1), (2), and (3) are all true.
3. Rank the following compounds in order of increasing reactivity in an SN1 reaction. Rank the
least reactive compound as 1, the most reactive compound as 4, and the compounds of
intermediate reactivity as 2 and 3.
Cl
Cl
OH
A
B
Cl
C
D
4. Consider the following two nucleophilic substitution reactions, labeled Reaction [1] and
Reaction [2]. (Only the starting materials are drawn.) Then answer True (T) or False (F) to each
of the following statements.
Reaction [1]
(CH3CH2)3CBr
CH3OH
Reaction [2]
CH3CH2CH2Br
OCH3
The rate equation for Reaction [1] is rate = k[(CH3CH2)3CBr][CH3OH].
Changing the leaving group from Br– to Cl– decreases the rate of both reactions.
Changing the solvent from CH3OH to (CH3)2S=O increases the rate of Reaction [2].
Doubling the concentration of both CH3CH2CH2Br and –OCH3 in Reaction [2] doubles the
rate of the reaction.
e. If entropy is ignored and Keq for Reaction [1] is < 1, then the reaction is exothermic.
f. If entropy is ignored and ΔH° is negative for Reaction [1], then the bonds in the product are
stronger than the bonds in the starting materials.
g. The energy diagram for Reaction [2] exhibits only one energy barrier.
a.
b.
c.
d.
5. Draw the organic products formed in the following reactions. Use wedges and dashes to show
stereochemistry in compounds with stereogenic centers.
a.
H
C
C
H
Br D
CH3OH
b.
(Consider substitution only.)
Cl
NaOH
c.
Br
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
171
Alkyl Halides and Nucleophilic Substitution 7–5
d.
Br
CH3COO
Answers to Practice Test
1. (7S)-7-chloro-3-ethyldecane
3. A–4
B–1
C–3
D–2
4. a. F
b. T
c. T
d. F
e. F
f. T
g. T
2. a. 4
b. 5
c. 5
d. 4
5. a.
c.
H
D C
OH
CH
b.
d.
O2CCH3
OCH3
CH3O
Answers to Problems
7.1 Classify the alkyl halide as 1°, 2°, or 3° by counting the number of carbons bonded directly
to the carbon bonded to the halogen.
C bonded to 2 C's
2° alkyl halide
C bonded to 1 C
1° alkyl halide
I
CH3
a.
F
b.
CH3CH2CH2CH2CH2 Br
c. CH3 C
CHCH3
d.
CH3 Cl
C bonded to 3 C's
3° alkyl halide
C bonded to 3 C's
3° alkyl halide
7.2 Use the directions from Answer 7.1.
3°, neither
2°, neither
2°, neither
a, b.
Cl
1°, neither
Br
Cl
Cl
3°, allylic
Cl
Br
Cl
3°, neither
telfairine
Br
vinyl
2°, neither
halomon
Cl
vinyl
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Chapter 7–6
7.3 Draw a compound of molecular formula C6H13Br to fit each description.
Br
a.
b.
Br
c.
Br
1° alkyl halide
one stereogenic center
2° alkyl halide
two stereogenic centers
3° alkyl halide
no stereogenic centers
7.4 To name a compound with the IUPAC system:
[1] Name the parent chain by finding the longest carbon chain.
[2] Number the chain so the first substituent gets the lower number. Then name and number
all substituents, giving like substituents a prefix (di, tri, etc.). To name the halogen
substituent, change the -ine ending to -o.
[3] Combine all parts, alphabetizing substituents, and ignoring all prefixes except iso.
a.
(CH3)2CHCH(Cl)CH2CH3
re-draw
[1]
2-methyl
[2]
Cl
5 carbon alkane = pentane
b.
[1]
[3] 3-chloro-2-methylpentane
2 Cl
Br
3-chloro
2-bromo
[2]
Br
[3] 2-bromo-5,5-dimethylheptane
2
7 carbon alkane = heptane
5,5-dimethyl
c.
2-methyl
[2]
[1]
Br
Br
[3] 1-bromo-2-methylcyclohexane
1-bromo
1
6 carbon cycloalkane =
cyclohexane
2-fluoro
d.
[1]
F
[2]
F
[3] 2-fluoro-5,5-dimethylheptane
5,5-dimethyl
1 2 3 4
7 carbon alkane = heptane
7
7.5 To work backwards from a name to a structure:
[1] Find the parent name and draw that number of carbons. Use the suffix to identify the
functional group (-ane = alkane).
[2] Arbitrarily number the carbons in the chain. Add the substituents to the appropriate carbon.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
173
Alkyl Halides and Nucleophilic Substitution 7–7
a. 3-chloro-2-methylhexane
[1]
6 carbon alkane
[2]
methyl at C2
1 2 3 4 5 6
chloro at C3
Cl
b. 4-ethyl-5-iodo-2,2-dimethyloctane
[1]
8 carbon alkane
[2]
ethyl at C4
1 2 3 4 5 6 7 8
I
2 methyls at C2
iodo at C5
c. cis-1,3-dichlorocyclopentane
[1]
5 carbon cycloalkane
[2]
chloro groups at C1 and C3, both on the same side
Cl
Cl
C3
C1
d. 1,1,3-tribromocyclohexane
[1]
6 carbon cycloalkane
3 Br groups
[2]
Br
C3
Br
Br
C1
e. propyl chloride
[1] 3 carbon alkyl group
CH3CH2CH2
[2] chloride on end
CH3CH2CH2 Cl
f. sec-butyl bromide
[1] 4 carbon alkyl group
CH3 CHCH2CH3
[2] bromide
CH3 CHCH2CH3
Br
7.6 a. Because an sp2 hybridized C has a higher percent s-character than an sp3 hybridized C, it
holds electron density closer to C. This pulls a little more electron density towards C, away
from Cl, and thus a Csp2–Cl bond is less polar than a Csp3–Cl bond.
b.
Cl
lowest boiling point
Cl
sp3 C–Cl bond
intermediate
boiling point
Br
larger halogen, sp3 C–Br bond
highest boiling point
7.7 a. Since chondrocole A has 10 C’s and only one functional group capable of hydrogen bonding
to water (an ether), it is insoluble in H2O. Because it is organic, it is soluble in CH2Cl2.
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Chapter 7–8
R
b.
H
Br
O
*
S
*
Three stereogenic centers are labeled with (*).
*
Cl
S
Br
c.
Br
O
O
Cl
Cl
enantiomer
constitutional isomer
7.8 To draw the products of a nucleophilic substitution reaction:
[1] Find the sp3 hybridized electrophilic carbon with a leaving group.
[2] Find the nucleophile with lone pairs or electrons in π bonds.
[3] Substitute the nucleophile for the leaving group on the electrophilic carbon.
+
a.
+ Br
OCH2CH3
Br
OCH2CH3
nonbonded e− pairs
nucleophile
leaving group
Cl
OH
b.
+
nonbonded e− pairs
nucleophile
leaving group
+
I
c.
+ Na+Cl–
Na+ −OH
N3
+ I
N3
nonbonded e− pairs
nucleophile
leaving group
Br
CN
d.
+ Na+Br–
Na+ −CN
+
nonbonded e− pairs
nucleophile
leaving group
7.9 Use the steps from Answer 7.8 and then draw the proton transfer reaction.
a.
+
Br
N(CH2CH3)3
substitution
+
N(CH2CH3)3 + Br
nucleophile
leaving group
b. (CH3)3C Cl
leaving group
+
H2O
substitution
nucleophile
(CH3)3C
O H
H
+ Cl
proton
(CH3)3C
transfer
O H
+ HCl
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Alkyl Halides and Nucleophilic Substitution 7–9
7.10 Draw the structure of CPC using the steps from Answer 7.8.
+
N
Cl
nucleophile
substitution
leaving group
+
N
+
Cl
CPC
7.11 Compare the leaving groups based on these trends:
• Better leaving groups are weaker bases.
• A neutral leaving group is always better than its conjugate base.
a. Cl–, I–
b. NH3, NH2–
farther down a column
of the periodic table
less basic
better leaving group
c. H2O, H2S
neutral compound
less basic
better leaving group
farther down a column
of the periodic table
less basic
better leaving group
7.12 Good leaving groups include Cl–, Br–, I–, and H2O.
a.
CH3CH2CH2 Br
c. CH3CH2CH2 OH2
b. CH3CH2CH2OH
Br− is a good
leaving group.
No good leaving group.
is too strong a base.
d. CH3CH3
No good leaving group.
H− is too strong a base.
H2O is a good
leaving group.
−OH
7.13 To decide whether the equilibrium favors the starting material or the products, compare the
nucleophile and the leaving group. The reaction proceeds towards the weaker base.
a.
CH3CH2 NH2
+
Br
nucleophile
better leaving group
weaker base
pKa (HBr) = –9
b.
I
+
+
CH3CH2 Br
CN
nucleophile
pKa (HCN) = 9.1
NH2
leaving group
Reaction favors
starting material.
pKa (NH3) = 38
CN
+
I
leaving group
better leaving group
weaker base
pKa (HI) = –10
Reaction favors product.
7.14 It is not possible to convert CH3CH2CH2OH to CH3CH2CH2Cl by nucleophilic substitution
with NaCl because –OH is a stronger base and poorer leaving group than Cl–. The equilibrium
favors the reactants, not the products.
CH3CH2CH2OH
Na Cl
weaker base
CH3CH2CH2Cl
Na
OH
stronger base
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Chapter 7–10
7.15 Use these three rules to find the stronger nucleophile in each pair:
[1] Comparing two nucleophiles having the same attacking atom the stronger base is a
stronger nucleophile.
[2] Negatively charged nucleophiles are always stronger than their conjugate acids.
[3] Across a row of the periodic table, nucleophilicity decreases when comparing species of
similar charge.
O
a. NH3, NH2
b. CH3NH2, CH3OH
A negatively charged
nucleophile is stronger
than its conjugate acid.
stronger nucleophile
c.
C
CH3
Across a row of the periodic
table, nucleophilicity decreases
with species of the same charge.
stronger nucleophile
O
CH3CH2O
same attacking
atom (O)
stronger base
stronger
nucleophile
7.16 Polar protic solvents are capable of hydrogen bonding, and therefore must contain a H
bonded to an electronegative O or N. Polar aprotic solvents are incapable of hydrogen
bonding, and therefore do not contain any O–H or N–H bonds.
a. HOCH2CH2OH
contains 2 O–H bonds
polar protic
b. CH3CH2OCH2CH3
no O–H bonds
polar aprotic
c. CH3COOCH2CH3
no O–H bonds
polar aprotic
7.17 • In polar protic solvents, the trend in nucleophilicity is opposite to the trend in basicity
down a column of the periodic table so that nucleophilicity increases.
• In polar aprotic solvents, the trend is identical to basicity so that nucleophilicity decreases
down a column.
a. Br− and Cl− in polar protic solvent
farther down the column
more nucleophilic
in protic solvent
b. −OH and Cl− in polar aprotic solvent
farther up the column
and to the left in the row
more basic
more nucleophilic
In polar aprotic solvents:
nucleophilicity increases
O F nucleophilicity
Cl
increases
c. HS− and F− in polar protic solvent
In polar protic solvents:
nucleophilicity increases
farther down the column
O F
nucleophilicity
and left in the row
S
increases
more nucleophilic
in protic solvent
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Alkyl Halides and Nucleophilic Substitution 7–11
7.18 The stronger base is the stronger nucleophile except in polar protic solvents when
nucleophilicity increases down a column. For other rules, see Answers 7.15 and 7.17.
a.
H2O
OH
no charge
weakest nucleophile
NH2
negatively charged
intermediate nucleophile
negatively charged
farther left in periodic table
strongest nucleophile
b.
Br
F
Basicity decreases down a
Basicity decreases
column in polar aprotic solvents.
across a row.
weakest nucleophile
intermediate nucleophile
c.
H2O
OH
strongest nucleophile
CH3COO
OH
weaker base than −OH
intermediate nucleophile
weakest nucleophile
strongest nucleophile
7.19 To determine what nucleophile is needed to carry out each reaction, look at the product to see
what has replaced the leaving group.
a. (CH3)2CHCH2CH2 Br
(CH3)2CHCH2CH2
c. (CH3)2CHCH2CH2 Br
SH
SH replaces Br.
HS− is needed.
b. (CH3)2CHCH2CH2 Br
(CH3)2CHCH2CH2
(CH3)2CHCH2CH2
OCOCH3
OCOCH3 replaces Br.
CH3COO− is needed.
OCH2CH3
d. (CH3)2CHCH2CH2 Br
OCH2CH3 replaces Br.
CH3CH2O− is needed.
(CH3)2CHCH2CH2
C CH
C≡CH replaces Br.
HC≡C− is needed.
7.20 The general rate equation for an SN2 reaction is rate = k[RX][:Nu–].
a. [RX] is tripled, and [:Nu–] stays the same: rate triples.
b. Both [RX] and [:Nu–] are tripled: rate increases by a factor of 9 (3 × 3 = 9).
c. [RX] is halved, and [:Nu–] stays the same: rate halved.
d. [RX] is halved, and [:Nu–] is doubled: rate stays the same (1/2 × 2 = 1).
7.21 The transition state in an SN2 reaction has dashed bonds to both the leaving group and the
nucleophile, and must contain partial charges.
a.
CH3CH2CH2 Cl
+
OCH3
CH3CH2CH2 OCH3
+
Cl
CH3CH2CH2
CH3O δ−
b.
Br
+
SH
SH
+
Br
δ−
SH
δ−
Br
Cl δ−
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Chapter 7–12
7.22 All SN2 reactions have one step.
Cl δ−
CH3CH2CH2
Energy
CH3O δ−
Ea
CH3CH2CH2 Cl
ΔH°
+ OCH3
CH3CH2CH2 OCH3 + Cl
Reaction coordinate
7.23 To draw the products of SN2 reactions, replace the leaving group by the nucleophile, and
then draw the stereochemistry with inversion at the stereogenic center.
CH3CH2 D
a.
C
D CH CH
2
3
Br +
CH3CH2O
OCH2CH3
b.
C
H
+
I
CN
CN
H
7.24 Increasing the number of R groups increases crowding of the transition state and decreases the
rate of an SN2 reaction.
Cl
or
a.
2° alkyl halide
b.
Cl
Br
or
2° alkyl halide
faster reaction
1° alkyl halide
faster reaction
Br
3° alkyl halide
7.25
+
H
N
N
H
CH3 SR2
H
N
A
+ SR2
N H
loss of
CH3
a proton
H
N
N
CH3
nicotine
7.26 In a first-order reaction, the rate changes with any change in [RX]. The rate is independent
of any change in [:Nu–].
a. [RX] is tripled, and [:Nu–] stays the same: rate triples.
b. Both [RX] and [:Nu–] are tripled: rate triples.
c. [RX] is halved, and [:Nu–] stays the same: rate halved.
d. [RX] is halved, and [:Nu–] is doubled: rate halved.
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Alkyl Halides and Nucleophilic Substitution 7–13
7.27 In SN1 reactions, racemization always occurs at a stereogenic center. Draw two products,
with the two possible configurations at the stereogenic center.
leaving group
nucleophile
CH3
a.
C
(CH3)2CH
CH3
H2O
Br
C
(CH3)2CH
CH2CH3
CH3
H
CH3COO
CH3
CH3CH2
Cl
(CH3)2CH
CH2CH3
+
HBr
OH
enantiomers
nucleophile
b.
C
+
OH
CH2CH3
H
CH3
CH3CH2
H
+
OOCCH3
CH3CH2
OOCCH3
CH3
diastereomers
leaving group
7.28 Carbocations are classified by the number of R groups bonded to the carbon:
0 R groups = methyl, 1 R group = 1°, 2 R groups = 2°, and 3 R groups = 3°.
a.
+
b. (CH3)3CCH2
+
1 R group
1° carbocation
2 R groups
2° carbocation
+
c.
d.
+
2 R groups
2° carbocation
3 R groups
3° carbocation
7.29 For carbocations: Increasing number of R groups = Increasing stability.
+
+
+
CH3CH2CH2CH2
CH3CHCH2CH3
CH3 C CH3
1° carbocation
least stable
2° carbocation
intermediate
stability
3° carbocation
most stable
CH3
7.30 For carbocations: Increasing number of R groups = Increasing stability.
+
CH2
+
+
1° carbocation
least stable
3° carbocation
most stable
2° carbocation
intermediate
stability
7.31 The rate of an SN1 reaction increases with increasing alkyl substitution.
CH3
a.
(CH3)3CBr
or
(CH3)3CCH2Br
1° alkyl halide
3° alkyl halide
faster SN1 reaction slower SN1 reaction
b.
Br
Br
or
3° alkyl halide
faster SN1 reaction
2° alkyl halide
slower SN1 reaction
+
Cl–
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Chapter 7–14
7.32 • For methyl and 1° alkyl halides, only SN2 will occur.
• For 2° alkyl halides, SN1 and SN2 will occur.
• For 3° alkyl halides, only SN1 will occur.
CH3 H
a. CH3 C
C
Br
b.
Br
CH3 CH3
Br
d.
2° alkyl halide
SN1 and SN2
1° alkyl halide
S N2
2° alkyl halide
SN1 and SN2
Br
c.
3° alkyl halide
SN1
7.33 • Draw the product of nucleophilic substitution for each reaction.
• For methyl and 1° alkyl halides, only SN2 will occur.
• For 2° alkyl halides, SN1 and SN2 will occur and other factors determine which mechanism
operates.
• For 3° alkyl halides, only SN1 will occur.
Strong nucleophile
favors SN2.
I
CH3OH
Cl
a.
OCH3
c.
+ HCl
3° alkyl halide
only SN1
Br
b.
OCH2CH3
CH3CH3O
+
I
+
HBr
2° alkyl halide
Both SN1 and SN2
are possible.
Weak nucleophile
favors SN1.
SH
SH
+ Br
CH3OH
d.
Br
1° alkyl halide
only SN2
OCH3
2° alkyl halide
Both SN1 and SN2
are possible.
7.34 First decide whether the reaction will proceed via an SN1 or SN2 mechanism. Then draw the
products with stereochemistry.
+
a.
H Br
Weak nucleophile
favors SN1.
2° alkyl halide
SN1 and SN2
Cl
b.
H D
1° alkyl halide
SN2 only
H 2O
+
C C H
+ HBr
+
H OH
HO H
SN1 = racemization at the
stereogenic C
enantiomers
HC C
D H
+
Cl
SN2 = inversion at the stereogenic C
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Alkyl Halides and Nucleophilic Substitution 7–15
7.35 Compounds with better leaving groups react faster. Weaker bases are better leaving groups.
a. CH3CH2CH2Cl
c. (CH3)3C OH or
CH3CH2CH2I
or
weaker base
better leaving group
b. (CH3)3CBr
+
OH2
weaker base
better leaving group
(CH3)3CI
or
(CH3)3C
d. CH3CH2CH2OH
weaker base
better leaving group
or
CH3CH2CH2 OCOCH3
weaker base
better leaving group
7.36 • Polar protic solvents favor the SN1 mechanism by solvating the intermediate carbocation
and halide.
• Polar aprotic solvents favor the SN2 mechanism by making the nucleophile stronger.
a. CH3CH2OH
polar protic solvent
contains an O–H bond
favors SN1
b. CH3CN
polar aprotic solvent
no O–H or N–H bond
favors SN2
c. CH3COOH
polar protic solvent
contains an O–H bond
favors SN1
d. CH3CH2OCH2CH3
polar aprotic solvent
no O–H or N–H bond
favors SN2
7.37 Compare the solvents in the reactions below. For the solvent to increase the reaction rate of
an SN1 reaction, the solvent must be polar protic.
+ CH3OH
a.
Cl
CH3OH
+
or
HCl
3o RX – SN1 reaction
Br
b.
OH
+
1o RX – SN2 reaction
c.
+
CH3O
H Cl
H2O
OH
+
Br
or
DMF [HCON(CH3)2]
Polar aprotic solvent
increases the rate of an
SN2 reaction.
DMF
CH3OH
+
or
HMPA
CH3OH
Polar protic solvent
increases the rate of an
SN1 reaction.
OCH3
DMSO
Cl
H OCH3
2o RX strong nucleophile
SN2 reaction
HMPA [(CH3)2N]3P=O
Polar aprotic solvent
increases the rate of an
SN2 reaction.
7.38 To predict whether the reaction follows an SN1 or SN2 mechanism:
[1] Classify RX as a methyl, 1°, 2°, or 3° halide. (Methyl, 1° = SN2; 3° = SN1; 2° = either.)
[2] Classify the nucleophile as strong or weak. (Strong favors SN2; weak favors SN1.)
[3] Classify the solvent as polar protic or polar aprotic. (Polar protic favors SN1; polar
aprotic favors SN2.)
a.
CH2Br
+
CH3CH2O
CH2OCH2CH3
+
Br
SN2 reaction
1° alkyl halide
SN2
b.
Br
2° alkyl halide
SN1 or SN2
+
N3
Strong nucleophile
favors SN2.
N3
+
Br
SN2 reaction = inversion at the stereogenic center
The leaving group was "up."
The nucleophile attacks from below.
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Chapter 7–16
I
c.
OCH3
CH3OH
+
SN1 reaction
HI
+
Weak nucleophile
favors SN1.
3° alkyl halide
SN1
d.
+
Cl
3° alkyl halide
SN1
+
H2O
Weak nucleophile
favors SN1.
+ HCl
OH
SN1 reaction
forms two enantiomers.
HO
7.39 Vinyl carbocations are even less stable than 1° carbocations.
CH3CH2CH2CH2CH=CH
CH3CH2CH2CH2CH2CH2
CH3CH2CH2CH2CHCH3
vinyl carbocation
least stable
1° carbocation
intermediate
stability
2° carbocation
most stable
7.40 Convert each ball-and-stick model to a skeletal or condensed structure and draw the reactants.
Na+ −CN
a.
CN
carbon
framework
Cl
CN
nucleophile
b. (CH3)3CCH2CH2
SH
(CH3)3CCH2CH2
Cl
Na+ −SH
(CH3)3CCH2CH2
SH
nucleophile
carbon
framework
OH
Cl
c.
OH
Na+ −OH
nucleophile
carbon
framework
Na+ −C
d.
CH3CH2 C C H
carbon
framework
CH3CH2 Cl
CH
CH3CH2 C C H
nucleophile
7.41
CH3O
Cl CH2CH3
CH3OCH2CH3
CH3CH2O
Cl CH3
CH3OCH2CH3
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Alkyl Halides and Nucleophilic Substitution 7–17
7.42 Use the directions from Answer 7.4 to name the compounds.
2
1
7 C chain = heptane
bromo at C2
ethyl at C5
one stereogenic center–R
(2R)-2-bromo-5-ethylheptane
a.
H
Br
5
R
3
5 C ring = cyclopentane
chloro at C1
isopropyl at C3
trans isomer–one substituent up, one down
(1R, 3R)-trans-1-chloro-3-isopropylcyclopentane
b.
2
1
R
Cl
R
7.43
H
CN (axial)
Br
a. (CH3)3C
CN
(eq) acetone
H
(CH3)3C
H
inversion (equatorial to axial)
H
polar aprotic solvent
SN2 reaction Large tert-butyl group is in
more roomy equatorial
position.
Br (axial)
H
b. (CH3)3C
H
CN
acetone
inversion (axial to equatorial)
CN (eq)
(CH3)3C
H
H
polar aprotic solvent
SN2 reaction
7.44 Use the directions from Answer 7.4 to name the compounds.
CH3
[1]
CH3 C CH2CH2 F
a.
[2]
[3] 1-fluoro-3,3-dimethylbutane
1
3 CH3
CH3
1-fluoro
3,3-dimethyl
4 carbon alkane = butane
b. [1]
CH3
CH3 C CH2CH2 F
I
[2]
3-ethyl
3
2-methyl
6 carbon alkane = hexane
2
1 I
1-iodo
[3]
3-ethyl-1-iodo-2-methylhexane
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Chapter 7–18
c. [1] (CH3)3CCH2Br
CH3
[2]
[3] 1-bromo-2,2-dimethylpropane
CH3 C CH2 Br
CH3
2 CH3 1
CH3 C CH2 Br
1-bromo
2,2-dimethyl
CH3
3 carbon alkane = propane
d. [1]
[2]
Br
6
2
Br
Cl
8 carbon alkane = octane
6-methyl
Br
[2]
2
I
1-bromo
[3] cis-1-bromo-3-iodocyclopentane
I
5 carbon cycloalkane =
cyclopentane
Cl
[1]
f.
Cl
6-bromo 2-chloro
Br
e. [1]
[3] 6-bromo-2-chloro-6-methyloctane
1
3-iodo
1
Cl
[2]
[3] trans-1,2-dichlorocyclohexane
trans-1,2-dichloro
Cl
Cl
6 carbon cycloalkane =
cyclohexane
2
g. [1] (CH3)3CCH2CH(Cl)CH2Cl
CH3
CH3
[2]
H
CH3
CH3 C CH2 C CH2 Cl
CH3
H
[3] 1,2-dichloro-4,4-dimethylpentane
CH3 C CH2 C CH2 Cl
Cl
Cl
4,4-dimethyl
1,2-dichloro
5 carbon alkane = pentane
h.
[1]
[2]
4
I H
I H
6 carbon alkane = hexane
(Indicate the R,S
designation also)
[3] (2R)-2-iodo-4,4-dimethylhexane
2 1
4,4-dimethyl
2
(2R)-2-iodo
3
I H4
Clockwise
R
1
7.45 To work backwards to a structure, use the directions in Answer 7.5.
c. 1,1-dichloro-2-methylcyclohexane
a. isopropyl bromide
Br
CH3 CHCH3
1 Cl
Bromine on middle C
makes it an isopropyl group.
2
b. 3-bromo-4-ethylheptane
3
Br
4-ethyl
1,1-dichloro
Cl
2-methyl
d. trans-1-chloro-3-iodocyclobutane
1-chloro
Cl
1
3
4
3-bromo
I
3-iodo
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Alkyl Halides and Nucleophilic Substitution 7–19
g. (1R,2R)-trans-1-bromo-2-chlorocyclohexane
1R
1-bromo
Br
1
e. 1-bromo-4-ethyl-3-fluorooctane
4-ethyl
Br
3
1
1-bromo
4
F
Cl
2-chloro
2R
h. (5R)-4,4,5-trichloro-3,3-dimethyldecane
5R
3,3-dimethyl
Cl H
1
4
5
3
3-fluoro
f. (3S)-3-iodo-2-methylnonane
2-methyl
4
3
1
I H
3S
3-iodo
Cl Cl
4,4,5-trichloro
7.46
H H
CH3
a.
CH3 C CH2CH2F
CH3
g.
d.
Br
1° halide
h.
I H
e.
2° halide
1° halide
2° halide
I
c.
1° halide
2° halide
2° halide
2° halide
Br
I
C C Cl
Cl H
Cl
3° halide
b.
(CH3)2CCH2
(CH3)3CCH2Br
Cl
1° halide
f.
Cl
Both are
2° halides.
7.47
1-chloro
1
Cl 1
3
3-chloropentane
Cl
1-chloropentane
1-chloro
Cl
Cl
1
1-chloro-2,2-dimethylpropane
2
2-chloro-2-methylbutane
Two stereoisomers
2-chloro
2
*
Cl
2-chloropentane
[* denotes stereogenic center]
2
3
Cl H 4
1
Clockwise
"4" in back =
R
2
3
1-chloro
1-chloro-3-methylbutane
2-chloro
2-methyl
2
3-methyl
Cl
3-chloro
3
4
1 Cl H
Clockwise
"4" in front =
S
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Chapter 7–20
Two stereoisomers
3-methyl
3
2
3
*
4H
2-chloro
Cl
2-chloro-3-methylbutane
[* denotes stereogenic center]
2
2
3
4H
Cl
1
Counterclockwise
"4" in front =
R
1Cl
Counterclockwise
"4" in back =
S
Two stereoisomers
1-chloro
2-methyl
*
3
4
3
H
4
H
Cl
Cl
2
1-chloro-2-methylbutane
[* denotes stereogenic center]
Cl
1
Clockwise
"4" in front =
S
2
1
Clockwise
"4" in back =
R
7.48
Br
a. (CH3)3CBr
or
CH3CH2CH2CH2Br
b.
larger surface area =
stronger intermolecular forces =
higher boiling point
I
or
Br
or
c.
more polar =
nonpolar
larger halide = more polarizable =
only VDW forces higher boiling point
higher boiling point
7.49
a.
CH3CH2CH2CH2 Br
b.
CH3CH2CH2CH2 Br
SH
c.
CH3CH2CH2CH2 Br
CN
d.
CH3CH2CH2CH2 Br
OCH(CH3)2
e.
CH3CH2CH2CH2 Br
C CH
f.
CH3CH2CH2CH2 Br
g.
CH3CH2CH2CH2 Br
NH3
h.
CH3CH2CH2CH2 Br
Na+ I−
i.
CH3CH2CH2CH2 Br
Na+ N3−
CH3CH2CH2CH2OH + Br
OH
H2O
CH3CH2CH2CH2SH + Br
CH3CH2CH2CH2CN + Br−
CH3CH2CH2CH2OCH(CH3)2 + Br−
CH3CH2CH2CH2C CH + Br−
CH3CH2CH2CH2OH2 + Br−
CH3CH2CH2CH2NH3 + Br−
CH3CH2CH2CH2I + Na+ Br−
CH3CH2CH2CH2N3 + Na+ Br−
CH3CH2CH2CH2OH + HBr
CH3CH2CH2CH2NH2 + HBr
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
187
Alkyl Halides and Nucleophilic Substitution 7–21
7.50 Use the steps from Answer 7.8 and then draw the proton transfer reaction, when necessary.
O
a.
+
Cl
CH3
C
CH3
nucleophile
I
leaving group
Cl
OH + HI
H2O
nucleophile
leaving group
d.
+ CH3CH2OH
OCH2CH3
Br
OCH3
+ Na+ –OCH3
f.
leaving group
+ NaBr
nucleophile
leaving group
Cl
+ HCl
nucleophile
leaving group
e.
+ NaI
nucleophile
+
I
c.
CN
O
Na+ –CN
+
Cl
O
leaving group
b.
+
O
CH3
+
SCH3 + Cl
CH3SCH3
nucleophile
7.51 A good leaving group is a weak base.
OH
c.
a.
bad leaving group
OH is a strong base.
b.
CH3CH2CH2CH2 Cl
Cl good leaving group
weak base
e.
This has only C–C
and C–H bonds.
No good leaving group.
OH2
d.
good leaving group
H2O is a weak base.
CH3CH2NH2
bad leaving group
NH2 is a strong base.
f.
CH3CH2CH2
I
I good leaving group
weak base
7.52 Use the rules from Answer 7.11.
a. increasing leaving group ability: −NH2 < −OH < F−
most basic
worst leaving
group
least basic
best leaving
group
b. increasing leaving group ability: −NH2 < −OH < H2O
most basic
least basic
worst leaving best leaving
group
group
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 7–22
c. increasing leaving group ability: Cl− < Br− < I−
d. increasing leaving group ability: NH3 < H2O < H2S
most basic
worst leaving
group
least basic
most basic
worst leaving best leaving
group
group
least basic
best leaving
group
7.53 Compare the nucleophile and the leaving group in each reaction. The reaction will occur if it
proceeds towards the weaker base. Remember that the stronger the acid (lower pKa), the
weaker the conjugate base.
I
NH2
+
+ I
a.
weaker base
pKa (HI) = –10
b.
CH3CH2I
+ CH3O
stronger base
pKa (NH3) = 38
CN
+ I
CH3CH2OCH3
stronger base
pKa (CH3OH) = 15.5
c.
Reaction will not occur.
NH2
+ I
weaker base
pKa (HI) = –10
Reaction will occur.
weaker base
pKa (HI) = –10
I
+
CN
Reaction will not occur.
stronger base
pKa (HCN) = 9.1
7.54 Use the directions in Answer 7.15.
a. Across a row of the periodic table
nucleophilicity decreases.
−OH < −NH <
2
d.
CH3−
b. • In a polar protic solvent (CH3OH), nucleophilicity
increases down a column of the periodic table,
so: −SH is more nucleophilic than −OH.
• Negatively charged species are more
nucleophilic than neutral species so −OH
is more nucleophilic than H2O.
Compare the nucleophilicity of N, S, and O.
In a polar aprotic solvent (acetone),
nucleophilicity parallels basicity.
CH3SH < CH3OH < CH3NH2
e.
In a polar aprotic solvent (acetone),
nucleophilicity parallels basicity. Across a row
and down a column of the periodic table
nucleophilicity decreases.
Cl− < F− < −OH
H2O < −OH < −SH
c. • In a polar protic solvent (CH3OH), nucleophilicity
increases down a column of the periodic table,
so: CH3CH2S− is more nucleophilic than CH3CH2O−.
• For two species with the same attacking atom, the more
basic is the more nucleophilic so CH3CH2O− is more
nucleophilic than CH3COO−.
CH3COO− < CH3CH2O− < CH3CH2S−
f.
Nucleophilicity decreases across a row so
−SH is more nucleophilic than Cl−.
In a polar protic solvent (CH3OH),
nucleophilicity increases down a column so Cl−
is more nucleophilic than F−.
F− < Cl− < −SH
189
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkyl Halides and Nucleophilic Substitution 7–23
7.55 Polar protic solvents are capable of hydrogen bonding, so they must contain a H bonded to
an electronegative O or N. Polar aprotic solvents are incapable of hydrogen bonding, so
they do not contain any O–H or N–H bonds.
a.
c.
(CH3)2CHOH
e.
no O–H or N–H bond
aprotic
contains O–H bond
protic
d.
CH3NO2
b.
CH2Cl2
no O–H or N–H bond
aprotic
N(CH3)3
no O–H or N–H bond
aprotic
f. HCONH2
contains an N–H bond
protic
NH3
contains N–H bond
protic
7.56
O
CH3
C
O
N
N
H
H
CH3
CH3
The amine N is more
nucleophilic since the electron
pair is localized on the N.
C
O
N
N
H
H
CH3
CH3
C
N
N
H
H
CH3
The amide N is less nucleophilic
since the electron pair is
delocalized by resonance.
7.57
1° alkyl halide
SN2 reaction
Br
CN
CN
+
+ Br
acetone
c. Transition state:
Ea
Br
y
b. Energy diagram:
+
Energy
a. Mechanism:
CN
² H°
CN
Br δ−
+ Br
δ−CN
Reaction coordinate
d. Rate equation: one-step reaction with both nucleophile and alkyl halide in the only step:
rate = k[R–Br][–CN]
e.
[1] The leaving group is changed from Br− to I−:
Leaving group becomes less basic → a better leaving group → faster reaction.
[2] The solvent is changed from acetone to CH3CH2OH:
Solvent changed to polar protic → decreases reaction rate.
[3] The alkyl halide is changed from CH3(CH2)4Br to CH3CH2CH2CH(Br)CH3:
Changed from 1° to 2° alkyl halide → the alkyl halide gets more crowded and the reaction
rate decreases.
[4] The concentration of −CN is increased by a factor of 5.
Reaction rate will increase by a factor of 5.
[5] The concentration of both the alkyl halide and −CN are increased by a factor of 5:
Reaction rate will increase by a factor of 25 (5 x 5 = 25).
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 7–24
7.58 Use the directions from Answer 7.24.
Br
Br
a.
Br
2° alkyl halide
intermediate
reactivity
3° alkyl halide
least reactive
b.
1° alkyl halide
most reactive
Br
Br
Br
vinyl halide
least reactive
2° alkyl halide
intermediate
reactivity
1° alkyl halide
most reactive
7.59
better leaving group
a. CH3CH2Br
CH3CH2Cl
+
OH
+
OH
faster reaction
stronger nucleophile
b.
Cl
+
NaOH
Cl
+
NaOCOCH3
I
+
OCH3
+
OCH3
c.
I
faster reaction
CH3OH
faster reaction
DMSO
polar aprotic
solvent
less steric hindrance
d.
Br
+
OCH2CH3
Br
+
OCH2CH3
faster reaction
7.60 All SN2 reactions proceed with backside attack of the nucleophile. When nucleophilic attack
occurs at a stereogenic center, inversion of configuration occurs.
CH3
a.
C
H
b.
CH3
Cl
* D
*
H
+
OCH3
Cl
+
OCH2CH3
C
H
+ Cl
D
* OCH3
inversion of configuration
OCH2CH3
*
H
+ Cl
No bond to the stereogenic center is broken,
since the leaving group is not bonded to
the stereogenic center.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
191
Alkyl Halides and Nucleophilic Substitution 7–25
c.
*
+
Br
*
CN
CN
* *
+ Br
inversion of configuration
[* denotes a stereogenic center]
7.61 Follow the definitions from Answer 7.28.
CH2CH3
a.
c.
CH3CH2CHCH2CH3
2° carbocation
e.
CH2
3° carbocation
1° carbocation
d.
b.
3° carbocation
2° carbocation
7.62 For carbocations: Increasing number of R groups = Increasing stability.
a.
CH2
b.
1° carbocation
least stable
2° carbocation
intermediate
stablity
1° carbocation
least stable
3° carbocation
most stable
2° carbocation
intermediate
stablity
3° carbocation
most stable
7.63 Both A and B are resonance stabilized, but the N atom in B is more basic and therefore more
willing to donate its electron pair.
A
B
O CH2
O CH2
N CH2
N
H
H
more basic
N atom
CH2
This resonance form stabilizes the carbocation
more than the equivalent resonance structure for A.
Thus, B is more stable then A.
7.64
Step [1]
CH3
a. Mechanism:
SN1 only
CH3 C CH2CH3
I
A
+ H2O
CH3
CH3
Step [2]
C CH2CH3 + H2O
B
+ I
CH3
CH3 C CH2CH3
OH2
C
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 7–26
b. Energy diagram:
Energy
192
Ea1
B
² H°1
Ea2
² H°2
² H°overall = 0
C CH3
A CH3
CH3 C CH2CH3
CH3 C CH2CH3
OH2
I
Reaction coordinate
c. Transition states:
CH3 δ
C CH2CH3
CH3
I δ–
+
CH3 δ+
C CH2CH3
CH3
OH2
δ+
d. rate equation: rate = k[(CH3)2CICH2CH3]
e. [1] Leaving group changed from I− to Cl−: rate decreases since I− is a better leaving group.
[2] Solvent changed from H2O (polar protic) to DMF (polar aprotic):
rate decreases since polar protic solvent favors SN1.
[3] Alkyl halide changed from 3° to 2°: rate decreases since 2° carbocations are less stable.
[4] [H2O] increased by factor of five: no change in rate since H2O is not in rate equation.
[5] [R–X] and [H2O] increased by factor of five: rate increases by a factor of five. (Only the concentration of
R–X affects the rate.)
7.65 The rate of an SN1 reaction increases with increasing alkyl substitution.
Br
Br
Br
1° alkyl halide
least reactive
a.
3° alkyl halide
most reactive
2° alkyl halide
intermediate
reactivity
Br
Br
Br
b.
aryl halide
least reactive
2° alkyl halide
intermediate
reactivity
3° alkyl halide
most reactive
7.66 The rate of an SN1 reaction increases with increasing alkyl substitution, polar protic solvents,
and better leaving groups.
a.
(CH3)3CCl
+ H2O
(CH3)3CI
+ H2O
b.
better leaving group
faster reaction
Br
CH3OH
3° halide
faster SN1 reaction
Br + CH OH
3
1° halide
slower SN1 reaction
+
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
193
Alkyl Halides and Nucleophilic Substitution 7–27
Cl
c.
aryl halide
slower reaction
+ H O
2
Cl
I
d.
I
2° halide
faster SN1 reaction
+ H2O
+ CH3CH2OH
CH3CH2OH
+ CH3CH2OH
DMSO
polar protic solvent
faster reaction
polar aprotic solvent
slower reaction
7.67
Br
a.
OCH2CH3
+
+ CH3CH2OH
Br
CH3CH2O
+ HBr
OH
b.
+
H2O
OH + HBr
+
7.68 The 1o alkyl halide is also allylic, so it forms a resonance-stabilized carbocation. Increasing the
stability of the carbocation by resonance, increases the rate of the SN1 reaction.
CH3CH2CH2CH CHCH2
CH3CH2CH2CH CH CH2
Br
CH3CH2CH2CH CH CH2
Br
resonance-stabilized carbocation
Use each resonance structure individually to continue the mechanism:
CH3CH2CH2CH CH CH2
CH3OH
CH3CH2CH2CH CHCH2
H OCH3
CH3CH2CH2CH CHCH2OCH3
HBr
Br
CH3CH2CH2CH CH CH2
CH3CH2CH2CH CH CH2
CH3CH2CH2CHCH CH2
H OCH3
HBr
OCH3
Br
CH3OH
7.69
H
H
Br
a.
+
acetone
1° alkyl halide
SN2 only
b.
CN + Br
CN
+
OCH3
DMSO
+
Br H
strong nucleophile
polar aprotic solvent
Both favor SN2.
2° alkyl halide
SN1 and SN2
c.
Br
H OCH3
+ CH3OH
CH2CH2CH3
3° alkyl halide
SN1 only
OCH3
+ HBr
CH2CH2CH3
Br
reaction at a stereogenic center
inversion of configuration
194
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Chapter 7–28
CH2CH3
d.
C
H
CH2CH3
C
+ CH3COOH
I
CH3
H
CH2CH3
C
CH3COO
H
CH3
OOCCH3
CH3
HI
2° alkyl halide Weak nucleophile
favors SN1.
SN1 and SN2
e.
+
OCH2CH3
Br
2° alkyl halide
SN1 and SN2
OCH2CH3
+ CH3CH2OH
2° alkyl halide
SN1 and SN2
reaction at a stereogenic center
inversion of configuration
+ Br
strong nucleophile
polar aprotic solvent
Both favor SN2.
Cl
f.
OCH2CH3
DMF
reaction at a stereogenic center
racemization of product
OCH2CH3
+
+ HCl
two products – diastereomers
Nucleophile attacks
from above and below.
Weak nucleophile
favors SN1.
7.70
Br
[2]
[1]
(CH3)2NCH2CH2O H
Na+H
C
OCH2CH2N(CH3)2
H
(CH3)2NCH2CH2O
Na+
C
+ NaBr
+ H2
H
diphenhydramine
7.71 First decide whether the reaction will proceed via an SN1 or SN2 mechanism (Answer 7.38),
and then draw the mechanism.
H
Br
OCH2CH3
Br
3° alkyl halide
SN1 only
CH3CH2OH
can attack from
above or below
OCH2CH3
H
+
Br
OCH2CH3
OCH2CH3
+
+ HBr
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Alkyl Halides and Nucleophilic Substitution 7–29
7.72
nucleophile
O
O
C O H
O
C O
C
OH
+ Br
O
+ H2O
Br
Br
C7H10O2
leaving group
7.73
Br
H
Br
H H
N
CH3
N
CO32–
Br–
N
CH3
N H
CO32–
CH3
N
N
Br
–
+ HCO3–
NaHCO3
+ NaBr
N
+
CH3
N
nicotine
7.74
a. Two diastereomers (C and D) are formed as products from the two enantiomers of A.
H CO2CH2CH3
R Br
O
H2N
+
OC(CH3)3
A
(CH3CH2)3N
nucleophile
leaving
group
C
B
CO2CH2CH3
H
S Br
CO2CH2CH3
H CH3
O
S N
H S
OC(CH3)3
H
+
B
(CH3CH2)3N
A
b.
CO2CH2CH3
H CH3
O
R N
H S
OC(CH3)3
H
H CH3
D
CH3CH2O
H
S
Both have the
S configuration
quinapril
O
H CH3
N S
H
O
N
CO2H
Since both stereogenic centers in D have
the S configuration and the corresponding
stereogenic centers in quinapril are also S,
D is needed to synthesize the drug.
196
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Chapter 7–30
7.75
CH3NH2
N
Cl
Cl
CH3NH2
+
N
Cl
+
N
H
N H
CH3
Cl
N CH3
+
N
Cl
N
H
+
CH3NH3
CH3
N
CH3NH3
N
H
CH3NH2
+
Cl
CH3
7.76 In the first reaction, substitution occurs at the stereogenic center. Since an achiral, planar
carbocation is formed, the nucleophile can attack from either side, thus generating a racemic
mixture.
3° alkyl halide
CH3OH
OCH3
two steps
SN1
Br
(6R)-6-bromo-2,6-dimethylnonane
Br
OCH3
achiral,
planar carbocation
racemic mixture
optically inactive
In the second reaction, the starting material contains a stereogenic center, but the nucleophile
does not attack at that carbon. Since a bond to the stereogenic center is not broken, the
configuration is retained and a chiral product is formed.
3° alkyl halide
Br
two steps
CH3OH
SN1
(5R)-2-bromo-2,5-dimethylnonane
OCH3
configuration
retained
Br
optically active
Reaction does not occur at
the stereogenic center.
7.77
The nucleophile has replaced the leaving group.
Missing reagent:
a.
O
I
O
Cl
b.
C CH
The nucleophile has replaced the leaving group.
Missing reagent:
C CH
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197
Alkyl Halides and Nucleophilic Substitution 7–31
N3
c.
N3
SH
SH
d.
The nucleophile has replaced the halide.
Starting material:
Cl
The nucleophile has replaced the halide.
Starting material:
Cl The leaving group must have the opposite
orientation to the position of the
nucleophile in the product.
7.78 To devise a synthesis, look for the carbon framework and the functional group in the product.
The carbon framework is from the alkyl halide and the functional group is from the
nucleophile.
SH
a.
SH
SH
functional
group
carbon
framework
Na
O
b.
Na
Cl
O
Cl
O
functional
carbon
group
framework
Na
c.
CH3CH2CN
carbon
framework
CH3CH2Cl
CN
CH3CH2CN
functional
group
O
Cl
d.
functional
group
carbon
framework
Na
2o halide
O Na
or
O
O
O
O
Cl
1o halide
carbon
functional framework
group
e.
CH3CH2 OCOCH3
carbon
framework
functional
group
CH3CH2Cl
Na
OCOCH3
CH3CH2OCOCH3
This path is preferred.
The strong nucleophile favors
an SN2 reaction so an unhindered
1o alkyl halide reacts faster.
198
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Chapter 7–32
7.79 Work backwards to determine the alkyl chloride needed to prepare benzalkonium chloride A.
CH3
a.
CH3(CH2)17Cl
CH2N(CH3)2
CH2
N
(CH2)17CH3
Cl–
CH3
B
A
b.
CH3
CH3(CH2)17N(CH3)2
CH2Cl
CH2
C
N
(CH2)17CH3
Cl–
CH3
A
7.80
OCH3
Na+ OCH3
I
C
B
very crowded 3° halide
E
CH3I
O Na
preferred method
The strong nucleophile favors SN2
reaction so the alkyl halide should
be unhindered for a faster reaction.
OCH3
A
D
E
unhindered methyl halide
7.81
H C C H
CH3(CH2)7CH2Br
NaH
H C C
CH3(CH2)7CH2C CH
A
B
NaH
CH3(CH2)7CH2C C
+ H2
C
+ H2
CH3(CH2)11CH2Br
H
H
addition of H2
C C
CH3(CH2)7CH2
CH3(CH2)7CH2C CCH2(CH2)11CH3
(1 equiv)
CH2(CH2)11CH3
D
muscalure
7.82
O
a.
H
[1] Na+H
[2] CH3–Br
O
O
CH3
+ H2
(Chapter 9)
b.
CH3CH2CH2 C C H
[1] Na+ NH2
[2] CH3CH2–Br
CH3CH2CH2 C C
CH3CH2CH2 C C CH2CH3
(Chapter 11)
c. H CH(CO2CH2CH3)2
(Chapter 23)
+ Na+ Br–
[1] Na+ –OCH2CH3
CH(CO2CH2CH3)2
[2] C6H5CH2–Br
+ Na+ Br–
+ NH3
C6H5CH2 CH(CO2CH2CH3)2
+ Na+ Br–
+ CH3CH2OH
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Alkyl Halides and Nucleophilic Substitution 7–33
7.83
quinuclidine
The three alkyl groups are "tied back"
in a ring, making the electron pair
more available.
triethylamine
CH3CH2
N
This electron pair is more hindered
by the three CH2CH3 groups.
N
These bulky groups around the N
CH2CH3
cause steric hindrance and this
CH2CH3
decreases nucleophilicity.
This electron pair on quinuclidine is much more available than the one on triethylamine.
less steric hindrance
more nucleophilic
7.84
O
O
H
O
[1]
H
+
H2
O
O
CH3 Br
CH3
[2]
minor product
O
+ NaBr
O
CH3 Br
CH3
[2]
major product
7.85
Br
Br
a.
OH
base
(CH3)3C
b.
O
(CH3)3C
OH
O
intramolecular
SN2
(CH3)3C
O
O
base
(CH3)3C
c.
(CH3)3C
(CH3)3C
Br
Br
Br
(CH3)3C
Br
base
OH
O
(CH3)3C
O
3° alkyl halide
harder reaction
OH
d.
Br
(CH3)3C
intramolecular
SN2
intramolecular
S N2
(CH3)3C
O
O
base
Br
(CH3)3C
3° alkyl halide
harder reaction
intramolecular
S N2
(CH3)3C
200
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Chapter 7–34
7.86
Cl bonded to sp2 C
cannot undergo SN1.
Cl
O
Cl
Cl
O
Cl
CH3OH
Cl bonded to sp3 C
no resonance stabilization possible
for the carbocation formed here
Cl
Cl
Cl bonded to sp3 C
Resonance-stabilized carbocation forms.
best for SN1
Cl
Cl
Cl
Cl
+
(1 equiv)
O
J
Cl
O
re-draw
Cl
Cl
Cl
Cl
+
CH3OH
O
O
O CH3
Cl
H
Cl
Cl
+
O
HCl
OCH3
K
7.87
a.
H OH
H OH
ee =
[α] mixture
[α] pure enantiomer
x 100%
+ 5.0
x 100% = 10.% excess of R isomer
+48
90% racemic mixture = 45% R and 45% S
Total R isomer = 45 + 10 = 55% R isomer
=
(S)-1-phenyl-1-propanol (R)-1-phenyl-1-propanol
[α] = –48
[α] = +48
b. The R product is the product of inversion and it predominates.
H Br
H OH
H2O
H OH
+
S
retention
R
inversion
c. The weak nucleophile favors an SN1 reaction, which occurs by way of an intermediate
carbocation. Perhaps there is more inversion than retention because H2O attacks the
intermediate carbocation while the Br– leaving group is still in the vicinity of the
carbocation. The Br– would then shield one side of the carbocation and backside attack
would be slightly favored.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
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Alkyl Halides and Elimination Reactions 8–1
Chapter 8 Alkyl Halides and Elimination Reactions
Chapter Review
A comparison between nucleophilic substitution and β-elimination
Nucleophilic substitution⎯A nucleophile attacks a carbon atom (7.6).
Nu
H
H Nu
+
C C
C C
X
substitution
product
X
good
leaving group
β-Elimination⎯A base attacks a proton (8.1).
B
H
C C
C C
+ H B+ +
elimination
product
•
•
X
good
leaving group
X
Similarities
In both reactions RX acts as an
•
electrophile, reacting with an electron-rich
reagent.
•
Both reactions require a good leaving
group X:– willing to accept the electron
density in the C–X bond.
Differences
In substitution, a nucleophile attacks a single
carbon atom.
In elimination, a Brønsted–Lowry base
removes a proton to form a π bond, and two
carbons are involved in the reaction.
The importance of the base in E2 and E1 reactions (8.9)
The strength of the base determines the mechanism of elimination.
• Strong bases favor E2 reactions.
• Weak bases favor E1 reactions.
strong base
OH
CH3
E2
CH3
C CH2
CH3 C CH3
Br
H2O
weak base
E1
+
H2O
+ Br
CH3
CH3
C CH2
CH3
same product
different mechanism
+
H3O
+
+ Br
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–2
E1 and E2 mechanisms compared
E2 mechanism
one step (8.4B)
•
E1 mechanism
two steps (8.6B)
[1] Mechanism
•
[2] Alkyl halide
•
rate: R3CX > R2CHX >
RCH2X (8.4C)
•
rate: R3CX > R2CHX > RCH2X
(8.6C)
[3] Rate equation
[4] Stereochemistry
•
•
•
•
•
•
[5] Base
•
rate = k[RX][B:]
second-order kinetics (8.4A)
anti periplanar arrangement of
H and X (8.8)
favored by strong bases (8.4B)
rate = k[RX]
first-order kinetics (8.6A)
trigonal planar carbocation
intermediate (8.6B)
favored by weak bases (8.6C)
[6] Leaving group
•
•
[7] Solvent
•
[8] Product
•
better leaving group → faster
reaction (8.4B)
favored by polar aprotic
solvents (8.4B)
more substituted alkene
favored (Zaitsev rule, 8.5)
•
•
•
better leaving group → faster
reaction (Table 8.4)
favored by polar protic solvents
(Table 8.4)
more substituted alkene favored
(Zaitsev rule, 8.6C)
Summary chart on the four mechanisms: SN1, SN2, E1, and E2
Alkyl halide type
1o RCH2X
2o R2CHX
3o R3CX
Conditions
strong nucleophile
strong bulky base
strong base and nucleophile
strong bulky base
weak base and nucleophile
weak base and nucleophile
strong base
Mechanism
SN2
E2
SN2 + E2
E2
SN1 + E1
SN1 + E1
E2
Zaitsev rule
•
•
β-Elimination affords the more stable product having the more substituted double bond.
Zaitsev products predominate in E2 reactions except when a cyclohexane ring prevents trans
diaxial arrangement.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
203
Alkyl Halides and Elimination Reactions 8–3
Practice Test on Chapter Review
1. Which of the following is true about an E1 reaction?
1. The reaction is faster with better leaving groups.
2. The reaction is fastest with 3° alkyl halides.
3. The reaction is faster with stronger bases.
4. Statements (1) and (2) are true.
5. Statements (1), (2), and (3) are all true.
2. Consider the SN2 and E1 reaction mechanisms. What effect on the rate of the reaction is observed
when each of the following changes is made? Fill in each box of the table with one of the
following phrases: increases, decreases, or remains the same.
Change
a. The alkyl halide is changed from (CH3)3CBr to
CH3CH2CH2CH2Br.
b. The solvent is changed from (CH3)2CO to
CH3CH2OH.
c. The nucleophile/base is changed from –OH to
H2O.
d. The alkyl halide is changed from CH3CH2Cl to
CH3CH2I.
e. The concentration of the base/nucleophile is
increased by a factor of five.
SN2 Mechanism
E1 Mechanism
3. Rank the following compounds in order of increasing reactivity in an E2 elimination reaction.
Rank the most reactive compound as 3, the least reactive compound as 1, and the compound of
intermediate reactivity as 2.
Br
A
Br
B
Br
C
4. Draw the organic products formed in the following reactions.
Cl
Br
KOH
a.
K+ –OC(CH3)3
c.
CH3
CH3
Br
b.
NaNH2
Br
(excess)
d.
Br
NaOCH3
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Chapter 8–4
5. a. Fill in the appropriate alkyl halide needed to synthesize the following compound as a single
product using the given reagents.
K+ –OC(CH3)3
C
CH3
CH3
b. What starting material is needed for the following reaction? The starting material must yield
product cleanly, in one step without any other organic side products.
K+ –OC(CH3)3
D
CH3CH2CH
CHCH3
(cis and trans mixture)
6. Draw all products formed in the following reaction.
CH3OH
Cl
CH3CH 2
CH 3
Answers to Practice Test
1. 4
5.
4.
6.
Cl
2.
SN2
a. increases
b. decreases
c. decreases
d. increases
e. increases
3. A–2
B–1
C–3
E1
decreases
increases
same
increases
same
a.
b.
CH3CH2
a.
CH 3 C
CH3
C
H
CH3
CH3CH2
CH3
C
CH3CH2
b.
Br
c.
CH3
d.
CH3
OCH3
CH 3
OCH3
(or Cl or I)
D
OCH3
CH3
CH3CH2
CH3
CH3CH2
CH2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
205
Alkyl Halides and Elimination Reactions 8–5
Answers to Problems
8.1 • The carbon bonded to the leaving group is the α carbon. Any carbon bonded to it is a β carbon.
• To draw the products of an elimination reaction: Remove the leaving group from the
α carbon and a H from the β carbon and form a π bond.
a.
b.
β
α
CH3CH2CH2CH2CH 2 Cl
β1
α
β2
K+ −OC(CH3)3
β1
CH3CH2CH2CH CH 2
(CH3CH2)2C
CH2
CH3CH=C(CH3)CH2CH3
Cl
β1
c.
K + −OC(CH3)3
α
β2
Br
K+ −OC(CH3)3
β1
8.2 Alkenes are classified by the number of carbon atoms bonded to the double bond. A
monosubstituted alkene has one carbon atom bonded to the double bond, a disubstituted alkene
has two carbon atoms bonded to the double bond, etc.
4 C's bonded to C=C
tetrasubstituted
2 C's bonded to each C=C
disubstituted
OH
a.
b.
vitamin D3
3 C's bonded to each C=C
trisubstituted
vitamin A
H
CH2
3 C's bonded to each C=C
trisubstituted
2 C's bonded to the C=C
disubstituted
HO
8.3 To have stereoisomers at a C=C, the two groups on each end of the double bond must be
different from each other.
two different groups
(CH3CH2 and H)
a.
two CH3 groups
no stereoisomers
possible
two different groups
(H and CH3)
b. CH3CH2CH CHCH3
stereoisomers possible
two different groups two different groups
(cyclohexyl and H) (cyclohexyl and H)
c.
CH CH
stereoisomers possible
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Chapter 8–6
8.4
a.
2 CH3's on
one end
2 H's on one end
Only this C=C exhibits stereoisomerism.
b. A diastereomer has a different 3-D arrangement of groups but the carbon skeleton and the
double bonds must stay in the original positions.
diastereomer
different arrangement of groups
around this double bond
8.5 Two definitions:
• Constitutional isomers differ in the connectivity of the atoms.
• Stereoisomers differ only in the 3-D arrangement of the atoms in space.
a.
and
c.
different connectivity of atoms
constitutional isomers
and
b.
and
cis
trans
different arrangement of atoms in space
stereoisomers
d.
trans
and
trans
identical
different connectivity of atoms
constitutional isomers
8.6 Two rules to predict the relative stability of alkenes:
[1] Trans alkenes are generally more stable than cis alkenes.
[2] The stability of an alkene increases as the number of R groups on the C=C increases.
a.
or
monosubstituted
b.
CH2CH3
CH3CH2
C C
cis
CH3CH2
H
H
CH2CH3
trans
more stable
CH3
or
c.
C C
or
H
H
CH3
disubstituted
more stable
trisubstituted
more stable
disubstituted
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Alkyl Halides and Elimination Reactions 8–7
8.7
A
B
Alkene A is more stable than alkene B because the double bond in A is in a six-membered ring.
The double bond in B is in a four-membered ring, which has considerable angle strain due to the
small ring size.
8.8 In an E2 mechanism, four bonds are involved in the single step. Use curved arrows to show
these simultaneous actions:
[1] The base attacks a hydrogen on a β carbon.
[2] A π bond forms.
[3] The leaving group comes off.
CH3CH2 H
CH3CH2 C
Br
OCH2CH3
CHCH3
(CH3CH2)2C=CHCH3 +
transition state:
δ−
HOCH2CH3
+
CH3CH2 C
new π bond
β carbon
CH3CH2 H
Br
OCH2CH3
CHCH3
δ− Br
8.9 In both cases, the rate of elimination decreases.
stronger base
faster reaction
a. CH CH Br
3
2
+
OC(CH3)3
CH3CH2 Br
+
OH
better leaving group
faster reaction
b. CH3CH2 Br
+
OC(CH3)3
CH3CH2 Cl
+
OC(CH3)3
8.10 As the number of R groups on the carbon with the leaving group increases, the rate of an E2
reaction increases.
a. (CH3)2CHCH2CH2CH2Br
(CH3)2CHCH2CH(Br)CH3
1° alkyl halide
least reactive
3° alkyl halide
most reactive
CH3
Cl
Cl
b.
Cl
1° alkyl halide
least reactive
(CH3)2C(Br)CH2CH2CH3
2° alkyl halide
intermediate reactivity
CH3
2° alkyl halide
intermediate reactivity
3° alkyl halide
most reactive
8.11 Use the following characteristics of an E2 reaction to answer the questions:
[1] E2 reactions are second order and one step.
[2] More substituted halides react faster.
[3] Reactions with strong bases or better leaving groups are faster.
[4] Reactions with polar aprotic solvents are faster.
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Chapter 8–8
Rate equation: rate = k[RX][Base]
a. tripling the concentration of the alkyl halide = rate triples
b. halving the concentration of the base = rate halved
c. changing the solvent from CH3OH to DMSO = rate increases (Polar aprotic solvent is better
for E2.)
d. changing the leaving group from I– to Br– = rate decreases (I– is a better leaving group.)
e. changing the base from –OH to H2O = rate decreases (weaker base)
f. changing the alkyl halide from CH3CH2Br to (CH3)2CHBr = rate increases (More substituted
halide reacts faster.)
8.12 The Zaitsev rule states: In a β-elimination reaction, the major product has the more substituted
double bond.
CH3 H
a.
α
CH3 C
C
H
Br
CH2CH3
(CH3)2C
loss of H and Br
CH3
b.
α
CHCH2CH3
+
trisubstituted
major product
Br
(CH3)2CHCH
CHCH3
disubstituted
minor product
CH3
CH3
CH2
CH3
CH3
CH3
loss of H and Br
CH3
trisubstituted
minor product
tetrasubstituted
major product
disubstituted
minor product
Cl
CH3
monosubstituted
minor product
CH3
disubstituted
major product
CH3
α Cl
d.
CH3CH2CH2CH2CH=CHCH3
loss of H and Cl
α
c.
trisubstituted
CH3 ONLY product
loss of H and Cl
CH3
CH3
8.13 An E1 mechanism has two steps:
[1] The leaving group comes off, creating a carbocation.
[2] A base pulls off a proton from a β carbon, and a π bond forms.
CH3
CH3 C
CH2CH3
[1]
+ CH3OH
CH3
CH3 C CH CH3 CH3OH + Cl
[2]
+
(CH3)2C=CHCH3 + CH3OH2 + Cl
H
Cl
transition state [1]:
transition state [2]:
CH3
CH3 +C
δ
CH2CH3
Cl δ−
CH3
CH3 C
δ+
CH CH3
H
OCH3
δ+
H
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Alkyl Halides and Elimination Reactions 8–9
8.14 The Zaitsev rule states: In a β-elimination reaction, the major product has the more substituted
double bond.
β2
CH3
CH3
β1
α
C CHCH3
CH3CH2
CH3CH2 C CH2CH3 + H2O
a.
Cl
CH3CH2
β1
trisubstituted
major product
CH2CH2CH3
β1
CH3
α
b.
β3
I
β2
CH2CH2CH3
+
CH3
CH3OH
+
C CH2
CH3CH2
disubstituted
CH2CH2CH3
+
tetrasubstituted
major product
CH2
CH2CH2CH3
CH3
+
trisubstituted
disubstituted
8.15 Use the following characteristics of an E1 reaction to answer the questions:
[1] E1 reactions are first order and two steps.
[2] More substituted halides react faster.
[3] Weaker bases are preferred.
[4] Reactions with better leaving groups are faster.
[5] Reactions in polar protic solvents are faster.
Rate equation: rate = k[RX]. The base doesn’t affect rate.
a. doubling the concentration of the alkyl halide = rate doubles
b. doubling the concentration of the base = no change (Base is not in the rate equation.)
c. changing the alkyl halide from (CH3)3CBr to CH3CH2CH2Br = rate decreases (More
substituted halides react faster.)
d. changing the leaving group from Cl– to Br– = rate increases (better leaving group)
e. changing the solvent from DMSO to CH3OH = rate increases (Polar protic solvent favors E1.)
8.16 Both SN1 and E1 reactions occur by forming a carbocation. To draw the products:
[1] For the SN1 reaction, substitute the nucleophile for the leaving group.
[2] For the E1 reaction, remove a proton from a β carbon and create a new π bond.
CH3
CH3
Br
a.
+ H2O
leaving
group nucleophile
and base
SN1 product
CH3
Cl
nucleophile
and base
CH3
E1 products
CH3
b. CH3 C CH2CH2CH3 + CH3CH2OH
leaving
group
CH2
OH
CH3 C CH2CH2CH3
OCH2CH3
SN1 product
CH3
CH3
C CH2
CH3CH2CH2
C CHCH2CH3
CH3
E1 products
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Chapter 8–10
8.17 E2 reactions occur with anti periplanar geometry. The anti periplanar arrangement uses a
staggered conformation and has the H and X on opposite sides of the C–C bond.
H
HO
C
CH3
CH3
H
C
CH3
H
Br
C
base
H
C
CH3
H
H and Br are on opposite sides =
anti periplanar
8.18 The E2 elimination reactions will occur in the anti periplanar orientation as drawn. To draw
the product of elimination, maintain the orientation of the remaining groups around the C=C.
CH3CH2O
H
CH3
a.
C6H5
C
CH3
C6H5
C H
Br
C6H5
C C
C6H5
H
The two benzene rings remain on
opposite sides of the newly formed
C=C. This makes them trans.
The two benzene rings are
anti in this conformation (one
wedge, one dash).
diastereomers
CH3CH2O
H
b.
C6H5
C6H5
C6H5 C
C H
CH3
Br
C6H5
C C
CH3
H
The two benzene rings are gauche
in this conformation
(both drawn on dashes, behind the plane).
The two benzene rings remain
on the same side of the newly formed
C=C. This makes them cis.
8.19 Note: The Zaitsev products predominate in E2 elimination except when substituents on a
cyclohexane ring prevent a trans diaxial arrangement of H and X.
axial H's
H
CH(CH3)2
two
conformations
a.
CH3
Cl
CH3
H
CH3
H
H
H
CH(CH3)2
Cl
A
Use this conformation.
It has Cl axial and
two axial H's.
H
H
H
H
B
H
Cl
CH(CH3)2
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211
Alkyl Halides and Elimination Reactions 8–11
CH3
H
H
H
H
β1
H
H
H
CH(CH3)2
−OH
CH3
Cl
β2
CH(CH3)2
H
H
CH3
H
H
H
[loss of H(β2) + Cl]
A
re-draw
CH(CH3)2
CH(CH3)2
CH3
CH3
trisubstituted
major product
disubstituted
H
CH(CH3)2
CH3
CH3
H
two
conformations
b.
CH3
Cl
Cl
H
H
CH(CH3)2
H
H
β1
β2
H
H
B
CH(CH3)2
CH3
−OH
H
H
H
H
H
Use this conformation.
It has Cl axial and
one axial H.
Cl
H
Cl
H
A
CH3
H
[loss of H(β1) + Cl]
re-draw
two different axial H's
CH(CH3)2
CH(CH3)2
B
only one axial H
on a β carbon
CH(CH3)2
H
H
H
H
CH(CH3)2
[loss of H(β1) + Cl]
=
CH3
disubstituted
only product
8.20 Draw the chair conformations of cis-1-chloro-2-methylcyclohexane and its trans isomer. For
E2 elimination reactions to occur, there must be a H and X trans diaxial to each other.
Two conformations of the cis isomer:
Cl
CH3
H
H
H
H
A
reacting conformation (axial Cl)
This reacting conformation has only one
group axial, making it more stable and
present in a higher concentration than B.
This makes a faster elimination reaction
with the cis isomer.
H
CH3
H
H
H
Two conformations of the trans isomer:
Cl
H
H
H
Cl
H
CH3
H
Cl
H
H
H
CH3
B
reacting conformation (axial Cl)
This conformation is less stable than A,
since both CH3 and Cl are axial.
This slows the rate of elimination
from the trans isomer.
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Chapter 8–12
8.21 E2 reactions are favored by strong negatively charged bases and occur with 1°, 2°, and 3°
halides, with 3° being the most reactive.
E1 reactions are favored by weaker neutral bases and do not occur with 1° halides since
they would have to form highly unstable carbocations.
CH3
a.
C(CH3)3
CH3 C CH3
+
Cl
+
strong negatively
charged base
E2
I
b.
Cl
c.
OCH3
+
CH3OH
weak neutral
base
E1
H2O
d. CH3CH2Br
weak neutral
base
E1
+
OC(CH3)3
strong negatively
charged base
E2
8.22 Draw the alkynes that result from removal of two equivalents of HX.
Br
Cl Cl
a.
NH2
C C CH2CH3
C C CH2CH3
c. CH3 C CH2CH3
NH2
CH3C CCH3
Br
H H
+ HC CCH2CH3
Br
KOC(CH3)3
b. CH3CH2CH2CHCl2
CH3CH2C
DMSO
NH2
d.
CH
C
C
Br
8.23
1° halide
SN2 or E2
H
b.
K+ −OC(CH3)3
Cl
a.
CH3 C CH2CH3
Cl
2° halide
any mechanism
strong bulky base
E2
OH
strong base
SN2 and E2
CH2CH3
I
c.
+
weak base
SN1 and E1
CH3 CH CHCH3
+
SN2 product
disubstituted
major E2 product
OCH2CH3
SN1 product
+
+
E1 product
CH3CH2O
Cl
3° halide
no SN2
CH3CH2OH
E2
major E2 product
monosubstituted
minor E2 product
CH2CH3
CHCH3
strong base
d.
CH2 CH CH2CH3
OH
CH2CH3
CH3CH2OH
3° halide
no SN2
H
CH3 C CH2CH3
minor E2 product
E1 product
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Alkyl Halides and Elimination Reactions 8–13
8.24
3° halide
no SN2
weak base
SN1 and E1
CH3
CH3
Br
CH3OH
CH3
overall
reaction
SN1
HBr
CH3
CH3
CH3OH
CH3
or
E1
+
+
CH3
The steps:
CH3
CH3
OCH3
O CH3
H
CH3
Br
+ Br
CH3
H
CH3
Br
8.25 More substituted alkenes are more stable. Trans alkenes are generally more stable than cis
alkenes. Order of stability:
<
B
least stable
<
C
A
most stable
8.26 The trans isomer reacts faster. During elimination, Br must be axial to give trans diaxial
elimination. In the trans isomer, the more stable conformation has the bulky tert-butyl group in
the more roomy equatorial position. In the cis isomer, elimination can occur only when both
the tert-butyl and Br groups are axial, a conformation that is not energetically favorable.
Br
Br
=
slow
This conformation must react,
cis
but it contains two axial groups.
D
cis-1-bromo-3-tert-butylcyclohexane
Br
=
Br
trans
preferred conformation
E
trans-1-bromo-3-tert-butylcyclohexane
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Chapter 8–14
8.27 Translate each model to a structure and arrange H and Br to be anti periplanar.
a.
CH3
H
C6H5
C
C
CH3
Br
H
CH3CH2
Br
CH3CH2
rotate
C
C6H5
C
CH3CH2
E2
H
H
C6H5
C
C
CH3
H
H and Br 180°
away from each other
Br
CH3CH2
b.
CH3
H
C
H
H
H
rotate
C
C
CH3CH2
CH3
C6H5
C
C6H5
CH3CH2
E2
H
C
Br
C
CH3
C6H5
H and Br 180°
away from each other
8.28
a. CH3CH2CH2CH2CH2CH2Br
CH3CH2CH2CH2CH CH2
Br
b.
CH3CH2CH2CH2CH CHCH2CH3 +
CH3CH2CH2CH2CH2CH CHCH3
CH3
c. CH3CH2CHCHCH3
CH3CH2CH C(CH3)2
+
CH3CH CHCH(CH3)2
Cl
CH2
I
d.
8.29 To give only one product in an elimination reaction, the starting alkyl halide must have only
one type of β carbon with H’s.
a.
CH2 CHCH2CH2CH3
β
β
α
CH2
CH2CH2CH2CH3
Cl
β
b. (CH3)2CHCH CH2
CH2
c.
(CH3)2CHCH2
CH3
CH3
α
β
Two β carbons
are identical.
CH2Cl
α
β
α Cl
d.
CH2Cl
e.
C(CH3)3
β Cl
α
β
C(CH3)3
Two β carbons
are identical.
8.30 To have stereoisomers, the two groups on each end of the double bond must be different from
each other.
2 H's on one end
a.
2 CH3's on
one end two different groups
at each end
can have
stereoisomers
OH
b.
2 H's on
one end
two different groups
at each end
can have
stereoisomers
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
215
Alkyl Halides and Elimination Reactions 8–15
8.31 Use the definitions in Answer 8.5.
CH3
a.
CH2
and
and
c.
trans trans
trans trans
different connectivity
constitutional isomers
identical
H
CH3CH2
CH3
CH3CH2
and
C C
b.
CH3
CH2CH3
CH2CH3
C C
CH3
CH3
C
CH3
CH3
d.
C
and
CH3
stereoisomers
H
CH3
stereoisomers
8.32
Double bond can be cis or trans.
OH
a. five sp3 stereogenic centers (four circled, one labeled)
b. Two double bonds can both be cis or trans.
c. 27 = 128 stereoisomers possible
CH2CH CH(CH2)3COOH
PGF2α
HO
CH CHCH(OH)(CH2)4CH3
sp3 stereogenic center
Double bond can be cis or trans.
8.33 Use the rules from Answer 8.6 to rank the alkenes.
CH2 CHCH(CH3)2
CH2 C(CH3)CH2CH3
monosubstituted
least stable
disubstituted
intermediate
stability
(CH3)2C
CHCH3
trisubstituted
most stable
8.34 A larger negative value for H° means the reaction is more exothermic. Since both
1-butene and cis-2-butene form the same product (butane), these data show that 1-butene was
higher in energy to begin with, since more energy is released in the hydrogenation reaction.
1-butene
+ H2
ΔHo = −127 kJ/mol
1-butene
CH3
CH3
+ H2
C C
H
H
cis-2-butene
cis-2-butene
CH3CH2CH2CH3
CH3CH2CH2CH3
ΔHo = −120 kJ/mol
Energy
CH2 CHCH2CH3
larger ² H° for
1-butene
higher in energy
butane
smaller ² H° for cis-2-butene
lower in energy, more stable
216
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Chapter 8–16
8.35
Cl
a.
β1
α
(CH3)3CO
CH3CH=CHCH2CH2CH(CH3)2
β2
(loss of β2 H)
major product
disubstituted
(loss of β1 H)
monosubstituted
DBU
b.
O
only product
O
O
β3
I
CH3
β1
c.
β α
O
Cl
α
OH
CH3CH2C(CH3)=C(CH3)CH2CH2CH3
β2
CH3CH2CH(CH3)C(CH3) CHCH2CH3
(loss of β1 H)
major product
tetrasubstituted
(loss of β2 H)
trisubstituted
CH2
(loss of β3 H)
disubstituted
β
d.
Cl
α
OC(CH3)3
only product
β2
e.
β1
β2
f.
α
OH
α I
Br
CH3CH2CH2CH2CH=CHCH3
(loss of β1 H)
major product
disubstituted
(loss of β2 H)
monosubstituted
OH
β1
(loss of β2 H)
major product
trisubstituted
(loss of β1 H)
disubstituted
8.36 To give only one alkene as the product of elimination, the alkyl halide must have either:
• only one β carbon with a hydrogen atom
• all identical β carbons, so the resulting elimination products are identical
CH3 H
a. CH3 C
H
C H
Cl
CH3
C
CH3
H
CH3 H
CH3 C
C
H
Cl
C H
H
Cl
b.
CH=CH2
Cl
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217
Alkyl Halides and Elimination Reactions 8–17
Cl
Cl
c.
8.37 Draw the products of the E2 reaction and compare the number of C’s bonded to the C=C.
β1
A
α
β2
β2
β1 α
major product
trisubstituted
Br
Br
β1
α
(CH3)2CHCH=CHCH3
α
disubstituted
α
β2
B
α
(CH3)2CHCH=CHCH3
β1
major product
disubstituted
β2
monosubstituted
A yields a trisubstituted alkene as the major product and a disubstituted alkene as minor product.
B yields a disubstituted alkene as the major product and a monosubstituted alkene as minor product.
Since the major and minor products formed from A have more alkyl groups on the C=C
(making them more stable) than those formed from B, A reacts faster in an elimination reaction.
8.38
a. Mechanism:
by-products
Br
H
OC(CH3)3
+ HOC(CH3)3 + Br−
(CH3)3COH
b. Rate = k[R–Br][–OC(CH3)3]
[1] Solvent changed to DMF (polar aprotic) = rate increases
[2] [–OC(CH3)3] decreased = rate decreases
[3] Base changed to –OH = rate decreases (weaker base)
[4] Halide changed to 2° = rate increases (More substituted RX reacts faster.)
[5] Leaving group changed to I– = rate increases (better leaving group)
8.39
K+ –OC(CH3)3
+
Cl
A
1-chloro-1-methylcyclopropane
B
The dehydrohalogenation of an alkyl halide usually forms the more stable alkene. In this case, A is
more stable than B even though A contains a disubstituted C=C whereas B contains a trisubstituted
C=C. The double bond in B is part of a three-membered ring, and is less stable than A because of
severe angle strain around both C’s of the double bond.
218
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–18
8.40
H
a.
KOH
CH3CH2CH2 C
NaOCH2CH3
b.
Cl
Br
trans isomer more stable
major product
trans isomer more stable
major product
8.41
CH3
a.
CH3
CH3
Cl
CH3
CH3
tetrasubstituted
major product
CH2
CH3
CH3
trisubstituted
disubstituted
Br
b.
trisubstituted
This isomer is more stable—
large groups farther away.
major product
disubstituted
trisubstituted
Cl
c.
disubstituted
trisubstituted
major product
8.42 Use the rules from Answer 8.21.
a.
OCH3
Br
2° halide
CH3CH CHCH3
strong base
E2
CH3OH
b.
Br
2° halide
weak base
E1
CH2CH2CH3
H2O
Cl
CH3
weak base
3° halide
E1
OH
Cl
e.
2° halide
strong base
E2
OH
f.
2° halide Cl
CH3CH2CH CH2
(cis and trans)
strong base
E2
1° halide
d.
CH3CH CHCH3
OC(CH3)3
I
c.
CH3CH2CH CH2
(cis and trans)
strong base
E2
CHCH2CH3
CH3
CH2CH2CH3
CH3
CH2CH2CH3
CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
219
Alkyl Halides and Elimination Reactions 8–19
8.43 The order of reactivity is the same for both E2 and E1: 1° < 2° < 3°.
CH3
Br
CH3
Cl
CH3
Cl
3° halide
2° halide
3° halide +
better leaving group
Increasing reactivity in E1 and E2
8.44
CH3
Cl
a.
Cl
OH
3° halide—faster reaction
CH3
Cl
H2O
b.
CH3 Cl
c. (CH3)3CCl
OH
H2O
strong base—E2
H2O
(CH3)3CCl
OH
DMSO
OH
3° halide—faster reaction
polar aprotic solvent
faster reaction
8.45
In a ten-membered ring, the cis isomer is more
stable and, therefore, the preferred elimination
product. The trans isomer is less stable because
strain is introduced when two ends of the double
bond are connected in a trans arrangement in this
medium-sized ring.
Br
cis-cyclodecene
bromocyclodecane
8.46 With the strong base –OCH2CH3, the mechanism is E2, whereas with dilute base, the
mechanism is E1. E2 elimination proceeds with anti periplanar arrangement of H and X. In
the E1 mechanism there is no requirement for elimination to proceed with anti periplanar
geometry. In this case the major product is always the most stable, more substituted alkene.
Thus, C is the major product under E1 conditions. (In Chapter 9, we will learn that additional
elimination products may form in the E1 reaction due to carbocation rearrangement.)
Cl
OCH2CH3
A
strong base
E2
B
Since this is an E2 mechanism,
dehydrohalogenation needs an anti
periplanar H to form the double bond.
There is only one H trans to Cl, so the
disubstituted alkene B must form.
Cl
CH3OH
A
weak base
E1
B
C
disubstituted alkene
trisubstituted alkene
more stable
220
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–20
8.47
Br
+ CH3CH=CHCH3
β1
β2
2-butene
1-butene
from loss of
β1 H
from loss of
β2 H
Na+ –OCH2CH3
19%
81%
K+ –OC(CH3)3
33%
67%
The H’s on the CH2 group of the β2 carbon are more sterically hindered than the H’s on the
CH3 group of the β1 carbon. Since K+ –OC(CH3)3 is a much bulkier base than Na+ –OCH2CH3,
it is easier to remove the more accessible H on β1, giving it a higher percentage of 1-butene.
8.48 H and Br must be anti during the E2 elimination. Rotate if necessary to make them anti; then
eliminate.
CH3
a.
C6H5
C6H5
Br
H
CH2CH3
CH3
CH3
E2
CH2CH3
CH3
Br
Br
H
CH3
C6H5
CH3
b.
CH3
rotate
CH2CH3
C6H5
E2
CH3
CH2CH3
CH3
CH3
CH2CH3
C6H5
CH2CH3
H
C6H5
c.
C6H5
C6H5
CH3
H
H
CH3
Br
rotate
CH2CH3
CH3
CH2CH3
E2
Br
CH3
CH3
CH3
8.49
Cl
a.
two chair
conformations
H
CH3
Cl
H
H
H
H
H
CH3
H
B
H
H
CH3
Choose this conformation.
axial Cl
H
H
Cl
B
(CH3)2CH
CH3
(CH3)2CH A H
CH(CH3)2
(CH3)2CH
H axial
Cl
H
H
H
one axial H
(CH3)2CH
=
CH3
H
only product
CH3
CH(CH3)2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
221
Alkyl Halides and Elimination Reactions 8–21
Cl
b.
H
two chair
conformations
H
(CH3)2CH
CH(CH3)2
H
H
H
H
c.
β1
re-draw
CH3
CH3
CH(CH3)2
CH(CH3)2
D
D
= D
D
H
(loss of β2 H)
(loss of β2 H)
D
Cl
H
Cl
D
H
H
(CH3)2CH
CH3
(loss of β1 H)
major product
trisubstituted
re-draw
D
H
H
H
(CH3)2CH
CH3
H
β1 H
two axial H's
B
B
Choose this conformation.
axial Cl
H
H
A
H
Cl
Cl
(CH3)2CH
CH3
β2
Cl
axial
H
H
Cl
H
CH3
(CH3)2CH
CH3
CH3
H
β2
D
H
H
=
D
H H
D
enantiomers
−OH
D
This conformation reacts.
axial Cl
D
D =
H
D
H
(loss of β1 H)
(loss of β2 D)
D
Cl
d.
=
D
H
Cl
H
H
β1
H
H
D
D
β2
This conformation reacts.
axial Cl
H
D
Cl
H
D
=
D
H
D
H
enantiomers
−OH
D
H
H
H
D
(loss of β1 D)
=
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–22
8.50
enantiomers
a.
H CH3
H
CH3 C C CH2CH3
CH3CH2
CH3
C3
C2
H
H
H
CH3 CH3
C C
Cl H
enantiomers
C C CH CH
2
3
CH3
Cl
Cl
–HCl
H and Cl are arranged anti in
each stereoisomer, for anti
periplanar elimination.
CH3CH2
Cl
–HCl
CH2CH3
H
CH3
CH3
–HCl
CH3
CH3
C C
C C
C
CH3
Cl
D
C
H
CH3
CH3
H
CH3
CH3 C C
CH3
CH2CH3 CH3CH2
–HCl
H
C
C C
B
A
2-chloro-3-methylpentane
H
H
H
CH3
H
C C
CH2CH3 CH3CH2
identical
CH3
identical
b. Two different alkenes are formed as products.
c. The products are diastereomers: Two enantiomers (A and B) give identical products. A and B
are diastereomers of C and D. Each pair of enantiomers gives a single alkene. Thus,
diastereomers give diastereomeric products.
8.51
CH2CHCl2
a.
C CH
NaNH2
(2 equiv)
CH 3
b.
CH3CH 2 C
NaNH2
CHCH 2Br
(2 equiv)
CH 3 Br
Cl
c.
CH3 C CH2CH3
Cl
NaNH2
NaNH2
C C
CH3 C C CH3
C C
(2 equiv)
Cl Cl
C CH
CH3
HC C CH2CH3
(excess)
H H
d.
CH3
CH 3CH2 C
8.52
H H
Br
a. CH3C CCH3
CH3 C CH2CH3
or
CH3 C C CH3
Br Br
Br
CH3
b. CH3 C C CH
CH3
CH3 Br
CH3 C
C CH3
CH3
CH3 Br
or
CH3 C
CH3 Br
CH CH2Br
CH3
CH3
Br H
c.
C C
or
CH3 C
C C
Br H
Br Br
or
C C
H H
CH2CHBr2
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223
Alkyl Halides and Elimination Reactions 8–23
8.53
H H
CH3 C C CH3
CH3 C C CH3
CH3 CH=C=CH2
CH2=CH CH=CH2
C
Br Br
sp sp
A
2,3-dibromobutane
sp
B
8.54 Use the “Summary chart on the four mechanisms: SN1, SN2, E1, or E2” on p. 8–2 to answer the
questions.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Both SN1 and E1 involve carbocation intermediates.
Both SN1 and E1 have two steps.
SN1, SN2, E1, and E2 all have increased reaction rates with better leaving groups.
Both SN2 and E2 have increased rates when changing from CH3OH (a protic solvent) to
(CH3)2SO (an aprotic solvent).
In SN1 and E1 reactions, the rate depends on only the alkyl halide concentration.
Both SN2 and E2 are concerted reactions.
CH3CH2Br and NaOH react by an SN2 mechanism.
Racemization occurs in SN1 reactions.
In SN1, E1, and E2 mechanisms, 3° alkyl halides react faster than 1° or 2° halides.
E2 and SN2 reactions follow second-order rate equations.
8.55
Br
a.
1° halide
SN2 or E2
OC(CH3)3
OCH2CH3
I
b.
E2
sterically
hindered base
OCH2CH3
SN2
strong
nucleophile
1° halide
SN2 or E2
Cl
NH2
c. CH3 C CH3
HC C CH3
(2 equiv)
Cl
strong base
dihalide
Br
d.
1° halide
SN2 or E2
DBU
sterically
hindered
base
CH2CH3
e.
Br
2° halide
SN1, SN2, E1, E2
E2
OC(CH3)3
sterically
hindered
base
CH2CH3
CH2CH3
E2
major product
224
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–24
OCH2CH3
Br
f.
CH3CH2OH
CH2CH3
weak base
3° halide
no SN2
g.
(CH3)2CH
SN1 product
CHCH2Br
CHCH3
CH2CH3
CH2CH3
2 NaNH2
(CH3)2CH
E1 products
C CH
dihalide Br
Cl Cl
h.
KOC(CH3)3
(2 equiv)
CH3
CH3 C C CH
CH3
DMSO
dihalide
OCH2CH3
I
CH3CH2OH
i.
2° halide
SN1, SN2, E1, E2
weak base
CH3CH CHCH3
SN1 product
E1 product
H2O
j.
Cl
CH3CH2C(CH3) CHCH3
OH
weak base
3° halide
no SN2
(cis and trans)
E1 product
(cis and trans)
E1 product
SN1 product
E1 product
8.56 [1] NaOCOCH3 is a good nucleophile and weak base, and substitution is favored. [3] KOC(CH3)3
is a strong, bulky base that reacts by E2 elimination when there is a β hydrogen in the alkyl
halide.
a.
CH3Cl
[1] NaOCOCH3
[2] NaOCH3
c.
CH3OCOCH3
[1] NaOCOCH3
Cl
OCOCH3
[2] NaOCH3
CH3OCH3
OCH3
SN2
[3] KOC(CH3)3
Cl
b.
[3] KOC(CH3)3
CH3OC(CH3)3
[1] NaOCOCH3
[2] NaOCH3
[3] KOC(CH3)3
OCOCH3
d.
OCH3
Cl
[1] NaOCOCH3
[2] NaOCH3
[3] KOC(CH3)3
OCOCH3
+
E2
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225
Alkyl Halides and Elimination Reactions 8–25
8.57
a. two enantiomers:
(CH3)3C
(CH3)3C
A
B
b. The bulky tert-butyl group anchors the cyclohexane ring and occupies the more roomy
equatorial position. The cis isomer has the Br atom axial, while the trans isomer has the Br
atom equatorial. For dehydrohalogenation to occur on a halo cyclohexane, the halogen must
be axial to afford trans diaxial elimination of H and X. The cis isomer readily reacts since
the Br atom is axial. The only way for the trans isomer to react is for the six-membered ring
to flip into a highly unstable conformation having both (CH3)3C and Br axial. Thus, the trans
isomer reacts much more slowly.
Br
trans diaxial
Br
(CH3)3C
(CH3)3C
H
H
cis-1-bromo-4-tert-butylcyclohexane
trans-1-bromo-4-tert-butylcyclohexane
OCH3
c. two products:
OCH3
(CH3)3C
C=
D = (CH3)3C
d. cis-1-Bromo-4-tert-butylcyclohexane reacts faster. With the strong nucleophile –OCH3,
backside attack occurs by an SN2 reaction, and with the cis isomer, the nucleophile can
approach from the equatorial direction, avoiding 1,3-diaxial interactions.
1,3-diaxial interactions
Br
OCH3
H
H
OCH3
(CH3)3C
equatorial approach preferred
cis-1-bromo-4-tert-butylcyclohexane
(CH3)3C
Br
axial approach
trans-1-bromo-4-tert-butylcyclohexane
e. The bulky base –OC(CH3)3 favors elimination by an E2 mechanism, affording a mixture of
two enantiomers, A and B. The strong nucleophile –OCH3 favors nucleophilic substitution
by an SN2 mechanism. Inversion of configuration results from backside attack of the
nucleophile.
8.58
Cl
a.
H
H
strong base
2° halide
SN2 and E2
SN1, SN2, E1, E2
Cl
H
b.
2° halide
SN1, SN2, E1, E2
OH
OH
H2O
weak base
SN1 and E1
SN2 product
major E2 product
inversion at
stereogenic center
HO
H
H
minor E2 product
OH
SN1 products
major E1 product
minor E1 product
minor E1 product
minor E2 product
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–26
CH3
CH3
c.
CH3
CH3
CH3OH
Cl
C6H5
OCH3
C6H5
weak base
SN1 and E1
3° halide
no SN2
CH3
CH3
CH3
CH3
C6H5
OCH3
C6H5
SN1 products
E1 product
NaOH
d.
strong base
SN2 and E2
Cl
OH
2° halide
SN1, SN2, E1, E2
CH3
Br
e.
Br
achiral
SN1 product
OH
KOH
f.
D
2° halide
SN1, SN2, E1, E2
strong base
SN2 and E2
minor E2 product
CH3
OOCCH3
CH3COO
weak base
good nucleophile
SN1
3° halide
no SN2
major E2 product
SN2 product
(trans diaxial elimination of D, Br)
D
SN2 product
inversion at
stereogenic center
E2 product
8.59
CH3OH
a.
Cl
weak base
SN1 and E1
OCH3
OCH3
S N1
SN1
E1
E1
E1
KOH
b.
Cl
strong base
E2
8.60
CH3
a.
Br
3° halide
CH3
OC(CH3)3
strong
bulky base
E2
OC(CH3)3
CH3
CH2
major product
more substituted alkene
No substitution occurs with a strong bulky base and a 3o RX. The C with the leaving
group is too crowded for an SN2 substitution to occur. Elimination occurs instead by an
E2 mechanism.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
227
Alkyl Halides and Elimination Reactions 8–27
Br
b.
1° halide
OCH3
strong nucleophile
SN2
OCH3
All elimination reactions are slow with 1° halides.
The strong nucleophile reacts by an SN2 mechanism instead.
CH3
Cl
c.
minor product only
OH
CH3
strong base
E2
3° halide
More substituted
alkene is favored.
I
d.
Cl
minor product only
good nucleophile,
weak base
SN2 favored
2° halide
I
major product
The 2o halide can react by an E2 or SN2 reaction with a negatively charged nucleophile or base.
Since I– is a weak base, substitution by an SN2 mechanism is favored.
8.61
3° halide, weak base:
SN1 and E1
CH3CH2OH
a.
overall
reaction
Cl
+
+
+ HCl
OCH2CH3
The steps:
+ HCl
SN1
OCH2CH3
CH3CH2OH
or
E1 H
Cl
Cl
H
+ HCl
or
E1
+ HCl
H
Cl
Any base (such as CH3CH2OH or Cl–) can
be used to remove a proton to form an
alkene. If Cl– is used, HCl is formed as a
reaction by-product. If CH3CH2OH is used,
(CH3CH2OH2)+ is formed instead.
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Chapter 8–28
CH3
3° halide
strong base
E2
CH3
OH
Cl
b.
CH2
+ H2O + Cl
+
overall
reaction
CH3
CH3
Cl
Each product:
H
one step
OH
or
+ H2O + Cl
OH
CH2 H
CH2
Cl
one step
8.62 Draw the products of each reaction with the 1° alkyl halide.
Cl
a.
NaOCH2CH3
OCH2CH3
H
H
H
CN
strong
nucleophile
SN2
H
DBU
sterically
hindered base
E2
H
KCN
Cl
b.
Cl
c.
strong
nucleophile
SN2
H
8.63
H
H
H
Br
H
H
H2O
H
O H
OH
H2O
Br
above
H
H
+
HBr
+
HBr
H
Br–
H
H
H
O H
re-draw
Br
below
H
H
H
H
+
H
OH
H2O
H
H
Br
HBr
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
229
Alkyl Halides and Elimination Reactions 8–29
8.64
good nucleophile
O
CH3
C
CH3 CHCH3
O
O
C
CH3COO− is a good nucleophile and a weak base
and so it favors substitution by SN2.
CH3
O
CH3 CHCH3
(only)
Br
CH3 CHCH3 + CH3CH CH2 The strong base gives both SN2 and
E2 products, but since the 2° RX is
OCH2CH3
CH3CH2O
strong base
20%
80%
somewhat hindered to substitution,
the E2 product is favored.
8.65
Cl
OCH3
CH3OH
+
+
+
+
OCH3
3° halide
weak base
SN1 and E1
Cl
CH3OH
Cl
H
OCH3
OCH3
Cl–
HCl
Cl
H
or
HCl
Cl–
CH3OH
HCl
OCH3
Cl
or
H
HCl
Cl
H
OCH3
HCl
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 8–30
8.66 E2 elimination needs a leaving group and a hydrogen in the trans diaxial position.
Cl
Cl
Two different
conformations:
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
This conformation has Cl's
axial, but no H's axial.
This conformation has no Cl's axial.
For elimination to occur, a cyclohexane must have a H and Cl in the trans diaxial arrangement.
Neither conformation of this isomer has both atoms—H and Cl—axial; thus, this isomer only
slowly loses HCl by elimination.
8.67
H Br CH
3
H and Br are anti periplanar.
Elimination can occur.
CH3O−
O
O
H
H
H (in the ring) and Br are NOT anti periplanar.
Elimination cannot occur using this H.
Instead elimination must occur with the
H on the CH3 group.
O
O
–HBr
CH3
Br
H major product
Elimination can occur here.
H
CH3O−
O
O
O
O
–HBr
H
H
Elimination cannot occur in the ring
because the required anti periplanar geometry is not present.
8.68
leaving group
H
N
C6H5O
O
S
N
C6H5O
DBN
N
O
B
H
H H
H H
N
O
overall reaction
S
O
O
H C
CH2Cl
O
SN2
E2
H
A sequence of two reactions forms the
final product: E2 elimination opens the
five-membered ring. Then the sulfur
nucleophile displaces the Cl– leaving group
to form the six-membered ring.
H H
N
C6H5O
Cl
S
O
N
CH2
C
O
O
8.69
a.
H
D
Δ
SeOC6H5
SeOC6H5 and H are on the
same side of the ring.
syn elimination
CH3
D
b.
C
H
Br
CH3
Br
rotate
C
H
Br
H
C
CH3
H
C
CH3
Zn
CH3
Br
Both Br atoms are on the opposite
sides of the C–C bond.
anti elimination
H
C C
H
CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkyl Halides and Elimination Reactions 8–31
8.70 One equivalent of NaNH2 removes one mole of HBr in an anti periplanar fashion from each
dibromide. Two modes of elimination are possible for each compound.
C1
Br
H Br
CH3
a.
H Br
CH3
H
C2
A
Rotate to make
CH3 H and Br anti
periplanar.
Br H
C1
CH3
Br
loss of H from C1
loss of Br from C2
C
H and Br are anti periplanar on
C1 and C2 as drawn.
H Br
H
NaNH2
loss of H from C2
loss of Br from C1
D
H Br
Br
H
Br
NaNH2
H
C2
B
Two different
rotations are needed.
Rotate to make
H and Br anti
periplanar.
C1
CH3
CH3
loss of H from C1
loss of Br from C2
F
C2
H Br
H
Br
NaNH2
b. C and F are diastereomers.
D and E are diastereomers.
C and D are constitutional isomers.
E and F are constitutional isomers.
Br
H
CH3
CH3
loss of H from C2
loss of Br from C1
E
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
233
Alcohols, Ethers, and Epoxides 9–1
Chapter 9 Alcohols, Ethers, and Epoxides
Chapter Review
General facts about ROH, ROR, and epoxides
• All three compounds contain an O atom that is sp3 hybridized and tetrahedral (9.2).
CH3
O
CH3
H
109o
an alcohol
•
O
60o
O
CH3
H
H
111o
an ether
H
H
an epoxide
All three compounds have polar C–O bonds, but only alcohols have an O–H bond for
intermolecular hydrogen bonding (9.4).
H
C
H
H
=
O
H
=
H
O
C
H H
H
hydrogen bond
•
Alcohols and ethers do not contain a good leaving group. Nucleophilic substitution can occur
only after the OH (or OR) group is converted to a better leaving group (9.7A).
+
R OH
+
R OH2
H Cl
+ Cl
strong acid
weak base
good leaving group
•
Epoxides have a leaving group located in a strained three-membered ring, making them reactive
to strong nucleophiles and acids HZ that contain a nucleophilic atom Z (9.15).
leaving group
With strong nucleophiles,
Nu
O
O
C C
OH
H OH
C C
[1]
C C
[2]
Nu
+ OH
Nu
Nu
A new reaction of carbocations (9.9)
• Less stable carbocations rearrange to more stable carbocations by shift of a hydrogen atom or an
alkyl group. Besides rearrangement, carbocations also react with nucleophiles (7.13) and bases (8.6).
1,2-shift
C C
C C
R
(or H)
R
(or H)
Preparation of alcohols, ethers, and epoxides (9.6)
[1] Preparation of alcohols
R X
+
OH
R OH
+ X
•
•
The mechanism is SN2.
The reaction works best for CH3X and 1o RX.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–2
[2] Preparation of alkoxides (a Brønsted–Lowry acid–base reaction)
R O H + Na+ H
R O
Na+ + H2
alkoxide
[3] Preparation of ethers (Williamson ether synthesis)
R X
+
OR'
R OR'
+
X
•
•
The mechanism is SN2.
The reaction works best for CH3X and 1o RX.
[4] Preparation of epoxides (intramolecular SN2 reaction)
•
B
H O
[1]
C C
O
C
X
+
halohydrin
O
C C
[2]
C
+X
X
H B+
A two-step reaction sequence:
[1] Removal of a proton with base
forms an alkoxide.
[2] Intramolecular SN2 reaction forms
the epoxide.
Reactions of alcohols
[1] Dehydration to form alkenes
[a] Using strong acid (9.8, 9.9)
C C
H OH
H2SO4
or
TsOH
C C
•
+
H2O
•
•
Order of reactivity: R3COH > R2CHOH >
RCH2OH.
The mechanism for 2o and 3o ROH is E1;
carbocations are intermediates and
rearrangements occur.
The mechanism for 1o ROH is E2.
The Zaitsev rule is followed.
•
•
The mechanism is E2.
No carbocation rearrangements occur.
•
Order of reactivity: R3COH > R2CHOH >
RCH2OH.
The mechanism for 2o and 3o ROH is SN1;
carbocations are intermediates and
rearrangements occur.
The mechanism for CH3OH and 1o ROH is SN2.
•
[b] Using POCl3 and pyridine (9.10)
C C
H OH
POCl3
C C
pyridine
+ H2O
[2] Reaction with HX to form RX (9.11)
R OH + H X
R X
+ H2O
•
•
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
235
Alcohols, Ethers, and Epoxides 9–3
[3] Reaction with other reagents to form RX (9.12)
R OH
R OH
+ SOCl2
+
R Cl
•
•
pyridine
PBr3
R Br
Reactions occur with CH3OH and 1o and 2o ROH.
The reactions follow an SN2 mechanism.
[4] Reaction with tosyl chloride to form alkyl tosylates (9.13A)
O
R OH + Cl S
CH3
O
pyridine
R O S
O
CH3
O
R OTs
• The C–O bond is not broken so the
configuration at a stereogenic center is
retained.
Reactions of alkyl tosylates
Alkyl tosylates undergo either substitution or elimination depending on the reagent (9.13B).
Nu
C C
+
OTs
•
Substitution is carried out with strong :Nu–,
so the mechanism is SN2.
•
Elimination is carried out with strong bases,
so the mechanism is E2.
H Nu
C C
H OTs
B
+ HB+
C C
–OTs
+
Reactions of ethers
Only one reaction is useful: Cleavage with strong acids (9.14)
R O R'
+
H X
R X
+
R' X
+
H2O
(2 equiv)
(X = Br or I)
•
•
With 2o and 3o R groups, the mechanism is
SN1.
With CH3 and 1o R groups the mechanism is
SN2.
Reactions of epoxides
Epoxide rings are opened with nucleophiles :Nu– and acids HZ (9.15).
•
OH
O
C
C
[1] Nu
[2] H2O
or
HZ
C
C
•
Nu
(Z)
•
The reaction occurs with backside attack,
resulting in trans or anti products.
With :Nu–, the mechanism is SN2, and
nucleophilic attack occurs at the less
substituted C.
With HZ, the mechanism is between SN1
and SN2, and attack of Z– occurs at the
more substituted C.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–4
Practice Test on Chapter Review
1. Give the IUPAC name for each of the following compounds.
a.
b.
O
HO
2. Draw the organic products formed in each reaction. Draw all stereogenic centers using wedges
and dashes.
O
HI
a.
(2 equiv)
[1] NaH
b.
H
D
[1] TsCl, pyr
c.
d.
e.
[2] CH3CH2Br
OH
OH
OH
O
[2] NaOCH3
HBr
[1] NaCN
[2] H2O
H2SO4
f.
OH
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alcohols, Ethers, and Epoxides 9–5
3. What starting material is needed for the following reaction?
[1] TsCl, pyridine
OCH3
[2] NaOCH3
CH3CH2CH2
4. What alkoxide and alkyl halide are needed to make the following ether?
O
CH2CH2CH(CH3)2
Answers to Practice Test
1. a. 3-ethoxy-2methylhexane
2.
a.
Br
d.
3.
Br
I
CH2I
OH
CH3CH2CH2
b. 4-ethyl-7-methyl3-octanol
4.
OH
e.
O
b.
CN
OCH2CH3
XCH2CH2CH(CH3)2
D
H
c.
f.
OCH3
Answers to Problems
9.1 • Alcohols are classified as 1°, 2°, or 3°, depending on the number of carbon atoms bonded to
the carbon with the OH group.
• Symmetrical ethers have two identical R groups, and unsymmetrical ethers have R groups
that are different.
OH
OH
1° alcohol
O
2° alcohol
OH
OH
3° alcohol
1° alcohol
symmetrical ether
CH3
O
unsymmetrical ether
CH3
O
unsymmetrical ether
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–6
9.2 Use the definitions in Answer 9.1.
O
HO
2°
CH2OH
1°
OH
3°
CH3
CH3
H
CH3
F
H
O
9.3 To name an alcohol:
[1] Find the longest chain that has the OH group as a substituent. Name the molecule as a
derivative of that number of carbons by changing the -e ending of the alkane to the suffix -ol.
[2] Number the carbon chain to give the OH group the lower number. When the OH
group is bonded to a ring, the ring is numbered beginning with the OH group, and the “1”
is usually omitted.
[3] Apply the other rules of nomenclature to complete the name.
1
OH
a. [1]
OH
[2]
[3] 3,3-dimethyl-1-pentanol
5 carbons = pentanol
b. [1]
CH3
CH3
[2]
OH
2-methyl
[3] cis-2-methylcyclohexanol
OH
1
6 carbon ring = cyclohexanol
6-methyl
c.
OH
[1]
OH
[2]
[3] 5-ethyl-6-methyl-3-nonanol
3
9 carbons = nonanol
5-ethyl
9.4 To work backwards from a name to a structure:
[1] Find the parent name and draw its structure.
[2] Add the substituents to the long chain.
OH
7
3
2
c. 2-tert-butyl-3-methylcyclohexanol
a. 7,7-dimethyl-4-octanol
4
OH
1
5
b. 5-methyl-4-propyl-3-heptanol
d. trans-1,2-cyclohexanediol
3
OH
OH
OH
or
OH
OH
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239
Alcohols, Ethers, and Epoxides 9–7
9.5 To name simple ethers:
[1] Name both alkyl groups bonded to the oxygen.
[2] Arrange these names alphabetically and add the word ether. For symmetrical ethers, name
the alkyl group and add the prefix di.
To name ethers using the IUPAC system:
[1] Find the two alkyl groups bonded to the ether oxygen. The smaller chain becomes the
substituent, named as an alkoxy group.
[2] Number the chain to give the lower number to the first substituent.
IUPAC name:
a. common name:
CH3 O CH2CH2CH2CH3
CH3 O CH2CH2CH2CH3
substituent:
methoxy
butyl
methyl
larger group – 4 C's
butane
butyl methyl ether
1-methoxybutane
IUPAC name:
b. common name:
OCH3
OCH3
substituent –
methoxy
methyl
larger group – 6 C's
cyclohexane
cyclohexyl
cyclohexyl methyl ether
methoxycyclohexane
IUPAC name:
c. common name:
CH3CH2CH2 O CH2CH2CH3
propyl
CH3CH2CH2 O CH2CH2CH3
propyl
propoxy
dipropyl ether
propane
1-propoxypropane
9.6 Name the ether using the rules from Answer 9.5.
1 CH3
CH3 C
O CH3
CH3
2-methyl
2-methoxy
propane
2-methoxy-2-methylpropane
9.7 Three ways to name epoxides:
[1] Epoxides are named as derivatives of oxirane, the simplest epoxide.
[2] Epoxides can be named by considering the oxygen as a substituent called an epoxy group,
bonded to a hydrocarbon chain or ring. Use two numbers to designate which two atoms the
oxygen is bonded to.
[3] Epoxides can be named as alkene oxides by mentally replacing the epoxide oxygen by a
double bond. Name the alkene (Chapter 10) and add the word oxide.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–8
H
Three possibilities:
[1] methyloxirane
[2] 1,2-epoxypropane
[3] propene oxide
O
a.
CH3
Three possibilities:
[1] cis-2-methyl-3-propyloxirane
[2] cis-2,3-epoxyhexane
[3] cis-2-hexene oxide
O
c.
H
1
b.
CH3
1-methyl
O
epoxy group
2
Two possibilities:
[1] 6 carbons = cyclohexane
1,2-epoxy-1-methylcyclohexane
[2] 1-methylcyclohexene oxide
9.8 Two rules for boiling point:
[1] The stronger the forces the higher the bp.
[2] Bp increases as the extent of the hydrogen bonding increases. For alcohols with the
same number of carbon atoms: hydrogen bonding and bp’s increase: 3° ROH < 2° ROH <
1° ROH.
CH3
OH
CH3
a.
b.
O
OH
OH
OH
VDW
lowest bp
VDW
DD
intermediate bp
VDW
DD
hydrogen
bonding
highest bp
3° ROH
lowest bp
2° ROH
intermediate bp
1° ROH
highest bp
9.9 Strong nucleophiles (like –CN) favor SN2 reactions. The use of crown ethers in nonpolar
solvents increases the nucleophilicity of the anion, and this increases the rate of the SN2
reaction. The nucleophile does not appear in the rate equation for the SN1 reaction. Nonpolar
solvents cannot solvate carbocations, so this disfavors SN1 reactions as well.
9.10 Draw the products of substitution in the following reactions by substituting OH or OR for X in
the starting material.
a. CH3CH2CH2CH2 Br
+ OH
b.
Cl
+
OCH3
c.
CH2CH2–I
+
OCH(CH3)2
CH2CH2–OCH(CH3)2
d.
Br +
OCH2CH3
OCH2CH3
alcohol
CH3CH2CH2CH2 OH + Br
OCH3
+
Cl
+
unsymmetrical ether
+
Br
I
unsymmetrical ether
unsymmetrical ether
9.11 Two possible routes to X are shown. Path [2] with a 1° alkyl halide is preferred. Path [1] cannot
occur because the leaving group would be bonded to an sp2 hybridized C, making it an unreactive
aryl halide.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
241
Alcohols, Ethers, and Epoxides 9–9
F
O
O
[1]
+
X
Br
O
aryl halide
N
CH3
F
O
O
[2]
+
Cl
X
preferred
O
N
CH3
1° alkyl halide
9.12 NaH and NaNH2 are strong bases that will remove a proton from an alcohol, creating a
nucleophile.
a. CH3CH2CH2 O H + Na+ H
CH3
b.
CH3CH2CH2 O Na+ + H2
CH3
+
+ Na NH2
C O H
C O
H
Na+ + NH3
+
H
c.
O
H
Na+ H
CH3CH2CH2–Br + Na+ + H2
O
O
O
d.
H
CH2CH2CH3 + Br–
O
Na+ H
+ Na+ + H2
Br
O
+ Br–
Br
C6H10O
9.13 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major
product.
H
a.
CH3 C CH3
TsOH
CH2 CH CH3
+ H2O
OH
CH3CH2
b.
TsOH
OH
+ H2O
C CHCH3
CH3
trisubstituted
major product
CH2
disubstituted
minor product
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Chapter 9–10
CH3
OH
c.
CH2
CH3
TsOH
+
+ H2O
trisubstituted
disubstituted
major product minor product
9.14 The rate of dehydration increases as the number of R groups increases.
CH3
OH
OH
CH3
HO
1° alcohol
slowest reaction
3° alcohol
fastest reaction
2° alcohol
intermediate
reactivity
9.15
transition state [1]:
transition state [2]:
H
H
CH3CH2 C H
δ+ OH
CH3 C
δ−
H
H
OSO3H
δ−
OSO3H
CH2
OH2
δ+
9.16
H H
HSO4
H
+
This alkene is also formed in addition
to Y from the rearranged carbocation.
β
+
rearranged 3° carbocation
H2SO4
The initially formed 2° carbocation gives two alkenes:
H
HSO4
H
or
H
H
+ CH
2
+
H
HSO4
9.17
CH3 H
a. CH3 C
H
C CH2CH3
+
rearrangement
CH3 H
CH3 C
+
1,2-H shift
2° carbocation
H
b.
+
2° carbocation
C CH2CH3
+
3° carbocation
more stable
+
rearrangement
1,2-methyl shift
H
3° carbocation
more stable
rearrangement
1,2-H shift
c.
2° carbocation
+
3° carbocation
more stable
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alcohols, Ethers, and Epoxides 9–11
9.18
CH3 H
CH3 C
H
CH3 H
H2O
C CH3
CH3 C
overall
reaction
Cl
CH3 H
C CH3
H
CH3 C
OH
C CH3
HCl
OH H
The steps:
CH3 H
CH3 C
Cl
CH3 H
C CH3
H
CH3 C
H2O
and
O H
H
CH3 H
CH3 C
C CH3
H
Cl
CH3 H
C CH3
CH3 C
CH3 H
C CH3
H
CH3 C
H
H2O
2° carbocation
H O
3° carbocation
C CH3
H
H
Rearrangement of H forms
a more stable carbocation.
Cl
9.19
CH3
CH3
HCl
a. CH3 C CH2CH3
CH3 C CH2CH3
OH
c.
HBr
Br
+ H 2O
Cl
OH
HI
b.
+ H2 O
I
OH
+ H2 O
9.20 • CH3OH and 1° alcohols follow an SN2 mechanism, which results in inversion of configuration.
• Secondary (2°) and 3° alcohols follow an SN1 mechanism, which results in racemization at
a stereogenic center.
H OH
a.
C
I H
HI
D
CH3CH2CH2
CH3
b.
HBr
Br
CH3
OH
HO
D
1° alcohol so inversion of configuration
CH3CH2CH2
achiral starting material
c.
C
HCl
3° alcohol, so Br− attacks from above and below.
The product is achiral.
achiral product
Cl
Cl
3° alcohol = racemization
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Chapter 9–12
9.21
OH
a.
HCl
Cl
HCl
c.
Cl
OH
(product formed after a 1,2-H shift)
Cl
HCl
b.
OH
(product formed after a 1,2-CH3 shift)
9.22 Substitution reactions of alcohols using SOCl2 proceed by an SN2 mechanism. Therefore, there
is inversion of configuration at a stereogenic center.
H OH
SOCl2
Reactions using SOCl2
proceed by an SN2 mechanism =
inversion of configuration.
Cl H
pyridine
S
R
9.23 Substitution reactions of alcohols using PBr3 proceed by an SN2 mechanism. Therefore, there is
inversion of configuration at a stereogenic center.
PBr3
H OH
Reactions using PBr3
proceed by an SN2 mechanism =
inversion of configuration.
Br H
S
R
9.24 Stereochemistry for conversion of ROH to RX by reagent:
[1] HX—with 1°, SN2, so inversion of configuration; with 2° and 3°, SN1, so racemization.
[2] SOCl2—SN2, so inversion of configuration.
[3] PBr3—SN2, so inversion of configuration.
OH
a.
SOCl2
c.
Cl
OH
PBr3
Br
pyridine
OH
HI
I
I
S N2 =
inversion
b.
3° alcohol, SN1 =
racemization
9.25 To do a two-step synthesis with this starting material:
[1] Convert the OH group into a good leaving group (by using either PBr3 or SOCl2).
[2] Add the nucleophile for the SN2 reaction.
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Alcohols, Ethers, and Epoxides 9–13
H
H
H
PBr3
CH3 C OH
N3
CH3
CH3 C Br
CH3
CH3 C N3
CH3CH2O
CH3
H
CH3 C OCH2CH3
CH3
bad leaving
group
good leaving
group
9.26
O
+ CH3
a. CH3CH2CH2CH2 OH
H OH
b.
CH3CH2CH2
C
CH3CH2CH2CH2 O S
SO2Cl
pyridine
CH3 + Cl
O
H OTs
TsCl
CH3
pyridine
CH3CH2CH2
C
+ Cl
CH3
9.27
OTs +
a.
1° tosylate
b.
CN
SN2
CN
+
CH3CH2CH2 OTs
1° tosylate
K+ −OC(CH3)3
E2
CH3CH CH2 + K+ −OTs + HOC(CH3)3
strong bulky
base
H OTs
c.
CH3
HS H
+
C
CH2CH2CH3
2° tosylate
+ OTs
strong
nucleophile
SH
SN2
strong
nucleophile
CH3
C
SN2 product
(inversion of configuration)
CH2CH2CH3
(Substitution is
favored over
elimination.)
9.28
HO H
S
TsCl
TsO H
NaOH
H OH
SN 2
pyridine
retention
S
enantiomers
One inversion from starting material
to product.
inversion
R
9.29 These reagents can be classified as:
[1] SOCl2, PBr3, HCl, and HBr replace OH with X by a substitution reaction.
[2] Tosyl chloride (TsCl) makes OH a better leaving group by converting it to OTs.
[3] Strong acids (H2SO4) and POCl3 (pyridine) result in elimination by dehydration.
246
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–14
H
a.
H
SOCl2
CH3 C OH
pyridine
CH3
H
b.
H
CH3 C Cl
CH3
CH3
H
TsCl
CH3 C OH
pyridine
CH3
H
CH3 C OH
CH3
H
CH3 C Br
CH3
H
e.
CH3 C OTs
CH3
H
c.
HBr
CH3 C OH
d.
[1] PBr3
[2] NaCN
CH3 C CN
CH3
H
H2SO4
CH3 C OH
CH2 CHCH3
CH3 C OH
f.
CH3
CH3
POCl3
pyridine
CH2 CHCH3
9.30
a. CH3CH2 O CH2CH3
CH3
b.
CH3 C O CH2CH3
HBr
HBr
H
2 CH3CH2 Br
c.
+ H2O
O CH3
HBr
CH3
Br
+ CH3Br
CH3 C Br
+
+ H2O
CH3CH2Br + H2O
H
9.31 Ether cleavage can occur by either an SN1 or SN2 mechanism, but neither mechanism can occur
when the ether O atom is bonded to an aromatic ring. An SN1 reaction would require formation
of a highly unstable carbocation on a benzene ring, a process that does not occur. An SN2
reaction would require backside attack through the plane of the aromatic ring, which is also not
possible. Thus, cleavage of the Ph–OCH3 bond does not occur.
OCH3
anisole
HBr
OH
+
CH3Br
Br
bromobenzene
phenol
NOT formed
SN1:
SN2:
O CH3
H
CH3
CH3OH
highly unstable
carbocation
O
H
Br
9.32 Two rules for reaction of an epoxide:
[1] Nucleophiles attack from the back side of the epoxide.
[2] Negatively charged nucleophiles attack at the less substituted carbon.
CH3
a.
O
[1] CH3CH2O
[2] H2O
Attack here:
less substituted C
backside attack
CH3
OH
OCH2CH3
b. CH3
H
O
C C
[1] H
H
C C
H [2] H O
2
Attack here:
less substituted C
backside attack
HO
CH3
H
H
H
C C
C
CH
247
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alcohols, Ethers, and Epoxides 9–15
9.33 In both isomers, –OH attacks from the back side at either C–O bond.
cis-2,3-dimethyloxirane
[1] –OH
O
H
CH3
C
C
H
CH3
CH3
OH
H
C
[2] H2O
C
HO
+
H
CH3
HO
H
C
CH3
C
H
CH3
OH
[1] –OH
[2] H2O
O
H
CH3
C
C
H
CH3
enantiomers
HO
HO
O
H
CH3
C
[1] –OH
C
CH3
H
CH3
OH
H
C
[2] H2O
HO
HO
+
C
H
CH3
CH3
H
C
C
H
CH3
OH
[1]
O
–OH
H
CH3
[2] H2O
C
C
CH3
H
identical
Rotate around the C–C bond to
see the plane of symmetry.
HO
trans-2,3-dimethyloxirane
HO
OH
C
CH3
H
HO
C
H
CH3
meso compound
9.34 Remember the difference between negatively charged nucleophiles and neutral nucleophiles:
• Negatively charged nucleophiles attack first, followed by protonation, and the nucleophile
attacks at the less substituted carbon.
• Neutral nucleophiles have protonation first, followed by nucleophilic attack at the more
substituted carbon.
But, trans or anti products are always formed regardless of the nucleophile.
CH3
a.
O
HBr
Br
CH3
OH
CH3CH2
CH3CH2
H
H
CH3
b.
[1] CN
[2] H2O
negatively charged
nucleophile:
attack at less
substituted C
CH3CH2OH
H2SO4
neutral nucleophile:
attack at more
substituted C
neutral nucleophile:
attack at more
substituted C
O
CH3CH2
CH3CH2
O
c.
CH3
OH
CN
O
d.
CH3CH2
CH3CH2
OH
C
C
CH3CH2O
H
HO
H
H
[1] CH3O
[2] CH3OH
H
C
CH3CH2
CH3CH2
negatively charged
nucleophile:
attack at less
substituted C
H
H
C
OCH3
248
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–16
9.35
2 3
HO
cyclohexanol
6 C ring
2 CH3's at C3
3,3-dimethylcyclohexanol
1
a.
b.
3
2
1
c.
O
2
1,2-epoxy-1-ethylcyclopentane
or
1-ethylcyclopentene oxide
ethyl isobutyl ether
or
1-ethoxy-2-methylpropane
O
1
9.36
a. (1R,2R)-2-isobutylcyclopentanol
f.
b. 2° alcohol
[1]
NaH
HO
O
A
H2SO4
HO
A
[2]
HO
c. stereoisomer
POCl3
[3]
pyridine
HO
HO
Cl
HCl
(1R,2S)-2-isobutylcyclopentanol
[4]
HO
SOCl2
d. constitutional isomer
[5]
pyridine
HO
Cl
OH
TsCl
(1S,3S)-3-isobutylcyclopentanol
[6]
pyridine
HO
TsO
e. constitutional isomer with an ether
O
butoxycyclopentane
9.37
HO
H
HBr
a.
Br
S
HO
H
b.
H
PBr3
HO
H
HCl
Cl
S
Br
2° alcohol
SN1 = racemization
R
Br
R
c.
H
H
H
PBr3 follows
SN2 = inversion.
H
Cl
R
2° alcohol
SN1 = racemization
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
249
Alcohols, Ethers, and Epoxides 9–17
HO
H
H
SOCl2
d.
pyridine
SOCl2 follows
SN2 = inversion.
Cl
R
9.38 Draw the structure of each alcohol, using the definitions in Answer 9.1.
OH
OH
a.
OH
1°
b.
OH
2°
3°
enol
9.39 Use the directions from Answer 9.3.
a.
(CH3)2CHCH2CH2CH2OH
[1]
[2]
H
CH3 C CH2CH2CH2OH
H
CH3
CH3
5 carbons = pentanol
b.
[3]
1
4-methyl-1-pentanol
CH3 C CH2CH2CH2OH
4-methyl
(CH3)2CHCH2CH(CH2CH3)CH(OH)CH2CH3
[1]
H
[2]
H OH
CH3 C CH2 C CH
CH3
CH2CH3
H
CH2CH3
CH3
7 carbons = heptanol
c.
[1]
[3]
4-ethyl-6-methyl-3-heptanol
CH2CH3
CH2CH3
6-methyl
4-ethyl
5-methyl
[2]
[3]
OH
OH
8 carbons = octanol
d.
[1]
3
H OH
CH3 C CH2 C CH
HO
OH
HO
OH
HO H
HO H
[2]
3
2
HO H
[3]
(2R,3R)-2,3-butanediol
2R,3R
4 carbons = butanediol
[2]
OH
4
[1]
OH
g.
[1]
[3]
5-methyl-2,3,4-heptanetriol
2
OH
5-methyl
[2]
CH(CH3)2
OH
OH
7 carbons = heptanetriol
HO
cis-1,4-cyclohexanediol
HO H
f.
OH
[3]
cis
cyclohexanediol
e.
[1]
4-ethyl
3
1
4
[2]
4-ethyl-5-methyl-3-octanol
[3]
1
HO
CH(CH3)2
5 carbons = cyclopentanol
3-isopropyl
trans-3-isopropylcyclopentanol
250
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–18
9.40 Use the rules from Answers 9.5 and 9.7.
a.
c.
O
O
1,2-epoxy-2-methylhexane
or 2-butyl-2-methyloxirane
or 2-methylhexene oxide
dicyclohexyl ether
4,4-dimethyl
CH3
b.
d.
OCH2CH2CH3
CH3
longest chain =
heptane
substituent =
3-propoxy
4,4-dimethyl-3-propoxyheptane
CH3
CH3 C O C CH3
CH3
tert-butyl tert-butyl
di-tert-butyl ether
9.41 Use the directions from Answer 9.4.
a. 4-ethyl-3-heptanol
3
f. diisobutyl ether
4
O
1
g. 1,2-epoxy-1,3,3-trimethylcyclohexane
OH
1
OH
b. trans-2-methylcyclohexanol
O
OH
or
3
CH3
HO
CH3
3
h. 1-ethoxy-3-ethylheptane
O 1
c. 2,3,3-trimethyl-2-butanol
3
2
i. (2R,3S)-3-isopropyl-2-hexanol
d. 6-sec-butyl-7,7-diethyl-4-decanol
OH
3
6
2
OH
4
1
e. 3-chloro-1,2-propanediol
HO
2
OH
7
3
j. (2S)-2-ethoxy-1,1-dimethylcyclopentane
1
Cl
2
O
1
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
251
Alcohols, Ethers, and Epoxides 9–19
9.42
Eight constitutional isomers of molecular formula C5H12O containing an OH group:
OH 1-pentanol
OH
2-pentanol
2° alcohol
OH
1 OH 3-methyl-1-butanol
1° alcohol
3
1° alcohol
OH
2 1
2,2-dimethyl-1-propanol
1° alcohol
OH
3-pentanol
2° alcohol
1 OH 2-methyl-1-butanol
1° alcohol
2
OH
3-methyl-2-butanol
2° alcohol
2-methyl-2-butanol
3° alcohol
2
9.43 Use the boiling point rules from Answer 9.8.
a.
CH3CH2OCH3
(CH3)2CHOH
ether
no hydrogen bonding
lowest bp
b.
2° alcohol
hydrogen bonding
intermediate bp
CH3CH2CH2OH
1° alcohol
hydrogen bonding
highest bp
CH3CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2OH
no OH group
lowest water solubility
one OH group
intermediate water solubility
HOCH2CH2CH2CH2CH2CH2OH
two OH groups
highest water solubility
9.44 Melting points depend on intermolecular forces and symmetry. (CH3)2CHCH2OH has a lower
melting point than CH3CH2CH2CH2OH because branching decreases surface area and makes
(CH3)2CHCH2OH less symmetrical so it packs less well. Although (CH3)3COH has the most
branching and least surface area, it is the most symmetrical, so it packs best in a crystalline
lattice, giving it the highest melting point.
OH
OH
–108 °C
lowest melting point
OH
–90 °C
intermediate melting point
26 °C
highest melting point
9.45 Stronger intermolecular forces increase boiling point. All of the compounds can hydrogen
bond, but both diols have more opportunity for hydrogen bonding since they have two OH
groups, making their bp’s higher than the bp of 1-butanol. 1,2-Propanediol can also
intramolecularly hydrogen bond. Intramolecular hydrogen bonding decreases the amount of
intermolecular hydrogen bonding, so the bp of 1,2-propanediol is somewhat lower.
Increasing boiling point
OH
HO
HO
OH
OH
1-butanol
118 °C
1,2-propanediol
187 °C
1,3-propanediol
215 °C
252
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–20
9.46
H2SO4
a. CH3CH2CH2OH
NaH
b. CH3CH2CH2OH
CH3CH2CH2Cl
ZnCl2
HBr
d. CH3CH2CH2OH
PBr3
f. CH3CH2CH2OH
[1] NaH
[1] TsCl
CH3CH2CH2OH
CH3CH2CH2Cl
CH3CH2CH2OTs
pyridine
h. CH3CH2CH2OH
+ H 2O
CH3CH2CH2Br
TsCl
g. CH3CH2CH2OH
+ H2
CH3CH2CH2Br + H2O
SOCl2
pyridine
e. CH3CH2CH2OH
+ H 2O
CH3CH2CH2O Na+
HCl
c. CH3CH2CH2OH
i.
CH3CH CH2
[2] CH3CH2Br
CH3CH2CH2O Na+
[2] NaSH
CH3CH2CH2OTs
CH3CH2CH2OCH2CH3
CH3CH2CH2SH
9.47
a.
OH
NaH
b.
OH
NaCl
c.
OH
HBr
d.
OH
HCl
e.
OH
H2SO4
f.
OH
NaHCO3
Br
g.
OH
[1] NaH
Cl
h.
OH
POCl3
pyridine
O
N.R.
Na+
+ H2
+ H2O
N.R.
O
[2] CH3CH2Br
O CH2CH3
9.48 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major
product.
TsOH
a.
OH
tetrasubstituted
major product
b.
CH2CH3
OH
c.
OH
disubstituted
CH2CH3
TsOH
TsOH
trisubstituted disubstituted
major product
CHCH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
253
Alcohols, Ethers, and Epoxides 9–21
d.
TsOH
CH 3CH 2CH2CH2OH
CH3CH 2CH CH2
OH
TsOH
e.
tetrasubstituted
major product
disubstituted
Two products formed
by carbocation rearrangement
9.49 The most stable alkene is the major product.
OH
H2SO4
monosubstituted
trans and disubstituted
major product
cis and disubstituted
9.50 OTs is a good leaving group and will easily be replaced by a nucleophile. Draw the products
by substituting the nucleophile in the reagent for OTs in the starting material.
a.
CH3CH2CH2CH2 OTs
b.
CH3CH2CH2CH2 OTs
c.
CH3CH2CH2CH2 OTs
d.
CH3CH2CH2CH2 OTs
CH3SH
CH3CH2CH2CH2 SCH3 + HOTs
S N2
NaOCH2CH3
CH3CH2CH2CH2 OCH2CH3
S N2
NaOH
CH3CH2CH2CH2 OH
SN2
K+
–
OC(CH3)3
+ Na+
+ Na+
–OTs
–OTs
CH3CH2CH CH2 + (CH3)3COH + K+ –OTs
E2
9.51
HBr
a.
b.
+
HO H
OH
H D
Br H
HCl
ZnCl2
Cl
HO H
H D
pyridine
H Cl
TsCl
pyridine
d.
HO H
2° Alcohol will undergo SN1.
racemization
1° Alcohol will undergo SN2.
inversion
SOCl2
c.
H Br
SOCl2 always implies SN2.
inversion
KI
TsO H
Configuration is maintained.
C–O bond is not broken.
SN2
inversion
H
I
254
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–22
9.52
NaH
CH3I
A=
(a)
B=
O H
(b)
TsCl
C=
pyridine
HO H
TsO H
PBr3
(c)
CH3O H
CH3O
D=
H OCH3
CH3O
F=
E=
(2R)-2-hexanol
H Br
CH3O H
Routes (a) and (c) given identical products, labeled B and F.
9.53 Acid-catalyzed dehydration follows an E1 mechanism for 2o and 3o ROH with an added step to
make a good leaving group. The three steps are:
[1] Protonate the oxygen to make a good leaving group.
[2] Break the C−O bond to form a carbocation.
[3] Remove a β hydrogen to form the π bond.
OH
CH3 H OSO3H
+
overall
reaction
The steps:
CH3
CH3
CH2
+
+ H2O
H
O H
CH3
H
H
+ HSO4
CH3
H
CH3
+ H2O
+ H2SO4
HSO4
2° carbocation
and
CH3
H
1,2-H shift
+ HSO4
CH2
CH2
H
2° carbocation
+ H2SO4
3° carbocation
and
H
+ HSO4
CH3
CH3
+ H2SO4
9.54 With POCl3 (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has a
β hydrogen, only one product is formed. With H2SO4, the mechanism of elimination is E1.
A 2° carbocation rearranges to a 3° carbocation, which has three pathways for elimination.
255
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alcohols, Ethers, and Epoxides 9–23
O
OH
Cl
P
Cl
H
POCl2
O
Cl
N
POCl2
N
O
H
+
N
H
V
+
W
Cl
+ –OPOCl2
+
N
H
H OSO3H
OH
+
OH2
+
+ H 2O
V
+
1,2–CH3 shift
2° carbocation
+ HSO4–
3° carbocation
HSO4–
H
H
+
+
+
HSO4–
3° carbocation
H
3° carbocation
X
3° carbocation
Y
+ H2SO4
HSO4–
+
Z
+
H2SO4
H2SO4
9.55 To draw the mechanism:
[1] Protonate the oxygen to make a good leaving group.
[2] Break the C−O bond to form a carbocation.
[3] Look for possible rearrangements to make a more stable carbocation.
[4] Remove a β hydrogen to form the π bond.
Dark and light circles are meant to show where the carbons in the starting material appear in the product.
OH
H OSO3H
O H
+ H2O
H
+ HSO4–
2o carbocation
H
+ HSO4–
3o carbocation
+ H2SO4
256
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–24
9.56
OH
[1] HBr
3-methyl-2-butanol
1,2-H shift
H
H
Br
Br
[2] –H2O
2° carbocation
2-methyl-1-propanol
HBr
HO
3° carbocation
H2 O
Br
Br
H Br
H2O
SN2
no carbocation
The 2° alcohol reacts by
an SN1 mechanism to
form a carbocation
that rearranges.
The 1° alcohol reacts with HBr
by an SN2 mechanism.
no carbocation intermediate =
no rearrangement possible
9.57
H Cl
A
Cl
OH
H2O
OH2
two resonance structures
for the carbocation
B
Cl
Cl
Cl
Cl
C
9.58
H2SO4, NaBr
CH3CH2CH2CH2OH
overall reaction
CH3CH2CH2CH2Br
CH3CH2CH CH2
CH3CH2CH2CH2OCH2CH2CH2CH3
Step [1] for all products: Formation of a good leaving group
CH3CH2CH2CH2
OH
H OSO3H
CH3CH2CH2CH2
O H
+ HSO4
H
Formation of CH3CH2CH2CH2Br:
CH3CH2CH2CH2
O H
SN2
H
CH3CH2CH2CH2Br + H2O
Br
Formation of CH3CH2CH=CH2:
CH3CH2CH CH2 O H
H
HSO4–
H
E2
CH3CH2CH CH2
+ H2 O
H2SO4
257
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Alcohols, Ethers, and Epoxides 9–25
Ether formation (from the protonated alcohol):
CH3CH2CH2CH2
CH3CH2CH2CH2
OH2
CH3CH2CH2CH2
S N2
O CH2CH2CH2CH3 + H2O
H
HSO4
O H
CH3CH2CH2CH2 O
CH2CH2CH2CH3
H2SO4
9.59
H OSO3H
OH
O
OH2
+
O
+ HSO4–
O
+ HSO4–
H
1,2-shift
O +
O
+ H2SO4
+ H 2O
H
O
+
9.60
HSO4–
H OSO3H
OH
OH2
OH
O
OH
+ HSO4
–
H
O
OH
+ H 2O
+ H2SO4
9.61
a.
OCH2CH2CH3
b.
O
OCH2CH2CH3
O
Br
O
Br
2° halide
Br
O
1° halide
less hindered RX
preferred path
O
OCH2CH2CH3
+ BrCH2CH2CH3
1° halide
less hindered RX
preferred path
2° halide
258
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–26
c.
CH3CH2OCH2CH2CH3
CH3CH2OCH2CH2CH3
CH3CH2O + BrCH2CH2CH3
CH3CH2Br + OCH2CH2CH3
1° halide
1° halide
Neither path preferred.
9.62 A tertiary halide is too hindered and an aryl halide too unreactive to undergo a Williamson
ether synthesis.
Two possible sets of starting materials:
CH3
Br
O C CH3
CH3
CH3
CH3
O C CH3
Br C CH3
CH3
aryl halide
unreactive in SN2
O
CH3
3° alkyl halide
too sterically
hindered for SN2
9.63
a.
(CH3)3COCH2CH2CH3
b.
HBr
HBr
O
(2 equiv)
2
c.
H2 O
(CH3)3CBr + BrCH2CH2CH3
(2 equiv)
OCH3
HBr
Br
(2 equiv)
CH3Br
H2O
H2 O
Br
9.64
overall reaction
CH3
a.
O
I
CH3
H
I
+ H2O
I
The steps:
I
CH3
O
CH3
+ I
H
+
CH3
O
H
H
CH3
O
I
CH3
CH3
H
b.
Cl
O
H
Na+ H
Cl
O
+ H2 + Na+
I
H
O
H
I
CH3
I
CH3
+ H2 + NaCl
O
259
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Alcohols, Ethers, and Epoxides 9–27
9.65
OC(CH3)3
O C(CH3)3
H–OCOCF3
CH3
OH
H
CH2
+ CF3CO2–
C
OH
CH3
+
CH3
C CH2
H
CH3
CF3CO2–
+ CF3CO2H
9.66
O
C
a. H
O
C
H
HBr
H
H
BrCH2CH2OH
C
d. H
b. H
C
H
C
H
HOCH2CH2OH
e. H C
H2SO4
H
H
HOCH2CH2OH
H
H
[2] H2O
O
CH3CH2OCH2CH2OH
[2] H2O
f. H C
CH2 CH2OH
[1] −OH
C
H
[1] CH3CH2O−
HC C
[2] H2O
O
H2O
H
H
O
c. H C
H
H
H
O
C
[1] HC C–
C
C
[1] CH3S−
H
H
H
CH3SCH2CH2OH
[2] H2O
9.67
O
a. CH
3
H
CH3
CH3
b.
CH3 C
H2SO4
H
H
Br
O
CH3
OH
[2] H2O
OH
HBr
c.
C H
CH3CH2O
[1] CH3CH2O− Na+
O
O
CH3 OH
CH3CH2OH
OH
CN
[1] NaCN
d.
[2] H2O
OCH2CH3
9.68
CH3
H
Cl
C
a.
CH3
H O
Cl
H
C
C
H
CH3
C
O
H
CH3
Cl
CH3
C
b.
H O
H
Cl
CH3
H
C
C
CH3
C
O
H
CH3
Na+ H
OH
Cl
C
c.
CH3
H
C
H
CH3
CH3
H
Cl
rotate
C
CH3
H
C
The 2 CH3 groups are anti in the
starting material, making them
trans in the product.
CH3
H
O
C4H8O
Na+ H
H
H
CH3 C
CH3
H C
C
O
C4H8O
C
Na+ H
CH3
The 2 CH3 groups are gauche in the
starting material, making them
cis in the product.
CH3
H
Cl
C
O H
CH3
H
H
C
O
backside
attack
CH3
H C
C
O
C4H8O
H
CH3
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Chapter 9–28
9.69 First, use the names to draw the structures of the starting material and both products. Because
the product has two OH groups, one OH must come from the epoxide oxygen, and one must
come from the nucleophile, either –OH or H2O.
With –OH, the nucleophile
attacks at the less
substituted end of the
epoxide to form the R
isomer of the product.
OH
[1] Na+ –OH
O
HO
[2] H2O
R
from the nucleophile
(2R)-2-ethyl-2-methyloxirane
(2R)-2-methyl-1,2-butanediol
With H2O and H2SO4,
H2O
O
H2O attacks at the
HO
stereogenic center, from
from the nucleophile
OH
H2SO4
the back side, and the S
(2R)-2-ethyl-2-methyloxirane
(2S)-2-methyl-1,2-butanediol
isomer is formed.
9.70
O
O
O
CH3CH2O
CH3CH2O
+
Cl
CH3CH2OCH2
Cl
Cl–
CH3
9.71
KOC(CH3)3
OTs
a.
b.
OH
*
Bulky base favors E2.
HBr
H CH3
c.
H CH3
O
CH3CH2 C C H
CH2CH3
H
[1] −CN
CH3CH2
H
[2] H2O
Keep the stereochemistry at the stereogenic
center [*] the same here since no bond is
broken to it.
+
C
H
CH3CH2
CH2CH3
H
NC
H
OH
e.
H
C
C
CH2CH3
CN
H
KCN
(CH3)3C
HO
OH
C
OTs
d.
SN2
inversion
PBr3
CH3
f.
Br
*
CN
(CH3)3C
Br
SN2
inversion
CH3
OH
H D
TsCl
pyridine
OTs
H D
CH 3CO2
CH3CO2
H D
identical
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
261
Alcohols, Ethers, and Epoxides 9–29
g.
h.
OH
Br
Br
OH
HBr
O
O
[2] H2O
OH
OCH3
OH
[1] NaOCH3
OH
OCH3
O
NaH
OCH2CH3
CH3CH2I
i.
CH2CH3
CH3
j.
CH2CH3
CH2CH3
CH3CH2 C O CH3
CH3
HI
(2 equiv)
CH3
CH3CH2 C
I
+
I
+ H2O
CH3
CH3
9.72
a.
HBr
or
PBr3
OH
b.
OH
c.
OH
Br
HCl, ZnCl2
or
SOCl2, pyridine
Cl
[1] Na+ H−
[2] CH3CH2Cl
O
O
[1] TsCl, pyridine
d.
OH
OTs
[2] N3−
N3
Make OH a good leaving
group (use TsCl); then add N3−.
9.73
a.
OH
HCl
or
SOCl2, pyridine
b.
OH
H2SO4
c.
OH [1] Na+ H−
d.
OH [1] TsCl, pyridine
Cl
O
[2] CH3Cl
OTs [2] –CN
OCH3
CN
Make OH a good
leaving group (use TsCl);
then add −CN.
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Chapter 9–30
9.74
Br
(a) = HBr
or PBr3
or other
strong base
OH
OTs
(c) =
(e) =
(d) = KOC(CH3)3
or other
strong base
TsCl, pyridine
H2SO4 or
TsOH
(b) = KOC(CH3)3
Br
(f) = KOC(CH3)3
NBS
or other
strong base
HOCl
OH
(g) = NaH
OH
(h) =
+ enantiomer
O
−OH,
Cl
H2O
OH
9.75
OH
O
O
Cl
O
O
O
NaH
(CH3)2CHNH2
N
H H
O
proton
transfer
O
OH
N
H
propranolol
9.76 With the cis isomer, –OH acts as a nucleophile to displace Br– from the back side, forming a
trans diol (A). With the trans isomer, the two functional groups are arranged in a manner that
allows an intramolecular SN2. –OH removes a proton to form an alkoxide, which can then
displace Br– by intramolecular backside attack to afford an ether (B). Such a reaction is not
possible with the cis isomer because the nucleophile and leaving group are on the same side.
HO
OH
Br
HO
OH
SN2
cis-4-bromocyclohexanol
A
O
OH
HO
Br
trans-4-bromocyclohexanol
proton
transfer
O
Br
+ H2O
SN2
B
263
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Alcohols, Ethers, and Epoxides 9–31
9.77 If the base is not bulky, it can react as a nucleophile and open the epoxide ring. The bulky base
cannot act as a nucleophile, and will only remove the proton.
–
N(CH2CH3)2
H
Li+
O
+
H OH
O
+
HN(CH2CH3)2
LiOH
OH
9.78 First form the 2° carbocation. Then lose a proton to form each product.
H
H
CH3CH2 C CH2 OH
CH3CH2 C CH2
H OH2
H
H
1o alcohol
no 1o carbocation
at this step
H
H
1,2-shift
OH2
CH3CH2 C CH2
H
+ H2O
2o carbocation
H
CH3CH2 C CH2
H
H
CH3CH2 C CH2
H2O
+
CH3
H
H
CH3CH2CH=CH2
H3 O
H
H
+
C C
CH3 CH C CH3
=
CH3
H
+
C C
CH3
CH3
H3O
H2O
9.79
HO
H OSO3H
H2O
+ H2O
2o carbocation
+ HSO4
H2O
1,2-shift
H
+ H3O+
3o and resonance-stabilized
allylic carbocation
+
H2O
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 9–32
9.80
H OSO3H
OH
a.
+ H2O
OH2
X
+
HSO4–
1,2-shift
1,2-shift
H
Y
+ H3O+
H2O
b. Other elimination products can form from carbocations X and Y.
+ H3O+
H
X
H2O
H
X
+ H3O+
H2O
+ H3O+
H
Y
H2O
9.81
OH OH
a. CH3 C
C CH3
CH3 CH3
pinacol
H OSO3H
OH
OH2 OH
CH3 C
CH3 C
C CH3
C CH3
shift
CH3 CH3
CH3 CH3
1,2-CH3
CH3
CH3 C
CH3
OH
C
CH3
+ H2O
+ HSO4
CH3
H2SO4 +
CH3 C
CH3
O
C
CH3
pinacolone
CH3 O H
CH3 C
CH3
C
CH3
HSO4
265
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alcohols, Ethers, and Epoxides 9–33
b. Two different carbocations can form. The carbocation with the (+) charge adjacent to the
benzene rings (A) is more stable, so it is preferred.
HO
HO
H2SO4
or
(two steps)
OH
OH
D
A
B
resonance-stabilized
preferred
1,2-CH3 shift
O
O
+ H3O+
H
H2O
9.82
O
O H
C
I
O
H
H
O
Na+ OH
C
I
O
O
O
O
O
H
H
O
C
I
O
H
H
O
O
H
C
C
H
I
O
H
H
O
I
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Chapter 9–34
9.83 The conversion of X to Y requires two operations. X contains both a nucleophile (NH2) and a
leaving group (OSO2CH3), so an intramolecular SN2 reaction forms an aziridine. Since the
aziridine is strained, the amine nucleophile (CH2=CHCH2NH2) opens the ring by backside
attack, resulting in the trans stereochemistry of the two N’s on the six-membered ring.
O
CO2CH2CH3
O
backside
attack
H2N
OSO2CH3
H
CO2CH2CH3
New C–N bond on opposite
side to C–OSO2CH3 bond
N
H
X
OSO2CH3
O
CO2CH2CH3
H2N
CO2CH2CH3
HN
Y
aziridine
H2N
proton transfer
H2N
CO2CH2CH3
backside
attack HN
HN
O
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
267
Alkenes 10–1
Chapter 10 Alkenes
Chapter Review
General facts about alkenes
• Alkenes contain a carbon–carbon double bond consisting of a stronger σ bond and a weaker
π bond. Each carbon is sp2 hybridized and trigonal planar (10.1).
• Alkenes are named using the suffix -ene (10.3).
• Alkenes with different groups on each end of the double bond exist as a pair of diastereomers,
identified by the prefixes E and Z (10.3B).
Two higher priority groups on
the same side
Two higher priority groups on
opposite sides
CH3
CH3
CH3
C C
H
CH2CH3
H
E isomer
(2E)-3-methyl-2-pentene
•
CH3
Z isomer
(2Z)-3-methyl-2-pentene
Alkenes have weak intermolecular forces, giving them low mp’s and bp’s, and making them
water insoluble. A cis alkene is more polar than a trans alkene, giving it a slightly higher boiling
point (10.4).
cis-2-butene
more polar isomer
trans-2-butene
CH3
CH3
CH3
C C
H
less polar isomer
C C
H
H
H
CH3
no net dipole
a small net dipole
higher bp
•
CH2CH3
C C
lower bp
Since a π bond is electron rich and much weaker than a σ bond, alkenes undergo addition
reactions with electrophiles (10.8).
Stereochemistry of alkene addition reactions (10.8)
A reagent XY adds to a double bond in one of three different ways:
• Syn addition—X and Y add from the same side.
C C
•
H
BH 2
C C
•
Syn addition occurs in hydroboration.
Anti addition—X and Y add from opposite sides.
C C
•
H BH2
X2
or
X2, H2O
X
•
C C
X(OH)
Anti addition occurs in halogenation and
halohydrin formation.
Both syn and anti addition occur when carbocations are intermediates.
H
X(OH)
H
• Syn and anti addition occur in
H X
C C
C C
C C
and
hydrohalogenation and
or
X(OH)
H2O, H+
hydration.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–2
Addition reactions of alkenes
[1] Hydrohalogenation—Addition of HX (X = Cl, Br, I) (10.9–10.11)
• The mechanism has two steps.
• Carbocations are formed as intermediates.
RCH CH2 + H X
R CH CH2
• Carbocation rearrangements are possible.
X
H
alkyl halide
• Markovnikov’s rule is followed. H bonds
to the less substituted C to form the more
stable carbocation.
• Syn and anti addition occur.
[2] Hydration and related reactions—Addition of H2O or ROH (10.12)
For both reactions:
• The mechanism has three steps.
+
H OH
RCH CH2
R CH CH2
H2SO4
• Carbocations are formed as intermediates.
OH H
• Carbocation rearrangements are possible.
alcohol
• Markovnikov’s rule is followed. H bonds
to the less substituted C to form the more
RCH CH2 + H OR
R CH CH2
stable carbocation.
H2SO4
OR H
• Syn and anti addition occur.
ether
[3] Halogenation—Addition of X2 (X = Cl or Br) (10.13–10.14)
• The mechanism has two steps.
• Bridged halonium ions are formed as
RCH CH2 + X X
R CH CH2
intermediates.
X
X
•
No rearrangements occur.
vicinal dihalide
• Anti addition occurs.
[4] Halohydrin formation—Addition of OH and X (X = Cl, Br) (10.15)
• The mechanism has three steps.
+
• Bridged halonium ions are formed as
X
X
RCH CH2
R CH CH2
H2O
intermediates.
OH X
• No rearrangements occur.
halohydrin
• X bonds to the less substituted C.
• Anti addition occurs.
• NBS in DMSO and H2O adds Br and OH
in the same fashion.
[5] Hydroboration–oxidation—Addition of H2O (10.16)
• Hydroboration has a one-step mechanism.
[1] BH3 or 9-BBN
• No rearrangements occur.
RCH CH2
R CH CH2
[2] H2O2, HO
• OH bonds to the less substituted C.
H OH
• Syn addition of H2O results.
alcohol
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
269
Alkenes 10–3
Practice Test on Chapter Review
1. Give the IUPAC name for the following compounds.
a.
b.
2. Draw the organic products formed in the following reactions. Draw all stereogenic centers using
wedges and dashes.
CH2
a.
b.
HCl
[1] BH3
[2] H2O2, –OH
Br2
c.
d.
H2O
H2SO4
3. Fill in the table with the stereochemistry observed in the reaction of an alkene with each reagent.
Choose from syn, anti, or both syn and anti.
Reagent
a. [1] 9-BBN; [2] H2O2, –OH
b. H2O, H2SO4
c. Cl2, H2O
d. HI
e. Br2
Stereochemistry
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–4
4. a. In which of the following reactions are carbocation rearrangements are observed?
1. hydrohalogenation
2. halohydrin formation
3. hydroboration–oxidation
4. Carbocation rearrangements are observed in reactions (1) and (2).
5. Carbocation rearrangements are observed in reactions (1), (2), and (3).
b. Which of the following products are formed when HCl is added to 3-methyl-1-pentene?
1. 2-chloro-2-methylpentane
2. 3-chloro-3-methylpentane
3. 1-chloro-3-methylpentane
4. Both (1) and (2) are formed.
5. Products (1), (2), and (3) are all formed.
Answers to Practice Test
1. a. (3E)-4-ethyl-2,5dimethyl-3-nonene
b. (3E)-4-isopropyl-2methyl-3-octene
2.
Cl
Cl
a.
b.
HO
HO
c. Br
Br
Br
Br
d.
3. a. syn
b. both
c. anti
d. both
e. anti
OH
Answers to Problems
10.1
Six alkenes of molecular formula C5H10:
trans
cis
diastereomers
10.2 To determine the number of degrees of unsaturation:
[1] Calculate the maximum number of H’s (2n + 2).
[2] Subtract the actual number of H’s from the maximum number.
[3] Divide by two.
4. a. 1
b. 2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
271
Alkenes 10–5
a. C2H2
[1] maximum number of H's = 2n + 2 = 2(2) + 2 = 6
[2] subtract actual from maximum = 6 – 2 = 4
[3] divide by two = 4/2 = 2 degrees of unsaturation
b. C6H6
[1] maximum number of H's = 2n + 2 = 2(6) + 2 = 14
[2] subtract actual from maximum = 14 – 6 = 8
[3] divide by two = 8/2 = 4 degrees of unsaturation
c. C8H18
[1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18
[2] subtract actual from maximum = 18 – 18 = 0
[3] divide by two = 0/2 = 0 degrees of unsaturation
d. C7H8O
Ignore the O.
[1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16
[2] subtract actual from maximum = 16 – 8 = 8
[3] divide by two = 8/2 = 4 degrees of unsaturation
e. C7H11Br
Because of Br, add one more H (11 + 1 H = 12 H's).
[1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16
[2] subtract actual from maximum = 16 – 12 = 4
[3] divide by two = 4/2 = 2 degrees of unsaturation
f. C5H9N
Because of N, subtract one H (9 – 1 H = 8 H's).
[1] maximum number of H's = 2n + 2 = 2(5) + 2 = 12
[2] subtract actual from maximum = 12 – 8 = 4
[3] divide by two = 4/2 = 2 degrees of unsaturation
10.3
One possibility for C6H10:
a. a compound that has 2 π bonds
c. a compound with 2 rings
b. a compound that has 1 ring and 1 π bond
d. a compound with 1 triple bond
10.4 To name an alkene:
[1] Find the longest chain that contains the double bond. Change the ending from -ane to -ene.
[2] Number the chain to give the double bond the lower number. The alkene is named by the
first number.
[3] Apply all other rules of nomenclature.
To name a cycloalkene:
[1] When a double bond is located in a ring, it is always located between C1 and C2. Omit the
“1” in the name. Change the ending from -ane to -ene.
[2] Number the ring clockwise or counterclockwise to give the first substituent the lower
number.
[3] Apply all other rules of nomenclature.
3-methyl
1
CH3
a. [1] CH2 CHCH(CH3)CH2CH3
[2] CH2 CHCHCH2CH3
5 C chain with double bond
pentene
[3] 3-methyl-1-pentene
1-pentene
3
CH3CH2
b. [1] (CH3CH2)2C CHCH2CH2CH3
7 C chain with double bond
heptene
C CHCH2CH2CH3
[2]
CH3CH2
3-heptene
3-ethyl
[3] 3-ethyl-3-heptene
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Chapter 10–6
2-ethyl
c. [1]
[2]
[3] 2-ethyl-4-methyl-1-pentene
4-methyl
1
1-pentene
5 C chain with double bond
pentene
1
[2]
d. [1]
4
[3] 3,4-dimethylcyclopentene
3
3,4-dimethyl
5 C ring with a double bond
cyclopentene
1-methyl
e. [1]
[2]
[3] 5-tert-butyl-1-methylcyclohexene
1
5-tert-butyl
6 C ring with a double bond
cyclohexene
10.5 Use the rules from Answer 10.4 to name the compounds. Enols are named to give the OH the
lower number. Compounds with two C=C’s are named with the suffix -adiene.
a.
1
3
[2]
[1]
[3] 4-ethyl-3-hexen-1-ol
OH
OH
4-ethyl
6 C chain with double bond
hexene
[1]
OH
1
4
[2]
[3] 5-ethyl-6-methyl-7-octen-4-ol
OH
b.
6-methyl
5-ethyl
8 C chain with double bond
octene
[1]
5
[2]
2
c.
[3] 2,6-dimethyl-2,5-heptadiene
2-methyl
6-methyl
7 C chain with two double bonds
heptadiene
10.6 To label an alkene as E or Z:
[1] Assign priorities to the two substituents on each end using the rules for R,S nomenclature.
[2] Assign E or Z depending on the location of the two higher priority groups.
• The E prefix is used when the two higher priority groups are on opposite sides.
• The Z prefix is used when the two higher priority groups are on the same side of the
double bond.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
273
Alkenes 10–7
higher priority
Cl
CH3
a.
higher priority
C C
higher priority
Two higher priority groups are
on opposite sides: E isomer.
OCH3
Br
H
higher priority
H
H
c.
higher priority
C6H5
O
H
higher priority
b.
higher priority
CH2CH3
CH3CH2
kavain
C C
In both double bonds, the two higher priority groups are on opposite
sides: E isomers.
CH3
H
higher priority
Two higher priority groups are
on the same side: Z isomer.
10.7
E
E
Z
E
Z
skeletal structure of 11-cis-retinal
CHO
10.8 To work backwards from a name to a structure:
[1] Find the parent name and functional group and draw, remembering that the double bond is
between C1 and C2 for cycloalkenes.
[2] Add the substituents to the appropriate carbons.
a. (3Z)-4-ethyl-3-heptene
The higher priority groups are
on the same side = Z.
c. (1Z)-2-bromo-1-iodo-1-hexene
6 carbons
7 carbons
The double bond is
between C1 and C2.
I
Br
4-ethyl
The double bond is
between C3 and C4.
b. (2E)-3,5,6-trimethyl-2-octene
The higher priority groups are
on the same side = Z.
The double bond is
between C2 and C3.
8 carbons
The higher priority groups
are on opposite sides = E.
3,5,6-trimethyl
10.9 Draw all of the stereoisomers and then use the rules from Answer 10.6 to name each diene.
E
E
(2E,4E)-2,4-hexadiene
E
Z
(2E,4Z)-2,4-hexadiene
Z
Z
(2Z,4Z)-2,4-hexadiene
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Chapter 10–8
10.10 To rank the isomers by increasing boiling point:
Look for polarity differences: small net dipoles make an alkene more polar, giving it a
higher boiling point than an alkene with no net dipole. Cis isomers have a higher boiling
point than their trans isomers.
CH3
CH3
CH3
H
CH3CH2
C C
CH3CH2
C C
CH3
H
All dipoles cancel.
smallest surface area
no net dipole
lowest bp
CH2CH3
C C
CH2CH3
H
Two dipoles cancel.
no net dipole
trans isomer
intermediate bp
H
Two dipoles reinforce.
net dipole
cis isomer
highest bp
10.11 Increasing number of double bonds = decreasing melting point.
O
O
OH
stearic acid
no double bonds
highest melting point
OH
stearidonic acid
O
4 double bonds
lowest melting point
OH
linolenic acid
3 double bonds
intermediate melting point
10.12
Br
a.
H2SO4
NaOCH2CH3
b.
OH
CH2=CHCH2CH2CH2CH3
+
CH3CH=CHCH2CH2CH3
10.13 To draw the products of an addition reaction:
[1] Locate the two bonds that will be broken in the reaction. Always break the π bond.
[2] Draw the product by forming two new σ bonds.
HCl
H
two new σ bonds
a.
CH3
c.
Cl
b.
CH3CH2CH2CH CHCH2CH2CH3
CH3
HCl
H
HCl
CH3
CH3
Cl
H H
CH3CH2CH2 C C CH2CH2CH3
H Cl
two new σ bonds
10.14 Addition reactions of HX occur in two steps:
[1] The double bond attacks the H atom of HX to form a carbocation.
[2] X– attacks the carbocation to form a C−X bond.
two new σ bonds
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Alkenes 10–9
transition state
step [1]:
H
+
H
H
H
[1]
H Cl
transition state
step [2]:
H
H
[2]
Cl
H
Cl
δ−
δ+
δ+ H
Cl
Cl
δ−
10.15 Addition to alkenes follows Markovnikov’s rule: When HX adds to an unsymmetrical
alkene, the H bonds to the C that has more H’s to begin with.
2 H's
H adds here.
no H's
Cl adds here.
CH3
CH3
Cl
HCl
a.
c.
H
H
H
CH2
HCl
Cl
CH3
no H's
Cl adds here.
one H
H adds here.
H
no H's
Cl adds here
CH3
b.
CH3
HCl
C CH2
Cl C
CH2
CH3 H
CH3
2 H's
H adds here.
10.16 To determine which alkene will react faster, draw the carbocation that forms in the ratedetermining step. The more stable, more substituted the carbocation, the lower the Ea to
form it and the faster the reaction.
CH3
CH3
H
CH3
CH3
C
C C
CH2
H
CH3
H
H X
H
C C
H
CH3CHCH2
H
H
2° carbocation
slower reaction
H X
3° carbocation
faster reaction
10.17 Look for rearrangements of a carbocation intermediate to explain these results.
H Cl
CH3
Cl
H
H
H
H
Cl
CH3
1-chloro-3methylcyclohexane
H
CH3
H
H
2° carbocation
CH3 H
+ Cl–
2° carbocation
Rearrangement would not further
stabilize this carbocation.
H
1,2-H shift
Cl
CH3
3° carbocation
CH3 Cl
1-chloro-1methylcyclohexane
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Chapter 10–10
10.18 Addition of HX to alkenes involves the formation of carbocation intermediates.
Rearrangement of the carbocation will occur if it forms a more stable carbocation.
H
a.
H CH3
CH3
C C
H
H CH3
H C C CH2CH3
CH2CH3
H C C CH2CH3
H
H Br
3° carbocation
no rearrangement
b.
CH3
H H
H
C C
H
CH2CH3
CH3
H
H H
CH3 C C CH2CH3
CH3 C C CH2CH3
H Br
H
Br H
H H
H H
H
C C
H
H H
CH3 C C CH2CH3
2° carbocation
2° carbocation
Rearrangement would not further
stabilize either carbocation.
no rearrangement
(cis or trans)
c.
H H
CH3 C C CH2CH3
CH(CH3)2
CH3 C C
H
H
(cis or trans)
H H
1,2-H shift
CH3 C C CH(CH3)2
CH3 C C CH(CH3)2
2° carbocation
C CH3
H H
2° carbocation
rearrangement
Rearrangement would not further
stabilize the carbocation.
CH3
3° carbocation
more stable
H H
H H
CH3 C C CH(CH3)2
CH3 C C
Br H
H H
CH3
C CH3
Br
10.19 To draw the products, remember that addition of HX proceeds via a carbocation intermediate.
Addition of H+ (from HBr)
from above and below gives
an achiral, trigonal planar carbocation.
a.
H
H
CH3
CH3
CH3
CH3
Addition of Br– from
above
and below.
CH3
CH3
CH3
Br
CH3
CH3
CH3
Br
CH3
enantiomers
Addition of H+ (from HCl) from above and
below by Markovnikov's rule forms
an achiral 3° carbocation.
b. CH3
CH3
CH3
CH3
Cl– attacks
from above
and below.
CH3
H
CH3
CH3
CH3
Cl
H
achiral, trigonal planar
3° carbocation
diastereomers
CH3
Cl
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Alkenes 10–11
10.20 The product of syn addition will have H and Cl both up or down (both on wedges or both dashes),
while the product of anti addition will have one up and one down (one wedge, one dash).
CH3
H
CH3
H
CH3
H
CH3
H
Cl
CH3
Cl
CH3
Cl
CH3
Cl
CH3
A
syn addition
B
anti addition
C
anti addition
D
syn addition
10.21
CH3
CH2CH2CH3
CH3
or
CH2 C
a. CH3 C CH2CH2CH3
OH
CH3
H+ would add here to form
a 3° carbocation.
CH3
OH
b.
CH3
H+ would add here to form
a 3° carbocation.
or
c.
OH
H+ would add here to form
a 3° carbocation.
CH2
or
H+ would add here to form
a 3° carbocation.
H+
C CHCH2CH3
CH3
CH3CH CHCH3
would add here to form
a 2° carbocation.
(cis or trans)
10.22
H
H2O
H2SO4
1-pentene
OH
HO
H
enantiomers
10.23 Halogenation of an alkene adds two elements of X in an anti fashion.
Br
Br2
a.
Br
Cl2
CH3
Cl
CH3
Cl
Cl
Cl
b.
Br
Br
10.24 To draw the products of halogenation of an alkene, remember that the halogen adds to both
ends of the double bond but only anti addition occurs.
a.
Cl
Cl
Cl
CH3
Cl
CH3
Cl2
enantiomers
H
b.
C
H
Br2
Br
H
C
C
C
H
Br
achiral meso compound
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Chapter 10–12
c.
Br2
CH3
CH3
CH3
Br
Br
Br
Br
diastereomers
10.25 The two steps in the mechanism for the halogenation of an alkene are:
[1] Addition of X+ to the alkene to form a bridged halonium ion
[2] Nucleophilic attack by X–
CH3
trans-2-butene
C C
H
Addition of Br+ can occur
from above or below:
Br Br
CH3
overall
CH3
H
H
Br
H
C
H
enantiomers
C
H
Attack of Br– can occur
from the left or right:
CH3
below
Br
C
CH3
C C
Br Br
above
CH3
H
Br
H
C
CH3
Br
CH3
right
left
left
right
+ Br
Br
CH3
C
H
H
C
CH3
CH3
H
Br
C
C
+ Br
CH3
H
C
Br
CH3
CH3
C
H
H
C
CH3
CH3
Br
C
H
C
Br
H
CH3
+ Br
Br
C
H
+ Br
Br
H
C
H
CH3
CH3
H
C
Br
CH3
H
C
C
CH3
H
Br
CH3
Br
CH3
H
C
Br
Br
C
H
CH3
All four compounds are identical—an achiral meso compound.
10.26 Halohydrin formation adds the elements of X and OH across the double bond in an anti
fashion. The reaction is regioselective so X ends up on the carbon that had more H’s to
begin with.
a.
Br
NBS
DMSO, H2O
HO
CH3
b.
HO
Cl2
H2O
Br
CH3
OH
CH3
OH
Cl
Cl
Cl bonds to the carbon
with more H's to begin with.
10.27
H
a. (CH3)2S
H
H CH2CH3
CH3
H B S
b. (CH3CH2)3N
CH3
H B N CH2CH3
H CH2CH3
H CH2CH2CH2CH3
c. (CH3CH2CH2CH2)3P
H B P CH2CH2CH2CH3
H CH2CH2CH2CH3
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Alkenes 10–13
10.28 In hydroboration the boron atom is the electrophile and becomes bonded to the carbon atom
that had more H’s to begin with.
CH3
C CH2
a.
CH3
BH3
CH3 C CH2
CH3
C with more H's.
B will add here.
CH2BH2
C with more H's.
B will add here.
BH3
b.
BH3
CH2
c.
H BH2
BH2
H
C with more H's.
B will add here.
10.29 The hydroboration–oxidation reaction occurs in two steps:
[1] Syn addition of BH3, with the boron on the less substituted carbon atom
[2] OH replaces the BH2 with retention of configuration.
a. CH3CH2CH CH2
BH3
CH3CH2CH CH2
H
CH2CH3
b.
c. CH3
CH3
H2O2, –OH
CH3CH2CH CH2
BH2
H
OH
CH2CH3
H
CH2CH3
H2O2, –OH
H
CH2CH3
H
CH2CH3
H
BH2
BH2
OH
OH
BH3
CH3
CH3
H
CH3
H
CH3
BH2
BH2
H2O2, –OH
CH3
H
CH3
CH3
CH3
H
OH
OH
10.30 Remember that hydroboration results in addition of OH on the less substituted C.
OH
HO
a.
c.
b.
(E or Z isomer can be used.)
OH
10.31
a.
CH2
CH2
H2O
H2SO4
[1] BH3
[2] H2O2, HO
OH
CH3
CH2OH
Hydration places the OH
on the more substituted carbon.
Hydroboration–oxidation places the OH
on the less substituted carbon.
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Chapter 10–14
OH
Hydration places the OH
on the more substituted carbon.
H2O
H2SO4
b.
OH
[1] BH3
[2] H2O2, HO
OH
H2 O
H2SO4
c.
Hydroboration–oxidation places the OH
on the less substituted carbon.
[1] BH3
[2] H2O2, HO
Hydration places the OH
on the more substituted carbon.
Hydroboration–oxidation places the OH
on the less substituted carbon.
HO
10.32 There are always two steps in this kind of question:
[1] Identify the functional group and decide what types of reactions it undergoes
(e.g., substitution, elimination, or addition).
[2] Look at the reagent and determine if it is an electrophile, nucleophile, acid, or base.
acid:
catalyzes loss of H2O
acid
CH2
a.
HBr
CH3
CH2CH2CH3
c.
Br
alkene:
addition reactions
H2SO4
CHCH2CH3
OH
CH2CH2CH3
alcohol:
substitution and elimination
nucleophile and base
Cl
OCH3
NaOCH3
b.
CH3CH2CH=CHCH3
2° alkyl halide:
substitution and elimination
(cis and trans)
10.33 To devise a synthesis:
[1] Look at the starting material and decide what reactions it can undergo.
[2] Look at the product and decide what reactions could make it.
CH3
?
Br
a.
Cl
alkyl halide
Can undergo
substitution and elimination.
Br
K+ –OC(CH3)3
halohydrin:
Can form from an
alkene with Cl2 and H2O.
Cl2
H2 O
Cl
OH
OH is added to
the more substituted C.
CH3
?
OH
b.
OH
OH
alcohol
Can undergo
substitution and elimination.
CH3
OH
H2SO4
alcohol
Can be formed by
substitution and addition.
CH3 [1] BH
3
[2] H2O2, HO–
(major
product)
CH3
OH
OH is added to
the less substituted C.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
281
Alkenes 10–15
10.34 Convert each ball-and-stick model to a skeletal structure and then name the molecule.
2
3
6 C chain with a C=C
hexene
C=C at C2
2-hexene
2 CH3's at C3 and C5
two higher priority groups on opposite sides
Answer: (2E)-3,5-dimethyl-2-hexene
5
a.
b.
E isomer
5 C ring with a C=C
cyclopentene
sec-butyl at C1
methyl at C2
Answer: 1-sec-butyl-2-methylcyclopentene
2
1
10.35
H
a.
The two higher priority groups are on the
same side of the C=C, making it a Z alkene.
higher
priority
H
higher priority
A
b. Add H2O in a Markovnikov fashion to form two products.
OH
OH
H
H
H2 O
H2SO4
H
10.36
C1
8-methyl-1-nonene
C8
a.
b.
Br2
Br2
Br
Br
OH
Br
H2O
c.
Br2
CH3OH
OCH3
Br
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Chapter 10–16
10.37 Use the directions from Answer 10.2 to calculate degrees of unsaturation.
a. C3H4
[1] maximum number of H's = 2n + 2 = 2(3) + 2 = 8
[2] subtract actual from maximum = 8 – 4 = 4
[3] divide by 2 = 4/2 = 2 degrees of unsaturation
f. C8H9Br
Because of Br, add one H (9 + 1 = 10 H's).
[1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18
[2] subtract actual from maximum = 18 – 10 = 8
[3] divide by 2 = 8/2 = 4 degrees of unsaturation
b. C6H8
[1] maximum number of H's = 2n + 2 = 2(6) + 2 = 14
[2] subtract actual from maximum = 14 – 8 = 6
[3] divide by 2 = 6/2 = 3 degrees of unsaturation
g. C8H9ClO
Ignore the O; count Cl as one more H (9 + 1 = 10 H's).
[1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18
[2] subtract actual from maximum = 18 – 10 = 8
[3] divide by 2 = 8/2 = 4 degrees of unsaturation
c. C40H56
[1] maximum number of H's = 2n + 2 = 2(40) + 2 = 82
[2] subtract actual from maximum = 82 – 56 = 26
[3] divide by 2 = 26/2 = 13 degrees of unsaturation
d. C8H8O
Ignore the O.
[1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18
[2] subtract actual from maximum = 18 – 8 = 10
[3] divide by 2 = 10/2 = 5 degrees of unsaturation
e. C10H16O2
Ignore both O's.
[1] maximum number of H's = 2n + 2 = 2(10) + 2 = 22
[2] subtract actual from maximum = 22 – 16 = 6
[3] divide by 2 = 6/2 = 3 degrees of unsaturation
h. C7H9Br
Because of Br, add one H (9 + 1 = 10 H's).
[1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16
[2] subtract actual from maximum = 16 –10 = 6
[3] divide by 2 = 6/2 = 3 degrees of unsaturation
i. C7H11N
Because of N, subtract one H (11 – 1 = 10 H's).
[1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16
[2] subtract actual from maximum = 16 – 10 = 6
[3] divide by 2 = 6/2 = 3 degrees of unsaturation
j. C4H8BrN
Because of Br, add one H, but subtract one for N
(8 + 1 – 1 = 8 H's).
[1] maximum number of H's = 2n + 2 = 2(4) + 2 = 10
[2] subtract actual from maximum = 10 – 8 = 2
[3] divide by 2 = 2/2 = 1 degree of unsaturation
10.38 First determine the number of degrees of unsaturation in the compound. Then decide which
combinations of rings and π bonds could exist.
C10H14
[1] maximum number of H's = 2n + 2 = 2(10) + 2 = 22
[2] subtract actual from maximum = 22 – 14 = 8
[3] divide by two = 8/2 = 4 degrees of unsaturation
possibilities:
4 π bonds
3 π bonds + 1 ring
2 π bonds + 2 rings
1 π bond + 3 rings
4 rings
10.39 The statement is incorrect because when naming isomers with more than two groups on a
double bond, one must use an E,Z label, rather than a cis, trans label.
higher
priority
Cl
Cl
higher
priority
higher
priority
higher
priority
O
enclomiphene
E isomer
N(CH2CH3)2
O
zuclomiphene
Z isomer
N(CH2CH3)2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
283
Alkenes 10–17
10.40 Name the alkenes using the rules in Answers 10.4 and 10.6.
CH3
a.
CH2=CHCH2CH(CH3)CH2CH3
CH2 CHCH2CHCH2CH3
6 C chain with a double bond =
hexene
1-hexene 4-methyl
4-methyl-1-hexene
5-ethyl
2-methyl
b.
5-ethyl-2-methyl-2-octene
2-octene
8 C chain with a double bond =
octene
2-isopropyl
2-isopropyl-4-methyl-1-pentene
c.
1-pentene
5 C chain with a double bond =
pentene
4-methyl
d.
1-ethyl
5-isopropyl
6 C ring with a double bond =
cyclohexene
E double bond (higher priority groups on
opposite sides with bold bonds)
4-isopropyl
e.
1-ethyl-5-isopropylcyclohexene
3-ol
OH
(4E)-4-isopropyl-4-hepten-3-ol
OH
7 C chain with a double bond =
heptene
4-heptene
2-cyclohexene 5-sec-butyl-2-cyclohexenol
1-ol
f.
OH
6 C ring with a double bond =
cyclohexene
OH
5-sec-butyl
10.41 Use the directions from Answer 10.8.
a. (3E)-4-ethyl-3-heptene
7 carbons
4-ethyl
3
Higher priority groups on
opposite sides = E.
b. 3,3-dimethylcyclopentene
5 carbon ring
3,3-dimethyl
1
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Chapter 10–18
c. cis-4-octene
8 carbons
1
f. cis-3,4-dimethylcyclopentene
4
Higher priority groups on
the same side = cis.
2
d. 4-vinylcyclopentene
5 carbon ring
5 carbon ring
or
3
4
3,4-dimethyl
g. trans-2-heptene
1
4
2 Higher priority groups on
opposite sides = trans.
7 carbons
2
h. 1-isopropyl-4-propylcyclohexene
e. (2Z)-3-isopropyl-2-heptene
3-isopropyl
7 carbons
6 carbon ring
1-isopropyl
Higher priority groups on
the same side = Z.
4-propyl
10.42
a.
(2E,4S)-4-methyl-2-nonene
(2E,4R)-4-methyl-2-nonene
(2Z,4S)-4-methyl-2-nonene
A
B
C
b. A and B are enantiomers. C and D are enantiomers.
c. Pairs of diastereomers: A and C, A and D, B and C, B and D.
(2Z,4R)-4-methyl-2-nonene
D
10.43
CH3
CH3
a.
CH3
H
c.
(1Z,4S)-1,4-dimethylcyclodecene
diastereomer
(1E,4R)-1,4-dimethylcyclodecene
CH3
CH3
CH3
b.
H
(1E,4S)-1,4-dimethylcyclodecene
enantiomer
(1Z,4R)-1,4-dimethylcyclodecene
diastereomer
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
285
Alkenes 10–19
10.44 Name the alkene from which the epoxide can be derived and add the word oxide.
1-ethyl
H
derived
a.
O
H
O
derived
c.
from
from
H
6 carbon ring
cyclohexene
1-ethylcyclohexene
1-ethylcyclohexene oxide
(3E)- 3-heptene oxide
or
trans-3-heptene oxide
2-methyl
O
b.
derived
derived
d.
O
C(CH3)3
C(CH3)3
from
from
2-methyl-2-hexene oxide
H
7 carbon chain
heptene
(3E)- 3-heptene
or
trans-3-heptene
5 carbon ring
cyclopentene
4-tert-butylcyclopentene oxide 4-tert-butylcyclopentene
6 carbon chain
hexene
2-methyl-2-hexene
10.45
2-sec-butyl
a. 2-butyl-3-methyl-1-pentene
As written, this is the parent chain,
but there is another longer chain
containing the double bond.
new name:
2-sec-butyl-1-hexene
b. (Z)-2-methyl-2-hexene
new name:
2-methyl-2-hexene
Two groups on one end of the C=C
are the same (2 CH3's), so no E and Z isomers are possible.
1
c. (E)-1-isopropyl-1-butene
new name:
(3E)-2-methyl-3-hexene
As written, this is the parent chain,
but there is another longer chain
containing the double bond.
1
d. 5-methylcyclohexene
4
As written the methyl
is at C5. Re-number
to put it at C4.
e. 4-isobutyl-2-methylcylohexene
new name:
4-methylcyclohexene
2
1
5
As written this methyl
is at C2. Re-number
to put it at C1.
1
new name:
5-isobutyl-1-methylcyclohexene
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Chapter 10–20
f.
1-sec-butyl-2-cyclopentene
3
1
1
3
2
new name:
3-sec-butylcyclopentene
This has the double bond between
C2 and C3. Cycloalkenes must
have the double bond between
C1 and C2. Re-number.
g.
OH
1-cyclohexen-4-ol
OH
3
1
The numbering is incorrect. When a compound
contains both a double bond and an OH group,
number the C skeleton to give the OH group the
lower number.
3-cyclohexenol (The "1" can be omitted.)
OH
h.
OH
6
3-ethyl-3-octen-5-ol
The numbering is incorrect. When a compound
contains both a double bond and an OH group,
number the C skeleton to give the OH group the
lower number.
4
5
6-ethyl-5-octen-4-ol
10.46 a, b. E,Z, and R,S designations are shown.
CH3O
E
E
R E
S
S
E
O
O
N
H
O
Z
H
N
CHO
OH
OCH3
E
S
E
S
S
c. Nine double bonds that can be E or Z
Six tetrahedral stereogenic centers
Maximum possible number of stereoisomers = 215
10.47
O
stearic acid
OH
hIghest melting point
no double bonds
OH
intermediate melting point
one E double bond
O
elaidic acid
O
oleic acid
OH
lowest melting point
one Z double bond
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Alkenes 10–21
10.48
O
a.
O
OH
all trans double bonds
higher melting point
OH
O
b.
eleostearic acid
OH
all cis double bonds
lower melting point
10.49 The more negative the H°, the larger the Keq assuming entropy changes are comparable.
Calculate the H° for each reaction and compare.
CH2 CH2 + HI
CH3CH2
[1] Bonds broken
I
[2] Bonds formed
ΔHo (kJ/mol)
C–C π bond
+ 267
I
+ 297
H
Total
ΔHo (kJ/mol)
CH2ICH2 H
− 410
I
− 222
C
+ 564 kJ/mol
CH2 CH2 + HCl
− 632 kJ/mol
Total
[3] Overall ΔHo =
sum in Step [1]
+
sum in Step [2]
+ 564 kJ/mol
− 632
kJ/mol
– 68
kJ/mol
CH3CH2 Cl
[1] Bonds broken
[2] Bonds formed
[3] Overall ΔHo =
C–C π bond
+ 267
CH2ClCH2 H
− 410
sum in Step [1]
+
sum in Step [2]
H Cl
+ 431
C Cl
− 339
+ 698 kJ/mol
+ 698 kJ/mol
Total
− 749 kJ/mol
ΔHo (kJ/mol)
Total
ΔHo (kJ/mol)
− 749
kJ/mol
– 51 kJ/mol
Compare the ΔH°:
Addition of HI: –68 kJ/mol = more negative ΔH°, larger Keq
Addition of HCl: –51 kJ/mol
288
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Chapter 10–22
10.50
H
HBr
a.
Br
Br2, H2O
f.
Br
H
HI
b.
NBS
g.
H
H2O
H2SO4
H
H2SO4
Br
OH
OH
H
[1] 9-BBN
[2] H2O2, HO–
OCH2CH3
Cl2
e.
[2] H2O2, HO–
i.
Br
H
OH
CH3CH2OH
d.
OH
OH
[1] BH3
h.
c.
OH
+
aqueous DMSO
I
Br
+
OH
Cl
+
Cl
Cl
Cl
10.51
CH3
a.
HBr
C CH2
HI
C CH2
c.
C CH2
C CH2
H2SO4
CH3CH2OH
C CH2
h.
OH
CH3
NBS
CH3 C CH2Br
aqueous DMSO
OH
CH3
[2] H2O2, HO–
CH3 C CH2OH
H
CH3
CH3 C CH3
C CH2
i.
OCH2CH3
CH3
CH3
Cl2
OH
[1] BH3
C CH2
CH3
CH3
CH3
CH3
CH3
CH3 C CH3
H2SO4
CH3
e.
C CH2
CH3
CH3
CH3
d.
g.
CH3
H2O
CH3 C CH2Br
CH3
CH3 C CH3
I
CH3
Br2, H2O
CH3
CH3
CH3
CH3
C CH2
f.
Br
CH3
CH3
b.
CH3
CH3
CH3 C CH3
CH3
[1] 9-BBN
[2] H2O2, HO–
CH3 C CH2OH
H
CH3 C CH2Cl
Cl
10.52
Br
a.
Br
d.
CH3CH2CH2CH=CHCH3
Br
Cl
b.
e.
CH3
c.
C CH3
Br
CH3
CH2 or
Cl
Cl
CH3
C CH2
CH3CH2
f.
(CH3CH2)3CBr
C CHCH3
CH3CH2
CH3
289
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkenes 10–23
10.53 Hydroboration–oxidation results in addition of an OH group on the less substituted carbon,
whereas acid-catalyzed addition of H2O results in the addition of an OH group on the more
substituted carbon.
a.
OH
c.
hydroboration–oxidation
and
acid-catalyzed addition
e.
OH
acid-catalyzed addition
OH
or
OH
Both methods would give
product mixtures.
OH
b.
d.
hydroboration–oxidation
hydroboration–oxidation
10.54
a.
Cl
(CH3CH2)2C=CHCH2CH3
CH3CH2
H
HCl
C C
CH3CH2
CH3CH2 H
CH3CH2 C
Cl
CH2CH3
Cl
C H
CH2CH3
e.
Br2
CH2Br
OH
Br adds here
to less substituted C.
(CH3CH2)2C=CH2
CH3CH2
CH3CH2
[1] 9-BBN
CH3CH2
H2O
C CH2
H2SO4
f.
CH3CH2 C CH3
OH
[1] BH3
(CH3)2C=CHCH3
CH2
CH2OH
[2] H2O2, HO−
OH adds here
to less substituted C.
H adds here
to less substituted C.
c.
CH2
H2O
H adds here
to less substituted C.
b.
Cl2
d.
H
(CH3)2C
DMSO, H2O
CHCH3
H
(CH3)2C
OH Br
Br adds here
to less substituted C.
[2] H2O2, HO−
BH3 adds here
to less substituted C.
NBS
g.
BH2
OH
h.
CHCH3
Br2
Br Br
10.55
CH3CH2
H
CH3CH2CH2
CH2CH2CH3
or
C C
H
or
CH3CH C
CH3
CH3
C CHCH2CH3
CH3CH2
Cl
CH3CH2 C CH2CH2CH3
CH3
290
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–24
10.56
H
a.
CH3
CH3
H
C C
C C
=
H
C
H
CH3
b.
CH3
CH3CH2
H
CH3CH2
CH3
=
CH3
CH3
C C
CH3
H
H2O
H
CH3CH2
NBS
HO
CH3
H
CH3
H
C
CH3CH2
CH3
CH2
H 2O
CH 3
(CH3)3C
H2SO 4
I H
CH3
c.
CH3
CH3
CH3
Cl
[1] BH3
CH3
OH
Br CH2CH3
CH3
H
CH3
only syn addition
OH
Br CH2CH3
HBr
e.
Cl
Cl2
OH
only anti addition
+
H2O
OH
NBS
h.
only anti addition
CH3
CH3
H
[2] H2O2, HO−
CH3
I
Cl
CH3
Cl
d.
H
Cl
Cl2
CH3
g.
OH
HI
b.
f.
CH3
(CH 3)3C
OH
DMSO, H2O
CH3CH=CHCH2CH3
H2O
H2SO4
Cl
Br
Br
HO CH3
H OH
HO CH3
HO H
OH
CH3
H
HO
10.57
a. (CH3)3C
C
HO
Br
C
Cl
CH3
CH3CH2
C
Cl
CH3CH2
CH3
C
C
Br
C
CH3
CH3CH2
C
CH3
CH3
H
C
DMSO, H2O
H
H
HO
Cl2
C C
CH3 CH3CH2
C C
c.
CH3
=
C C
H
C
Br
CH3
H
Br
Br
H
Br2
C
Br
291
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkenes 10–25
10.58 Draw each reaction. (a) The cis isomer of 4-octene gives two enantiomers on addition of Br2.
(b) The trans isomer gives a meso compound.
H
H
H
Br
Br2
a.
H
Br
H
H
Br
cis-4-octene
Br
(4R,5R)-4,5-dibromooctane
(4S,5S)-4,5-dibromooctane
enantiomers
H
H
Br
Br2
b.
H
H
Br
trans-4-octene
(4R,5S)-4,5-dibromooctane
meso compound
10.59
CH3CH2
CH3CH2
CH2CH3
C C
H
H
H
H Cl
CH3CH2
CH3CH2
CH2CH3
H
C
Cl
H
H
Cl
H
By protonation of the alkene, the cis and
trans isomers produce identical
carbocation intermediates.
CH2CH3
Cl–
CH3CH2
CH2CH2CH3
H
C C H
Cl–
CH3CH2
CH2CH3
H Cl
C C H
H
H
C C
C
CH3CH2
C
CH2CH2CH3
H
Cl
CH3CH2
C
CH2CH2CH3
CH2CH2CH3
Cl
H
Both cis- and trans-3-hexene give the same
racemic mixture of products, so the
reaction is not stereospecific.
10.60
a.
Cl
H
H
H Cl
H
Cl
H
H
H
H
and
O
H
H
H
HO
C
Cl
O C
H
H
CH3
H
O
O
O C
CH3
H
H
H
CH3
+ HCl
292
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–26
H
H Br
b.
H
Br
H
1,2-H shift
Br
C H
CH3
C
CH3
2o carbocation
3o carbocation
+ Br–
10.61
a.
H
CH3
CH3
CH3
1,2-shift
H2O
CH3
CH3
CH3
O H
–
CH3 HSO4
OH
CH3
H2SO4
HSO4–
H OSO3H
H2O
2o carbocation
3o carbocation
+ H2SO4
H OSO3H
b.
OH
OH
+ H2SO4
H2SO4
O
HSO4–
CH3
O
CH3
H
–OSO
3H
10.62
H
H OH2
H
+ H3O+
O H
H2O
OH
H2O
H
H
O H
1,2-shift
H2O
H2O
H
H
OH
+ H3O+
H2O
1,2-shift
O H
H2O
OH
H
H
+ H3O+
293
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkenes 10–27
10.63 The isomerization reaction occurs by protonation and deprotonation.
CH3
H
CH2
OSO3H
CH3
C
C H
CH3
CH3
C
CH3
CH3
C H
CH3
CH3
CH3
C C
HSO4–
CH3
2,3-dimethyl-1-butene
CH3
2,3-dimethyl-2-butene
10.64
H
H Br
H
C C
H
H
H
CH3 C
C C
H
H
CH3CH CHCH2
C C
H
H
Br
CH3CH=CH–CH2Br
H
Br
Since two resonance structures can be drawn
for the intermediate carbocation, two different
products result from attack by Br–.
Br
CH3CHCH CH2
10.65
Br
HBr
A
OCH3
OCH3
B
H Br
COOCH3
D
H Br
Br
O
OCH3
OCH3
H
Br
HBr
COOCH3
C
H
This carbocation is resonance
stabilized by the O atom, and
therefore preferentially forms and
results in B.
C
H
O
CH3
O
C
CH3
δ+ O
H
This carbocation is destabilized
by the δ+ on the adjacent C,
so it does not form.
This carbocation is formed
preferentially and results in product
D. It is not destabilized by an
adjacent electron-withdrawing
COOCH3 group.
10.66
Br Br
H
OH
Br
OH
O
O
Br
+ HBr
+ Br
+ Br
Br
294
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–28
10.67
OH PBr3
OH
K+ –OC(CH3)3
Br
H2O + H2SO4
a.
OH adds to more substituted C.
POCl3, pyridine
Br
b.
CH3 C CH3
K+ –OC(CH3)3
Br
Br2
CH3 CH CH2
CH3 C CH2Br
H
H
OH
H2SO4
c.
[1] BH3
[2] H2O2,
d.
CH3 CH CH2I
NaH
OH
HO–
+ H2
CH3
K+ –OC(CH3)3
(CH3)2C
CH2
HCl
CH3 C Cl
CH3
CH3
Cl
e.
Br
Br2
K+ –OC(CH3)3
H2O
f.
CH3I
O–
CH3CH CH2
Br2
Br
OH
Br
CH3CH CH2
NaNH2
(2 equiv)
CH3C CH
10.68
Cl2
a.
H2 O
HBr
b.
Br
Cl
O
O
OH
+ H2
Br
CN
NaCN
OH
NaH
NaOH
c.
Cl
Na+ H–
(from b.)
[1] –SH
d.
O
[2] H2O
OH
+ enantiomer
SH
(from a.)
e.
O
(from a.)
[1] –C≡CH
OH
[2] H2O
C
+ enantiomer
CH
O
O
CH3CH2CH2I
OCH3
295
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkenes 10–29
10.69
Br
Br K+ –OC(CH3)3
a.
d.
Br
NaNH2
C CH
(2 equiv)
Br
Br2
b.
Br
(from b.)
OH
OH
Br2
c.
O
NaH
Br
e.
(from a.)
Br
(from c.)
H2O
(from a.)
10.70
OH
Br
OH
POCl3
Br2
pyridine
A
O
NaH
H2O
B
major product
H2SO4
OH
Br2
NaH
O
H2O
Br
C
10.71 Having two rings joined together as in A and B creates a very rigid ring system and
constrains bond angles. Evidently the bond angles around the C=C in A are close enough to
the trigonal planar bond angle of 120° so that A is stable. With B, however, the C=C is
located at a carbon shared by both rings and the bond angles around the C=C deviate greatly
from the desired angle, so that B is not a stable compound.
296
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–30
10.72 a. Br2 adds in an anti fashion to form a meso dibromide. Rotate around the C–C bond to
place H and Br anti periplanar in the second step. HBr can be eliminated in two ways, but
both give the same product.
C6H5
H
C C
C6H5
H
Br
Br2
C
H
C6H5
anti
C
Br
C6H5
H
rotate
C
C
H
C6H5
Br
Br
C6H5
H
C
or
C
C6H5
Br
H
C6H5
H
Br
H and Br must be anti.
meso compound
KOH
(–HBr)
Br
H
C C
C6H5
C6H5
Two higher priority groups are
on opposite sides = E.
b. Br2 addition forms two enantiomers. Anti periplanar elimination of H and Br gives the
same alkene from both compounds.
Br
H
H
Br2
C C
C6H5
H
C6H5
C6H5 anti
H
C
C6H5
C
or
C
H
C6H5
Br
Br
H
C6H5
C
Br
C6H5
two enantiomers
rotate
Br
H
C6H5
Br
H
C
C
C
Br
rotate
H
H
C6H5
Br
C6H5
C
C
C6H5
Br
KOH
(–HBr)
H
Br
C
H
C6H5
Br
C6H5
H
C C
C6H5
C6H5
Br
C
KOH
(–HBr)
C6H5
H
or
C
Br
H
H
C6H5
C C
identical
Z isomer
Br
C6H5
Z isomer
c. The products in (a) and (b) are diastereomers.
10.73
TsO H
H
TsO–
+ TsOH
A
+ TsO–
isocomene
297
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkenes 10–31
10.74
Na+ HCO3–
O
H O
O
O
O
O
O
O
I
I
I
OCH3
HO
B
HO
HO
Na+
I
OCH3
OCH3
OCH3
HO
I
C
H2CO3
10.75
OH
OH2
H OSO3H
H
2° carbocation
+ HSO4–
H2O
1,2-H shift
H
HSO4–
H
3° carbocation
+ H2SO4
10.76
TsO H
OH
H2O
OH2
nerol
OTs
+ TsOH
H2O
O H
H
H
OH
OH
H OSO2Cl
OTs
OSO2Cl
OH
OH
α-cyclogeraniol
nerol
OSO2Cl
OH
α-terpineol
HSO3Cl
298
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 10–32
10.77
O
H OSO3H
O H
OH
+ HSO4
–
H2 O
H O
H
HSO4–
OH
HO
OH
+ H2SO4
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
299
Alkynes 11–1
Chapter 11 Alkynes
Chapter Review
General facts about alkynes
• Alkynes contain a carbon–carbon triple bond consisting of a strong σ bond and two weak π bonds.
Each carbon is sp hybridized and linear (11.1).
180o
H C C H
=
acetylene
sp hybridized
•
•
•
Alkynes are named using the suffix -yne (11.2).
Alkynes have weak intermolecular forces, giving them low mp’s and low bp’s, and making them
water insoluble (11.3).
Since its weaker π bonds make an alkyne electron rich, alkynes undergo addition reactions with
electrophiles (11.6).
Addition reactions of alkynes
[1] Hydrohalogenation—Addition of HX (X = Cl, Br, or I) (11.7)
•
X H
H X
R C C H
R C C H
(2 equiv)
X H
Markovnikov’s rule is followed. H bonds to
the less substituted C in order to form the
more stable carbocation.
geminal dihalide
[2] Halogenation—Addition of X2 (X = Cl or Br) (11.8)
•
X X
X X
R C C H
R C C H
(2 equiv)
•
X X
Bridged halonium ions are formed as
intermediates.
Anti addition of X2 occurs.
tetrahalide
[3] Hydration—Addition of H2O (11.9)
R C C H
H2O
H2SO4
HgSO4
H
R
C C
H
HO
enol
•
O
R
C
CH3
ketone
•
Markovnikov’s rule is followed.
H bonds to the less substituted C in
order to form the more stable
carbocation.
The unstable enol that is first formed
rearranges to a carbonyl group.
300
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–2
[4] Hydroboration–oxidation—Addition of H2O (11.10)
R C C H
R
C C
[2] H2O2, HO
H
•
O
H
R
[1] R2BH
OH
C
C
H
H H
enol
The unstable enol, first formed after
oxidation, rearranges to a carbonyl
group.
aldehyde
Reactions involving acetylide anions
[1] Formation of acetylide anions from terminal alkynes (11.6B)
+
R C C H
B
R C C
+
+
HB
•
Typical bases used for the reaction are
NaNH2 and NaH.
[2] Reaction of acetylide anions with alkyl halides (11.11A)
H C C
+
R X
H C C R + X
•
•
The reaction follows an SN2 mechanism.
The reaction works best with CH3X and
RCH2X.
[3] Reaction of acetylide anions with epoxides (11.11B)
[1] O
H C C CH2CH2OH
H C C
[2] H2O
•
•
The reaction follows an SN2 mechanism.
Ring opening occurs from the back side at
the less substituted end of the epoxide.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
301
Alkynes 11–3
Practice Test on Chapter Review
1. Draw the structure for the compound with the following IUPAC name: 5-tert-butyl-6,6-dimethyl3-nonyne.
2. a. Which of the following compounds is an enol tautomer of compound A?
O
OH
OH
1.
2.
OH
3.
A
4. A can be a tautomer of both compounds (1) and (2).
5. A can be a tautomer of compounds (1), (2), and (3).
b. Which of the following bases is strong enough to deprotonate CH3C≡CH (propyne, pKa = 25)?
The pKa’s of the conjugate acids of the bases are given in parentheses.
1.
2.
3.
4.
5.
CH3Li (pKa = 50)
NaOCH3 (pKa = 15.5)
NaOCOCH3 (pKa = 4.8)
The bases in (1) and (2) are both strong enough.
The bases in (1), (2), and (3) are all strong enough.
3. Draw the organic products formed in the following reactions.
[1] R2BH
a.
[2] H2O2, –OH
b.
[1] NaH
C
C
H
[2] O
[3] H2O
HO H
c.
d.
D
C
C
H
[1] TsCl, pyridine
[2]
C
CH
H2O
H2SO4
HgSO4
e.
C
C
H
2 HBr
302
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–4
4. Draw two different enol tautomers for the following compound.
O
OH
5. What acetylide anion and alkyl halide are needed to make the following alkyne?
CH3
CH3 C C
C
CH2CH3
CH3
Answers to Practice Test
1.
H
CH3CH2 C C C
(CH3)3C
2. a. 2
b. 1
CH3
C
4.
3.
HO
a.
CH2CH2CH3
H
CH3
O
b.
OH
C
C
(E + Z)
CH2CH2OH
C D
HC
c.
HO
H
OH
(E + Z)
CH3
d.
5.
C
CH3
O
Br H
e.
C
C
Br
H
CH3X
H
C C
C
CH3
CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
303
Alkynes 11–5
Answers to Problems
11.1 • An internal alkyne has the triple bond somewhere in the middle of the carbon chain.
• A terminal alkyne has the triple bond at the end of the carbon chain.
HC C CH2CH 2CH 3
CH3 C C CH2CH 3
HC C CH CH3
CH3
terminal alkyne
internal alkyne
terminal alkyne
11.2
Csp–Csp3
(c)
Csp2–Csp
(b)
OH
(a)
Csp3–Csp2
11.3
O
santalbic acid
To name an alkyne:
[1] Find the longest chain that contains both atoms of the triple bond, change the -ane
ending of the parent name to -yne and number the chain to give the first carbon of the
triple bond the lower number.
[2] Name all substituents following the other rules of nomenclature.
CH2CH2CH3
a.
Increasing bond strength: a < c < b
H C C CH2CCH2CH2CH3
CH2=CHCH2CHC
CH2CH2CH3
CCCH2CH2CH3 number to the first site of
unsaturation.)
CH3
4,4-dipropyl-1-heptyne
b. CH3C CCClCH2CH3
(Number to give the lower
CH2CH3 CH3
c.
4-ethyl-7,7-dimethyl-1-decen-5-yne
d.
1
5
CH3
3
(The longest chain must contain
both functional groups.)
4-chloro-4-methyl-2-hexyne
3-isopropyl-1,5-octadiyne
11.4
To work backwards from a name to a structure:
[1] Find the parent name and the functional group.
[2] Add the substituents to the appropriate carbon.
a. trans-2-ethynylcyclopentanol
5 C ring with OH at C1
b. 4-tert-butyl-5-decyne
OH
OH
OH on C1
C CH
ethynyl at C2
tert-butyl at C4
10 C chain with
a triple bond
triple bond at C5
or
C CH
304
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–6
c. 3-methylcyclononyne
9 C ring with a
triple bond at C1
3-methyl
triple bond at C1
11.5
Two factors cause the boiling point increase. The linear sp hybridized C’s of the alkyne
allow for more van der Waals attraction between alkyne molecules. Also, since a triple bond
is more polarizable than a double bond, this increases the van der Waals forces between two
molecules as well.
11.6
To convert an alkene to an alkyne:
[1] Make a vicinal dihalide from the alkene by addition of X2.
[2] Add base to remove two equivalents of HX and form the alkyne.
a. Br2CH(CH2)4CH3
Na+ –NH2
Na+ –NH2
BrCH CHCH2CH2CH2CH3
HC CCH2CH2CH2CH3
not isolated
b. CH2=CCl(CH2)3CH3
Na+ –NH2
Cl2
c. CH2=CH(CH2)3CH3
HC CCH2CH2CH2CH3
CH2CHCH2CH2CH2CH3
Cl Cl
11.7
HC CCH2CH2CH2CH3
(2 equiv)
Acetylene has a pKa of 25, so bases having a conjugate acid with a pKa above 25 will be
able to deprotonate it.
a. CH3NH– [pKa (CH3NH2) = 40]
pKa > 25 = Can deprotonate acetylene.
b. CO32– [pKa (HCO3–) = 10.2]
pKa < 25 = Cannot deprotonate acetylene.
11.8
Na+ –NH2
c. CH2=CH– [pKa (CH2=CH2) = 44]
pKa > 25 = Can deprotonate acetylene.
d. (CH3)3CO– {pKa [(CH3)3COH] = 18}
pKa < 25 = Cannot deprotonate acetylene.
To draw the products of reactions with HX:
• Add two moles of HX to the triple bond, following Markovnikov’s rule.
• Both X’s end up on the more substituted C.
a. CH3CH2CH2CH2 C C H
Br
2 HBr
CH3CH2CH2CH2 C CH3
Br
b. CH3 C C CH2CH3
2 HBr
Br
CH3 CH2 C CH2CH3
Br
Br
c.
C CH
2 HBr
C CH3
Br
Br
CH3 C CH2 CH2CH3
Br
305
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkynes 11–7
11.9
a.
+
Cl
c.
Cl
+
+
CH3
b. CH3 O CH2
O
NH
NH
CH2
11.10 Addition of one equivalent of X2 to alkynes forms trans dihalides.
Addition of two equivalents of X2 to alkynes forms tetrahalides.
2 Br2
CH3CH2 C C CH2CH3
Br Br
CH3CH2 C C CH2CH3
Br Br
CH3CH2 C C CH2CH3
Cl
Cl2
CH2CH3
C C
CH3CH2
trans dihalide
Cl
11.11
Cl2
CH3 C C CH3
Cl
CH3
The two Cl atoms are electron withdrawing,
making the π bond less electron rich and
therefore less reactive with an electrophile.
C C
CH3
Cl
11.12 To draw the keto form of each enol:
[1] Change the C–OH to a C=O at one end of the double bond.
[2] At the other end of the double bond, add a proton.
H H
C
H
CH2
a.
OH
c.
O
OH
H
H
new C–H bond
OH
O
H
new C–H bond
O
b.
H
new C–H bond
H H
11.13 The treatment of alkynes with H2O, H2SO4, and HgSO4 yields ketones.
H2O
CH3CH=C(OH)CH2CH3
H2SO4, HgSO4
CH3C(OH)=CHCH2CH3
CH3 C C CH2CH3
O
O
Two enols form.
two ketones after
tautomerization
11.14
OH
OH
a.
OH
OH
b.
enol tautomers
constitutional isomers,
but not tautomers
306
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Chapter 11–8
11.15 Reaction with H2O, H2SO4, and HgSO4 adds the oxygen to the more substituted carbon.
Reaction with [1] R2BH, [2] H2O2, –OH adds the oxygen to the less substituted carbon.
a. (CH3)2CHCH2 C C H
(CH3)2CHCH2
b.
C C H
CH3
H2O
H2SO4, HgSO4
C CH
O
CH2 C
CH3
H
Forms a ketone. H2O is added with the
O atom on the more substituted carbon.
CH3
[1] R2BH
[2] H2O2, HO
C CH
CH3 C
CH3 C
O
CH2 CH2 C
H
H
Forms an aldehyde. H2O is added with the
O atom on the less substituted carbon.
O
H2O
C
H2SO4, HgSO4
CH3
Forms a ketone. H2O is added with the
O atom on the more substituted carbon.
H H
Forms an aldehyde. H2O is added with the
C
O atom on the less substituted carbon.
C O
H
[1] R2BH
[2] H2O2, HO
11.16
[1] NaH
a. H C C H
[1] NaH
H C C
+ H2
C C
+ H2
[2] (CH3)2CHCH2–Cl
(CH3)2CHCH2 C C H
[2] CH3CH2–Br
C C CH2CH3
+ NaBr
b.
C CH
[1] NaNH2
C C
+ NH3
[2] (CH3)3CCl
CH3
C CH2
+ NaCl
+ NaCl
1° alkyl halide
substitution product
3° alkyl halide
elimination product
CH3
+
C CH
11.17
a.
(CH3)2CHCH2C
CH3
CH3 C CH2
H
CH
C C H
CH3
CH3 C CH2Cl
H
C C H
terminal alkyne
only one possibility
1° RX
b. CH3C CCH2CH2CH2CH3
CH3
C C
[1]
CH2CH2CH2CH3
[2]
[1]
CH3Cl
[2] CH3 C C
C C CH2CH2CH2CH3
Cl CH2CH2CH2CH3
1° RX
internal alkyne
two possibilities
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
307
Alkynes 11–9
c. (CH3)3CC CCH2CH3
CH3
CH3 C C C
CH3
CH3
CH3 C C C
CH3
CH2CH3
internal alkyne
only one possibility
Cl CH2CH3
1° RX
CH3
CH3 C Cl
CH3
The 3° alkyl halide
would undergo elimination.
C CCH2CH3
3° RX
too crowded for SN2 reaction
11.18
HC C H
Na+ H
Na+ H
H C C CH2CH3
CH3CH2–Br
HC C
C C CH2CH3 + H2
H
+ H2
+ Na+Br–
+ Na+
(CH3)2CHCH2C
Br
H
(CH3)2CHCH2CH2
–
C C CH2CH3 + Br
11.19
CH3
CH3
CH3
CH3 C C C C CH3
CH3
CH3
+
CH3 C C C
CH3
X C CH3
CH3
CH3
The 3° alkyl halide is too crowded for nucleophilic
substitution. Instead, it would undergo elimination
with the acetylide anion.
2,2,5,5-tetramethyl-3-hexyne
11.20
CH3
a.
[1]
CH3
OH
C C H
O
[2] H2O
Epoxide is drawn up, so the
acetylide anion attacks from below
at less substituted C.
C
CH
O
b.
[1]
CH
C
OH
C C H
+
[2] H2O
C
OH
Backside attack of the nucleophile
(–C≡CH) at either C since both
ends are equally substituted
CH
enantiomers
11.21
a. CH3CH2CH2Br
CH3CH2C C–Na+
b. (CH3)2CHCH2CH2Cl
c. (CH3CH2)3CCl
CH3CH2CH2C CCH2CH3
CH3CH2C C–Na+
CH3CH2C C–Na+
d. BrCH2CH2CH2CH2OH
(CH3)2CHCH2CH2C
(CH3CH2)2C
CH3CH2C C–Na+
CHCH3
CH3CH2C CH
CCH2CH3
CH3CH2C CH
BrCH2CH2CH2CH2O–Na+
308
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–10
e.
f.
O
CH3CH2C C–Na+
O
CH3CH2C C–Na+
H 2O
CH3CH2C CCH2CH2OH
H 2O
CH3CH2C CCH2CHCH3
OH
11.22 To use a retrosynthetic analysis:
[1] Count the number of carbon atoms in the starting material and product.
[2] Look at the functional groups in the starting material and product.
• Determine what types of reactions can form the product.
• Determine what types of reactions the starting material can undergo.
[3] Work backwards from the product to make the starting material.
[4] Write out the synthesis in the synthetic direction.
?
CH3CH2C CCH2CH3
HC CH
6 C's
CH3CH2C CCH2CH3
HC C H
Na+ H
CH3CH2C C
CH3CH2–Br
HC C
2 C's
+ CH3CH2Br
CH3CH2C C H
Na+ H
HC C
CH3CH2C C
+ CH3CH2Br
CH3CH2–Br
CH3CH2C CCH2CH3
11.23
O
HC CH
CH3CH2CH2 C
H
product:
4 carbons, aldehyde functional group
(can be made by hydroboration–oxidation of
a terminal alkyne)
starting material:
2 carbons, C≡C functional group
(can form an acetylide
anion by reaction with NaH)
O
Retrosynthetic
analysis:
CH3CH2CH2 C
CH3CH2C C H
H C C H
H
Forward direction: H C C H
Na+ H
C C H
CH3CH2–Br
CH3CH2C C H
O
[1] R2BH
[2] H2O2,
HO–
11.24
a.
4
6
3
2
5
1
6 C alkyne
hexyne
C≡C at C1
1-hexyne
CH3 and CH2CH3 at C3
3-ethyl-3-methyl-1-hexyne
1-hexyne
3
b.
5
11
1
12 C chain with 2 C C's
dodecadiyne
C≡C's at C3 and C5
3,5-dodecadiyne
CH3 at C11
11-methyl-3,5-dodecadiyne
CH3CH2CH2 C
H
309
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkynes 11–11
11.25
O
OH
b.
a.
O
OH
keto form
enol form
H
enol form
keto form
11.26
a, b.
N
O
O
O
N
O
c.
d. * = sp hybridized C
e. 3 < 1 < 2
H
(1)
(2)
*
* *
O
*
C*
HN
(3)
HO
erlotinib
most acidic
C–H proton
phomallenic acid C
most acidic
proton
shortest C–C single bond
Csp–Csp2
11.27 Use the rules from Answer 11.3 to name the alkynes.
3-hexyne
1-hexyne
2-hexyne
3-methyl-1-pentyne
3,3-dimethyl-1-butyne
4-methyl-2-pentyne
4-methyl-1-pentyne
11.28 Use the rules from Answer 11.3 to name the alkynes.
5
3
a. CH3CH2CH(CH3)C CCH2CH3
5-methyl
5-methyl-3-heptyne
3-heptyne
e.
CH3CHC CCHCH3
CH3
HC C CH(CH2CH3)CH2CH2CH3
1-hexyne
3-hexyne
b.
d.
3-ethyl
CH3CH2C CCH2C
5
2,5-dimethyl-3-hexyne
2
CH3
3-ethyl-1-hexyne
CCH3
2,5-octadiyne
2
5-ethyl
f.
2,5-dimethyl
4-nonyne
c.
6
4-ethyl
CH3
7-methyl
CH3CH2CHC CCHCHCH2CH3
CH3CH2
CH2CH3
3,6-diethyl
3,6-diethyl-7-methyl-4-nonyne
6-methyl
g.
(2E)-4,5-diethyl-2-decen-6-yne
1
ethynyl
1-ethynyl-6-methylcyclohexene
310
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–12
11.29 Use the directions from Answer 11.4 to draw each structure.
a. 5,6-dimethyl-2-heptyne
d. cis-1-ethynyl-2-methylcyclopentane
1-ethynyl
2-heptyne
or
C CH
C CH
2-methyl
b. 5-tert-butyl-6,6-dimethyl-3-nonyne
6,6-dimethyl
e. 3,4-dimethyl-1,5-octadiyne
3,4-dimethyl
3-nonyne
5-tert-butyl
diyne
c. (4S)-4-chloro-2-pentyne
Cl
H
diyne
f. (6Z)-6-methyl-6-octen-1-yne
4-chloro
S configuration
6-methyl
Z
2-pentyne
1
11.30 Keto–enol tautomers are constitutional isomers in equilibrium that differ in the location of a
double bond and a hydrogen. The OH in an enol must be bonded to a C=C.
O
a.
CH3
C
OH
CH3
and
CH2
C
• C=O
• C=C
• one more CH bond
• OH on C=C
keto–enol tautomers
H
and
O
• C=O
• one more CH bond
• C=C
• OH on C=C
keto–enol tautomers
O
OH
O
b.
OH
c.
CH3
OH
and
d.
and
OH is not bonded
to the C=C.
NOT keto–enol tautomers
OH is not bonded
to the C=C.
NOT keto–enol tautomers
11.31 To draw the enol form of each keto form: [1] Change the C=O to a C–OH. [2] Change
one single C–C bond to a double bond, making sure the OH group is bonded to the C=C.
Use the directions from Answer 11.12 to draw each keto form.
a.
O
H
O
b.
OH
OH
CH3CH2CHO =
H
OH
E/Z isomers
possible
H
OH
d.
+
CH2CH3
O
c.
CH2CH3
CH2CH3
OH
O
311
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkynes 11–13
11.32 Tautomers are constitutional isomers that are in equilibrium and differ in the location of a
double bond and a hydrogen atom.
O
O
OH
O
O
a.
A
OH
OH
tautomer
O
O
c.
b.
constitutional isomer
OH
d.
constitutional isomer
neither
11.33
O
OH
OH
(E and Z)
2-butanone
C=C has one
C bonded to it.
C=C has two C's bonded to it.
The more substituted double
bond is more stable.
11.34
O H
O
HO
O
H OH
HO
O
H OH
11.35
NHCH3
NHCH3
X
H O H
enamine 2
+ H 2O
H
NCH3
NCH3
+ H 3O
H
+ H2 O
Y
imine
11.36
HC CCH2CH2CH2CH3
a.
HCl
(2 equiv)
b.
HBr
O
Cl
CH3 C CH2CH2CH2CH3
e.
Cl
Br
CH3 CCH2CH2CH2CH3
(2 equiv)
f.
Cl2
(2 equiv)
d.
HC CCH2CH2CH2CH3
g.
Cl Cl
O
H2O
H2SO4, HgSO4
[2] H2O2, HO–
NaH
H
C
CH2CH2CH2CH2CH3
Na+ C CCH2CH2CH2CH3
Br
Cl Cl
c.
[1] R2BH
CH3
C
h.
CH2CH2CH2CH3
[1] –NH2
CH3CH2C CCH2CH2CH2CH3
[2] CH3CH2Br
[1] –NH2
[2] O
[3] H2O
HOCH2CH2 C CCH2CH2CH2CH3
312
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–14
11.37
Br Br
HBr
a.
H2O
c.
(2 equiv)
b.
H2SO4
O
Br Br
Br2
O
[1] R2BH
d.
(2 equiv)
[2] H2O2, HO–
Br Br
11.38
O
a.
(CH3CH2)3C
H2O
C CH
(CH3CH2)3C
H2SO4, HgSO4
C
CH3
[1] R2BH
b.
(CH3CH2)3C
C CH
c.
(CH3CH2)3C
C CH
d.
(CH3CH2)3C
C CH
(CH3CH2)3C
[2] H2O2, HO–
HCl
CH2CHO
(CH3CH2)3CCCl2CH3
(2 equiv)
[1] NaH
(CH3CH2)3C
[2] CH3CH2Br
C CCH2CH3
11.39 Reaction rate (which is determined by Ea) and enthalpy (H°) are not related. More
exothermic reactions are not necessarily faster. Since the addition of HX to an alkene forms
a more stable carbocation in an endothermic, rate-determining step, this carbocation is
formed faster by the Hammond postulate.
H
R C C R'
HX
H X
R C C R'
H
R C C R'
R C C R'
HX
H H
sp hybridized carbocation
less stable
slower reaction
R C C R'
H H
H H
sp2 hybridized carbocation
more stable
faster reaction
11.40
O
OH
C
a.
CH3
C
CH3
O
C
c.
CH3
CH3
C
CH3 C CH
CH2
OH
C
d.
C
CH2
CH3CH2
O
CH
OH
O
b.
OH
CH
C C
H X
R C C R'
CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
313
Alkynes 11–15
11.41 To determine what two alkynes could yield the given ketone, work backwards by drawing
the enols and then the alkynes.
H
HC C CH2CH3
OH
C C
H
HO
O
C
CH3
CH2CH3
H
CH3 C C CH3
C C
CH2CH3
CH3
2-butanone
CH3
11.42
CH2CHO
a.
C CH
b.
(CH3)2CHC
CCH(CH3)2
O
11.43
Cl
2 Cl2
b.
(CH3)3CC
Br Br
2 HBr
C CH
a.
(CH3)3CC
CH
CHCl2
Cl
c.
CH
Cl Cl
[1] Cl2
CH
[2] NaNH2
C C
C C
(2 equiv)
H H
e.
[1] R2BH
C CH
d.
C
[2] H2O2, HO
HC C + D2O
HC CD
O
+
DO–
O
O
f.
g.
C C CH3
CH2
CH3CH2C C + CH3CH2CH2–OTs
CH3
h.
H2O
H2SO4
H
C
C
CH3
CH2CH3
CH3CH2C CCH2CH2CH3
CH3
[1] HC C
O
[2] HO–H
+ OTs
CH3
O
OH
C CH
C CH
CH2–I
i.
CH3CH2C C–H
[1] Na+ –NH2
C C–H [1] Na+ H–
j.
CH3CH2C C
C C
[2] O
[2]
CH3CH2C C
CH2
C CCH2CH2O– [3] HO–H
C C CH2CH2OH
314
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–16
11.44
CH2CH2Br
KOC(CH3)3
H H
Br2
CH=CH2
C C H
KOC(CH3)3
A
B
C CH
DMSO
(2 equiv)
Br Br
C
NaNH2
D
CH3I
C CCH3
C C
E
11.45
most acidic H
[1] NaNH2
[2] CH3I
H C C CH2CH2CH2OH
CH3 C C CH2CH2CH2OH
NaNH2 will remove the
proton from the
OH since it is more acidic.
A
H C C CH2CH2CH2OCH3 = B
H C C CH2CH2CH2O
CH3
I
11.46
stereogenic center at
the site of reaction
identical
Cl HC C
a.
C CH
D H
H D
stereogenic center NOT
at the site of reaction
CH3 H
O
H
CH3
[1] HC C
H
CH3 [2] H2O
C
H
CH3
CH3
C
C
OH
H
C
C
C
CH
inversion
H
CH3
HC
Configuration is retained.
O
Cl HC C
b.
c.
CH3
H
HO
C CH
CH3 H
d.
H
CH3
[1] HC C
H
CH3
[2] H2O
HO
H
C
H
CH3
CH3
CH3
OH
H
C
C
C
C
CH
HC
enantiomers
C
H
CH3
315
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkynes 11–17
11.47
H D
H D
TsCl
OH pyridine
D H
HC C
OTs
SN2
B
A
retention
H D
D H
PBr3
OH
inversion
Br
SN2
H D
HC C
SN2
C
A
inversion
inversion
11.48
H–C CCH2OR'
new C–C bond
CH3 Li+
OH
a.
Br
PBr3
C CCH2OR'
C CCH2OR'
B
OCOR
OCOR
OCOR
A
C
OH
O
[1] D = HC C
OR
b.
[2] H2O
OH
OR
H C C
OH
H C C
E = TsCl,
pyridine
These 2 C's are added.
OCH2CH3
OCH2CH3
[1] NaH
CH3 C C
[2] CH3I
OH
OH
[1] NaH
H C C
H C C
H C C
[2] F = CH3CH2Br
G
OTs
new C–C bond
11.49
Br H
CH3CH2 C C
CH3CH2
H
Br H
H
CH3–C
H
H
C C
Br
NH2
Br
C CH3
H
CH3CH2 C C H
1-butyne
H2N
H
CH3
CH3 C C CH3
C C
CH3
Br
NH2
Br
(E and Z isomers possible)
2-butyne
NH2
Reaction by-products:
H
Br H
CH3CH2 C C
major product
more substituted alkyne
H
Br H
NH2
NH2 H
C H
CH3 C C
Br
H
CH3CH C CH2
1,2-butadiene
2 NH3 + 2 Br–
316
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–18
11.50 A carbanion is more stable when its lone pair is in an orbital with a higher percentage of the
smaller s orbital. A carbocation is more stable when its positive charge is due to a vacant
orbital with a lower percentage of the smaller s orbital. In HC≡C+, the positively charged C
uses two p orbitals to form two π bonds. If the σ bond is formed using an sp hybrid orbital,
the second hybrid orbital would have to remain vacant, a highly unstable situation.
HC C
CH2 CH
sp hybridized
higher % s-character
more stable
HC C
sp2 hybridized
lower % s-character
less stable
CH2 CH
sp hybridized
sp hybridized
Vacant orbital has 50% s-character. Vacant orbital is a p orbital.
less stable
more stable
11.51
CH3
Cl
CH3C CCH3
H Cl
CH3
H
+
C
Cl
C C
CH3
CH3
C CH3
CH3
Cl– attack on the same side as
the H yields the E isomer.
C C
H
Cl
Cl– attack on the opposite side
to the H yields the Z isomer.
H
Cl
11.52
a.
C
C
C
C
C
C
C
C
H
H
C O
H
H
CH3CH2 Li+
OH
H OH
+ CH3CH3
+
C H
O
Li+
H OSO3H
OH
HSO4
OH2
b. CH3 C C CH
CH3 C C CH
H
+ OH
CH3 C C CH
H2O
C C CH
CH3
H
H
H
H O H
CH3 C C C H
H
resonance structures
HSO4
HSO4
CH3 CH CH
O H
C O
H2SO4
H
CH3 CH C C
H
OH
CH3 CH C C
H
H
O H
CH3 CH C C H
H
H OSO3H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
317
Alkynes 11–19
11.53
H
C C OCH3
CH3CH2CH2
H OH2
OH
C C
OCH3
CH3CH2CH2
H
CH3CH2CH2C C OCH3
OH
C C
OCH3
CH3CH2CH2
H
C C OCH3
CH3CH2CH2
H OH2
H2O
H
H
H2O
O
CH3CH2CH2CH2 C
OCH3
O H
H2O
CH3CH2CH2CH2 C
OCH3
OH
CH3CH2CH2CH2 C
OCH3
H2O
H OH2
11.54
a. C6H5CH2CHBr2
KOC(CH3)3
(2 equiv)
DMSO
KOC(CH3)3
b. C6H5CHBrCH3
H2SO4
c. C6H5CH2CH2OH
C6H5C CH
C6H5CH=CH2
C6H5CH=CH2
Br2
Br2
NaNH2
C6H5CHBrCH2Br
excess
NaNH2
C6H5CHBrCH2Br
excess
C6H5C CH
C6H5C CH
11.55 The alkyl halides must be methyl or 1°.
a. HC C
CH2CH2CH(CH3)2
HC C
Cl CH2CH2CH(CH3)2
CH3
b. CH3
CH3
C C C CH2CH3
CH3 Cl
C C C CH2CH3
CH3
c.
C C
1° RX
CH3
CH2CH2CH3
C C
Cl CH2CH2CH3 1° RX
11.56
a.
HC C–H
b.
HC C–H
Na+ H–
Na+ H–
HC C
HC C
(CH3)2CHCH2–Cl
CH3CH2CH2–Cl
(CH3)2CHCH2C
CH
NaH
CH3CH2CH2C CH
CH3CH2CH2C C
CH3CH2CH2–Cl
CH3CH2CH2C CCH2CH2CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–20
c.
[1] R2BH
CH3CH2CH2C CH
[2] H2O2, HO–
(from b.)
d.
O
H2O
CH3CH2CH2C CH
H2SO4
HgSO4
2 HCl
(from b.)
e.
CH3CH2CH2CH2CHO
CH3CH2CH2C CH
CH3CH2CH2
C
CH3
CH3CH2CH2CCl2CH3
(from b.)
f.
O
H 2O
CH3CH2CH2C CCH2CH2CH3
H2SO4, HgSO4
CH3CH2CH2
C
CH2CH2CH2CH3
(from b.)
11.57
HBr
b. CH3CH2C CH
(from a.)
(2 equiv)
Cl2
c. CH3CH2C CH
(from a.)
(2 equiv)
Br2
d. CH3CH2CH CH2
e.
Cl
Cl2
a. CH3CH2CH CH2
NaH
CH3CH2C CH
Cl
2 –NH2
CH3CH2CH CH2
CH3CH2C CH
CH3CH2CBr2CH3
CH3CH2CCl2CHCl2
CH3CH2CHBrCH2Br
CH3CH2C C
CH3CH2CH2 Br
CH3CH2C C CH2CH2CH3
(from a.)
[1]
f.
CH3CH2C C
O
CH3CH2C C CH2CH2OH
[2] H2O
(from e.)
O
g. CH3CH2C C
(from e.)
OH
[1]
[2] H2O
C CCH2CH3
(+ enantiomer)
11.58
NaH
a.
HC CH
b.
CH3CH2CH2CH2CH2CH2C CH
HC C
CH3CH2CH2CH2CH2CH2Br
NaH
CH3CH2CH2CH2CH2CH2C CH
CH3CH2CH2CH2CH2CH2C C
CH3CH2Br
CH3CH2CH2CH2CH2CH2C CCH2CH3
(from a.)
c.
CH3CH2CH2CH2CH2CH2C C
(from b.)
d.
[1]
O
[2] H2O
CH3CH2CH2CH2CH2CH2C CCH2CH2OH
(from c.)
CH3CH2CH2CH2CH2CH2C CCH2CH2OH
[1] NaH
[2] CH3CH2Br
CH3CH2CH2CH2CH2CH2C CCH2CH2OCH2CH3
319
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkynes 11–21
11.59
K+ –OC(CH3)3
Br
Br2
CH2 CH2
Br
NaNH2
NaH
HC CH
HC C
Br (2 equiv)
Br
H2 O
Br
C C
NaH
C C
HC C
H2SO4, HgSO4
O
11.60
H2SO4
a.
2 –NH2
Br2
OH
NaH
CH3 C CH
Br
CH3 C C
Br
CH3 C CCH2CH2CH3
SOCl2
OH
Cl2
b.
(from a.)
[1] CH3 C C
NaH
Cl
H2O
(from a.)
[2] H2O
O
OH
Cl
OH
CH3C CCH2CHCH3
11.61
a.
H2SO4
OH
CH2 CH2
Br
Br2
NaNH2
(2 equiv)
Br
NaH
HC CH
HC C
Br
PBr3
OH
[1] NaH
C C
HO
HC C
[2] O
[3] H2O
OH
b.
H2SO4
C C
CH2 CH2
Br
Br2
H2O
O
OH
C C
NaH
HO
NaH
O
(from a.)
C C
CH3CH2Br
(from a.)
CH3CH2 O
11.62 Two resonance structures can be drawn for an enol.
OH
OH
negative charge on C that
is part of the enol C=C
Since the second resonance structure places an electron pair (and therefore a negative charge)
on an enol carbon, this makes the C=C more nucleophilic than the C=C of an alkene for which
no additional resonance forms can be drawn. Thus, the OH group donates electron density to
the C=C by a resonance effect.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–22
11.63
O
H
H
O
H
O
H
O
H2O
H
O
OH2
+ H3O
H
H
H
H
H
H
H
11.64
Br Br
Br
CH3 C C H
CH3
CH3
Br
C C
CH3 C C H
H
H O
H2O
C C H
H
HO
H
H O
H
Br–
Br
CH3
Br
C C
+ H2O
H OH2
H2O
CH3
C
Br
CH3
O
C C H
H O
CH2Br
H3O+
H
H2O
11.65
Only this carbocation forms because
it is resonance stabilized. The
positive charge is delocalized on oxygen.
H
TsOH
O
TsO H
H
O
O
re-draw
O
+ TsO–
H
not resonance stabilized
(not formed)
H
OCH3
NOT
O
O
CH3OH
O
OCH3
H
CH3OH
OCH3
X
O
Y
+ CH3OH2
11.66 A more stable internal alkyne can be isomerized to a less stable terminal alkyne under these
reaction conditions because when CH3CH2C≡CH is first formed, it contains an sp hybridized
C–H bond, which is more acidic than any proton in CH3–C≡C–CH3. Under the reaction
conditions, this proton is removed with base. Formation of the resulting acetylide anion
drives the equilibrium to favor its formation. Protonation of this acetylide anion gives the
less stable terminal alkyne.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Alkynes 11–23
H
H
CH3 C C CH2
NH2
CH3 C C CH2
CH3 C C CH2
H
H
2-butyne
NH3
NH2
H
H2O
NH2
NH2
NH3
CH3CH2 C C
CH3 C C CH
CH3CH2 C C H
acetylide anion
1-butyne
H
A
C C C(CH3)2
(CH3)2CH
NH2
NH3
CH3 C C CH
H
H
(CH3)2CH
NH2
CH3 C C CH
B
C C C(CH3)2
(CH3)2CH
C C C(CH3)2
H
2,5-dimethyl-3-hexyne
NH3
NH2
(CH3)2CH
NH2
C C C(CH3)2
H
2,5-dimethyl-2,3-hexadiene
In this case the reaction stops with formation of 2,5-dimethyl-2,3-hexadiene because a terminal
alkyne (with an acidic sp hybridized C–H bond) is not formed. Removal of the circled H in the
diene re-forms the anion shown in resonance structures A and B.
11.67
CH3
C
C
O
H
C
O
H
CH3
CH3
OH
OH2 C
C
C
C
+
H
C
CH3
H2O
C
CH3
O
C
C
O
H2O
CH3
O
C OH
H
O
C
O
CH3
C O
CH3
C O H
H
C
O
CH3
C OH
CH3
H
HCOOH
resonance structures
C OH
H
enol
O
O
C
H
H
322
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 11–24
11.68 In the presence of acid, (R)-α-methylbutyrophenone enolizes to form an achiral enol.
H
O
OH2
OH
H
OH
H
H2 O
(R)-α-methylbutyrophenone
(+ 1 resonance
structure)
(E and Z isomers)
achiral enol
The achiral enol can then be protonated from above or below the plane to form a racemic
mixture that is optically inactive.
O H
H2O
O
H3O+
below
OH
H
O H
H
H2O
O
above
H
OH2
H3O+
H
H
racemic mixture
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
323
Oxidation and Reduction 12–1
Chapter 12 Oxidation and Reduction
Chapter Review
Summary: Terms that describe reaction selectivity
• A regioselective reaction forms predominately or exclusively one constitutional isomer
(Section 8.5).
CH3
CH3
I
OH
major product
trisubstituted alkene
•
minor product
disubstituted alkene
A stereoselective reaction forms predominately or exclusively one stereoisomer (Section 8.5).
H Br
H
Na+ –OCH2CH3
C C
+
C C
H H
C C
H
H
trans alkene
major product
•
CH2
+
H
cis alkene
minor product
An enantioselective reaction forms predominately or exclusively one enantiomer (Section 12.15).
C C
CH2OH
allylic alcohol
O
C C
Sharpless
reagent
or
CH2OH
C C
O
CH2OH
One enantiomer is favored.
Definitions of oxidation and reduction
Oxidation reactions result in:
• an increase in the number of C–Z bonds,
or
• a decrease in the number of C–H bonds.
Reduction reactions result in:
• a decrease in the number of C–Z bonds, or
• an increase in the number of C–H bonds.
[Z = an element more electronegative than C]
Reduction reactions
[1] Reduction of alkenes—Catalytic hydrogenation (12.3)
R
CH CH R
H2
Pd, Pt, or Ni
H H
R C C R
H H
alkane
•
•
Syn addition of H2 occurs.
Increasing alkyl substitution on the C=C
decreases the rate of reaction.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–2
[2] Reduction of alkynes
[a]
H H
2 H2
R C C R
•
Two equivalents of H2 are added and four
new C–H bonds are formed (12.5A).
R
•
H
•
Syn addition of H2 occurs, forming a cis
alkene (12.5B).
The Lindlar catalyst is deactivated so that
reaction stops after one equivalent of H2
has been added.
R C C R
Pd-C
H H
alkane
[b]
R
H2
R C C R
C C
Lindlar
catalyst
H
cis alkene
[c]
R
Na
R C C R
H
•
C C
NH3
H
R
Anti addition of H2 occurs, forming a
trans alkene (12.5C).
trans alkene
[3] Reduction of alkyl halides (12.6)
R X
[1] LiAlH4
[2] H 2O
R H
alkane
•
•
The reaction follows an SN2 mechanism.
CH3X and RCH2X react faster than more
substituted RX.
•
•
The reaction follows an SN2 mechanism.
In unsymmetrical epoxides, H– (from
LiAlH4) attacks at the less substituted
carbon.
•
•
•
The mechanism has one step.
Syn addition of an O atom occurs.
The reaction is stereospecific.
•
Ring opening of an epoxide intermediate
with –OH or H2O forms a 1,2-diol with two
OH groups added in an anti fashion.
[4] Reduction of epoxides (12.6)
O
C C
OH
[1] LiAlH4
[2] H2O
C C
H
alcohol
Oxidation reactions
[1] Oxidation of alkenes
[a] Epoxidation (12.8)
C C
+
O
C C
RCO3H
epoxide
[b] Anti dihydroxylation (12.9A)
[1] RCO3H
C C
HO
C
C
[2] H2O ( H+ or HO–)
OH
1,2-diol
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
325
Oxidation and Reduction 12–3
[c] Syn dihydroxylation (12.9B)
HO
[1] OsO4; [2] NaHSO3, H2O
C C
OH
C
or
[1] OsO4, NMO; [2] NaHSO3, H2O
or
KMnO4, H2O, HO–
C
•
Each reagent adds two new C–O
bonds to the C=C in a syn fashion.
•
Both the σ and π bonds of the alkene
are cleaved to form two carbonyl
groups.
1,2-diol
[d] Oxidative cleavage (12.10)
R'
R
C C
R
H
[1] O3
R
[2] Zn, H2O or
CH3SCH3
R
R'
C O
O C
+
H
ketone
aldehyde
[2] Oxidative cleavage of alkynes (12.11)
[a]
[1] O3
R C C R'
internal alkyne
[2] H2O
R
R'
C O
+
•
O C
HO
The σ bond and both π bonds of the
alkyne are cleaved.
OH
carboxylic acids
[b]
R C C H
terminal alkyne
R
[1] O3
[2] H2O
+
C O
CO2
HO
[3] Oxidation of alcohols (12.12, 12.13)
H
[a]
[b]
PCC
or
HCrO4–
H
A-26
1o alcohol Amberlyst
resin
H
Oxidation of a 1o alcohol with PCC or
HCrO4– (Amberlyst A-26 resin) stops at the
aldehyde stage. Only one C–H bond is
replaced by a C–O bond.
•
Oxidation of a 1o alcohol under harsher
reaction conditions—CrO3 (or Na2Cr2O7 or
K2Cr2O7) + H2O + H2SO4—affords a
RCOOH. Two C–H bonds are replaced by
two C–O bonds.
Since a 2o alcohol has only one C–H bond
on the carbon bearing the OH group, all
Cr6+ reagents—PCC, CrO3, Na2Cr2O7,
K2Cr2O7, or HCrO4– (Amberlyst A-26
resin)—oxidize a 2o alcohol to a ketone.
H
aldehyde
R
C O
H2SO4, H2O
1o alcohol
•
C O
CrO3
R C OH
H
HO
carboxylic acid
H
[c]
R
R C OH
PCC or CrO3
or
R
HCrO4–
o
2 alcohol Amberlyst A-26
resin
R C OH
•
R
C O
R
ketone
[4] Asymmetric epoxidation of allylic alcohols (12.15)
H
CH2 OH
C C
R
H
(CH3)3C
OOH
Ti[OCH(CH3)2]4
O
H C C
R
CH2OH
H
with (–)-DET
or
H
CH2OH
R C C
H
O
with (+)-DET
326
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–4
Practice Test on Chapter Review
1.a. Compound X has a molecular formula of C9H12 and contains no triple bonds. X is hydrogenated
to a compound of molecular formula C9H14 with excess H2 and a palladium catalyst. What can
be said about X?
1. X has four rings.
2. X has three rings and one double bond.
3. X has two rings and two double bonds.
4. X has one ring and three double bonds.
5. X has four double bonds.
b. Syn addition to an alkene occurs exclusively with which reagents?
1.
2.
3.
4.
5.
OsO4
KMnO4, H2O, –OH
mCPBA, then H2O, –OH
Both reagents (1) and (2) give syn addition exclusively.
Reagents (1), (2), and (3) give syn addition exclusively.
c. Which of the following reagents adds to an alkene exclusively in an anti fashion?
1. Br2
2. H2, Pd-C
3. BH3, then H2O2, –OH
4. Reagents (1) and (2) both add in an anti fashion.
5. Reagents (1), (2), and (3) all add in an anti fashion.
2. Label each statement as True (T) or False (F).
a.
b.
c.
d.
e.
f.
g.
h.
PCC oxidizes 1° alcohols to aldehydes.
CrO3 oxidizes 2° alcohols to ketones.
Treatment of 2-hexyne with Na in NH3 forms cis-2-hexene.
Reduction of propene oxide with LiAlH4 forms 1-propanol.
Ozonolysis of 2-methyl-2-octene forms one ketone and one aldehyde.
mCPBA is an oxidizing agent that converts alkenes to trans diols.
1-Octen-5-yne reacts with H2 and Pd-C, but does not react with H2 and Lindlar catalyst.
Treatment of cyclohexene with OsO4 affords an optically inactive product mixture that
contains two enantiomers.
3. Label each reagent as an oxidizing agent, reducing agent, or neither.
a.
b.
c.
d.
e.
f.
O3
LiAlH4
mCPBA
H2O, H2SO4
PCC
Na, NH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
327
Oxidation and Reduction 12–5
4. Draw the organic products formed in each reaction and indicate stereochemistry when necessary.
Na
a.
NH3
H2
b.
Pd–C
OH
c.
PCC
HO
d.
OH
[1] SOCl2
[2] LiAlH4
5. a. Fill in the appropriate starting material (including any needed stereochemistry) in the following
reaction.
[1] OsO4
[2] H2O, NaHSO3
HO
H
C
H
C6H5
C
C6H5
+
OH
b. Fill in the appropriate reagent in the following reaction.
O
H
OH
c. What starting material is needed for the following reaction?
[1] O3
CHO
[2] (CH3)2S
O
OH
enantiomer
328
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–6
Answers to Practice Test
1.a. 2
b. 4
c. 1
2.a. T 3.a. oxidizing
b. T
b. reducing
c. F
c. oxidizing
d. F
d. neither
e. T
e. oxidizing
f. F
f. reducing
g. F
h. F
4.
5.
H
H
a.
a.
H
H
b.
b. Sharpless reagent
(+)-DET
c.
CHO
c.
HO
d.
Chapter 12: Answers to Problems
12.1 Oxidation results in an increase in the number of C–Z bonds (usually C–O bonds) or a
decrease in the number of C–H bonds.
Reduction results in a decrease in the number of C–Z bonds (usually C–O bonds) or an
increase in the number of C–H bonds.
O
O
oxidation
a.
c.
reduction
b.
d.
CH3
C
O
CH3
CH2 CH2
CH3
C
OCH3
CH3CH2Cl
oxidation
neither
1 new C–H bond
and 1 new C–Cl bond
12.2 Hydrogenation is the addition of hydrogen. When alkenes are hydrogenated, they are
reduced by the addition of H2 to the π bond. To draw the alkane product, add a H to each C of
the double bond.
CH3
CH3
b.
CH2CH(CH3)2
C C
a.
H
CH3 CH2CH(CH3)2
CH3 C
H
C H
H
c.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–7
12.3 Draw the alkenes that form each alkane when hydrogenated.
a.
or
H2
Pd-C
or
c.
or
or
or
H2
Pd-C
or
or
H2
Pd-C
or
b.
12.4 Cis alkenes are less stable than trans alkenes, so they have larger heats of hydrogenation.
Increasing alkyl substitution increases the stability of a C=C, thus decreasing the heat of
hydrogenation.
cis alkane
less stable
larger heat of hydrogenation
or
b.
or
a.
trans alkane
trisubstituted
disubstituted
less stable
larger heat of hydrogenation
12.5 Hydrogenation products must be identical to use hydrogenation data to evaluate the relative
stability of the starting materials.
Different products are formed.
Hydrogenation data can't
be used to determine the
relative stability of the starting
materials.
2-methyl-2-pentene
3-methyl-1-pentene
12.6
Two enantiomers are formed in equal amounts:
new stereogenic center
H
a.
b.
CH3 C CH2CH2CH3
C C
H
CH3
CH2CH3
CH2CH3
CH2CH3
CH2CH2CH3
CH2
H
CH3
CH3
H
=
C
CH3
+ CH3
CH2CH3
+
CH2CH2CH3
H
H
CH3
diastereomers
c.
C(CH3)3
C(CH3)3
+
diastereomers
C(CH3)3
CH3CH2CH2 C CH
3
H
330
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–8
12.7
A
Molecular formula
before hydrogenation
C10H12
Molecular formula
after hydrogenation
C10H16
Number
of rings
3
Number of
π bonds
2
B
C
C4H8
C6H8
C4H10
C6H12
0
1
1
2
Compound
12.8
H2, Pd-C
CH2OCO(CH2)16CH3
CHOCO(CH2)6(CH2CH=CH)2(CH2)4CH3
CH2OCO(CH2)16CH3
CHOCO(CH2)6(CH2CH2CH2)2(CH2)4CH3 B
excess
CH2OCO(CH2)16CH3
H2, Pd-C
CH2OCO(CH2)16CH3
CH2OCO(CH2)16CH3
CHOCO(CH2)6CH2CH=CH(CH2)7CH3 CHOCO(CH2)9CH2CH=CH(CH2)4CH3
1 equiv
A
CH2OCO(CH2)16CH3
CH2OCO(CH2)16CH3
CH2OCO(CH2)16CH3
or
C
A has 2 double bonds.
lowest melting point
C has 1 double bond.
intermediate melting point
C
B has 0 double bonds.
highest melting point
12.9 Hydrogenation of HC≡CCH2CH2CH3 and CH3C≡CCH2CH3 yields the same compound. The
heat of hydrogenation is larger for HC≡CCH2CH2CH3 than for CH3C≡CCH2CH3 because
internal alkynes are more stable (lower in energy) than terminal alkynes.
12.10
O
CH2 C
O
C CH2CH3
H2
CH2CH3
C
C
Lindlar catalyst
CH3
H H
C
CH3
A
H
cis-jasmone
(perfume component
isolated from jasmine flowers)
H
12.11 In the presence of Pd-C, H2 adds to alkenes and alkynes to form alkanes. In the presence of
the Lindlar catalyst, only alkynes react with H2 to form cis alkenes.
H2
a.
C6H10
Pd-C
H2
b.
Lindlar
catalyst
no reaction
C6H10
Pd-C
Lindlar
catalyst
12.12 Use the directions from Answer 12.11.
a. CH3OCH2CH2C CCH2CH(CH3)2
H2 (excess)
Pd-C
CH3OCH 2CH2CH2CH2CH2CH(CH3)2
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–9
b. CH3OCH2CH2C CCH2CH(CH3)2
CH3OCH2CH2
H2 (1 equiv)
CH2CH(CH3)2
C C
Lindlar catalyst
H
c.
CH3OCH2CH2C CCH2CH(CH3)2
H
CH3OCH2CH2
H2 (excess)
CH2CH(CH3)2
C C
Lindlar catalyst
H
H
CH3OCH2CH2
Na, NH3
d. CH3OCH2CH2C CCH2CH(CH3)2
H
C C
H
CH2CH(CH3)2
12.13
H2
Lindlar
catalyst
A
H
R
B
H
12.14 LiAlH4 reduces alkyl halides to alkanes and epoxides to alcohols.
CH3
[1] LiAlH4
Cl
a.
H
b.
[2] H2O
O
[1] LiAlH4
H
[2] H2O
H replaces Cl.
CH3
OH
H
H
12.15 To draw the product, add an O atom across the π bond of the C=C.
O
mCPBA
a.
(CH3)2C
CH2
b.
(CH3)2C
C(CH3)2
(CH3)2C
c.
CH2
CH2
mCPBA
O
mCPBA
(CH3)2C
C(CH3)2
12.16 For epoxidation reactions:
• There are two possible products: O adds from above and below the double bond.
• Substituents on the C=C retain their original configuration in the products.
H
CH3
mCPBA
C C
a.
H
H
CH3CH2
b.
CH2CH3
mCPBA
C C
H
O
CH3 C C H
H
H
H
CH3
H
H C C H enantiomers
O
CH3CH2
CH2CH3
O
H C C H
CH3CH2 C C CH2CH3
O
H
H
identical
CH3
c.
mCPBA
CH3
CH3
O
O
H
H
enantiomers
O
CH2
332
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–10
12.17 Treatment of an alkene with a peroxyacid followed by H2O, HO– adds two hydroxy groups in
an anti fashion. cis-2-Butene and trans-2-butene yield different products of dihydroxylation.
cis-2-Butene gives a mixture of two enantiomers and trans-2-butene gives a meso compound.
The reaction is stereospecific because two stereoisomeric starting materials give different
products that are also stereoisomers of each other.
CH3 C C
H
HO
[1] RCO3H
CH3
H
[2] H2O, HO−
C
CH3
cis-2-butene
CH3 C C
H
C
CH3
H
CH3
H
OH
H
OH
C
C
HO
H
CH3
enantiomers
H
[1] RCO3H
CH3
[2] H2O, HO−
HO
H
C
CH3
CH3
H
CH3
C
OH
H
trans-2-butene
OH
C
C
H
CH3
HO
identical
meso compound
12.18 Treatment of an alkene with OsO4 adds two hydroxy groups in a syn fashion. cis-2-Butene
and trans-2-butene yield different stereoisomers in this dihydroxylation, so the reaction is
stereospecific.
HO
CH3 C C
H
CH3
[1] OsO4
H
[2] NaHSO3, H2O
OH
C
CH3
H
H
[1] OsO4
HO
H
CH3
[2] NaHSO3, H2O
trans-2-butene
C
HO
OH
C
CH3
CH3
C
CH3
H
C
H
CH3
H
HO
CH3
H
OH
identical
meso compound
cis-2-butene
CH3 C C
H
CH3
H
C
H
C
CH3
C
OH
enantiomers
12.19 To draw the oxidative cleavage products:
• Locate all the π bonds in the molecule.
• Replace all C=C’s with two C=O’s.
Replace this π bond with two C=O's.
[1] O3
a. (CH3)2C
CHCH2CH2CH2CH3
[2] Zn, H2O
[1] O3
[2] Zn, H2O
+
O CHCH2CH2CH2CH3
aldehyde
+
O
One ketone and
one aldehyde
are formed.
Two aldehydes are formed.
O
[2] Zn, H2O
c.
O
ketone
H
[1] O3
b.
(CH3)2C
H
O
O
H
A dicarbonyl compound is formed.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–11
12.20 To find the alkene that yields the oxidative cleavage products:
• Find the two carbonyl groups in the products.
• Join the two carbonyl carbons together with a double bond. This is the double bond that
was broken during ozonolysis.
CH3
a. (CH3)2C O
+ (CH3CH2)2C
O
(CH3)2C
c.
C(CH2CH3)2
CH3
O only
C
C
CH3
CH3
O
CHCH3
+
CH3
With only one product,
the alkene must be symmetrical
around the double bond. Join
this C to the same C in another
identical molecule.
Join these two C's.
b.
CH3
C
CH3CHO
Join these two C's.
12.21
O
a.
[1] O3
O
[2] CH3SCH3
H
O
[2] CH3SCH3 H
c.
[2] CH3SCH3
O O
H
O
O
[1] O3
H
[1] O3
b.
O
O
O
H
H
H
O
O
H
H
12.22 To draw the products of oxidative cleavage of alkynes:
• Locate the triple bond.
• For internal alkynes, convert the sp hybridized C to COOH.
• For terminal alkynes, the sp hybridized C–H becomes CO2.
a.
[2] H2O
O
O
[1] O3
CH3CH2 C C CH2CH2CH3
CH3CH2
C
OH
+
HO
C
CH2CH2CH3
internal alkyne
O
[1] O3
b.
C
C C
[2] H2O
internal alkyne
terminal alkyne
c.
O
+
OH
HO
C
identical compounds
internal alkyne
H C C CH2 CH2 C C CH3
[1] O3
[2] H2O
O
CO2 +
HO
C
O
C
O
OH
+
HO
C
CH3
H
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–12
12.23
a.
c.
CH3CH2CH(CH3)CO2H
b.
CO2 + CH3(CH2)8CO2H
CH3CH2CO2H, HO2CCH2CO2H, CH3CO2H
CH3CH2CH C C CHCH2CH3
CH3(CH2)8 C CH
CH3
CH3
CH3CH2 C C CH2 C C CH3
d.
HO2C(CH2)14CO2H
12.24 For the oxidation of alcohols, remember:
• 1° Alcohols are oxidized to aldehydes with PCC.
• 1° Alcohols are oxidized to carboxylic acids with oxidizing agents like CrO3 or Na2Cr2O7.
• 2° Alcohols are oxidized to ketones with all Cr6+ reagents.
O
OH
a.
PCC
CrO3
OH
H
c.
OH
H2SO4, H2O
O
OH
OH
O
O
PCC
b.
CrO3
d.
H2SO4, H2O
12.25 Upon treatment with HCrO4––Amberlyst A-26 resin:
• 1° Alcohols are oxidized to aldehydes.
• 2° Alcohols are oxidized to ketones.
H
OH
HCrO4
a.
O
–
c.
Amberlyst A-26 resin
b. HO
OH
HCrO4–
O
OH
OH
H
HCrO4
–
Amberlyst A-26 resin
O
O
O
Amberlyst A-26 resin
12.26
NaOCl
a.
OH
b.
O
+ NaCl + H2O
The by-products of the reaction with sodium
hypochlorite are water and table salt (NaCl), as
opposed to the by-products with HCrO4––
Amberlyst A-26 resin, which contain
carcinogenic Cr3+ metal.
Oxidation with NaOCl has at least two advantages over oxidation with CrO3, H2SO4 and
H2O. Since no Cr6+ is used as oxidant, there are no Cr by-products that must be disposed of.
Also, CrO3 oxidation is carried out in corrosive inorganic acids (H2SO4) and oxidation with
NaOCl avoids this.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
335
Oxidation and Reduction 12–13
12.27 To draw the products of a Sharpless epoxidation:
• With the C=C horizontal, draw the allylic alcohol with the OH on the top right of the
alkene.
• Add the new oxygen above the plane if (−)-DET is used and below the plane if (+)-DET
is used.
OH (CH3)3C
a.
OH
OOH
Ti[OCH(CH3)2]4
(+)-DET
O H
(+)-DET adds O
below the plane.
OH
b.
OH
re-draw
(CH3)3C
H O
OOH
OH
Ti[OCH(CH3)2]4
(–)-DET
(−)-DET adds O
above the plane.
12.28 Sharpless epoxidation needs an allylic alcohol as the starting material. Alkenes with no
allylic OH group will not undergo reaction with the Sharpless reagent.
This alkene is part of an allylic alcohol
and will be epoxidized.
OH
geraniol
This alkene is not part of an
allylic alcohol and will not be epoxidized.
12.29
OH
a.
OH
H2
Pd-C
A
OH
b.
mCPBA
O
O
OH
A
OH
c.
PCC
CHO
A
OH
d.
A
CrO3
H2SO4
H2O
COOH
OH
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–14
e.
OH
O
Sharpless reagent
(+)-DET
A
OH
12.30
O
[1] O3
CH3
[2] CH3SCH3
H
O
H
H
H
O
O
12.31
[1] NaH
HC CH
[2] CH3CH2Br
CH3CH2C CH
[1] NaH
[2]
OH
Na
O
OH
NH3
[3] H2O
12.32 Use the rules from Answer 12.1.
OH
a.
b.
CH3CH2Br
c.
CH2CH3 reduction
C CH
CH2 CH2 neither
1 C–H and 1 C–Br
bond are removed.
O
reduction
d. HO
OH
O
O oxidation
12.33 Use the principles from Answer 12.2 and draw the products of syn addition of H2 from above
and below the C=C.
H2
Pd-C
a.
CH3
Pd-C
CH3
CH2CH3
c.
CH3
C CH2
(CH3)2CH
+
CH3
CH3
CH2CH3
H2
H2
Pd-C
CH2CH3
+
Pd-C
CH3CH2
d.
CH3
CH3
H2
b.
CH3
CH3
CH2CH3
H C
(CH3)2CH
CH2CH3
+
CH3
C
(CH3)2CH
H
CH3
12.34 Increasing alkyl substitution increases alkene stability, thus decreasing the heat of
hydrogenation.
2-methyl-2-butene
trisubstituted
smallest ΔHo = –112 kJ/mol
2-methyl-1-butene
disubstituted
intermediate ΔHo = –119 kJ/mol
3-methyl-1-butene
monosubstituted
largest ΔHo = –127 kJ/mol
337
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–15
12.35
A possible structure:
a. Compound A: molecular formula C5H8: hydrogenated to C5H10.
2 degrees of unsaturation, 1 is hydrogenated.
1 ring and 1 π bond
b. Compound B: molecular formula C10H16: hydrogenated to C10H18.
3 degrees of unsaturation, 1 is hydrogenated.
2 rings and 1 π bond
c. Compound C: molecular formula C8H8: hydrogenated to C8H16.
5 degrees of unsaturation, 4 are hydrogenated.
1 ring and 4 π bonds
12.36
c.
a. monosubstituted
largest heat of hydrogenation
b. fastest reaction rate
O
[1] O3
H
[2] Zn, H2O
A
c.
a. tetrasubstituted
smallest heat of hydrogenation
b. slowest reaction rate
+
C
H
[1] O3
O
[2] Zn, H2O
B
+ O
identical
a. trisubstituted
intermediate heat of hydrogenation
b. intermediate reaction rate
c.
O
[1] O3
[2] Zn, H2O
O
+
H
C
12.37
O
O
OH
a.
H2 (excess)
OH
Pd-C
stearidonic acid
stearic acid
O
b.
H
O
H2 (1 equiv)
O
OH
OH
Pd-C
O
O
OH
O
OH
c.
trans
one possibility
OH
338
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–16
O
d.
O
O
OH
OH
<
OH
<
stearidonic acid
4 cis C=C's
product in (b)
3 cis C=C's
product in (c)
2 cis and one trans C=C's
12.38
H2
a.
[1] LiAlH4
h.
Pd-C
H2
b.
no reaction
Lindlar catalyst
Na
NH3
c.
H
no reaction
(CH3)3COOH
[1] CH3CO3H
no reaction
Ti[OCH(CH3)2]4
(−)-DET
O
[2] H2O,
O
H
[2] CH3SCH3
CH3CO3H
e.
O
[1] O3
i.
j.
d.
no reaction
[2] H2O
OH
OH
mCPBA
k.
HO−
OH
O
OH
anti addition
f.
l.
syn addition
[2] NaHSO3, H2O
O
OH
[2] H2O
OH
OH
KMnO4
g.
[1] LiAlH4
OH
[1] OsO4 + NMO
syn addition
H2O, HO−
OH
12.39
a. CH3CH2CH2 C C CH2CH2CH3
H2 (excess)
Pd-C
CH3CH2CH2
H2
b. CH3CH2CH2 C C CH2CH2CH3
Lindlar catalyst
c. CH3CH2CH2 C C CH2CH2CH3
d. CH3CH2CH2 C C CH2CH2CH3
CH2CH2CH3
C C
Na
NH3
[1] O3
[2] H2O
H
H
H
CH2CH2CH3
trans alkene
C C
CH3CH2CH2
H
O
CH3CH2CH2
cis alkene
C
O
+
OH
CH3CH2CH2
C
OH
identical
12.40
a.
CH3CH2
C
CH3
H
C
H2
Pd-C
CH2OH
H C * C CH2OH
CH3 H
[* = new stereogenic center]
CH3
CH3
CH3CH2 H
C
CH3CH2
CH2CH2OH
H
+
HOCH2CH2
H
Two enantiomers are formed.
C
CH2CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–17
b.
CH3CH2
H
C
O
mCPBA
C
CH3
CH3CH2
CH3
CH2OH
CH3CH2
CH3
H
CH2OH
H
CH2OH
e.
f.
CH3 C C
CH3CH2
CH2OH
CH3 C C
CH3CH2
CH2OH
(CH3)3COOH
Ti[OCH(CH3)2]4
(+)-DET
H
O
c. CH3CH2
C
C
CH2OH
d. CH3CH2
CH3
CH3CH2
PCC
C
CH3
C
CH3CH2
H
CrO3
CH2OH
H
C
H
[1] PBr3
O
CH3
CH3CH2
CH3CH2
CH2OH
C
CH3
CH2Br
[2] LiAlH4
[3] H2O
CHO
CH3CH2
H2SO4, H2O
CH3CH2
H
H
C
C C
CH3
CH2OH
H
H
C
C
CH3
C
CH3
H
C
g.
CH2OH
H
O
(CH3)3COOH
Ti[OCH(CH3)2]4
(−)-DET
H
CH3
CH3CH2
COOH
C
CH3
h.
CH3CH2
C
H
C
CH3
CH3
CH3CH2
HCrO4–
Amberlyst A-26
CH2OH
C
CH3
H
C
CHO
12.41
OH
O
PCC
a.
OH
c.
CrO3
OH
H2SO4, H2O
O
b. CH3CH2CH2CH2OH
PCC
CH3CH2CH2
C
H
12.42
OH
a.
b.
c.
d.
CH2
[1] SOCl2
Cl
pyridine
[2] LiAlH4
[3] H2O
[1] OsO4
OH
[2] NaHSO3, H2O
CH2OH
[1] mCPBA
CH2
H2
Lindlar
catalyst
O
[2] LiAlH4
OH
[3] H2O
CH3
O
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–18
12.43
a. H2, Pd-C
d. [1] LiAlH4
b. mCPBA
O
c. KMnO4
H2O, HO–
or [1] OsO4
[2] NaHSO3, H2O
OH f. PBr3
[2] H2O
g. CrO3,
or PCC or HCrO4–,
H2SO4, H2O
Amberlyst
A-26 resin
O
OH
OH
h. [1] LiAlH4
or HBr
[2] H2O
e. H2O (–OH)
OH
Br
OH
(+ enantiomer)
12.44 LiAlH4 attacks at the less substituted end of an unsymmetrical epoxide to form an alcohol
with an OH on the more substituted carbon.
O
or
OH
O
2-methyl-2-pentanol
Hydride attacks here.
12.45 The two sides of the C=C of A are different. Since D2 adds only from above, this must mean
that this side is less sterically hindered. Other reagents will also add from the same side.
O
mCPBA
a.
A
Br
b.
Br
H2O
Br2
H H
H2O
A
H
Br OH
OH
H
base
Since the OH of the bromohydrin
is down, the resulting epoxide
must also be down.
O
12.46 Alkenes treated with [1] OsO4 followed by NaHSO3 in H2O will undergo syn addition,
whereas alkenes treated with [2] CH3CO3H followed by –OH in H2O will undergo anti
addition.
a.
[1]
(CH3)2CH
CH(CH3)2
C C
H
[2] (CH3)2CH
H
H
H
C C
CH(CH3)2
[1] OsO4
[2] NaHSO3, H2O
[1] CH3CO3H
[2] –OH, H2O
anti addition
HO
(CH3)2CH
H
OH
C
H
HO
(CH3)2CH
H
C
CH(CH3)2
H CH(CH )
3 2
C
C
OH
HO
OH
rotate
(CH3)2CH
H
C
C
H
CH(CH3)2
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–19
b.
C6H5
[1]
H
C C
H
C6H5
H
H
H
C
[2] –OH, H2O
H
C
HO
[1] CH3CO3H
C C
rotate
H
C6H5
syn addition
HO
OH
C
[2] NaHSO3, H2O
C6H5
C6H5
[2]
HO
[1] OsO4
H
C6H5
C
C6H5
C6H5
C
H
C6H5
C6H5
C
H
+ enantiomer
OH
+ enantiomer
OH
OH
CH3CH2CH2
c. [1]
CH2CH2CH3
C C
H
HO
[1] OsO4
CH3CH2CH2
[2] NaHSO3, H2O
H
CH3CH2CH2
OH
C
H
H
rotate
C
H
CH2CH2CH3
H
CH2CH2CH3
syn addition
OH
OH
CH3CH2CH2
[2]
CH3CH2CH2
H
[1] CH3CO3H
C C
H
H
CH2CH2CH3
H
[2] –OH, H2O
CH2CH2CH3
OH
12.47
OH
O
H
A
O
OH
H OH
+ AlH3 + Li+
H3Al H
Li
D– (from LiAlD4) opens the epoxide ring from the back
side, so it is oriented on a wedge in the final product.
D
OH
12.48
Cl
H OH
[1]
C
mCPBA
O
[2]
O
O
O
O
Br
OH
Br
Br
OH
H
O
OH
Br
H
Na+ H
[3]
Cl
OH
C
OH
Br
Na+ + H2 +
O
O
Br–
O
+
HO
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–20
12.49 Use the directions from Answer 12.19.
a. (CH3CH2)2C
CHCH2CH3
[1] O3
(CH3CH2)2C
[2] CH3SCH3
O CHCH2CH3
CH3
[1] O3
b.
+
O
+ O C
O
[2] Zn, H2O
CH3
O
C
[1] O3
C CH
c.
OH + CO2
[2] H2O
O
d.
[1] O3
C C
C
[2] H2O
O
OH
+
HO
C
identical
12.50
a. (CH3)2C
b.
O and CH2 O
O and
(CH3)2C
CH2
O
c.
d.
CH3CH2CH2CH CHCH2CH2CH3
CH3CH2CH2CHO only
Join this C to the same C
in another identical molecule.
O
O
and 2 equivalents of
Join both of these C's
to a C from formaldehyde.
CH2 O
formaldehyde C
12.51 Use the directions from Answer 12.20.
Join these two C's.
a. C10H18
O
[1] O3
H
[2] CH3SCH3
O
2 degrees of unsaturation
one ring + one π bond
O
b. C10H16
[1] O3
[2] CH3SCH3
3 degrees of unsaturation
O
two rings + one π bond
12.52
Join these two C's.
a. CH3CH2CH2CH2COOH and
CO2
Join these two C's.
CH3CH2CH2CH2C CH c.
Join these two C's.
b. CH3CH2COOH and CH3CH2CH2COOH
CH3CH2C CCH2CH2CH3
COOH
and
C CCH3
CH3COOH
343
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–21
12.53
a.
squalene
A
B
B
C
B
B
A
[1] O3
[2] Zn, H2O
H
+ O
O
+
O
H
2 equiv
(from portion A)
O
O
H
1 equiv
(from portion C)
4 equiv
(from portion B)
COOH
b.
c.
linolenic acid
[1] O3
O
H
+
[2] Zn, H2O
[1] O3
[2] Zn, H2O
O
O
+
H
O
O
O
O
zingiberene
H
+
H
H
O
H
H
H
+
COOH
O
O
2 equiv
12.54
O
[1] O3
A
a.
H
H
[2] CH3SCH3
O
C8H12
[1] NaNH2
H2 (excess)
b.
[2] CH3I
Pd-C
B
C6H10
C7H12
C
12.55
The hydrogenation reaction tells you that both
oximene and myrcene have 3 π bonds (and no rings).
Use this carbon backbone and add in the
double bonds based on the oxidative cleavage products.
H2
C10H16
Pd-C
2,6-dimethyloctane
3 degrees of unsaturation
O
Oximene:
(CH3)2C
O
CH2 O
CH2(CHO)2
CH3 C CHO
O
O
Myrcene:
(CH3)2C
O
CH2 O
(2 equiv)
H
C
C
CH2CH2
CHO
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–22
12.56
OH
H2SO4
H2
Pd-C
decalin
C10H18O
A
C10H16
B
C10H16
C
ozonolysis
O
O
C10H16O2
E
CHO
D
O
12.57 Since hydrogenation of DHA forms CH3(CH2)20COOH, DHA is a 22-carbon fatty acid. The
ozonolysis products show where the double bonds are located.
CO2H
A
B
B
B
B
DHA
C
B
[1] O3
[2] Zn, H2O
+
CH3CH2CHO
OHCCH2CHO
5 equiv
(from portion B)
(from portion A)
+
OHCCH2CH2CO2H
(from portion C)
12.58 The stereogenic center (labeled with *) in both structures can be R or S.
*
O
O
H
O
H
H
H
Pd-C
or
H
O
O
H2
ozonolysis
*
butylcyclopheptane
O
possible structures for
dictyopterene D'
H
12.59
OH
OH
a.
re-draw
(CH3)3COOH
Ti[OC(CH3)2]4
(−)-DET
b.
H C C
(CH3)3C
CH2OH
H
(CH3)3COOH
Ti[OC(CH3)2]4
(+)-DET
H
(CH3)3C
C
C
O
CH2OH
H
H
O
OH
H
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–23
12.60
OH
re-draw
OH
O
(CH3)3COOH
Ti[OC(CH3)2]4
(−)-DET
CH2OH
enantiomeric excess =
% one enantiomer – % second enantiomer
CH2OH
+
H
H
O
major product
87%
ee = 87% – 13% = 74%
minor product
13%
12.61
H
a.
Replace this O
to make an alkene.
O
H
H
H
CH2
H
HOCH2
HOCH2
CH3
CH3 C C CH3
O
H
CH2OH
H
(+)-DET
CH2OH
CH3
C
(–)-DET
Replace this O
to make an alkene.
H
O
c.
(–)-DET HO
HO
b.
Replace this O
to make an alkene.
C
CH3
CH3
12.62 Use retrosynthetic analysis to devise a synthesis of each hydrocarbon from acetylene.
a.
CH3CH2CH CH2
CH3CH2CH CH
NaH
HC CH
C CH
CH3CH2Cl
C CH
HC CH
H2
CH3CH2C CH
CH3CH2CH CH2
Lindlar catalyst
CH3
CH3
C C
b.
H
CH3 C C CH3
C CH
CH3 C CH
HC CH
H
NaH
HC CH
C CH
CH3Cl
CH3 C CH
NaH
CH3 C C
CH3Cl
CH3
H2
CH3 C C CH3
Lindlar
catalyst
CH3
d.
CH3 C C CH3
NaH
C CH
CH3Cl
C CH
CH3 C CH
CH3 C CH
(CH3)2CHCH2CH2CH2CH2CH(CH3)2
HC CH
H
HC CH
CH3
H
HC CH
H
H
C C
c.
NaH
Cl
C CH
HC C
NaH
CH3 C C
CH3Cl
CH3 C C CH3
Na
CH3
NaH
C C
C CH
Cl
C C
HC CH
H2
(2 equiv)
Pd-C
H
C C
NH3
H
HC CCH2CH(CH3)2
CH3
C C
CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–24
12.63
Cl
NaH
HC CH
HC C
NaH
Cl
H2
Lindlar catalyst
12.64
Br
–NH
Br2
Na
2
NH3
(2 equiv)
cis
trans
Br
12.65
CH3
O
a.
CH3 C
H
C
CH3
H
NaH
HC CH
CH3
C
H
CH3 C C CH3
C
H
CH3Cl
HC C
HC C CH3
HC C CH3
NaH
C C CH3
O
CH3
H
C
C
CH3Cl
HC CH
CH3 C C CH3
mCPBA
CH3
CH3
H
C
H
H2
Lindlar catalyst
CH3
C
H
O
b. H C
C
CH3
H
CH3
H
CH3
CH3
(+ enantiomer)
CH3 C C CH3
c.
OH
C
H
CH3
H
H
H
CH3
C C
CH3
CH3
H2O, HO−
C
C
CH3
H
+ enantiomer
H
CH3
KMnO4
HC CH
O
mCPBA
CH3 C C CH3
C C
H
CH3
(from a.)
CH3
C C
CH3
H
H
C
HC C CH3
H
H
Na, NH3
(from a.)
HO
CH3 C C CH3
C C
CH3
HO
C
H
CH3
OH
C
H
CH3
HC C CH3
HC CH
347
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–25
HO
d.
OH
C
H
C
H
CH3
CH3
H
CH3
CH3 C C CH3
C C
CH3
HC C CH3
HC CH
H
(+ enantiomer)
H
CH3
KMnO4
C C
CH3
H2O, HO−
H
(from b.)
HO
OH
C
H
CH3
(+ enantiomer)
C
H
CH3
12.66
[1] BH3
a. C6H5CH CH2
[2] H2O, –OH
PCC
C6H5CH2 CH2OH
C6H5CH2 CHO
HO
H2O
b. C6H5CH CH2
c. C6H5CH CH2
H2SO 4
[1] BH3
[2] H2O, –OH
O
PCC
C6H5CH CH3
C6H5
CrO3
C6H5CH2 CH2OH
C
CH3
C6H5CH2 COOH
H2SO4, H2O
HC CH
NaH
O
mCPBA
d. C6H5CH CH2
C C H
H H
C6 H5
HO
[1] C CH
C6H5CH CH2C CH
[2] H2O
12.67
HO
CHO
O
NaH
CH3O
Br
CHO
Br
Br
O
CHO
CH3O
CH3O
NaH
CH3OH
CH3O
CH3O
O
Br
CH3O
O
PBr3
CH3O–
O
H2
OH
CHO
Pd-C
CH3O
CH3O
A
CH3O
12.68
Br
Br2
Br
NaH
2 NaNH2
1-pentene
CH3Cl
Na
(2E)-2-hexene
NH3
348
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–26
12.69
a.
b.
[1] 9-BBN or BH3
CH3CH2CH CH2
[2] H2O2,
HO–
POCl3
CH3CH2CH2CH2OH
CrO3
CH3CH2CH2CH2OH
CH3CH2CH CH2
pyridine
mCPBA
CH3CH2CH2COOH
H2SO4, H2O
HO
O
[1] LiAlH4
CH3CH2CH CH3
CH3CH2CH CH2
[2] H2O
H2O, H2SO4
CrO3
H2SO4, H2O
CH3CH2
C
CH3
O
O
mCPBA
c.
O
OH
H2 O
HC C
C CH
NaH
O
PCC
C CH
C CH
HC CH
O
H
NaH
C CH
d.
C C
CH3Cl
C C CH3
H
mCPBA
Na
CH3
NH3
(+ enantiomer)
12.70
NaH
a. HC CH
CH3CH2Br
HC C
NaH
HC C CH2CH3
C C CH2CH3
O
H2
OH
Lindlar catalyst
OH
b.
Na
OH
NH3
(from a.)
OH
c.
HO CH2CH2 C C CH2CH3
Na
O CH2CH2 C C CH2CH3
PBr3
Br
PCC
CHO
OH
NH3
(from a.)
H2O
12.71 Determine which 2 C’s can come from HC≡CH and two alkyl halides that are unhindered
(1° or CH3X).
C6H5
C6H5
HC
CH
C6H5
X
X
1-phenyl-5-methylhexane
In synthetic direction:
HC
CH
NaH
Br
HC
C
H
NaH
H2
+ H2
C6H5
2 H2
C6H5
Pd-C
C6H5
Br
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
349
Oxidation and Reduction 12–27
12.72
?
O
[1] LiAlH4
[2] H2O
H2, Pd-C
H2SO4
OH
major
product
12.73
NaH
HC CH
NaH
Br
C CH
CH3CH2 C CH
Br
CH3CH2 C C
Na, NH3
Cl
H
Cl2
anti
addition
Cl
H
(3R,4S)-3,4-dichlorohexane
12.74
a.
K+ –OC(CH3)3
PBr3
OH
HC CH
NaH
C CH
Br2
CH2 CH2
Br
Br
CH3CH2 C CH
NaH
2 NaNH2
Br
Br
CH3CH2 C C
HC CH
Br
Na, NH3
H2
b.
(from a.)
(from b.)
H
Lindlar catalyst
[1] OsO4
c.
[2] NaHSO3, H2O
O
mCPBA
HO
H
C
CH3CH2
H
C
CH2CH3
OH
H
350
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–28
12.75
OCH2CH3
H OCH2CH3
H
OCH3
OCH3
OCH3
OCH3
Li
Li
Li
H
OCH2CH3
OCH3
Li
OCH3
H OCH2CH3
12.76
The favored conformation
for both molecules places
the tert-butyl group equatorial.
OH
(CH3)3C
OH
H
H
= (CH ) C
3 3
A
A
This OH is axial and
will react faster because
the OH group is more
hindered.
OH
(CH3)3C
=
(CH3)3C
OH
B
This OH is equatorial
and will react more slowly
because the OH group
is less hindered.
B
351
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Oxidation and Reduction 12–29
12.77
O
H O
Cl
O
O
O C
O
O
O
mCPBA
OSiR3
R = alkyl group
OSiR3
O
C
O
H
Cl
O
O
O
H
O
O
O
OH
O
Cl
O
O
O
H
O
C
O
O
C
Cl
C
Cl
O
O
OSiR3
O
H O
C
Cl
OSiR3
O
O
O
OH
O
O
C
Cl
C
HO
O
Cl
OSiR3
12.78
O
mCPBA
"down" bond
a.
O comes in from below.
X
"up" bond
OH
Br2
H2O
Br
b.
Br
NaH
Br+ comes in from below.
X
H2O attacks from the back side at the more substituted
C. This places the OH group axial, on an "up" bond.
12.79 The two OH’s are added to opposite faces of the C=C, so anti addition occurs.
H2O2
HCO2H
trans-2-butene
CH3
H
HO
S
C
OH
H
C
R
CH3
meso
H
H2O2
HCO2H
cis-2-butene
HO
C
HO
H
C
CH3
CH3
C
R R
CH3
OH
H
S
two enantiomers
H
OH
C
S
CH3
O
352
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 12–30
12.80
H
OSO3H
OH
OH2
H2O
HSO4–
OH
H
O
H2O
H
+ H2SO4
HSO4–
O
Cr
O
O
H2O
H
O
O
O
O
Cr
H
O
proton
transfer
O
O
Cr
Cr
O
OH
O
O
12.81
Sharpless
epoxidation
O
O
O
O
(+)-DET
OH
O
H
O
X
Y
Na+ –OH
inversion here
O
O
O
O
O
O
backside
attack
O
O
H OH
Sharpless epoxidation
deteremines the stereochemistry
of this C–O bond.
C6H5S H
NaOH
O
O
O
OH
C6H5S
O
O
O
SC6H5
OH
OH
H OH
O
OH
O
SC6H5
OH
Z
OH
OH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
353
Mass Spectrometry and Infrared Spectroscopy 13–1
Chapter 13 Mass Spectrometry and Infrared Spectroscopy
Chapter Review
Mass spectrometry (MS)
• Mass spectrometry measures the molecular weight of a compound (13.1A).
• The mass of the molecular ion (M) = the molecular weight of a compound. Except for isotope
peaks at M + 1 and M + 2, the molecular ion has the highest mass in a mass spectrum (13.1A).
• The base peak is the tallest peak in a mass spectrum (13.1A).
• A compound with an odd number of N atoms gives an odd molecular ion. A compound with an
even number of N atoms (including zero) gives an even molecular ion (13.1B).
• Organic chlorides show two peaks for the molecular ion (M and M + 2) in a 3:1 ratio (13.2).
• Organic bromides show two peaks for the molecular ion (M and M + 2) in a 1:1 ratio (13.2).
• The fragmentation of radical cations formed in a mass spectrometer gives lower molecular
weight fragments, often characteristic of a functional group (13.3).
• High-resolution mass spectrometry gives the molecular formula of a compound (13.4A).
Electromagnetic radiation
• The wavelength and frequency of electromagnetic radiation are inversely related by the
following equations: λ = c/ν or ν = c/λ (13.5).
• The energy of a photon is proportional to its frequency; the higher the frequency the higher the
energy: E = hν (13.5).
Infrared spectroscopy (IR, 13.6 and 13.7)
• Infrared spectroscopy identifies functional groups.
• IR absorptions are reported in wavenumbers:
wavenumber = ν~ = 1/λ
•
•
•
The functional group region from 4000–1500 cm–1 is the most useful region of an IR spectrum.
C–H, O–H, and N–H bonds absorb at high frequency, ≥ 2500 cm–1.
As bond strength increases, the wavenumber of an absorption increases; thus triple bonds absorb
at higher wavenumber than double bonds.
C C
~1650 cm–1
C C
~2250 cm–1
Increasing bond strength
Increasing ν~
•
The higher the percent s-character, the stronger the bond, and the higher the wavenumber of an
IR absorption.
C
C H
C H
H
Csp3–H
25% s-character
3000–2850 cm–1
Csp2–H
33% s-character
3150–3000 cm–1
Csp–H
50% s-character
3300 cm–1
Increasing percent s-character
Increasing ~
ν
354
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–2
Practice Test on Chapter Review
1. a. Which compound has a molecular ion at 112 and a peak at 1720 cm–1 in its IR spectrum?
O
O
1.
O
2.
3.
4. Compounds (1) and (2) both fit these criteria.
5. Compounds (1), (2), and (3) all fit these criteria.
b. Which of the following compounds has peaks at 3300, 3000, and 2250 cm–1 in its IR spectrum?
1.
C
CH
2.
C
N
3.
C
CCH3
4. Compounds (1) and (2) have these peaks in their IR spectra.
5. Compounds (1), (2), and (3) all contain these peaks in their IR spectra.
c. What is the base peak in a mass spectrum?
1. the peak due to the radical cation formed when a molecule loses an electron
2. the tallest peak in the mass spectrum
3. the peak due to the fragment with the largest m/z ratio
4. Both (1) and (2) describe the base peak.
5. Statements (1), (2), and (3) describe the base peak.
d. Which compounds are possible structures for a molecule that has a molecular ion at 150 in its
mass spectrum?
O
O
O
1.
O
2.
H
3.
CH3O
OCH3
4. Both (1) and (2) are possible structures.
5. Compounds (1), (2), and (3) are all possible structures.
e. Which compounds exhibit prominent M + 2 peaks in their mass spectra?
Br
1.
2.
3.
Cl
4. Both (1) and (2) show M + 2 peaks.
5. Compounds (1), (2), and (3) all show M + 2 peaks.
F
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
355
Mass Spectrometry and Infrared Spectroscopy 13–3
2. Answer each question with the number that corresponds to one of the following regions of an IR
spectrum.
1.
2.
3.
4.
a.
b.
c.
d.
e.
f.
4000–2500 cm–1
2500–2000 cm–1
2000–1500 cm–1
< 1500 cm–1
This region is called the fingerprint region of an IR spectrum.
The OH group of 1-propanol absorbs in this region.
The C=N of C6H5CH2CH=NCH3 absorbs in this region.
An unsymmetrical C≡C absorbs in this region.
An sp hybridized C–H bond absorbs in this region.
Ethyl benzoate (C6H5CO2CH2CH3) absorbs in all regions of the IR except this one.
3. Answer True (T) or False (F).
a. IR spectroscopy is useful for determining the molecular weight of a compound.
b. A C–H bond that absorbs at 3140 cm–1 is stronger than a C–H bond that absorbs at 2950 cm–1.
c. A compound with a molecular ion at 109 contains a N atom.
d. A compound with a base peak at 57 must contain a N atom.
e. 2-Butyne shows an IR absorption at 2250 cm–1.
f. 1-Propanol shows an IR absorption at 3200–3600 cm–1.
g. An ether shows no IR absorptions at 3200–3600 or 1700 cm–1.
h. In its mass spectrum, a compound that has a molecular ion with two peaks of approximately
equal intensity at 124 and 126 contains chlorine.
Answers to Practice Test
1.a. 4
b. 1
c. 2
d. 4
e. 4
2.a. 4
b. 1
c. 3
d. 2
e. 1
f. 2
3.a. F
b. T
c. T
d. F
e. F
f. T
g. T
h. F
356
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–4
Answers to Problems
13.1
The molecular ion formed from each compound is equal to its molecular weight.
a. C3H6O
molecular weight = 58
molecular ion (m/z) = 58
b. C10H20
molecular weight = 140
molecular ion (m/z) = 140
c. C8H8O2
molecular weight = 136
molecular ion (m/z) = 136
d. C10H15N
molecular weight = 149
molecular ion (m/z) = 149
13.2
Some possible formulas for each molecular ion:
a. Molecular ion at 72: C5H12, C4H8O, C3H4O2
b. Molecular ion at 100: C8H4, C7H16, C6H12O, C5H8O2
c. Molecular ion at 73: C4H11N, C2H7N3
13.3
To calculate the molecular ions you would expect for compounds with Cl, calculate the
molecular weight using each of the two most common isotopes of Cl (35Cl and 37Cl).
13.4
a. C4H935Cl = 92
C4H937Cl = 94
Two peaks in 3:1 ratio at m/z 92 and 94
c. C4H11N = 73
One peak at m/z 73
b. C3H7F = 62
One peak at m/z 62
d. C4H4N2 = 80
One peak at m/z 80
Convert the ball-and-stick model to a skeletal structure and determine the molecular formula.
Calculate the molecular weight using each of the two common isotopes for Br (79Br and 81Br).
Br
C6H11Br
13.5
C6H1179Br = 162
C6H1181Br = 164
Two peaks in a 1:1 ratio at m/z 162 and 164
After calculating the mass of the molecular ion, draw the structure and determine which C–C
bond is broken to form fragments of the appropriate mass-to-charge ratio.
e–
Cleave bond [1].
[1]
CH3 CH3
CH3 C
C
H
H
m/z = 100
CH3 C CH3
+
CH3CHCH2CH3
H
m/z = 43
CH2CH3
e–
Cleave bond [1].
CH3 C CH2CH3
H
m/z = 57
+
(CH3)2CH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
357
Mass Spectrometry and Infrared Spectroscopy 13–5
13.6
Break this bond.
CH3 H
CH3 C
CH3
C
e–
C CH3
H
CH3 H
+
CH3 C CH3
CH3
m/z = 114
H
CH3
C
C CH3
H
H
m/z = 57
This 3° carbocation is more
stable than others that can
form, and is therefore the
most abundant fragment.
13.7
a.
OH
Cleave bond [1].
+
CH3
C CH2 CH3
OH
H
m/z = 59
CH3 C CH2 CH3
H
[1]
OH
[2]
Cleave bond [2].
CH2CH3
+
CH3 C
H
m/z = 45
b.
OH
– H 2O
CH2=CHCH2CH3
CH3 C CH2 CH3
+
CH3CH=CHCH3
m/z = 56
H
m/z = 56
13.8
O
a.
C
O C
CH3CH2 C O
CH3CH2
(from cleavage of bond [2])
[1]
b.
CH3CH2CH2CH2CH2CH2OH
13.9
(from cleavage of bond [1])
[2]
CH3CH2CH2C O
c. CH3CH2CH2CHO
+ CH2OH
HC O
Use the exact mass values given in Table 13.1 to calculate the exact mass of each compound.
C7H5NO3
mass: 151.0270
C8H9NO2
C10H17N
mass: 151.0634
compound X
mass: 151.1362
358
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–6
13.10
CH3
benzene
C6H6 m/z = 78
toluene
C7H8 m/z = 92
CH3
CH3
p-xylene
C8H10 m/z = 106
GC–MS analysis:
Three peaks in the gas chromatogram.
Order of peaks: benzene, toluene, p-xylene,
in order of increasing bp.
Molecular ions observed in the three mass spectra:
78, 92, 106.
13.11 Wavelength and frequency are inversely proportional. The higher frequency light will
have a shorter wavelength.
a. Light having λ = 102 nm has a higher ν than light with λ = 104 nm.
b. Light having λ = 100 nm has a higher ν than light with λ = 100 μm.
c. Blue light has a higher ν than red light.
13.12 The energy of a photon is proportional to its frequency, and inversely proportional to its
wavelength.
a. Light having ν = 108 Hz is of higher energy than light having ν = 104 Hz.
b. Light having λ = 10 nm is of higher energy than light having λ = 1000 nm.
c. Blue light is of higher energy than red light.
13.13 Higher wavenumbers are proportional to higher frequencies and higher energies.
a. IR light with a wavenumber of 3000 cm–1 is higher in energy than IR light with a
wavenumber of 1500 cm–1.
b. IR light having λ = 10 μm is higher in energy than IR light having λ = 20 μm.
13.14 Stronger bonds absorb at a higher wavenumber. Bonds to lighter atoms (H versus D) absorb
at higher wavenumber.
a.
CH3 C C CH2CH3 or CH2 C(CH3)2
stronger bond
higher wavenumber
b.
CH3 H
or CH3 D
lighter atom H
higher wavenumber
13.15 Cyclopentane and 1-pentene are both composed of C–C and C–H bonds, but 1-pentene also has
a C=C bond. This difference will give the IR of 1-pentene an additional peak at 1650 cm–1
(for the C=C). 1-Pentene will also show C–H absorptions for sp2 hybridized C–H bonds at
3150–3000 cm–1.
13.16 Look at the functional groups in each compound below to explain how each IR is different.
O
CH3
C
A
CH3
C=O peak at ~1700 cm–1
OH
CH3OCH CH2
B
C=C peak at 1650 cm–1
Csp2–H at 3150–3000 cm–1
C
O–H peak at 3200–3600 cm–1
13.17 a. Compound A has peaks at ~3150 (sp2 hybridized C–H), 3000–2850 (sp3 hybridized C–H),
and 1650 (C=C) cm–1.
b. Compound B has a peak at 3000–2850 (sp3 hybridized C–H) cm–1.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
359
Mass Spectrometry and Infrared Spectroscopy 13–7
13.18 All compounds show an absorption at 3000–2850 cm–1 due to the sp3 hybridized C–H bonds.
Additional peaks in the functional group region for each compound are shown.
a.
d.
O
C=O bond at ~1700 cm–1
no additional peaks
O
e.
b.
CH3O
N
OH
O–H bond at 3600–3200
H
HO
cm–1
O–H at 3600–3200 cm–1
N–H at 3500–3200 cm–1
Csp2–H at 3150–3000 cm–1
C=O at ~1700 cm–1
C=C at 1650 cm–1
aromatic ring at 1600, 1500 cm–1
c.
Csp2–H at 3150–3000 cm–1
C=C bond at 1650 cm–1
13.19
O
OH
Csp2–H at 3150–3000 cm–1
Csp3–H at 3000–2850 cm–1
C=O at ~1700 cm–1
C=C at 1650 cm–1
O–H above 3000 cm–1
The OH of a COOH is much broader
than the OH of an alcohol and occurs
at 3500–2500 cm–1 (see Chapter 19).
13.20 Possible structures are (a) CH3COOCH2CH3 and (c) CH3CH2COOCH3. Compounds (b) and
(d) also have an OH group that would give a strong absorption at ~3600–3200 cm–1, which is
absent in the IR spectrum of X, thus excluding them as possibilities.
13.21
a. Hydrocarbon with a molecular ion at m/z = 68
IR absorptions at 3310 cm–1 = Csp–H bond
3000–2850 cm–1 = Csp3–H bonds
2120 cm–1 = C≡C bond
Molecular formula: C5H8
H C C CH2CH2CH3
or H C C CHCH3
CH3
b. Compound with C, H, and O with a molecular
ion at m/z = 60
IR absorptions at 3600–3200 cm–1 = O–H bond
3000–2850 cm–1 = Csp3–H bonds
Molecular formula: C3H8O
CH3CH2CH2 O H
or
CH3CH O H
CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–8
13.22
Csp2–H at 3000–3150 cm–1
Csp3–H at 2850–3000 cm–1
C=C at 1650 cm–1
a.
b.
OH
O–H at 3200–3600 cm–1
Csp2–H at 3000–3150 cm–1
Csp3–H at 2850–3000 cm–1
C=C at 1650 cm–1
13.23
Cleave bond [1].
[1]
CH3
[2]
CH3
+
CH3CH2 C CH2CH2CH2CH3 m/z = 127
CH2CH3
CH3CH2 C CH2CH2CH2CH3
CH3
CH2CH3
[3]
Cleave bond [2].
m/z = 142
CH2CH2CH2CH3
+
m/z = 85
CH3CH2 C
CH2CH3
CH3
Cleave bond [3].
CH2CH3
+
C CH2CH2CH2CH3
m/z = 113
CH2CH3
13.24
c.
a.
e.
(CH3)3CCH(Br)CH(CH3)2
O
molecular formula: C5H10O
molecular ion (m/z): 86
molecular formula: C6H6
molecular ion (m/z): 78
d.
b.
molecular formula: C10H16
molecular ion (m/z): 136
molecular formula: C8H17Br
molecular ions (m/z): 192, 194
Cl
molecular formula: C5H11Cl
molecular ions (m/z): 106, 108
13.25
O
CH2CH2CH3
C9H12
molecular weight = 120
C
CH2CH3
C9H10O
molecular weight = 134
13.26 Examples are given for each molecular ion.
a. molecular ion 102: C8H6, C6H14O, C5H10O2, C5H14N2
b. molecular ion 98: C8H2, C7H14, C6H10O, C5H6O2
c. molecular ion 119: C8H9N, C6H5N3
d. molecular ion 74: C6H2, C4H10O, C3H6O2
OCH2CH3
C8H10O
molecular weight = 122
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
361
Mass Spectrometry and Infrared Spectroscopy 13–9
13.27 Likely molecular formula, C8H16 (one degree of unsaturation—one ring or one π bond).
Four structures with m/z = 112
13.28
CH3
CH3 CH CO2CH3
Cl
B
C4H7O2Cl
molecular weight: 122, 124
should show 2 peaks for the
molecular ion with a 3:1 ratio
C
C8H10O
OCH3
A
molecular weight: 122
Mass spectrum [1]
CH3CH2CH2Br
Mass spectrum [2]
C3H7Br
molecular weight: 122, 124
should show 2 peaks for the
molecular ion with a 1:1 ratio
Mass spectrum [3]
13.29
H2
or
or
or
or
Pd-C
Possible structures
C7H12
(exact mass 96.0940)
13.30
[1] [2]
a.
OH
OH
OH
(from cleavage of bond [1])
[1]
O
[2]
O
(from cleavage of bond [2])
O
b.
(from cleavage of bond [1])
(from cleavage of bond [2])
[1] [2]
c.
O
O
(from cleavage of bond [1])
O
(from cleavage of bond [2])
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–10
13.31
[1]
a.
– H2O
C6H5 CH CH2
H H
Cleave bond [1].
H H
m/z = 104
H
e–
C6H5 C C OH
(resonance-stabilized
carbocation)
C6H5 C
m/z = 122
H
m/z = 91
e–
CH2 C
Cleave bond [1].
[3]
b.
CH3 H H
CH2 C C C
– H2O
[1]
CH3 H H
CH2 C C C OH
H
[2]
m/z = 68
H H
H H
C C OH
H H
m/z = 71
e–
Cleave bond [2].
m/z = 86
e–
Cleave bond [3].
CH2 C CH3
m/z = 41
CH2 OH
m/z = 31
13.32
Ketone A
Ketone B
O
O
m/z = 128
m/z = 128
α cleavage
α cleavage
O
CH2CH3
+
O
+
CH3
+
+
m/z = 99
m/z = 113
This is ketone A since α
cleavage gives a fragment
with m/z of 99.
This is ketone B since α
cleavage gives a fragment
with m/z of 113.
13.33 One possible structure is drawn for each set of data:
a. A compound that contains a benzene ring
and has a molecular ion at m/z = 107
c. A compound that contains a carbonyl group
and gives a molecular ion at m/z = 114
O
NH2
CH3
CH2CH2CH2CH2CH3
C7H14O
C7H9N
b. A hydrocarbon that contains only sp3 hybridized
carbons and a molecular ion at m/z = 84
C
d. A compound that contains C, H, N, and O and has
an exact mass for the molecular ion at 101.0841
O
CH3
C6H12
C
NHCH2CH2CH3
C5H11NO
13.34 Use the values given in Table 13.1 to calculate the exact mass of each compound. C8H11NO2
(exact mass 153.0790) is the correct molecular formula.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Mass Spectrometry and Infrared Spectroscopy 13–11
13.35 Two isomers such as CH2=CHCH2CH2CH2CH3 and (CH3)2C=CHCH2CH3 have the same
molecular formulas and therefore give the same exact mass, so they are not distinguishable
by their exact mass spectra.
13.36 Alpha cleavage of a 1° alcohol (RCH2OH) forms an alkyl radical (R•) and a resonancestabilized carbocation with m/z = 31.
CH2OH
m/z = 31
CH2 OH
resonance-stabilized carbocation
13.37 An ether fragments by α cleavage because the resulting carbocation is resonance stabilized.
[1]
(CH3)2CH
CH2
Cleave
bond [1].
[2]
O
CH2 CH3
Cleave
bond [2].
(CH3)2CH
CH2
O
CH2 CH3
CH2 O
CH2 CH3
resonance-stabilized carbocations
(CH3)2CH
CH3
CH2
O
CH2
(CH3)2CH
CH2
O
CH2
13.38
a. (CH3)2C O or (CH3)2CH OH
stronger bond
higher ~
ν absorption
b. (CH3)2C
NCH3
or (CH3)2CH NCH3
H or
c.
stronger bond
higher ~
ν absorption
H
stronger bond
higher ~
ν absorption
13.39 Locate the functional groups in each compound. Use Table 13.2 to determine what IR
absorptions each would have.
a.
C CH
Csp–H at 3300 cm–1
Csp3–H at 2850–3000 cm–1
C–C triple bond at 2250 cm–1
c.
O
Csp3–H at 2850–3000 cm–1
C=O at 1700 cm–1
O
O–H at > 3000 cm–1
Csp2–H at 3000–3150 cm–1
C=O at ~1700 cm–1
phenyl group at 1600, 1500 cm–1
The OH of the RCOOH is even broader than
the OH of an alcohol (3500–2500 cm–1), as
we will learn in Chapter 19.
C
b.
OH
O–H at 3200–3600 cm–1
Csp3–H at 2850–3000 cm–1
d.
OH
13.40
O
a.
and
C=C bond
1650 cm–1
Csp2–H at 3150–3000 cm–1
HC CCH2CH2CH3
C≡C bond
2250 cm–1
Csp–H at 3300 cm–1
b. CH CH C OH
3
2
O
and
CH3
C
OCH3
O–H bond
no O–H bond
> 3000 cm–1
[See note on OH in Answer 13.39d.]
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–12
O
c.
CH3CH2
and
C
CH3
O–H bond
3200–3600 cm–1
Csp2–H at 3150–3000 cm–1
C=C bond at 1650 cm–1
C=O bond
1700 cm–1
d.
OCH3
OCH3
e.
CH3CH CHCH2OH
f.
CH3(CH2)5
no C=O bond
C
CH3CH2C CH
Csp–H bond
3300 cm–1
C≡C bond at ~2250 cm–1
no C≡C absorption
due to symmetry
HC CCH2N(CH2CH3)2
O
and
and
CH3C CCH3
and
CH3(CH2)5C N
Csp–H bond
3300 cm–1
OCH3
C=O bond
~1700 cm–1
13.41 The IR absorptions above 1500 cm–1 are different for each of the narcotics.
CH3
HO
C
O
CH3O
O
O
O
H
H
HO
O
N
CH3
CH3
morphine
C
O
H
H
O
heroin
• O–H bond at
~3200–3600 cm–1
• no C=O bond
• C=O bond at
~1700 cm–1
• no O–H bond
N
CH3
H
N
OH
O
CH3
oxycodone
• C=O bond at
~1700 cm–1
• O–H bond at
~3200–3600 cm–1
13.42 Look for a change in functional groups from starting material to product to see how IR
could be used to determine when the reaction is complete.
Loss of the C=C will be visible in the IR
by disappearance of the peak at 1650 cm–1.
H2
a.
Pd
OH
O
PCC
b.
[1] O3
c.
CH3
O
[2] CH3SCH3
d.
OH
[1] NaH
[2] CH3Br
Loss of the O–H group will be visible in the IR
by disappearance of the peak at 3200–3600 cm–1
and appearance of the C=O at ~1700 cm–1.
O C
CH3
O
Loss of the C=C will be visible in the IR
by disappearance of the peak at 1650 cm–1
and appearance of the C=O at ~1700 cm–1.
Loss of the O–H will be visible in the IR
by disappearance of the peak at 3200–3600 cm–1.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
365
Mass Spectrometry and Infrared Spectroscopy 13–13
13.43 In addition to Csp3–H at ~3000–2850 cm–1:
Spectrum [1]:
Spectrum [2]:
CH2=C(CH3)CH2CH2CH2CH3 (B)
C=C peak at 1650 cm–1
Csp2–H at ~3150 cm–1
(CH3CH2)3COH (F)
OH at 3600–3200 cm–1
Spectrum [3]:
Spectrum [4]:
(CH3)2CHOCH(CH3)2 (D)
No other peaks above 1500 cm–1
CH(CH3)2
(C)
Csp –H at ~3150 cm–1
Phenyl peaks at 1600 and 1500 cm–1
2
Spectrum [5]:
Spectrum [6]:
CH3CH2CH2CH2COOH (A)
OH at ~3500–2500 cm–1
C=O at ~1700 cm–1
CH3COOC(CH3)3 (E)
C=O at ~1700 cm–1
13.44
a. Compound with a molecular ion at m/z = 72
IR absorption at 1725 cm–1 = C=O bond
Molecular formula: C4H8O
c. Compound with a molecular ion at m/z = 74
IR absorption at 3600–3200 cm–1 = O–H bond
Molecular formula: C4H10O
OH
O
or
b. Compound with a molecular ion at m/z = 55
The odd molecular ion means an odd number
of N’s present. Molecular formula: C3H5N
IR absorption at 2250 cm–1 = C≡N bond
OH
OH
OH
or
CH3 CH 2C N
13.45
Chiral hydrocarbon with a molecular ion at m/z = 82
Molecular formula: C6H10
IR absorptions at 3300 cm–1 = Csp–H bond
3000–2850 cm–1 = Csp3–H bonds
2250 cm–1 = C≡C bond
*
HC CCHCH2CH3
stereogenic center
CH3
Two possible enantiomers: CH3
C
CH3CH2
C CH
H
CH3
or
C
HC C
H
CH2CH3
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–14
13.46 The chiral compound Y has a strong absorption at 2970–2840 cm–1 in its IR spectrum due to
sp3 hybridized C–H bonds. The two peaks of equal intensity at 136 and 138 indicate the
presence of a Br atom. The molecular formula is C4H9Br. Only one constitutional isomer of
this molecular formula has a stereogenic center:
Br
Br
and
Y=
two possible enantiomers
13.47 The molecular ion of 192 suggests C12H16O2 as a possible molecular formula. IR absorption
at 1721 cm–1 is due to a C=O, and the absorptions around 3000 cm–1 are due to Csp2–H and
Csp3–H. The compound is an ester, formed in the following manner.
O
O
OH
+ NaOH
O
I
O
O
H
C12H16O2
13.48
O
H
CH3
Zn(Hg)
HCl
m/z = 92; molecular formula C7H8
IR absorptions at:
3150–2950 cm–1 = Csp3–H and Csp2–H bonds
1605 cm–1 and 1496 cm–1 due to phenyl group
Z
13.49
H
O
Br
O
H–Br
H
+ Br
H Br
C CH2
CH3
H
CH3
CH3
HBr
B
O
CH3
C
H O
CH2
H
Br
Br
H2O
13.50 The molecular ion of 144 suggests C8H16O2 as a possible molecular formula for X. The IR
absorption at 1739 cm–1 is due to a C=O, and the absorptions at less than 3000 cm–1 are due to
Csp3–H.
OH
O
HO
O
O
2-methyl-1-propanol
C8H16O2
X
possible structure
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Mass Spectrometry and Infrared Spectroscopy 13–15
13.51
O
fragments:
J
C6H12O
m/z = 100
IR absorption at 2962 cm–1 = Csp3–H bonds
1718 cm–1 = C=O bond
O
O
α cleavage product
α cleavage product
m/z = 43
m/z = 85
The fragment at m/z = 57 could be due to (C4H9)+ or (C3H5O)+.
13.52
O
C
NH2
CHO
K
L
C7 H6 O
m/z = 106
C7 H9 N
m/z = 107
IR absorptions at 3373 and 3290 cm–1 = N–H
3062 cm–1 = Csp2–H bonds
2920 cm–1 = Csp3–H bonds
1600 cm–1 = benzene ring
m/z = 105
m/z = 77
IR absorption at 3068 cm–1 = Csp2–H bonds on ring
2850 cm–1 = Csp3–H bond
–1
2820 cm and 2736 cm–1 = C–H of RCHO (Appendix E)
1703 cm–1 = C=O bond
1600 cm–1 = aromatic ring
The odd molecular ion indicates
the presence of a N atom.
13.53
Possible structures of P:
Cl
Cl2
CH3O
CH3O
Cl
or
Cl
or
CH3O
CH3O
FeCl3
C7H7ClO
m/z = 142, 144
IR absorption at 3096–2837 cm–1 = Csp3–H bonds and Csp2–H bonds
1582 cm–1 and 1494 cm–1 = benzene ring
The peak at M + 2 shows the presence of Cl or Br. Since
Cl2 is a reactant, the compound presumably contains Cl.
13.54 The mass spectrum has a molecular ion at 71. The odd mass suggests the presence of an odd
number of N atoms; likely formula, C4H9N. The IR absorption at ~3300 cm–1 is due to N–H
and the 3000–2850 cm–1 is due to sp3 hybridized C–H bonds.
H
N
Br
H
N
H
Na+ H
Br
N H
W
+ Na+ + H2
+
Na+ Br– + H2
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 13–16
13.55 Because the carbonyl absorption of an amide is at lower wavenumber than the carbonyl
absorption of an ester, the C=O of the amide must be weaker and have more single bond
character. This can be explained by resonance. Although both an ester and amide are
resonance stabilized, the N atom of the amide is more basic, making it more willing to donate
its electron pair.
R
O
O
O
C
C
C
NH2
R
NH2
R
amide
O
OR
R
C
OR
ester
This form contributes more to
the hybrid, so there is more
single bond character.
This form contributes
less to the hybrid.
Since the amide carbonyl has more single bond character, the bond is weaker and it absorbs
at lower wavenumber.
13.56 The α,β-unsaturated carbonyl compound has three resonance structures, two of which place a
single bond between the C and O atoms. This means that the C–O bond has partial single
bond character, making it weaker than a regular C=O bond, and moving the absorption to
lower wavenumber.
+
O
O
O
+
three resonance structures for 2-cyclohexenone
13.57 If a ketone carbonyl absorbs at lower wavenumber than an aldehyde carbonyl, the ketone
carbonyl is weaker and has more single bond character. This can be explained by the fact
that R groups are electron donating and stabilize an adjacent (+) charge.
R
O
O
O
C
C
C
R
R
R
R
O
H
R
C
H
ketone
The two R groups stabilize the (+)
charge, so this form contributes more to
the hybrid than the charge-separated
resonance form of an aldehyde.
One R group stabilizes
the (+) charge less.
As a result, the charge-separated resonance form of a ketone, which contains a C–O single
bond, contributes more to the hybrid of a ketone, making the C=O weaker and shifting the
absorption to lower wavenumber.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Mass Spectrometry and Infrared Spectroscopy 13–17
13.58
PCC
a, b.
H
O
PCC
NaOCOCH3
H
A
O
+
PCC
+
O Cr OH
O
H
OH
O
+
B
+
OH
OH
H
+
+H B
B
a, c.
+
O
H
O
HO
isopulegone
+
+
H B
+ Cr4+
H B
O
isopulegone
O
B
citronellol
molecular ion at 154
C10H18O
IR at 1730 cm–1 (C=O)
+ Cr4+ + H B+
O
O Cr OH
OH
+
H2O
O
O
H OH
B
OH
B
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
371
Nuclear Magnetic Resonance Spectroscopy 14–1
Chapter 14 Nuclear Magnetic Resonance Spectroscopy
Chapter Review
1
H NMR spectroscopy
[1] The number of signals equals the number of different types of protons (14.2).
CH3 O CH3
Ha
CH3CH2
CH3 O CH2CH3
Cl
Ha
Ha Hb
Ha
all equivalent H's
1 NMR signal
2 types of H's
2 NMR signals
Hb Hc
3 types of H's
3 NMR signals
[2] The position of a signal (its chemical shift) is determined by shielding and deshielding effects.
• Shielding shifts an absorption upfield; deshielding shifts an absorption downfield.
• Electronegative atoms withdraw electron density, deshield a nucleus, and shift an absorption
downfield (14.3).
C H
•
This proton is shielded.
Its absorption is upfield,
0.9–2 ppm.
C H
X
This proton is deshielded.
Its absorption is farther downfield,
2.5–4 ppm.
Loosely held π electrons can either shield or deshield a nucleus. Protons on benzene rings
and double bonds are deshielded and absorb downfield, whereas protons on triple bonds are
shielded and absorb upfield (14.4).
H
C
C H
H
deshielded H
downfield absorption
shielded H
upfield absorption
[3] The area under an NMR signal is proportional to the number of absorbing protons (14.5).
[4] Spin–spin splitting tells about nearby nonequivalent protons (14.6–14.8).
• Equivalent protons do not split each other’s signals.
• A set of n nonequivalent protons on the same carbon or adjacent carbons split an NMR signal
into n + 1 peaks.
• OH and NH protons do not cause splitting (14.9).
• When an absorbing proton has two sets of nearby nonequivalent protons that are equivalent
to each other, use the n + 1 rule to determine splitting.
• When an absorbing proton has two sets of nearby nonequivalent protons that are not
equivalent to each other, the number of peaks in the NMR signal = (n + 1)(m + 1). In flexible
alkyl chains, peak overlap often occurs, resulting in n + m + 1 peaks in an NMR signal.
13
C NMR spectroscopy (14.11)
[1] The number of signals equals the number of different types of carbon atoms. All signals are
single lines.
[2] The relative position of 13C signals is determined by shielding and deshielding effects.
• Carbons that are sp3 hybridized are shielded and absorb upfield.
• Electronegative elements (N, O, and X) shift absorptions downfield.
• The carbons of alkenes and benzene rings absorb downfield.
• Carbonyl carbons are highly deshielded, and absorb farther downfield than other carbon types.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–2
Practice Test on Chapter Review
1. a. Which of the following statements is true about 1H NMR absorptions?
1. A signal that occurs at 1800 Hz on a 300 MHz NMR spectrometer occurs at 3000 Hz on a
500 MHz NMR spectrometer.
2. A signal that occurs at 3.3 ppm on a 60 MHz NMR absorbs at 198 Hz upfield from TMS.
3. A signal that occurs at 600 Hz is downfield from a signal that occurs at 800 Hz.
4. Statements (1) and (2) are both true.
5. Statements (1), (2), and (3) are all true.
b. Which of the following statements is true about 1H NMR spectroscopy?
1. Electronegative elements shield a nucleus so an absorption shifts downfield.
2. A triplet is due to a proton that has four adjacent nonequivalent protons.
3. Circulating π electrons create a magnetic field that reinforces the applied field in the
vicinity of the protons in benzene.
4. Statements (1) and (2) are both true.
5. Statements (1), (2), and (3) are all true.
2. How many different types of protons does each of the following molecules contain?
a.
c.
(CH3)2C
e.
CHCH3
g.
Cl
Cl
O
b.
O
Cl
d.
Br
f.
Cl
Cl
Br
CH3
CH2CH3
C C
H
CH3
h.
Cl H
3. Into how many peaks will each of the circled protons be split in a proton NMR spectrum?
H
O
H
a.
c.
e.
O
Br
H
H
d.
b.
HO
H
H
OH
f.
CH3
g.
O
O
H
H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
373
Nuclear Magnetic Resonance Spectroscopy 14–3
4. How many lines are presents in the 13C NMR spectrum of each compound?
a.
c.
CH3O
Cl
e.
O
O
b. CH3CH2CH2CH2CH2CH2CH2CH3
d.
f.
O
5. With reference to the 1H NMR absorptions in the following compound, (a) which proton absorbs
farthest upfield; (b) which proton absorbs farthest downfield?
Ha
Hb
O
O
Hd
Hc
6. With reference to the 13C NMR absorptions in the following compound, (a) which carbon
absorbs farthest downfield; (b) which carbon absorbs farthest upfield?
Cb
Ca
O
F
Cc
Cd
Answers to Practice Test
1. a. 1
b. 3
2. a. 5
b. 5
c. 4
d. 5
e. 4
f. 5
g. 3
h. 9
3. a. 8
b. 3
c. 7
d. 4
e. 4
f. 3
g. 8
4. a. 6
b. 4
c. 6
d. 4
e. 4
f. 5
5. a. Hb
b. Hc
6. a. Cc
b. Ca
Answers to Problems
14.1
Use the formula δ = [observed chemical shift (Hz)/ν of the NMR (MHz)] to calculate the
chemical shifts.
a. CH3 protons:
δ = [1715 Hz] / [500 MHz]
= 3.43 ppm
14.2
OH proton:
δ = [1830 Hz] / [500 MHz]
= 3.66 ppm
b. The positive direction of the δ scale is downfield
from TMS. The CH3 protons absorb upfield
from the OH proton.
Calculate the chemical shifts as in Answer 14.1.
a. one signal:
δ = [1017 Hz] / [300 MHz]
= 3.39 ppm
second signal:
δ = [1065 Hz] / [300 MHz]
= 3.55 ppm
b. one signal:
3.39 = [x Hz] / [500 MHz]
x = 1695 Hz
second signal:
3.55 = [x Hz] / [500 MHz]
x = 1775 Hz
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–4
14.3
To determine if two H’s are equivalent replace each by an atom X. If this yields the same
compound or mirror images, the two H’s are equivalent. Each kind of H will give one NMR
signal.
2 kinds of H's
2 NMR signals
e. CH3CH2CO2CH2CH3
4 kinds of H's
4 NMR signals
g. CH3(CH2)7Cl
8 kinds of H's
8 NMR signals
d. (CH3)2CHCH(CH3)2
2 kinds of H's
2 NMR signals
f. CH3OCH2CH(CH3)2
4 kinds of H's
4 NMR signals
h. CH3CH2CH2OH
4 kinds of H's
4 NMR signals
c. CH3CH2CH2CH3
a. CH3CH3
1 kind of H
1 NMR signal
b. CH3CH2CH3
2 kinds of H's
2 NMR signals
14.4
Draw in all of the H’s and compare them. If two H’s are cis and trans to the same group,
they are equivalent.
H
H
CH3
Ha
a. 4 identical H's
H
Hb
Ha
b.
CH3
Hb
H
Hc
H
Ha
CH3
Ha
H
Ha
H
Hb
H
c.
H
H
H
2 NMR signals
14.5
CH3
Hb
CH3
Hc
Hb
H
CH3
Hd
Hc
3 NMR signals
4 NMR signals
If replacement of H with X yields enantiomers, the protons are enantiotopic.
If replacement of H with X yields diastereomers, the protons are diastereotopic. In general, if the
compound has one stereogenic center, the protons in a CH2 group are diastereotopic.
a.
CH3CH2CH2CH2CH2CH3
c.
replacement
of H with X
CH3
C
CH3CH2CH2CH2
CH3
H
X
C
CH3CH2CH2CH2
Pick one configuration at the
existing stereogenic center.
X
H
CH3
C
HO
enantiomers =
enantiotopic H's
b.
CH3CH(OH)CH2CH2CH3
H
H
X
replacement
of H with X
CH3
C
C
CH2CH3
HO
H
X
H
C
CH2CH3
diastereomers =
diastereotopic H's
CH3CH2CH2CH2CH3
replacement
of H with X
CH3CH2CHCH2CH3
X
14.6
no stereogenic center
neither enantiotopic nor diastereotopic
The two protons of a CH2 group are different from each other if the compound has one
stereogenic center. Replace one proton with X and compare the products.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
375
Nuclear Magnetic Resonance Spectroscopy 14–5
a. The stereogenic center makes the H's in
the CH2 group diastereotopic and
therefore different from each other.
stereogenic
center
Cl H Hb
Hd
CH3 C C CH3
Hc
c.
b.
Hc
H H
Hd
Cl C C O CH3
He
Ha
H H He
Hb Hc
stereogenic
center
H H H Hd
Ha CH3 C C C CH3
stereogenic
center
Ha
H CH3 Hb
Hf Hg
5 NMR signals
7 NMR signals
5 NMR signals
14.7
He
Br H H
Decreased electron density deshields a nucleus and the absorption goes downfield.
Absorption also shifts downfield with increasing alkyl substitution.
a. FCH2CH2CH2Cl
F is more electronegative than Cl. The CH2 group
adjacent to the F is more deshielded and the H's will
absorb farther downfield.
b. CH3CH2CH2CH2OCH3
The CH2 group adjacent to the O will absorb farther
downfield because it is closer to the electronegative
O atom.
c. CH3OC(CH3)3
The CH3 group bonded to the O atom
will absorb farther downfield.
14.8
O
a. ClCH2CH2CH2Br
b. CH3OCH2OC(CH3)3
Ha H b H c
3 types of protons:
Hb < Hc < Ha
c. CH3
Ha Hb
Hc
3 types of protons:
Hc < Ha < Hb
C
CH2CH3
Ha
Hb Hc
3 types of protons:
Hc < Ha < Hb
14.9
O
a.
CH3 C C H
CH3CH CH2
CH3CH2CH3
Hc
Hb
Ha
Hc protons are shielded because they are
bonded to an sp3 C.
Ha is shielded because it is bonded to an sp C.
Hb protons are deshielded because they are
bonded to an sp2 C.
Hc < Ha < Hb
b.
CH3
Ha
C
OCH2CH3
Hb Hc
Hc protons are shielded because they are bonded to an sp3
C.
Ha protons are deshielded slightly because the CH3 group
is bonded to a C=O.
Hb protons are deshielded because the CH2 group is
bonded to an O atom.
Hc < Ha < Hb
14.10 An integration ratio of 2:3 means that there are two types of hydrogens in the compound, and
that the ratio of one type to another type is 2:3.
a. CH3CH2Cl
2 types of H's
3:2 - YES
b. CH3CH2CH3
2 types of H's
6:2 or 3:1 - no
c. CH3CH2OCH2CH3
2 types of H's
6:4 or 3:2 - YES
d. CH3OCH2CH2OCH3
2 types of H's
6:4 or 3:2 - YES
376
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–6
14.11 To determine how many protons give rise to each signal:
• Divide the total number of integration units by the total number of protons to find the
number of units per H.
• Divide each integration value by this value and round to the nearest whole number.
C8H14O2
total number of integration units = 14 + 12 + 44 = 70 units
total number of protons = 14 H's
70 units/14 H's = 5 units per H
Signal [A] = 14/5 = 3 H
Signal [B] = 12/5 = 2 H
Signal [C] = 44/5 = 9 H
14.12
downfield absorption
closer to O
downfield absorption
closer to O
CH3O2CCH2CH2CO2CH3
CH3CO2CH2CH2O2CCH3
A
ratio of absorbing signals 2:3
Signal [1] = 4 H = 2.64
Signal [2] = 6 H = 3.69
6 H's with
downfield absorption
B
ratio of absorbing signals 3:2
Signal [1] = 6 H = 2.09
Signal [2] = 4 H = 4.27
4 H's with
downfield absorption
14.13 To determine the splitting pattern for a molecule:
• Determine the number of different kinds of protons.
• Nonequivalent protons on the same C or adjacent C’s split each other.
• Apply the n + 1 rule.
O
a.
CH3CH2
C
O
Cl
c.
Ha Hb
Ha: 3 peaks - triplet
Hb: 4 peaks - quartet
H
b.
Br
Ha
Ha: 2 peaks - doublet
Hb: 4 peaks - quartet
C
Ha
CH3CH2
CH2CH2Br
e.
d.
Ha
H
Ha
H
Hb
Ha: 2 peaks - doublet
Hb: 2 peaks - doublet
Cl
f.
C C
Br
H
C C
CH3
Hb Hc
Ha: 1 peak - singlet
Hb: 3 peaks - triplet
Hc: 3 peaks - triplet
Hb
CH3 C Br
CH3
H
ClCH2CH(OCH3)2
Hb
Ha: 2 peaks - doublet
Hb: 2 peaks - doublet
Ha Hb
Ha: 2 peaks - doublet
Hb: 3 peaks - triplet
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
377
Nuclear Magnetic Resonance Spectroscopy 14–7
14.14 Use the directions from Answer 14.13.
Ha
Ha
Ha O
a.
O
Hb
c. CH3 C H
Ha: quartet
Hb: triplet
2 NMR signals
Hb
Hb
Ha: doublet
Hb: quartet
2 NMR signals
O
b. CH3
C
OCH2CH2OCH3 Ha and Hd are both singlets.
Ha
Hb Hc
Hb: triplet
Hc: triplet
4 NMR signals
Hd
Ha: triplet
Hb: doublet
Hc
Hc: singlet
3 NMR signals
d. Cl2CHCH2CO2CH3
Ha Hb
14.15 CH3CH2Cl
3 units
2 units
3
There are two kinds of protons, and they can split
each other. The CH3 signal will be split by the CH2
protons into 2 + 1 = 3 peaks. It will be upfield from
the CH2 protons since it is farther from the Cl. The
CH2 signal will be split by the CH3 protons into
3 + 1 = 4 peaks. It will be downfield from the CH3
protons since the CH2 protons are closer to the Cl.
The ratio of integration units will be 3:2.
1
chemical shift (ppm)
14.16
Cl
a.
(CH3)2CHCO2CH3
b.
CH3CH2CH2CH2CH3
c.
CH2Br
Ha
C C
d.
H
Hb
Ha H
Ha Hb Hc
Ha: split by 2 H's
Ha: split by 1 H
3 peaks
2 peaks
Hc: split by 4 equivalent H's
Hb: split by 2 sets of H's
5 peaks
(1 + 1)(2 + 1) = 6 peaks
Hb: split by 2 sets of H's
(3 + 1)(2 + 1) = 12 peaks (maximum)
Since this is a flexible alkyl chain, the signal
due to Hb will have peak overlap, and
3 + 2 + 1 = 6 peaks will likely be visible.
split by 6 equivalent H's
6 + 1 = 7 peaks
H
H
Hb
(all H's)
Br
H Hc
Ha: split by 2 different H's
(1+1)(1+1) = 4 peaks
Hb: split by 2 different H's
(1+1)(1+1) = 4 peaks
Hc: split by 2 different H's
(1+1)(1+1) = 4 peaks
C C
14.17
CH3CH2
O
a. CH3OCH2CH3
Ha
H b Hc
Ha: singlet at ~3 ppm
Hb: quartet at ~3.5 ppm
Hc: triplet at ~1 ppm
b.
CH3CH2
H a Hb
C
c. CH3OCH2CH2CH2OCH3
OCH(CH3)2
Hc Hd
Ha: triplet at ~1 ppm
Hb: quartet at ~2 ppm
Hc: septet at ~3.5 ppm
Hd: doublet at ~1 ppm
Ha
Hb Hc Hb Ha
Ha: singlet at ~3 ppm
Hb: triplet at ~3.5 ppm
Hc: quintet at ~1.5 ppm
d.
Ha
Hb
CH2CH3
Ha
C C
H
H
Hc
Ha: triplet at ~1 ppm
Hb: multiplet (8 peaks) at ~2.5 ppm
Hc: triplet at ~5 ppm
378
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–8
14.18
Hb
Splitting diagram for Hb
Hb
Cl
Jab = 13.1 Hz
Cl
Ha
Hc
1 trans Ha proton splits Hb into
1 + 1 = 2 peaks
a doublet
Hc
2 Hc protons
2 Hc protons split Hb into
2 + 1 = 3 peaks
Now it's a doublet of triplets.
trans-1,3-dichloropropene
Jbc = 7.2 Hz
14.19
Cl
Hb
H
ClCH2
C3H4Cl2
Cl
CH3
Ha
A
Ha: 1.75 ppm, doublet, 3 H, J = 6.9 Hz
Hb: 5.89 ppm, quartet, 1 H, J = 6.9 Hz
H
doublet
singlet Cl
H
doublet
B
signal at 4.16 ppm, singlet, 2 H
signal at 5.42 ppm, doublet, 1 H, J = 1.9 Hz
signal at 5.59 ppm, doublet, 1 H, J = 1.9 Hz
14.20 Remember that OH (or NH) protons do not split other signals, and are not split by adjacent
protons.
triplet
singlet
doublet
triplet
b. CH3CH2CH2OH
a. (CH3)3CCH2OH
singlet
c. (CH3)2CHNH2
singlet
singlet
singlet
3 NMR signals
12 peaks (maximum)
6 peaks (more likely,
resulting from peak overlap)
4 NMR signals
7 peaks
3 NMR signals
14.21
A
Ha: doublet at ~1.4 due to the CH3 group, split into two peaks
by one adjacent nonequivalent H (Hc).
Hb: singlet at ~2.7 due to the OH group. OH protons are not
split by nor do they split adjacent protons.
Hc: quartet at ~4.7 due to the CH group, split into four peaks
by the adjacent CH3 group.
Hd: Five protons on the benzene ring.
O
Hd
H
Ha: one adjacent nonequivalent H, so two peaks
Hb: one adjacent nonequivalent H, so two peaks
Hc: Hc is located on a N atom so there is no splitting and it
appears as one peak.
Hd: Hd has one nonequivalent H on the same carbon and one
on an adjacent carbon, so it is split into (1 + 1)(1 + 1) = 4
peaks (a doublet of doublets).
H
Hc
C CH3
Hd
OH
5 H's on
benzene ring
Ha
Hb
14.22
H H
N
N
H2N
H2N
HO H N
b
HN
N
N
H
Cl H
a
H
N
Hc
palau'amine
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
379
Nuclear Magnetic Resonance Spectroscopy 14–9
14.23 Use these steps to propose a structure consistent with the molecular formula, IR, and NMR
data.
• Calculate the degrees of unsaturation.
• Use the IR data to determine what types of functional groups are present.
• Determine the number of different types of protons.
• Calculate the number of H’s giving rise to each signal.
• Analyze the splitting pattern and put the molecule together.
• Use the chemical shift information to check the structure.
• Molecular formula C7H14O2
2n + 2 = 2(7) + 2 = 16
16 – 14 = 2/2 = 1 degree of unsaturation
1 π bond or 1 ring
• IR peak at 1740 cm-1
C=O absorption is around 1700 cm–1 (causes the degree of unsaturation).
No signal at 3200–3600 cm–1 means there is no O–H bond.
• NMR data: absorption ppm integration
singlet
26 units/3 units per H = 9 H's
1.2
26
1.3
10
triplet
10 units/3 units per H = 3 H's (probably a CH3 group)
4.1
6
quartet
6 units/3 units per H = 2 H's (probably a CH2 group)
• 3 kinds of H's
• number of H's per signal
total integration units: 26 + 10 + 6 = 42 units
42 units / 14 H's = 3 units per H
• look at the splitting pattern
CH3
C CH3
The singlet (9 H) is likely from a tert-butyl group:
CH3
The CH3 and CH2 groups split each other: CH3 CH2
• Join the pieces together.
O
CH3CH2O
C
O
CH3
C CH3
or
CH3CH2
CH3
Pick this structure due to the chemical shift data.
The CH2 group is shifted downfield (4 ppm), so it
is close to the electron-withdrawing O.
C
CH3
O C CH3
CH3
380
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–10
14.24
•
Molecular formula: C3H8O
•
•
IR peak at 3200–3600 cm–1
NMR data:
• doublet at ~1.2 (6 H)
• singlet at ~2.2 (1 H)
• septet at ~4 (1 H)
¾ Calculate degrees of unsaturation
2n + 2 = 2(3) + 2 = 8
8 – 8 = 0 degrees of unsaturation
¾ Peak at 3200–3600 cm–1 is due to an O–H bond.
3 types of H's
septet from 1 H
singlet from 1 H
doublet from 6 H's
from the O–H proton
split by 6 H's
split by 1 H
¾ Put information together:
CH3
HO C CH3
H
14.25 Identify each compound from the 1H NMR data.
singlet
at 2.5
a.
CH2 CHCOCH3
O
HCl
H H
CH3
Cl
b.
triplet
at 3.6
(CH3)2C
O
base
O
H2O
H H
singlet
A
singlet
at 1.3
singlet
at 3.8
B
singlet
at 2.2
triplet
at 3.05
OH
14.26 Each different kind of carbon atom will give a different 13C NMR signal.
O
b.
a. CH3CH2CH2CH3
CH3CH2
C a Cb Cb Ca
2 kinds of C's
2 13C NMR signals
C
c. CH3CH2CH2 O CH2CH2CH3
OCH3
Each C is different.
4 kinds of C's
4 13C NMR signals
Ca Cb C c
Cc Cb Ca
same groups on both sides of O
3 kinds of C's
3 13C NMR signals
CH3CH2
d.
H
Ha Hb Ha
Hd
H H H
Hc
a.
Ha
H H H
Hb
Ha
H C C C H
H C C C H
H Cl Cl
Cl H Cl
Hc
Hb
4 1H NMR signals
2 1H NMR signals
H H Cl
Hc
Ha
H Cl H
H C C C H
H C C C H
H H Cl
H Cl H
Hb
3 1H NMR signals
all H's identical
1 1H NMR signal
Hc
H
Each C is different.
4 kinds of C's
4 13C NMR signals
14.27
Hb Ha
H
C C
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
381
Nuclear Magnetic Resonance Spectroscopy 14–11
b.
H H H
H H H
H H Cl
H Cl H
H C C C H
H C C C H
H C C C H
H C C C H
H Cl Cl
Ca Cl H Cl Ca
Cb
Each C is different.
3 kinds of C's
3 13C NMR signals
2 kinds of C's
2 13C NMR signals
H H Cl
Each C is different.
3 kinds of C's
3 13C NMR signals
Ca H Cl H Ca
Cb
2 kinds of C's
2 13C NMR signals
c. Although the number of 13C signals cannot be used to distinguish these isomers, each
isomer exhibits a different number of signals in its 1H NMR spectrum. As a result, the
isomers are distinguishable by 1H NMR spectroscopy.
14.28
These 2 C's are different because they
are cis and trans to different groups.
H
CO2H
H
H
chrysanthemic acid
Every carbon is different so
there are 10 lines for the 10 C
atoms.
14.29 Electronegative elements shift absorptions downfield. The carbons of alkenes and benzene
rings, and carbonyl carbons are also shifted downfield.
O
b.
CH3CH2OCH2CH3
a.
The CH2 group is closer
to the electronegative O
and will be farther downfield.
c.
BrCH2CHBr2
The C of the CHBr2 group has two
bonds to electronegative Br atoms
and will be farther downfield.
H
C
d. CH3CH CH2
OCH3
The carbonyl carbon is
highly deshielded and
will be farther downfield.
The CH2 group is part of
a double bond and will
be farther downfield.
14.30
a. In order of lowest to highest chemical shift:
Ca C b C c Cd
CH3CHCH2CH3
OH
Cd < Ca < Cc < Cb
b. In order of lowest to highest chemical shift:
(CH3CH2)2C=O
Ca Cb
Cc
Ca < Cb < Cc
14.31
• molecular formula C4H8O2
2n + 2 = 2(4) + 2 = 10
10 – 8 = 2/2 = 1 degree of unsaturation
O
• no IR peaks at 3200–3600 or 1700 cm–1
no O–H or C=O
O
• 1H NMR spectrum at 3.69 ppm
only one kind of proton
•
13C
NMR spectrum at 67 ppm
only one kind of carbon
This structure satisifies all the data.
One ring is one degree of
unsaturation. All carbons and
protons are identical.
382
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–12
14.32
O
OH
• molecular formula C4H8O
2n + 2 = 2(4) + 2 = 10
10 – 8 = 2/2 = 1 degree of unsaturation
•
• molecular formula C4H8O
2n + 2 = 2(4) + 2 = 10
10 – 8 = 2/2 = 1 degree of unsaturation
13C
• all 13C NMR signals at < 160 ppm
NO C=O
NMR signal at > 160 ppm due to
C=O
14.33
Hb
Ha
H
H CH3
Hb
CH3
H
Hd
H
H H
O
O
Cl
O
CH3
H
4° C
CH3
Ha
Hd
H
Hc
CH3
O
Hf
H H
H H
Hc
He
A
B
a. 6 1H NMR signals
b. 7 13C NMR signals (including the carbonyl C)
a. 4 1H NMR signals
b. 5 13C NMR signals (including the 4° C)
14.34
Ha
Hc
H H
H H
O
Br
H
Hb
C
CH3
Ha
Hd
H H Hc
O
CH3
HO
H H
Hb
H H
Cl
H
Hd
H
He
D
Hc
a. 4 1H NMR signals
b. Ha: 1 adjacent H, so 2 peaks
Hb: 2 adjacent H's, so 3 peaks
Hc: 3 adjacent H's, so 4 peaks
Hd: 2 adjacent H's, so 3 peaks
a. 5 1H NMR signals
b. Ha: singlet
Hb: 2 adjacent H's, so 3 peaks
Hc: 2 adjacent H's, so 3 peaks
Hd: 1 nonequivalent H on the same C, so 2 peaks
He: 1 nonequivalent H on the same C, so 2 peaks
14.35 Use the directions from Answer 14.3.
a.
(CH3)3CH
2 kinds of H's
b.
(CH3)3CC(CH3)3
1 kind of H
c. CH3CH2OCH2CH2CH2CH2CH3
7 kinds of H's
e.
O
5 kinds of H's
d.
Ha CH3
H
Hc
f.
CH3CH2
C C
C C
Hb
H
H
4 kinds of H's
Hd
H
H
3 kinds of H's
CH2CH3
383
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Nuclear Magnetic Resonance Spectroscopy 14–13
g.
CH3CH2CH2OCH2CH2CH3
3 kinds of H's
h.
Ha
CH3
Hb
CH3
i.
CH2CH3
C C
Br
Hb
Hd
H H
j.
Hf
CH3 C C CH3
Hd
Ha
Hc
Ha
H
Hb
H
He
HO H
Hc
CH3
O
H
Hd
4 kinds of H's
Hc
6 kinds of H's
4 kinds of H's
14.36
CH3
H
H
Hb
H
a.
H
Hc
Ha
CH3
H
Ha
Hb
b.
Hd
Hf
Hd
H
H
H
H
H
H
Hb
CH2CH3
Hg
Hf
H
H
H
Hd
Hd H
Ha
H
H
H
Hb
H
H
CH3
Hd
H
Ha CH3
d.
Hc
H
CH3
c.
H
Hb
3 kinds of protons
He
Hc
4 kinds of protons
Hb
Hc
Hb
H
H H
c
Ha
4 kinds of protons
Hd CH3
Ha
Hc
7 kinds of protons
14.37
equivalent
O
O
a.
CH3
O
b.
CH3
CHO
c.
d.
N
N
OH
H
N
N
CH3
caffeine
4 NMR signals
OCH3
OH
CH3O
N
H
HO
thymol
7 NMR signals
capsaicin
15 NMR signals
equivalent
vanillin
6 NMR signals
14.38
δ (in ppm) = [observed chemical shift (Hz)] / ν of the NMR (MHz)]
14.39
a. 2.5 = x Hz/300 MHz
x = 750 Hz
b. ppm = 1200 Hz/300 MHz
= 4 ppm
c. 2.0 = x Hz/300 MHz
x = 600 Hz
2.16 = x Hz/500 MHz
x = 1080 Hz (chemical shift of acetone in Hz)
1080 Hz + 1570 Hz = 2650 Hz
2650 Hz/500 MHz = 5.3 ppm (chemical shift of the CH2Cl2 signal)
384
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–14
14.40 Use the directions from Answer 14.7.
a. CH3CH2CH2CH2CH3 or
c.
CH3CH2CH2OCH3
CH3OCH2CH3
or
Adjacent O deshields the H's.
farther downfield
b. CH3CH2CH2I
Increasing alkyl substitution
farther downfield
or CH3CH2CH2F
d. CH3CH2CHBr2 or
More electronegative F
deshields the H's.
farther downfield
CH3CH2CH2Br
Two electronegative
Br's deshield the H.
farther downfield
14.41 Use the directions from Answer 14.11.
Signal of 13 units is from 1 H.
Signal of 33 units is from 3 H's.
Signal of 73 units is from 6 H's.
[total number of integration units] / [total number of protons]
[13 + 33 + 73] / 10 = ~12 units per proton
14.42
Hb Hb
Hb
Ha
CH3
a. CH3CO2C(CH3)3 and CH3CO2CH3
c.
Ha
Hb
Ha
CH3
Hb
Ha
Ha
Ha : Hb = 1:3
Ha : Hb = 1:1
different ratio of peak areas
Hb in CH3CO2CH3 is farther downfield
than all H's in CH3CO2C(CH3)3.
and
CH3
Hb
CH3
Ha
Hb Hb
Hb
CH3
Ha
Ha : Hb = 3:1
Ha : Hb = 3:2
different ratio of peak areas
b. CH3OCH2CH2OCH3 and CH3OCH2OCH3
Ha
Hb Hb
Ha
Ha
Hb
Ha
Ha : Hb = 3:2
Ha : Hb = 3:1
different ratio of peak areas
14.43 The following compounds give one singlet in a 1H NMR spectrum:
CH3
CH3CH3
CH3 C C CH3
Cl
Cl
Br Br
CH3
O
C C
CH3
CH3
(CH3)3C
C
C(CH3)3
14.44
O
a. CH3CH(OCH3)2
CH3 protons split by 1 H = doublet
CH proton split by 3 H's = quartet
b. CH3OCH2CH2
C
OCH3
both CH2 groups split
each other = triplets
c.
CH2CH3
CH3 protons split by 2 H's = triplet
CH2 protons split by 3 H's = quartet
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Nuclear Magnetic Resonance Spectroscopy 14–15
O
g. CH3CH2CH2CH2OH
d. CH3OCH2CHCl2
CH2 protons split by 1 H = doublet
CH proton split by 2 H's = triplet
O
e. (CH3)2CH C OCH2CH3
Ha Hb
Hc Hd
Ha protons split by 1 H = doublet
Hb proton split by 6 H's = septet
Hc protons split by 3 H's = quartet
Hd protons split by 2 H's = triplet
Ha protons split by 2 H's = triplet
Hb protons split by CH3 + CH2 protons = 12
peaks (maximum)
Hc protons split by 2 different CH2 groups = 9
peaks (maximum)
Hd protons split by 2 H's = triplet
Since Hb and Hc are located in a flexible alkyl
chain, it is likely that peak overlap occurs, so
that the following is observed: Hb (3 + 2 + 1 =
6 peaks), Hc (2 + 2 + 1 = 5 peaks).
Ha Hb
CH3
j.
Hb
H
Ha
H
Hb
C C
CH3CH2
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
CH3
k.
H
Ha
H
Hb
C C
Br
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
Ha Hb Hc
Ha protons split by 2 CH2 groups =
quintet
Hb protons split by 2 H's = triplet
H
Ha: split by CH3 group + Hb
= 8 peaks (maximum)
Hb: split by 2 H's = triplet
h. CH CH CH C OH
3
2
2
HOCH2CH2CH2OH
C
Ha
Ha Hb Hc Hd
O
f.
CH3CH2
i.
Ha protons split by 2 H's = triplet
Hc protons split by 2 H's = triplet
Hb protons split by CH3 + CH2 protons = 12
peaks (maximum)
Since Hb is located in a flexible alkyl chain,
it is likely that peak overlap occurs, so that
only 3 + 2 + 1 = 6 peaks will be observed.
CH3
l.
H
Ha
C C
Hc
H
H
Hb
Ha: split by Hb + Hc doublet of doublets (4 peaks)
Hb: split by Ha + Hc doublet of doublets (4 peaks)
Hc: split by CH3, Ha + Hb - 16
peaks
14.45
Ha
Br
H
Hb
H
H
Br
C C
Ha
C C
CO2CH3
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
Ha and Hb are geminal.
Hb H
CO2CH3
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
Ha and Hb are trans.
Both compounds exhibit two doublets for the H's on the C=C, but the
coupling constants (Jgeminal and Jtrans) are different. Jgeminal is much smaller
than Jtrans (0–3 Hz versus 11–18 Hz).
14.46
Hb
Ha
C C
Hc
CN
Jab = 11.8 Hz
Jbc = 0.9 Hz
Jac = 18 Hz
Ha: doublet of doublets at 5.7 ppm. Two large J values are seen for the H’s cis
(Jab = 11.8 Hz) and trans (Jac = 18 Hz) to Ha.
Hb: doublet of doublets at ~6.2 ppm. One large J value is seen for the cis H
(Jab = 11.8 Hz). The geminal coupling (Jbc = 0.9 Hz) is hard to see.
Hc: doublet of doublets at ~6.6 ppm. One large J value is seen for the trans H
(Jac = 18 Hz). The geminal coupling (Jbc = 0.9 Hz) is hard to see.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–16
Ha
Splitting diagram for Ha
1 trans Hc proton splits Ha into
1 + 1 = 2 peaks
a doublet
Jac = the coupling constant between Ha and Hc
1 cis Hb proton splits Ha into
1 + 1 = 2 peaks
Now it's a doublet of doublets.
Jab = the coupling constant between Ha and Hb
14.47
Four constitutional isomers of C4H9Br:
Br
Br
Br
Br
4 different C's
4 different C's
2 different C's
3 different C's
14.48 Only two compounds in Problem 14.43 give one signal in their 13C NMR spectrum:
CH3CH 3
14.49
O
R
C
O
OR'
C
R
OR'
The O atom of an ester donates electron density, so the carbonyl carbon has
less δ+, making it less deshielded than the carbonyl carbon of an aldyhyde or
ketone. Therefore, the carbonyl carbon of an aldehyde or ketone is more
deshielded and absorbs farther downfield.
14.50
a. HC(CH3)3
2 signals
d.
g.
O
5 signals
7 signals
e. CH3CH2
b.
CH2
CH2CH3
h.
O
C C
5 signals
H
4 signals
H
3 signals
c. CH3OCH(CH3)2
3 signals
f.
OH
7 signals
i.
3 signals
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Nuclear Magnetic Resonance Spectroscopy 14–17
14.51
O
a. CH3CH2
Ca Cb
O
OH
C
OH
c.
b. CH3CH2CHCH2CH3
Cc
Ca
Ca Cb Cc
Ca < Cb < C c
C
Ca < C b < Cc
d. CH2 CHCH2CH2CH2Br
CH2CH3
Ca
Cb Cc
Cc < Cb < Ca
Cb Cc
Cb < Cc < Ca
14.52
19 ppm
62 ppm
16 ppm
a. CH3CH2CH2CH2OH
205 ppm
b. (CH3)2CHCHO
14 ppm 35 ppm
143 ppm
23 ppm
c. CH2=CHCH(OH)CH3
41 ppm
113 ppm
69 ppm
14.53
Ca
Cb Ca
Cb
O
Ca
Cd
a.
CH3O2C
H
H
H
b.
Cc
C C
Cc
CO2CH3
H
CH3O2C
Cb Ca
Ca
3 different C's
3 signals
Cb
H
Cc
C C
Cc
CO2CH3
Cb Ca
3 signals
Cd
CO2CH3
C C
H
O
Cc
O
CO2CH3
Ce O
Cb Ca
4 signals
C b Cc
Ca
5 signals
14.54 Use the directions from Answer 14.23.
a. C4H8Br2: 0 degrees of unsaturation
IR peak at 3000–2850 cm–1: Csp3–H bonds
NMR: singlet at 1.87 ppm (6 H) (2 CH3 groups)
singlet at 3.86 ppm (2 H) (CH2 group)
CH3
CH3 C CH2Br
c. C5H10O2: 1 degree of unsaturation
IR peak at 1740 cm–1: C=O
NMR: triplet at 1.15 ppm (3 H) (CH3 split by 2 H's)
triplet at 1.25 ppm (3 H) (CH3 split by 2 H's)
quartet at 2.30 ppm (2 H) (CH2 split by 3 H's)
quartet at 4.72 ppm (2 H) (CH2 split by 3 H's)
O
Br
CH3CH2
b. C3H6Br2: 0 degrees of unsaturation
IR peak at 3000–2850 cm–1: Csp3–H bonds
NMR: quintet at 2.4 ppm (split by 2 CH2 groups)
triplet at 3.5 ppm (split by 2 H's)
Br
Br
C
O
CH2CH3
d. C6H14O: 0 degrees of unsaturation
IR peak at 3600–3200 cm–1: O–H
NMR: triplet at 0.8 ppm (6 H) (2 CH3 groups
split by CH2 groups)
singlet at 1.0 ppm (3 H) (CH3)
quartet at 1.5 ppm (4 H) (2 CH2 groups split
by CH3 groups)
singlet at 1.6 ppm (1 H) (O–H proton)
CH3
CH3CH2 C CH2CH3
OH
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–18
e. C6H14O: 0 degrees of unsaturation
IR peak at 3000–2850 cm–1: Csp3–H bonds
NMR: doublet at 1.10 ppm (integration = 30 units)
(from 12 H's)
septet at 3.60 ppm (integration = 5 units)
(from 2 H's)
H
f. C3H6O: 1 degree of unsaturation
IR peak at 1730 cm–1: C=O
NMR: triplet at 1.11 ppm
multiplet at 2.46 ppm
triplet at 9.79 ppm
O
H
CH3 C O C CH3
CH3
CH3CH2
CH3
C
H
14.55
Two isomers of C9H10O: 5 degrees of unsaturation (benzene ring likely)
Compound B:
IR absorption at 1688 cm–1: C=O
NMR data:
Absorptions:
triplet at 1.22 (3 H) (CH3 group split by 2 H's)
quartet at 2.98 (2 H) (CH2 group split by 3 H's)
multiplet at 7.28–7.95 (5 H)
(likely a monosubstituted benzene ring)
Compound A:
IR absorption at 1742 cm–1: C=O
NMR data:
Absorptions:
singlet at 2.15 (3 H) (CH3 group)
singlet at 3.70 (2 H) (CH2 group)
broad singlet at 7.20 (5 H)
(likely a monosubstituted benzene ring)
O
CH2
C
O
C
CH3
14.56 IR absorptions:
3088–2897 cm–1: sp2 and sp3 hybridized C–H
1740 cm–1: C=O
1606 cm–1: benzene ring
triplet at 2.91
H H
O
5H
multiplet
7.20–7.35
H H
CH3
singlet at 2.02
O
triplet at 4.25
C10H12O2
W
14.57 IR absorption at 1713 cm–1 is due to C=O.
triplet at 2.43
doublet
at 1.09
CH3 H H
CH3
CH3
H
O
septet at 2.60
C7H14O
V
H H
multiplet at 1.6
triplet at 0.91
CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
389
Nuclear Magnetic Resonance Spectroscopy 14–19
14.58
Compound C:
molecular ion 146 (molecular formula C6H10O4)
IR absorption at 1762 cm–1: C=O
1H NMR data:
Absorptions:
Ha: doublet at 1.47 (3 H) (CH3 group adjacent to CH)
Hb: singlet at 2.07 (6 H) (2 CH3 groups)
Hc: quartet at 6.84 (1 H adjacent to CH3)
Hc
O
CH3
C
H
O C O
O
C
CH3
Hb
CH3
Hb
Ha
14.59
Hb
O
[1] LiC CH
[2] H2O
OH
Hc
CH3 C C CH
Ha
CH3
Ha
D
Compound D:
molecular ion 84 (molecular formula C5H8O)
IR absorptions at 3600–3200 cm–1: OH
3303 cm–1: Csp–H
2938 cm–1: Csp3–H
2120 cm–1: C C
1H NMR data:
Absorptions:
Ha: singlet at 1.53 (6 H) (2 CH3 groups)
Hb: singlet at 2.37 (1 H)
alkynyl CH and OH
Hc: singlet at 2.43 (1 H)
14.60
Compound E:
C4H8O2:
1 degree of unsaturation
IR absorption at 1743 cm–1: C=O
NMR data:
Compound F:
C4H8O2:
1 degree of unsaturation
IR absorption at 1730 cm–1: C=O
NMR data:
total integration units/# H's
(23 + 29 + 30)/8 = ~10 units per H
total integration units/# H's
(18 + 30 + 31)/8 = ~10 units per H
Ha: quartet at 4.1 (23 units - 2 H)
Hb: singlet at 2.0 (29 units - 3 H)
Hc: triplet at 1.4 (30 units - 3 H)
Ha: singlet at 4.1 (18 units - 2 H)
Hb: singlet at 3.4 (30 units - 3 H)
Hc: singlet at 2.1 (31 units - 3 H)
O
CH3
Hb
C
O
OCH2CH3
CH3
Ha Hc
Hc
C
CH2OCH3
Ha
Hb
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–20
14.61
Compound H:
C8H11N:
4 degrees of unsaturation
IR absorptions at 3365 cm–1: N–H
3284 cm–1: N–H
3026 cm–1: Csp2–H
2932 cm–1: Csp3–H
1603 cm–1: due to benzene
1497 cm–1: due to benzene
Compound I:
C8H11N:
4 degrees of unsaturation
IR absorptions at 3367 cm–1: N–H
3286 cm–1: N–H
3027 cm–1: Csp2–H
2962 cm–1: Csp3–H
1604 cm–1: due to benzene
1492 cm–1: due to benzene
NMR data:
NMR data:
multiplet at 7.2–7.4 ppm, 5 H on a benzene ring
Ha: triplet at 2.9 ppm, 2 H, split by 2 H's
Hb: triplet at 2.8 ppm, 2 H, split by 2 H's
Hc: singlet at 1.1 ppm, 2 H, no splitting (on NH2)
multiplet at 7.2–7.4 ppm, 5 H on a benzene ring
Ha: quartet at 4.1 ppm, 1 H, split by 3H's
Hb: singlet at 1.45 ppm, 2 H, no splitting (NH2)
Hc: doublet at 1.4 ppm, 3 H, split by 1 H
Ha
CH3
CH2CH2NH2
H
Hb
Hc
C NH2
Hc
Hb
Ha
14.62
a. C9H10O2:
5 degrees of unsaturation
IR absorption at 1718 cm–1: C=O
NMR data:
multiplet at 7.4–8.1 ppm, 5 H on a benzene ring
quartet at 4.4 ppm, 2 H, split by 3 H's
triplet at 1.3 ppm, 3 H, split by 2 H's
b. C9H12:
4 degrees of unsaturation
IR absorption at 2850–3150 cm–1:
C–H bonds
NMR data:
O
C
CH3
OCH2CH3
C CH3
H
downfield due to the O atom
14.63 IR absorption at 1717 cm–1 is due to a C=O.
singlet at 1.02
singlet at 7.1–7.4 ppm, 5 H, benzene
septet at 2.8 ppm, 1 H, split by 6 H's
doublet at 1.3 ppm, 6 H, split by 1 H
CH3 CH3 O
CH3
CH3
singlet at 2.13
H H
C7H14O
R
singlet at 2.33
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Nuclear Magnetic Resonance Spectroscopy 14–21
14.64 IR absorption at 1730 cm–1 is due to a C=O. Eight lines in the 13C NMR spectrum means
there are eight different types of C.
triplet at
4.20
Hb
(CH3)2N
singlet at
2.32
O
Ha
H H
H
O
H H
Hb
Ha
singlet at 9.97
Ha and Hb appear as two doublets.
triplet at
3.05
14.65
a. Compound J has a molecular ion at 72: molecular formula C4H8O
1 degree of unsaturation
O
IR spectrum at 1710 cm–1: C=O
1H NMR data (ppm):
C
CH3
CH2CH3
1.0 (triplet, 3 H), split by 2 H's
2.1 (singlet, 3 H)
2.4 (quartet, 2 H), split by 3 H's
b. Compound K has a molecular ion at 88: molecular formula C5H12O
0 degrees of unsaturation
IR spectrum at 3600–3200 cm–1: O–H bond
1
OH
H NMR data (ppm):
CH3 C CH3
0.9 (triplet, 3 H), split by 2 H's
1.2 (singlet, 6 H), due to 2 CH3 groups
CH2CH3
1.5 (quartet, 2 H), split by 3 H's
1.6 (singlet, 1 H), due to the OH proton
14.66
Compound L has a molecular ion at 90: molecular formula C4H10O2
0 degrees of unsaturation
total integration units/# H's
IR absorptions at 2992 and 2941 cm–1: Csp3–H
(25 + 46 + 7)/10 = ~8 units per H
1H NMR data (ppm):
O
Ha: 1.2 (doublet, 3 H), split by 1 H
H+
CH3 OH
Hb: 3.3 (singlet, 6 H), due to 2 CH3 groups
H
Hc: 4.8 (quartet, 1 H), split by 3 adjacent H's
H
Hc
CH3 C OCH3
Ha
OCH3
L
Hb
Hb
392
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–22
14.67
TsOH
multiplet at ~2.0 singlet at ~1.7 multiplet at ~1.5
triplet at ~0.95
triplet at ~0.95
2 singlets
at ~1.6
M
OH
OH2
H
N
triplet at ~5.1
H OTs
two signals
at ~4.6
triplet at ~2.0
OTs
1,2-H shift
H
TsOH
N
H H
H2O
OTs
or
TsOH
H
OTs
M
14.68
Compound O has a molecular formula C10H12O.
5 degrees of unsaturation
IR absorption at 1687 cm–1
1H NMR data (ppm):
Ha: 1.0 (triplet, 3 H), due to CH3 group, split by 2 adjacent H's
Hb: 1.7 (sextet, 2 H), split by CH3 and CH2 groups
Hc: 2.9 (triplet, 2 H), split by 2 H's
7.4–8.0 (multiplet, 5 H), benzene ring
14.69
Compound P has a molecular formula C5H9ClO2.
1 degree of unsaturation
13C NMR shows 5 different C's, including a C=O.
O
1H NMR data (ppm):
C
ClCH2CH2
OCH2CH3
Ha: 1.3 (triplet, 3 H), split by 2 H's
Hc Hb
Hd Ha
Hb: 2.8 (triplet, 2 H), split by 2 H's
Hc: 3.7 (triplet, 2 H), split by 2 H's
P
Hd: 4.2 (quartet, 2 H), split by CH3 group
O
C CH2CH2CH3
Hc H b Ha
O
393
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Nuclear Magnetic Resonance Spectroscopy 14–23
14.70
Compound Q: Molecular ion at 86.
Molecular formula: C5H10O:
O
1 degree of unsaturation
C
CH3
CH2CH3
IR absorption at ~1700 cm–1: C=O
NMR data:
Ha: doublet at 1.1 ppm, 2 CH3 groups split by 1 H
Hb: singlet at 2.1 ppm, CH3 group
Hc: septet at 2.6 ppm, 1 H split by 6 H's
Hb O
[1] strong base
[2] CH3I
C
CH3
Hc Ha
CH(CH3)2
= Q
MW = 86
14.71
a. Compound R, the odor of banana: C7H14O2
1 degree of unsaturation
1H NMR (ppm):
Ha: 0.93 (doublet, 6 H)
Hb: 1.52 (multiplet, 2 H)
Hc: 1.69 (multiplet, 1 H)
Hd: 2.04 (singlet, 3 H)
He: 4.10 (triplet, 2 H)
Hc
O
CH3
C
b. Compound S, the odor of rum: C7H14O2
1 degree of unsaturation
1H NMR (ppm):
OCH2CH2CHCH3
CH3
Hd
He Hb
Ha
Hc
O
Ha: 0.94 (doublet, 6 H)
Hb: 1.15 (triplet, 3 H)
Hc: 1.91 (multiplet, 1 H)
Hd: 2.33 (quartet, 2 H)
He: 3.86 (doublet, 2 H)
CH3CH2
C
OCH2CHCH3
CH3
He
Hb Hd
Ha
Ha
Ha
14.72
C6H12:
1 degree of unsaturation
K+ –OC(CH3)3
T
1H
Br
H+
U
1H
NMR of T (ppm):
Ha: 1.01 (singlet, 9 H)
Hb: 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz)
Hc: 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz)
Hd: 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz)
NMR of U: 1.60 (singlet) ppm.
CH3
C
OH
CH3
CH3
C
CH3
14.73
a.
C6H12O2:
1 degree of unsaturation
IR peak at 1740 cm–1: C=O
1H NMR 2 signals: 2 types of H's
13C NMR: 4 signals: 4 kinds of C's,
including one at ~170 ppm due a C=O
O
CH3
C
C
O
CH3 CH3
b.
C6H10:
2 degrees of unsaturation
IR peak at 3000 cm–1: Csp3–H bonds
peak at 3300 cm–1: Csp–H bond
peak at ~2150 cm–1: C≡C bond
13
C NMR: 4 signals: 4 kinds of C's
CH3
HC C C CH3
CH3
CH3
Hc
Ha
(CH3)3C
H
C C
Hd
H
H
Hb
All H's are identical,
so there is only one
singlet in the NMR.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–24
14.74 a. Since A has no absorptions at 1700 cm–1 or 3600–3200 cm–1, it has no C=O or OH. An
oxygen-containing compound without these functional groups is an ether (or an epoxide).
Since B is formed from a reaction with HCl, A must contain an epoxide, because ethers
are unreactive with HCl.
doublet at 3.5
H
O
doublet at 1.4
CH3
quartet of doublets at 3.0
H
singlet at 3.8
CH3O
2 doublets at 6.9 and 7.2
A
b. Epoxide A is equally substituted by R groups on both C’s. With HCl, the epoxide is
protonated first and then backside attack by Cl– forms the chlorohydrin. Attack at the C
adjacent to the benzene ring is preferred because the δ+ in the transition state at this carbon
can be delocalized on the benzene ring.
H
O
H
CH3
H
O
H
CH3
H
H Cl
CH3O
H
Cl
CH3O
CH3O
A
S
R
OH
CH3
H
B
Cl
The benzene ring and CH3 group must be trans in the epoxide to give the correct
configuration at the two stereogenic centers in the product.
14.75
O
H
C
O
N(CH3)2
N,N-dimethylformamide
H
C + CH3
N
CH3
cis to the O atom
cis to the H atom
A second resonance structure for N,N-dimethylformamide places the two CH3 groups in different
environments. One CH3 group is cis to the O atom, and one is cis to the H atom. This gives rise
to two different absorptions for the CH3 groups.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
395
Nuclear Magnetic Resonance Spectroscopy 14–25
14.76
Ho
Ho
Ho
Ho
Ho
Hi
Hi
Hi
Ho
Hi
Ho
Hi
Hi
Ho
Ho
Ho
Ho
Ho
18-Annulene has 18 π electrons that create an induced magnetic
field similar to the 6 π electrons of benzene. 18-Annulene has
12 protons that are oriented on the outside of the ring (labeled Ho),
and 6 protons that are oriented inside the ring (labeled Hi). The
induced magnetic field reinforces the external field in the vicinity of
the protons on the outside of the ring. These Ho protons are
deshielded and so they absorb downfield (8.9 ppm). In contrast, the
induced magnetic field is opposite in direction to the applied
magnetic field in the vicinity of the protons on the inside of the ring.
This shields the Hi protons and the absorption is therefore very far
upfield, even higher than TMS (–1.8 ppm).
14.77
Ca
stereogenic center
H
CH3 H
CH3 C
H
HO C CH3
C CH3 Replace a CH3 group
X C CH3
with X.
OH
Cb
H
Replace Ca.
3-methyl-2-butanol
H
or
HO C CH3
CH3 C X
H
Replace Cb.
The CH3 groups are not equivalent to each other,
since replacement of each by X forms two diastereomers.
Thus, every C in this compound is different
and there are five 13C signals.
14.78
O
CH3 P OCH3
OCH3
One P atom splits each nearby CH3 into a
doublet by the n + 1 rule, making two doublets.
Ha
Hb
Ha
All 6 Ha protons are equivalent.
14.79 a. Splitting pattern:
Jab = 11 Hz
Jbc = 4 Hz
3 Hz
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 14–26
b. Three resonance structures can be drawn for 2-cyclohexenone.
O
A
O
B
O
C
Resonance structure C places a (+) charge on one C of the C=C, deshielding the H
attached to it and shifting the absorption downfield.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
397
Radical Reactions 15–1
Chapter 15 Radical Reactions
Chapter Review
General features of radicals
• A radical is a reactive intermediate with an unpaired electron (15.1).
• A carbon radical is sp2 hybridized and trigonal planar (15.1).
• The stability of a radical increases as the number of C’s bonded to the radical carbon
increases (15.1).
least stable
most stable
RCH2
CH3
1o
R2CH
R3C
2o
3o
Increasing alkyl substitution
Increasing radical stability
•
Allylic radicals are stabilized by resonance, making them more stable than 3o radicals (15.10).
CH2 CH CH2
CH2 CH CH2
two resonance structures for the allyl radical
Radical reactions
[1] Halogenation of alkanes (15.4)
R H
X2
hν or Δ
X = Cl or Br
•
•
R X
alkyl halide
•
•
The reaction follows a radical chain mechanism.
The weaker the C–H bond, the more readily the H
is replaced by X.
Chlorination is faster and less selective than
bromination (15.6).
Radical substitution results in racemization at a
stereogenic center (15.8).
[2] Allylic halogenation (15.10)
CH2 CH CH3
NBS
hν or ROOR
CH2 CHCH2Br
allylic halide
•
The reaction follows a radical chain mechanism.
[3] Radical addition of HBr to an alkene (15.13)
H H
• A radical addition mechanism is followed.
HBr
RCH CH2
R C C H
• Br bonds to the less substituted carbon atom to
hν, Δ, or
H Br
form the more substituted, more stable radical.
ROOR
alkyl bromide
[4] Radical polymerization of alkenes (15.14)
CH2 CHZ
ROOR
Z
Z
polymer
Z
•
A radical addition mechanism is followed.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–2
Practice Test on Chapter Review
1. a. Which alkyl halide(s) can be made in good yield by radical halogenation of an alkane?
Br
1.
Cl
2.
3.
Br
4. Both (1) and (2) can be made in good yield.
5. Compounds (1), (2), and (3) can all be made in good yield.
b. In which of the following reactions will rearrangement not occur?
1. halogenation of an alkane with Cl2 and heat
2. addition of Cl2 to an alkene
3. addition of HCl to an alkene
4. Rearrangements do not occur in reactions (1) and (2).
5. Rearrangements do not occur in reactions (1), (2), and (3).
c. Which labeled H is most easily abstracted in a radical halogenation reaction?
Ha
He
Hd
Hc
Hb
1.
2.
3.
4.
5.
Ha
Hb
Hc
Hd
He
d. Which of the labeled C–H bonds in the following compound has the smallest bond dissociation
energy?
Hc
Ha
Hb
1.
2.
3.
4.
5.
CH3
Hd
CH2 He
C–Ha
C–Hb
C–Hc
C–Hd
C–He
2. (a) Which radical is the most stable? (b) Which radical is the least stable?
A
B
C
D
3. Draw all of the organic products formed in each reaction. Indicate stereochemistry in part (c).
a.
HBr
ROOR
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–3
b.
NBS
hν
c.
Br2
hν
4. What monomer is needed to make the following polymer?
Cl
Cl
Cl
5. In each box, fill in the appropriate reagents needed to carry out the given reaction. This question
involves reactions from Chapter 15, as well as previous chapters.
Br
OH
OH
O
CN
O
Answers to Practice Test
1.a. 2
b. 4
c. 2
d. 3
2.a. A
b. B
3.a.
4.
3.c.
Br
Br
Br
3.b.
BrCH2CH=CHCH3
Br
(cis and trans)
Cl
400
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–4
5.
[1] BH3
Br2
KOC(CH3)3
hν or Δ
[2] H2O2, –OH
Br
OH
PCC or
CrO3 or
Na2Cr2O7
mCPBA
[1] –CN
[2] H2O
OH
O
CN
O
Answers to Problems
15.1
1° Radicals are on C’s bonded to one other C; 2° radicals are on C’s bonded to two other C’s;
3° radicals are on C’s bonded to three other C’s.
a. CH3CH2 CHCH2CH3
c.
b.
2° radical
15.2
3° radical
1° radical
b. (CH3)3CCHCH3
c. (CH3)3CCH2
d.
Reaction of a radical with:
• an alkane abstracts a hydrogen atom and creates a new carbon radical.
• an alkene generates a new bond to one carbon and a new carbon radical.
• another radical forms a bond.
a. CH3 CH3
b. CH2 CH2
15.4
2° radical
The stability of a radical increases as the number of alkyl groups bonded to the radical carbon
increases. Draw the most stable radical.
a. (CH3)2CCH2CH3
15.3
d.
Cl
H Cl
CH3 CH2
Cl
c.
Cl
d.
Cl
Cl Cl
Cl
Cl
CH2 CH2
+
O O
Cl O O
Monochlorination is a radical substitution reaction in which a Cl replaces a H, thus
generating an alkyl halide.
Cl2
a.
b. CH3CH2CH2CH2CH2CH3
c. (CH3)3CH
Cl
Cl2
Cl2
H
ClCH2CH2CH2CH2CH2CH3
Cl
CH3 C CH3
CH3
CH3 C CH2CH2CH2CH3
Cl
H
CH3 C CH2Cl
CH3
H
CH3CH2 C CH2CH2CH3
Cl
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
401
Radical Reactions 15–5
15.5
Cl2
Δ
A
Cl
Cl
Cl2
Cl
Δ
B
Cl
15.6
Br
Initiation:
Propagation:
CH3
H
Br
+
Br
+
Br
+
Br
CH3
Br Br
CH3
Termination:
hν
or Δ
Br
H Br
CH3 Br
Br
Br
Br Br
or
CH3
CH3
CH3 CH3
or
CH3
Br
CH3 Br
15.7
Step 1:
CH3 H
+
Br
CH3
1 bond broken
+435 kJ/mol
CH3
Step 2:
15.8
1 bond formed
–368 kJ/mol
Br Br
CH3 Br +
1 bond broken
+192 kJ/mol
1 bond formed
–293 kJ/mol
ΔH° = –101 kJ/mol
Br
The rate-determining step for halogenation reactions is formation of CH3• + HX.
CH3 H
+
I
CH3
1 bond broken
+435 kJ/mol
15.9
ΔH° = +67 kJ/mol
H Br
H
I
1 bond formed
–297 kJ/mol
ΔH° = +138 kJ/mol
This reaction is more endothermic
and has a higher Ea than a similar
reaction with Cl2 or Br2.
The weakest C–H bond in each alkane is the most readily cleaved during radical
halogenation.
a.
b.
H
H
3°
most reactive
c.
CH3CHCH2CH3
H
3°
most reactive
2°
most reactive
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–6
15.10 To draw the product of bromination:
• Draw out the starting material and find the most reactive C–H bond (on the most
substituted C).
• The major product is formed by cleavage of the weakest C–H bond.
Br2
a.
Br2
c.
Br
Δ
Δ
Br
Br
Br2
b.
Br2
Br
Δ
d.
Δ
15.11 If 1o C–H and 3o C–H bonds were equally reactive there would be nine times as much
(CH3)2CHCH2Cl as (CH3)3CCl since the ratio of 1o to 3o H’s is 9:1. The fact that the ratio is
only 63:37 shows that the 1o C–H bond is less reactive than the 3o C–H bond.
(CH3)2CHCH2Cl is still the major product, though, because there are nine 1o C–H bonds and
only one 3o C–H bond.
15.12
CH3
a.
CH3 C Br
Δ
CH3
CH3
CH3
Br2
CH3 C H
c.
CH3
CH3
H2O
C CH2
CH3 C OH
H2SO4
CH3
CH3
(from b.)
CH3
b.
(CH3)3CO–K+
CH3 C Br
CH3
CH3
CH3
C CH2
d.
CH3
CH3 C Cl
CCl4
CH3
(from a.)
CH2Cl
Cl2
C CH2
CH3
(from b.)
15.13
Br K+ –OC(CH )
3 3
Br2
a.
Br
Br2
+ enantiomer
hν
Br
mCPBA
O
b.
(from a.)
15.14 Since the reaction does not occur at the stereogenic center, leave it as is.
C1
H Br
C4
CH3
H Br
Cl2
C
CH2CH3
(2R)-2-bromobutane
hν
H Br
C
C
CH2CH3
ClCH2
S
CH3
CH2CH2Cl
R
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–7
15.15
Cl
Cl
Cl2
a. CH3CH2CH2CH2CH3
CH3CH2CH2CH2CH2Cl
Δ
CH3CH2CHCH2CH3
Cl
b.
Cl2
CH3
H
c.
H Cl
H Cl
CH3
ClCH2CH2CH(CH2CH3)2
CH3
CH3
C
CH3
Cl H
CH(CH2CH3)2
H Cl
H Cl
hν
(Consider attack at
C2 and C3 only.)
CH3
H
Cl
CH3
CH2CH3
Cl Cl
Cl2
d.
C
Cl H
CH3CH2 C CH2CH3
hν
CH2CH3
Cl
Cl
Cl2
CH3CH2 C CH2CH3
CH3
CH3
Cl
H
CH3 CH3CH2CH2
CH3CH2CH2
Cl
Cl
CH2Cl
Δ
C
C
CH(CH2CH3)2
H Cl
15.16
O N
Chain propagation:
O N O
+
O3
+
O
O N O + O2
O N
+
O2
The radical is re-formed.
15.17 Draw the resonance structure by moving the π bond and the unpaired electron. The hybrid is
drawn with dashed lines for bonds that are in one resonance structure but not another. The
symbol δ• is used on any atom that has an unpaired electron in any resonance structure.
a.
CH3 CH CH CH2
CH3 CH CH CH2
δ
c.
δ
hybrid: CH3 CH CH CH2
hybrid:
δ
δ
d.
b.
δ
hybrid:
hybrid:
δ
δ
δ
404
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–8
15.18 Reaction of an alkene with NBS or Br2 + hν yields allylic substitution products.
Br
NBS
a.
c. CH2 CH CH3
hν
Br2
CH2CHCH3
Br Br
NBS
b. CH2 CH CH3
hν
CH2 CH CH2Br
15.19
CHBrCH3
NBS
NBS
a. CH3 CH CH CH3
c. CH2 C(CH2CH3)2
CH3 CH CH CH2Br
hν
+
hν
CH2 C
CH2CH3
CH3CH(Br)CH CH2
CH3
CH3
b.
+
BrCH2C(CH2CH3) CHCH3
NBS
CH3
CH3
Br
hν
CH3
CH3
+
Br
15.20
CH3
CH3
CH3
NBS
CH3
Br
CH3
Br
hν
Br
Br
15.21 Reaction of an alkene with NBS + hν yields allylic substitution products.
a.
one possible product: high yield
Br
b.
CH3CH2CH CHCH2Br
c.
Br
Cannot be made in high yield by allylic halogenation.
Any alkene starting material would yield a mixture of allylic halides.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–9
15.22 The weakest C–H bond is most readily cleaved. To draw the hydroperoxide products, add
OOH to each carbon that bears a radical in one of the resonance structures.
O
C
O
C
OH
OH
linoleic acid
H
This allylic C–H bond is most readily cleaved.
O
C
O
C
OH
hydroperoxide products:
HO
O
O
O
C
C
OH
O
O
C
HO
OH
OH
OH
(E/Z isomers are possible.)
O
15.23 Only (b) has an OH on a benzene ring, so it may be an antioxidant.
HO
b.
phenol
15.24
a.
CH2 CHCH2CH2CH2CH3
HBr
CH3CHCH2CH2CH2CH3
Br
CH2 CHCH2CH2CH2CH3
HBr
CH2BrCH2CH2CH2CH2CH3
ROOR
b.
HBr
c. CH3CH CHCH2CH2CH3
HBr
ROOR
Br
HBr
or HBr, ROOR
CH3CH CH2CH2CH2CH 3
Br
15.25 In addition of HBr under radical conditions:
• Br• adds first to form the more stable radical.
• Then H• is added to the carbon radical.
Br
CH3CH2 CHCH2CH2CH3
Br
OH
406
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–10
CH3 H
2 radical possibilities:
Br C
CH3
or
C
C
CH3 H
CH3
CH3 H
H
C Br
H
H
C
C Br
CH3 H
3° radical
more stable
This radical forms.
1° radical
less stable
15.26
Br
HBr
ROOR
a.
Br2
c.
Br
Br
Br
HBr
b.
15.27
Transition state 1:
H
CH3 C
Energy
δ
Transition state 2:
CH2
Br
H
δ
CH3 C CH2
δ
H Br
Br
CH3 CH CH2
+
δ
Br
–18 kJ/mol
[Use the bond dissociation
energies in Appendix C.]
H
CH3 C CH2
Br
–29 kJ/mol
H
CH3 C CH2 + Br
H Br
Reaction coordinate
15.28
H
a.
CH2 C
H
H
CH2 C CH2 C CH2
C6 H5
C6 H5
C6 H5
polystyrene
H
C
C6 H5
b.
CH2 CH OCOCH3
O
O
O
COCH3 COCH3 COCH3
poly(vinyl acetate)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
407
Radical Reactions 15–11
15.29
Initiation:
RO
[1]
OR
Cl
+ CH2 C
2 RO
Cl
[2]
ROCH2 C
H
H
Cl
Cl
Propagation:
ROCH2 C
Cl
[3]
ROCH2 C
CH2 C
H
Cl
Cl
CH2 C
H
new C–C bond
Cl Cl
[4]
C CH2
H
Cl
CH2 C
H
H
Repeat Step [3] over and over.
Termination:
carbon radical
CH2 C C CH2
H
[one possibility]
H H
15.30 With Cl2, each H of the starting material can be replaced by Cl. With Br2, cleavage of the
weakest C–H bond is preferred.
Cl
Cl
a. Cl2
hν
Cl
A
Cl
Cl
b.
Br2
Δ
Br
Cl
a. Cl2
hν
Cl
B
b. Br2
Δ
Br
Cl
Cl
408
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–12
15.31
(CH3)3C
H O
OCH3
Abstraction of the phenol H
produces a resonance-stabilized radical.
BHA
(CH3)3C
(CH3)3C
O
OCH3
O
(CH3)3C
OCH3
(CH3)3C
(CH3)3C
O
OCH3
O
OCH3
O
OCH3
15.32
a. increasing bond strength: 2 < 3 < 1
b. and c.
CH3
CH2 C CHCH3
CH3
CH3
H CH2 C CHCH3
H H
H
1° radical
least stable
2° radical
intermediate stability
H CH2 C CHCH3
H
3° radical
most stable
d. increasing ease of H abstraction: 1 < 3 < 2
15.33 Use the directions from Answer 15.2 to rank the radicals.
a.
(CH3)2CHCH2CH(CH3)CH2
1° radical
least stable
(CH3)2CHCHCH(CH3)2
(CH3)2CCH2CH(CH3)2
2° radical
intermediate stability
3° radical
most stable
b.
2° radical
least stable
3° radical
intermediate stability
allylic radical
most stable
15.34 Draw the radical formed by cleavage of the benzylic C–H bond. Then draw all of the
resonance structures. Having more resonance structures (five in this case) makes the radical
more stable, and the benzylic C–H bond weaker.
CH2 H
benzylic C–H bond
bond dissociation energy = 356 kJ/mol
CH2
CH2
CH2
CH2
CH2
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–13
15.35
Hb
CH2 CHCHCHC(CH3)CH2 H
Ha
Ha = bonded to an sp3 3° carbon
Hb = bonded to an allylic carbon
Hc = bonded to an sp3 1° carbon
Hd = bonded to an sp3 2° carbon
Hd
H H
Hc
H
Increasing ease of abstraction:
Hc < Hd < Ha < Hb
15.36
Cl
a.
Cl
b.
(CH3)3CCH2CH2CH2CH2Cl
(CH3)3CCH2CH2CH2CH3
Cl
(CH3)3CCH2CH2CHCH3
(CH3)3CCH2CHCH2CH3
CH2Cl
Cl
(CH3)2CCH2CH2CH2CH3
(CH3)3CCHCH2CH2CH3
Cl
Cl
Cl
c.
Cl
Cl
Cl
d.
CH3
CH3
Cl
CH3
CH3
CH2Cl
15.37 To draw the product of bromination:
• Draw out the starting material and find the most reactive C–H bond (on the most
substituted C).
• The major product is formed by cleavage of the weakest C–H bond.
Br
a.
c.
Br
CH3
CH3
Br
b.
(CH3)3CCH2CH(CH3)2
(CH3)3CCH2C(CH3)2
d.
(CH3)3CCH2CH3
(CH3)3CCHCH3
Br
15.38 Draw all of the alkane isomers of C6H14 and their products on chlorination. Then determine
which letter corresponds to which alkane.
A
Cl
Cl
Cl2
Cl
*
*
hν
Cl
*
Cl
Cl
Cl2
hν
B
*
Cl
* *
Cl
Cl
410
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–14
Cl2
*
hν
Cl
Cl
C
Cl
Cl2
*
Cl
hν
D
Cl
Cl2
Cl
hν
*
Cl
Cl
*
[* = stereogenic center]
E
15.39 Halogenation replaces a C–H bond with a C–X bond. To find the alkane needed to make
each of the alkyl halides, replace the X with aV H.
Br
a.
c.
Cl
Br
b.
d.
(CH3)3CCH3
(CH3)3CCH2Cl
15.40 For an alkane to yield one major product on monohalogenation with Cl2, all of the hydrogens
must be identical in the starting material. For an alkane to yield one major product on
bromination, it must have a more substituted carbon in the starting material.
a.
Cl
c.
d.
Cl
b.
Br
Br on 2° carbon
The product with Br on 3° carbon
will form predominantly.
three different
C–H bonds
Br
These two compounds can be formed
in high yield from an alkane.
These two compounds cannot be
formed in high yield from an alkane.
15.41 A single constitutional isomer is formed in both halogenations in (a), (c), and (d).
Bromination often forms a single product by cleavage of the weakest C–H bond. For
chlorination to form a single product, the starting material must have only one kind of H that
reacts. In (c), a single chlorination product is formed because there is only one type of sp3
hybridized C–H bond.
a.
Cl2
Cl
hν
Br2
Cl2
hν
Br2
Br
Δ
Cl
Br2
Δ
Cl
hν
Br
Δ
b.
Cl2
c.
Br
Cl
d.
Cl2
Cl
hν
Br2
Δ
Br
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–15
Cl
Cl
Cl2
e.
hν
Br
Br2
Δ
15.42 In bromination, the predominant (or exclusive) product is formed by cleavage of the weaker
C–H bond to form the more stable radical intermediate.
CH3
Br2
CH2 H
Δ
CH3
CH2Br
CH3
weaker bond
H
o
C
stronger bond ΔH = 356 kJ/mol
As usual, more product is formed
ΔHo = 460 kJ/mol
by homolysis of the weaker bond.
CH3
Br
D
NOT formed
15.43 Chlorination is not selective so a mixture of products results. Bromination is selective, and
the major product is formed by cleavage of the weakest C–H bond.
Cl
Cl
Cl
Cl2
a.
Δ
Cl
Y
Cl
Br2
b.
Δ
Br
Y
K+–OC(CH3)3
Br2
c.
Δ
Y
Br
15.44 Draw the resonance structures by moving the π bonds and the radical.
CH2
a.
b.
Cl
CH2
c.
Z
412
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–16
15.45 Reaction of an alkene with NBS + hν yields allylic substitution products.
Br
NBS
a.
hν
Br
NBS
b. CH3CH2CH CHCH2CH3
hν
NBS
c. (CH3)2C CHCH3
(CH3)2C
hν
Br
CH3CH2CH CHCHCH3
CHCH2Br
(CH3)2C(Br)CH
Br
CH3CH2CHCH CHCH3
CH2 C(CH3)CH(Br)CH3
BrCH2C(CH3) CHCH3
CH2
Br
NBS
d.
hν
Br
Br
15.46 It is not possible to form 5-bromo-1-methylcyclopentene in good yield by allylic bromination
because several other products are formed. 1-Methylcyclopentene has three different types
of allylic hydrogens (labeled with *), all of which can be removed during radical
bromination.
Br
*
*
NBS
Br
Br
hν
*
Br
Br
5-bromo-1-methylcyclopentene
15.47
Br
NBS
hν
Br
X
Br
Br
Br
Br
15.48
Cl
+ Cl2
a.
hν
Cl
Cl
Cl
CH3
+ Br2
b.
CH3
CH3
Δ
CH3
Br
CH3
CH3
(major product)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
413
Radical Reactions 15–17
Br
c.
hν
+ Br2
Br
Br
(minor product)
(two major products)
Br
CH2
d.
hν
+ NBS
CH2
Br
Br
Br2
HBr
e.
Br
h.
HBr
ROOR
f.
Br
Br
Br
g.
CH2
hν
HBr
ROOR
CH=CHCH2Br
NBS
i.
Br
15.49
Br
Br
HBr
a.
Br2
b.
Br
NBS
c.
hν
Br
+
enantiomer
15.50
KOC(CH3)3
Br2
hν
Br
A
B
ozonolysis
O
O
Br
acetone
cyclohexanone
C
D
15.51
CH3
a.
CH3
b.
Br2
hν
Cl2
hν
CH3
Br
CH2Cl
CH3
CH3
CH3
Cl
Cl
CH3
CH3
Cl
Cl
Cl
CH3
Cl
CH3
Cl
414
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–18
Cl
Cl2
c.
Cl
Cl
Cl
hν
Cl
Cl
Cl
Cl
CH3
d.
Br2
CH3
CH3
Br
Br
hν
CH3
CH3
CH3
CH3
Br2
Δ
e.
H
CH3
+
CH3 Br
CCl3
f.
C
CH3CH2
CH3
Br CH3
CCl3
Cl2
Δ
F
H
C
ClCH2CH2
CCl3
+
F
H
C
CH3CH2
CCl3
+
F
Cl
F
Cl
+
C
CH2CH3
F H
F H
CH3
C
C
CCl3
CH3
+
H Cl
C
C
CCl3
Cl H
15.52
Cl
Cl
Cl2
a.
hν
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
A
B
C
D
E Cl
(2R)-2-chloropentane
Cl
Cl Cl
G
F
Cl
b. There would be seven fractions, since each molecule drawn has different physical properties.
c. Fractions A, B, D, E, and G would show optical activity.
15.53
Cl2
a.
H
Cl
Cl
Cl
hν
Cl
1
2
3
Cl
4
Cl
Cl
Cl
5
6
9
Cl
7
Cl
10
11
8
Cl
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
415
Radical Reactions 15–19
a. There would be 10 fractions, since 4 and 5 (two enantiomers) would be in the same fraction.
b. All fractions except the one that contains 4 and 5 would be optically active.
15.54
Br
NBS
CH2
Br
CH2
hν
CH2
CH2
A
CH2
Br
CH2Br
Br
CH2Br
15.55
CH3
a.
CH3
hν
CH3 C CH3 + Br2
CH3 C CH3 + HBr
H
Br
C–H bond broken
+381 kJ/mol
Br–Br bond broken
+192 kJ/mol
C–Br bond formed
–272 kJ/mol
total bonds broken = +573 kJ/mol
Br
b. Initiation:
Propagation:
hν
or Δ
Br
(CH3)3C
H
(CH3)3C
Termination:
(one possibility)
Br
+
+
Br
Br
total bonds formed = –640 kJ/mol
+
ΔH° = –67 kJ/mol
Br
H Br
(CH3)3C
Br Br
Br
H–Br bond formed
–368 kJ/mol
(CH3)3C
Br
Br
c. ΔH° = (bonds broken) – (bonds formed)
= (+381 kJ/mol) + (–368 kJ/mol)
= +13 kJ/mol
ΔH° = (bonds broken) – (bonds formed)
= (+192 kJ/mol) + (–272 kJ/mol)
= –80 kJ/mol
Br Br
d, e.
Transition state 1:
Transition state 2:
+13
kJ/mol
Energy
Transition state 1:
(CH3)3C
H
CH3
(CH3)3C
CH3
+ Br
δ
C CH
3
H
δ
–80
kJ/mol
(CH3)3C
Reaction coordinate
Br
Br
Br
Transition state 2:
δ
CH3
CH3 C CH3
Br
Br
δ
416
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–20
15.56
O
Initiation:
O
N
Br
N
hν
O
NBS
+
Br
O
Propagation:
+
Br
H
Br
Br
Br
+
Br (from NBS)
H Br
(from NBS)
+
Br
+
Br
Br
Br
Termination:
(one possibility)
+
Br Br
Br
15.57 Calculate the H° for the propagation steps of the reaction of CH4 with I2 to show why it
does not occur at an appreciable rate.
CH3 H
+
I
CH3
+435 kJ/mol
ΔHo = +138 kJ/mol
I
–297 kJ/mol
I
CH3
H
I
CH3
+151 kJ/mol
I
This step is highly endothermic, making it
difficult for chain propagation to occur
over and over again.
ΔHo = –83 kJ/mol
I
–234 kJ/mol
15.58 Calculate ΔH° for each of these steps, and use these values to explain why this alternate
mechanism is unlikely.
[1]
CH4
+
Cl
C–H bond broken
+435 kJ/mol
[2]
H
+
Cl2
Cl–Cl bond broken
+242 kJ/mol
CH3Cl + H
C–Cl bond formed
–351 kJ/mol
HCl +
ΔH° = +84 kJ/mol
The ΔH° for Step [1] is very endothermic,
making this mechanism unlikely.
Cl
H–Cl bond formed ΔH° = –189 kJ/mol
–431 kJ/mol
15.59
H Br
Br–
1,2-CH3 shift
Br
Br–
3,3-dimethyl-1-butene
2o carbocation
3o carbocation
2-bromo-2,3-dimethylbutane
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–21
Br
3,3-dimethyl-1-butene
HBr
Br
HBr
peroxide
Br
The 2o radical does
NOT rearrange.
Br
1-bromo-3,3-dimethylbutane
Addition of HBr without added peroxide occurs by an ionic mechanism and forms a 2o carbocation,
which rearranges to a more stable 3o carbocation. The addition of H+ occurs first, followed by Br–.
Addition of HBr with added peroxide occurs by a radical mechanism and forms a 2o radical that does
not rearrange. In the radical mechanism Br• adds first, followed by H•.
15.60
H
Step 1:
CH3 CH CH2
+
Cl
CH3
ΔH1° = –72 kJ/mol
C CH2
Cl
1 bond broken
+267 kJ/mol
1 bond formed
–339 kJ/mol
H
Step 2:
H
CH3 C CH2
+ H–Cl
CH3 C CH2
Cl
H Cl
1 bond broken
+431 kJ/mol
1 bond formed
–397 kJ/mol
ΔH2° = +34 kJ/mol
This step of propagation is endothermic. It
prohibits chain propagation from occurring over
and over.
15.61
Cl2
a.
Δ
K+ –OC(CH3)3
Cl
mCPBA
e.
O
(from a.)
NBS
b.
ROOR
Br
f.
(from a.)
(from a.)
g.
H2SO4
Br
Br
−OH
OH
Cl
OH
h.
O
Cl
(from a.)
(from e.)
[1] NaCN
[2] H2O
CN
15.62
a.
Br
(from b.)
Cl2
d.
OH
Br
Br2
(from b.)
H2O
c.
Br
Br2
Br
H 2O
K+–OC(CH3)3
hν
H2SO4
major
product
OH
418
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–22
mCPBA
b.
Cl2
c.
O
Cl
+ enantiomer
Cl
(from a.)
(from a.)
15.63
Br2
a. CH3CH2CH3
Br
b.
CH3CHCH3
hν
K+ –OC(CH3)3
CH3CH CH2
Br Br
Br2
K+ –OC(CH3)3
CH3 C C H
K+ –OC(CH3)3
HBr
CH3C CH
(2 equiv)
DMSO
H H
Br
Br
ROOR
c. HC CH
NaH
H2
HBr
Lindlar catalyst
ROOR
CH3CH2Br
HC C
Br
15.64
a. CH3 CH3
b.
Br2
hν
NaH
HC CH
HC C
(from a.)
O
c. CH2 CH2
(from a.)
mCPBA
d. HC CCH2CH3
(from b.)
K+ –OC(CH3)3
CH3 CH2Br
CH3 CH2Br
C
(from b.)
Br2
2 NaNH2
BrCH2 CH2Br
HC CCH2CH3
(from a.)
[1] HC
CH2 CH2
HC CCH2CH2OH
[2] H2O
NaH
C CCH2CH3
CH3 CH2Br
(from a.)
CH3CH2 C C CH2CH3
Na
NH3
H2 O
e. CH3CH2 C C CH2CH3
H2SO4
HgSO4
(from d.)
O
15.65
Cl2
hν
Cl
K+ –OC(CH3)3
[1] O3
OHC
[2] (CH3)2S
CHO
HC CH
419
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Radical Reactions 15–23
15.66
O2 abstracts aN H here.
O O
COOH
HH
COOH
COOH
O O
C5H11
C5H11
C5H11
arachidonic acid
+ HOO
R H
OOH
COOH
COOH
+
C5H11
O O
COOH
R H
C5H11
C5H11
another molecule of
arachidonic acid
5-HPETE
This process is repeated.
15.67
[1]
O
H
O O
O O
[2]
H
O
O
O O
O
O
[3]
+ HOO
O OH
+
O
Then, repeat
Steps [2] and [3].
15.68
HOO
a.
(cis and trans)
OOH
O2 abstracts aN H here.
O O
b.
O O
[1]
H H
+ HOO
[2a]
[3a]
R H
O O
O O
[2b]
O O
[RH = 1-hexene]
OOH
[3b]
O O
+ R
HOO
+ R
R H
Repeat Steps [2] and [3] again and again.
15.69 For resonance structures A–F, an additional resonance form can be drawn that moves the
position of the three π bonds in the ring bonded to two OH groups.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–24
O
O
HO
O
HO
HO
a.
OH
A
OH
B
OH
O
C
O
HO
O
HO
HO
F
OH
OH
E
OH
O
D
O
HO
HO
O
HO
OH
OH
OH
b. Homolysis of the indicated OH bond is preferred because it allows the resulting radical to
delocalize over both benzene rings. Cleavage of one of the other OH bonds gives a radical
that delocalizes over only one of the benzene rings.
15.70 Abstraction of the labeled H forms a highly resonance-stabilized radical. Four of the
possible resonance structures are drawn.
OH
HO
OH
OH
O
HO
O
OH
vitamin C
O
HO
–O
O
OH
O
HO
O
OH
O
O
HO
O
O
O
O
X
OH
OH
O
HO
O
O
O
HO
O
O
O
O
15.71 The monomers used in radical polymerization always contain double bonds.
a.
b.
CH2 CHCO2Et
COOEt COOEt COOEt
polyisobutylene
poly(ethyl acrylate)
(used in Latex paints)
Et = CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
421
Radical Reactions 15–25
15.72
a.
CH3
CH3
CH2 C
CH2 C
COOCH3
CH2
CH3
C CH2
C
COOCH3 COOCH3 COOCH3
PMMA
methyl methacrylate
b.
CH3
CH3
CH2 C
HO
CO2CH2CH2OH
O
hydroxyethyl methacrylate
poly-HEMA
O
O
O
OH
15.73 Polystyrene has H atoms bonded to benzylic carbons, carbons bonded directly to a benzene
ring. These C–H bonds are unusually weak because the radical that results from homolysis is
resonance stabilized.
H
–H
No such resonance stabilization is possible for the radical
that results from C–H bond cleavage in polyethylene.
15.74
Overall reaction:
CH2 CHCN
ROOR
CN
Initiation:
RO
OR
[1]
CN
CN
+ CH2 C
2 RO
CN
CN
[2]
ROCH2 C
H
H
CN
CN
Propagation:
ROCH2 C
CN
[3]
ROCH2 C
CH2 C
H
CN
NC
CH2 C
C CH2
H
H
CN
CH2 C
H
H
H
Repeat Step [3] over and over.
Termination:
carbon radical
[4]
new C–C bond
CN CN
CH2 C C CH2
H H
[one possibility]
422
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–26
15.75
a. CH3O
CH=CH2
A
OCH3
OCH3
OCH3
b. The OCH3 group stabilizes an intermediate carbocation by resonance. This makes A
react faster than styrene in cationic polymerization.
OCH3
OCH3
OCH3
three of the possible resonance structures
15.76
Cl
C6H5
Cl
Cl Cl
C6H5
Cl Cl
C6H5
Cl Cl
C6H5
alternating copolymer
15.77
singlet singlet
at 2.23 at 4.04
Cl2
Cl
Cl
Cl
hν
Cl
doublet
at 1.69
Cl Cl
A
C
doublet at 5.85
Cl
multiplet Cl B
at 4.34
15.78
triplet
Cl2
hν
Cl
Cl
quintet
minor product
molecular formula C3H6Cl2
Integration:
(57 units + 29 units)/6 H's = 14 units per H
one signal is 57 units/14 units per H = 4 H's
second signal is 29 units/14 units per H = 2 H's
1H NMR data:
quintet at 2.2 (2 H's) split by 4 H's
triplet at 3.7 (4 H's) split by 2 H's
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
423
Radical Reactions 15–27
15.79
Cl2
a. CH3CH3
equivalent H's
singlet
doublet quartet
ClCH2CH2Cl
CH3CHCl2
hν
Cl
b. Initiation:
Propagation:
X
hν
Cl
Y
Cl
or Δ
+
Cl
H
H
+
CH3 C H
Cl
CH3 C
H
H Cl
H
H
H
+
CH3 C
Cl
Cl
CH3 C Cl
Cl
H
H
H
Cl
Formation of Y: CH3 CH Cl
Cl
Cl
Cl
H CH2 CH2Cl
Cl
Cl
CH3CHCl2
Y
CH3 CH Cl
Formation of X:
H Cl
CH3 CH Cl
H Cl
CH2 CH2Cl
Cl
CH2 CH2Cl
Cl
ClCH2CH2Cl
X
Termination:
Cl
Cl
+
Cl Cl
15.80
RO
OR
RO
O
O
CH3
C
OR
O
[1]
H
OR
CH3
[2]
C
O
H
C
H OR
CH3 [3]
O
+
O
CH3
C
(Repeat Steps [2]
and [3].)
15.81 a. The triphenylmethyl radical is highly resonance stabilized, because the radical can be
delocalized on each of the benzene rings. As an example using one ring:
C6H5
C6H5
C6H5
C6H5
C6H5
C6H5
In addition, the radical is very sterically hindered,
making it difficult to undergo reactions.
C6H5
C6H5
C6H5
C6H5
424
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 15–28
b. First, draw the resonance form of the radical that places the unpaired electron on the C that
forms the new C–C bond.
A
c. Hexaphenylethane formation would require two very crowded 3° radicals to combine.
The formation of A results from a radical on one of the six-membered rings, which is
much more accessible for reaction.
d. The 1H NMR spectrum of hexaphenylethane should show signals only in the aromatic
region (7–8 ppm), whereas the 1H NMR spectrum of A will also have signals for the sp2
hybridized C–H bonds (4.5–6.0 ppm) of the alkenes, as well as the single H on the sp3
hybridized carbon. The 13C NMR spectrum of hexaphenylethane should consist of lines
due to the 4° C’s and the aromatic C’s. For A, the 13C NMR spectrum will also have lines
for the sp3 and sp2 hybridized C’s that are not contained in the aromatic rings.
15.82
Initiation:
R3SnH
R3Sn
+ Z
Propagation:
+ HZ
CH2
Br
+ R3SnBr
R3Sn
R3SnH
CH2
R3SnH
CH2
CH2
R3SnH
+
+
R3Sn
+
R3Sn
R3Sn
15.83
CO2H
O
O
COOH
O
O
O
COOH
O
O
A
O
COOH
O
O
COOH
O
O
OOH
+ R
O O
R H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
425
Conjugation, Resonance, and Dienes 16–1
Chapter 16 Conjugation, Resonance, and Dienes
Chapter Review
Conjugation and delocalization of electron density
• The overlap of p orbitals on three or more adjacent atoms allows electron density to delocalize,
thus adding stability (16.1).
• An allyl carbocation (CH2=CHCH2+) is more stable than a 1o carbocation because of p orbital
overlap (16.2).
• In a system X=Y–Z:, Z is generally sp2 hybridized to allow the lone pair to occupy a p orbital,
making the system conjugated (16.5).
Four common examples of resonance (16.3)
[1] The three atom “allyl” system:
X
Y
Z
*
X Y
*
* = +, –, •, or ••
Z
[2] Conjugated double bonds:
[3] Cations having a positive charge
adjacent to a lone pair:
X Y
[4] Double bonds having one atom more
electronegative than the other:
X Y
+
+
X Y
+
X Y
Electronegativity of Y > X
Rules on evaluating the relative “stability” of resonance structures (16.4)
[1] Structures with more bonds and fewer charges are more stable.
CH3
more stable
resonance structure
CH3
C O
+
C O
CH3
CH3
all neutral atoms
one more bond
charge separation
[2] Structures in which every atom has an octet are more stable.
+
+
CH3 O CH2
CH3 O CH2
more stable
resonance structure
All 2nd row elements have an octet.
[3] Structures that place a negative charge on a more electronegative element are more stable.
The (–) charge is on the more
electronegative O atom.
O
CH3
C
O
CH2
CH3
C
CH2
more stable
resonance structure
426
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 16–2
The unusual properties of conjugated dienes
[1] The C–C σ bond joining the two double bonds is unusually short (16.8).
[2] Conjugated dienes are more stable than similar isolated dienes. ΔHo of hydrogenation is smaller
for a conjugated diene than for an isolated diene converted to the same product (16.9).
[3] The reactions are unusual:
• Electrophilic addition affords products of 1,2-addition and 1,4-addition (16.10, 16.11).
• Conjugated dienes undergo the Diels–Alder reaction, a reaction that does not occur with
isolated dienes (16.12–16.14).
[4] Conjugated dienes absorb UV light in the 200–400 nm region. As the number of conjugated
π bonds increases, the absorption shifts to longer wavelength (16.15).
Reactions of conjugated dienes
[1] Electrophilic addition of HX (X = halogen) (16.10, 16.11)
CH2 CH CH CH2
HX
(1 equiv)
CH2 CH CH
H
X
1,2-product
kinetic product
•
•
•
•
CH2
+
CH2 CH CH
CH2
H
X
1,4-product
thermodynamic product
The mechanism has two steps.
Markovnikov’s rule is followed. Addition of H+ forms the more stable allylic
carbocation.
The 1,2-product is the kinetic product. When H+ adds to the double bond, X– adds
to the end of the allylic carbocation to which it is closer (C2 not C4). The kinetic
product is formed faster at low temperature.
The thermodynamic product has the more substituted, more stable double bond.
The thermodynamic product predominates at equilibrium. With 1,3-butadiene, the
thermodynamic product is the 1,4-product.
[2] Diels–Alder reaction (16.12–16.14)
Z
1,3-diene
•
•
•
•
•
•
•
Δ
Z
The three new bonds
are labeled in bold.
dienophile
The reaction forms two σ and one π bond in a six-membered ring.
The reaction is initiated by heat.
The mechanism is concerted: all bonds are broken and formed in a single step.
The diene must react in the s-cis conformation (16.13A).
Electron-withdrawing groups in the dienophile increase the reaction rate (16.13B).
The stereochemistry of the dienophile is retained in the product (16.13C).
Endo products are preferred (16.13D).
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
427
Conjugation, Resonance, and Dienes 16–3
Practice Test on Chapter Review
1. a. Which of the following statements is true about the Diels–Alder reaction?
1. The reaction is faster with electron-donating groups in the dienophile.
2. The reaction is endothermic.
3. The diene must adopt the s-cis conformation.
4. Statements (1) and (2) are true.
5. Statements (1), (2), and (3) are all true.
b. Which of the following statements is true about the absorption of ultraviolet light by unsaturated
systems?
1. 1,4-Pentadiene requires light having a wavelength < 200 nm for electron promotion.
2. 1,3-Cyclohexadiene absorbs ultraviolet light with a wavelength > 200 nm.
3. As the number of conjugated π bonds increases, the energy difference between the excited
state and ground state decreases.
4. Statements (1) and (2) are true.
5. Statements (1), (2), and (3) are all true.
c. Which of the following compounds contains a labeled carbon atom that is sp2 hybridized?
CH2
A
1.
2.
3.
4.
5.
CH2
B
CH2
C
A only
B only
C only
A and B
A, B, and C
d. Which of the following represent valid resonance structures for A?
NH2
NH2
1.
A
NH2
2.
NH2
3.
4. Both (1) and (2) are valid resonance structures.
5. Structures (1), (2), and (3) are all valid.
2. Name the following compounds and indicate the conformation around the σ bond that joins the
two double bonds.
a.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 16–4
b.
3. a. Consider the four hydrocarbons (A–D) drawn below. [1] Which compound absorbs the shortest
wavelength of ultraviolet light? [2] Which compound absorbs the longest wavelength of
ultraviolet light?
A
B
C
D
b. Consider the four dienes (A–D) drawn below. [1] Which diene is most reactive in the
Diels–Alder reaction? [2] Which diene is the least reactive in the Diels–Alder reaction?
A
B
C
D
4. Draw the organic products formed in each reaction. In part (b), label the kinetic and
thermodynamic products.
a.
O
Δ
O
indicate
stereochemistry
OCH3
HBr
b.
(1 equiv)
[1]
c.
[2]
CH3O2C
Δ
CO2CH3
indicate
stereochemistry
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
429
Conjugation, Resonance, and Dienes 16–5
5. What diene and dienophile are needed to synthesize the following Diels–Alder adducts?
O
Cl
a.
b.
O
Cl
Cl
CH3O
CN
CH3O
Answers to Practice Test
1.a. 3
b. 5
c. 4
d. 1
4.a.
5.a.
H
CH3O
b. (2Z,4E)-3-ethyl-6,6-dimethyl2,4-nonadiene, s-trans
CH3O
CH3O
H
2.a. (4Z,6E)-6,7-diethyl-2methyl-4,6-decadiene, s-trans
O
O
O
(both H's up or both H's down)
O
b.
b.
Cl Cl
Cl
3.a. [1] A; [2] C
b. [1] D; [2] A
Br
kinetic
Br
thermodynamic
c.
CO2CH3
CO2CH3
CN
430
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 16–6
Answers to Problems
16.1 Isolated dienes have two double bonds separated by two or more σ bonds.
Conjugated dienes have two double bonds separated by only one σ bond.
a.
b.
One σ bond separates
two double bonds =
conjugated diene
c.
Two σ bonds separate
two double bonds =
isolated diene
One σ bond separates
two double bonds =
conjugated diene
Four σ bonds separate
two double bonds =
isolated diene
Conjugation occurs when there are overlapping p orbitals on three or more adjacent atoms.
Double bonds separated by two σ bonds are not conjugated.
16.2
a.
CH2 CH CH CH CH CH2
e.
c.
O
All of the carbon atoms are sp2
hybridized. Each π bond is
separated by only one σ bond.
conjugated
The two π bonds are
separated by only one σ bond.
conjugated
d.
b.
+
This carbon is not sp2 hybridized.
NOT conjugated
+
Three adjacent carbon atoms are sp2 hybridized
and have an unhybridized p orbital.
conjugated
The two π bonds are
separated by three σ bonds.
NOT conjugated
16.3
d.
Two resonance structures differ only in the placement of electrons. All σ bonds stay in the
same place. Nonbonded electrons and π bonds can be moved. To draw the hybrid:
• Use a dashed line between atoms that have a π bond in one resonance structure and not
the other.
• Use a δ symbol for atoms with a charge or radical in one structure but not the other.
resonance hybrid:
a.
+
δ+
+
δ
+
The + charge is delocalized
on two carbons.
resonance hybrid:
b.
+
+
δ+
The + charge is delocalized
on two carbons.
δ+
The + charge is delocalized
on two carbons.
δ+
resonance hybrid:
c.
16.4
+
+
δ+
SN1 reactions proceed via a carbocation intermediate. Draw the carbocation formed on loss
of Cl and compare. The more stable the carbocation, the faster the SN1 reaction.
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Conjugation, Resonance, and Dienes 16–7
CH3CH2CH2Cl is a 1o halide, which does not react by
an SN1 reaction because cleavage of the C–Cl bond
forms a highly unstable 1o carbocation.
3-chloro-1-propene CH2 CHCH2Cl
more reactive
CH2
CH CH2
CH2 CH
CH2
1-chloropropane
less reactive
resonance-stabilized carbocation
Two resonance structures delocalize
the positive charge on 2 C's making
3-chloro-1-propene more reactive.
CH3CH2CH2Cl
CH3CH2CH2
only one Lewis structure
very unstable
16.5
a. CH2 CH CH CH CH2
CH2 CH CH CH
CH2
CH2
CH CH CH CH2
Move the charge
and the double bond.
O
b.
CH3CH2
C
O
CH3
C
CH3CH2
C
H
+
+
C
c. CH3 CH Cl
CH3
CH3 CH Cl
Move the lone pair.
H
Move the charge
and the double bond.
d.
Move the charge
and the double bond.
16.6 To compare the resonance structures remember:
• Resonance structures with more bonds are better.
• Resonance structures in which every atom has an octet are better.
• Resonance structures with neutral atoms are better than those with charge separation.
• Resonance structures that place a negative charge on a more electronegative atom are
better.
no octet
one more bond
+
a. CH3 C NH2
δ+
+
NH2
NH2
CH3 C
CH3 C
CH3
δ+
c.
hybrid CH3
least stable
most stable
one more bond
All atoms have an octet.
better resonance structure
intermediate stability
CH3
b.
CH3
C
NH
least stable
CH3
C
least stable
δ+
+
CH3(CH2)3C O
CH3(CH2)3C+ O
δ
one more bond
better resonance structure
intermediate stability
most stable
δ−
O
O
O
CH3(CH2)3C O
+
NH
negative charge on the
more electronegative atom
better resonance structure
intermediate stability
CH3
C
δ−
NH
hybrid
most stable
d.
N
least stable
N
one more bond
better resonance structure
intermediate stability
δ+
δ–
N
most stable
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 16–8
16.7
OCH3
OCH3
OCH3
OCH3
largest contribution
more bonds and all
atoms have octets
16.8 Remember that in any allyl system, there must be p orbitals to delocalize the lone pair.
CH2
O
a.
b.
O
CH3 C
c.
O
sp2 hybridized
trigonal planar geometry
sp2 hybridized
trigonal planar geometry
sp2 hybridized
trigonal planar geometry
The s-cis conformation has two double bonds on the same side of the single bond.
The s-trans conformation has two double bonds on opposite sides of the single bond.
16.9
a. (2E,4E)-2,4-octadiene in the s-trans conformation
c. (3Z,5Z)-4,5-dimethyl-3,5-decadiene in
both the s-cis and s-trans conformations
Z
double bonds on opposite sides
s-trans
b. (3E,5Z)- 3,5-nonadiene in the s-cis conformation
s-cis
Z
s-trans
Z
Z
Z
E
double bonds on the same side
s-cis
16.10
s-cis
conjugated
s-trans
OH
conjugated
E E
conjugated
isolated
Z
Z
HO
isolated
Z
O
isolated
OH
Z
isolated
16.11 Bond length depends on hybridization and percent s-character. Bonds with a higher percent
s-character have smaller orbitals and are shorter.
HC C C CH
sp hybridized carbons
50% s-character
shortest bond
CH2 CH CH CH2
CH3 CH3
sp2 hybridized carbons
33% s-character
intermediate length
sp3 hybridized carbons
25% s-character
longest bond
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Conjugation, Resonance, and Dienes 16–9
16.12 Two equivalent resonance structures delocalize the π bond and the negative charge.
O
O
CH3 C
CH3
CH3 C
O
O δ−
hybrid:
O
C
O δ−
These bond lengths are equal
because they are identical.
16.13 The less stable (higher energy) diene has the larger heat of hydrogenation. Isolated dienes
are higher in energy than conjugated dienes, so they will have a larger heat of hydrogenation.
or
a.
Double bonds separated by
two σ bonds = isolated diene
larger heat of hydrogenation
Double bonds separated by
one σ bond = conjugated diene
smaller heat of hydrogenation
or
b.
Double bonds separated by
two σ bonds = isolated diene
larger heat of hydrogenation
Double bonds separated by
one σ bond = conjugated diene
smaller heat of hydrogenation
16.14 Isolated dienes are higher in energy than conjugated dienes. Compare the location of the
double bonds in the compounds below.
*
*
0 conjugated double bonds
least stable
*
*
3 conjugated double bonds
most stable
*
2 conjugated double bonds
intermediate stability
16.15 Conjugated dienes react with HX to form 1,2- and 1,4-products.
a.
HCl
CH3 CH CH CH CH CH3
CH3 CH CH
H
Cl
CH CH CH3
Cl
H
isolated diene
c.
Cl
HCl
Cl
Cl
d.
Cl
HCl
A
B
Cl
1,4-product
1,2-product
HCl
b.
CH3 CH CH CH CH CH3
C
This double bond is more reactive, so C is probably a minor product
because it results from HCl addition to the less reactive double bond.
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Chapter 16–10
16.16 The mechanism for addition of DCl has two steps:
[1] Addition of D+ forms a resonance-stabilized carbocation.
[2] Nucleophilic attack of Cl– forms 1,2- and 1,4-products.
[1]
D
D Cl
+ Cl
D
Cl
[2]
[2]
Cl
Cl
D
D
16.17 Label the products as 1,2- or 1,4-products. The 1,2-product is the kinetic product, and the
1,4-product, which has the more substituted double bond, is the thermodynamic product.
This is C1.
The H added here.
The H added here.
CH3
CH3
HCl
CH3
Cl
+
Cl
This is C4.
1,2-product
kinetic product
1,4-product
thermodynamic product
16.18 To draw the products of a Diels–Alder reaction:
[1] Find the 1,3-diene and the dienophile.
[2] Arrange them so the diene is on the left and the dienophile is on the right.
[3] Cleave three bonds and use arrows to show where the new bonds will be formed.
COOH
+
a.
Δ
re-draw
COOH
COOH
dienophile
diene
CH3
Δ
re-draw
+
b.
COOCH3
COOCH3
diene
Rotate to
make it s-cis.
CH3
+
O
dienophile
CH3
dienophile
re-draw
c.
COOCH3
diene
Rotate to
make it s-cis.
O
Δ
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
435
Conjugation, Resonance, and Dienes 16–11
16.19 For a diene to be reactive in a Diels–Alder reaction, a diene must be able to adopt an s-cis
conformation.
rotate
s-trans
cannot rotate
unreactive
s-cis
most reactive
The diene is always in the
s-cis conformation.
s-cis
reactive
16.20 Zingiberene reacts much faster than β-sesquiphellandrene as a Diels–Alder diene because its
diene is constrained in the s-cis conformation. The diene in β-sesquiphellandrene is
constrained in the s-trans conformation, so it is unreactive in the Diels–Alder.
16.21 Electron-withdrawing substituents in the dienophile increase the reaction rate.
H
H
CH2 CH2
HOOC
COOH
no electron-withdrawing groups
least reactive
H
C C
CH2 C
one electron-withdrawing group
intermediate reactivity
COOH
two electron-withdrawing groups
most reactive
16.22 A cis dienophile forms a cis-substituted cyclohexene.
A trans dienophile forms a trans-substituted cyclohexene.
CH3OOC
a.
COOCH3
H
COOCH3
Δ
C C
+
H
COOCH3
cis dienophile
CH3OOC
b.
COOCH3
Δ
+
COOCH3
COOCH3
COOCH3
trans-substituted products
O
Δ
+
COOCH3
+
trans dienophile
c.
COOCH3
cis-substituted products
H
H
COOCH3
+
H
O
H
O
+
O
cis dienophile
H
O
identical
cis-substituted product
H
O
436
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Chapter 16–12
16.23 The endo product (with the substituents under the plane of the new six-membered ring) is the
preferred product.
+
a.
Δ
CH2 CHCOOCH3
COOCH3
COOCH3
Δ
+
b.
endo substituent
COOCH3 both groups endo
COOCH3
COOCH3
16.24 To find the diene and dienophile needed to make each of the products:
[1] Find the six-membered ring with a C–C double bond.
[2] Draw three arrows to work backwards.
[3] Follow the arrows to show the diene and dienophile.
COOCH2CH3
COOCH2CH3
COOCH2CH3
+
a.
CH3O
CH3O
COOCH3
COOCH3
CH3O
COOCH3
+
b.
COOCH3
COOCH3
Cl
Cl
Cl
H
H
c.
CH3OOC
Cl
H
H
O
O
Cl
O
Cl
O
O
+
O
O
O
O
16.25
CH3O
CH3O
+
CH3O
CH3O
NC
NC
H
(+ enantiomer)
O
O
H O
A
O
16.26 Conjugated molecules absorb light at a longer wavelength than molecules that are not
conjugated.
a.
or
conjugated
longer wavelength
b.
not conjugated
or
all double bonds conjugated
longer wavelength
one set of
conjugated dienes
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
437
Conjugation, Resonance, and Dienes 16–13
16.27 Sunscreens contain conjugated systems to absorb UV radiation from sunlight. Look for
conjugated systems in the compounds below.
O
O
a.
O
b.
CH3O
O
c.
CH3O
OH
conjugated system
could be a sunscreen
not a conjugated system
conjugated system
could be a sunscreen
16.28
2
a.
3
b.
s-cis conformation
3
2
4
(2Z,4E)-3,4-dimethyl-2,4-heptadiene
4
s-trans conformation
(2E,4Z)-3,4-dimethyl-2,4-octadiene
16.29
O
a.
CO2CH3
H OCH3
H
O
CH3O2C
OCH3
H
O
O
O
O
b.
H
16.30 Use the definition from Answer 16.1.
CH2=CHC N
3 π bonds with only
1 σ bond between
conjugated
2 multiple bonds with only
1 σ bond between
conjugated
CH2
3 π bonds with 2 or
more σ bonds between
NOT conjugated
This C is sp2.
1 π bond with
an adjacent sp2 hybridized atom
The lone pair occupies a p orbital,
so there are p orbitals on
three adjacent atoms.
conjugated
1 π bond with
no adjacent sp2 hybridized atoms
NOT conjugated
CH2OCH3
1 π bond with
no adjacent sp2 hybridized atoms
NOT conjugated
438
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Chapter 16–14
16.31
isolated isolated
isolated
O
OOH
C
OH
5-HPETE
conjugated
16.32
b.
CH3O
d.
a.
CH CH CH2
CH3O
CH CH CH2
CH3O CH CH CH2
CH2
CH2
e.
O
O
O
c.
O
O
O
O
O
O
f.
16.33
CH2
CH2
CH2
CH2
CH2
a.
OH
OH
OH
OH
OH
b.
CH2
c.
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
OCH3
OCH3
OCH3
OCH3
OCH3
d.
CH2
OCH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
439
Conjugation, Resonance, and Dienes 16–15
16.34 No additional resonance structures can be drawn for compounds (a) and (d).
OCH3
OCH3
b.
c.
N
N
CH3
CH3
16.35
resonance hybrid:
δ−
Five resonance structures delocalize the negative charge
on five C's, making them all equivalent.
δ−
δ−
δ−
All of the carbons are identical in the anion.
δ−
16.36
CH2
a.
CH CH2
H
CH3CH2CH2 H
more acidic
CH2 CH CH2
less acidic
CH2 CH
CH3CH2CH2
CH2
only one Lewis structure
Resonance stabilization delocalizes the
negative charge on 2 C's after loss of a proton.
This makes propene more acidic than propane.
b. Draw the products of cleavage of the bond.
ethane CH3 CH3
CH3
+
CH3
1-butene CH3 CH2CH=CH2
CH3
Two unstable radicals form.
+
CH2 CH CH2
CH2 CH CH2
One resonance-stabilized radical forms.
This makes the bond dissociation
energy lower because a more stable radical is formed.
16.37 Use the directions from Answer 16.9.
a. (3Z)-1,3-pentadiene in the s-trans conformation
double bonds on opposite sides
s-trans
c. (2E,4E,6E)-2,4,6-octatriene
d. (2E,4E)-3-methyl-2,4-hexadiene in the s-cis conformation
b. (2E,4Z)-1-bromo-3-methyl-2,4-hexadiene
Br
s-cis
440
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 16–16
16.38
E
Z
C
1,4-pentadiene
(3E)-1,3-pentadiene
(3Z)-1,3-pentadiene
2-methyl-1,3-butadiene
1,2-pentadiene
C
C
2,3-pentadiene
3-methyl-1,2-butadiene
16.39
2E,4E
2E,4Z
2Z,4Z
2Z,4E
16.40
and
a.
Z
and
c.
E
(3E)-1,3,5-hexatriene
(3E)-1,3,5-hexatriene
both s-cis
both s-trans
different conformations
b.
(3E)-1,3,5-hexatriene
(3Z)-1,3,5-hexatriene
different stereoisomers
and
(3Z)-1,3,5-hexatriene
(3Z)-1,3,5-hexatriene
both s-cis
both s-trans
different conformations
16.41 Use the directions from Answer 16.13 and recall that more substituted double bonds are more
stable.
Increasing heat of hydrogenation
conjugated diene
one tetra-, one disubstituted
double bond
smallest
heat of hydrogenation
conjugated diene
one di-, one trisubstituted
double bond
smaller intermediate
heat of hydrogenation
isolated diene
one di-, one trisubstituted
double bond
larger intermediate
heat of hydrogenation
isolated diene
both disubstituted
double bonds
largest
heat of hydrogenation
16.42 Conjugated dienes react with HX to form 1,2- and 1,4-products.
Br
a.
HBr
(1 equiv)
isolated diene
major product, formed by addition of HBr to
the more substituted C=C
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Conjugation, Resonance, and Dienes 16–17
b.
Br
Br
HBr
(E and Z isomers can form.)
(1 equiv)
1,2-product
1,4-product
Br
Br
Br
HBr
c.
(1 equiv)
Br
1,2-product
1,4-product
1,2-product
(E and Z isomers)
1,4-product
16.43
Br
H Br
H
Br
Br
Br
Br
Br
16.44
This cation forms because it is benzylic and resonance stabilized.
CH CHCH3
A
CH CH2CH3
H Br
Br
CH CH2CH3
CH CH2CH3
CH CH2CH3
CH CH2CH3
Br
CHCH2CH3
C
H
CH2CH CH2
C–CH
CH3
H
B
H Br
2o
carbocation
1,2H shift
CH CH2CH3
Br
This 2o carbocation
is also benzylic, making it
resonance stabilized, as above.
Br
CHCH2CH3
C
16.45 To draw the mechanism for reaction of a diene with HBr and ROOR, recall from Chapter 15
that when an alkene is treated with HBr under these radical conditions, the Br ends up on the
carbon with more H’s to begin with.
442
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Chapter 16–18
RO OR
RO
HOR +
H Br
OR
Br
Br
Use each resonance structure to react with HBr.
Br
H Br
Br
Br
Br
Br
H Br
Br
Br
16.46
C2
CH2
a, b.
HCl
CH3
CH3
Cl
X
C4
Y
H adds here at C1.
Z
Cl
Cl added at C4.
1,4-product
thermodynamic
product
Cl added at C2.
1,2-product
kinetic product
Y is the kinetic product because of the proximity effect. H and Cl add across two adjacent atoms.
Z is the thermodynamic product because it has a more stable trisubstituted double bond.
If addition occurred at the other C=C,
the following allylic carbocation would form:
Addition occurs at the labeled double bond due to the
stability of the carbocation intermediate.
c.
CH2
CH3
CH3
CH2
The two resonance structures for this allylic
cation are 3o and 2o carbocations.
more stable intermediate
Addition occurs here.
CH2
The two resonance structures for this allylic
cation are 1o and 2o carbocations.
less stable
16.47 Addition of HCl at the terminal double bond forms a carbocation that is highly resonance
stabilized since it is both allylic and benzylic. Such stabilization does not occur when HCl is
added to the other double bond. This gives rise to two products of electrophilic addition.
Cl
H Cl
Cl
1,2-product
Cl
Cl
(+ three more resonance structures that delocalize
the positive charge onto the benzene ring)
1,4-product
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
443
Conjugation, Resonance, and Dienes 16–19
16.48 There are two possible products:
disubstituted C=C
(CH3)2C
H
CH CH C(CH3)2
(CH3)2C
Br
CH CH C(CH3)2
H
Br
trisubstituted C=C
1,2-product
1,4-product
The 1,2-product is always the kinetic product because of the proximity effect. In this case, it is also
the thermodynamic (more stable) product because it contains a more highly substituted C=C
(trisubstituted) than the 1,4-product (disubstituted). Thus, the 1,2-product is the major product at
high and low temperature.
16.49 The electron pairs on O can be donated to the double bond through resonance. This increases
the electron density of the double bond, making it less electrophilic and therefore less
reactive in a Diels–Alder reaction.
CH2 CH OCH3
CH2
+
CH OCH3
methyl vinyl ether
This C now bears a net negative charge.
16.50 Use the directions from Answer 16.18.
Δ
a.
re-draw
dienophile
diene
COOCH3
Δ
b.
Cl
diene
COOCH3
Cl
Cl
trans-substituted products
trans dienophile
COOCH3
c.
COOCH3
COOCH3
COOCH3
Cl
Cl
Δ
Cl
diene
cis dienophile
cis-substituted products
O
H O
Δ
d.
H
diene
dienophile
endo ring
Δ
e.
O
re-draw
O
H
endo substituent
diene
dienophile
O
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Chapter 16–20
O
f.
O
H
H
Δ
+
O
H
Both C=C's of the
dienophile react so two
Diels–Alder reactions
take place.
or
excess
H
O
O
diene
H
H
H
O
H
dienophile
16.51 Use the directions from Answer 16.24.
CH3
a.
CH3
COOCH3
COOCH3
COOCH3
COOCH3
COOCH3
b.
+
CH3
CH3
CH3
O
c.
CH3
COOCH3
O
H
H
COOCH3
COOCH3
O +
COOCH3
+
d.
identical
Cl
e.
Cl
O
O C
O
CH3
O
f.
O
O
O
+
O C
Cl
CH3
O
O
O
O
+
O
O
O
C
CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
445
Conjugation, Resonance, and Dienes 16–21
16.52
O
O
This pathway is preferred because
the dienophile has electronwithdrawing C=O groups that
make it more reactive.
O
O
O
O
CH2
+
CH2
O
O
no electron-withdrawing groups
less reactive
16.53
COOCH3
COOCH3
Δ
+ HC C COOCH3
a.
dienophile
diene
CO2CH3
Δ
+ CH3O2C C C CO2CH3
b.
dienophile
diene
CO2CH3
CO2CH3
CO2CH3
16.54
OCH3
OCH3
+
CHO
OCH3
CHO
Δ
+
CHO
C
OCH3
H
O
C
OCH3
H
1,2-disubstituted product
major
1,3-disubstituted product
minor
The major product is formed when the circled
carbons with a δ+ and δ– react.
O
resonance hybrids:
δ+OCH3
−
Oδ
δ−
δ+ H
δ−
δ+
For the 1,2-product, carbons with unlike charges would react.
This is favored because the electron-rich and the electronpoor C's can bond.
δ+OCH3
δ+
δ−
δ+H
δ−
O δ−
For the 1,3-product, there are no partial
charges of opposite sign on reacting
carbons. This arrangement is less attractive.
446
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Chapter 16–22
16.55
COOH
Δ
+
COOH
COOH
COOH
These are the only two double bonds that are
conjugated and have the s-cis conformation
needed for a Diels–Alder reaction.
16.56
H
C
+
Δ
Δ
C
Cl
H
Cl
Cl
Cl
Cl
Cl
Cl
Cl
mCPBA
Cl
1 equiv
Cl
Cl
Cl
Cl
Cl
Cl
X
Z
aldrin
Cl
Cl
O
dieldrin
Y Cl
This double bond is more electron rich,
so it is epoxidized more readily.
16.57
In each problem, the synthesis must
begin with the preparation of
cyclopentadiene from dicyclopentadiene.
Δ
2
dicyclopentadiene
a.
HO
[1] OsO4
COOCH3
Δ
cyclopentadiene
HO
[2] NaHSO3, H2O
COOCH3
COOCH3
O
O
b.
Δ
O
mCPBA
O
O
O
c.
O
H2 (excess)
CHO
Δ
O
O
O
CHO
[1] NaH
Pd-C
[2] CH3CH2CH2CH2Br
OH
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
447
Conjugation, Resonance, and Dienes 16–23
16.58
O
Δ
re-draw
O
a.
O
dienophile
diene
COOCH3
Δ
re-draw
b.
COOCH3
COOCH3
dienophile
diene
16.59 A transannular Diels–Alder reaction forms a tricyclic product from a monocyclic starting
material.
Δ
O
O
O
O
16.60
CH3CH CHCH2OH
CH3CH CHCH2 OH2
+ Br
CH3CH CHCH2
two resonance
structures
H Br
+ H2O
CH3CH CHCH2Br
+ Br
CH3CHCH CH2
CH3CHCH CH2
Br
+ Br
16.61
a.
HCl
(CH3)2C=CHCH2CH2CH=CH2
(CH3)2C(Cl)CH2CH2CH2CH=CH2
isolated diene
major product
I
b.
HI
I
conjugated diene
1,4-product
1,2-product
O
O
H
Δ
c.
O
+
H
O
O
O
diene
dienophile
O
O
(CH3)2C=CHCH2CH2CHClCH3
minor product
448
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Chapter 16–24
d.
diene
COOCH3
Δ
COOCH3
+
COOCH3
+
trans dienophile
COOH
Δ
+
e.
+
COOH
COOH
diene
cis dienophile
HBr
f.
COOH
COOH
COOH
Br
Br
conjugated diene
1,2-product
1,4-product
16.62 The more stable the carbocation, the faster the SN1 reaction. The carbocation from A is more
stable because the lone pairs on the O atom of the OCH3 afford additional resonance
stabilization.
more stable carbocation
Br
CH3O
CH3O
CH2
A
CH3O
CH2
CH3O
CH2
CH3O
CH2
Br
CH3O
CH2
CH3O
CH2
additional resonance structure
with all atoms having an octet
less stable carbocation
O
Br
O
O
O
CH2
B
CH2
δ+
CH2
Br
O
CH2
This resonance structure is
destabilized because the (+) charge is
adjacent to the δ+ on the C=O atom.
O
CH2
16.63 The mechanism is E1, with formation of a resonance-stabilized carbocation.
OH
OH2
H
H A
A
H2O
A
H A
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
449
Conjugation, Resonance, and Dienes 16–25
16.64
Cl
β2
a.
β1
loss of H (from β1 C ) + Cl
conjugated
more stable
major product
loss of H (from β2 C) + Cl
more substituted
b. Dehydrohalogentaion generally forms the more stable product. In this reaction, loss of H from the β1
carbon forms a more stable conjugated diene, so this product is preferred even though it does not
contain the more substituted C=C.
16.65
singlet at 1.42 ppm
H
O
mCPBA
Each H is a doublet of doublets in the 5.2–5.4 ppm region.
H
H
isoprene
H
H
doublet of doublets at 5.5–5.7 ppm
two doublets at 2.7–2.9 ppm
16.66
1 H: doublet of doublets at 6.0 ppm
The IR shows an OH absorption at
3200–3600 cm–1.
H
Br
H2O
H
B
H
OH
Each H is a doublet
of doublets in the
4.9–5.2 ppm region.
OH peak at 1.5 ppm
6 H: singlet at 1.3 ppm
16.67
isolated diene
shortest wavelength
1
2 conjugated bonds
intermediate
wavelength
2
3 conjugated bonds
intermediate
wavelength
3
4 conjugated bonds
longest wavelength
4
16.68
O
The phenol makes ferulic acid an
antioxidant. Loss of H forms a highly
stabilized phenoxy radical that inhibits
radical formation during oxidation.
C
HO
OCH3
ferulic acid
OH
The highly conjugated π system
makes it a sunscreen.
450
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 16–26
16.69 There are two possible modes of addition of HBr to allene. When H+ adds to the terminal
carbon, a 2° vinyl carbocation is formed, which affords 2-bromo-1-propene after nucleophilic
attack.
[1]
CH2 C CH2
CH2
C CH2
CH3 C CH2
Br
H
H Br
2-bromo-1-propene
Br
2° vinyl carbocation
When H+ adds to the middle carbon, an intermediate carbocation with a (+) charge adjacent to the
C=C is formed. This carbocation is not resonance stabilized (at least initially), because the two C=C’s
of allene are oriented 90° to each other, a geometry that does not allow for overlap of the C=C with the
empty p orbital of the carbocation.
These C=C's are oriented 90° to each other.
[2]
CH2 C CH2
CH2
C CH2
H
CH2 CH CH2
Br
1o carbocation
H Br
Br
As a result, path [1] forms a more stable carbocation and is preferred.
16.70
cyclohexanamine
aniline
NH2
NH2
NH2
N is surrounded by 3 atoms and
1 lone pair so it is sp3 hybridized.
N has a lone pair on an atom
adjacent to a C=C. N must be
sp2 hybridized to delocalize the
lone pair by resonance.
+ other resonance structures
Basicity is a measure of how willing an atom is to donate an electron pair. Since the lone pair on the
N in aniline is delocalized on the benzene ring, it is less available for donation to a proton. This
makes aniline much less basic.
16.71
O
H
O
+
O
O
H O
Δ
H
O3 cleaves
the C=C.
O
O
Endo product formed.
H
[1] O3
O
O
[2] (CH3)2S
H
H
O
O
The Diels–Alder reaction establishes the
stereochemistry of the four carbons on the sixmembered ring. All four carbon atoms bonded to the
six-membered ring are on the same side.
451
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Conjugation, Resonance, and Dienes 16–27
16.72
O
O
O
CH3
CH3
Δ
+
O
Δ
+
H
O
H
COOCH3
CH3O2C
COOCH3
Diels–Alder
reaction
A
O
O C O
O
B
loss of CO2
16.73
diene
1st Diels–Alder
reaction
Δ
CO2CH3
C
CO2CH3
CO2CH3
CO2CH3
+
CH3O2C C C CO2CH3
Either of these alkenes
becomes the dienophile
2nd Diels–Alder
for an intramolecular Diels–
reaction
Alder reaction in a second step.
D
CO2CH3
+
CO2CH3
two products
C16H16O4
16.74 Retro Diels–Alder reaction forms a conjugated diene. Intramolecular Diels–Alder reaction
then forms N.
CO2CH3
CO2CH3
CO2CH3
NOCH3
HN
Δ
NOCH3
NOCH3
retro
Diels–Alder
HN
M
intramolecular
Diels–Alder
reaction
HN
N
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
453
Benzene and Aromatic Compounds 17–1
Chapter 17 Benzene and Aromatic Compounds
Chapter Review
Comparing aromatic, antiaromatic, and nonaromatic compounds (17.7)
•
Aromatic compound
•
A cyclic, planar, completely conjugated compound that
contains 4n + 2 π electrons (n = 0, 1, 2, 3, and so forth).
An aromatic compound is more stable than a similar acyclic
compound having the same number of π electrons.
•
•
•
•
•
A cyclic, planar, completely conjugated compound that
contains 4n π electrons (n = 0, 1, 2, 3, and so forth).
An antiaromatic compound is less stable than a similar acyclic
compound having the same number of π electrons.
A compound that is not •
aromatic
A compound that lacks one (or more) of the requirements to be
aromatic or antiaromatic.
Antiaromatic
compound
Properties of aromatic compounds
• Every carbon has a p orbital to delocalize electron density (17.2).
• They are unusually stable. ΔHo for hydrogenation is much less than expected, given the number
of degrees of unsaturation (17.6).
• They do not undergo the usual addition reactions of alkenes (17.6).
• 1H NMR spectra show highly deshielded protons because of ring currents (17.4).
Examples of aromatic compounds with 6 π electrons (17.8)
benzene
N
N
H
pyridine
pyrrole
+
cyclopentadienyl
anion
tropylium
cation
Examples of compounds that are not aromatic (17.8)
not cyclic
not planar
not completely
conjugated
454
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–2
Practice Test on Chapter Review
1. Give the IUPAC name for each of the following compounds.
O2N
OH
a.
CH3
b.
(CH3)2CHCH2
OH
c.
O2N
2. Label each compound as aromatic, nonaromatic, or antiaromatic. Choose only one possibility.
Assume all completely conjugated rings are planar.
NH2
a.
c.
g.
e.
+
i.
+
O
N
b.
O
f.
d.
j.
h.
HN
N
3. Answer the following questions about compounds A–E drawn below.
H
Na
[4]
[2]
N
O
N
N
[1]
[3]
B
Hd
Ha
H
Hc
Hb
Nb
A
H
C
D
E
N
NH2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
455
Benzene and Aromatic Compounds 17–3
a.
b.
c.
d.
e.
f.
g.
How is nitrogen Na in compound A hybridized?
In what type of orbital does the lone pair on Na reside?
How is nitrogen Nb in compound B hybridized?
In what type of orbital does the lone pair on Nb reside?
Which of the labeled bonds in compound C is the shortest?
Which of the labeled bonds in compound C is the longest?
When considering both compounds D and E, which of the labeled hydrogen atoms
(Ha, Hb, Hc, or Hd) is the most acidic? Give only one answer.
h. When considering both compounds D and E, which of the labeled hydrogen atoms
(Ha, Hb, Hc, or Hd) is the least acidic? Give only one answer.
Answers to Practice Test
1.a. 2-sec-butyl-5-nitrophenol
b. o-isobutyltoluene
c. 2-tert-butyl-4-nitrophenol
3.a. sp2
b. p
c. sp2
d. sp2
e. 4
f. 3
g. Ha
h. Hc
2.a. nonaromatic
b. aromatic
c. nonaromatic
d. antiaromatic
e. aromatic
f. nonaromatic
g. aromatic
h. aromatic
i. antiaromatic
j. aromatic
Answers to Problems
17.1
Move the electrons in the π bonds to draw all major resonance structures.
N(CH3)2
O
O
N(CH3)2
O
N(CH3)2
O
N(CH3)2
diphenhydramine
17.2
Look at the hybridization of the atoms involved in each bond. Carbons in a benzene ring are
surrounded by three groups and are sp2 hybridized.
Csp2–Csp3
a.
H
Csp2–H1s
b.
Csp2–Csp2
Cp–Cp
shortest of all
the indicated bonds
in (a) and (b)
Csp2–Csp2
Csp2–Csp2
Cp–Cp
456
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–4
17.3
•
•
•
To name a benzene ring with one substituent, name the substituent and add the word
benzene.
To name a disubstituted ring, select the correct prefix (ortho = 1,2; meta = 1,3; para = 1,4)
and alphabetize the substituents. Use a common name if it is a derivative of that
monosubstituted benzene.
To name a polysubstituted ring, number the ring to give the lowest possible numbers and
then follow other rules of nomenclature.
isopropyl group
a.
OH
1
1 CH2CH3
phenol
m-butylphenol
CH3
1
2 Br
ethyl
2-bromo
d.
2
Cl 5
3
iodo Two groups are 1,4 = para.
p-ethyliodobenzene
17.4
butyl group
3
isopropylbenzene
I 4
2
c.
PhCH(CH3)2
b.
Two groups are 1,3 = meta.
toluene (CH3 group must be at the "1" position,
if the molecule is named as a toluene derivative.)
5-chloro
2-bromo-5-chlorotoluene
Work backwards to draw the structures from the names.
c. cis-1,2-diphenylcyclohexane
a. isobutylbenzene
isobutyl
group
b. o-dichlorobenzene
Cl
d. m-bromoaniline
f. 3-tert-butyl-2-ethyltoluene
Br
Cl
Cl
NH2
17.5
OH
1
6
2
propofol
e. 4-chloro-1,2-diethylbenzene
aniline
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
457
Benzene and Aromatic Compounds 17–5
17.6
Molecular formula C10H14O2: 4 degrees of unsaturation
IR absorption at 3150–2850 cm–1: sp2 and sp3 hybridized C–H bonds
NMR absorptions (ppm):
1.4 (triplet, 6 H)
O
O
4.0 (quartet, 4 H)
6.8 (singlet, 4 H)
17.7
Count the different types of carbons to determine the number of 13C NMR signals.
Ce Cf
Ca
Cb
a.
Cc
CH2CH3
Cl
Cd
Cc Cb
C C
Ca b c
4 signals
All C's are different.
7 signals
The less stable compound has a larger heat of hydrogenation.
CH3
CH2
A
benzene ring, more stable
smaller ΔH°
17.9
B
no benzene ring, less stable
larger ΔH°
The protons on sp2 hybridized carbons in aromatic hydrocarbons are highly deshielded and
absorb at 6.5–8 ppm whereas hydrocarbons that are not aromatic show an absorption at
4.5–6 ppm, typical of protons bonded to the C=C of an alkene.
H
H
H
H
H
H
H
H
not aromatic
alkene H's ~4.5–6 ppm
H
H
H
aromatic ring
H's ~6.5–8 ppm
aromatic ring
H's ~6.5–8 ppm
17.10 To be aromatic, a ring must have 4n + 2 π electrons.
16 π e−
4n
4(4) = 16
antiaromatic
H
c.
b.
a.
17.11
Cb Cc
c. Cd
b.
Cb
4 types of C's in the benzene ring
6 signals
17.8
Cc Cb
CH3
Cd
Ca
20 π e−
4n
4(5) = 20
antiaromatic
22 π e−
4n + 2
4(5) + 2 = 22
aromatic
458
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–6
17.12 In determining if a heterocycle is aromatic, count a nonbonded electron pair if it makes the
ring aromatic in calculating 4n + 2. Lone pairs on atoms already part of a multiple bond
cannot be delocalized in a ring, and so they are never counted in determining aromaticity.
O
a.
b.
c.
+
O
O
Count one lone pair from O.
4n + 2 = 4(1) + 2 = 6
aromatic
d.
O
no lone pair from O
4n + 2 = 4(1) + 2 = 6
aromatic
N
N
With one lone
Both N atoms are part of
a double bond, so the lone
pair from each O
there would be 8 electrons. pairs cannot be counted:
there are 6 electrons.
If O's are sp3 hybridized, the
4n + 2 = 4(1) + 2 = 6
ring is not completely
aromatic
conjugated.
not aromatic
17.13
H
quinine
(antimalarial drug)
N
HO
H
N is sp3 hybridized and the
lone pair is in an sp3 hybrid orbital.
CH3O
N
N is sp2 hybridized and the lone pair is
not part of the aromatic ring.
This means it occupies an sp2 hybrid orbital.
17.14
a. The five-membered ring is aromatic
because it has 6 π electrons, two from
each π bond and two from the N atom
that is not part of a double bond.
F
F
N
2
N
N
NH2
CF3
N
CF3
N
O
sp2 hybridized N
lone pair in p orbital
F
2
F
sp2 hybridized N
lone pair in p orbital
b, c.
N
2
NH2 O
F
F
sp3 hybridized N
lone pair in sp3 orbital
N
N
sp2 hybridized N
lone pair in sp2 orbital
sp2 hybridized N
lone pair in sp2 orbital
sitagliptin
17.15 Since a neutral oxygen atom forms only two bonds, it must always donate its electron pair to
a delocalized π electron system (like the N atom of pyrrole). It can never be part of a double
bond, because it would carry a net (+) charge. With nitrogen, however, a N atom can either
donate an electron pair or be part of a double bond, since N forms three bonds. In a neutral
five-membered aromatic ring, there are two π bonds (where the O atom cannot be located),
and only one atom that can donate its electron pair (where the O atom may reside).
17.16 Compare the conjugate base of 1,3,5-cycloheptatriene with the conjugate base of
cyclopentadiene. Remember that the compound with the more stable conjugate base will
have a lower pKa.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
459
Benzene and Aromatic Compounds 17–7
B
B
H H
H H
H
1,3,5-cycloheptatriene
pKa = 39
8 π Electrons make this
conjugate base
especially unstable
(antiaromatic).
Since the conjugate base is unstable,
the pKa of 1,3,5-cycloheptatriene is high.
H
6 π electrons
aromatic conjugate base
very stable anion
cyclopentadiene
Since the conjugate base is very stable, the pKa
of cyclopentadiene is much lower.
17.17 The compound with the most stable conjugate base is the most acidic.
Conjugate bases:
no resonance
delocalization
2 resonance
structures
aromatic conjugate base
most stable
most unstable base so
least acidic acid
The acid is
intermediate in acidity.
The acid is the
most acidic.
17.18
17.19 To be aromatic, the ions must have 4n + 2 π electrons. Ions in (b) and (c) do not have the
right number of π electrons to be aromatic.
a.
2 π electrons
4(0) + 2 = 2
aromatic
d.
10 π electrons
4(2) + 2 = 10
aromatic
17.20
absorbs at 7.6 ppm
H
A
=
The NMR indicates that A is aromatic. The C’s of the triple
bond are sp hybridized. Each triple bond has one set of
electrons in p orbitals that overlap with other p orbitals on
adjacent atoms in the ring. This overlap allows electrons to
delocalize. Each C of the triple bonds also has a p orbital in
the plane of the ring. The electrons in these p orbitals are
localized between the C’s of the triple bond, not delocalized
in the ring. Although A has 24 π e– total, only 18 e– are
delocalized around the ring.
460
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–8
17.21 In using the inscribed polygon method, always draw the vertex pointing down.
+
2 antibonding MOs
2 π electrons
All bonding MOs are filled.
aromatic
1 bonding MO
17.22 Draw the inscribed pentagons with the vertex pointing down. Then draw the molecular
orbitals (MOs) and add the electrons.
Cation:
Radical:
2 antibonding MOs
3 bonding MOs
4 π electrons
Not all bonding MOs are filled.
not aromatic
5 π electrons
Not all bonding MOs are filled.
not aromatic
17.23 C60 would exhibit only one 13C NMR signal because all the carbons are identical.
17.24
g
d
f
c
e
a.
a
b
c
2
Br
b.
d
6
5
Answer: p-isopropyltoluene
seven lines in 13C NMR (lettered Ca–Cg)
Name as a derivative of toluene.
g
OH
1
3
4
isopropyl group
Answer: 3-bromo-5-nitrophenol
All C's are different, so there are six lines in the 13C NMR.
Name as a derivative of phenol.
NO2
17.25
not sp2 hybridized
H
a.
NH2
H
not completely
conjugated
not aromatic
H
aromatic ring
N
b.
N
N
c.
O
H
6 π electrons
aromatic
The six-membered ring is aromatic. The bicyclic
ring system with the O atom is not completely
conjugated, so it is nonaromatic.
17.26
a. If the Kekulé description of benzene was accurate, only one product would form in
Reaction [1], but there would be four (not three) dibromobenzenes (A–D), because
adjacent C–C bonds are different—one is single and one is double. Thus, compounds
A and B would not be identical. A has two Br’s bonded to the same double bond, but
B has two Br’s on different double bonds.
b. In the resonance description, only one product would form in Reaction [1], since all C’s are
identical, but only three dibromobenzenes (ortho, meta, and para isomers) are possible.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
461
Benzene and Aromatic Compounds 17–9
A and B are identical because each C–C bond is identical and intermediate in bond length
between a C–C single and C–C double bond.
Br
Br
[1]
Br
Br
[2]
Br
Br
Br
Br
Br
A
B
C
D
17.27
C8H10:
Br
Br
Br
C8H9Br:
Br
Br
Br
Br
Br
Br
3 isomers
2 isomers
3 isomers
1 isomer
17.28 To name the compounds use the directions from Answer 17.3.
NH2
d.
a.
CH3CH2
aniline
CH2CH2CH3
g.
Cl
sec-butylbenzene
Br
b.
e.
m-chloroethylbenzene
Cl
toluene
p-chlorotoluene
h.
Br
NH2
CH2CH3
CH3
1-ethyl-3-isopropyl-5-propylbenzene
Br
Cl
c.
CH(CH3)2
o-chloroaniline
aniline
2,3-dibromoaniline
OH
f.
NO2
NO2
2,5-dinitrophenol
phenol (OH at C1)
Ph
cis-1-bromo-2-phenylcyclohexane
462
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–10
17.29
d. o-bromonitrobenzene
a. p-dichlorobenzene
Cl
g. 2-phenyl-2-propen-1-ol
Br
Cl
OH
NO2
b. m-chlorophenol
e. 2,6-dimethoxytoluene
h. trans-1-benzyl-3-phenylcyclopentane
OCH3
Cl
CH3
OCH3
OH
c. p-iodoaniline
or
f. 2-phenyl-1-butene
I
H2N
17.30
a, b. constitutional isomers of molecular formula C8H9Cl and names of the trisubstituted benzenes
Cl
Cl
Cl
Cl
Cl
stereoisomers
for this isomer
Cl
Cl
Cl
Cl
Cl
Cl
2-chloro-1,3-dimethylbenzene
1-chloro-2,3-dimethylbenzene
4-chloro-1,2-dimethylbenzene
Cl
Cl
Cl
1-chloro-2,4-dimethylbenzene
c. stereoisomers
Cl
1-chloro-3,5-dimethylbenzene
2-chloro-1,4-dimethylbenzene
Cl
17.31 Count the electrons in the π bonds. Each π bond holds two electrons.
a.
b.
10 π electrons
7 π electrons
c.
d.
10 π electrons
e.
14 π electrons
12 π electrons
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
463
Benzene and Aromatic Compounds 17–11
17.32 To be aromatic, the compounds must be cyclic, planar, completely conjugated, and have
4n + 2 π electrons.
Circled C's are not sp2.
not completely conjugated
not aromatic
a.
Circled C is not sp2.
not completely conjugated
not aromatic
b.
12 π electrons
does not have 4n + 2
π electrons
not aromatic
c.
12 π electrons
does not have 4n + 2
π electrons
not aromatic
d.
17.33 In determining if a heterocycle is aromatic, count a nonbonded electron pair if it makes the
ring aromatic in calculating 4n + 2. Lone pairs on atoms already part of a multiple bond
cannot be delocalized in a ring, and so they are never counted in determining aromaticity.
S
O
a.
c.
O
6 π electrons
counting a lone pair from S
4(1) + 2 = 6
aromatic
not aromatic
not aromatic
N
N
b.
g.
e.
O
6 π electrons
counting the lone
pair from N
4(1) + 2 = 6
aromatic
H
N
O
d.
f.
h.
10 π electrons
4(2) + 2 = 10
aromatic
6 π electrons,
counting a lone pair from O
4(1) + 2 = 6
aromatic
17.34
Circled C's are
not sp2.
not aromatic
a.
c.
Circled C is
not sp2.
not aromatic
d.
4 π electrons
4(1) = 4
antiaromatic
O
10 π electrons
in 10-membered ring
4(2) + 2 = 10
aromatic
b.
17.35
6 π electrons
in this ring
+
A
6 π electrons
in this ring
Count
these 2e–.
N
N
N
6 π electrons
counting a lone pair from O
4(1) + 2 = 6
aromatic
N
10 π electrons
4(2) + 2 = 10
aromatic
These lone
pairs are on
doubly
bonded N
atoms, so they
can't be
counted.
464
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–12
A resonance structure can be drawn for A that places a negative charge in the five-membered
ring and a positive charge in the seven-membered ring. This resonance structure shows that
each ring has six π electrons, making it aromatic. The molecule possesses a dipole such that
the seven-membered ring is electron deficient and the five-membered ring is electron rich.
17.36 Each compound is completely conjugated. A compound with 4n + 2 π electrons is especially
stable, while a compound with 4n π electrons is especially unstable.
pentalene
azulene
10 π electrons
4(2) + 2 = 10
aromatic
very stable
8 π electrons
4(2) = 8
antiaromatic
unstable
17.37
N
N
N
N
H
heptalene
sp2 hybridized but
with lone pair in p orbital
purine
12 π electrons
6(2) = 12
antiaromatic
unstable
a. Each N atom is sp2 hybridized.
b. The three unlabeled N atoms are sp2 hybridized with
lone pairs in one of the sp2 hybrid orbitals. The
labeled N has its lone pair in a p orbital.
c. 10 π electrons
d. Purine is cyclic, planar, completely conjugated, and
has 10 π electrons [4(2) + 2] so it is aromatic.
17.38
a. 16 total π electrons
b. 14 π electrons delocalized in the ring. [Note: Two of the electrons in
the triple bond are localized between two C’s, perpendicular to the π
electrons delocalized in the ring.]
c. By having two of the p orbitals of the C–C triple bond co-planar with
the p orbitals of all the C=C’s, the total number of π electrons
delocalized in the ring is 14. 4(3) + 2 = 14, so the ring is aromatic.
C
17.39 A second resonance structure can be drawn for the six-membered ring that gives it three π
bonds, thus making it aromatic with six π electrons.
O
HO
N3
O
HO
N
O
NH
O
N
N3
O
NH
O
6 π electrons
465
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Benzene and Aromatic Compounds 17–13
17.40 The rate of an SN1 reaction increases with increasing stability of the intermediate carbocation.
Increasing reactivity
Cl
Cl
+
Cl
+
The aromatic carbocation is
delocalized over the whole ring,
making it a very stable intermediate
and most easily formed in an SN1
reaction.
+
4 π electrons
6 π electrons
2o carbocation
antiaromatic
aromatic
very unstable intermediate
very stable intermediate
Increasing stability
17.41
Na+ H –
H OH
D
D
D
D
+ Na+ OH
+H D
HO H
Two additional resonance
structures are not drawn.
D
D
D
D
H OH
17.42 α-yrone reacts like benzene because it is aromatic. A second resonance structure can be
drawn showing how the ring has six π electrons. Thus, α-pyrone undergoes reactions
characteristic of aromatic compounds—that is, substitution rather than addition.
O
O
O
α-pyrone
+
O
6 π electrons
17.43
a.
cyclopropenyl radical
b.
N
N
N
H
H
H
and
pyrrole
N
N
N
H
H
H
466
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–14
c.
phenanthrene
17.44
Naphthalene can be drawn as three resonance structures:
(a)
(a)
(a)
(b)
(b)
(b)
In two of the resonance structures bond (a) is a double bond, and
bond (b) is a single bond. Therefore, bond (b) has more single
bond character, making it longer.
17.45
a.
N
N
N
H
H
H
and
pyrrole
N
N
N
H
H
H
O
O
Pyrrole is less resonance stabilized than benzene
because four of the resonance structures have
charges, making them less good.
b.
O
O
O
and
O
furan
Furan is less resonance stabilized than pyrrole because its O atom is
less basic, so it donates electron density less "willingly." Thus,
charge-separated resonance forms are more minor contributors to
the hybrid than the charge-separated resonance forms of pyrrole.
17.46 The compound with the more stable conjugate base is the stronger acid. Draw and compare
the conjugate bases of each pair of compounds.
conjugate bases
a.
or
or
more acidic
resonance-stabilized
but not aromatic
or
b.
more acidic
6 π electrons, aromatic
more stable conjugate base
Its acid is more acidic.
or
6 π electrons, aromatic
more stable conjugate base
Its acid is more acidic.
antiaromatic
highly destabilized
conjugate base
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
467
Benzene and Aromatic Compounds 17–15
17.47
+ NaNH2
+ NH3
Na+
indene
The conjugate base of indene has 10 π electrons, making it aromatic
and very stable. Therefore, indene is more acidic than many hydrocarbons.
17.48
Ha
Ha
Hb
Hc
– Hb
CH3
– Hc
CH3
CH3
CH3
A
Hd
B
Hb is most acidic because its
conjugate base is aromatic (6 π electrons).
Hd
Hc is least acidic because its
conjugate base has 8 π electrons,
making it antiaromatic.
17.49
conjugate base
pyrrole
N
Both pyrrole and the conjugate base of pyrrole have 6 π
electrons in the ring, making them both aromatic. Thus,
deprotonation of pyrrole does not result in a gain of aromaticity
since the starting material is aromatic to begin with.
N
H
conjugate base
cyclopentadiene
more acidic
H H
Cyclopentadiene is not aromatic, but the conjugate base has 6 π
electrons and is therefore aromatic. This makes the C–H bond in
cyclopentadiene more acidic than the N–H bond in pyrrole, since
deprotonation of cyclopentadiene forms an aromatic conjugate base.
H
17.50
H H
+ H+
a.
N H
pyrrole
or
+ H+
N
C2
H
A
N
B
H
H
H H
N
H H
H
N
H
Protonation at C2 forms conjugate acid A
because the positive charge can be
delocalized by resonance. There is no
resonance stabilization of the positive
charge in B.
468
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–16
b. H H
Loss of a proton from A (which is not aromatic)
gives two electrons to N, and forms pyrrole,
which has six π electrons that can then
delocalize in the five-membered ring, making it
aromatic. This makes deprotonation a highly
favorable process, and A more acidic.
H
N
A
pKa = 0.4
Both C and its conjugate base pyridine are
aromatic. Since C has six π electrons, it is
already aromatic to begin with, so there is less to
pKa = 5.2 be gained by deprotonation, and C is thus less
acidic than A.
C
N H
17.51
a.
3 antibonding MOs
2K
2 nonbonding MOs
3 bonding MOs
cyclooctatetraene
dianion of
cyclooctatetraene
cyclooctatetraene and its 8 π electrons
b. Even if cyclooctatetraene were flat, it has two unpaired electrons in its HOMOs
(nonbonding MOs) so it cannot be aromatic.
c. The dianion has 10 π electrons.
d. The two additional electrons fill the nonbonding MOs; that is, all the bonding and
nonbonding MOs are filled with electrons in the dianion.
e. The dianion is aromatic since its HOMOs are completely filled, and it has no electrons in
antibonding MOs.
17.52
4 antibonding MOs
4 antibonding MOs
5 bonding MOs
+
cyclononatetraenyl
cation
4 antibonding MOs
5 bonding MOs
5 bonding MOs
cyclononatetraenyl
radical
8 π electrons
antiaromatic
cyclononatetraenyl
anion
9 π electrons
not aromatic
10 π electrons
aromatic
All bonding MOs are filled.
17.53 The number of different types of C’s = the number of signals.
3
4
1
2 CH3
a.
CH3
1
3
3
2
1
c.
b.
5
2
d.
1
4
CH2CH3
5 different C's
all unique
9 different C's
1
2
3
1
2
3 different C's
2
1
2
3
4
3
2
3 4 3
2
4 different C's
1
1
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
469
Benzene and Aromatic Compounds 17–17
17.54 Draw the three isomers and count the different types of carbons in each. Then match the
structures with the data.
5
4
5
4
3
2
3 2
1
ortho isomer
5 types of C
5 lines in spectrum
Spectrum [B]
17.55
6
1
5
4
5
2 3
1
1
2 3 4 3 2
meta isomer
6 types of C
6 lines in spectrum
Spectrum [A]
4
3
1
2
4 4
para isomer
4 types of C
4 lines in spectrum
Spectrum [C]
1
a. C10H14: IR absorptions at 3150–2850 (sp2 and sp3 hybridized C–H), 1600, and 1500
(due to a benzene ring) cm–1
1
H NMR data:
Absorption
ppm
# of H’s Explanation
doublet
1.2
6
6 H’s adjacent to 1 H
(CH3)2CH group
singlet
2.3
3
CH3
septet
3.1
1
1 H adjacent to 6 H’s
multiplet
7–7.4
4
a disubstituted benzene ring
You can’t tell from these data where the two groups are on the benzene ring. They are not
para, since the para arrangement usually gives two sets of distinct peaks (resembling two
doublets) so there are two possible structures—ortho and meta isomers.
or
b. C9H12: 13C NMR signals at 21, 127, and 138 ppm → means three different types of C’s.
1
H NMR shows two types of H’s: 9 H’s probably means 3 CH3 groups; the other 3 H’s
are very deshielded so they are bonded to a benzene ring.
Only one possible structure fits:
CH3
CH3
CH3
c. C8H10: IR absorptions at 3108–2875 (sp2 and sp3 hybridized C–H), 1606, and 1496
(due to a benzene ring) cm–1
1
H NMR data:
Absorption
ppm # of H’s Explanation
Structure:
triplet
1.3
3
3 H’s adjacent to 2 H’s
CH2CH3
quartet
2.7
2
2 H’s adjacent to 3 H’s
multiplet
7.3
5
a monosubstituted benzene ring
470
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–18
17.56
a. Compound A: Molecular formula C8H10O
IR absorption at 3150–2850 (sp2 and sp3 hybridized C–H) cm–1
1
H NMR data:
Absorption ppm
# of H’s Explanation
triplet
1.4
3
3 H’s adjacent to 2 H’s
quartet
3.95
2
2 H’s adjacent to 3 H’s
multiplet
6.8–7.3
5
a monosubstituted benzene ring
b. Compound B: Molecular formula C9H10O2
IR absorption at 1669 (C=O) cm–1
1
H NMR data:
Absorption ppm # of H’s Explanation
singlet
2.5
3
CH3 group
singlet
3.8
3
CH3 group
doublet
6.9
2
2 H’s on a benzene ring
doublet
7.9
2
2 H’s on a benzene ring
Structure:
OCH2CH3
Structure:
O
C
CH3
CH3O
or
CH3
O
C
OCH3
It would be hard to distinguish
these two compounds with
the given data.
17.57
3 equivalent H's
singlet at ~2.3 ppm
* 3 other H's on benzene ring
(arrows with *)
at ~6.9 ppm
CH3
*
H
*
H
H
thymol
IR absorptions:
3500–3200 cm–1 (O–H)
3150–2850 cm–1 (C–H bonds)
1621 and 1585 cm–1 (benzene ring)
OH
H
CH3 CH3
1H
septet at ~3.2 ppm
6 equivalent H's
doublet at ~1.2 ppm
X
Y
Z
basic structure of
thymol
Thymol must have this basic structure, given the NMR and IR data
since it is a trisubstituted benzene ring with one singlet and two
doublets in the NMR at ~6.9 ppm. However, which group [OH, CH3,
or CH(CH3)2] corresponds to X, Y, and Z is not readily distinguished
with the given data. The correct structure for thymol is given.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
471
Benzene and Aromatic Compounds 17–19
17.58
C1
C2 C3
13C
C4
CH3
NMR has four lines that are located in the aromatic region
(~110–155 ppm), corresponding to the four different types of
carbons in the aromatic ring of the para isomer. The ortho and
meta isomers have six different C's, and so six lines would be
expected for each of them.
O
N
CH3
OCH2CH3
C2 C3
17.59 Because tetrahydrofuran has a higher boiling point and is more water soluble, it must be more
polar and have stronger intermolecular forces than furan. There are two contributing factors.
One lone pair on furan’s O atom is delocalized on the five-membered ring to make it aromatic.
This makes it less available for H-bonding with water and other intermolecular interactions.
Also, the C–O bonds in furan are less polar than the C–O bonds in tetrahydrofuran because of
hybridization. The sp2 hybridized C’s of furan pull a little more electron density towards them
than do the sp3 hybridized C’s of tetrahydrofuran. This counteracts the higher electronegativity
of O compared to C to a small extent.
sp2 C
sp3 C
O
O
more polar C–O bond
less polar C–O bond
delocalized
17.60
a, b.
N
electron pair in an sp2 hybridized orbital
The ring system is aromatic with 10 π electrons, 8 π
electrons from the double bonds and 2 π electrons
from the N atoms common to both rings.
N
zolpidem
O
electron pair in a p orbital
so it can delocalize
N(CH3)2
472
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–20
N
N
N
c.
N
N
O
N
O
O
N(CH3)2
N(CH3)2
N(CH3)2
N
N
N
N
N
N
O
O
O
N(CH3)2
N(CH3)2
N
N(CH3)2
N
N
N
N
O
N
O
N(CH3)2
O
N(CH3)2
N
N(CH3)2
N
N
N
O
O
N(CH3)2
N(CH3)2
17.61
H
O
O
O
O
a.
HO
OH
OCH3
OCH3
curcumin
The enol form is more stable because the enol
double bond makes a highly conjugated
system. The enol OH can also intramolecularly
hydrogen bond to the nearby carbonyl O atom.
sp3
HO
OCH3
keto form
OH
OCH3
473
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Benzene and Aromatic Compounds 17–21
O
O
O
O
b.
HO
OH
OCH3
HO
OCH3
The enol O–H proton is more acidic than an
alcohol O–H proton because the conjugate
base is resonance stabilized.
OH
OCH3
O
O
OCH3
HO
OH
OCH3
OCH3
c. Curcumin is colored because it has many conjugated π electrons, which shift absorption of
light from the UV to the visible region.
d. Curcumin is an antioxidant because it contains a phenol. Homolytic cleavage affords a
resonance-stabilized phenoxy radical, which can inhibit oxidation from occurring, much
like vitamin E and BHT in Chapter 15.
(+ other resonance structures)
O
O
O
O
phenoxy radical
Resonance delocalizes the radical on the ring and C chain of curcumin.
17.62 a. Pyrazole rings are aromatic because they have six π electrons—two from the lone pair on
the N atom that is not part of the double bond and four from the double bonds.
b.
OH
H
sp2
H
N
N
H
p
H
H
474
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–22
OH
OH
OH
c.
H
H
N
N
H
H
H
H
N
N
H
H
H
H
N
H
H
OH
H
N
H
H
OH
H
H
H
H
N
N
H
H
H
OH
H
N
H
N
H
N
H
N
H
H
H
H
OH
H
H
N
N
H
H
H
d. The N atom in the NH bond in the pyrazole ring is sp2 hybridized with 33% s-character,
increasing the acidity of the N–H bond. The N–H bond of CH3NH2 contains an sp3
hybridized N atom.
17.63 Both A and B are cyclic, and if the lone pair of electrons on N is in a p orbital, they are
completely conjugated with 10 π electrons, a number that satisfies Hückel’s rule. To be
aromatic, A and B must be planar, and the internal bond angles of A and B would be much
larger than 120°, the theoretical bond angle of sp2 hybridized C’s. The fact that A is aromatic
means that the lone pair on N occupies a p orbital, so it can delocalize on the nine-membered
ring. The stabilization gained by being aromatic is greater than any angle strain. With B, the
lone pair on N is also delocalized on the C=O, making it less available for donation to the
ring, so the ring is not aromatic.
O
NH
O
+N
N
OCH2CH3
A
2 electrons in a p orbital
10 π electrons
B
Electron pair is less available so
the ring is not aromatic.
OCH2CH3
475
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Benzene and Aromatic Compounds 17–23
17.64 With 14 π electrons in the double bonds, the system is aromatic [4(3) + 2 = 14 π electrons].
The ring current generated by the circulating π electrons deshields the protons on the C=C’s,
so they absorb downfield (8.14–8.67 ppm). The CH3 groups, however, are very shielded
since they lie above and below the plane, so they absorb far upfield (–4.25 ppm). The
dianion of C now has 16 π electrons, making it antiaromatic, so the position of the
absorptions reverses. The C=C protons are now shielded (–3 ppm), and the CH3 protons are
now deshielded (21 ppm).
–4.75 ppm
CH3
CH3
21 ppm
~ 8 ppm
CH3
CH3
C
2–
–3 ppm
dianion
17.65 A second resonance structure for A shows that the ring is completely conjugated and has six
π electrons, making it aromatic and especially stable. A similar charge-separated resonance
structure for B makes the ring completely conjugated, but gives the ring four π electrons,
making it antiaromatic and especially unstable.
O
O
O
O
+
+
B
6 π electrons
aromatic
stable
A
4 π electrons
antiaromatic
not stable
17.66 The conversion of carvone to carvacrol involves acid-catalyzed isomerization of two double
bonds and tautomerization of a ketone to an enol tautomer. In this case the enol form is part
of an aromatic phenol. Each isomerization of a C=C involves Markovnikov addition of a
proton, followed by deprotonation.
O
O
HSO4–
H OSO3H
H
O
O
HSO4–
H OSO3H
O
H
H OSO3H
(R)-carvone
OH
H2SO4 +
OH
H
HSO4–
carvacrol
476
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 17–24
17.67 Resonance structures for triphenylene:
A
B
C
D
I
H
G
E
F
Resonance structures A–H all keep three double and three single bonds in the three six-membered
rings on the periphery of the molecule. This means that each ring behaves like an isolated benzene
ring undergoing substitution rather than addition because the π electron density is delocalized within
each six-membered ring. Only resonance structure I does not have this form. Each C–C bond of
triphenylene has four (or five) resonance structures in which it is a single bond and four (or five)
resonance structures in which it is a double bond.
Resonance structures for phenanthrene:
A
B
C
D
E
With phenanthrene, however, four of the five resonance structures keep a double bond at the labeled
C’s. (Only C does not.) This means that these two C’s have more double bond character than other
C–C bonds in phenanthrene, making them more susceptible to addition rather than substitution.
17.68
X
Y
O
N(CH3)2
N(CH3)2
C
H
113 ppm
O
C
O
C
H
H
130 ppm
The negative charge and
increased electron density make
the carbon more shielded and
shift the absorption upfield.
The positive charge and
decreased electron density make
the carbon deshielded and shift
the absorption downfield.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
477
Reactions of Aromatic Compounds 18–1
Chapter 18 Reactions of Aromatic Compounds
Chapter Review
Mechanism of electrophilic aromatic substitution (18.2)
• Electrophilic aromatic substitution follows a two-step mechanism. Reaction of the aromatic ring
with an electrophile forms a carbocation, and loss of a proton regenerates the aromatic ring.
• The first step is rate-determining.
• The intermediate carbocation is stabilized by resonance; a minimum of three resonance structures
can be drawn. The positive charge is always located ortho or para to the new C–E bond.
H
E
H
E
H
E
(+) ortho to E
(+) para to E
(+) ortho to E
Three rules describing the reactivity and directing effects of common substituents (18.7–18.9)
[1] All ortho, para directors except the halogens activate the benzene ring.
[2] All meta directors deactivate the benzene ring.
[3] The halogens deactivate the benzene ring.
Summary of substituent effects in electrophilic aromatic substitution (18.6–18.9)
Substituent
Inductive effect
Resonance
effect
Reactivity
Directing effect
donating
none
activating
ortho, para
withdrawing
donating
activating
ortho, para
withdrawing
donating
deactivating
ortho, para
withdrawing
withdrawing
deactivating
meta
R
[1]
R = alkyl
Z
[2]
Z = N or O
X
[3]
X = halogen
[4]
Y (δ+ or +)
478
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–2
Five examples of electrophilic aromatic substitution
[1] Halogenation—Replacement of H by Cl or Br (18.3)
H
Cl
X2
•
Br
or
FeX3
[X = Cl, Br]
aryl chloride
aryl bromide
Polyhalogenation occurs on
benzene rings substituted by
OH and NH2 (and related
substituents) (18.10A).
[2] Nitration—Replacement of H by NO2 (18.4)
H
NO2
HNO3
H2SO4
nitro compound
[3] Sulfonation—Replacement of H by SO3H (18.4)
H
SO3H
SO3
H2SO4
benzenesulfonic
acid
[4] Friedel–Crafts alkylation—Replacement of H by R (18.5)
R
H
RCl
AlCl3
alkyl benzene
(arene)
•
•
•
•
Variations:
Rearrangements can occur.
Vinyl halides and aryl halides are unreactive.
The reaction does not occur on benzene rings
substituted by meta deactivating groups or NH2
groups (18.10B).
Polyalkylation can occur.
R
H
ROH
[1] with alcohols
H2SO4
R
H
[2] with alkenes
CH2 CHR
H2SO4
CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
479
Reactions of Aromatic Compounds 18–3
[5] Friedel–Crafts acylation—Replacement of H by RCO (18.5)
O
H
C
RCOCl
AlCl3
•
R
ketone
The reaction does not occur on benzene rings
substituted by meta deactivating groups or
NH2 groups (18.10B).
Nucleophilic aromatic substitution (18.13)
[1] Nucleophilic substitution by an addition–elimination mechanism
•
•
Nu
X
Nu
A
•
A
X = F, Cl, Br, I
A = electron-withdrawing group
•
The mechanism has two steps.
Strong electron-withdrawing groups at the
ortho or para position are required.
Increasing the number of electronwithdrawing groups increases the rate.
Increasing the electronegativity of the halogen
increases the rate.
[2] Nucleophilic substitution by an elimination–addition mechanism
•
•
•
Reaction conditions are harsh.
Benzyne is formed as an intermediate.
Product mixtures may result.
•
A benzylic C–H bond is needed for reaction.
Nu
X
Nu
X = halogen
Other reactions of benzene derivatives
[1] Benzylic halogenation (18.14)
Br
R
Br2
hν or Δ
or
NBS
hν or ROOR
R
benzylic bromide
[2] Oxidation of alkyl benzenes (18.15A)
R
KMnO4
COOH
benzoic acid
480
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–4
[3] Reduction of ketones to alkyl benzenes (18.15B)
O
C
R
Zn(Hg), HCl
or
NH2NH2, –OH
R
alkyl benzene
[4] Reduction of nitro groups to amino groups (18.15C)
NO2
H2, Pd-C
or
Fe, HCl
or
Sn, HCl
NH2
aniline
Practice Test on Chapter Review
1. a. Which of the following statements is true about an ethoxy substituent (–OCH2CH3) on a
benzene ring?
1. OCH2CH3 increases the rates of both electrophilic substitution and nucleophilic substitution.
2. OCH2CH3 decreases the rates of both electrophilic substitution and nucleophilic substitution.
3. OCH2CH3 increases the rate of electrophilic substitution and decreases the rate of nucleophilic
substitution.
4. OCH2CH3 decreases the rate of electrophilic substitution and increases the rate of nucleophilic
substitution.
5. None of these statements is true.
b. Which of the following statements is true about a –CO2CH3 group on a benzene ring?
1. CO2CH3 increases the rates of both electrophilic substitution and nucleophilic substitution.
2. CO2CH3 decreases the rates of both electrophilic substitution and nucleophilic substitution.
3. CO2CH3 increases the rate of electrophilic substitution and decreases the rate of nucleophilic
substitution.
4. CO2CH3 decreases the rate of electrophilic substitution and increases the rate of nucleophilic
substitution.
5. None of these statements is true.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
481
Reactions of Aromatic Compounds 18–5
c. Which of the following is not a valid resonance structure for the carbocation that results from
ortho attack of an electrophile on C6H5C(CH3)=CH2?
1.
3.
5.
E
E
2.
E
4.
E
E
2. Draw the organic products formed in the following reactions.
[1] CH3CH2CH2COCl, AlCl3
a. CH3
OCH3
e.
[2] Zn(Hg), HCl
SO3
H2SO4
Br
b.
CN
HNO3
Fe
H2SO4
HCl
O
f.
Br2 (1 equiv)
N
H
FeBr3
KMnO4
c.
g. O2N
d. CH3O
CH3
HNO3
H2
H2SO4
Pd-C
h.
NaOCH2CH3
Cl
NaOH
Cl
Δ
3. (a) Considering the compound drawn below, which ring is most reactive in electrophilic
aromatic substitution? Which ring is the least reactive in electrophilic aromatic substitution?
C
B
H
N
A
O
N
D
N
H
NO2
4. Classify each substituent as [1] ortho, para activating, [2] ortho, para deactivating, or [3] meta
deactivating.
a. –Br
c. –COOH
e. –N(CH2)6COCH3
b. –CH2CH2CH2Br
d. –NHCOCH2CH3
f. –CCl3
5. What reagents are needed to convert toluene (C6H5CH3) to each compound?
a. C6H5COOH
c. p-bromotoluene
e. p-ethyltoluene
b. C6H5CH2Br
d. o-nitrotoluene
f.
O
CH3
482
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–6
Answers to Practice Test
1. a. 3
b. 4
c. 4
2.
3. a. A
b. C
a. CH3
e.
OCH3
Br
4. a. 2
b. 1
c. 3
d. 1
e. 1
f. 3
SO3H
O
CH3
N
H
f.
H2N
b.
Br
CN
5. a. KMnO4
b. Br2, hν
c. Br2, FeBr3
d. HNO3, H2SO4
e. CH3CH2Cl, AlCl3
f. CH3CH2COCl, AlCl3
Br
O
N
H
CO2H
c.
CO2H
g. O2N
OCH2CH3
H2N
d. CH3O
CH3
OH
h.
OH
Answers to Problems
18.1
The π electrons of benzene are delocalized over the six atoms of the ring, increasing
benzene’s stability and making them less available for electron donation. With an alkene, the
two π electrons are localized between the two C’s, making them more nucleophilic and thus
more reactive with an electrophile than the delocalized electrons in benzene.
18.2
H
E
+
B
E
+
H
E
B
E
483
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–7
18.3
Reaction with Cl2 and FeCl3 as the catalyst occurs in two parts. First is the formation of an
electrophile, followed by a two-step substitution reaction.
+
[1]
+ FeCl3
Cl Cl
Cl Cl FeCl3
Lewis base Lewis acid
H
electrophile
H
Cl
+
[2]
H
Cl
H
Cl
Cl Cl FeCl3
FeCl4
resonance-stabilized carbocation
H
Cl
[3]
18.4
Cl FeCl3
Cl
+ HCl + FeCl3
There are two parts in the mechanism. The first part is formation of an electrophile. The
second part is a two-step substitution reaction.
R
=
A
H
O
O
[1]
O
S
O
H OSO3H
+
S
O + O H
H
SO3H
H
+
[2]
=
SO3H
R
R
+
SO3H
HSO4–
+
electrophile
H
SO3H
H
SO3H
R
R
resonance-stabilized carbocation
–
HSO4
H
SO3H
[3]
SO3H
H2SO4
R
R
B
18.5
Friedel–Crafts alkylation results in the transfer of an alkyl group from a halogen to a benzene
ring. In Friedel–Crafts acylation an acyl group is transferred from a halogen to a benzene ring.
CH(CH3)2
a.
+
(CH3)2CHCl
c.
Cl
b.
+
AlCl3
O
O
AlCl3
+ CH3CH2
C
C
Cl
AlCl3
CH2CH3
484
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–8
18.6
Remember that an acyl group is transferred from a Cl atom to a benzene ring. To draw the
acid chloride, substitute a Cl for the benzene ring.
O
C
a.
O
CH2CH2CH(CH3)2
Cl
O
Cl
b.
C
c.
CH2CH2CH(CH3)2
C
To be reactive in a Friedel–Crafts alkylation reaction, the X must be bonded to an sp3
hybridized carbon atom.
Br
a.
sp2
Br
Br
b.
unreactive
18.8
O
C
O
C
18.7
C
O
sp
c.
3
reactive
Br
d.
sp2
unreactive
sp
3
reactive
The product has an “unexpected” carbon skeleton, so rearrangement must have occurred.
CH3
[1] CH3 C CH2 Cl
AlCl3
CH3
CH3 C
+
_
CH2 Cl AlCl3
CH3
1,2-H shift
CH3 C
H
H
Rearrangement
C(CH3)3
+
C(CH3)3
+ AlCl4–
H
H
H
[2]
CH3
H
C(CH3)3
C(CH3)3
_
H
C(CH3)3
[3]
18.9
Cl AlCl3
C(CH3)3
HCl
AlCl3
Both alkenes and alcohols can form carbocations for Friedel–Crafts alkylation reactions.
a.
+
b.
+ (CH3)2C
OH
H2SO4
CH2
H2SO4
c.
H2SO4
+
C(CH3)3
OH
d.
+
H2SO4
Cl
485
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–9
18.10
[1] Cl
Cl
Cl
O
A
Cl
Cl
Cl
Cl
Cl
O
Cl
Cl3Al Cl
Al Cl
+
O
AlCl4–
Cl
O
Cl
H
[2] Cl
H
Cl
O
Cl
H
O
Cl
H
O
Cl
Cl
Cl
O
Cl
_
Cl AlCl3
H
[3] Cl
O
Cl
Cl
+
O
Cl
HCl
+
AlCl3
B
18.11 In parts (b) and (c), a 1,2-shift occurs to afford a rearrangement product.
a.
Cl
c.
OH
b.
18.12
a.
b.
CH2CH2CH2CH3
alkyl group
electron donating
c. OCH2CH3
electronegative O
electron withdrawing
Br
halide
electron withdrawing
18.13 Electron-donating groups place a negative charge in the benzene ring. Draw the resonance
structures to show how –OCH3 puts a negative charge in the ring. Electron-withdrawing
groups place a positive charge in the benzene ring. Draw the resonance structures to show how
–COCH3 puts a positive charge in the ring.
O
a.
CH3
O
CH3
O
CH3
O
CH3
O
CH3
486
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–10
b.
O
O
O
C
C
C
CH3
CH3
O
O
C
C
CH3
O
C
CH3
CH3
O
C
CH3
CH3
18.14 To classify each substituent, look at the atom bonded directly to the benzene ring. All R
groups and Z groups (except halogens) are electron donating. All groups with a positive
charge, δ+, or halogens are electron withdrawing.
I
OCH3
a.
C(CH3)3
c.
b.
lone pair on O
electron donating
halogen
electron withdrawing
R group
electron donating
18.15 Electron-donating groups make the compound react faster than benzene in electrophilic
aromatic substitution. Electron-withdrawing groups make the compound react more
slowly than benzene in electrophilic aromatic substitution.
O
C
a.
O
CH3
O2N
C
Cl
CH3
O2N
+
halogen
electron withdrawing
reacts slower
electron withdrawing
reacts slower
CN
Cl
d.
CN
NO2
Cl
O2N
b.
CH2CH3
electron withdrawing
reacts slower
CH2CH3
e.
NO2
OH
OH
c.
R group
electron donating
reacts faster
OH
+
NO2
lone pairs on O
electron donating
reacts faster
O2N
+
CH2CH3
O2N
18.16 Electron-donating groups make the compound more reactive than benzene in electrophilic
aromatic substitution. Electron-withdrawing groups make the compound less reactive than
benzene in electrophilic aromatic substitution.
C(CH3)3
a.
OH
b.
+
COOCH2CH3
c.
N(CH3)3
d.
OH
R group
electron donating
more reactive
two OH's
electron donating
more reactive
C with 2 electronegative O's
electron withdrawing
less reactive
electron withdrawing
less reactive
487
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–11
18.17
Cl
OCH3
halogen
electron withdrawing
least reactive
lone pairs on O
electron donating
most reactive
intermediate
reactivity
18.18 Chlorine inductively withdraws electron density and decreases the rate of electrophilic
aromatic substitution. The closer the Cl is to the ring, the larger the effect it has. The larger
the number of Cl’s, the larger the effect.
Cl
Cl
Cl
Cl
least reactive
intermediate
reactivity
most reactive
18.19 Especially stable resonance structures have all atoms with an octet. Carbocations with
additional electron donor R groups are also more stable structures. Especially unstable
resonance structures have adjacent like charges.
C(CH3)3
C(CH3)3
a.
NO2
C(CH3)3
C(CH3)3
H
H
H
NO2
NO2
NO2
especially stable with additional R group
stabilized carbocation
OH
OH
b.
NO2
OH
OH
OH
H
H
H
H
NO2
NO2
NO2
NO2
especially stable
All atoms have an octet.
O
CHO
CHO
CHO
C
H
H
H
NO2
NO2
NO2
c.
NO2
O
C
H
H
H
NO2
especially unstable
2 adjacent (+) charges
18.20 Polyhalogenation occurs with highly activated benzene rings containing OH, NH2, and
related groups with a catalyst.
OH
OH
Cl2
a.
Cl
OH
Cl
b.
FeCl3
Cl
OH
Cl2
OH
CH3
Cl
c.
Cl
CH3
Cl2
CH3
Cl
FeCl3
Cl
488
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–12
18.21 Friedel–Crafts reactions do not occur with strongly deactivating substituents including NO2,
or with NH2, NR2, or NHR groups.
a.
CH3Cl
SO3H
no reaction
c.
AlCl3
CH3Cl
N(CH3)2
no Friedel–Crafts
reaction
AlCl3
strongly deactivating
CH3
CH3
b.
Cl
CH3Cl
d.
Cl
AlCl3
NHCOCH3
Cl
CH3
CH3Cl
NHCOCH3
AlCl3
+
Cl is an o,p
director.
CH3
NHCOCH3
18.22 To draw the product of reaction with these disubstituted benzene derivatives and HNO3,
H2SO4 remember:
• If the two directing effects reinforce each other, the new substituent will be on the
position reinforced by both.
• If the directing effects oppose each other, the stronger activator wins.
• No substitution occurs between two meta substituents.
o,p
o,p
OCH3
OCH3
a.
CH3
NO2
HNO3
c.
H2SO4
COOCH3
CH3
NO2 HNO3 O2N
meta
CH3
NO2
COOCH3
NO2
o,p
meta
o,p (strong)
OCH3
OCH3
Br
b.
HNO3
O2N
Cl
OCH3
Br
Cl
Br
HNO3
d.
H SO
Cl
O2 N
Br H2SO4
2
4
o,p
oppose
products due to
OCH3 directing effects
NO2
H2SO4
Br
Br
NO2
NO2
o,p
18.23
Cl
a.
OH
Cl2, FeCl3
SO3, H2SO4
SO3H
c.
OH
CH3Cl, AlCl3
CH3
(+ ortho isomer)
O
ClCOCH3
C
AlCl3
Br
Br2
SO3H
Put meta director on first.
b.
OH
O
HNO3, H2SO4
C
CH3
CH3
O2N
CH3
Br goes ortho to
the stronger activator.
489
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–13
18.24
O2N
O2N
O
NaOCH3
a.
Cl
O
–OH
b. F
HO
OCH3
NO2
NO2
18.25
N(CH3)2
N(CH3)2
N(CH3)2
F3C
H O
O
Cl
CF3
HOC(CH3)3
OC(CH3)3
N(CH3)2
F3C
O
O
(+ 2 additional
resonance structures)
Cl
Cl
18.26
a.
Cl
NaNH2
NH2
NH3
NaOH
b.
CH3O
Cl
CH3O
H2O
Δ
c.
Cl
OH
CH3O
OH
KNH2
NH2
NH3
NH2
18.27
KNH2
CH3
CH3
NH3
Cl
CH3
Two different benzynes are possible.
CH3
NH2
CH3
CH3
NH2
H2N
Ortho, meta, and para products are formed.
18.28 This reaction proceeds via a radical bromination mechanism and two radicals are possible:
A (2o and benzylic) and B (1o). Since B (which leads to C6H5CH2CH2Br) is much less stable,
this radical is not formed so only C6H5CH(Br)CH3 is formed as product.
490
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–14
CH2CH3
CHCH3
CH2CH2
or
A
2o and benzylic
1o
B
Br
CH2CH2Br
CHCH3
only product
not formed
18.29
Br
Br2
hν
a.
Br
Br2
K+ −OC(CH3)3
FeBr3
Br
Br
(+ para isomer)
Br
NaOH
Br2
hν
b.
OH
Br
c.
O
mCPBA
K+ −OC(CH3)3
Br2
hν
OH
[1] BH3
d.
[2] H2O2, HO–
(from c.)
18.30 First use an acylation reaction, and then reduce the carbonyl group to form the alkyl benzenes.
O
a.
O
C
Cl
C
CH2CH2CH2CH3
CH2CH2CH2CH3
Zn(Hg) + HCl
CH2CH2CH2CH2CH3
AlCl3
O
b.
Cl
C
O
C
C(CH3)3
C(CH3)3
CH2C(CH3)3
Zn(Hg) + HCl
AlCl3
18.31
O
O
Cl
Zn(Hg)
O
AlCl3
HCl
O
Cl
AlCl3
p-isobutylacetophenone
(+ ortho isomer)
491
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–15
18.32
CH3
CH3Cl
COOH
KMnO4
a.
AlCl3
NO2
HNO3
b.
H2SO4
Pd-C
CH3
CH3Cl
c.
NH2
H2
CH3
Br2
FeBr3
AlCl3
COOH
KMnO4
Br
Br
(+ para isomer)
18.33
CH3
O
ClCH2CH3
a.
C
Cl
C
O
CH3
CH2CH3
O
SO3
CH3
AlCl3
AlCl3
C
H2SO4
HO3S
CH2CH3
CH2CH3
(+ ortho isomer)
Br
o,p director
o,p director
b.
O
Cl
C
Br
CH3
AlCl3
CH2CH3
Br2
C
Both are o,p directors, but they are meta to each
other. The alkyl group must be obtained by
reduction of a carbonyl.
Br
CH3
Zn(Hg)
FeBr3
C
O
CH3
HCl
CH2CH3
O
Br
c.
Cl2
FeCl3
CH3CH2Cl
Cl
AlCl3
K+ –OC(CH3)3
Br2
Cl
hν
Cl
Cl
(+ ortho isomer)
CHO
Cl
[1] BH3
[2] H2O2, –OH
OH
PCC
Cl
492
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–16
18.34
Br
a. Br2
O
O
O
FeBr3
Br
CHO
CHO
CHO
A
NO2
O
b. HNO3
O
H2SO4
O 2N
CHO
CHO
O
O
c. CH3CH2COCl
O
AlCl3
CHO
O
CHO
Br
H
N
Br
O
a. Br2
H
N
Br
O
FeBr3
B
NO2
b. HNO3
H
N
Br
O
H2SO4
O
c. CH3CH2COCl
Br
H
N
O
AlCl3
18.35 Intramolecular Friedel–Crafts acylation occurs on the more activated aromatic ring.
OCH3
OCH3 O
O
Cl
CH3O
CH3O groups activate this ring.
CH3O
493
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–17
18.36 OH is an ortho, para director.
OH
OH
OH
HNO3
a.
OH
OH
NO2
Cl
Cl2
g.
H2SO4
FeCl3
NO2
OH
OH
Cl
OH
OH
OH
SO3H
SO3
b.
h.
OH
OH
NO2
NH2
OH
OH
OH
AlCl3
O
O
d.
CH(CH2CH3)2
AlCl3
C
OH
NH2NH2
(+ ortho
isomer)
O
OH
(CH3CH2)2CH
C
OH
Br
(+ ortho
isomer)
k.
FeBr3
Br2
Br
OH
OH
(+ ortho
isomer)
hν
Br
OH
OH
Br
Br2
f.
CH2CH(CH2CH3)2
CH(CH2CH3)2
OH
Br
OH
(+ ortho
isomer)
OH
OH
Br2
e.
CH2CH(CH2CH3)2
CH(CH2CH3)2
j.
O
OH
C
(+ ortho
isomer)
HCl
OH
C
(CH3CH2)2CHCOCl
Zn(Hg)
(+ ortho
isomer)
i.
OH
(+ ortho
isomer)
HCl
SO3H
CH3CH2Cl
OH
Sn
(+ ortho
isomer)
H2SO4
c.
Cl
l.
KMnO4
(+ ortho
isomer)
(+ ortho
isomer)
Br
O
C
OH
18.37
CH(CH3)2 CH3CH2COCl
a.
AlCl3
CH(CH3)2
C
CH2CH3
CH(CH3)2
+
CH3CH2
O
O
b.
C
CH3CH2COCl
CH(CH3)2
N(CH3)2
c.
AlCl3
CH3CH2COCl
AlCl3
No reaction
No Friedel–Crafts reaction
C
O
494
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–18
Br
Br
CH3CH2COCl
d.
AlCl3
Br
CH2CH3
C
O
CH3CH2
H
N
e.
C
CH3
CH3CH2
C
O
C
O
H
N
CH3CH2COCl
O
+
C
H
N
CH3
+
CH3CH2
O
AlCl3
C
O
18.38
CN
CN
SO3
a.
H2SO4
SO3H
NO2
HNO3
b.
NO2
HO
CH3
c.
O2N
NO2
H2SO4
SO3
NO2
CH3
OH
H2SO4
OH
HO
HO
CH3
HO3S
OH
SO3H
CH2CH3
d. Cl
CH3CH2Cl
OCOCH3
CHO
Cl
AlCl3
CHO
Br2
e.
FeBr3
CH3
OCOCH3
CHO
Br
CH3
CH3
Br
COCH3
O
f.
CH3COCl
NHCOCH3
NHCOCH3
AlCl3
CH3O
CH3O
O2N
NO2
O
NO2
NO2
HNO3
g.
H2SO4
NO2
NO2
NO2
NO2
Cl
h.
COOCH3
CH3O
OCH3
Br
i.
SO3
Cl2
CH3O
FeCl3
Br
OCH3
Br
SO3H
HO3S
H2SO4
j.
F
NO2
NaSCH2CH3
SCH2CH3
NO2
COOCH3
OCH3
C
O
CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
495
Reactions of Aromatic Compounds 18–19
18.39 Watch out for rearrangements.
a.
Cl
b.
c.
Cl
d.
Cl
Cl
rearrangement
2° carbocation
rearrangement
3° carbocation
18.40
a.
CH3
KMnO4
C(CH3)3
HOOC
C(CH3)3
Br
Br2
b.
hν
CH3
Zn(Hg), HCl
c.
C
CH3
CH2CH2CH3
C
H
O
CH2CH2CH3
H
Br2
d.
FeBr3
Br
Br
O
NH2NH2
e.
–OH
NO2
f.
O2N
Cl
NO2
CH3NH2
O2N
NO2
NHCH3
NO2
18.41
C bonded to 2 H's
must use acylation
followed by reduction.
O
O
Cl
Zn(Hg)
a.
AlCl3
C bonded to 1 H
can be added directly
by alkylation.
Cl
b.
AlCl3
HCl
496
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–20
O
O
CH2CH3
C
Cl
c.
C
CH3
AlCl3
Ethyl group can be introduced
by two methods.
C(CH3)3
CH3CH2Cl
HCl
AlCl3
Method [1]
Method [2]
C(CH3)3
(CH3)3CCl
d.
CH2CH3
CH3 Zn(Hg)
AlCl3
no H's
18.42
NC
N
CN
NaH
OH
F
A
O
N
B
C
18.43
CHO
N
N
N
O
[1]
N
Path [1]
SN2
[2]
CHO
Cl
O
D
Path [2]
nucleophilic aromatic
substitution
N
CHO
HO
CHO
N
O
+ NaH
F
N
N
OH
+ NaH
18.44
SO3H
SO3H
[1] CH3COCl, AlCl3
a.
[2] Cl2, FeCl3
Cl
O
Alternate synthesis:
Cl
AlCl3
SO3H
SO3
CH3COCl
Cl2
FeCl3
Step [1] won't work because a Friedel–Crafts reaction
= A can't be done on a deactivated benzene ring, as is the
case with the SO3H substituent. Even if Step [1] did
work, the second step would introduce Cl meta to
CH3
SO3H, not para as drawn.
Cl
H2SO4
O
CH3
(+ para isomer)
(+ isomer)
Cl
O
CH3
497
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–21
OCH3
Step [1] involves a Friedel–Crafts alkylation
o
= B using a 1 alkyl halide that will undergo
rearrangement, so that a butyl group will not be
NO2
introduced as a side chain.
OCH3
[1] CH3CH2CH2CH2Cl, AlCl3
b.
[2] HNO3, H2SO4
CH3CH2CH2CH2
Alternate synthesis:
OH
OCH3
[1] NaH
OCH3
CH3(CH2)2COCl
AlCl3
[2] CH3Cl
(+ ortho isomer)
CH3CH2CH2
O
OCH3
CH3CH2CH2CH2
OCH3
HNO3
H2SO4
NO2
Zn(Hg), HCl
CH3CH2CH2CH2
B
18.45 Use the directions from Answer 18.16 to rank the compounds.
OH
NO2
NH2
NO2
a.
c.
least reactive
intermediate
reactivity
CHO
most reactive
least reactive
Cl
intermediate
reactivity
CH2NH2
b.
most reactive
CH3
NH2
d.
least reactive
intermediate
reactivity
most reactive
least reactive
intermediate
reactivity
most reactive
18.46
Br
[1]
a.
b.
c.
d.
C N
O
[2]
withdraw
donate
less
deactivate
a.
b.
c.
d.
[3]
withdraw
withdraw
less
deactivate
a.
b.
c.
d.
C
CH3
O
withdraw
donate
more
activate
18.47
O
O
a.
C
O
more electron rich
due to O atom
more reactive
b.
CH3
more electron rich
due to CH3 group
more reactive
c.
more electron rich
due to C atom
more reactive
498
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–22
18.48
E
a.
N
O2N
c.
N
more electron rich
due to N atom
faster
E
electron
withdrawing
O2N
N(CH3)3
N(CH3)3
less electron rich
due to (+) charge on N and
electron-withdrawing NO2 group
slower
N
E
E
b.
NH(CH3)2
E
NH(CH3)2
d. O2N
less electron rich
due to (+) charge on N
slower
N
O2N
N
electron electron donating
withdrawing
Effects cancel out.
similar in reactivity to benzene
18.49
E
a.
E+
+
E
Ortho and para products are isolated.
ortho, para
director
A benzene ring is an electron-rich
substituent that stabilizes an
intermediate positive charge by an
electron-donating resonance effect.
As a result, it activates a benzene
ring toward reaction with
electrophiles.
With ortho and para attack there is additional resonance stabilization that delocalizes the
positive charge onto the second benzene ring. Such additional stabilization is not possible with
meta attack.
Ortho attack:
H
E H
E H
E H
E H
E H
E+
E H
Meta attack:
H E
H
E+
H E
H E
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–23
Para attack:
E+
H
E
H
H
E
H
E
H
E
H
E
H
E
E
b.
E+
O N
O N
+
O N
E
Ortho and para products are isolated.
ortho, para
director
With ortho and para attack there is additional resonance stabilization that delocalizes the
positive charge onto the nitroso group. Such additional stabilization is not possible with
meta attack. This makes –NO an ortho, para director. Since the N atom bears a partial (+)
charge (because it is bonded to a more electronegative O atom), the –NO group inductively
withdraws electron density, thus deactivating the benzene ring towards electrophilic attack.
In this way, the –NO group resembles the halogens. Thus, the electron-donating resonance
effect makes –NO an o,p director, but the electron-withdrawing inductive effect makes it a
deactivator.
Ortho attack:
E H
H
E H
E H
E H
E+
O N
O N
O N
O N
O N
especially stable
Meta attack:
H E
H
H E
H E
E+
O N
O N
O N
O N
Para attack:
E+
O N
H
O N
H
E
O N
H
E
O N
H
E
especially stable
O N
H
E
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–24
18.50
O
OCH2CH3
alkyl group on the benzene ring
R stabilizes (+) charges on the o,p positions by an
electron-donating inductive effect. This group behaves
like any other R group so that ortho and para products
are formed in electrophilic aromatic substitution.
O
O
OCH2CH3
OCH2CH3
(+) charge on atom bonded to the benzene ring
Drawing resonance structures in electrophilic aromatic substitution
results in especially unstable structures for attack at the o,p positions—
two (+) charges on adjacent atoms. This doesn't happen with meta
attack, so meta attack is preferred. This is identical to the situation
observed with all meta directors.
18.51 Under the acidic conditions of nitration, the N atom of the starting material gets protonated,
so the atom directly bonded to the benzene ring bears a (+) charge. This makes it a meta
director, so the new NO2 group is introduced meta to it.
H
N(CH3)2
N(CH3)2
HNO3
N(CH3)2
HNO3
+
H2SO4
H2SO4
acts as a base
now a meta director
NO2
18.52 Increasing the number of electron-withdrawing groups (especially at the ortho and para
positions to the leaving group) increases the rate of nucleophilic aromatic substitution.
Increasing the electronegativity of the halogen increases the rate.
a.
Cl
F
O2N
p-fluoronitrobenzene
most reactive
O2N
chlorobenzene
least reactive
F
m-fluoronitrobenzene
O2N
b.
F
O2N
F
O2N
O2N
1-fluoro-3,5-dinitrobenzene
least reactive
c. CH3
O2N
Cl
NO2
4-chloro-3-nitrotoluene
least reactive
F
NO2
1-fluoro-3,4dinitrobenzene
CH3
F
NO2
4-fluoro-3nitrotoluene
1-fluoro-2,4-dinitrobenzene
most reactive
O 2N
F
NO2
1-fluoro-2,4-dinitrobenzene
most reactive
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–25
18.53 A CH3O group has an electron-donating resonance effect. This stabilizes a (+) charge, so it
increases stability of the carbocation intermediate in electrophilic aromatic substitution. This
destabilizes a (–) charge, so it decreases the stability of the carbanion intermediate in
nucleophilic aromatic substitution.
18.54
Cl
AlCl3
+
_
+
Cl AlCl3
+
+ AlCl4–
resonance-stabilized carbocation
Use both resonance forms to show how two products are formed.
_
Cl AlCl3
H
H
H
H
+
+
+
+
+
H
HCl
+ AlCl3
+
+
+
H
+
H
H
_
Cl AlCl3
18.55
OCH3
H OSO3H
H
OCH3
+
OCH3
OCH3
H
H
OCH3
HSO4–
HSO4–
OCH3
H2SO4
+
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–26
18.56
OH
H OSO3H
CH3O
OH2
CH3O
CH3O
+
CH3O
CH3O
H
HSO4–
1,2-CH3 shift
CH3O
CH3O
H
CH3O
+ H2O
H
CH3O
+
H2SO4
H
HSO4–
18.57
a. The product has one stereogenic center.
stereogenic center
b. The mechanism for Friedel–Crafts alkylation with this 2° halide involves formation of a trigonal
planar carbocation. Since the carbocation is achiral, it reacts with benzene with equal
probability from two possible directions (above and below) to afford an optically inactive,
racemic mixture of two products.
H Cl
AlCl3
(2R)-2-chlorobutane
H
Cl
AlCl3
H
H
AlCl4
trigonal planar
achiral carbocation
–
racemic mixture
optically inactive
H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
503
Reactions of Aromatic Compounds 18–27
18.58 The reaction follows the two-step addition–elimination mechanism for nucleophilic aromatic
substitution.
Na+ –OCH3
N
Cl
OCH3
N
2-chloropyridine
OCH3
N
Cl
OCH3
N
Cl
Cl
A
Cl–
+
N
OCH3
Resonance structure A is stabilized because the negative charge is located on an
electronegative N. This makes nucleophilic aromatic substitution on 2-chloropyridine faster
than a similar reaction with chlorobenzene, which has no N atom to stabilize the intermediate
negative charge.
18.59 Since there is no electron-withdrawing group on the benzene ring, the mechanism likely
proceeds via elimination–addition.
N
Cl
CH3
N CH3
H
Cl
B
H
N CH3
Cl
+
+ HB
B
N CH3
H
N
N
CH3
CH3
B
H B
+ Cl–
504
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–28
18.60
E
E
E+
A
This product is formed.
B
This product is not formed.
H E
H E
H E
H E
E+
Attack to form A proceeds via a
carbocation for which 7 resonance
structures can be drawn. Four
resonance structures contain an intact A
benzene ring.
E
E+
H E
E
H
Attack to form B proceeds via a
carbocation for which 6 resonance
structures can be drawn. Only two
resonance structures contain an
intact benzene ring.
H E
E
H
H E
E
H
E
E
H
B
A reaction that occurs by way of the more stable carbocation is preferred, so product A is formed.
18.61
O
CCl3
C
H OSO3H
H
H
CCl3
H
O
C
H
CCl3
Cl
O
C
OH
OH
H C
H
Cl
H C
CCl3H
HSO4–
H OSO3H
Cl
CCl3
HSO4–
(+ 3 resonance
structures)
OH2
H C
HSO4–
Cl
CCl3
HSO4–
CCl3
Cl
H CCl3
C
Cl
H
Cl
C
H
DDT
H2SO4
(+ 3 resonance
structures)
Cl
Cl
H C
Cl
CCl3
H2O
(+ 5 resonance
structures)
H
H
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–29
18.62
O
O
O
O
O
O
O
O
O AlCl3
O
O
O AlCl3
H
resonance-stabilized
acylium ion
AlCl3
AlCl3
O
O
O
O AlCl3
O
O
O AlCl3
H
H
O
O AlCl3
+ HB+
B
H OH2
O
(any base in the reaction mixture)
O
OH
H2 O
+ AlCl3
18.63 Benzyl bromide forms a resonance-stabilized intermediate that allows it to react rapidly
under SN1 conditions.
Formation of a resonance-stabilized carbocation:
CH2
CH2
CH2
CH2
CH2
CH2
Br
benzyl bromide
resonance-stabilized carbocation
CH2OCH3
CH2
CH2OCH3
+ HBr
H
+ Br
+ CH3OH
O2N
CH2Br
CH2Br
electron-withdrawing group
destabilizing
slower reaction
benzyl methyl ether
CH3O
CH2Br
electron-donor group
stabilizing
faster reaction
The electron-withdrawing NO2 group will destabilize the carbocation so the benzylic halide will be
less reactive, while the electron-donating OCH3 group will stabilize the carbocation, so the benzylic
halide will be more reactive.
18.64
O
a.
Cl2
FeCl3
Cl
Cl
Zn(Hg), HCl
Cl
AlCl3
O
(+ para isomer)
Cl
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–30
b.
NO2
NO2
HNO3
Br2
FeBr3
H2SO4
Br
CH3
CH3Cl
c.
CH3
SO3
AlCl3
H2SO4
COOH
KMnO4
SO3H
SO3H
(+ para isomer)
CH3
CH3Cl
d.
COOH
KMnO4
COOH
HNO3
AlCl3
H2SO4
NO2
CH3Cl
e.
AlCl3
KMnO4
Br2
CH3
Br
FeBr3
Br
COOH
(+ ortho isomer)
CH3
f.
CH3
CH3
CH3
CH3Cl
SO3
Cl2 (excess)
AlCl3
H2SO4
FeCl3
COOH
Cl
Cl
SO3H
KMnO4
Cl
SO3H
Cl
SO3H
(+ ortho isomer)
Br
g.
(CH 3)2CHCl
HNO3
AlCl3
H2SO4
Cl2
NO2
Br2
FeCl3
(+ para isomer)
Cl
hν
NO2
K+ –OC(CH3)3
NO2
Cl
(+ isomer)
18.65
NO2
a.
NH2
Br2
HNO3
Sn
FeBr3
H2SO4
HCl
Br
Br
Br
(+ ortho isomer)
b.
CH3
CH3
CH3
CH3Cl
HNO3
Br2
AlCl3
H2SO4
FeBr3
NO2
(+ ortho isomer)
Br
NO2
Cl
NO2
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–31
NO2
Cl
Cl
HNO3
O
c.
AlCl3
AlCl3
(+ isomer)
H2SO4
O
O
(+ para isomer)
d.
CH3Cl
HNO3
AlCl3 CH3
H2SO4
NO2
NO2
KMnO4
HCl
HOOC
CH3
NH2
Sn
HOOC
(+ ortho isomer)
O
O
O
Cl
e.
Br
Br2
AlCl3
NH2NH2
FeBr3
–
Br
HNO3
Br
H2SO4
OH
NO2
(+ isomer)
H2
Pd-C
Br
NH2
O
f.
CH3CH2Cl
AlCl3
AlCl3
Br2 Br
Cl2
Cl
O
FeCl3
O
hν
O
Cl
(+ ortho isomer)
Cl
K+ –OC(CH3)3
NH2NH2
O
–OH
Cl
Cl
18.66
CH3
CH2Br
Br2
a.
hν
CH3
CHO
CH2Br
Br2
b.
PCC
hν
c.
Cl2
CH3
CH3
FeCl3
Cl
CH3
KMnO4
COOH
(+ para isomer)
CH3
HNO3
d.
CH2OC(CH3)3
OC(CH3)3
Cl
CH3
NO2
H2SO4
HNO3
COOH
NO2
KMnO4
NO2
H2SO4
(+ para isomer)
NO2
NO2
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–32
O
CH3 Br2
e.
FeBr3
CH3
CH3
Cl
AlCl3
Br
Br2
hν
Br
(+ ortho isomer)
Br
O
O
O
O
O2N
HNO3
Cl
CH3
O2N
NH2NH2
H2SO4
AlCl3
OH
Br
O
O
f.
–OH
Br
–OH
H2
Pd-C
(+ ortho isomer)
H2N
O
Cl
CH3
g.
O
AlCl3
Br
Br2
Cl2
CH3
CH3
hν
O
FeCl3
O
Cl
(+ ortho isomer)
Cl
–
CHO
OH
PCC
CH2OH
O
O
Cl
Cl
18.67
Cl
[1] NaH
a.
[2] CH3Br
HO
AlCl3
CH3O
CH3O
(+ ortho isomer)
b.
HO
CH3
[1] NaH
CH3Cl
[2] CH3CH2Br CH3CH2O
AlCl3 CH3CH2O
hν
[2] CH3I
[2] CH3I
HO
CH3
CH3Cl
AlCl3
CH3O
–OH
CH3O
(+ ortho isomer)
HNO3
H2SO4
(excess)
CH2OH
[1] NaH
CH3CH2O
[1] NaH
CH3CH2O
(+ ortho isomer)
OCH3
c.
CH2Br
Br2
O2N
CH3CH2O
CH3
H2
H2 N
CH3
Pd-C CH O
3
CH3O
NO2
NH2
18.68
HO
Br
SOCl2
a.
Cl
AlCl3
Br2
hν
Br
K+ −OC(CH3)3
Br2
Br
2 NH2
C CH
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–33
C CH
O
C C
NaH
b.
C CCH2CH2O
C CCH2CH2OH
H2O
(from a.)
HNO3
c.
Br2
O2N
hν
H2SO4
K+ –OC(CH3)3
O2N
(+ ortho isomer)
(from a.)
HC CH
NaH
O2N
C CH
d.
HBr
ROOR
Br
C CH
CH2CH2 C CH
O2N
Na
C CCH2CH3
[1] NaH
NH3
[2] CH3CH2Cl
(from a.)
O2N
Br
(from a.)
Br
Br2
e.
FeBr3
(from a.)
K+ −OC(CH3)3
Br2
hν
Br
Br
OH
[1] OsO4
Br
[2] H2O, NaHSO3
Br
OH
(+ ortho isomer)
HNO3
Cl2
f.
FeCl3
(from a.)
Br2
Cl
H2SO4
Cl
K+ –OC(CH3)3
Cl
hν
NO2
(+ ortho isomer)
Cl
Br
NO2
mCPBA
NO2
Cl
O
NO2
18.69 One possibility:
Cl
Br
Br2
O
Cl
AlCl3
hν
AlCl3
O
O
K+ –OC(CH3)3
(+ ortho isomer)
HO
Na2Cr2O7
O
HO
[1] BH3
[2] H2O, –OH
H2O, H2SO4
O
Zn(Hg), HCl
HO
O
ibufenac
O
O
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–34
18.70
HO
[2] CH3CH2CH2Br
CH3
CH3Cl
[1] NaH
O
AlCl3
COOH
KMnO4
O
O
(+ ortho isomer)
HNO3
H2SO4
COOH
O2N
X
O
18.71 Use integration data and the molecular formula to determine the number of H’s that give rise
to each signal (Section 14.5, How To).
1
H NMR data of compound A (C8H9Br):
Absorption
ppm
# of H’s
triplet
1.2
3
quartet
2.6
2
two signals
7.1 and 7.4
2+2
Explanation
3 H’s adjacent to 2 H’s
2 H’s adjacent to 3 H’s
para disubstituted benzene
Structure:
Br
1
H NMR data of compound B (C8H9Br):
Absorption
ppm
# of H’s
triplet
3.1
2
triplet
3.5
2
multiplet
7.1–7.4
5
Explanation
2 H’s adjacent to 2 H’s
2 H’s adjacent to 2 H’s
monosubstituted benzene
Structure:
Br
18.72 IR absorption at 1717 cm–1 means compound C has a C=O.
1
H NMR data of compound C (C10H12O):
Absorption
ppm
# of H’s
singlet
2.1
3
triplet
2.8
2
triplet
2.9
2
multiplet
7.1–7.4
5
Explanation
3 H’s
2 H’s adjacent to 2 H’s
2 H’s adjacent to 2 H’s
monosubstituted benzene
Structure:
Explanation
6 H’s adjacent to 1 H
1 H adjacent to 6 H’s
monosubstituted benzene
Structure:
O
18.73
1
H NMR data of compound X (C10H12O):
Absorption
ppm
# of H’s
doublet
1.3
6
septet
3.5
1
multiplet
7.4–8.1
5
O
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Reactions of Aromatic Compounds 18–35
1
H NMR data of compound Y (C10H14):
Absorption
ppm
# of H’s
doublet
0.9
6
multiplet
1.8
1
doublet
2.5
2
multiplet
7.1–7.3
5
Explanation
6 H’s adjacent to 1 H
1 H adjacent to many H’s
2 H’s adjacent to 1 H
monosubstituted benzene
Structure:
18.74
a
CH3
CH3
OH
+
C CH2
d
b
H2SO4
OH c
CH3
CH3
p-cresol
C(CH3)3
d
2-methyl-1-propene
(2 equiv)
C(CH3)3
a
1H
NMR spectral data:
1.4 (singlet, 18 H) (a)
2.27 (singlet, 3 H) (b)
5.0 (singlet, 1 H) (c)
7.0 (singlet, 2 H) (d) ppm
BHT (C15H24O)
H OSO3H
CH3
CH3 C CH3
C CH2
CH3
CH3
OH
OH
HSO4–
OH
C(CH3)3
H
CH3 C CH3
OH
C(CH3)3
H
OH
C(CH3)3
H
C(CH3)3
H
CH3
CH3
CH3
OH
CH3
CH3
CH3
OH
C(CH3)3
H
C(CH3)3
Repeat to add the second C(CH3)3 group.
H2SO4
HSO4–
CH3
CH3
18.75 Five resonance structures can be drawn for phenol, three of which place a negative charge on
the ortho and para carbons. These illustrate that the electron density at these positions is
increased, thus shielding the protons at these positions and shifting the absorptions to lower
chemical shift. Similar resonance structures cannot be drawn with a negative charge at the
meta position, so it is more deshielded and absorbs farther downfield, at higher chemical shift.
OH
OH
OH
(–) charges on the ortho and para positions
OH
OH
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–36
18.76 a. Pyridine: The electron-withdrawing inductive effect of N makes the ring electron poor.
Also, electrophiles E+ can react with N, putting a positive charge on the ring. This makes
the ring less reactive with another positively charged species.
To understand why substitution occurs at C3, compare the
stability of the carbocation formed by attack at C2 and C3.
Electrophilic attack on N:
Electrophilic attack at C2:
Electrophilic attack at C3:
E
H
E+
N
N
E+
E
N
E+
N
H
E
N
N
less reactive
than benzene
E
H
N
H
E
N
E
H
N
H
E
N does not have an octet.
(+) charge on an electronegative N atom
poor resonance structure
attack at C2 does not occur
N
better resonance structures
Since attack at C3 forms a more stable carbocation, attack at C3 occurs. Attack at C4
generates a carbocation of similar stability to attack at C2, so attack at C4 does not occur.
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Reactions of Aromatic Compounds 18–37
b. Pyrrole is more reactive than benzene because the C’s are more electron rich. The lone pair on N
has an electron-donating resonance effect.
Attack at C2:
+
N
H
N
H
N
H
N
H
Attack at C3:
E+
H
E
N
H
E
H
+ E+
N
H
N
H
more reactive than benzene
E
N
H
H
H
E
N
H
E
H
E
N
H
N
H
3-position
fewer resonance structures
Attack at C3 does not occur.
E
N
H
2-position
more resonance structures
attack at C2
Since attack at C2 forms a more stable carbocation, electrophilic substitution occurs at C2.
18.77
O
H OSO3H
OH
OH
OH
OH
1,2-CH3
shift
HSO4–
OH
HSO4–
H
514
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–38
18.78 Draw a stepwise mechanism for the following intramolecular reaction, which was used in the
synthesis of the female sex hormone estrone.
overall reaction
Lewis acid
OH
or HA
RO
RO
A
H A
The steps:
+
+
+
OH2
RO
RO
+ A
+ H2O
RO
(+ 1 resonance structure)
A
H
+
H A +
RO
A
RO
(+ 3 resonance structures)
18.79 a. In quinoline the lone pair on N occupies an sp2 hybrid orbital, so it can never be donated to
the ring by resonance. The N atom decreases the electron density of the ring in which it is
located by an electron-withdrawing inductive effect, so substitution occurs on the other
ring. In indole, the N atom donates its electron pair (which is contained in a p orbital) to
the five-membered ring, increasing its electron density, so substitution occurs on the fivemembered ring with the N atom.
b. In the presence of acid, the N atom is protonated prior to electrophilic attack. For
substitution to occur at C8 rather than C7, the carbocation that results from electrophilic
addition at C8 must be more stable. Attack at C8 generates a carbocation with more
resonance structures and four structures keep one ring aromatic (1–4). Attack at C7
generates a carbocation with fewer resonance structures and only two have an intact
aromatic ring (5 and 6).
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Reactions of Aromatic Compounds 18–39
attack
7
8
N
N
H
N
H
H Br
Br Br
H OSO3H
H Br
at C8
N
H
8
H Br 1
N
H
+ Br–
N
H
H Br
H Br
4
N
H
H Br
3
N
H
2
N
H
H Br
attack
7
at C7
N
H
H
Br
N
H
H
Br
N
H
H
Br
H
Br
N
H
H
Br
N
5 H
Br Br
6
c. With indole, attack at C3 forms the more highly resonance-stabilized carbocation.
H Br
3
H Br
H Br
attack
2
at C3
N
H
N
H
N
H
N
H
all octets
Br Br
attack
H Br
at C2
N
H
H
H
H
Br
Br
Br
N
H
all octets
N
H
H
H
Br
Br
N
H
N
H
H
N
H
Br
N
H
all octets
N
H
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 18–40
Attack at C3 forms resonance structures, all of which have an intact aromatic ring, and two of
which have all atoms with octets. Attack at C2 forms a cation with more resonance
structures, but only two have an intact aromatic ring, and only one has complete octets.
18.80
OH
N
H
O
BF3
H
BF3
N
H
HOBF3
N
H
(+ 3 additional
resonance structures)
1,2-shift
N
H
N
H
H2O
BF3
H
HO BF3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
517
Carboxylic Acids and the Acidity of the O–H Bond 19–1
Chapter 19 Carboxylic Acids and the Acidity of the O–H Bond
Chapter Review
General facts
x Carboxylic acids contain a carboxy group (COOH). The central carbon is sp2 hybridized and
trigonal planar (19.1).
x Carboxylic acids are identified by the suffixes -oic acid, carboxylic acid, or -ic acid (19.2).
x Carboxylic acids are polar compounds that exhibit hydrogen bonding interactions (19.3).
Summary of spectroscopic absorptions (19.4)
IR absorptions
C=O
O–H
~1710 cm–1
3500–2500 cm–1 (very broad and strong)
1
O–H
C–H D to COOH
10–12 ppm (highly deshielded proton)
2–2.5 ppm (somewhat deshielded Csp3–H)
C=O
170–210 ppm (highly deshielded carbon)
H NMR absorptions
13
C NMR absorption
General acid–base reaction of carboxylic acids (19.9)
x Carboxylic acids are especially acidic
O
O
+
R C
B
R C
because carboxylate anions are
+
H B
O H
O
resonance stabilized.
carboxylate
anion
pKa | 5
x For equilibrium to favor the products,
the base must have a conjugate acid
with a pKa > 5. Common bases are
listed in Table 19.3.
Factors that affect acidity
Resonance effects. A carboxylic acid is more acidic than an alcohol or phenol because its conjugate
base is more effectively stabilized by resonance (19.9).
OH
ROH
pKa = 16–18
O
R C
OH
pKa = 10
pKa | 5
Increasing acidity
Inductive effects. Acidity increases with the presence of electron-withdrawing groups (like the
electronegative halogens) and decreases with the presence of electron-donating
groups (like polarizable alkyl groups) (19.10).
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–2
Substituted benzoic acids.
x Electron-donor groups (D) make a substituted benzoic acid less acidic than benzoic acid.
x Electron-withdrawing groups (W) make a substituted benzoic acid more acidic than
benzoic acid.
COOH
COOH
COOH
D
W
less acidic
higher pKa
more acidic
lower pKa
pKa < 4.2
pKa = 4.2
pKa > 4.2
Increasing acidity
Other facts
x Extraction is a useful technique for separating compounds having different solubility properties.
Carboxylic acids can be separated from other organic compounds by extraction, because aqueous
base converts a carboxylic acid into a water-soluble carboxylate anion (19.12).
x A sulfonic acid (RSO3H) is a strong acid because it forms a weak, resonance-stabilized conjugate
base on deprotonation (19.13).
x Amino acids have an amino group on the D carbon to the carboxy group [RCH(NH2)COOH].
Amino acids exist as zwitterions at pH | 6. Adding acid forms a species with a net (+1)
charge [RCH(NH3)COOH]+. Adding base forms a species with a net (–1) charge
[RCH(NH2)COO]– (19.14).
Practice Test on Chapter Review
1. a. Give the IUPAC name for the following compound.
CO2H
b. Draw the structure corresponding to the following name: sodium m-bromobenzoate.
2. a. Which of the labeled atoms is least acidic?
Ha
O
O
O
O Hd
He
1. Ha
3. Hc
2. Hb
5. He
4. Hd
Hc
Hb
O
b. Which of the following carboxylic acids has the lowest pKa?
COOH
1.
COOH
2.
CH3
COOH
3.
Cl
COOH
COOH
4.
5.
CH3O
O2 N
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
519
Carboxylic Acids and the Acidity of the O–H Bond 19–3
c. Which compound(s) can be converted to A by an oxidation reaction?
OH
1.
CO2H
4. Both [1] and [2] can be converted to A.
O
O
5. Compounds [1], [2], and [3] can all
be converted to A.
OH
A
2.
OH
3.
OH
3. Rank the following compounds in order of increasing basicity. Label the least basic compound
as 1, the most basic compound as 3, and the compound of intermediate basicity as 2.
COO–
COO–
O2N
A
O–
B
C
4. Draw the organic products formed in each of the following reactions.
+
NH3
a.
CH3
H
OH
NaOH
C
COOH
(1 equiv)
N
HO
[2] CH3I
NaH
COOH
b.
[1] NaH
c.
(1 equiv)
Answers to Practice Test
1.a. cis-2-methylcyclopentanecarboxylic acid
2.a. 1
b. 5
c. 5
3. A–2
B–1
C–3
4.a.
OCH3
NH3
CH3
H
b.
c.
+
C
N
COO
b.
COO–Na+
COO
HO
Br
520
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–4
Answers to Problems
To name a carboxylic acid:
[1] Find the longest chain containing the COOH group and change the -e ending to -oic acid.
[2] Number the chain to put the COOH carbon at C1, but omit the number from the name.
[3] Follow all other rules of nomenclature.
19.1
3
a.
2
4
H
CH3
c.
CH3CH2CH2 C CH2COOH
CH3 2
H
CH3CH2 C CH2
1
CH3CH2
1
C COOH
CH2CH3
Number the chain to put COOH at C1.
6 carbon chain = hexanoic acid
2,4-diethylhexanoic acid
Number the chain to put COOH at C1.
6 carbon chain = hexanoic acid
3,3-dimethylhexanoic acid
O
4
b.
6
H
CH3 C CH2CH2COOH
d.
1
Cl
4
1 OH
8
9
Number the chain to put COOH at C1.
5 carbon chain = pentanoic acid
4-chloropentanoic acid
Number the chain to put COOH at C1.
9 carbon chain = nonanoic acid
4-isopropyl-6,8-dimethylnonanoic acid
19.2
c. 3,3,4-trimethylheptanoic acid
a. 2-bromobutanoic acid
O
e. 3,4-diethylcyclohexanecarboxylic
acid
O
O
HO
OH
HO
Br
b. 2,3-dimethylpentanoic acid
d. 2-sec-butyl-4,4-diethylnonanoic acid
O
O
HO
f. 1-isopropylcyclobutanecarboxylic acid
COOH
OH
19.3
O
D
aD-methoxyvaleric acid
c. DE-dimethylcaproic acid
OH
E
OCH3
O
O
b. E-phenylpropionic acid
OH
D
E
d. D-chloro-E-methylbutyric acid
OH
Cl
E
OH
D
O
19.4
O
O
a.
O
Li+
lithium benzoate
b.
Na+
O
O
O
H
sodium formate
or
sodium methanoate
c.
O
K+
d.
O
Na+
Br
potassium 2-methylpropanoate
sodium 4-bromo-6-ethyloctanoate
521
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and the Acidity of the O–H Bond 19–5
19.5
COO–Na+
COOH
C2
2-propylpentanoic acid
19.6
19.7
sodium 2-propylpentanoate
More polar molecules have a higher boiling point and are more water soluble.
COOCH3
CH2CH2CH2OH
least polar
lowest boiling point
least H2O soluble
intermediate polarity
intermediate boiling point
CH2COOH
most polar
highest boiling point
most H2O soluble
Look for functional group differences to distinguish the compounds by IR. Besides sp3
hybridized C–H bonds at 3000–2850 cm–1 (which all three compounds have), the following
functional group absorptions are seen:
O
CH3CH2CH2CH2
C
OH
O
OH
CH3CH2CH2CH2
C
O
OCH3
carboxylic acid
2 strong absorptions
~1710 (C=O)
~2500–3500 (OH) cm–1
ester
1 strong absorption
~1700 (C=O) cm–1
Molecular formula: C4H8O2
one degree of unsaturation
1
alcohol
1 strong absorption
~3600–3200 (OH) cm–1
19.8
H NMR data (ppm):
0.95 (triplet, 3 H)
1.65 (multiplet, 2 H)
2.30 (triplet, 2 H)
11.8 (singlet, 1 H)
O
HO
19.9
HO
HO
COOH
HO
COOH
HO
OH
PGF2D
a prostaglandin
OH
enantiomer
There are five tetrahedral stereogenic
centers. Both double bonds can
exhibit cis–trans isomerism.
Therefore, there are
27 = 128 stereoisomers.
19.10 1° Alcohols are converted to carboxylic acids by oxidation reactions.
O
a.
OH
OH
O
b.
(CH3)2CH
C
OH
(CH3)2CH
CH2 OH
c.
COOH
CH2OH
522
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–6
19.11
a.
Na2Cr2O7
CH2OH
COOH
H2SO4, H2O
A
b.
c. O2N
KMnO4
CH3
O2 N
COOH
C
(Any R group with benzylic H's
can be present para to NO2.)
OH
[1] O3
CH3C CCH3
[2] H2O
B
2° OH
d.
CH3COOH
(2 equiv)
O
CrO3
OH
CO2H
H2SO4, H2O
D
1° OH
19.12
a.
b.
COOH
CH3
CH3
NaOH
OH
COO
NaOCH3
CH3
Na+
+ H2 O
O
CH3
NaH
c. CH3 C OH
CH3
CH3
Na+
d.
C O Na+ + H2
CH3
COOH
NaHCO3
+ HOCH3
COO
Na+
+ H2CO3
19.13 CH3COOH has a pKa of 4.8. Any base having a conjugate acid with a pKa higher than 4.8 can
deprotonate it.
a. F– pKa (HF) = 3.2 not strong enough
b. (CH3)3CO– pKa [(CH3)3COH]= 18 strong enough
c. CH3– pKa (CH4) = 50 strong enough
d. –NH2 pKa (NH3) = 38 strong enough
e. Cl– pKa (HCl) = –7.0 not strong enough
19.14
Ha
H OHb
Increasing acidity: Ha < Hb < Hc
OHc
O
H OHb
– Ha
OHc
negative charge on C
unstable conjugate base
O
mandelic acid
Ha
H O
– Hb
OHc
O
Ha
negative charge on O
more stable conjugate base
Ha
H OHb
– Hc
H OHb
O
O
O
O
negative charge on O,
resonance stabilized
most stable conjugate base
19.15 Electron-withdrawing groups make an acid more acidic, lowering its pKa.
CH3CH2 COOH
least acidic
pKa = 4.9
ICH2
COOH
one electron-withdrawing group
intermediate acidity
pKa = 3.2
CF3 COOH
three electron-withdrawing F's
most acidic
pKa = 0.2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
523
Carboxylic Acids and the Acidity of the O–H Bond 19–7
19.16
a. CH3COOH
least acidic
HSCH2COOH
HOCH 2COOH
intermediate
acidity
most acidic
b. ICH2CH2COOH
least acidic
ICH2COOH
I2CHCOOH
intermediate
acidity
most acidic
19.17
a.
CH3
COOH
least acidic
COOH
Cl
intermediate
acidity
COOH
most acidic
O
b. CH3O
COOH
CH3
C
COOH
COOH
CH3
least acidic
intermediate
acidity
most acidic
19.18
OH
Phenol A has a higher pKa than phenol
because of its substituents. Both the OH
and CH3 are electron-donating groups,
which make the conjugate base less stable.
Therefore, the acid is less acidic.
HO
A
19.19 To separate compounds by an extraction procedure, they must have different solubility
properties.
a. CH3(CH2)6COOH and CH3CH2CH2CH2CH=CH2: YES. The acid can be extracted into
aqueous base, while the alkene will remain in an organic layer.
b. CH3CH2CH2CH2CH=CH2 and (CH3CH2CH2)2O: NO. Both compounds are soluble in
organic solvents and insoluble in water. Neither is acidic enough to be extracted into
aqueous base.
c. CH3(CH2)6COOH and NaCl: one carboxylic acid, one salt: YES. The carboxylic acid is
soluble in an organic solvent while the salt is soluble in water.
d. NaCl and KCl: two salts: NO.
19.20
weaker conjugate base
better leaving group
stronger conjugate base
worse leaving group
CF3SO3–
CF3SO3H
CH3SO3–
CH3SO3H
CF3 is electron withdrawing.
stronger acid
lower pKa
CH3 is electron donating.
weaker acid
higher pKa
19.21
phenylalanine
COOH
C
H2N
H
methionine
COOH
H
COOH
C
C
NH2
H2N
H
CH2CH2SCH3
R
R
S
COOH
H
CH3SCH2CH2
C
NH2
S
524
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–8
19.22 Since amino acids exist as zwitterions (i.e., salts), they are too polar to be soluble in organic
solvents like diethyl ether. Thus, they are soluble in water.
19.23
COOH
H 3N C H
COO
COO
H3 N C H
H2N C H
H
H
H
pH = 1
glycine
pH = 11
neutral form
19.24
COO
pKa(COOH) + pKa(NH3+)
pI =
=
(2.58) + (9.24)
= 5.91
H3N C H
2
2
CH2
19.25
+
+
H3N CH COO–
H3N CH COOH
H
H
electron-withdrawing group
The nearby (+) stabilizes the conjugate base by an
electron-withdrawing inductive effect, thus making
the starting acid more acidic.
19.26
6
5
4
2
1
3
OH
A
a. 2,5-dimethylhexanoic acid
2
B
a. 3-ethyl-3-methylcyclohexanecarboxylic acid
NaOH
NaOH
O
b.
COOH
1
O
3
COO Na+
b.
O Na+
c. sodium 2,5-dimethylhexanoate
c. sodium 3-ethyl-3-methylcyclohexanecarboxylate
d. An alcohol or ether would have a much
higher pKa than a carboxylic acid.
d.
OH
OH
OH
O
H
19.27
COOH
COOH
CH3
least acidic
COOH
CF3
intermediate acidity
most acidic
525
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and the Acidity of the O–H Bond 19–9
19.28 Use the directions from Answer 19.1 to name the compounds.
a.
(CH3)2CHCH2CH2CO2H
b.
BrCH2COOH
COOH
4-methylpentanoic acid
g.
o-bromobenzoic acid
2-bromoacetic acid
or 2-bromoethanoic acid
Br
O
h.
c.
CH3CH2
COOH p-ethylbenzoic acid
OH
4,4,5,5-tetramethyloctanoic acid
d.
–
+
CH3CH2CH2COO Li
O
lithium butanoate
O– Na+
i.
sodium 2-methylhexanoate
1-ethylcyclopentanecarboxylic acid
e.
COOH
COOH
10
j.
3
7
2,4-dimethylcyclohexanecarboxylic acid
f.
5
7-ethyl-5-isopropyl-3-methyldecanoic acid
COOH
19.29
OH
f. o-chlorobenzoic acid
a. 3,3-dimethylpentanoic acid
COOH
O
Cl
OH
b. 4-chloro-3-phenylheptanoic acid
CH3
O
c. (2R)-2-chloropropanoic acid
Cl
Cl
Cl
e. m-hydroxybenzoic acid
C
O
K+
O
h. sodium D-bromobutyrate
OH
O
Na+
Br
O
d. EE-dichloropropionic acid
Cl
g. potassium acetate O
i. 2,2-dichloropentanedioic acid
O
O
O
HO
HOOC
OH
Cl Cl
OH
j. 4-isopropyl-2-methyloctanedioic acid
OH
O
HO
OH
O
19.30
O
O
OH
pentanoic acid
OH
3-methylbutanoic acid
O– Na+
sodium pentanoate
O– Na+
sodium 3-methylbutanoate
OH
OH
2-methylbutanoic acid
2,2-dimethylpropanoic acid
O
O
O
O
O
O
O– Na+
sodium 2-methylbutanoate
O– Na+
sodium
2,2-dimethylpropanoate
526
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–10
19.31
C5 or G carbon
OH
a.
OH
C2 or D carbon
b.
CO2H
CO2H
HO
C3 or E carbon
IUPAC: 2-hydroxypropanoic acid
common:D-hydroxypropionic acid
IUPAC: 3,5-dihydroxy-3-methylpentanoic acid
common: EG-dihydroxy-E-methylvaleric acid
19.32
O
a.
OH
OH
O
lowest boiling point
intermediate boiling point
O
highest boiling point
O
OH
b.
HO
lowest boiling point
intermediate boiling point
highest boiling point
19.33
O
OH
a.
CrO3
OH
[1] O3
c.
CH3
CO2
[2] H2O
H2SO4, H2O
b. (CH3)2CH
COOH +
C C H
KMnO4
COOH d. CH3(CH2)6CH2OH
HOOC
Na2Cr2O7
CH3(CH2)6COOH
H2SO4, H2O
19.34
[1] BH3
a.
CH2
[2] H2O2, OH
HC CH
[2] CH3I
B
[1] NaNH2
HC CCH3
C
[2] CH3CH2I
CH(CH3)2
(CH3)2CHCl
c.
COOH
H2SO4, H2O
A
[1] NaNH2
b.
CrO3
CH2OH
–
CH3CH2C CCH3
D
[1] O3
[2] H2O
CH3CH2COOH
CH3COOH
E + F
COOH
KMnO4
AlCl3
G
H
19.35
Bases: [1] –OH pKa (H2O) = 15.7; [2] CH3CH2– pKa (CH3CH3) = 50; [3] –NH2 pKa (NH3) = 38;
[4] NH3 pKa (NH4+) = 9.4; [5] HC{C– pKa (HC{CH) = 25.
a.
CH3
COOH
pKa = 4.3
All of the bases
can deprotonate this.
b.
–OH,
Cl
OH
pKa = 9.4
CH3CH2–, –NH2, and HC{C–
can deprotonate this.
c. (CH3)3COH
pKa = 18
CH3CH2–, –NH2, and HC{C–
can deprotonate this.
527
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and the Acidity of the O–H Bond 19–11
19.36
a.
K+
COOH
OC(CH
COO K+
3)3
+ HOC(CH3)3
pKa = 18
Reaction favors products.
pKa = 4.2
OH
b.
NH4+
O
NH3
pKa | 16
+
OH
c.
Na+ NH2
+ NH3 + Na+
O
CH3–Li+
COOH
COO Li+
CH3
Na+H
OH
Na+
O
Reaction favors products.
H2
Reaction favors products.
pKa = 35
pKa | 16
CH3
f.
CH4
pKa = 50
CH3
pKa | 4
e.
Reaction favors products.
pKa = 38
pKa = 10
d.
Reaction favors reactants.
pKa = 9.4
OH
Na2CO3
O– Na+
CH3
With the same pKa for the starting
acid and the conjugate acid, an
equal amount of starting materials
and products is present.
Na+ HCO3
pKa = 10.2
pKa = 10.2
19.37 The stronger acid has a lower pKa and a weaker conjugate base.
COOH
a.
CH2OH
or
carboxylic acid
stronger acid
lower pKa
weaker conjugate base
c.
or
COOH
or
ClCH2COOH
FCH2COOH
F is more electronegative.
weaker acid
stronger acid
higher pKa
lower pKa
stronger conjugate base
weaker conjugate base
d.
Cl
or
NCCH2COOH
CN is electron withdrawing.
stronger acid
lower pKa
weaker conjugate base
CH3COOH
weaker acid
higher pKa
stronger conjugate base
19.38
Br
a.
COOH
least acidic
Br is electronegative
intermediate acidity
OH
b.
CH3
least acidic
Cl
COOH
COOH
Cl more electronegative
most acidic
OH
Cl
intermediate acidity
COOH
Cl is electron withdrawing.
stronger acid
lower pKa
weaker conjugate base
CH3 is electron donating.
weaker acid
higher pKa
stronger conjugate base
alcohol
weaker acid
higher pKa
stronger conjugate base
b.
CH3
OH
O2N
most acidic
528
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–12
19.39
O
a. BrCH2COO–
BrCH2CH2COO–
weakest base
(CH3)3CCOO–
intermediate basicity
O2N
strongest base
weakest base
b.
O
NH
weakest base
O
O
c.
intermediate basicity
strongest base
CH2
intermediate basicity
strongest base
19.40
Increasing acidity
ICH2COOH
pKa values
BrCH2COOH
least acidic
3.12
FCH2COOH
2.86
2.66
F2CHCOOH
F3CCOOH
most acidic
0.28
1.24
19.41 The OH of the phenol group in morphine is more acidic than the OH of the alcohol (pKa 10
versus pKa 16). KOH is basic enough to remove the phenolic OH, the most acidic proton.
most acidic proton
HO
The OH is part of a phenol.
Methylation occurs here.
O
O
an alcohol
H
HO
N
H
CH3
CH3O
[2] CH3I
H
HO
H
CH3
HO
CH3
O
H
H
HO
N
CH3
O
O
O
N
H
N
Many resonance structures
stabilize the conjugate base.
O
O
O
HO
morphine
H
O
[1] KOH
H
O
H
N
H
CH3
HO
O
H
N
CH3
H
HO
H
N
CH3
codeine
19.42 The closer the electron-withdrawing CH3CO– group is to the carboxylic acid, the more it will
stabilize the conjugate base, making the acid stronger.
O
C
CH3 G+ C
O
O
OH
pyruvic acid
stronger acid
O
C
CH3 G+ CH2 OH
C
farther away
acetoacetic acid
weaker acid
529
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and the Acidity of the O–H Bond 19–13
19.43
a. The negative charge on the conjugate base of p-nitrophenol is delocalized on the NO2 group,
stabilizing the conjugate base, and making p-nitrophenol more acidic than phenol (where the
negative charge is delocalized only around the benzene ring).
OH
O
O
O +
N
O +
N
O +
N
O
O
O
p-nitrophenol
pKa = 7.2
OH
O
phenol
pKa = 10
two of the possible resonance structures for the
conjugate base [See part (b) for all the possible
resonance structures.]
b. In the para isomer, the negative charge of the conjugate base is delocalized over both the benzene
ring and onto the NO2 group, whereas in the meta isomer it cannot be delocalized onto the NO2
group. This makes the conjugate base from the para isomer more highly resonance stabilized, and
the para substituted phenol more acidic than its meta isomer.
OH
O2N
O
O2N
O
O
O
O2N
pKa = 7.2
p-nitrophenol
O
N
O
N
O
O2N
O
negative charge on
two O atoms
very good resonance structure
more stable conjugate base
O2N
stronger acid
OH
NO2
O
O
O
NO2
O
NO2
O
NO2
NO2
O
O
NO2
pKa = 8.3
m-nitrophenol
19.44 A CH3O group has an electron-withdrawing inductive effect and an electron-donating
resonance effect. In 2-methoxyacetic acid, the OCH3 group is bonded to an sp3 hybridized C,
so there is no way to donate electron density by resonance. The CH3O group withdraws
electron density because of the electronegative O atom, stabilizing the conjugate base, and
making CH3OCH2COOH a stronger acid than CH3COOH.
– H+
O
CH3O
OH
more acidic acid
O
CH3O
O
more stable conjugate base
530
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–14
In p-methoxybenzoic acid, the CH3O group is bonded to an sp2 hybridized C, so it can donate
electron density by a resonance effect. This destabilizes the conjugate base, making the
starting material less acidic than C6H5COOH.
O
– H+
CH3O
O
O
CH3O
CH3O
OH
O
O
less acidic acid
like charges nearby
less stable conjugate base
19.45 Phenol has a pKa of 10, making p-methylthiophenol (pKa = 9.53) the stronger acid. A
substituent that increases the acidity of a phenol must withdraw electron density to stabilize
the negative charge of the conjugate base. An electron-withdrawing substituent deactivates a
benzene ring towards electrophilic aromatic substitution, making p-methylthiophenol less
reactive than phenol.
OH
OH
CH3S
stronger acid
less reactive in
electrophilic substitution
19.46
The O in A is more electronegative than the N in C so there is a
stronger electron-withdrawing inductive effect. This stabilizes
the conjugate base of A, making A more acidic than C.
CO2H
O
A
pKa = 3.2
CO2H
O
B
pKa = 3.9
N
H
CO2H
C
pKa = 4.4
Since the O in A is closer to the COOH group than the O atom in
B, there is a stronger electron-withdrawing inductive effect. This
makes A more acidic than B.
531
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and the Acidity of the O–H Bond 19–15
19.47
CO2H
CO2H
O2N
O2N
D
CO2H
C
E
– H+
– H+
O
– H+
O
O
O
O
O2N
O2N
O
Since the benzene ring is bonded
to the D carbon (not the carbonyl
carbon), this compound is not
much different than any alkylsubstituted carboxylic acid.
least acidic
The electron-withdrawing
Since the NO2 group is bonded to a
inductive effect of the NO2 benzene ring that is bonded directly to the
–
group helps stabilize the COO
carbonyl group, inductive effects and
group.
resonance effects stabilize the conjugate
intermediate acidity
base. For example, a resonance
structure can be drawn that places a (+)
charge close to the COO– group.
Two of the resonance structures for the conjugate base of C:
most acidic
O
O
O
N
O
N
O
O
O
O
unlike charges nearby
stabilizing
19.48
O
CH3
C *
OH
NaOH
CH3
O
O
C *
O
C *
O
H3O
labeled O atom
CH3
+
H3O+
OH
O
CH3
The resonance-stabilized carboxylate
anion can now be protonated on either O
atom, the one with the label and the one
without the label.
C *
OH
CH3
C *
O
The label is now in two different locations.
19.49
O
a.
Ha
O
Hc
Hb
loss of Hb:
O
Hb
Hc
O
one Lewis structure
least stable conjugate base
O
Hc
O
2 resonance structures
intermediate stability
O
Hc
Hb
The most acidic proton forms the
most stable conjugate base.
O
1,3-cyclohexanedione
increasing acidity: Hb < Ha < Hc
loss of Ha:
Ha
O
O
loss of Hc: H
a
Hb
O
Ha
O
Hb
Ha
O
Hb
3 resonance structures
most stable conjugate base
O
532
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–16
Hb
N
N
b.
Hc
O
Ha
N
Hc
loss of Hb:
O
Ha
O
Ha
acetanilide
N
Ha
Hc
N
Hc
O
Hb
N
Hc
O
Ha
7 resonance structures
most stable conjugate base
increasing acidity: Ha < Hc < Hb
loss of Ha:
N
Hc
O
Ha
Hb
Hb
N
N
N
Hc
Ha
N
loss of Hc:
O
O
Ha
one Lewis structure
least stable conjugate base
O
Ha
Hc
O
Ha
Hc
O
2 resonance structures that
delocalize the negative charge
intermediate stability
19.50
O
O
O
H
H
H
H
H H
HO
O
O
H
H
H
OH
HO
O
O
O
H
H
H
H
The conjugate base is resonance
stabilized. Two of the structures place a
negative charge on an O atom.
weaker conjugate base
stronger acid
The conjugate base has only one Lewis structure.
stronger conjugate base
weaker acid
O
19.51
The negatively charged C is more nucleophilic
than the negatively charged O atom.
CH3COOH
CH2COO–
strong base
(2 equiv)
CH3CH2CH2CH2 Br
X
CH3CH2CH2CH2CH2COO–
H3O+
Two equivalents of strong base remove
both the O–H and C–H protons.
CH3CH2CH2CH2CH2COOH
hexanoic acid
19.52
CH3
O
O
C
C
NH2
acetamide
CH3
CH3
N
CH3
C
N
O
C
C
CH3
O
O
H
O is more electronegative than N, making the
conjugate base of CH3COOH more stable
than the conjugate base of acetamide. Therefore,
acetamide is less acidic.
somewhat less stable
with the (–) charge on N
O
OH
O
H
CH3
C
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
533
Carboxylic Acids and the Acidity of the O–H Bond 19–17
19.53
COOH
A
x
x
B
x
Dissolve both compounds in CH2Cl2.
Add 10% NaHCO3 solution. This makes a carboxylate
anion (C10H7COO–) from B, which dissolves in the
aqueous layer. The other compound (A) remains in
the CH2Cl2.
Separate the layers.
19.54
OH
OH
and
x
x
x
Dissolve both compounds in CH2Cl2.
Add 10% NaOH solution. This converts C6H5OH into
a phenoxide anion, C6H5O–, which dissolves in the
aqueous solution. The alcohol remains in the organic
layer (neutral) since it is not acidic enough to be
deprotonated to any significant extent by NaOH.
Separate the layers.
19.55 To separate two compounds in an aqueous extraction, one must be water soluble (or be able
to be converted into a water-soluble ionic compound by an acid–base reaction), and the other
insoluble. 1-Octanol has greater than 5 C’s, making it insoluble in water. Octane is an
alkane, also insoluble in water. Neither compound is acidic enough to be deprotonated by a
base in aqueous solution. Since their solubility properties are similar, they cannot be
separated by an extraction procedure.
19.56
O
C
one double bond or ring
a. Molecular formula: C3H5ClO2
ClCH2CH2
OH
C=O and O–H
IR: 3500–2500 cm–1, 1714 cm–1
NMR data: 2.87 (triplet, 2 H), 3.76 (triplet, 2 H), and 11.8 (singlet, 1 H) ppm
b. Molecular formula: C8H8O3
5 double bonds or rings CH3O
COOH
IR: 3500–2500 cm–1, 1688 cm–1
C=O and O–H
NMR data: 3.8 (singlet, 3 H), 7.0 (doublet, 2 H), 7.9 (doublet, 2 H), and 12.7 (singlet, 1 H) ppm
para disubstituted benzene ring
5 double bonds or rings
c. Molecular formula: C8H8O3
OCH2COOH
IR: 3500–2500 cm–1, 1710 cm–1
C=O and O–H
NMR data: 4.7 (singlet, 2 H), 6.9–7.3 (multiplet, 5 H), and 11.3 (singlet, 1 H) ppm
monosubstituted benzene ring
534
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–18
19.57
Compound A: Molecular formula C4H8O2 (one degree of unsaturation)
IR absorptions at 3600–3200 (O–H), 3000–2800 (C–H), and 1700 (C=O) cm–1
1
H NMR data:
Absorption
singlet
singlet
triplet
triplet
ppm
2.2
2.55
2.7
3.9
# of H’s
3
1
2
2
Explanation
a CH3 group
1 H adjacent to none or OH
2 H’s adjacent to 2 H’s
2 H’s adjacent to 2 H’s
Structure:
O
CH3
C
A
CH2CH2OH
Compound B: Molecular formula C4H8O2 (one degree of unsaturation)
IR absorptions at 3500–2500 (O–H) and 1700 (C=O) cm–1
1
H NMR data:
Absorption
doublet
septet
singlet (very broad)
ppm
1.6
2.3
10.7
# of H’s
6
1
1
Explanation
6 H’s adjacent to 1 H
1 H adjacent to 6 H’s
OH of RCOOH
Structure:
CH3
CH3 C COOH
H
B
19.58
Compound C: Molecular formula C4H8O3 (one degree of unsaturation)
IR absorptions at 3600–2500 (O–H) and 1734 (C=O) cm–1
1
H NMR data:
Absorption
triplet
quartet
singlet
singlet
ppm
1.2
3.6
4.1
11.3
# of H’s
3
2
2
1
Explanation
a CH3 group adjacent to 2 H’s
2 H’s adjacent to 3 H’s
2 H’s
OH of COOH
Structure:
O
O
OH
C
19.59
Compound D: Molecular formula C9H9ClO2 (five degrees of unsaturation)
13
C NMR data: 30, 36, 128, 130, 133, 139, 179 = 7 different types of C’s
1
H NMR data:
Absorption
triplet
triplet
two signals
singlet
ppm
2.7
2.9
7.2
11.7
# of H’s
2
2
4
1
Explanation
2 H’s adjacent to 2 H’s
2 H’s adjacent to 2 H’s
on benzene ring
OH of COOH
Structure:
COOH
Cl
D
19.60
Molecular formula: C8H6O4: 6 degrees of unsaturation
IR 1692 cm–1 (C=O)
1H NMR 8.2 and 10.0 ppm (singlets)
aromatic H
COOH
O
HO
O
OH
535
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and the Acidity of the O–H Bond 19–19
19.61
A
COOH
COOH
B
O
C
OH
3 different C's
Spectrum [2]: peaks at 27, 39, 186 ppm
5 different C's
Spectrum [1]: peaks at 14, 22, 27, 34, 181 ppm
4 different C's
Spectrum [3]: peaks at 22, 26, 43, 180 ppm
19.62
GBL: Molecular formula C4H6O2 (two degrees of unsaturation)
IR absorption at 1770 (C=O) cm–1
1
H NMR data:
Absorption
multiplet
triplet
triplet
ppm
2.28
2.48
4.35
# of H’s
2
2
2
Explanation
2 H’s adjacent to several H’s
2 H’s adjacent to 2 H’s
2 H’s adjacent to 2 H’s
Structure:
O
O
GBL
19.63
HOOC
H
C
threonine
OH
HOOC
H
C
H
H2N
C
H2N
CH3
OH
HOOC
C
CH3
H
HOOC
C
H
H2N
2S,3S
2R,3S
OH
H
C
C
CH3
H2N
H
C
OH
H
CH3
2S,3R
naturally
occurring
2R,3R
19.64
NH
NH
COOH
NH2
COOH
proline
enantiomer
O
O
COO
zwitterion
19.65
a. methionine
O
H3N CH C OH
b. serine
O
O
H2N CH C O
CH2
CH2
CH2
CH2
CH2
CH2
CH2SCH3
CH2SCH3
CH2SCH3
OH
OH
OH
pH = 1
pH = 6
form at isoelectric point
pH = 11
H3N CH C OH
pH = 1
H3N CH C O
O
H3N CH C O
pH = 6
form at isoelectric point
19.66
a. cysteine pI =
b. methionine pI =
pKa(COOH) + pKa(NH3+)
= (2.05) + (10.25) / 2 = 6.15
2
pKa(COOH) + pKa(NH3+)
2
H2N CH C O
= (2.28) + (9.21) / 2 = 5.75
pH = 11
536
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–20
19.67
O
O
H2N
C
C
OH
N
H
H2N H
lysine
This lone pair is localized
on the N atom, making it a
base.
OH
H2N H
tryptophan
This lone pair is delocalized in the S system to give 10S
electrons, making it aromatic. This is similar to pyrrole
(Chapter 17). Since these electrons are delocalized in the
aromatic system, this N atom in tryptophan is not basic.
19.68 The first equivalent of NH3 acts as a base to remove a proton from the carboxylic acid.
A second equivalent then acts as a nucleophile to displace X to form the ammonium salt of
the amino acid.
O
R
OH
X
O
O
NH3
R
acid–base reaction
NH3
O– NH4+
SN2 reaction
X
R
O– NH4+
NH3
19.69
a. At pH = 1, the net
charge is (+1).
NH3
b. increasing pH: As base is added, the
most acidic proton is removed first,
then the next most acidic proton, and
so forth.
NH3
C
HOOCCH2CH2
COOH
H
NH3
C
OOCCH2CH2
COO– Na+
H
C
HOOCCH2CH2
c. monosodium glutamate
COOH
H
base (1 equiv)
NH3
C
HOOCCH2CH2
COO–
H
base (2nd equiv)
NH3
C
OOCCH2CH2
COO–
H
base (3rd equiv)
NH2
C
OOCCH2CH2
COO–
H
19.70 The first equivalent of NaH removes the most acidic proton—that is, the OH proton on the
phenol. The resulting phenoxide can then act as a nucleophile to displace I to form a
substitution product. With two equivalents, both OH protons are removed. In this case the
more nucleophilic O atom is the stronger base—that is, the alkoxide derived from the alcohol
(not the phenoxide), so this negatively charged O atom reacts first in a nucleophilic
substitution reaction.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
537
Carboxylic Acids and the Acidity of the O–H Bond 19–21
nucleophile
most acidic proton
NaH
–
O
(CH2)4OH
–O
(CH2)4O–
(1 equiv)
[1] CH3I
CH3O
(CH2)4OH
[2] H2O
(CH2)4OH
HO
NaH
(2 equiv)
[1] CH3I
HO
(CH2)4OCH3
[2] H2O
nucleophile
19.71
O
HO
COOH
HO
C
O
p-hydroxybenzoic acid
less acidic than benzoic acid
+
HO
O
C
O
like charges on nearby atoms
destabilizing
The OH group donates electron density by its resonance
effect and this destabilizes the conjugate base, making the
acid less acidic than benzoic acid.
O H
OH
O
C
COOH
O
o-hydroxybenzoic acid
more acidic than benzoic acid
Intramolecular hydrogen bonding stabilizes the
conjugate base, making the acid more acidic than
benzoic acid.
19.72
O
Hd
OHe
HaO
HbO Hc O
2-hydroxybutanedioic acid
increasing acidity:
Hd < Hc < Hb < He < Ha
Ha and He must be the two most acidic protons since they are part
of carboxylic acids. Loss of a proton forms a resonance-stabilized
carboxylate anion that has the negative charge delocalized on two
O atoms. Ha is more acidic than He because the nearby OH group
on the D carbon increases acidity by an electron-withdrawing
inductive effect. Hb is the next most acidic proton because the
conjugate base places a negative charge on the electronegative
O atom, but it is not resonance stabilized.
The least acidic H’s are Hc and Hd since these H’s are bonded to
C atoms. The electronegative O atom further acidifies Hc by an
electron-withdrawing inductive effect.
538
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 19–22
19.73
most acidic H
OH
O
O
O
O
O
base
O
O
O
O
O
A
O
B
O
O
O
O
C
The conjugate base has three resonance structures, two of which place a negative charge on the
oxygens. In this way the conjugate base resembles a carboxylate anion. In addition, the C=C’s in
A and C are conjugated.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
539
Introduction to Carbonyl Chemistry 20–1
Chapter 20 Introduction to Carbonyl Chemistry
Chapter Review
Reduction reactions
[1] Reduction of aldehydes and ketones to 1o and 2o alcohols (20.4)
O
R
C
OH
NaBH4, CH3OH
H(R')
R C H(R')
or
H
[1] LiAlH4 [2] H2O
1o or 2o alcohol
or
H2, Pd-C
[2] Reduction of α,β-unsaturated aldehydes and ketones (20.4C)
OH
NaBH4
CH3OH
• reduction of the C=O only
R
O
O
H2 (1 equiv)
R
Pd-C
• reduction of the C=C only
R
OH
H2 (excess)
Pd-C
• reduction of both π bonds
R
[3] Enantioselective ketone reduction (20.6)
[1] (S)- or (R)CBS reagent
O
C
R
HO H
C
[2] H2O
H OH
R
(R) 2o alcohol
C
or
R
(S) 2o alcohol
•
A single
enantiomer is
formed.
[4] Reduction of acid chlorides (20.7A)
[1] LiAlH4
[2] H2O
O
R
C
RCH2OH
1o alcohol
•
LiAlH4, a strong reducing agent,
reduces an acid chloride to a
1o alcohol.
•
With LiAlH[OC(CH3)3]3, a milder
reducing agent, reduction stops at the
aldehyde stage.
Cl
O
[1] LiAlH[OC(CH3)3]3
[2] H2O
R
C
H
aldehyde
540
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–2
[5] Reduction of esters (20.7A)
[1] LiAlH4
RCH2OH
[2] H2O
1o alcohol
O
R
C
•
LiAlH4, a strong reducing agent,
reduces an ester to a 1o alcohol.
•
With DIBAL-H, a milder reducing
agent, reduction stops at the aldehyde
stage.
OR'
O
[1] DIBAL-H
C
R
[2] H2O
H
aldehyde
[6] Reduction of carboxylic acids to 1o alcohols (20.7B)
O
R
C
[1] LiAlH4
OH
[2] H2O
RCH2OH
1o alcohol
[7] Reduction of amides to amines (20.7B)
O
R
C
[1] LiAlH4
[2] H2O
N
RCH2 N
amine
Oxidation reactions
Oxidation of aldehydes to carboxylic acids (20.8)
•
O
R
C
O
CrO3, Na2Cr2O7, K2Cr2O7, KMnO4
H
or
Ag2O, NH4OH
C
R
•
OH
carboxylic acid
Preparation of organometallic reagents (20.9)
[1] Organolithium reagents:
R X
[2] Grignard reagents:
All Cr6+ reagents except PCC
oxidize RCHO to RCOOH.
Tollens reagent (Ag2O +
NH4OH) oxidizes RCHO only.
Primary (1°) and secondary (2°)
alcohols do not react with
Tollens reagent.
R X
+
2 Li
R
Li
+ Mg
(CH3CH2)2O
[3] Organocuprate reagents:
R X
2 R Li
+
2 Li
+ CuI
+
LiX
R Mg X
R Li + LiX
R2Cu Li+
+
LiI
541
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–3
[4] Lithium and sodium acetylides:
Na+ –NH2
R C C H
R C C
Na+
+
NH3
a sodium acetylide
R Li
R C C H
R C C Li
+ R H
a lithium acetylide
Reactions with organometallic reagents
[1] Reaction as a base (20.9C)
R M
+
H O R
+
R H
M+
•
•
O R
RM = RLi, RMgX, R2CuLi
This acid–base reaction occurs with H2O,
ROH, RNH2, R2NH, RSH, RCOOH,
RCONH2, and RCONHR.
[2] Reaction with aldehydes and ketones to form 1o, 2o, and 3o alcohols (20.10)
O
R
C
OH
[1] R"MgX or R"Li
R C H (R')
[2] H2O
H (R')
R"
1o, 2o, or 3o alcohol
[3] Reaction with esters to form 3o alcohols (20.13A)
[1] R"Li or R"MgX
(2 equiv)
O
R
C
OR'
[2] H2O
OH
R C R"
R"
3o alcohol
[4] Reaction with acid chlorides (20.13)
[1] R"Li or R"MgX
(2 equiv)
[2] H2O
O
R
C
OH
•
More reactive organometallic reagents—R"Li
and R"MgX—add two equivalents of R" to an
acid chloride to form a 3o alcohol with two
identical R" groups.
•
Less reactive organometallic reagents—
R'2CuLi—add only one equivalent of R' to an
acid chloride to form a ketone.
R C R"
R"
3o alcohol
Cl
O
[1] R'2CuLi
[2] H2O
R
C
R'
ketone
542
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–4
[5] Reaction with carbon dioxide—Carboxylation (20.14A)
O
[1] CO2
R MgX
R C
[2] H3O+
OH
carboxylic acid
[6] Reaction with epoxides (20.14B)
OH
O
C
[1] RLi, RMgX, or R2CuLi
C
C
[2] H2O
C
R
alcohol
[7] Reaction with α,β-unsaturated aldehydes and ketones (20.15B)
[1] R'Li or R'MgX
OH
R
[2] H2O
R
C
C
C
•
More reactive organometallic reagents—
R'Li and R'MgX—react with α,βunsaturated carbonyls by 1,2-addition.
•
Less reactive organometallic reagents—
R'2CuLi— react with α,β-unsaturated
carbonyls by 1,4-addition.
R'
O
C
C
allylic alcohol
C
O H
C
C
R
[1] R'2CuLi
[2] H2O
C
R'
ketone
Protecting groups (20.12)
[1] Protecting an alcohol as a tert-butyldimethylsilyl ether
CH3
R O H
+ Cl
CH3
Si C(CH3)3
CH3
R O Si C(CH3)3
N
CH3
NH
R O TBDMS
Cl TBDMS
tert-butyldimethylsilyl ether
[2] Deprotecting a tert-butyldimethylsilyl ether to re-form an alcohol
CH3
R O Si C(CH3)3
CH3
(CH3CH2CH2CH2)4N+ F–
R O H
+
F Si C(CH3)3
CH3
R O TBDMS
CH3
F
TBDMS
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
543
Introduction to Carbonyl Chemistry 20–5
Practice Test on Chapter Review
1. Which compounds undergo nucleophilic addition and which undergo substitution?
O
a.
CH3
C
O
b.
CH3
CH3CH2CH2
C
O
O
c.
Cl
CH3
C
C
d.
OCH3
H
2. What product is formed when CH3CH2CH2Li reacts with each compound, followed by
quenching with water and acid?
a. CH3CH2CHO
b. (CH3)2CO
c. CH3CH2CO2CH3
d. CH3CH2COCl
e. CO2
f. CH2=CHCOCH3
g. ethylene oxide
h. CH3COOH
3. What product is formed when HO(CH2)4CHO is treated with each reagent?
a. NaBH4, CH3OH
b. PCC
c. Ag2O, NH4OH
d. Na2Cr2O7, H2SO4, H2O
4. What reagent is needed to convert (CH3CH2)2CHCOCl into each compound?
a. (CH3CH2)2CHCOCH2CH3
b. (CH3CH2)2CHCHO
c. (CH3CH2)2CHC(OH)(CH2CH3)2
d. (CH3CH2)2CHCH2OH
5. Draw the organic products formed in the following reactions.
OH
N
a.
[1] LiAlH4
[2] CH3Li
[3] H2O
[Indicate stereochemistry.]
c. O
[2] H2O
[1] TBDMS–Cl, imidazole
O
O
[2]
b.
[1] Mg
OCH3
d.
Br
[3] H3O+
O
O
NaBH4
CH3OH
544
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–6
6. What starting materials are needed to synthesize each compound using the indicated reagent or
functional group?
OH
OH
c. Synthesize:
a. Synthesize:
using a Grignard reagent
from an ester
O
b. Synthesize:
using an organocuprate reagent
Answers to Practice Test
1. a. addition
4.a. (CH3CH2)2CuLi
b. substitution
b. DIBAL-H
c. substitution
c. CH3CH2MgBr
d. addition
d. LiAlH4
5.
6.
CO2R
a.
N
a.
+ CH3MgBr
O
2. a. CH3CH2CH(OH)CH2CH2CH3
b. (CH3)2C(OH)CH2CH2CH3
c. (CH3CH2)3COH
d. (CH3CH2)3COH
e. CH3CH2CH2CO2H
f. CH2=CHC(OH)(CH3)CH2CH2CH3
g. CH3(CH2)4OH
h. CH3CH2CH3 + CH3CO2H
OH
b.
b.
CuLi
2
OTBDMS
c.
HO
3. a. HO(CH2)5OH
b. OHC(CH2)3CHO
c. HO(CH2)4CO2H
d. HO2C(CH2)3CO2H
c.
OTBDMS
MgBr
+ CH3CHO
HO
OCH3
d.
OH
O
Answers to Problems
20.1
[1] Csp2–Csp2
a.
H
[3]
Csp2–Csp2
[2] σ: C –O
π: Cp–Op
sp2
α-sinensal
O
sp2
b. The O is sp2 hybridized.
Both lone pairs occupy sp2 hybrid orbitals.
545
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–7
20.2
A carbonyl compound with a reasonable leaving group (NR2 or OR bonded to the C=O)
undergoes substitution reactions. Those without good leaving groups undergo addition.
O
no good leaving group
addition reactions
O
O
O
O
OH
N
O
H
All other C=O's have leaving groups.
substitution reactions
OH
HO
O
O
H
O
O
O
20.3
Aldehydes are more reactive than ketones. In carbonyl compounds with leaving groups, the
better the leaving group, the more reactive the carbonyl compound.
O
a.
CH3CH2CH2
O
C
H
or
CH3CH2CH2
C
O
c.
CH3
O
CH3CH2
C
O
or
CH3
CH3CH(CH3)
C
CH3
C
OCH3
CH2CH3
CH3
C
O
or
OCH3
CH3
C
NHCH3
better leaving group
more reactive
NaBH4 reduces aldehydes to 1° alcohols and ketones to 2° alcohols.
O
a.
Cl
O
d.
less hindered carbonyl
more reactive
20.4
O
or
better leaving group
more reactive
less hindered carbonyl
more reactive
b.
CH3CH2
C
CH3CH2CH2
C
OH
NaBH4
H
CH3OH
CH3CH2CH2 C H
NaBH4
c.
H
CH3OH
O
OH
NaBH4
b.
20.5
O
OH
CH3OH
1° Alcohols are prepared from aldehydes and 2° alcohols are from ketones.
H
OH
O
a.
b.
OH
O
OH
c.
20.6
OH
1-methylcyclohexanol
3° Alcohols cannot be made by reduction of a carbonyl group,
because they do not contain a H on the C with the OH.
O
546
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–8
20.7
O
a.
H2 (excess)
d.
[2] H2O
O
OH
O
OH
[1] LiAlH4
Pd-C
OH
NaBH4
O
NaBH4 (excess)
e.
b.
CH3OH
CH3OH
O
O
O
H2 (1 equiv)
c.
OH
D OH
NaBD4
f.
CH3OH
Pd-C
20.8
O
OH
CH3OH
NaBH4
CHO
b.
CH3OH
c. (CH3)3C
20.9
OH
NaBH4
a.
OH
NaBH4
O
(CH3)3C
CH3OH
OH
(CH3)3C
OH
The 2° alcohol comes from a carbonyl group. Since hydride was delivered from the back
side, the (R)-CBS reagent must be used.
O
O
O
H OH
H drawn back
F
N
N
F
2° OH O
O
F
X
+
(R)-CBS
F
20.10
Part [1]: Nucleophilic substitution of H for Cl
O
CH3
C
O
Cl
[1]
H3Al H
CH3 C Cl
H
O
[2]
CH3
C
+
H
aldehyde
+ AlH3
Cl–
H replaces Cl.
Part [2]: Nucleophilic addition of H– to form an alcohol
O
O
CH3
C
H3Al H
H
CH3 C H
[3]
H
+ AlH3
H OH
[4]
OH
CH3 C H
+ –OH
H
1o alcohol
20.11 Acid chlorides and esters can be reduced to 1° alcohols. Keep the carbon skeleton the same
in drawing an ester and acid chloride precursor.
547
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–9
O
O
CH2OH
a.
C
C
or
Cl
OH
CH2OH
OCH3
Cl
b.
C
O
or O
CH3O
or
O
CH3O
c.
OCH3
Cl
C
O
OCH3
CH3O
20.12
O
a.
O
[1] LiAlH4
OH
C
c.
OH
[2] H2O
N(CH3)2
[1] LiAlH4
CH2N(CH3)2
[2] H2O
O
O
b.
[1] LiAlH4
NH2
NH2
[2] H2O
NH
d.
[1] LiAlH4
NH
[2] H2O
20.13
O
CH2NH2
C
a.
O
NH2
c.
O
N
b.
CH2CH3
C
CH2CH3
N
H
O
N
CH2CH3
CH2CH3
N
H
N
or
CH2CH3
20.14
O
COOCH3
a.
[1] LiAlH4
O
b. CH3O
OH
NaBH4
COOCH3
CH3OH
[1] LiAlH4
OH [2] H2O
NaBH4
CH3OH
HO
OH
+ HOCH3
Neither functional group reduced
20.15
CO2CH2CH3
a.
O
b.
CO2CH2CH3
O
H2
(1 equiv)
Pd-C
CO2CH2CH3
O
H2
(2 equiv)
Pd-C
[1] LiAlH4
CO2CH2CH3
OH
CH3O
OH
CH3O
OH
[2] H2O
+ HOCH3
CH3OH
O
O
c.
OH
[2] H2O
NaBH4
CH3O
OH
548
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–10
CO2CH2CH3
c.
[1] LiAlH4
+ CH3CH2OH
OH
[2] H2O
O
OH
d.
NaBH4
CO2CH2CH3
CO2CH2CH3
CH3OH
O
OH
20.16 Tollens reagent reacts only with aldehydes.
Ag2O, NH4OH
a.
C
O
COOH
Na2Cr2O7
H2SO4, H2O
O
CHO
b.
CH2OH
OH
Ag2O, NH4OH
OH
No reaction
OH
O
Na2Cr2O7
H2SO4, H2O
C
OH
20.17
O
OH
CHO
c. B
CHO
PCC
O
HO
B
H
OH
OH
a. B
CH2OH
NaBH4
d. B
O
Ag2O, NH4OH
C
HO
CH3OH HO
O
O
OH
b. B
[1] LiAlH4
[2] H2O
OH
CH2OH
e. B
C
CrO3
H2SO4, H2O
HO
OH
O
OH
20.18
a. CH3CH2Br + 2 Li
c. CH3CH2Br + 2 Li
CH3CH2Li + LiBr
b. CH3CH2Br + Mg
CH3CH2MgBr
CH3CH2Li
2 CH3CH 2Li + CuI
+ LiBr
LiCu(CH2CH3)2 + LiI
20.19
HC CCH2CH2CH2CH2CH2CH3 + NaH
Na+ C CCH2CH2CH2CH2CH2CH3 + H2
hydrogen gas
HC CCH2CH2CH2CH2CH2CH3
BrMgC CCH2CH2CH2CH2CH2CH3 + CH4
methane gas
+ CH3MgBr
20.20
a.
b.
Li + H2O
CH3
CH3 C MgBr + H2O
CH3
+ LiOH
c.
MgBr
CH3
CH3 C H
CH3
+ HOMgBr
d. CH3CH2C C Li
+ H2O
+ H2O
+ HOMgBr
CH3CH2C CH + LiOH
549
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–11
20.21 To draw the products, add the alkyl or phenyl group to the carbonyl carbon and protonate the
oxygen.
[1] CH3CH2CH2Li
a.
[2] H2O
O
b.
H
C
Li
[1]
H
O
[2] H2O
HO CH2CH2CH3
OH
O
OH
[1] C6H5Li
c.
d.
[1] CH2=O
Na+
C C
C C CH2OH
[2] H2O
H C H
[2] H2O
20.22 Addition of RM always occurs from above and below the plane of the molecule.
H OH
O
a.
CH3
C
H
CH3
b.
H OH
[1] CH3CH2MgBr
+
[2] H2O
O
[1] CH3CH2Li
CH2CH3
CH3
+
OH
CH3
OH
[2] H2O
CH2CH3
20.23
OH
OH
a.
O
CH3 C CH3
CH3MgBr +
H
b.
c.
CH2OH
H
MgBr
C
CH3
+
C
+ CH3CH2MgBr
H
+
H
C
O
OH
O
MgBr
O
OH
O
C CH2CH3
MgBr
+
d.
H
OH
O
+ CH3MgBr
CH2CH3
H
OH
or
O
C CH2CH3
+ CH3CH2MgBr
H
H
20.24
OH
O
a.
+ CH3Li
linalool
(three methods)
OH
Li
O
Li
lavandulol
O
+ CH2=O
Li
550
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–12
b.
c. Linalool is a 3° ROH. Therefore, it has no H on the
carbon with the OH group, and cannot be prepared by
reduction of a carbonyl compound.
CHO
OH
NaBH4
CH3OH
20.25
N(CH3)2
N(CH3)2
OH
O
BrMg
OCH3
OCH3
venlafaxine
20.26
CH3 O
CH3 O
TBDMS–Cl
CH3OH
C CH
[1] Li C CH
imidazole
[2] H2O
HO
TBDMSO
TBDMSO
estrone
(CH3CH2CH2CH2)4NF
CH3OH
C CH
ethynylestradiol
HO
20.27
O
a.
CH3CH2
C
Cl
b.
CH3CH2 C CH2CH2CH2CH3
CH2CH2CH2CH3
[2] H2O
O
C
OH
[1] CH3CH2CH2CH2MgBr
(2 equiv)
OCH3
[1] CH3CH2CH2CH2MgBr
(2 equiv)
OH
CH3CH2CH2CH2 C CH2CH2CH2CH3
[2] H2O
O
c.
OCH2CH3
[1] CH3CH2CH2CH2MgBr
(2 equiv)
OH
CH3CH2CH2CH2 C CH2CH2CH2CH3
[2] H2O
CH2CH2CH2CH2CH3
20.28
O
OH
a.
C
CH3
CH3O
C
+
MgBr
CH3
(2 equiv)
O
b. (CH3CH2CH2)3COH
CH3O
C
CH2CH2CH3 + CH3CH2CH2MgBr
(2 equiv)
551
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–13
O
OH
c. CH3 C CH2CH(CH3)2
C
CH3O
CH2CH(CH3)2
CH3
+ CH3MgBr
(2 equiv)
20.29 The R group of the organocuprate has replaced the Cl on the acid chloride.
O
a. CH3CH2
Cl
C
CH3CH2
[2] H2O
O
[1] (Ph)2CuLi
c. CH3CH2 C Cl
CH3
[2] H2O
Ph
O
[1] [CH3CH2CH(CH3)]2CuLi
O
b.
O
O
[1] (CH3)2CuLi
C
[2] H2O
Cl
20.30
O
a. (CH3)2CHCH2
Cl [2] H O
2
O
O
[1] LiAlH[OC(CH3)3]3
C
(CH3)2CHCH2
C
H
c. (CH3)2CHCH2
O
b.
(CH3)2CHCH2
C
[2] H2O
O
C
Cl
[1] CH3CH2Li (2 equiv)
HO
d.
(CH3)2CHCH2
O
[1] (CH3CH2)2CuLi
Cl
C
[1] LiAlH4
Cl
[2] H2O
(CH3)2CHCH2CH2OH
[2] H2O
20.31
O
C
a.
O
2CuLi
CH3
Cl
C
CH3
Cl + [(CH3)3C]2CuLi
b.
O
or
O
C
O
or
O
C
CH3
(CH3)2CuLi
Cl
Cl
O
+ [(CH3)2CH]2CuLi
O
20.32
O
[1] Mg
Br
MgBr
[2] CO2
a.
b.
c. CH3O
Cl
[1] Mg
CH2Br
MgCl
[1] Mg
CH3O
C
O
O
[2] CO2
[3] H3O+
COO–
CH2MgBr
[2] CO2
C
OH
[3] H3O+
CH3O
COOH
CH2COO–
[3] H3O+
CH3O
CH2COOH
552
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–14
20.33
OH
a.
O
c.
O
O
OH
b.
OH
OH
O
d.
(+ enantiomer)
20.34 The characteristic reaction of α,β-unsaturated carbonyl compounds is nucleophilic addition.
Grignard and organolithium reagents react by 1,2-addition and organocuprate reagents react
by 1,4-addition.
O
O
CH3
O
a.
[1] (CH3)2CuLi
c.
[2] H2O
CH3
O
CH3
[1] (CH3)2CuLi
[2] H2O
CH3
CH3
HO C CH
CH3
[1] H C C Li
[2] H2O
HO C CH
[1] H C C Li
[2] H2O
CH3
[1] (CH3)2CuLi
b.
[2] H2O
O
O
CH3
[1] H C C Li
[2] H2O
HO C CH
20.35
a. CH3CH2OH
HBr or
PBr3
PCC
OH
Mg
CH3CH2Br
O
CH3CH2MgBr
[1] CH3CH2MgBr
OH
[2] H2O
b.
HBr
OH
Br
(from a.)
c.
OH
H2SO4
H2
+
Pd-C
(from a.)
d.
OH
HBr or
PBr3
Br
O
MgBr
MgBr
H2O
CH3CH2OH
e.
Mg
PCC
CH3CHO
OH
H2O
(from d.)
CH3CH2OH
H2SO4
CH2=CH2
mCPBA
O
PCC
OH
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
553
Introduction to Carbonyl Chemistry 20–15
20.36
O
OH
a. NaBH4
CH3OH
A
OH
b. [1] LiAlH4
[2] H2O
OH
c. [1] CH3MgBr (excess)
[2] H2O
HO C6H5
d. [1] C6H5Li (excess)
[2] H2O
e. Na2Cr2O7
No reaction
H2SO4
H2O
O
CH3O
OCH3
a. NaBH4
No reaction
CH3OH
B
CH3O
b. [1] LiAlH4
OH
[2] H2O
CH3O
c. [1] CH3MgBr (excess)
OH
[2] H2O
d. [1] C6H5Li (excess)
C6H5 C6H5
CH3O
OH
[2] H2O
e. Na2Cr2O7
H2SO4
H2O
No reaction
20.37
PBr3
a.
OH
Mg
Br
O
PCC
H
[1] BrMg CH2CH3
[2] H2O
OH
CH3OH
PBr3
CH3Br
Mg
O
b.
[1] CH3MgBr
H
(from a.)
[2] H2O
OH
PCC
O
[1] CH3MgBr
[2] H2O
OH
554
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–16
OH
Br
PBr3
[1] Mg
c.
(from b.)
CH3OH
d.
OH
OH
[2] H2C=O
[3] H2O
H2SO4
PCC
CH2 CH2
O
mCPBA
[1] BrMg CH2CH3
OH
(from a.)
[2] H2O
20.38
O
O
NaBH4
a.
H
CH3OH
OH
O
H
[1] LiAlH4
OH
h.
H
[2] H2O
O
H2
H
OH
[1] (CH3)2CuLi
i.
H
Pd-C
No reaction
[2] H2O
O
d.
O
PCC
H
No reaction
O
H
OH
H2SO4, H2O
[2] H2O
OH
NH4OH
OH
[1] CH3C≡CLi
H
[2] H2O
O
Ag2O
H
H
k.
OH
[1] HC≡CNa
O
O
f.
j.
O
Na2Cr2O7
e.
OH
[1] C6H5Li
[2] H2O
O
c.
H
[2] H2O
O
b.
OH
[1] CH3MgBr
g.
TBDMSCl
l.
OH
O–TBDMS
imidazole
20.39
Li (2 equiv)
a.
Br
b.
Br
c.
Mg
Br
+
Li
LiBr
MgBr
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
(CH3CH2CH2CH2)2CuLi
H2O
d.
Li
e.
MgBr
f.
+ LiOH
D2 O
D
+ DOMgBr
CH3C≡CH
+
Li
LiC≡CCH3
20.40
a.
MgBr
CH2 O
H2 O
OH
d.
b.
MgBr
O
MgBr CH3CH2COOCH3
OH
H2 O
HO
H2O
e.
MgBr CH3COOH
f.
MgBr
+ CH3COO
OH
c.
MgBr
CH3CH2COCl
H2O
HC CH
HC C
555
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–17
g.
MgBr
h.
MgBr
O
H3O+
CO2
OH
i.
MgBr
j.
MgBr
D2O
O
OD
O
OH
H2O
D +
H2O
OH
20.41
O
C
O
Cl
a.
C
(CH3CH2CH2CH2)2CuLi
CH2CH2CH2CH3
O
C
OCH3
b.
(CH3CH2CH2CH2)2CuLi
No reaction
O
O
CH3
c.
CH3
[1] (CH3CH2CH2CH2)2CuLi
[2] H2O
O
d.
CH3
[1] (CH3CH2CH2CH2)2CuLi
CH3
[2] H2O
CH2CH2CH2CH3
HO
20.42 Arrange the larger group [(CH3)3C–] on the left side of the carbonyl.
HO H
H OH
a.
NaBH4, CH3OH
S
O
R
HO H
b.
[1] (S)-CBS reagent; [2] H2O
c.
[1] (R)-CBS reagent; [2] H2O
R
H OH
S
20.43
HO
OH
CH3
a. NaBH4, CH3OH
C
d. [1] CH3Li; [2] H2O
O
O
CH3
C
CH3
HO
CH2CH3
C
CH3 b. H2 (1 equiv), Pd-C
CH3
e. [1] CH3CH2MgBr; [2] H2O
OH
O
A
c. H2 (excess), Pd-C
CH3
CH3
C
f.
[1] (CH2=CH)2CuLi; [2] H2O
CH3
CH=CH2
556
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–18
20.44
O
a. (CH3)2CHCH2CH2
C
Cl
C
c. (CH3)2CHCH2CH2
C
H
O
[1] (CH2=CH)2CuLi
Cl
(CH3)2CHCH2CH2
[2] H2O
[1] C6H5MgBr
(2 equiv)
O
C
(CH3)2CHCH2CH2
[2] H2O
O
b. (CH3)2CHCH2CH2
O
[1] LiAlH[OC(CH3)3]3
Cl
O
CH=CH2
OH
(CH3)2CHCH2CH2
C C6H5
[2] H2O
C6H5
OH
[1] LiAlH4
d. (CH ) CHCH CH C Cl
3 2
2
2
C
(CH3)2CHCH2CH2
[2] H2O
C H
H
20.45
O
a.
CH3CH2
C
[1] LiAlH4
OCH2CH2CH3
O
b.
CH3CH2
C
[1] CH3CH2CH2MgCl
(2 equiv)
OCH2CH2CH3
O
c.
CH3CH2
C
CH3CH2CH2OH
[2] H2O
OH
CH3CH2 C CH2CH2CH3
[2] H2O
CH2CH2CH3
O
[1] DIBAL-H
OCH2CH2CH3
CH3CH2
[2] H2O
C
H
20.46
a.
NaBH4
O
COOCH3
OH
O
[1] LiAlH4
c.
CH3OH
COOCH3
OH
(CH3)2N
[2] H2O
OH
(CH3)2N
O
[1] LiAlH4
b. O
OH
d.
COOCH3 [2] H2O
CH2OH
[1] LiAlH[OC(CH3)3]3
O
OH
Cl
[2] H2O
O
OH
H
O
O
20.47
CH3
a. CH3
CH3
MgBr
[1] CO2
CH3
COCl
COOH
c.
C
CH3
O [1] CH3CH2MgBr
b.
OH
[2] H2O
[2] H3O+
CH3
[1] C6H5MgBr
(excess)
[2] H2O
OH
d.
COOCH2CH3
[1] CH3MgCl
(excess)
CH3
[2] H2O
OH
C CH3
557
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–19
e.
OH
[1] CH2=O
MgBr
O
CH2OH
g.
[2] H2O
O
f.
CH3
[2] H2O
OH
O
[1] (CH3)2CuLi
[1] CH3MgBr
O
h.
[2] H2O
[1] (CH3)2CuLi
C6H5
[2] H2O
20.48
[1] CH3Li
a. (CH3)3C
O
[2] H2O
b.
(CH3)3C
[1] CH3CH2MgBr
O
OH
OH
OH
CH3
CH3
CH=CH2
OH
c.
O
OH
CH2CH3
+
CH2CH3
[2] H2O
+ (CH3)3C
[1] (CH2=CH)2CuLi
+
[2] H2O
OH
CH=CH2
[1] Mg
d.
Br
COOH
[2] CO2
H
H
[3] H3O+
O
H OH
[1] (S)-CBS reagent
e.
[2] H2O
O
f.
OCH2CH3
H
[1] LiAlH4
CH2OH
+ HOCH2CH3
[2] H2O
H
20.49 Three carbons bear a δ+ because they are bonded to electronegative O atoms.
C3
O
C2
O
O
C1
C1 is an sp3 hybridized C bonded to an O so it bears a δ+. There are no additional resonance
structures that affect C1. C2 is part of a carbonyl that has three resonance structures, only
one of which places a (+) charge on C. The O atom of the ester donates its electron pair,
making the carbonyl C less electrophilic than C3. C3 is most electrophilic. Its carbonyl is
stabilized by two resonance structures, one of which places a (+) charge on carbon.
558
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–20
O
O
O
C2
O
O
O
O
O
O
O donates electron density.
C3
O
O
O
O
O
full (+) charge
most electrophilic site
O
20.50 The organolithium reagent is a nucleophile and a base. As a base it can remove the most acidic
proton (between the benzene ring and C=O) to form an enolate that is protonated by D3O+.
OCH3 H
OCH3
OCH3 D
O
O
O
D OD2
+ D2O
HC C Li
OCH3
OCH3
OCH3
HC CH
B
C12H13DO3
Li
A
OCH3
O
OCH3
20.51 Both ketones are chiral molecules with carbonyl groups that have one side more sterically
hindered than the other. In both reductions, hydride approaches from the less hindered side.
The CH3 groups on the bridgehead carbon
make the top more hindered. H– attacks
from below to afford an exo OH group.
Attack comes from below.
2 H's
less hindered
Attack comes from above.
H
H
CH3
CH3
[1] LiAlH4
H
CH3
H
[2] H2O
O
[1] LiAlH4
from above
OH
endo OH group
The concave shape of the six-membered ring makes the
bottom face of the C=O more sterically hindered. Addition
of the H– occurs from above to place the new C–H bond
exo, making the OH endo.
O
OH
[2] H2O
H
from
exo OH group below
559
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–21
20.52 Since a Grignard reagent contains a carbon atom with a partial negative charge, it acts as a
base and reacts with the OH of the starting halide, BrCH2CH2CH2CH2OH. This acid–base
reaction destroys the Grignard reagent so that addition cannot occur. To get around this
problem, the OH group can be protected as a tert-butyldimethylsilyl ether, from which a
Grignard reagent can be made.
basic site
Br
OH
Mg
acidic site
δ−
OH
BrMg
proton transfer
CH3
O
+ HOMgBr
These will react.
INSTEAD: Use a protecting group.
Br
OH
TBDMS–Cl
imidazole
OTBDMS
Br
Mg
ether
BrMg
OTBDMS
O
[1]
protected OH
[2] H2O
OH
(CH3CH2CH2CH2)4NF
OH
OH
OTBDMS
A
20.53 Compounds F, G, and K are all alcohols with aromatic rings so there will be many
similarities in their proton NMR spectra. These compounds will, however, show differences
in absorptions due to the CH protons on the carbon bearing the OH group. F has a CH2OH
group, which will give a singlet in the 3–4 ppm region of the spectrum. G is a 3o alcohol that
has no protons on the C bonded to the OH group so it will have no peak in the 3–4 ppm
region of the spectrum. K is a 2o alcohol that will give a doublet in the 3–4 ppm region of
the spectrum for the CH proton on the carbon with the OH group.
560
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–22
[1] O3
[2] CH3SCH3
C
O
CHO
O
OH
[1] Mg
H2SO4
[1] C6H5MgBr
HCl
Cl
[2] CH2=O
D
[2] H2O
CH2OH F
[3] H2O
B
A
mCPBA
[1] LiAlH4
[1] (CH3)2CuLi
O
E
[2] H2O
OH
Br
PBr3
G
OH
CH3
[2] H2O
OH
MgBr
[1] C6H5CHO
Mg
[2] H2O
I
H
J
K
20.54
O
Cl
O
R
Si
R
H OH
[1] (R)-CBS
reagent
R
Si
[2] H2O
R
O
H OH
Cl
O
NaI
R
Si
A
O
I
O
R
B
O
(CH3CH2)3SiCl
imidazole
H OH
HO
H
N
KF
H OSi(CH2CH3)3
NHCH(CH3)2
O
R
Si
R
HO
O
O
R
Si
(CH3)2CHNH2
D
H OSi(CH2CH3)3
I
R
C
O
20.55
H OH
O
a.
O
O
CH3 MgBr
O
+
+ MgBr
O
O
O
O
H OH
+
+ MgBr
CH3 MgBr
OH
HOMgBr
+
OH
561
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–23
H OH
H OH
O
O
b.
O
O
O
O
CH2CH2CH2CH2
MgBr
O
BrMg
O
+
+
+ MgBr
BrMg CH2CH2CH2CH2 MgBr
+ MgBr
OH
HOMgBr
OH
+
20.56
O
MgBr
O
+ CH O C OCH
3
3
O
CH3O C OCH3
CH3O
C
+ CH3O
MgBr
MgBr
HO H
O
O
O
C
C
CH3O C
MgBr
MgBr
+ CH3O
MgBr
H2O
OH
C
+ CH3OH + HOMgBr
20.57
O
O
O
O
H
H AlH3
O
O
O
O
H AlH3
H
O
O
H
O
O
O
O
OH
HO
OH
H AlH3
O
H2O
Protonate
four alkoxides.
O
O
O
O
O
O
+ AlH3
O
AlH3
OH
H
O
+ AlH3
+ AlH3
O
O
H
O
O H
O
H AlH3
562
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–24
20.58
MgBr
CH2OH
a.
O
H
C
BrMg
d.
H
HO
O
or
b.
H
MgBr
OH
or
H
O
MgBr
O
or
MgBr
CH3MgBr
O
OH
c.
(C6H5)2C=O
(C6H5)3COH
O
BrMg
O
MgBr
e.
20.59
(a)
(a)
N
N
(c)
(b)
MgBr
O
OH
(b)
BrMg
N
O
(c)
N
MgBr
O
20.60
O
OH
a.
C
CH3O
O
C
MgBr
(2 equiv)
CH3
CH3O
C
CH3O
C
CH3
BrMg
(2 equiv)
O
OH
b. CH3 C CH2CH2CH(CH3)2
c. (CH3CH2CH2CH2)2C(OH)CH3
CH2CH2CH(CH3)2
+ CH3–MgBr
(2 equiv)
563
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–25
20.61
OH
Li
O
OH
+
a.
H
(two ways)
C CH2CH3
O
or
(three ways)
H
+
Li
O
OH
b.
c.
O
C
+
or
Li
+
O
or
(three ways)
O
Li
CH2CH3 +
Li
+
Li CH2CH3
O
or
or
+
Li
CH3CH2
O
C
Li
OCH3
+
(2 equiv)
20.62
OH
a.
+
MgBr
O
HO
H
H
c.
O
C6H5
b.
O
H
+ BrMg C6H5
H
+
OH
MgBr
20.63
C6H5
C6H5
C6H5
Br
Mg
Br
KOC(CH3)3
Br
C6H5
H2O
MgBr
C6H5
H2
C6H5
C6H5
Pd-C
LiAlH4
C6H5
20.64
Br
Mg
O
MgBr
OH
H2O
H
CHO
MgBr
Mg
Br
564
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–26
20.65
OH
Br
PBr3
a.
MgBr
Mg
CH2OH
[1] CH2=O
[2] H2O
MgBr
COOH
[1] CO2
b.
[2] H3O+
(from a.)
OH
c.
O
CrO3
[1] CH3MgBr
H2SO4, H2O
H2SO4
OH
[2] H2O
mCPBA
O
major product
OH
d.
mCPBA
H2SO4
O
OH
[1] C6H5MgBr
[2] H2O
O [1] CH3CH2CH2MgBr
e.
Pd-C
+
(from c.)
O
OH
[1] CH3MgBr
H2SO4
[1] O3
CH3
[2] H2O
O
CHO
[2] (CH3)2S
(from c.)
C6H5
H2
[2] H2O
f.
C6H5
H2SO4
OH
O
PCC
major product
OH
[1]
O
MgBr
OH
O
(from a.)
H
PCC
g.
PCC
[2] H2O
(from a.)
[1]
MgBr
(from a.)
O
OH
H2SO4
h.
[2] H2O
major product
(from c.)
20.66
a.
OH
Br
PBr3
MgBr
[1] O
OH
b.
[2] H2O
(from a.)
Mg
OH
[1] CH3CHO
[2] H2O
MgBr
OH
c.
PCC
O
[1]
[2] H2O
MgBr
OH
565
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–27
MgBr
[1]
O
d.
(from a.)
OH
[1] 2 Li
e.
[(CH3)2CH]2CuLi
Br
[2] CuI (0.5 equiv)
[2] H2O
(from a.)
[1] O
[2] H2O
O
20.67
OH
H
O
H
O
H
H
PCC
H
[1] NaH
H
H
HO
H
H
[2] CH3I
HO
H
CH3O
estradiol
[1] HC CLi
[2] H2O
OH
C CH
H
H
H
CH3O
mestranol
20.68
Br
a.
CH3CH2Cl
Br2
AlCl3
hν
Br2
Br
K+ –OC(CH3)3
OH
[1] BH3
[2] H2O2, –OH
MgBr
Mg
FeBr3
O
OH
[1]
[2] H2O
CH3Cl
Br2
AlCl3
hν
OH
Br Mg
MgBr [1] H2C=O
[2] H2O
566
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–28
O
OH
MgBr [1] CH3CH=O
b.
PCC
[2] H2O
(from a.)
O
O
CH3
Cl
CH3Br
AlCl3
O
Br NaOH
OH PCC
O
OH
Mg
[1] CH3MgBr
H
PCC
[2] H2O
(from a.)
20.69
Br
Br2
a.
MgBr
Mg
[2] H3O+
FeBr3
MgBr
b.
COOH
[1] CO2
CH3OH
PCC
OH
CH2OH
[1] CH2=O
CHO [1] PhMgBr
PCC
(from a.)
[2] H2O
O
PCC
[2] H2O
(from a.)
PBr3
c. CH3CH2CH2OH
Mg
CH3CH2CH2Br
PCC
CH3CH2CH2OH
CH3CH2CH2MgBr
[1] CH3CH2CH2MgBr
CH3CH2CHO
HO
[1] PhMgBr
PCC
[2] H2O
OH
O
(from a.)
[2] H2O
O
d.
PCC
OH
H
OH
O
O
H2O
O
Br2
PCC
Br
FeBr3
MgBr
(from a.)
20.70
PCC
a. HO
[1] CH3CH2MgBr
H
O
OH
b.
PCC
OH
Mg
PBr3
CH3CH2OH
PCC
[2] H2O
CH3CH2Br
O
CH3CH2MgBr
[1] CH3CH2CH2MgBr
OH
[2] H2O
CH3CH2CH2OH
PBr3
Mg
CH3CH2CH2Br
CH3CH2CH2MgBr
O
567
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Introduction to Carbonyl Chemistry 20–29
PCC
OH
c.
[1] CH3CH2CH2CH2MgBr
H
PCC
[2] H2O
O
PBr3
CH3CH2CH2CH2OH
CH3CH2CH2CH2Br
O
CH3CH2CH2CH2MgBr
[1] (CH3)2CHCH2MgBr
H
d.
OH
Mg
PBr3
Mg
[2] H2O
O
OH
Br
MgBr
(from c.)
PBr3
(CH3)2CHCH2OH
[1] CO2
Mg
(CH3)2CHCH2Br
[2] H3O+
(CH3)2CHCH2MgBr
CO2H
e.
HBr
OH
Mg
Br
MgBr
H2O
K+ –OC(CH3)3
PBr3
+
OH
H
(from c.)
Br
O
20.71
a.
O
[1] (CH3)2CuLi
[1] CH3MgBr
[2] H2O
O
[2] H2O
OH
H2SO4
H2, Pd-C
+
b.
PCC
CHO
OH
OH
[1] A
OTs
TsCl
[2] H2O
pyridine
PBr3
CN
–CN
Mg
Br
MgBr
A
O
c.
OCH3
[1] LiAlH4
OH
Br2
FeBr3
[2] H2O
Br
Br
(+ ortho isomer)
HBr
d.
ROOR
[1] Mg
Br
[2] O
[3] H2O
O CH3CH2I
OH NaH
OH
Br
O
568
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–30
HO
[1] NaH
e. HC CH
TBDMS–Cl
CH3CHC CH
[2] CH3CH=O
[3] H2O
imidazole
OTBDMS
NaH
CH3CHC CH
OTBDMS
CH3CHC C
[1] O
[2] H2O
OTBDMS
OH
CH3CHC C
OH
(CH3CH2CH2CH2)4NF
OH
20.72
Hb
Hb
O
CH3
Ha
H
CH3
Hb
cm–1 (C=O)
IR peak: 1716
1H NMR: 2 signals (ppm)
CH3
doublet 1.2 (Hb)
H
CH3
Ha
septet 2.7 (Ha)
Hb
Hd
Hd
H
NaBH4
OHa
CH3
CH3
CH3OH
H
Hc
H
CH3 CH3
Hd
Hd
Hb
Hc
C7H16O
C7H14O
IR peak: 3600–3200 cm–1 (OH)
1H NMR: 4 signals (ppm)
doublet 0.9 (Hd)
singlet 1.5 (Ha)
multiplet 1.7 (Hc)
triplet 3.0 (Hb)
B
A
20.73
Hb
H
Hb
H
O
CH3
CH3
C4H8O
Ha
Ha
Hc
Hc
H
H
[2] H2O
C
1H NMR: 2 signals (ppm)
singlet (6 H) 1.3 (Ha)
singlet (2 H) 2.4 (Hb)
IR peak 3600–3200 cm–1 (OH)
NMR: 4 signals (ppm)
singlet (6 H) 1.2 (Ha)
singlet (1 H) 1.6 (Hb)
singlet (2 H) 2.7 (Hc)
multiplet (5 H) 7.2 (benzene ring)
OHb
[1] C6H5MgBr
1H
CH3 CH3
Ha Ha
C10H14O
D
20.74
O
CH3
C
Hc
OCH2CH3
Ha
Hb
C4H8O2
E
IR peak 1743 cm–1 (C=O)
1H NMR: 2 signals (ppm)
triplet (3 H) 1.2 (Hc)
singlet (3 H) 2.0 (Ha)
quartet (2 H) 4.1 (Hb)
[1] CH3CH2MgBr
(excess)
[2] H2O
Hb OH Hb
d
CH3CH2 C CH2CH3
Ha
CH3
Hc
C6H14O
F
Ha
IR peak 3600–3200 cm–1 (OH)
1H NMR: 4 signals (ppm)
triplet (6 H) 0.9 (Ha)
singlet (3 H) 1.1 (Hc)
quartet (4 H) 1.5 (Hb)
singlet (1 H) 1.55 (Hd)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
569
Introduction to Carbonyl Chemistry 20–31
20.75 Molecular ion at m/z = 86: C5H10O (possible molecular formula).
Determine the number of integration units per H:
Total number of integration units: 25 + 17 + 24 + 17 = 83
83 units/10 H's = 8.3 units per H
Divide each integration value by 8.3 to determine the number of H's per signal:
25 units/ 8.3 = 3 H's
24 units/ 8.3 = 3 H's
17 units/ 8.3 = 2 H's
H
Hb
O
[1] CH3MgBr
CH3CH2CH2C N
b
H H
CH3
+
[2] H3O
CH3
H H
Hc
Ha
Hd Hd
G
IR peak 1721 cm–1 (C=O)
1H NMR: 4 signals (ppm)
triplet (3 H) 0.9 (Ha)
sextet (2 H) 1.6 (Hb)
singlet (3 H) 2.1 (Hc)
triplet (2 H) 2.4 (Hd)
20.76 Molecular ion at m/z = 86: C5H10O (possible molecular formula).
Hb Hd Hd
CH3 H H
[1] (CH3)3CLi
H
[2] CH2=O
[3] H2O
OH
He H
Hf
H H
Ha
Hc Hc
H
IR peaks: 3600–3200 cm–1 (OH)
1651 cm–1 (C=C)
1H NMR: 6 signals (ppm)
singlet (1 H) 1.7 (Ha)
singlet (3 H) 1.8 (Hb)
triplet (2 H) 2.2 (Hc)
triplet (2 H) 3.8 (Hd)
two signals at 4.8 and 4.9
due to 2 H's: He and Hf
20.77
Hb
O
He
[1] CH3MgBr
[2] H2O
O
IR peak 3600–3200 cm–1 (OH)
1H NMR: 5 signals (ppm)
triplet (3 H) 0.94 (Ha)
multiplet (2 H) 1.39 (Hb)
multiplet (2 H) 1.53 (Hc)
singlet (1 H) 2.24 (Hd)
triplet (2 H) 3.63 (He)
OH
Ha
Hc
Hd
CH3 MgBr
O
H OH
OH
HOMgBr
570
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–32
20.78
O
OH
[1] (R)-CBS
reagent
Br
[2] H2O
HO
OTBDMS
Br
imidazole
(2 equiv)
HO
COOCH3
COOCH3
[1] NaH
HO
Br
TBDMSCl
C6H5
COOCH3
Br
[2] Br
A
TBDMS O
C6H5
O
Br
NH3
H2N
C6H5
O
B
A
+
OTBDMS
H
N
B
C6H5
O
TBDMS O
[1] LiAlH4
COOCH3
[2] H2O
OTBDMS
H
N
C6H5
O
TBDMS O
(CH3CH2CH2CH2)4NF
OH
(R)-salmeterol
20.79
larger group
equatorial
axial
OH
[1] L-selectride
(CH3)3C
O
[2] H2O
(CH3)3C
H
= (CH3)3C
OH
equatorial
cis-4-tert-butylcyclohexanol
major product
4-tert-butylcyclohexanone
L-Selectride adds H– to a C=O group. There are two possible reduction products—cis and trans isomers—
but the cis isomer is favored. The key element is that the three sec-butyl groups make L-selectride a large, bulky
reducing agent that attacks the carbonyl group from the less hindered direction.
O–
O
(CH3)3C
(CH3)3C
R3B H
OH axial
H
equatorial
H2O
(CH3)3C
H equatorial
cis product
When H– adds from the equatorial direction, the product has an axial OH and a new equatorial
H. Since the equatorial direction is less hindered, this mode of attack is favored with large
bulky reducing agents like L-selectride. In this case, the product is cis.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
571
Introduction to Carbonyl Chemistry 20–33
R3B H
Axial H's hinder
H
axial attack.
H
H
O
(CH3)3C
(CH3)3C
H axial
axial
O–
H2O
OH equatorial
(CH3)3C
trans product
The axial H's hinder H– attack from the axial direction. As a result, this mode of attack is more difficult with
larger reducing agents. In this case the product is trans. This product is not formed to any appreciable extent.
20.80 The β carbon of an α,β-unsaturated carbonyl compound absorbs farther downfield in the 13C
NMR spectrum than the α carbon, because the β carbon is deshielded and bears a partial
positive charge as a result of resonance. Since three resonance structures can be drawn for an
α,β-unsaturated carbonyl compound, one of which places a positive charge on the β carbon,
the decrease of electron density at this carbon deshields it, shifting the 13C absorption
downfield. This is not the case for the α carbon.
122.5 ppm
O
150.5 ppm
O
O
Oδ−
hybrid:
β α
mesityl oxide
δ+
δ+
20.81
CH2CN
OH
CN
OH
[1] Li+ –N[CH(CH3)2]2
[2]
N(CH3)2
CH2NH2
OH
[1] LiAlH4
O
OCH3
[2] H2O
OCH3
OCH3
OCH3
W
H
CN
H C
[3] H2O
–
N[CH(CH3)2]2
X
C15H19NO2
H C CN
O
H OH
O
OCH3
venlafaxine
Y
OCH3
HN[CH(CH3)2]2
CN
OH
CN
OCH3
OCH3
OH
572
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 20–34
20.82
H
MgCl
O
O
H
OH
H OH
P
MgCl
OH
MgCl
MgCl
H OH
H
O
H
H
HB
O
O
H OH
B
H2O
Any base (such as the
alkoxide) can deprotonate
the intermediate.
CH3
OH
N
OH
20.83
O
NH2OH
O
O
H2NOH
H2NOH
proton
transfer
H
O
OH
H
N
OH
OH
N
N
=
proton
transfer
H NOH
O
O
O
(+ 1 resonance structure)
20.84
O
N
H A
O
CH2CH3
CH3CH2 MgBr
N
CH3
CH3
(any proton
source)
OH
CH2CH3
N
+ A
CH3
+ MgBr+
CH2CH3
MgBr+
CH2CH3
+
N
CH3
CH2CH3
CH3CH2 MgBr
N
+
CH3
OH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
573
Aldehydes and Ketones—Nucleophilic Addition 21–1
Chapter 21 Aldehydes and Ketones—Nucleophilic Addition
Chapter Review
General facts
• Aldehydes and ketones contain a carbonyl group bonded to only H atoms or R groups. The
carbonyl carbon is sp2 hybridized and trigonal planar (21.1).
• Aldehydes are identified by the suffix -al, while ketones are identified by the suffix -one (21.2).
• Aldehydes and ketones are polar compounds that exhibit dipole–dipole interactions (21.3).
Summary of spectroscopic absorptions of RCHO and R2CO (21.4)
C=O
~1715 cm–1 for ketones
IR absorptions
• increasing frequency with decreasing ring size
~1730 cm–1 for aldehydes
• For both RCHO and R2CO, the frequency
decreases with conjugation.
Csp2–H
of CHO
CHO
C–H α
to C=O
C=O
1
H NMR
absorptions
13
C NMR
absorption
~2700–2830 cm–1 (one or two peaks)
9–10 ppm (highly deshielded proton)
2–2.5 ppm (somewhat deshielded Csp3–H)
190–215 ppm
Nucleophilic addition reactions
[1] Addition of hydride (H–) (21.8)
O
R
C
OH
NaBH4, CH3OH
R C H(R')
or
H(R')
H
[1] LiAlH4 [2] H2O
•
•
The mechanism has two steps.
H:– adds to the planar C=O from
both sides.
•
•
The mechanism has two steps.
R:– adds to the planar C=O from
both sides.
1o or 2o alcohol
[2] Addition of organometallic reagents (R–) (21.8)
O
R
C
OH
[1] R"MgX or R"Li
R C H(R')
[2] H2O
H(R')
R"
1o, 2o, or 3o alcohol
[3] Addition of cyanide (–CN) (21.9)
O
R
C
NaCN
H(R')
HCl
OH
R C H(R')
CN
cyanohydrin
•
•
The mechanism has two steps.
–
CN adds to the planar C=O from both
sides.
574
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–2
[4] Wittig reaction (21.10)
R
R
C O
+ Ph3P
C C
•
alkene
•
C
(R')H
(R')H
Wittig reagent
The reaction forms a new C–C σ bond and a new
C–C π bond.
Ph3P=O is formed as by-product.
[5] Addition of 1o amines (21.11)
R
R
R"NH2
C O
C N
mild acid
(R')H
(R')H
R"
•
•
The reaction is fastest at pH 4–5.
The intermediate carbinolamine is unstable, and
loses H2O to form the C=N.
•
•
The reaction is fastest at pH 4–5.
The intermediate carbinolamine is unstable, and
loses H2O to form the C=C.
•
The reaction is reversible. Equilibrium favors the
product only with less stable carbonyl
compounds (e.g., H2CO and Cl3CCHO).
The reaction is catalyzed with either H+ or –OH.
imine
[6] Addition of 2o amines (21.12)
O
NR2
R2NH
H
(R')H
(R')H
mild acid
enamine
[7] Addition of H2O—Hydration (21.13)
H2O
H+ or –OH
O
R
C
H(R')
OH
R C H(R')
OH
•
gem-diol
[8] Addition of alcohols (21.14)
O
R
C
H+
H(R')
OR"
R C H(R')
+ R"OH
(2 equiv)
+
OR''
acetal
H2O
•
•
•
The reaction is reversible.
The reaction is catalyzed with acid.
Removal of H2O drives the equilibrium to
favor the products.
Other reactions
[1] Synthesis of Wittig reagents (21.10A)
RCH2X
[1] Ph3P
[2] Bu
Li
Ph3P CHR
•
•
Step [1] is best with CH3X and RCH2X since the reaction
follows an SN2 mechanism.
A strong base is needed for proton removal in Step [2].
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
575
Aldehydes and Ketones—Nucleophilic Addition 21–3
[2] Conversion of cyanohydrins to aldehydes and ketones (21.9)
OH
O
–OH
R C H(R')
CN
C
H(R')
+ H2 O
aldehyde or
ketone
+ –CN
R
•
This reaction is the reverse of cyanohydrin
formation.
[3] Hydrolysis of nitriles (21.9)
OH
R C H(R')
CN
OH
H2 O
H+ or –OH
Δ
R C H(R')
COOH
α-hydroxy
carboxylic acid
[4] Hydrolysis of imines and enamines (21.12)
NR
(R')H
O
NR2
H
or
H2O, H+
(R')H
imine
(R')H
H
+ RNH2 or R2NH
aldehyde or
ketone
enamine
[5] Hydrolysis of acetals (21.14)
OR"
R C H(R') + H2O
OR"
•
O
H+
R
C
H(R')
aldehyde or
ketone
+ R"OH
(2 equiv)
•
The reaction is acid catalyzed and is
the reverse of acetal synthesis.
A large excess of H2O drives the
equilibrium to favor the products.
576
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–4
Practice Test on Chapter Review
1. Give the IUPAC name for the following compounds.
O
O
a.
b.
H
c.
O
2. (a) Considering compounds A–D, which compound forms the smallest amount of hydrate?
(b) Which compound forms the largest amount of hydrate?
CH3O
CHO
O2N
Cl
CHO
CH3O
O
O
A
B
C
D
3. (a) Considering compounds A–D, which compound absorbs at the lowest wavenumber in its IR
spectrum? (b) Which compound absorbs at the highest wavenumber in its IR spectrum?
O
O
O
A
B
O
C
D
4. Fill in the lettered reagents (A–G) in the following reaction scheme.
HO
CHO
B
O
HO
COOH
C
HO C CH
A
O
O
HO
D
TBDMSO
E
TBDMSO
OH
F
G
HO
OH
577
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–5
5. Draw the organic products formed in the following reactions.
a. CH3CH2Br
NH
[1] Ph3P
d.
[2] BuLi
[3]
b.
mild acid
O
O
NH2
CHO
+
O
e.
HO
OH
TsOH
mild acid
O
c.
O
[1] NaCN, HCl
f.
HO
OH
CH3CH2OH
H+
[2] H2O, H+, Δ
[Indicate stereochemistry.]
Answers to Practice Test
1.a. 54.
isopropyl-2,4- A = [1] LiC≡CH; [2] H2O
dimethylB = [1] R2BH; [2] H2O2, –OH
cyclohexanone C = Ag2O, NH4OH
b. 3,3D = H2O, H2SO4, HgSO4
dimethyl-5E = TBDMS–Cl, pyridine
phenyl-2F = [1] CH3Li; [2] H2O
pentanone
G = (CH3CH2CH2CH2)4N+F–
c. 4-ethyl-2methylcyclohexanecarbaldehyde
2. a. D
b. C
3. a. B
b. C
5.
O
a.
e.
O
b.
N
O
f.
O
+
OH
c.
O
CO2H
d.
N
O
(E + Z)
Answers to Problems
21.1
As the number of R groups bonded to the carbonyl C increases, reactivity towards
nucleophilic attack decreases. Steric hindrance decreases reactivity as well.
O
O
Increasing reactivity
decreasing steric hindrance
OH
O
OH
578
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–6
21.2
More stable aldehydes are less reactive towards nucleophilic attack.
CHO
CHO
benzaldehyde
Several resonance structures delocalize the partial
positive charge on the carbonyl carbon, making
it more stable and less reactive towards
nucleophilic attack.
21.3
cyclohexanecarbaldehyde
This aldehyde has no added
resonance stabilization.
O
O
O
C
C
C
H
H
O
O
C
C
H
O
C
H
O
C
H
(CH3)3CC(CH3)2CH2CHO
3 2 1 H
5
O
5 C chain = pentanal
O
Cl
Cl
CHO
Cl
3 4
Cl
3,3-dichlorocyclobutane4 C ring =
carbaldehyde
cyclobutanecarbaldehyde
3,3,4,4-tetramethylpentanal
8
1
7
CHO
b.
6
8 C chain = octanal
2 1
c.
4
H
21.4
H
• To name an aldehyde with a chain of atoms: [1] Find the longest chain with the CHO group
and change the -e ending to -al. [2] Number the carbon chain to put the CHO at C1, but
omit this number from the name. Apply all other nomenclature rules.
• To name an aldehyde with the CHO bonded to a ring: [1] Name the ring and add the
suffix-carbaldehyde. [2] Number the ring to put the CHO group at C1, but omit this
number from the name. Apply all other nomenclature rules.
CHO
a.
H
CHO
5
4
3
2
2,5,6-trimethyloctanal
Work backwards from the name to the structure, referring to the nomenclature rules in
Answer 21.3.
a. 2-isobutyl-3-isopropylhexanal
6 C chain
O
b. trans-3-methylcyclopentanecarbaldehyde
5 carbon ring
CHO
CHO
or
H
CH3
CH3
579
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–7
d. 3,6-diethylnonanal
c. 1-methylcyclopropanecarbaldehyde
9 C chain
CHO
3 carbon ring
H
CH3
O
21.5 • To name an acyclic ketone: [1] Find the longest chain with the carbonyl group and change
the -e ending to -one. [2] Number the carbon chain to give the carbonyl C the lower
number. Apply all other nomenclature rules.
• To name a cyclic ketone: [1] Name the ring and change the -e ending to -one. [2] Number
the C’s to put the carbonyl C at C1 and give the next substituent the lower number. Apply
all other nomenclature rules.
a.
1
3
O
4
c.
5
O
8 C chain =
octanone
1
8
O
5-ethyl-4-methyl-3-octanone
CH3
(CH3)3C
(CH3)3CCOC(CH3)3
b.
3
O
4
5
2,2,4,4-tetramethyl-3-pentanone
2
1
5 C ring =
cyclopentanone
3
O
5 C chain =
pentanone
CH3
(CH3)3C
2
O
3-tert-butyl-2-methylcyclopentanone
Most common names are formed by naming both alkyl groups on the carbonyl C, arranging
them alphabetically, and adding the word ketone.
21.6
a. sec-butyl ethyl ketone
d. 3-benzoyl-2-benzylcyclopentanone
benzyl group:
benzoyl group:
e. 6,6-dimethyl-2-cyclohexenone
6 C ketone
5 C ketone
O
O
O
CH2
b. methyl vinyl ketone
C
f. 3-ethyl-5-hexenal
O
O
CHO
c. p-ethylacetophenone
O
CH2
O
21.7
Compounds with both a C–C double bond and an aldehyde are named as enals.
a. (2Z)-3,7-dimethyl-2,6-octadienal
3
7
6
2
1 CHO
neral
b. (2E,6Z)-2,6-nonadienal
3
7
6
1
CHO
2
cucumber aldehyde
580
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–8
Even though both compounds have polar C–O bonds, the electron pairs around the sp3 hybridized
O atom of diethyl ether are more crowded and less able to interact with electron-deficient sites in
other diethyl ether molecules. The O atom of the carbonyl group of 2-butanone extends out from
the carbon chain, making it less crowded. The lone pairs of electrons on the O atom can more
readily interact with the electron-deficient sites in the other molecules, resulting in stronger forces
and a higher boiling point.
21.8
O
O
diethyl ether
2-butanone
For cyclic ketones, the carbonyl absorption shifts to higher wavenumber as the size of the
ring decreases and the ring strain increases. Conjugation of the carbonyl group with a
C=C or a benzene ring shifts the absorption to lower wavenumber.
21.9
CHO
a.
CHO
or
conjugated C=O
lower wavenumber
b.
or
O
higher wavenumber
O
smaller ring
higher wavenumber
21.10 The number of lines in their 13C NMR spectra can distinguish the constitutional isomers.
O
O
2-pentanone
5 lines
O
3-pentanone
3 lines
3-methyl-2-butanone
4 lines
21.11
[1] DIBAL-H
a. CH3CH2CH2COOCH3
PCC
b. CH3CH2CH2CH2OH
CH3CH2CH2CHO
[2] H2O
CH3CH2CH2CHO
[1] R2BH
c.
HC CCH2CH3
[2] H2O2, HO–
d. CH3CH2CH2CH CHCH2CH2CH3
CH3CH2CH2CHO
[1] O3
[2] Zn, H2O
CH3CH2CH2CHO
21.12
O
Cl
a.
O
CH3
c.
AlCl3
Cl
[1] (CH3)2CuLi
[2] H2O
C CH
H2O
H2SO4
HgSO4
O
O
b.
O
581
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–9
21.13
CH3O
OCH3
CH3O OCH3
[1] O3
OHC
CHO
[2] (CH3)2S
Cleave this C=C with O3.
21.14 Addition of hydride or R–M occurs at a planar carbonyl C, so two different configurations at
a new stereogenic center are possible.
O
a.
Add
stereochemistry:
H OH
NaBH4
CH3OH
new stereogenic
center
O
Add
CH=CH2
CH=CH2
OH
OH
[2] H2O
(CH3)3C
H OH
OH
CH=CH2 stereochemistry:
[1] CH2 CHMgBr
b.
H OH
(CH3)3C
(CH3)3C
(CH3)3C
21.15 Treatment of an aldehyde or ketone with NaCN, HCl adds HCN across the double bond.
Cyano groups are hydrolyzed by H3O+ to replace the three C–N bonds with three C–O bonds.
OH
CHO
CHCN
NaCN
a.
OH
b.
HCl
OH
H3 O+, Δ
CN
COOH
21.16
HO
HO
HO
O
O
OH
HO
HO
O
O
OH
CN
HO
enzyme
C
O
CN
enzyme
H
C
21.17
CH3
CH3
C O
+
CH3
b.
Ph3P=CH2
C CH2
CH3
O +
+
HCN
toxic
by-product
amygdalin
a.
H
Ph3P=CHCH2CH2CH2CH3
CHCH2CH2CH2CH3
582
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–10
21.18
BuLi
a. Ph3P + Br−CH2CH3
Ph3P−CH2CH3
Br–
b. Ph3P + Br−CH(CH3)2
Ph3P−CH(CH3)2
Ph3P=CHCH3
BuLi
Ph3P=C(CH3)2
Br–
c. Ph3P + Br−CH2C6H5
Ph3P−CH2C6H5
BuLi
Ph3P=CHC6H5
Br–
21.19
CHO
a.
CH2CH3
Ph3P CHCH2CH3
+
CH2CH3
CHO
Ph3P CHC6H5
+
b.
CHO
+
c.
COOCH3
Ph3P CHCOOCH3
COOCH3
21.20
Ph3P=CHCO2CH2CH3
N
N
CHO
A
CO2CH2CH3
(+ cis isomer)
B
H2, Pd–C
N
CO2CH2CH3
C
21.21 To draw the starting materials of the Wittig reactions, find the C=C and cleave it. Replace it
with a C=O in one half of the molecule and a C=PPh3 in the other half. The preferred
pathway uses a Wittig reagent derived from a less hindered alkyl halide.
a.
CH3
CH2CH3
C C
CH3
H
CH3
C PPh3
CH3
CH2CH3
+
O C
or
H
2° halide precursor
(CH3)2CHX
b.
CH2CH3
CH3CH2
H
CH2CH3
C
H
(cis or trans)
H
PPh3
CH2CH3
C
H
1° halide precursor
XCH2CH2CH3
preferred pathway
CH3CH2
C C
CH3
C O + Ph3P
CH3
+
(only one route possible)
O C
H
583
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–11
C6H5
c.
CH3
C
C6H5
C
H
H
C6H5
CH3
C
PPh3
+
O
or
C
H
H
(cis or trans)
CH3
C O
+
Ph3P
H
C
H
(both routes possible)
1° halide precursor
C6H5CH2X
1° halide precursor
XCH2CH3
21.22
a. Two-step sequence:
O
[1] CH3MgBr
HO CH3
H2SO4
[2] H2O
One-step sequence:
O
minor product
Ph3P=CH2
tetrasubstituted
major product
(E and Z
isomers)
only product
b. Two-step sequence:
O
[1] C6H5CH2MgBr
OH
[2] H2O
CH2C6H5
H2SO4
CHC6H5
trisubstituted
conjugated C=C
One-step sequence:
O
Ph3P=CHC6H5
CH2C6H5
trisubstituted
CHC6H5 only product
21.23 When a 1o amine reacts with an aldehyde or ketone, the C=O is replaced by C=NR.
a.
CHO
O
CH3CH2CH2CH2NH2
NCH2CH2CH2CH3
NCH2CH2CH2CH3
CH3CH2CH2CH2NH2
b.
CH
c.
O
CH3CH2CH2CH2NH2
NCH2CH2CH2CH3
21.24 Remember that the C=NR is formed from a C=O and an NH2 group of a 1° amine.
CH3
a.
CH3
C
C NCH2CH2CH3
H
b.
CH3
O + NH2CH2CH2CH3
H
N
CH3
O
NH2
21.25
O
N(CH3)2
N(CH3)2
CH3
N
CH3
H
+
584
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–12
21.26 • Imines are hydrolyzed to 1° amines and a carbonyl compound.
• Enamines are hydrolyzed to 2° amines and a carbonyl compound.
O
H2O
CH N
a.
+
C
H+
H2N
H
imine
1° amine
H2O
b.
CH2 N
(CH3)2NCH
O
CH3
enamine
c.
+
CH2 N H
H+
CH3
2° amine
O
H2O
C(CH3)2
H+
enamine
+
(CH3)2NH
C CH(CH3)2
H
2° amine
21.27 • A substituent that donates electron density to the carbonyl C stabilizes it, decreasing the
percentage of hydrate at equilibrium.
• A substituent that withdraws electron density from the carbonyl C destabilizes it,
increasing the percentage of hydrate at equilibrium.
a. CH3CH2CH2CHO
or
b.
CH3CH2COCH3
one R group on C=O
higher percentage of hydrate
2 R groups
on C=O
CH3CF2CHO
or
CH3CH2CHO
F atoms are electron withdrawing.
higher percentage of hydrate
21.28
H
O H
H OH2
O H
O H
O H
O H
+
O
+ H2O
+ H3O
(+ 1 resonance
structure)
H2O
21.29 Treatment of an aldehyde or ketone with two equivalents of alcohol results in the formation
of an acetal (a C bonded to two OR groups).
O
a.
O
+
2 CH3OH
TsOH
OCH3
+ HO
b.
OH
TsOH
O
O
OCH3
21.30
OCH3
OCH3
a.
b.
OCH3
O
OCH3
2 OR groups
on different C's
2 ethers
O
c.
2 OR groups
on same C
acetal
CH3
CH3
2 OR groups
on same C
acetal
d.
OCH3
OH
1 OR group and
1 OH group
on same C
hemiacetal
585
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–13
21.31 The mechanism has two parts: [1] nucleophilic addition of ROH to form a hemiacetal;
[2] conversion of the hemiacetal to an acetal.
O
O
O
TsO−H
+ HOCH CH OH
2
2
+ H2O
overall reaction
TsO
H
O H
O H
HO O CH2CH2OH
HO O CH2CH2OH
+ TsO−H
+ TsO
hemiacetal
HOCH2CH2OH
H
HO O CH2CH2OH
H O
O CH2CH2OH
O CH2CH2OH
O CH2CH2OH
TsO−H
+ H2O
+ TsO–
O CH2CH2OH
O
TsO
O H
O
O
+ TsO−H
carbocation
re-drawn
acetal
21.32
CH3O OCH3
a.
+
O
H2SO4
H2O
CH3
C
O
CH3
+
b.
+ 2 CH3OH
H 2O
CH3
O
OH
O
H2SO4
C
+
CH3
OH
21.33
HO
O
O
H2 O
O
H2SO4
safrole
+
H
H
HO
formaldehyde
21.34 Use an acetal protecting group to carry out the reaction.
O
O
O
[1] CH3Li (2 equiv)
HOCH2CH2OH
O
O
H3
O+
[2] H2O
TsOH
COOCH2CH3
O
COOCH2CH3
OH
OH
586
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–14
21.35
OH
O
C4
C1
OH
O
a. HO
O
H
b.
C1
O
H
C4
OH
C5
C1
C5
C1
21.36
acetal
O
HO
H
O
O
H
OH
O
O
H
OCH3
O
acetal
H
HO
O
H
O
HO
COOH
HO
acetal
O
H
O
HO
O
H
O
OH
H
OH
acetal
OH
monensin
O
digoxin
hemiacetal
Ether O atoms are indicated in bold.
21.37 The hemiacetal OH is replaced by an OR group to form an acetal.
OH
OCH2CH3
O
a.
H+
+ CH3CH2OH
O
OH
HO
O
b.
OCH2CH3
+ CH3CH2OH
H+
HO
HO
O
HO
21.38
HO OH
a.
HO OH
O
*
*
*
HO
*
OH *
OH
5 stereogenic centers
(labeled with *)
HO
HO OH
b.
OH
d.
HO OH
O
HO
hemiacetal C
OH
OH
HO OH
O
OH
HO
OH
β-D-galactose
e.
HO OH
O
OCH3
HO
OH
α-D-galactose
c.
CHO
OH
+
O
HO
OH
OCH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
587
Aldehydes and Ketones—Nucleophilic Addition 21–15
21.39
2
O
a.
3
4
O 1
1
cis-5-isopropyl-2methylcyclohexanone
B
3,3-dimethylbutanal
A
b.
[1] A
NaBH4
OH
CH3OH
[2] A
[3] A
[4] A
[5] A
[1] B
CH3MgBr
OH
Ph3P=CHOCH3
+
OH
OCH3 (+ cis isomer)
CH3CH2CH2NH2
NCH2CH3
mild H+
O
HOCH2CH2CH2OH
H+
O
NaBH4
HO
HO
H2O
CH3MgBr
HO
[3] B
H
H
H2O
CH3OH
[2] B
4
5
CH
2
3
HO
Ph3P=CHOCH3
CH3O
OCH3
[4] B
[5] B
CH3CH2CH2NH2
mild H+
HOCH2CH2CH2OH
H+
CH3CH2CH2N
O
O
21.40 The least hindered carbonyl group is the most reactive.
O
O
H
O
least reactive
intermediate
reactivity
most reactive
588
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–16
21.41
O
CHO
HO
O
a.
O
OH
O
OH
b.
O
OH
21.42 Use the rules from Answers 21.3 and 21.5 to name the aldehydes and ketones.
O
a. Ph
O
8 C = octanone
8-phenyl-3-octanone
e.
CH3
o-nitroacetophenone
NO2
b.
CHO
5 C ring
2-methylcyclopentanecarbaldehyde
f.
6 C ring = cyclohexanone
5-ethyl-2-methylcyclohexanone
g.
CHO
6 C = hexanal
3,4-diethylhexanal
O
c.
O
E
8 C = octenone
(5E)-2,5-dimethyl-5-octen-4-one
CHO
d.
trans-2-benzylcyclohexanecarbaldehyde
CH2Ph
h.
CHO
6 C = hexenal
3,4-diethyl-2-methyl-3-hexenal
21.43
f. 2-formylcyclopentanone
a. 2-methyl-3-phenylbutanal
O
O
CHO
H
O
O
b. dipropyl ketone
CHO
g. (3R)-3-methyl-2-heptanone
c. 3,3-dimethylcyclohexanecarbaldehyde
h. m-acetylbenzaldehyde
CHO
O
d. α-methoxypropionaldehyde
O
H
i. 2-sec-butyl-3-cyclopentenone
O
OCH3
O
j. 5,6-dimethyl-1-cyclohexenecarbaldehyde
e. 3-benzoylcyclopentanone
CHO
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
589
Aldehydes and Ketones—Nucleophilic Addition 21–17
21.44
O
O
H
O
H
H
2-ethylbutanal
hexanal
O
H
2,2-dimethylbutanal
O
O
H
(2S)-2,3-dimethylbutanal
O
3,3-dimethylbutanal
O
H
(2R)-2,3-dimethylbutanal
H
4-methylpentanal
O
H
O
(2R)-2-methylpentanal
H
(2S)-2-methylpentanal
O
H
H
(3R)-3-methylpentanal
(3S)-3-methylpentanal
21.45
H
= C6H5CH2CHO
phenylacetaldehyde
O
a. NaBH4, CH3OH
g.
C6H5CH2CH2OH
b. [1] LiAlH4; [2] H2O
C6H5CH2CH2OH
h.
C6H5CH2CH(OH)CH3
NaCN, HCl
OCH2CH3
H
NH
C6H5CH2CH=CHCH3
i.
(E and Z isomers)
f.
OCH2CH3
CH3CH2OH (excess), H+
C6H5CH2CH(OH)CN
e. Ph P CHCH
3
3
(E and Z isomers)
N(CH2CH3)2
c. [1] CH3MgBr; [2] H2O
d.
H
(CH3CH2)2NH, mild acid
mild acid
H
(CH3)2CHNH2
mild acid
(E and Z isomers)
N
NCH(CH3)2
j.
CH3CH=C(CH3)2
c.
O
OH
HO
O
H+
21.46
a. CH3CH2Cl
[1] Ph3P
[1] Ph3P
CH2Cl
CH=CHCH2CH2CH3
[2] BuLi
[3] CH3CH2CH2CHO
[2] BuLi
[3] (CH3)2C=O
(E and Z isomers)
[1] Ph3P
b.
CH2Br
[2] BuLi
[3] C6H5CH2CH2CHO
CH=CHCH2CH2C6H5
(E and Z isomers)
21.47
a.
Ph3P CHCH2CH2CH3
b.
Ph3P C(CH2CH2CH3)2
BrCH2CH2CH2CH3
BrCH(CH2CH2CH3)2
c. Ph3P CHCH=CH2
BrCH2CH=CH2
590
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–18
21.48
Ph3P
CHO
a.
H
b. CH3CH2CHO +
c.
mild
H2N
O
CH3CH2CH N
acid
HOCH2CH2OH
O
H+
O
H3O+
d.
H2N
N
O
O
e.
mild
+
C6H5
N
H
N
acid
(E and Z isomers)
C6H5
HO
H3O+, Δ
CN
C
f.
C6H5
HO
COOH
C6H5
C6H5
C6H5
CH3CH2OH
OH
g.
i.
OCH3
CH3O
O
H3O+
N
h.
OCH2CH3
H+
O
O
H3O+
OCH3
O
j.
+
O + HOCH3
CH3O
Ph3P CH(CH2)5COOCH3
HN
CH(CH2)5COOCH3
21.49
CH3CH2O
O
OCH2CH3
+ HOCH2CH3
a.
CH3O
O
OCH3
OCH3
OCH3
c.
b.
HO
O
OCH2CH3
OCH3
OCH3
+ HOCH3
H
O
+ HOCH2CH3
591
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–19
21.50 Consider para product only, when an ortho, para mixture can result.
Br
Br2
O
CH3COCl
FeBr3
O
HOCH2CH2OH
Br
AlCl3
O
Br
B
A
O
Mg
O
H+
MgBr
D
C
[1] CH3CHO
[2] H2O
O
O
O
H2 O
O
PCC
O
H+
G
O
OH
O
F
E
21.51
Ph3P=CHCH2CH2CH3
a. CH3CH2CH2CHO
CH3CH2CH2
CH2CH2CH3
H
O
HO CN
NaCN
b.
+
+
C C
H
CH3CH2CH2
H
C C
H
CH2CH2CH3
NC OH
HCl
NaBH4
O
c.
CH3CH2
d. HO
+
OH
OH
CH3OH
CH3CH2
CH3OH
O
HO
CH3CH2
+
O
O
HO
HCl
OCH3
OH
OCH3
21.52
new stereogenic center
O
OH
O
OH
O
OH
CHO
HO
A
achiral
S
new stereogenic center
H
CHO
S
HO H B
chiral
An equal mixture of enantiomers
results, so the product is optically
inactive.
H
O
H
O
OH
O
OH
A mixture of diastereomers results.
Both compounds are chiral and
OH they are not enantiomers, so the
mixture is optically active.
592
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–20
21.53
OH
H
HO
a.
acetal O
acetal
O
O
O
H
H
O
acetal
etoposide
O
O
H
O
CH3O
OCH3
OH
b. Lines of cleavage are drawn in.
OH
H
HO
O
H
O
O
O
O
O
CH3O
HO
H+
O
H
OH
H
H
O
H2O
O
+
HO
OCH3
H
CH3O
O
HO
OH
H
O CH
HO H
OH
OH
+ CH3CHO
+ CH2=O
OCH3
OH
OH
21.54 Use the rules from Answer 21.1.
O
O
O
Increasing reactivity
decreasing steric hindrance
21.55 The less stable carbonyl compound forms the higher percentage of hydrate.
O
H
δ+
O
H
O
The aldehyde is destabilized since there is a δ+ on its adjacent carbon, which is part of a C=O.
Thus, PhCOCHO has the higher concentration of hydrate.
21.56 Electron-donating groups decrease the amount of hydrate at equilibrium by stabilizing the
carbonyl starting material. Electron-withdrawing groups increase the amount of hydrate at
equilibrium by destabilizing the carbonyl starting material. Electron-donating groups make
the IR absorption of the C=O shift to lower wavenumber because they stabilize the chargeseparated resonance form, giving the C=O more single bond character.
593
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–21
O2N
COCH3
CH3O
COCH3
a.
p-nitroacetophenone
NO2 withdrawing group
less stable
b.
higher percentage of hydrate
p-methoxyacetophenone
CH3O donating group
more stable
lower percentage of hydrate
c.
higher wavenumber
lower wavenumber
21.57 Use the principles from Answer 21.21.
CH3CH2
a.
CH3CH2
CH2CH2CH3
C C
CH3CH2
H
CH3CH2
CH2CH2CH3
C O
Ph3P C
CH3CH2
or
H
CH2CH3
CH3
C C
b.
H
H
CH3
or
Ph3P C
H
O C
H
H
(Both routes possible.)
1° alkyl halide precursor
(XCH2CH3)
O
PPh3
c.
Ph3P CH2
methyl halide precursor
(CH3X)
preferred pathway
C6H5
C6H5
O
CH2CH3
C PPh3
H
1° alkyl halide precursor
(XCH2CH2CH3)
d.
H
2° alkyl halide precursor
[(CH3CH2)2CHX]
CH2CH3
C O
O C
CH3CH2
1° alkyl halide precursor
(XCH2CH2CH2CH3)
preferred pathway
CH3
CH2CH2CH3
C PPh3
or
Ph3P
or
CH2 O
2° alkyl halide precursor
(CH3CH2CH2CHXCH3)
C6H5
Ο
PPh3
H
1° alkyl halide precursor
(C6H5CH2X)
preferred pathway
2° alkyl halide precursor
X
21.58
a.
N
O
O
O
H
+ HOCH2CH2OH
c.
O
+
NH2
b.
N
O
+
HN
d. CH3O
OCH3
OCH3
CH3O
+ HOCH3
H
O
594
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–22
21.59
PCC
a. C6H5–CH2OH
b. C6H5–COCl
C6H5–CHO
[1] LiAlH[OC(CH3)3]3
[2] H2O
c. C6H5–COOCH3
[1] DIBAL-H
f. C6H5–CH=CH2
[1] O3
C6H5–CHO
g. C6H5–CH=NCH2CH2CH3
[1] LiAlH4
e. C6H5–CH3
KMnO4
PCC
C6H5–CH2OH
[2] H2O
C6H5–COOH
C6H5–CHO
H+
C6H5–CHO
H+
C6H5–CHO
[1] LiAlH4
C6H5–CH2OH
[2] H2O
H2O
H2O
h. C6H5–CH(OCH2CH3)2
C6H5–CHO
[2] H2O
d. C6H5–COOH
C6H5–CHO
[2] Zn, H2O
PCC
C6H5–CHO
21.60
PCC
a.
OH
Cl
b.
c. CH3COCl (CH3CH2)2CuLi
e. CH3C CCH3
O
O
(CH3)2CuLi
O
H2 O
d. CH3CH2C CH
O
H2O
H2SO4
HgSO4
H2SO4
HgSO4
O
O
21.61
O
a. One-step sequence:
preferred route
only one product formed
Ph3P
or
O
Two-step sequence:
OH
H2SO4
[1] CH3CH2CH2MgBr
[2] H2O
b. One-step sequence:
Ph3P
CHO
preferred route
only one product formed
or
OH
CHO [1] (CH ) CHMgBr
3 2
H2SO4
Two-step sequence:
[2] H2O
+ other alkenes that result from
carbocation rearrangement
21.62
a.
CH3CH2CH2CH CHCH3
PCC
One possibility:
CHO
OH
PBr3
OH
Br
[1] Ph3P
[2] BuLi
CHO
Ph3P CHCH3
CH3CH2CH2CH CHCH3
(E and Z)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
595
Aldehydes and Ketones—Nucleophilic Addition 21–23
b.
C6H5CH CHCH2CH2CH3
CH3OH
One possibility:
SOCl2
CH3Cl
Br2
AlCl3
hν
CHO
Br [1] Ph3P
[2] BuLi
C6H5CH PPh3
(from a.)
C6H5CH CHCH2CH2CH3
(E and Z)
c.
PCC
OH
OH
O
Br
PBr3
O
[1] Ph3P
Ph3P CHCH(CH3)2
[2] BuLi
21.63
H2 O
a.
PCC
OH
H2SO4
O
Ph3P CHCH2CH3
PBr3
CH3CH2CH2OH
O
b.
H2NCH2CH2CH3
CHCH2CH3
CH3CH2CH2Br
[1] Ph3P
[2] BuLi
Ph3P CHCH2CH3
NCH2CH2CH3
mild acid
NH3 (excess)
(from a.)
BrCH2CH2CH3
PBr3
HOCH2CH2CH3
OH
c.
Br
HBr
MgBr
Mg
CH3OH
PCC
[1] H2C=O
OH
O
H
PCC
MgBr
[2] H2O
[1]
(from a.)
C(OCH2CH3)2
O
CH3CH2OH
(2 equiv)
PCC
TsOH
2
d.
[1] OsO4
[2] NaHSO3, H2O
OH
OH
O
TsOH
PCC
OH
OH
O
O
[2] H2O
596
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–24
21.64
CH3OH
SOCl2
CH3Cl
a.
KMnO4
CH3
AlCl3
OH
COOH
Br
PBr3
OH
[1] LiAlH4
PCC
CHO
[2] H2O
PPh3
[1] PPh3
[2] BuLi
PPh3
CH CH
CHO
(E and Z isomers)
OH
b.
CH3Cl
(from a.)
AlCl3
OH
OCH3
[1] NaH
[2] CH3Br
CH3
OCH3
P(C6H5)3
hν
CH3
PBr3
(+ ortho isomer)
OCH3
Br2
Br
P(C6H5)3
CH3OH
Br
BuLi
OCH3
CH3O
CH CH
(CH3)3COH
C(CH3)3
P(C6H5)3
(E and Z isomers)
HCl
CH3Cl
(from a.)
C(CH3)3
(CH3)3CCl
AlCl3
AlCl3
C(CH3)3
C(CH3)3
KMnO4
CH3
C(CH3)3
[1] LiAlH4
[2] H2O
[3] PCC OHC
HOOC
(+ ortho isomer)
21.65
O
a.
OH
HOCH2CH2OH
O
O
OH
H+
PBr3
O
[1] Mg
O
PCC
OH
O
O
Br [2] (CH3)2CO
[3] H2O
OH
H2O, H+
O
[1] LiAlH4
OH
O
[2] H2O
OH
OH
CH3CH2CH2OH
PCC
b.
O
[1] Ph3P
O
Br
(from a.)
[2] BuLi
O
O
PPh3
CH3CH2CHO O
H3O+
O
CH=CHCH2CH3
(E/Z mixture)
O
CH=CHCH2CH3
597
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–25
21.66
O
a. CH3CH2OH
PCC
CH3
C
H
PBr3
CH3CH2Br
Mg
CH3CH2MgBr
H2SO4
b. CH3CH2OH
O
[1]
C
CH3
H
[2] H2O
CH2=CH2
O
H OH
PCC
C
CH3
CH2CH3
Br2
Br
Br
CH3
2 NaNH2
C
CH3CH2OH
CH2CH3
TsOH
NaH
HC CH
[1]
CH3
O
HO
O
O
OH
O
C
[2] H2O
H
(from a.)
OH
PCC
CH2CH3
HC C Na+
[1] OsO4
[2] NaHSO3, H2O
OCH2CH3
CH3 C OCH2CH3
TsOH
21.67
O
O
O
O
mCPBA
O
(CH3)3CNH2
OH
NHC(CH3)3
O
NHC(CH3)3
O
H2O
O
O
O
H3O+
X
OH
NHC(CH3)3
HO
+ (CH3)2C=O
HO
albuterol
21.68
[1]
HOCH2CH2OH
Br
O
Br
CHO
TsOH
Mg
O
O
O
O
OH
O
BrMg
O
[2] H2O
H2O
H+
OH
A
21.69
a.
O H
Na+H–
O
Cl
+ H2 + Na+
b.
O
O
acetal
O
CH3
O
+ Cl–
O
methoxy methyl ether
CHO
598
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–26
H
O
O
c.
O
H OH2
H
H2O
O
O
+ H2O
O
O
O
O
H2O
H
+ HOCH3
(The three organic products are boxed in.)
O CH2
H
+ H3O
H
O
O
H
H OH2
O
O
OH
H
CH2 OH
H2O
CH2 O H
CH2 O
H2O
+ H3O
21.70
H
H
N
N
H2O
H2O
H
OH
H
HO
N
H
N
HO
H OH2
H
H OH2
H2O
(+ 1 resonance
structure)
O
O
H2O
H H
N
NH3
O
NH2
H OH2
H
H2O
NH2
(+ 1 resonance
structure)
599
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–27
21.71
O
O
O
CF3
Ar
O
O
CF3
Ar'NHNH2
Ar
proton
transfer
H Cl
OH
O
CF3
Ar'NHNH
Ar
OH2
CF3
Ar'NHNH
+ Cl–
Ar
NHNH2
CF3
H2NSO2
O
N
O
Ar
Ar'
CF3
Ar
N
Ar
CF3
H2O
H
N
Ar'NH
H
+ H3O+
proton
transfer
H Cl HO
O
N
N
Ar Ar' H
Ar'
CF3
H2O
CF3
N
N
H2O
N
N
N
Ar
Ar'
CF3
H
Ar
Ar'
+
N
N
Ar'
+ H3O+
H2NSO2
Cl–
CF3
N
celecoxib
21.72 The OH groups react with the C=O in an intramolecular reaction, first to form a hemiacetal,
and then to form an acetal.
O
O OH
HO
OH
O O
OH
OH adds here to form a hemiacetal.
Then, the acetal is formed by a second
intramolecular reaction.
hemiacetal
acetal
C9H16O2
21.73
a.
H H OH
O
OCH2CH3
O
OCH2CH3
O
H
H OH2
+
HO
O
O
OH
H
H2O
H OH
H OH2
+ H OCH2CH3
H2O
O
H3O+
O
H
HO
H O
H
O
H
OH + H2O
600
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–28
O
b.
H OH2
H
OH
OH
+
H
H
O
O
OH
H
H OH2
O
OH
H
OH
O
H OH
O
H OH
OH2
O
H
H OH
H
O
H2O
H
O
O
O
O
H OH
+ H3O+
O
21.74
HSO4–
H–OSO3H
O
H
OH
O
O
HO
O
HO
O
H–OSO3H
+ HSO4–
HSO4–
H
H2SO4 +
O
O
+
H2O
O
+ HSO4–
H2O
21.75
O
H Cl
O
H
O
OH
OH
O
O CH2
O
CH2
+ Cl–
NH2
S
N
S
O
OH
N
S
H
O
OH
H H
H Cl
Cl
Cl
H
S
N
O
H
H
OH
N
S
H
HO
NH
CH2
OH
S
NH
S
H Cl
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–29
21.76
H O
HO
O
NH2
+ CH3
C
HO
H CH3
H
HO
HO
H OH2
H OH
proton
N C H
HO
N C H
CH3
transfer
HO
dopamine
H OH2
HO
HO
HO
NH
HO
H
CH3
H3 O
N H
HO
CH3
CH3
H2O
salsolinol
N C H
CH3
NH
HO
HO
(+ 3 more resonance
structures)
C
HO
H2O
H
H2 O
21.77
O
O
O
(CH3)2S
CH2 H
Bu Li
(CH3)2S
CH2
+ (CH3)2S
S(CH3)2
sulfur ylide
X
sulfonium salt
X
+ Bu H + LiX
21.78 Hemiacetal A is in equilibrium with its acyclic hydroxy aldehyde. The aldehyde can undergo
hydride reduction to form 1,4-butanediol and a Wittig reaction to form an alkene.
OH
O
a.
OH
O
OH
C
A
OH
H
This can now be reduced with NaBH 4.
OH
O
O
b.
OH
A
C
Ph3P=CHCH2CH(CH3)2
H
reacts with the Wittig reagent
1,4-butanediol
(CH3)2CHCH2CH=CHCH2CH2CH2OH
(E and Z isomers)
602
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–30
+
c.
PPh3
Y
NaOCH2CH3
HO
O
X
PPh3
O +
H
O
O HO
aldehyde for
Wittig reaction
HO
O
New C=C could be cis or
trans. The more stable
trans C=C is drawn.
isotretinoin
21.79
O
H OH2
CH3–MgI
O
HO
OCH3
HO
OCH3
OCH3
OCH3
OCH3
OCH3
H2O
5,5-dimethoxy-2-pentanone
OCH3
OCH3
H
H OH2
HO
H
H OH2
HO
OCH3
HO
HO
OCH3
OH
OCH3
OH
H
O
OCH3
(+ 1 resonance
structure)
H
H2O
H2O
H2O
HO
H O CH3
O
OH
(+ 1 resonance
structure)
H O CH3
OH
O
H
H2O
Y
OH
H3O
H2O
603
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–31
21.80
cyclopropenone
(1640 cm–1)
O
O
2-cyclohexenone
(1685 cm–1)
O
O
O
These three resonance structures include an
aromatic ring; 4n + 2 = 2 π electrons. Although
they are charge separated, the stabilized aromatic
ring makes these three structures contribute to the
hybrid more than usual. Since these three
resonance contributors have a C–O single bond,
the absorption is shifted to a lower wavenumber.
O
O
There are three resonance structures for
2-cyclohexenone, but the charge-separated
resonance structures are not aromatic so they
contribute less to the resonance hybrid. The
C=O absorbs in the usual region for a
conjugated carbonyl.
21.81
A. Molecular formula C5H10O
IR absorptions at 1728, 2791, 2700 cm–1
NMR data: singlet at 1.08 (9 H)
singlet at 9.48 (1 H) ppm
B. Molecular formula C5H10O
IR absorption at 1718 cm–1
NMR data: doublet at 1.10 (6 H)
singlet at 2.14 (3 H)
septet at 2.58 (1 H) ppm
C. Molecular formula C10H12O
IR absorption at 1686 cm–1
NMR data: triplet at 1.21 (3 H)
singlet at 2.39 (3 H)
quartet at 2.95 (2 H)
doublet at 7.24 (2 H)
doublet at 7.85 (2 H) ppm
D. Molecular formula C10H12O
IR absorption at 1719 cm–1
NMR data: triplet at 1.02 (3 H)
quartet at 2.45 (2 H)
singlet at 3.67 (2 H)
multiplet at 7.06–7.48 (5 H) ppm
1 degree of unsaturation
C=O, CHO
3 CH3 groups
CHO
CHO
CH3 C CH3
CH3
1 degree of unsaturation
C=O
2 CH3's adjacent to H
CH3
CH adjacent to 2 CH3's
O
5 degrees of unsaturation (4 due to a benzene ring)
C=O
CH3 adjacent to 2 H's
CH3
CH2 adjacent to 3 H's
2 H's on benzene ring
2 H's on benzene ring
O
CH3
5 degrees of unsaturation (4 due to a benzene ring)
C=O
CH3 adjacent 2 H's
O
2 H's adjacent to 3 H's
CH2
a monosubstituted benzene ring
21.82
C7H16O2: 0 degrees of unsaturation
IR: 3000 cm–1: C–H bonds
NMR data (ppm):
Ha: quartet at 3.5 (4 H), split by 3 H's
Hb: singlet at 1.4 (6 H)
Hc: triplet at 1.2 (6 H), split by 2 H's
CH3
Hb
CH3CH2 O C O CH2CH3
Hc Ha
CH3
Hb
Ha Hc
604
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–32
21.83
B. Molecular formula C9H10O
A. Molecular formula C9H10O
5 degrees of unsaturation
5 degrees of unsaturation
IR absorption at 1720 cm–1 → C=O
IR absorption at 1700 cm–1 → C=O
IR absorption at ~2700 cm–1 → CH of RCHO
IR absorption at ~2700 cm–1 → CH of RCHO
NMR data (ppm):
NMR data (ppm):
2 triplets at 2.85 and 2.95 (suggests –CH2CH2–)
triplet at 1.2 (2 H's adjacent)
multiplet at 7.2 (benzene H's)
quartet at 2.7 (3 H's adjacent)
signal at 9.8 (CHO)
doublet at 7.3 (2 H's on benzene)
doublet at 7.7 (2 H's on benzene)
O
singlet at 9.9 (CHO)
CH2 CH2 C
H
O
C
CH2 CH3
H
21.84
C. Molecular formula C6H12O3
1 degree of unsaturation
IR absorption at 1718 cm–1 → C=O
To determine the number of H's that give rise to each signal, first find the number of integration units per H by
dividing the total number of integration units (7 + 40 + 14 + 21 = 82) by the number of H's (12); 82/12 = 6.8. Then
divide each integration unit by this number (6.8).
O
OCH3
NMR data (ppm):
singlet at 2.2 (3 H's)
OCH3
doublet at 2.7 (2 H's)
singlet at 3.2 ( 6 H's – 2 OCH3 groups)
triplet at 4.8 (1 H)
21.85
D. Molecular ion at m/z = 150: C9H10O2 (possible molecular formula)
5 degrees of unsaturation
IR absorption at 1692 cm–1 → C=O
O
NMR data (ppm):
triplet at 1.5 (3 H's – CH3CH2)
quartet at 4.1 (2 H's – CH3CH2)
doublet at 7.0 (2 H's – on benzene ring)
doublet at 7.8 (2 H's – on benzene ring)
singlet at 9.9 (1 H – on aldehyde)
21.86
OH
OH
OH
a. HO
CHO
OH
b.
HO
HO
HO
OH
CHO
OH
CHO
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Aldehydes and Ketones—Nucleophilic Addition 21–33
21.87
OH
OH
O
HO
HO
O
HO
HO
H Cl
OH
OH
OH
β-D-glucose
OH
above
H
OH
Cl
OCH3
CH3
OH
+
HCl
acetal
OH
O
HO
HO
below
O
HO
HO
O
OH
CH3OH
OH
OH
Cl
OH
O
O
H2O
O
HO
HO
CH3OH
HO
HO
OH
Cl
OH
HO
HO
O
HO
HO
OH2
OH
OH
O
HO
HO
OH
O
CH3
H
+
acetal
OH
HCl
OCH3
The carbocation is trigonal planar, so CH3OH attacks from two
different directions, and two different acetals are formed.
21.88
H+
H
O
H
O
OH
O
OH
O
OH
O
O
O
OH
OH
O H
O
O
H2O
H3O
21.89
O
a.
O
brevicomin
O
HO
b.
brevicomin
Br
H3O+
OH
O
O
O
[1] Ph3P
O
O
Br [2] BuLi
H+
OH
PPh3
CH3CH2CHO
H3O+ O
O
OH
[1] OsO4
[2] NaHSO3, H2O
OH
OH
O
O
PCC
CH3CH2CH2OH
CH=CHCH2CH3
(E and Z isomers)
606
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 21–34
21.90
acetal carbon
OH
a.
OCH3
O
HO
HO
c.
OH
HO
O
HO
CH3O
CH3O
O
OCH3
O
CH3O
HO
CH3O
OH
O
CH3O
OH
OH
(OH can be up or down in both
products.)
HO
hemiacetal carbon
OH
+
b. [1] H3O HO
O
HO
HO
(OH can be up or down.)
OH
OH
[2] CH3OH, HCl
O
HO
HO
OH
HO
O
HO
O
OCH3
HO
(OCH3 can be up or down.)
OCH3
[3] NaH (excess)
O
CH3O
CH3O
CH3O
O
CH3O
CH3I (excess)
OCH3
O
OCH3
OCH3
21.91
OH
OH
[1] O3
O
O
CHO
[2] (CH3)2S
CH3O
CH3O
H OSO2R
RSO3H
H
H
OH
CH3O
OH
O
R
S
C6H10O3
OH
OH
O
OH
O
S:
OH
O
O
CH3OH
CH3O
Mechanism of R
OH
OH
NaBH4
O
O
CH3O H
OSO2R
OH
OH
H
H
H
H
O
H
CH3O H
OH
O
O
S
OSO2R
RSO3H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
607
Aldehydes and Ketones—Nucleophilic Addition 21–35
21.92 The mechanism involves SN2 displacement of Br, followed by intramolecular enamine
formation.
H2N
+ H
H2N
N
N
+
SN2
H
H2N
+
HCO3–
CO32–
O
Br
H
H
N
H2N
N
O
CO32–
N
R
N
+
H N
H
CO32–
H2O
+
(any acid)
N
H A
H
H
HO
N
HN
+ HCO3– + H2O
O
NH
conivaptan
proton
transfer
N
R
N
R
Br–
N
R
O
N
R
N
H N
+ HCO3–
O
N+
N
O
N
R
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609
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–1
Chapter 22 Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution
Chapter Review
Summary of spectroscopic absorptions of RCOZ (22.5)
IR absorptions
• All RCOZ compounds have a C=O absorption in the region
1600–1850 cm–1.
• RCOCl: 1800 cm–1
• (RCO)2O: 1820 and 1760 cm–1 (two peaks)
• RCOOR': 1735–1745 cm–1
• RCONR'2: 1630–1680 cm–1
• Additional amide absorptions occur at 3200–3400 cm–1 (N–H stretch)
and 1640 cm–1 (N–H bending).
• Decreasing the ring size of a cyclic lactone, lactam, or anhydride
increases the frequency of the C=O absorption.
• Conjugation shifts the C=O to lower wavenumber.
1
H NMR absorptions
• C–H α to the C=O absorbs at 2–2.5 ppm.
• N–H of an amide absorbs at 7.5–8.5 ppm.
•
13
C NMR absorption
C=O absorbs at 160–180 ppm.
Summary of spectroscopic absorptions of RCN (22.5)
IR absorption
• C≡N absorption at 2250 cm–1
13
C NMR absorption
• C≡N absorbs at 115–120 ppm.
Summary: The relationship between the basicity of Z– and the properties of RCOZ
• Increasing basicity of the leaving group (22.2)
• Increasing resonance stabilization (22.2)
R
O
O
O
C
C
C
Cl
R
acid chloride
O
anhydride
O
R
R
O
OH
carboxylic acid
~
R
C
O
OR'
ester
R
C
NR'2
amide
• Increasing leaving group ability (22.7B)
• Increasing reactivity (22.7B)
• Increasing frequency of the C=O absorption in the IR (22.5)
General features of nucleophilic acyl substitution
• The characteristic reaction of compounds having the general structure RCOZ is nucleophilic acyl
substitution (22.1).
• The mechanism consists of two steps (22.7A):
[1] Addition of a nucleophile to form a tetrahedral intermediate
[2] Elimination of a leaving group
• More reactive acyl compounds can be used to prepare less reactive acyl compounds. The reverse
is not necessarily true (22.7B).
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Chapter 22–2
Nucleophilic acyl substitution reactions
[1] Reaction that synthesizes acid chlorides (RCOCl)
From RCOOH
(22.10A):
O
R
C
O
+
SOCl2
OH
R
+
C
+
SO2
Cl
HCl
[2] Reactions that synthesize anhydrides [(RCO)2O]
[a] From RCOCl
(22.8):
O
R
C
+–
Cl
O
O
O
C
C
R'
R
O
C
O
Cl–
+
R'
O
Δ
OH
[b] From dicarboxylic
acids (22.10B):
O
O
+
OH
H2O
O
O
cyclic anhydride
[3] Reactions that synthesize carboxylic acids (RCOOH)
O
O
[a] From RCOCl
(22.8):
[b] From (RCO)2O
(22.9):
R
R
C
+
Cl
O
O
C
C
O
H2O
pyridine
R
R
C
+
OH
R
H 2O
2
R
C
Cl–
OH
O
+
OR'
H2 O
(H+ or –OH)
R
C
O
OH
(with acid)
or
R
R
C
C
O–
+
(with base)
O
H2O, H+
[d] From RCONR'2 (R' =
H or alkyl, 22.13):
+
N
H
O
+
O
[c] From RCOOR'
(22.11):
C
+
+
R'2NH2
+
R'2NH
+
+
N
H
OH
O
R
C
NR'2
R' = H or alkyl
O
H2O, –OH
R
C
O–
[4] Reactions that synthesize esters (RCOOR')
O
[a] From RCOCl
(22.8):
R
C
O
+
Cl
R'OH
pyridine
R
C
OR'
Cl–
R'OH
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Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–3
[b] From (RCO)2O
(22.9):
R
O
O
C
C
O
O
+
R'OH
R
O
[c] From RCOOH
(22.10C):
R
C
OH
+
+
RCOOH
OR'
O
H2SO4
R'OH
C
R
R
C
+ H 2O
OR'
[5] Reactions that synthesize amides (RCONH2) [The reactions are written with NH3 as the
nucleophile to form RCONH2. Similar reactions occur with R'NH2 to form RCONHR', and with
R'2NH to form RCONR'2.]
O
[a] From RCOCl
(22.8):
R
C
O
+
NH3
Cl
R
(2 equiv)
[b] From (RCO)2O
(22.9):
R
O
O
C
C
O
R
C
R
NH3
R
(2 equiv)
C
OH
R
[2] Δ
C
NH2
R
C
C
NH2
+
RCOO– NH4+
+
H2O
O
OH
R'NH2
+
C
R
DCC
O
[d] From RCOOR'
(22.11):
NH4+Cl–
O
[1] NH3
O
R
+
NH2
O
+
O
[c] From RCOOH
(22.10D):
C
+ H2O
NHR'
O
+
NH3
OR'
R
C
NH2
+
R'OH
Nitrile synthesis (22.18)
Nitriles are prepared by SN2 substitution using unhindered alkyl halides as starting materials.
R X
R = CH3,
+
–
CN
R C N
X–
+
SN 2
1o
Reactions of nitriles
[1] Hydrolysis (22.18A)
O
H 2O
R C N
(H+
or
–OH)
R
C
O
OH
(with acid)
or
R
C
O–
(with base)
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–4
[2] Reduction (22.18B)
[1] LiAlH4
R CH2NH2
[2] H2O
1o amine
R C N
O
[1] DIBAL-H
[2] H2O
R
C
H
aldehyde
[3] Reaction with organometallic reagents (22.18C)
R C N
O
[1] R'MgX or R'Li
[2] H2O
R
C
R'
ketone
Practice Test on Chapter Review
1. Give the IUPAC name for each of the following compounds.
CN
a.
O
O
b.
c.
O
N
CH3
2. (a) Which compound absorbs at the lowest wavenumber in the IR? (b) Which compound absorbs
at the highest wavenumber?
O
O
H
N
O
NH
O
O
O
A
B
C
D
3. a. Which of the following reaction conditions can be used to synthesize an ester?
1.
2.
3.
4.
5.
RCOCl + R'OH + pyridine
RCOOH + R'OH + H2SO4
RCOOH + R'OH + NaOH
Both methods [1] and [2] can be used to synthesize an ester.
Methods [1], [2], and [3] can all be used to synthesize an ester.
b. Which of the following compounds is most reactive in nucleophilic acyl substitution?
1. CH3COCl
2. CH3COOCH3
3. CH3CON(CH3)2
4. (CH3CO)2O
5. CH3COOH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
613
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–5
4. What reagent is needed to convert (CH3CH2)2CHCOOH into each compound?
a. (CH3CH2)2CHCOO–Na+
b. (CH3CH2)2CHCOCl
c. (CH3CH2)2CHCON(CH3)2
d. (CH3CH2)2CHCO2CH2CH3
e. [(CH3CH2)2CHCO]2O
5. What reagent is needed to convert CH3CH2CH2CN to each compound?
a. CH3CH2CH2COOH
b. CH3CH2CH2CH2NH2
c. CH3CH2CH2COCH2CH3
d. CH3CH2CH2CHO
6. Draw the organic products formed in the following reactions.
H D
a.
Br
CO2H
[1] NaCN
[1] SOCl2
d.
[2] LiAlH4
[2] (CH3CH2)2NH
[3] H2O
[Indicate stereochemistry.]
b.
O
O
O
O CH2CH3
O
OH
c.
OH
O
NaOH, H2O
+
O
O
e.
+ CH3CH2NH2
(excess)
O
[1] NaCN
f.
Br
O
[2] CH3CH2MgBr
[3] H2O
O
Answers to Practice Test
1. a. 5-ethyl-2-methylheptanenitrile
b. cyclohexyl 2methylbutanoate
c. N-cyclohexyl-Nmethylbenzamide
2. a. C
b. A
3. a. 4
b. 1
4. a. NaOH
b. SOCl2
c. HN(CH3)2, DCC
d. CH3CH2OH, H2SO4
e. heat
5. a. H3O+
b. [1] LiAlH4; [2] H2O
c. [1] CH3CH2Li; [2] H2O
d. [1] DIBAL-H; [2] H2O
6.
D H
a.
b.
O
CH2NH2
O
O
O
O
N(CH2CH3)2
d.
O
NHCH2CH3
e.
O
O– H3NCH2CH3
HOCH2CH3
OCOCH3
c.
OH
O
O
f.
+
OH
O
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–6
Answers to Problems
22.1
The number of C–N bonds determines the classification as a 1o, 2o, or 3o amide.
NH2 1° amide
O
O
O
1° amide H2N
O
NH
oxytocin
All seven others are 2° amides.
NH
O
N
H
O
O
O
HN
N
H
N
O
N
H
H2N
NH2 S
H
N
O
1° amide
S
3° amide
O
HO
22.2
As the basicity of Z increases, the stability of RCOZ increases because of added resonance
stabilization.
R
O
O
C
C
O
R + Br
Br
R
C
+
Br
The basicity of Z determines how much
this structure contributes to the hybrid.
Br– is less basic than –OH, so RCOBr
is less stable than RCOOH.
22.3
O
CH3 Cl
CH3
C
O
Cl
CH3
H
C
Cl
O
O
CH3 NH2
C
NH2
H
C
NH2
This resonance structure contributes little to the hybrid
since Cl– is a weak base. Thus, the C–Cl
bond has little double bond character, making it similar in
length to the C–Cl bond in CH3Cl.
This resonance structure contributes more to the hybrid
since –NH2 is more basic. Thus, the C–N bond
in HCONH2 has more double bond character, making
it shorter than the C–N bond in CH3NH2.
22.4
a. (CH3CH2)2CH
COCl
b.
C6H5COOCH3
re-draw
re-draw
O
O
Cl 2-ethylbutanoyl chloride
2-ethyl
OCH3
methyl benzoate
alkyl group = methyl
acyl group =
benzoate
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Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–7
O
e.
c. CH3CH2CON(CH3)CH2CH3
CH3CH2
O
C
C
O
benzoic propanoic anhydride
re-draw
O
N
acyl group =
propanoic
N-ethyl-N-methyl
acyl group =
benzoic
N-ethyl-N-methylpropanamide
acyl group =
propanamide
1
O
d.
H
CN
f.
C
OCH2CH3
3
6 carbon chain =
hexanenitrile
alkyl group = ethyl
acyl group =
formate
3-ethylhexanenitrile
ethyl formate
22.5
d. N-isobutyl-N-methylbutanamide
a. 5-methylheptanoyl chloride
g. sec-butyl 2-methylhexanoate
O
O
O
N
Cl
b. isopropyl propanoate
O
h. N-ethylhexanamide
e. 3-methylpentanenitrile
O
O
CN
O
f. o-cyanobenzoic acid
c. acetic formic anhydride
O
O
O
H
O
N
H
OH
CH3
CN
22.6
CH3CONH2 has two H’s bonded to N that can hydrogen bond. CH3CON(CH3)2 does not
have any H’s capable of hydrogen bonding. This means CH3CONH2 has much stronger
intermolecular forces, which leads to a higher boiling point.
22.7
a. CH3
O
O
O
C
C
C
OCH2CH3
and CH3
N(CH2CH3)2
c. CH3CH2CH2
b.
O
O
O
and
O
d.
O
O
smaller ring:
C=O at a higher
wavenumber
NHCH3
and
CH3CH2CH2
2° amide: 1 N–H
absorption at
3200–3400 cm–1
amide: C=O at
lower wavenumber
O
O
anhydride:
2 C=O peaks
NH2
1° amide: 2
N–H absorptions
O
and
C
Cl
616
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–8
22.8
Hb
O
Hc signal from the 5 H's
on the aromatic ring
Ha
O
H H
CH3
Hb
Ha
A
Molecular formula C9H10O2
5 degrees of unsaturation
IR: 1743 cm–1 from ester C=O
3091–2895 cm–1 from sp2 and sp3 C–H
1
H NMR: Ha = 2.06 ppm (singlet, 3 H) – CH3
Hb = 5.08 ppm (singlet, 2 H) – CH2
Hc = 7.33 ppm (broad singlet, 5 H)
22.9
H H
O
O
Hb signal from the
10 H's on the two
B
aromatic rings
Molecular formula C14H12O2
9 degrees of unsaturation
IR: 1718 cm–1 from conjugated ester C=O
3091–2953 cm–1 from sp2 and sp3 C–H
1H NMR:
Ha = 5.35 ppm (singlet, 2 H)
Hb = 7.26–8.15 ppm (multiplets, 10 H)
More reactive acyl compounds can be converted to less reactive acyl compounds.
a.
CH3COCl
CH3COOH
more reactive
YES
b. CH3CONHCH3
less reactive
NO
c.
CH3COOCH3
less reactive
CH3COOCH3
d.
more reactive
(CH3CO)2O
more reactive
CH3COCl
NO
more reactive
YES
less reactive
less reactive
CH3CONH2
22.10 The better the leaving group is, the more reactive the carboxylic acid derivative. The
weakest base is the best leaving group.
O
O
C
C
a.
NH2
−NH
2
CH3CH2
C
OCH3
−OCH
3
strongest base
least reactive
b.
O
intermediate
−Cl
weakest base
most reactive
O
O
O
C
C
C
CH3CH2
NHCH3
−NHCH
3
OH
−OH
strongest base
least reactive
intermediate
Cl
CH3CH2
O
O
C
CH2CH3
−OOCCH
2CH3 weakest base
most reactive
22.11
CH3
O
O
O
O
C
C
C
C
O
CH3
Cl3C
O
CCl3
acetic anhydride
trichloroacetic anhydride
The Cl atoms are electron withdrawing,
which makes the conjugate base (the
leaving group, CCl3COO–) weaker
and more stable.
617
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–9
22.12
O
O
H2O
a.
O
OH
+
N
–
H Cl
pyridine
Cl
O
NH2 + NH4+Cl–
NH3
excess
O
O
CH3COO−
b.
c.
CH3 +
O
Cl–
(CH3)2NH
N(CH3)2
d.
excess
+ (CH3)2NH2 Cl–
22.13 The mechanism has three steps: [1] nucleophilic attack by O; [2] proton transfer;
[3] elimination of the Cl– leaving group to form the product.
OCH3
OCH3
O
OH
+
OCH3
O Cl
O
Cl
O Cl
O
H
N
H
OCH3
OCH3
OCH3
N
OCH3
O
Cl
O
OCH3
A
22.14
O
O
O
O
O
OH
H2O
O
NH3
HO
a.
c.
O
b.
CH3OH
O
NH2
O
NH4
excess
O
CH3
O
O
(CH3)2NH
HO
d.
excess
O
N(CH3)2
O
(CH3)2NH2
22.15 Reaction of a carboxylic acid with thionyl chloride converts it to an acid chloride.
O
a. CH3CH2
C
O
SOCl2
OH
CH3CH2
O
b.
C
C
Cl
O
OH
[1] SOCl2
C
O
[2] (CH3CH2)2NH (excess)
Cl
C
N(CH2CH3)2
(CH3CH2)2NH2 Cl–
618
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–10
22.16
H2SO4
COOH + CH3CH2OH
a.
OCH2CH3 + H2O
C
O
O
COOH
b.
+
C
H2SO4
OH
+ H2O
O
O
COOH
c.
+
C
O
Na+
+ CH3OH
O
H2SO4
d. HO
O
NaOCH3
+ H2O
OH
O
22.17
O
C
O
CH318OH
OH
C
18
OCH3 + H2O
22.18
O
C
HO
OH
O
H
H−OSO3H
C
HO
O
HO OH
H
O
HO OH
H
+ HSO4–
O
–OSO
O
O H
O
+ H2SO4
3H
HSO4–
HO OH2
O
O
H2O +
H−OSO3H
+ HSO4–
22.19
O
a. O
CH3NH2
+
O− H3NCH3
OH
b.
O
O
[1] CH3NH2
OH
[2] Δ
NHCH3 + H2O
c.
O
O
CH3NH2
OH
DCC
NHCH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
619
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–11
22.20
O
[1] H3O+
OH
+
HO
O
+
HO
O
a.
CH3O
O
CH3O
O
[2] H2O, –OH
octinoxate
CH3O
O
[1] H3O+
O
b.
OH
O
+
HO
+
HO
OH
O
OH
[2] H2O, –OH
octyl salicylate
O
OH
22.21 Bonds broken during hydrolysis are indicated.
O
HO
HO
H
C(CH3)3
O
O
O
CH3 OH
COOH
OH
HO
HO
O
H3O+
HO
HOOC
O
H
C(CH3)3
CH3 OH
HO H
HOOC
O H
O
ginkgolide B
22.22
HOCH2
O
OH
HOCH2
HO
RCOOH, H2SO4
O
O
HO
RCOOCH2
OH
OH
RCOO
CH2OH
a long-chain fatty acid
sucrose
RCOO OOCR
OOCR
CH2
O
O
CH2OOCR
OOCR
O
RCOO
olestra
22.23
CH2OCO(CH2)15CH3
CHOCO(CH2)15CH3
CH2OCO(CH2)7CH=CH(CH2)7CH3
cis
CH2 OH
Na+ −OOC(CH2)15CH3
CH OH
Na+ −OOC(CH2)15CH3
CH2 OH
Na+ −OOC(CH2)7CH=CH(CH2)7CH3
hydrolysis
glycerol
soap
cis
620
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–12
22.24
O
O
NH
OH
H2O, OH
proton
transfer
O
N H
H
O
O
O
NH
NH2
22.25 Aspirin has an ester, a more reactive acyl group, whereas acetaminophen has an amide, a less
reactive acyl group.
a. The ester makes aspirin more easily hydrolyzed with water from the air than
acetaminophen. Therefore, Tylenol can be kept for many years, whereas aspirin
decomposes.
b. Similarly, aspirin will be hydrolyzed and decompose in the aqueous medium of a liquid
medication, but acetaminophen is stable due to the less reactive amide group, allowing it
to remain unchanged while dissolved in H2O.
ester
O
C
amide
H
N
CH3
O
COOH
CH3
O
HO
acetylsalicylic acid
C
acetaminophen
22.26
"Regular" amide is not hydrolyzed.
H H H
N
S
R
O
H H
N
R
H3O+
O
N
O
COOH
O
H
S
HN
OH
COOH
22.27
O
H
N
O
H
N
N
H
N
H
O
O
nylon 6,10
O
Cl
Cl
O
+
H2N
NH2
621
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–13
22.28
OH
+
HOOC
COOH
In the polyester Kodel, most of the
bonds in the polymer backbone are
part of a ring, so there are fewer
degrees of freedom. Fabrics made
from Kodel are stiff and crease
resistant, due to these less flexible
polyester fibers.
HO
1,4-dihydroxymethylcyclohexane
terephthalic acid
O
O
O
O
O
O
O
O
Kodel
22.29
O
O
O
OH
+
OH
OH
O
O
O
OH
PLA
polyl(lactic acid)
O
O
O
O
Reaction occurs here (–H2O).
22.30 Acetyl CoA acetylates the NH2 group of glucosamine, since the NH2 group is the most
nucleophilic site.
HO
HO
O
O
HO
+
OH
CH3
C
O
HO
NH2
HO
OH
SCoA
HO
glucosamine
HN
NAG
O
22.31
a. CH3CH2CH2 Br
NaCN
CH3CH2CH2 CN
H2O, –OH
c.
CN
CN
H2O, H+
b.
COO
COOH
CN
COOH
22.32
a.
CH3
O
OH
NH
C
C
C
NH2
O
b.
C
CH3
c.
NH
OH
NHCH3
C
NCH3
CH3CH2
OH
NH2
CH3CH2
C
O
622
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–14
22.33
[1] NaCN
a. CH3CH2 Br
b. CH3CH2CH2 CN
CH3CH2 CH2NH2
[2] LiAlH4
[3] H2O
[1] DIBAL-H
O
CH3CH2CH2 C
[2] H2O
H
22.34
OCH3
[1] CH3CH2MgCl
a.
CN
[2] H2O
OCH3
O
C
CH2CH3
O
CN
C
[1] C6H5Li
b.
[2] H2O
22.35
O
O
CH2 CN
a.
[1] CH3MgBr
CH2 CN
CH2 C
CH3
[2] H2O
c.
[1] DIBAL-H
CH2 C
H
[2] H2O
O
CH2 CN
b.
[1] (CH3)3CMgBr
O
CH2 C
CH2 CN
C(CH3)3
CH2 C
d.
H 3 O+
OH
[2] H2O
22.36
CH3 C N
O
O
[1] CH3CH2MgBr
[1] CH3MgBr
CH3CH2 C N
[2] H2O
[2] H2O
22.37 The better the leaving group, the more reactive the acyl compound.
O
O
OH
O
SH
worst leaving group
least reactive
intermediate
reactivity
22.38
O
a.
3
2
1
1 CN
O
A
isobutyl 2,2-dimethylpropanoate
6
5
4
3
2
B
2-ethylhexanenitrile
Cl
best leaving group
most reactive
623
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–15
b.
A
A
[1] H3O+
COOH
[2] –OH
COO
HO
HO
H2O
OH
[3] CH3CH2CH2MgBr
A
HO
COO
[2] –OH
B
OH
COOH
[1] H3O+
B
HO
H2O
[4] LiAlH4
A
H2O
H2O
O
[3] CH3CH2CH2MgBr
B
H2O
NH2
H2O
[4] LiAlH4
B
22.39 Better leaving groups make acyl compounds more reactive. C has an electron-withdrawing
NO2 group, which stabilizes the negative charge of the leaving group, whereas D has an
electron-donating OCH3 group, which destabilizes the leaving group.
O
CH3
C
O
O
NO2
C
CH3
C
O
OCH3
D
an electron-withdrawing substituent an electron-donating substituent
O
O
N
O
O
O
leaving group from C
N
O
OCH3
O
OCH3
O
one possible
resonance structure
Delocalizing the negative charge on the
NO2 stabilizes the leaving group,
making C more reactive than D.
one possible
resonance structure
leaving group from D
Adjacent negative charges destabilize
the leaving group.
624
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–16
22.40
Cl
a.
f.
2,2-dimethylpropanoyl chloride
m-chlorobenzonitrile
Cl
O
O
O
b.
CN
cyclohexyl pentanoate
g.
Cl
O
O
c.
3-phenylpropanoyl chloride
O
O
h.
O
cis-2-bromocyclohexanecarbonyl chloride
C
Cl
Br
cyclohexanecarboxylic anhydride
O
i.
O
d.
N
phenyl phenylacetate
N,N-diethylcyclohexanecarboxamide
O
O
O
e.
N
N-cyclohexylbenzamide
H
j.
cyclopentyl
cyclohexanecarboxylate
O
22.41
e. isopropyl formate
a. propanoic anhydride
O
i. benzoic propanoic anhydride
O
O
O
H
O
b. α-chlorobutyryl chloride
O
O
O
f. N-cyclopentylpentanamide
O
j. 3-methylhexanoyl chloride
NH
O
Cl
Cl
O
Cl
c. cyclohexyl propanoate
g. 4-methylheptanenitrile
k. octyl butanoate
O
O
CN
O
d. cyclohexanecarboxamide
h. vinyl acetate
O
C
O
l. N,N-dibenzylformamide
O
O
NH2
O
H
CH3
N
22.42 Rank the compounds using the rules from Answer 22.10.
O
a.
O
NH2
−NH
2 strongest base
least reactive
O
OCH2CH2CH3
−OCH CH CH
2
2
3
intermediate
Cl
Cl– weakest base
most reactive
625
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–17
O
b.
O
O
O
O
O
F3C
ester
least reactive
anhydride
intermediate
O
O
CF3
anhydride with electronwithdrawing F's
most reactive
22.43
resonance structures for the leaving group
O
O
C N
N
Nu
O
N
R C N
R
R
Nu
C
N
N
Nu
imidazolide
N
N
N
N
N
N
N
N
The leaving group is both resonance stabilized and aromatic
(6 π electrons), making it a much better leaving group than
exists in a regular amide.
22.44
Reaction as an acid:
O
CH3
C
O
B
NH2
CH3
C
O
NH
CH3
C
CH3CH2 NH2
NH
B
CH3CH2 NH
no resonance stabilization
of the conjugate base
These two resonance structures
make the conjugate base more stable,
and therefore CH3CONH2 a stronger acid.
Reaction as a base:
O
CH3
C
O
NH2
CH3
C
This electron pair
is localized on N.
CH3CH2 NH2
NH2
This electron pair is delocalized by resonance,
making it less available for electron donation.
Thus, CH3CONH2 is a much weaker base.
22.45
CH3CH2CH2CH2COCl
a.
pyridine
b.
NH3
H2O
CH3CH2OH
d.
CH3CH2CH2CH2COOH
N Cl–
H
c.
O
CH3COO−
CH3CH2CH2CH2
O
excess
N Cl–
H
O
CH3
NH4 Cl–
(CH3CH2)2NH
e.
CH3CH2CH2CH2COOCH2CH3
pyridine
CH3CH2CH2CH2CONH2
excess
Cl–
CH3CH2CH2CH2CON(CH2CH3)2
(CH3CH2)2NH2 Cl–
C6H5NH2
f.
excess
CH3CH2CH2CH2CONHC6H5
C6H5NH3 Cl–
626
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–18
22.46
O
O
O
a.
b.
SOCl2
d.
no reaction
O
e.
2
no reaction
(CH3CH2)2NH
O
H2O
NaCl
OH
excess
N(CH2CH3)2
O
O H2N(CH2CH3)2
O
CH3OH
O
OCH3
c.
f.
O
CH3CH2NH2
NHCH2CH3
O
excess
OH
O
H3NCH2CH3
22.47
C6H5CH2COOH
a.
O
NaHCO3
+ H2CO3 g.
O Na
b.
NaOH
h.
O
+ H2O
i.
O
SOCl2
OCH3
O
CH3OH
–OH
O Na
c.
O
CH3OH
H2SO4
O
[1] NaOH
O
Cl
d.
NaCl
no reaction
j.
O
CH3NH2
DCC
e.
NH3
O
k.
(1 equiv)
O NH4
f.
O
[1] NH3
[2] Δ
l.
NH2
O
O
[2] CH3COCl
NHCH3
[1] SOCl2
O
[2] CH3CH2CH2NH2
NHCH2CH2CH3
O
[1] SOCl2
[2] (CH3)2CHOH
OCH(CH3)2
22.48
c.
CH3CH2CH2CO2CH2CH3
H2O, –OH
O
O
HO
O
a.
SOCl2
no reaction
d.
NH3
O
b.
NH2
HO
O
H3O+
OH
HO
e. CH3CH2NH2
NHCH2CH3
HO
627
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–19
22.49
NH2
H3 O +
a.
CH2COOH
CH2COO
H2O, –OH
b.
O
22.50
H3O+
a.
COOH
d.
[1] CH3CH2Li
O
[2] H2O
H2O, –OH
b.
C6H5CH2CN
COO
C
[2] H2O
O
[1] LiAlH4
CH2NH2
CH3
c. [1] CH3MgBr
f.
[2] H2O
H
e. [1] DIBAL-H
O
[2] H2O
22.51
O
O
COOH
C
SOCl2
a.
OH
O
f.
Cl
O
H3O+
OH
OH
O
O
[1] NaCN
O
b. C6H5COCl
+
+
C6H5
N
H
(excess)
c. C6H5CN
OH
N
N
Cl–
CH3CH2CH2CH2Br
O
[2] H2O, –OH
C6H5
[2] H2O
h. C6H5CH2COOH
C
[1] SOCl2
O
C6H5(CH2)2NH(CH2)3CH3
[2] CH3CH2CH2CH2NH2
[3] LiAlH4
[4] H2O
CH2CH2CH3
O
O
H2SO4
(CH3)2CH
CH3CH2CHOH
C
OCH(CH3)CH2CH3 i.
Δ
HOOC
O
COOH
CH3
e.
CH3CH2CH2CH2
H H
O
[1] CH3CH2CH2MgBr
d. (CH3)2CHCOOH +
g.
O
NHCOCH3
H2 O
–OH
O
j.
+
(CH3CO)2O +
NH2
NHCOCH3
NH2
O
(excess)
+ CH3COO– H3N
22.52 Both lactones and acetals are hydrolyzed with aqueous acid, but only lactones react with
aqueous base.
O
a.
O
O
O
X
H3O+
O
OH
O
OH
COOH
OH
b.
NaOH
O
H 2O
O
O
O
O
X
CO2 Na+
OH
628
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–20
22.53
Br NaCN
CN
H3O+
COOH SOCl2
A
COCl [1] (CH3)2CuLi
[1] LiAlH4
CH3
O
[2] H2O
D
B
C
E
CH3OH, H+
[2] H2O
PCC
CH2NH2
CO2CH3 [1] DIBAL-H
CHO [1] CH3Li
[2] H2O
C
G
F
OH
[2] H2O
H
[1] LiAlH4
(CH3CO)2O
[2] H2O
CH2OTs NaCN
CH2OH
CH2NHCOCH3
CH2CN
TsCl, pyridine
I
K
J
L
[1] CH3MgBr
[2] H2O
O
M
22.54
O
H D
C CH3
a.
CH3COCl
OH
c.
O
pyridine
CH3
CH3
NaCN
Br
b.
CH3
H+
COOH
O
CN
H D
C
H D
CH3CH2OH
D H
d.
CH3
C
+
Cl C H
6 5
CH3
C
CH3
CH3
C
C
H
NH2
(2 equiv)
C6H 5
COOCH2CH3
CH3
C
H +
H
C6H5
NH
NH3
+
C
Cl–
CH3
O
22.55 Hydrolyze the amide and ester bonds in both starting materials to draw the products.
a.
O
O
CO2CH2CH3
H2 O
O
CO2H
O
HO
N
H
OH
H 2N
NH2
oseltamivir
NH2
629
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–21
NH2
HOOC
b.
H
N
NH2
H2 O
O
CH3O
HOOC
H2 N
+ CH3OH
OH
COOH
O
O
aspartame
phenylalanine
22.56
F
F
F
(–H+)
CH3SO2Cl
O
(CH3CH2)3N
O
O
OH
CH3O
OSO2CH3
CH3O
N
H
CH3O
PhCH2NH2
intramolecular
amide formation
F
O
N
Ph
F
C18H18FNO
22.57
O
C
a.
Cl
O
C Cl
C Cl
H
b.
HO
O
O
O
proton
O
O
OH
NHNH2 + Cl
+ HN
N
O
C
NH2NH
NH2NH
NH2NH2
O
O
OH
transfer HO
O
O
Ph
630
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–22
22.58
H+
O
CH3
C
OH
OH
CH3
C
OH
OH
OH
CH3 C
18
O H
H
18
OH
CH3 C OH2
18
18
O H
+ H2O
OH
H+
CH3 C OH
CH3 C OH
18
O H
O H
A
+ H3O+
+ H2O
+ H2O
Two possibilities for A:
OH
O H
CH3 C
CH3
18
O H
C
O
H2O
18
C
CH3
OH
+ H3O+
18
OH
A
OH
OH
CH3 C
CH3
18
O H
OH
C 18
O H
C
CH3
H2O
+ H3O+
18
O
A
22.59
H2O
H
H2O
O H
O
O
OH
OH
OH
H−OH2
O
O
OH
O
OH
O
H2O
OH
H
γ-butyrolactone
H−OH2
O
H3O
O
HO
HO
OH
H
H2O
OH
4-hydroxybutanoic acid
GHB
22.60
enzyme
CH2OH
CH2
O H
O
O
CO2H
H A
aspirin
O
O
OH
CO2H
OH
CO2H
CH2
A
H A
O
O
OH
CO2H
A
CH2
H O
O
HA
CH2O C
CH3
inactive enzyme
A
O H
CH2O C
OH
O
CO2H
OH
CO2H
+
CH3
salicylic acid
A
631
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–23
22.61
O
Ar
Ar
F
O
RO
O
Ar'
N
CH3CH3
Ar
RO
N
O
MgBr+
Ar'
N
OR
Ar'
H
CH3CH2 MgBr
O
N
RO–
O
F
22.62
CH3O
CH3O
O
C
O
O
OH
OH
O
O
proton
O
OH
O
O
OCH3
transfer
HO
OCH3
CO2H
HO
OCH3
D
H OH2
632
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–24
22.63
H OCH2CH3
O
O CH2CH3
OH
O
O
H–Cl
A
Cl
H
OCH2CH3
OH
O
OH
O
Cl
H–Cl
OCH2CH3
OH2
OCH2CH3
O
OH
OCH2CH3
O
OH
O
OCH2CH3
OH
O
H
Cl
H
H–Cl
Cl
Cl
CO2CH2CH3
OCH2CH3
OCH2CH3
O
Cl
+
Cl
Cl
=
O
H2O
B
22.64 The mechanism is composed of two parts: hydrolysis of the acetal and intramolecular Fischer
esterification of the hydroxy carboxylic acid.
H2O
H
O
O
OH
H OH2
O
O
H2O
O
HO
O
O
OH
O
O
H2O
H
O
O
OH
OH
O
+ H2O
HO
H
HO HO
H OH2
O
HO
HO
HO HO
O
OH
O
OH
HO
OH
O
O
+ H3O+
+ H2O
H
H OH2
H2O
HO
H
O
OH
OH
OH
HO
HO
O
OH
H OH2
OH
OH
HO
+ H 2O
O
O
O
O
HO
O H
HO
HO
+ H3O+
H2O
OH2
OH
+ H 2O
633
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–25
22.65
O
H3O+
HO
Ha
CN
A
C6H10O2: 2 degrees of unsaturation
O
IR: 1770 cm–1 from ester C=O in a five-membered ring
NMR: Ha = 1.27 ppm (singlet, 6 H) – 2 CH3 groups
Hb = 2.12 ppm (triplet, 2 H) – CH2 bonded to CH2
Hc = 4.26 ppm (triplet, 2 H) – CH2 bonded to CH2
1H
Hb Hc
A
proton
C N
HO
HO
C N
OH2
HO
transfer
C NH
H OH2
OH
imidic acid
H2O
C NH2
HO
C NH2
HO
OH
C NH2
HO
O
H OH2
O H
amide
H2O
H2O
H2O
HO
NH2
HO
O H
NH2
H O
H OH2
O
H O
NH3
O
O
NH3
O
H3O
H2O
H2O
O
22.66 Fischer esterification is the treatment of a carboxylic acid with an alcohol in the presence of
an acid catalyst to form an ester.
a. (CH3)3CCO2CH2CH3
(CH3)3CCOOH
O
c.
+ HOCH2CH3
b.
O
OH
HO
O
OH
O
O
O
d.
HO
OH
HO
O
O
O
22.67
a.
NH3
Br excess
NH2
H3O+
NaCN
Br
H
N
OH
NH2
CN
O
DCC
O
634
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–26
[1]
b.
(from a.)
CN
MgBr
Mg
[2] H2O
Br
MgBr
H2SO4
O
O
O
CrO3
NaOH
c.
OH
Br
OH
H2SO4, H2O
OH
O
(from b.)
[1]
O
d.
MgBr
OH
(2 equiv)
[2] H2O
O
(from c.)
22.68
a.
Br
b.
COOH
(from a.)
c.
–CN
CN
H2O, H+
HOCH2CH3
(from a.)
C
[2] H2O
CN
(from a.)
COCl
CH3
O
[1] LiAlH4
d.
SOCl2
CO2CH2CH3
H2SO4
[1] CH3MgBr
CN
COOH
CH2NH2
[2] H2O
CH3COCl
CH2NHCOCH3
22.69
a. CH3Cl
+ NaCN
CH3 CN
H3 O +
CH3 COOH
[1] CO2
CH3 Cl + Mg
Br
+ NaCN
b.
sp2
CH3 MgCl
MgBr
+ Mg
(CH3)3C
Cl + Mg
d. HOCH2CH2CH2CH2Br
CH3 COOH
This method can't be used because an SN2 reaction
can't be done on an sp2 hybridized C.
Br
c. (CH3)3CCl + NaCN
[2] H3O+
COOH
[1] CO2
[2] H3
O+
This method can't be used because an SN2 reaction
can't be done on a 3° C.
(CH3)3C
+ NaCN
HOCH2CH2CH2CH2 Br + Mg
MgCl
[1] CO2
[2] H3O+
(CH3)3C
HOCH2CH2CH2CH2 CN
COOH
H3O+
HOCH2CH2CH2CH2 COOH
This method can't be used because you can't make
a Grignard reagent with an acidic OH group.
635
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–27
22.70
CH3OH
O
SOCl2
CH3
CH3Cl
CH3
HNO3
H2SO4
AlCl3
COOH
KMnO4
O2N
C
CH3CH2OH
H2SO4
O 2N
OCH2CH3
O2N
H2
Pd-C
(+ ortho isomer)
O
C
OCH2CH3
H2N
22.71
O
NH2
HO
HO
CH3COCl
N
H
N
H
NaH
O
N
H
SOCl2
serotonin
O
N
H
N
H
CH3Cl
CH3COOH
SOCl2
CH3OH
O
CrO3
H2SO4, H2O
N
H
CH3O
CH3CH2OH
N
H
melatonin
22.72
O
O
CH3CH2Cl
a.
AlCl3
OH
COH
KMnO4
NO2
H2, Pd-C
HO
HO
(+ ortho isomer)
c.
HO
H
N
C
CH3
O
acetaminophen
(from b.)
H
N
NaH
O
OH
salicylamide
NH2
H
N
CH3COCl
H2SO4
OH
CNH2
NH3
OH
OH
(+ para isomer)
b.
Cl
(2 equiv)
OH
HNO3
SOCl2
O
C
CH3
acetaminophen
H
N
CH3CH2Br
O
CH3CH2O
CH3
O
HO
(More nucleophilic
NH2 reacts first.)
C
C
CH3
O
p-acetophenetidin
636
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–28
22.73
HO
[1] CrO3, H2SO4, H2O
[2] SOCl2
O
a.
O
O
Cl
Cl2
Cl
AlCl3
FeCl3
CH3OH
O
SOCl2
b.
CH3Cl
CH3
AlCl3
CH3
HNO3
H2SO4
COOH
KMnO4
O2N
C
CH3OH
H+
O2N
O2N
(+ ortho isomer)
H2, Pd-C
O
Br
FeBr3
H2SO4
Cl
[1] CrO3, H2SO4,
H2O
[2] SOCl2
Br
HNO3
C
OCH3
N
H
Br2
O
O
C
O
c.
OCH3
O2N
(+ ortho isomer)
H2N
CH3CH2CH2OH
O
Br
Cl
H2, Pd-C
Br
[1] CrO3, H2SO4,
H2O
[2] SOCl2
H2 N
OCH3
H
N
O
CH3CH2OH
(CH3)3COH
CH3OH
HCl
d.
SOCl2
(CH3)3CCl
CH3Cl
AlCl3
KMnO4
SOCl2
HO
AlCl3
(+ ortho isomer)
Br2
FeBr3
Br
CH3Cl
Br
Cl
O
O
Br
KMnO4
HO
AlCl3
(+ ortho isomer)
Br
NaOH
O
O
O
O
O
O
(CH3)3C
Br
637
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–29
22.74
CH3Cl
a.
CH3OH
Br2
CH3
AlCl3
SOCl2
NaCN
CH2Br
hν
CH2CN
H2O
H+
CH2COOH
HOCH2CH3
H2SO4
CH3Cl
CH2COOCH2CH3
ethyl phenylacetate
NO2
(from a.)
CH3
CH3Cl
b.
NO2
CH3
HNO3
NH2
COOH
KMnO4
H2SO4
AlCl3
NH2
COOH
H2
HOCH3
Pd-C
H2SO4
COOCH3
methyl anthranilate
(+ para isomer)
(from a.)
O
CH3 KMnO
4
CH3Cl
c.
COOH
AlCl3
CH3CH2OH
OH CH3COOH
H2SO4
[1] LiAlH4
[2] H2O
CrO3
O
benzyl acetate
CH3COOH
H2SO4, H2O
22.75
O
C
a. CH3
OH
c. CH3CH2Br
d. CH3 CH2OH
13
OCH2CH3
O
O
CrO3
CH3CH2OH
13
H2SO4, H2O
H218O
CH3
C
–
13
CH3 CH2Br
OCH2CH3
O
CH3COCl
CH3
C
18
OCH2CH3
18
18
OH
13C
CH3
H2SO4
OH
CH3CH218OH
+ base (H18O–)
PBr3
C
CH3
H2SO4
b. CH313CH2OH
13
O
CH313CH2OH
CH3
13
CH218OH
CrO3
H2SO4, H218O
18
O
13 C 18
OH
CH3
CH3CH2OH
H2SO4
O
13 C
CH3
+
OCH2CH3
H218O
22.76
a. HO
OH
and
HO
C
C
O
O
O
O
O
OH
O
O
O
O
O
O
NH
b. ClOC
COCl
and
H2N
NH2
O
NH
HN
HN
O
O
638
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–30
22.77
O
O
a.
O
O
O
O
O
O
O O
O O
b.
O
OH
HO
O
HO
OH
O
O O
O
O
O
HO
OH
HO
OH
22.78
a. Docetaxel has fewer C’s and one more OH group than taxol. This makes docetaxel more
water soluble than taxol.
OH
O
b.
O
O
O
OH
O
N
O
H
carbamate
=
OH
docetaxel
HO
O
O
RO
NHR
carbamate
H
O
O
O
O
RO
O
NHR
RO
1
most stable
O
NHR
O
RO
4
least stable
NHR
RO
3
Increasing stability: 4 < 3 < 2 < 1
O
c.
H OH2
OH
O
O
NHR
O
NHR
H2O
NHR
2
More basic N atom allows N to
donate electron density more
than O, so this structure
contributes more than 3 to the
hybrid.
H
O
C
H
H2O
NHR
H2O
H OH2
O
C
+ NH2R
H OH2
O
H3O
H3NR
H2O
639
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–31
OH
O
d.
O
O
OH
O
N
O
H
H3O+
OH
HO
O
docetaxel
O
COOH
O
O
O
CO2
H
OH
H3N
OH
OH
O
O
OH
CH3CO2H
HO
HO
H
HO
HO
O
22.79
O
a. CH3
C
O
and
OCH3
CH3CH2
C
c.
OH
contains a broad, strong OH
absorption at 3500–2500 cm–1
O
N(CH3)2
and
NH2
2 NH absorptions at 3200–3400 cm–1
C=O at < 1700 cm
due to the stabilized amide C=O absorption at higher wavenumber
–1
O
O
and
b.
O
O
d. O
Cl
OH
O
and
CH3
Acid chloride CO absorbs at
much higher wavenumber.
Cl
OH absorption at
3500–3200 cm–1 + C=O
ketone
only C=O
22.80
O
a. C6H5COOCH2CH3
CH3CH2COOCH2CH3
O
b.
most resonance
stabilized
CH3CONH2
Increasing wavenumber
least resonance
stabilized
CH3COOCH3
CH3COCl
Increasing wavenumber
22.81
a.
C6H12O2 → one degree of unsaturation
IR: 1738 cm–1 → C=O
NMR: 1.12 (triplet, 3 H), 1.23 (doublet, 6 H),
2.28 (quartet, 2 H), 5.00 (septet, 1 H) ppm
O
O
O
b.
c. C8H9NO
IR: 3328 (NH), 1639 (conjugated amide C=O) cm–1
NMR: 2.95 (singlet, 3 H), 6.95 (singlet, 1 H),
7.3–7.7 (multiplet, 5 H) ppm
C4H7N
IR: 2250 cm–1 → triple bond
NMR: 1.08 (triplet, 3 H), 1.70 (multiplet, 2 H),
2.34 (triplet, 2 H) ppm
CH3CH2CH2C N
N
H
CH3
d. C4H7ClO → one degree of unsaturation
IR: 1802 cm–1 → C=O (high wavenumber, RCOCl)
NMR: 0.95 (triplet, 3 H), 1.07 (multiplet, 2 H),
2.90 (triplet, 2 H) ppm
O
Cl
640
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–32
e. C5H10O2 → one degree of unsaturation
IR: 1750 cm–1 → C=O
NMR: 1.20 (doublet, 6 H), 2.00 (singlet, 3 H),
4.95 (septet, 1 H) ppm
f. C10H12O2 → five degrees of unsaturation
IR: 1740 cm–1 → C=O
NMR: 1.2 (triplet, 3 H), 2.4 (quartet, 2 H),
5.1 (singlet, 2 H), 7.1–7.5 (multiplet, 5 H) ppm
O
O
O
O
g. C8H14O3 → two degrees of unsaturation
IR: 1810, 1770 cm–1 → 2 absorptions due to
C=O (anhydride)
NMR: 1.25 (doublet, 12 H), 2.65 (septet, 2 H) ppm
O
O
O
22.82
A. Molecular formula C10H12O2 → five degrees of unsaturation
IR absorption at 1718 cm–1 → C=O
NMR data (ppm):
triplet at 1.4 (CH3 adjacent to 2 H's)
singlet at 2.4 (CH3)
quartet at 4.4 (CH2 adjacent to CH3)
doublet at 7.2 (2 H's on benzene ring)
doublet at 7.9 (2 H's on benzene ring)
O
O
B. IR absorption at 1740 cm–1 → C=O
NMR data (ppm):
singlet at 2.0 (CH3)
triplet at 2.9 (CH2 adjacent to CH2)
triplet at 4.4 (CH2 adjacent to CH2)
multiplet at 7.3 (5 H's, monosubstituted benzene)
O
O
22.83
Molecular formula C10H13NO2 → five degrees of unsaturation
IR absorptions at 3300 (NH) and 1680 (C=O, amide or conjugated) cm–1
NMR data (ppm):
triplet at 1.4 (CH3 adjacent to CH2)
singlet at 2.2 (CH3C=O)
quartet at 3.9 (CH2 adjacent to CH3)
CH3CH2O
doublet at 6.8 (2 H's on benzene ring)
singlet at 7.2 (NH)
doublet at 7.4 (2 H's on benzene ring)
H
N
CH3
O
phenacetin
22.84
Molecular formula C11H15NO2 → five degrees of unsaturation
IR absorption 1699 (C=O, amide or conjugated) cm–1
NMR data (ppm):
triplet at 1.3 (3 H) (CH3 adjacent to CH2)
singlet at 3.0 (6 H) (2 CH3 groups on N)
quartet at 4.3 (2 H) (CH2 adjacent to CH3)
doublet at 6.6 (2 H) (2 H's on benzene ring)
doublet at 7.9 (2 H) (2 H's on benzene ring)
CH3
O
N
CH3
OCH2CH3
C
641
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–33
22.85
a. Molecular formula C6H12O2 → one degree of unsaturation
IR absorption at 1743 cm–1 → C=O
1
O
H NMR data (ppm):
triplet at 0.9 (3 H) – CH 3 adjacent to CH2
O
multiplet at 1.35 (2 H) – CH2
D
multiplet at 1.60 (2 H) – CH2
singlet at 2.1 (3 H – from CH3 bonded to C=O)
triplet at 4.1 (2 H) – CH 2 adjacent to the electronegative O atom and another CH 2
b. Molecular formula C6H12O2 → one degree of unsaturation
IR absorption at 1746 cm–1 → C=O
O
1
H NMR data (ppm):
doublet at 0.9 (6 H) – 2 CH3's adjacent to CH
O
E
multiplet at 1.9 (1 H)
singlet at 2.1 (3 H) – CH3 bonded to C=O
doublet at 3.85 (2 H) – CH2 bonded to electronegative O and CH
22.86 The extent of resonance stabilization affects the position of the C=O absorption in the IR of
an amide.
N
A
N
O
O
NH
NH
O
O
This resonance structure does not contribute significantly
to the hybrid because it places a double bond at the
bridgehead N, an impossible geometry in small rings.
The C=O has more double bond character, so the
absorption is at a higher wavenumber.
This resonance structure contributes significantly
to the hybrid, so the C=O has more single bond
character. The absorption shifts to a lower
wavenumber.
B
22.87
O
O
ClCH2 C
ClCH2 C
NH2
N H
4.02 ppm
H
There is restricted rotation around the
amide C–N bond. The 2 H's are in
different environments (one is cis to an O
atom, and one is cis to CH2Cl), so they
give different NMR signals.
different
environments
7.35 and 7.60 ppm
This resonance structure makes a significant
contribution to the resonance hybrid.
22.88
18
18
O
C6H5
C
–OH
OCH2CH3
ethyl benzoate
18
O
C6H5 C OCH2CH3
OH
H2O
OH
C6H5 C OCH2CH3
18
–OH
OH
OH
O
C6H5 C OCH2CH3
O
C6H5
C
OCH2CH3
18
–
Two OH groups are now equivalent
and either can lose H2O to form
labeled or unlabeled ethyl benzoate.
OH
Unlabeled starting material
was recovered.
642
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–34
22.89
NH2
abbreviate as:
R
N
H
CH3O
NH2
R
H
NH2
C
RCH2CH2NH2
H C O
O
CH3OOC
CH3OOC
CH3OOC
OCOCH3
CH3OOC
OCOCH3
OCH3
OCH3
proton transfer
H
RCH2CH2N
H2O
RCH2CH2NH OH2
H C
CH3OOC
any proton CH3OOC
source
CH3OOC
CH3OOC
OCOCH3
CH3OOC
OCH3
RCH2CH2NH OH
H C
OCOCH3
CH3OOC
OCH3
OCOCH3
OCH3
R
RCH2CH2N
H BH3
RCH2CH2N
CH3O
CH3OOC
CH3OOC
OCOCH3
OCH3
H3O
CH3O
CH3O
N
N O
H
O
C
O
OCOCH3
CH3OOC
OCH3
CH3OOC
H
H
OCOCH3
OCH3
N
CH3OOC
OCOCH3
OCH3
BH3
CH3O
22.90 Both acetals are hydrolyzed by the usual mechanism for acetal hydrolysis (Steps [1 –[6]),
forming four new OH groups. Intramolecular esterification forms a lactone (Steps [10]–[15]),
followed by conversion of a carbonyl tautomer to an enol (Steps [16]–[17]). Both acetals are
hydrolyzed at once in the given mechanism.
643
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–35
H
O H
H2O
H Cl
O
OO
O
CO2H
O
O H
OO
[1]
O
OH
OO
CO2H
O
[2]
H
H Cl
CO2H
O HO
Cl
OH
OO
[3]
CO2H
O HO
O H
+ 2 Cl–
H2O
H
Cl
[4]
H
OH
OH
O
O
[7]
CO2H
HO HO
+
Cl–
Cl
O
O
[5]
CO2H
O HO
H Cl
O H
OH
O H
+ 2 HCl
Cl
+
H
O H
H Cl
CO2H
O HO
H
O
O
H
[6]
CO2H
HO HO
[8]
OH
O H
OH
OH
O
OH
OH
H Cl
(2 equiv)
OH
HO
CO2H
OH
O
+
+ 2 HCl
[9]
HO
(2 equiv)
OH
H Cl
O
O
OH
HO
OH
C +
HO
[10]
OH
OH
OH
OH
C
HO
HO
O
[12]
HO
HO
O H
O
Cl
HO
HO
HO
+OH2
O
[14]
HO
HO
H Cl
OH
O
+ H2O
O
+ Cl–
[15]
HO
HO
HO
HO
HO
O
O
O
HO
O
[17]
H
H Cl
O
H
Cl
+ Cl–
OH
O
O
[16]
OH
O
[13]
OH
O
HO
OH
O
+ Cl–
HO
Cl
OH
[11]
OH
OH
HO
H
O
HO
O
OH
vitamin C
+ HCl
644
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 22–36
22.91
O
O
N
N
O
Ph
O
O
Ar N
Ph
O
Ar N
O
Ph
O
O
H
F
R Li
O
Ar
O
Ar N
Ar N
O
O
O
O
H
+ RH +
Li+
O
O
O H
O
Ph
O
O
O
O
O O
O
Ph
O
O
Ar N
O
O
O
O
H OH
Ph
Ph
O
O
Ar N
Ar N
O
+ –OH
OH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
645
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–1
Chapter 23 Substitution Reactions of Carbonyl Compounds at the α Carbon
Chapter Review
Kinetic versus thermodynamic enolates (23.4)
O
R
kinetic enolate
Kinetic enolate
• The less substituted enolate
• Favored by strong base, polar aprotic solvent, low temperature:
LDA, THF, –78 oC
O
R
thermodynamic
enolate
Thermodynamic enolate
• The more substituted enolate
• Favored by strong base, protic solvent, higher temperature:
NaOCH2CH3, CH3CH2OH, room temperature
Halogenation at the α carbon
[1] Halogenation in acid (23.7A)
O
R
C
O
C
H
X2
R
CH3COOH
C
X
C
•
•
The reaction occurs via enol intermediates.
Monosubstitution of X for H occurs on the
α carbon.
•
The reaction occurs via enolate
intermediates.
Polysubstitution of X for H occurs on the
α carbon.
α-halo aldehyde
or ketone
X2 = Cl2, Br2, or I2
[2] Halogenation in base (23.7B)
O
R
C
C
R
O
X2 (excess)
–OH
R
C
C
R
•
X X
H H
X2 = Cl2, Br2, or I2
[3] Halogenation of methyl ketones in base—The haloform reaction (23.7B)
O
R
C
CH3
–OH
X2 = Cl2, Br2, or I2
•
O
X2 (excess)
R
C
O–
+ HCX3
haloform
The reaction occurs with methyl ketones
and results in cleavage of a carbon–carbon
σ bond.
646
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–2
Reactions of α-halo carbonyl compounds (23.7C)
[1] Elimination to form α,β-unsaturated carbonyl compounds
O
R
α
Li2CO3
Br
LiBr
DMF
O
R
•
Elimination of the elements of Br and H forms
a new π bond, giving an α,β-unsaturated
carbonyl compound.
•
The reaction follows an SN2 mechanism,
generating an α-substituted carbonyl
compound.
β
α
[2] Nucleophilic substitution
O
O
R
Nu
α Br
α Nu
R
Alkylation reactions at the α carbon
[1] Direct alkylation at the α carbon (23.8)
O
•
O
C α H
C
[1] Base
[2] RX
C α R
C
X–
+
•
•
The reaction forms a new C–C bond to the
α carbon.
LDA is a common base used to form an
intermediate enolate.
The alkylation in Step [2] follows an SN2
mechanism.
[2] Malonic ester synthesis (23.9)
[1] NaOEt
α
[2] RX
[3] H3O+, Δ
H
•
R CH2COOH
•
H Cα COOEt
COOEt
diethyl malonate
[1] NaOEt
[1] NaOEt
[2] RX
[2] R'X
[3] H3O+, Δ
α
The reaction is used to prepare
carboxylic acids with one or two alkyl
groups on the α carbon.
The alkylation in Step [2] follows an
SN2 mechanism.
R CHCOOH
R'
[3] Acetoacetic ester synthesis (23.10)
O
CH3
C α H
C H
[1] NaOEt
[2] RX
[3] H3O+, Δ
•
O
CH3
C
CH2 R
•
COOEt
ethyl acetoacetate
O
[1] NaOEt
[2] RX
[1] NaOEt
[2] R'X
[3] H3O+, Δ
CH3
C
CH R
R'
The reaction is used to prepare
ketones with one or two alkyl
groups on the α carbon.
The alkylation in Step [2]
follows an SN2 mechanism.
647
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–3
Practice Test on Chapter Review
1. a. Which of the following compounds can be prepared using either the acetoacetic ester or malonic
ester syntheses?
COOH
O
1.
O
3.
2.
4. Both [1] and [2] can be prepared by one of these routes.
5. Compounds [1], [2], and [3] can all be prepared by these routes.
b. Which of the following compounds is not an enol form of dicarbonyl compound A?
O
O
OH
O
OH
1.
O
O
OH
3.
2.
A
5. Compounds [1][4] are all enols of A.
2. a. Which proton in compound A has the lowest pKa?
b. Which proton in compound A has the highest pKa?
Hb
Hc
Hd
O
Ha
A
c. What proton in compound B is the least acidic?
d. What proton in compound B is the most acidic?
Hb
O
O
Hd H
c
O
Ha
B
3. What reagents are needed to convert 4-heptanone to each compound?
a. 3-bromo-4-heptanone
b. 3-methyl-4-heptanone
c. 3,5-dimethyl-4-heptanone
d. 2-hepten-4-one
e. 2-methyl-4-heptanone
O
4.
OH
648
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–4
4. Draw the organic products formed in each of the following reactions.
O
I2 (excess)
a.
O
d.
O
[1] NaOEt
OEt
–OH
[2] CH3CH2CH2CH2Br
[3] H3O+, Δ
O
CO2Et
O
[2] CH3CH2Br
[3] H3O+, Δ
b.
c.
[1] LDA
CH2(CO2Et)2
[1] LDA
e.
[2] CH3CH2Br
[1] NaOEt
[1] NaOEt
[2] CH3CH2Br
[2] C6H5CH2Cl
[3] H3O+, Δ
5. a. What starting materials are needed to synthesize carboxylic acid A by a malonic ester
synthesis?
O
OH
A
OCH3
b. What starting materials are needed to synthesize ketone B by the acetoacetic ester synthesis?
O
B
Answers to Practice Test
1. a. 1
b. 4
2. a. Hb
b. Hd
c. Hb
d. Hc
3. a. Br2, CH3CO2H
b. [1] LDA; [2] CH3I
c. []1 LDA; [2] CH3I;
[3] LDA; [4] CH3I
d. [1] Br2, CH3CO2H;
[2] Li2CO3, LiBr, DMF
e. [1] Br2, CH3CO2H;
[2] Li2CO3, LiBr, DMF;
[3] (CH3)2CuLi; [4] H2O
O
4.
CO2
a.
5.
a.
d.
+ HCI3
CH3O
O
CH2(CO2Et)2
O
e.
b.
Br
Br
Br
b.
O
c.
Br
OH
O
C6H5
CO2Et
649
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–5
Answers to Problems
23.1 • To convert a ketone to its enol tautomer, change the C=O to C–OH, make a new double bond
to an α carbon, and remove a proton at the other end of the C=C.
• To convert an enol to its keto form, find the C=C bonded to the OH. Change the C–OH to a
C=O, add a proton to the other end of the C=C, and delete the double bond.
[In cases where E and Z isomers are possible, only one stereoisomer is drawn.]
HO
a.
O
OH
O
b.
O
d.
C6H5
OH
C6H5
H
C6H5
H
c.
C6H5
O
O
O
OH
O
e.
O
O
O
OH
OH
C6H5
f.
C6H5
OH
C6H5
Draw mono enol
tautomers only.
O
OH
OH
O
(Conjugated enols are preferred.)
23.2
The mechanism has two steps: protonation followed by deprotonation.
OH
OH
O H
H2O
H
H
H H OH2
O
H
H
H
H
H 3O +
23.3
Cl
H
H
H
O
O D
D Cl
D Cl
H
O D
Cl
D H
D
D Cl
H
H
O D
O D
H Cl
H Cl
D D
D D
H
H
D Cl
O
O D
Cl
650
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–6
23.4
a.
CH3CH2O
O
O
O
O
O
C
C
C
C
C
C
OCH2CH3
CH3CH2O
C
H
b.
CH3
OCH2CH3
CH3CH2O
H
O
O
O
O
C
C
C
C
C
OCH2CH3
CH3
C
c.
CH3
O
C
C
CH3
CHC N
C
OCH2CH3
OCH2CH3
CH3
O
C
C
H
O
C
H
O
CH
O
OCH2CH3
H
O
CH3
CHC N
C
C C N
H
23.5
The indicated H’s are α to a C=O or C≡N group, making them more acidic because their
removal forms conjugate bases that are resonance stabilized.
O
a.
CH3
C
O
b.
CH2CH2CH3
c.
CH3CH2CH2 CN
CH3CH2CH2
C
d. O
OCH2CH3
O
CH3
23.6
O
O
O
–H+
O
O
–H+
O
O
–H+
O
O
O
no resonance stabilization
least acidic
23.7
O
O
O
O
O
O
O
O
Two resonance structures
stabilize the conjugate base.
intermediate acidity
Three resonance structures
stabilize the conjugate base.
most acidic
In each of the reactions, the LDA pulls off the most acidic proton.
O
O
O
LDA
a.
c.
THF
CHO
b.
LDA
THF
O
LDA
CH3 C
OCH2CH3
CHO
CN
d.
THF
LDA
THF
CH2 C
OCH2CH3
CN
651
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–7
23.8
the gas
O
O
O
O
O
CH4
OCH2CH3
H3
OCH2CH3
O
O+
OCH2CH3
+ MgBr+
H
CH3–MgBr
The CH2 between the two C=O’s contains acidic H’s, so CH3MgBr reacts as a base to
remove a proton. Thus, proton transfer (not nucleophilic addition) occurs.
23.9
In addition to being strong bases, organolithiums are good nucleophiles that can add to a
carbonyl group instead of pulling off a proton to generate an enolate.
23.10 • LDA, THF forms the kinetic enolate by removing a proton from the less substituted C.
• Treatment with NaOCH3, CH3OH forms the thermodynamic enolate by removing a proton
from the more substituted C.
O
O
LDA, THF
O
a.
O
O
NaOCH3
LDA, THF
O
c.
CH3OH
NaOCH3
O
CH3OH
LDA, THF
O
b.
O
NaOCH3
CH3OH
23.11
a. This acidic H is removed with base to form an achiral enolate.
O
O
H
CH3
NaOH
(2R)-2-methylcyclohexanone
b.
O
α
O
H2O
NaOH
H
CH3
CH3
H
H
achiral
Protonation of the planar achiral enolate occurs with
equal probability from two sides, so a racemic mixture is
formed. The racemic mixture is optically inactive.
O
α
O
CH3
O
H
CH3
or
O
H
CH3
H2O
α
α
H
CH3
(3R)-3-methylcyclohexanone
This stereogenic center is not located at the α carbon,
so it is not deprotonated with base. Its configuration is retained in
the product, and the product remains optically active.
652
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–8
23.12
O
Cl
a.
O
H2O, HCl
O
b.
CH3CH2CH2
C
Br
Br2, CH3CO2H
Cl2
O
O
c.
O
Br2
H
CH3CO2H
CH3CH2CH
C
H
Br
23.13
O
O
O
Br
Br
–
Br2, OH
a.
O
I2, –OH
b.
I
I
O
I
I
O
I2, –OH
c.
O
+ HCI3
23.14
O
O
Br
O
O
Li2CO3
LiBr
DMF
a.
CH3SH
c.
Br
O
SCH3
O
CH3CH2NH2
b.
Br
NHCH2CH3
23.15 Bromination takes place on the α carbon to the carbonyl, followed by an SN2 reaction with
the nitrogen nucleophile.
CH3
Br
O
O
O
N
O
O
O
O
NHCH3
Br2
LSD
CH3CO2H
N
N
COC6H5
(Section 18.5)
N
COC6H5
M
COC6H5
23.16
O
a. CH3CH2
C
CH2CH3
[2] CH3CH2I
O
O
[1] LDA, THF
CH3CH2
C
CHCH3
c.
CH2CH3
O
b.
O
[1] LDA, THF
O
O
[2] CH3CH2I
O
[1] LDA, THF
[1] LDA, THF
d.
[2] CH3CH2I
CH2CH2CN
[2] CH3CH2I
CH2CHCN
CH2CH3
653
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–9
23.17
O
O
[1] LDA, THF
a.
[2] CH3I
O
O
O
+
[1] LDA, THF
b.
[2] CH3I
COOCH3
c.
[1] LDA, THF
C
[2] CH3I
O
CH3 +
C
O
O
CH3
O
23.18 Three steps are needed: [1] formation of an enolate; [2] alkylation; [3] hydrolysis of the ester.
CO2CH2CH3
CO2CH2CH3
[1] LDA
CH3O
CO2CH2CH3
[2] CH3I
CH3O
CH3O
A
[3] H3O+
The product is racemic because the new
stereogenic center is formed by
alkylation of a planar enolate with equal
probability from above and below. CH3O
CO2H
naproxen
23.19
LDA
O
a.
O
O
CH3CH2Br
THF
O
KOC(CH3)3
b.
O
O
CH3Br
(CH3)3COH
O
(CH3)3COH
O
O
CH3I
LDA
c.
(from a.)
O
d.
THF
LDA
THF
(from b.)
KOC(CH3)3
O
O
CH3Br
O
CH3Br
O
654
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–10
23.20
LDA
THF
O
O
CH3I
O
CH2
Br
Br2
O
O
O
A
Li2CO3
LiBr
DMF
O
CH3CO2H
O
C
B
O
O
α-methyleneγ-butyrolactone
23.21 Decarboxylation occurs only when a carboxy group is bonded to the α C of another carbonyl
group.
O
COOH
a.
COOH
α
β
b.
COOH
O
α
c.
COOH
α
β
d.
COOH
COOH
α
YES
NO
YES
NO
23.22
H3O+
[1] NaOEt
a. CH2(CO2Et)2
[2]
Br
CH2 CH2COOH
Δ
b. CH2(CO2Et)2 [1] NaOEt
[1] NaOEt
H3O+
[2] CH3Br
[2] CH3Br
Δ
CH3
CH3 C COOH
H
23.23
COOH
a.
Cl
b.
Cl
Br
O
O
Br
COOH
23.24 Locate the α C to the COOH group, and identify all of the alkyl groups bonded to it. These
groups are from alkyl halides, and the remainder of the molecule is from diethyl malonate.
a. (CH3)2CHCH2CH2CH2CH2 CH2COOH
α
H3O+
[1] NaOEt
CH2(CO2Et)2
[2] (CH3)2CHCH2CH2CH2CH2Br
b.
(CH3)2CHCH2CH2CH2CH2CH2COOH
Δ
COOH
α
H3O+
[1] NaOEt
[1] NaOEt
[2] CH3CH2CH2Br
[2] CH3CH2CH(CH3)CH2CH2Br
CH2(CO2Et)2
c.
Δ
(CH3CH2CH2CH2)2 CHCOOH
α
[1] NaOEt
[1] NaOEt
[2] CH3CH2CH2CH2Br
[2] CH3CH2CH2CH2Br
CH2(CO2Et)2
H3O+
Δ
(CH3CH2CH2CH2)2CHCOOH
COOH
655
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–11
23.25 The reaction works best when the alkyl halide is 1° or CH3X, since this is an SN2 reaction.
(CH3)3C–CH2COOH
a.
α
COOH
b.
(CH3)3CX
3° alkyl halide
(too crowded)
c.
X
aryl halide
(leaving group on an
sp2 hybridized C)
Aryl halides are unreactive
in SN2 reactions.
(CH3)3C–COOH
This compound has 3 CH3 groups on the α
carbon to the COOH. The malonic ester
synthesis can be used to prepare mono- and
disubstituted carboxylic acids only: RCH2COOH
and R2CHCOOH, but not R3CCOOH.
23.26
O
O
a.
CH3
C
[1] NaOEt
CH2CO2Et
CH3
[2] CH3I
[3] H3O+, Δ
C
O
O
b.
CH2CH3
CH3
C
[1] NaOEt
CH2CO2Et
CH3
[2] CH3CH2CH2Br
[3] NaOEt
[4] C6H5CH2I
[5] H3O+, Δ
C
CH CH2
CH2CH2CH3
23.27 Locate the α C. All alkyl groups on the α C come from alkyl halides, and the remainder of
the molecule comes from ethyl acetoacetate.
O
a.
C
CH3
O
CH2 CH2CH3
CH3
C
CH2
COOEt
O
H3O+
[1] NaOEt
Δ
[2] CH3CH2Br
CH3
C
CH2CH2CH3
α
O
O
b. CH3
C
CH3
CH(CH2CH3)2
α
α
CH2
COOEt
[1] NaOEt
[2] CH3CH2Br
[2] CH3CH2Br
O
CH3
C
CH2
[1] NaOEt
[1] NaOEt
[2] CH3CH2Br
[2] CH3(CH2)3Br
COOEt
23.28
O
CH3
C
CH2CO2Et
O
H3O+
Δ
CH3
C
CH(CH2CH3)2
O
O
c.
C
[1] NaOEt
+
Br
Br
O
NaOEt
(2 equiv)
CH3
X
CO2Et
H3O+
Δ
656
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–12
23.29
O
O
O
a.
CH3
C
H3
[1] NaOEt
CH2CO2Et
[2]
Δ
CO2Et
CH3O
Br
O+
CH3O
nabumetone
CH3O
O
b. CH3
C
O
[1] LDA, THF
[2]
CH3
Br
CH3O
CH3O
nabumetone
23.30 Use the directions from Answer 23.1 to draw the enol tautomer(s). In cases where E and Z
isomers can form, only one isomer is drawn.
O
a.
OH
H
O
O
OH
O
OH
O
b.
H
OEt
OEt
OEt
unconjugated enol
(less stable)
O
OH
OH
OEt
OH
OEt
23.31
a.
O
(CH3)3C
NaOH
H2O
b.
O
(CH3)3C
A
(CH3)3C
B
COCH3
H
(CH3)3C
one axial and one equatorial
group, less stable
C
O
H
COCH3
(CH3)3C
Both groups are equatorial.
more stable
This isomerization will occur since it makes
a more stable compound.
(CH3)3C
COCH3
Both groups are equatorial = cis.
Compound C will not isomerize since it
already has the more stable arrangement
of substituents.
23.32 Use the directions from Answer 23.1 to draw the enol tautomer(s). In cases where E and
Z isomers can form, only one isomer is drawn.
a.
O
O
OH
conjugated enol
(more stable)
b.
O
HO
O
(mono enol form)
OH
O
OH
O
O
HO
O
c.
O
OH
d.
OH
OH
conjugated enol
(more stable)
657
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–13
23.33
O
O
O
OCH2CH3
O
OCH2CH3
ethyl acetoacetate
The ester C=O is resonance stabilized, and is therefore
less available for tautomerization. Since the carbonyl
form of the ester group is stabilized by electron
delocalization, less enol is present at equilibrium.
23.34
a. CH3CH2CH2CO2CH(CH3)2
c.
O
C
CH2CH3
e. NC
O
OH
O
b.
d. CH3O
O
f.
CH2CN
O
HOOC
23.35
CH2
a.
C
H
Hb O
CH3
O
O
O
b.
c.
CH3
Ha
H
Hc
H
HO
Ha
Hc
H
H
Hb
H
c
Hc is bonded to an sp2 hybridized C = least acidic.
Ha Hb
Ha is bonded to an α C = intermediate acidity.
Hb is bonded to an sp3 hybridized
Hc is bonded to an α C =
Hb is bonded to an α C, and is adjacent
C = least acidic.
least acidic.
to a benzene ring = most acidic.
Hc is bonded to an α C =
Ha is bonded to an α C, and is
intermediate acidity.
adjacent to a benzene ring =
Ha is bonded to O = most acidic.
intermediate acidity.
Hb is bonded to an α C between two
C=O groups = most acidic.
23.36
CN
LDA
a.
d.
THF
O
CN
LDA
THF
O
O
OCH3
b.
O
O
OCH3
THF
O
O
LDA
e.
THF
O
LDA
c.
LDA
THF
f.
LDA
O
O
THF
O
O
23.37 Enol tautomers have OH groups that give a broad OH absorption at 3600–3200 cm–1, which
could be detected readily in the IR.
658
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–14
23.38
O
O
Hb
Ha
Ha
remove Ha
O
Hb
Ha
Hb
O
Hb
Ha
Hb
Hb
O
Hb
Ha
Ha
remove Hb
O
Ha
Ha
Hb
Removal of Ha gives two
resonance structures. The
negative charge is never on O.
O
Ha
Ha
Hb
Hb
Ha
Ha
Hb
Removal of Hb gives three resonance structures. The negative
charge is on O in one resonance structure, making the
conjugate base more stable and Hb more acidic (lower pKa).
23.39
O
O
HO
O
5,5-Dimethyl-1,3-cyclohexanedione exists predominantly
in its enol form because the C=C of the enol is conjugated
with the other C=O of the dicarbonyl compound.
Conjugation stabilizes this enol.
5,5-dimethyl-1,3cyclohexanedione
O
O
HO
O
2,2-dimethyl-1,3cyclohexanedione
The enol of 2,2-dimethyl-1,3-cyclohexanedione is not
conjugated with the other carbonyl group. In this way it
resembles the enol of any other carbonyl compound,
and thus it is present in low concentration.
23.40 a.
O
N
HN
H2N
OH
N
N
N
N
OH
H2N
OH
N
N
O
O
keto form
acyclovir
enol form
In the enol form, the bicyclic ring system has four π bonds (eight π electrons) and a lone
pair on N, for a total of 10 π electrons. This makes it aromatic by Hückel’s rule.
b. The keto form of acyclovir can also be drawn in a resonance form that gives it 10 π
electrons, making it aromatic as well.
O
HN
H2N
O
N
OH
N
N
H2N
O
N
HN
OH
N
N
O
aromatic
completely conjugated
10 π electrons
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
659
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–15
23.41
O
CH3
C
O
CH3
O
CH2
C
O
O
CH3
CH2
C
O
O
CH3
CH2
C
O
CH3
ester
The O atom of the ester OR group donates electron density by a resonance effect. The
resulting resonance structure keeps a negative charge on the less electronegative C end
of the enolate. This destabilizes the resonance hybrid of the conjugate base, and makes
the α H's of the ester less acidic.
CH3
O
O
C
C
CH3
CH2
O
CH3
CH2
C
CH3
no additional
resonance structures
ketone
This structure, which places a negative charge on the O
atom, is the major contributor to the hybrid, stabilizing it, and
making the α H's of the ketone more acidic.
23.42
O
O
2,4-pentanedione
base (1 equiv)
O
O
[1] CH3I
O
O
base
O
O
(2nd equiv)
[2] H2O
A
[1] CH3I
O
O
[2] H2O
B
more nucleophilic site
With a second equivalent of base a dianion is
formed. Since the second enolate is less
resonance stabilized, it is more nucleophilic
and reacts first in an alkylation with CH3I,
forming B after protonation with H2O.
One equivalent of base removes
the most acidic proton between
the two C=O's, to form A on
alkylation with CH3I.
23.43 The mechanism of acid-catalyzed halogenation consists of two parts: tautomerization of the
carbonyl compound to the enol form and reaction of the enol with halogen.
A higher percentage of the more stable enol is present.
O
CH3COOH
OH
OH
Br2
O
O
Br
2-pentanone
C=C has
1 bond to C
C=C has
2 bonds to C
more stable
(E and Z isomers)
A
Br
B
major product formed
from the more stable enol
660
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–16
23.44 • The mechanism of acid-catalyzed halogenation [Part (a)] consists of two parts:
tautomerization of the carbonyl compound to the enol form and reaction of the enol with
halogen.
• In the haloform reaction [Part (b)], the three H’s of the CH3 group are successively replaced
by X, to form an intermediate that is oxidatively cleaved with base.
O
O
O
H
O H
O
H
H
a.
O
O H
O
H
H
O H
O
O
Br
O
H Br
O
C
b.
H
C
H H
OH
H
I
H
Br
[2]
I
C
I
C
H
+
H
+ I
O
O
O
O
H
Repeat
[1] and [2]
two times.
O
+
Br
H
+ H2O
O
O
O H
H
–
H
O
C
H
Br Br
Br
O
[1]
H
Br
O H
H
O
H
–CI
CI3
CI3
OH
3
CHI3
–OH
23.45 Use the directions from Answer 23.24.
a.
α
CH3OCH2Br
CH3OCH2 CH2COOH
α
b.
C6H5
COOH
Br
COOH
c.
and
Br
Br
C6H5
α
23.46
a. CH3CH2CH2CH2CH2 CH2COOH
[2] CH3CH2CH2CH2CH2Br
α
b.
H3O+
[1] NaOEt
CH2(CO2Et)2
COOH
[1] NaOEt
[1] NaOEt
[2] CH3CH2CH2CH2CH2Br
[2] CH3Br
CH2(CO2Et)2
Δ
CH3CH2CH2CH2CH2CH2COOH
H3O+
Δ
COOH
α
c.
COOH
α
[1] NaOEt
[1] NaOEt
[2] (CH3)2CHCH2CH2CH2CH2Br
[2] CH3Br
CH2(CO2Et)2
H3O+
Δ
COOH
661
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–17
23.47
O
O
O
O
[1] NaOEt
O
[2] CH3CH2CH2Br
O
O
[1] NaOEt
O
O
O
O
[2] CH3CH2CH2Br
O
H3 O +
Δ
O
H
O
valproic acid
23.48
H 3O +
[1] NaOEt
a.
CH2(CO2Et)2
[2] BrCH2CH2CH2CH2CH2Br
[3] NaOEt
b.
COOH
CH2OH
[2] H2O
(from a.)
c.
[1] LiAlH4
COOH
Δ
COOH
CH3OH
CH3
[1] CH3MgBr (2 equiv)
CO2CH3
H2SO4
C OH
[2] H2O
CH3
(from a.)
COOH
d.
CH3CH2OH
H2SO4
COOCH2CH3
CH3
[1] LDA
COOCH2CH3
[2] CH3I
(from a.)
23.49
a.
O
CH3
[1] Na+ –CH(COOEt)2 HO
[2] H2O
CH3CHCH2 CH(COOEt)2
O
c.
CH3
C
O
[1] Na+ –CH(COOEt)2
C
[2] H2O
CH3
CH(COOEt)2
Cl
nucleophilic attack here
b.
CH2 O
[1] Na+ –CH(COOEt)2
[2] H2O
HOCH2CH(COOEt)2
d.
CH3
O
O
C
C
O
CH3
[1] Na+ –CH(COOEt)2
[2] H2O
CH
O
C
3
CH(COOEt)2
+ CH3COOH
23.50 Use the directions from Answer 23.27.
O
α
a.
CH3
O
b.
O
α
C
[1] NaOEt
CH2
COOEt
O
CH3
C
CH2
[2] Br
H3O+
Δ
[1] NaOEt
[1] NaOEt
[2] CH3CH2Br
[2] Br
COOEt
O
H3O+
Δ
O
662
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–18
α
O
O
c.
CH3
O
H3O+
[1] NaOEt
C
CH2
COOEt
O
Δ
[2] Br
α
O
d.
C
CH3
[1] NaOEt
CH2
COOEt
O
H3O+
[2]
Δ
Br
Br
[3] NaOEt
23.51
O
a.
CH3
[1] NaOEt
C
COOEt
CH2
b.
CH3
C
Δ
CH3
[1] NaOEt
[1] NaOEt
H3O+
[2] CH3Br
[2] CH3Br
[2] CH3CH2Br
O
CH2
COOEt
O
c.
CH3
C
O
H3O+
THF
CH(CH3)2
C
CH2
CH2CH2CH3
O
Δ
O
LDA
C
CH3
C
CH(CH3)2
O
CH3I
CH3CH2
CH(CH3)2
C
CH(CH3)2
(from b.)
O
CH3
d.
C
CH(CH3)2
O
KOC(CH3)3
(CH3)3COH
C
CH3
O
CH3I
C(CH3)2
CH3
C
C(CH3)3
(from b.)
23.52
O
O
O
Li2CO3
a.
LiBr
DMF
Br
O
f.
[2] Li2CO3, LiBr, DMF
O
COOH
b.
(E and Z)
O
[1] Br2, CH3CO2H
O
Δ
O
I2 (excess)
g.
O
+ CHI3
–OH
COOH
c. CH3CH2CH2CO2Et
[1] LDA
COOH
CH3CH2
CH3CH2CHCO2Et
h.
Cl
NaH
CN
C N
[2] CH3CH2I
O
O
Br
d.
O
O
NHCH(CH3)2
(CH3)2CHNH2
O
Br2 (excess)
i.
–OH
O
O
Br Br
O
[1] LDA
e.
[2] CH3CH2I
NaI
j.
Cl
I
663
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–19
23.53
O
O
O
O
O
[1] LDA
[1] LDA
a.
c.
Cl
[2]
[2] CH3I
H
H
O
D H
O
[1] LDA
C
b.
CH3
H D
[2]
C
CH3
I
23.54
CHO
a.
OH
NaBH4
CH3OH
[1] PBr3
CN
[2] CH3I
A
p-isobutylbenzaldehyde
CN
[1] LDA
[2] NaCN
B
C
H3O+
Δ
COOH
ibuprofen
COOH
b.
COO
[1] LDA
removal of the
most acidic H
D
[2] CH3I
COOCH3
substitution
reaction
+ I–
Removal of the most acidic proton with LDA forms a carboxylate anion that reacts as a
nucleophile with CH3I to form an ester as substitution product.
23.55
CO2CH3
CO2CH3
Cl
a.
NH
+
sp2
Cl on
hybridized
C does not react.
Cl on sp3
hybridized C reacts.
H COOH
H2SO4
Cl
A
pyridine
Cl
B
Cl
clopidogrel
(racemic)
H COOCH3
TsCl
HO
S
The N atom acts as a
nucleophile to displace Cl–.
H COOCH3
CH3OH
b. HO
K2CO3
S
Cl
N
NH
S isomer
S
TsO
Cl
C
inversion of
configuration
SN2
CO2CH3
N
S
Cl
clopidogrel
(single enantiomer)
664
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–20
23.56
O
O
O
O
LDA
KOC(CH3)3
THF
Br
Br
–78 oC
A
B
(CH3)3COH
room temperature
A
C
23.57
O
O
O
α
a.
Δ
β
+
CO2
c.
O
[1] LDA
H
COOH
[2] CH3CH2I
In order for decarboxylation to occur readily, the COOH
group must be bonded to the α C of another carbonyl
group. In this case, it is bonded to the β carbon.
[1] NaOEt
(CH3CH2)3CCH(CO2Et)2
b. CH2(CO2Et)2
[2] (CH3CH2)3CBr
LDA removes a H from the less substituted
C, forming the kinetic enolate. This
product is from the thermodynamic enolate,
which gives substitution on the more
substituted α C.
The 3° alkyl halide is too crowded to react with the
strong nucleophile by an SN2 mechanism.
23.58 Protonation in Step [3] can occur from below (to re-form the R isomer) or from above to
form the S isomer as shown.
H A
A
H
H
O
H
O
[1]
H
O
O
H
OH
R isomer
inactive enantiomer
O
[2]
OH
H
OH
achiral enol
H A
A
[3]
H
H
O
HA
H
+
H
O
[4]
H
O
O
S isomer
active enantiomer
O
H
OH
H
A
(+ one more resonance structure)
23.59
O
O
CH3
LDA
O
O
O
I
+
I–
O
CH3
I
+
I–
H
665
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–21
23.60
O
O
O
O
O
O
O
EtO O
O
OEt
EtO
OEt
EtO
OEt
H H
H
EtO
OEt
O
O
+ HOEt
–
OEt
–
OEt
O
O
CO2Et H3O+
O
O
H
O
OEt
O
O
OEt
O
+ HOEt
23.61 Protons on the γ carbon of an α,β-unsaturated carbonyl compound are acidic because of
resonance.
There is no H on this C, so a planar
enolate cannot form and this
stereogenic center cannot change.
γ
O
α
β
γ
X
Remove the H on this γ C.
O
O
H
OH
H OH
Removal of this proton forms a
resonance-stabilized anion.
One resonance structure places
a negative charge on O.
O
O
Protonation of the
O
planar enolate can
occur from below
(to re-form starting
material X), or from
above to form Y.
H
Y
666
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–22
23.62
LDA = B
O
C6H5
Br
B
+
Br
C6H5
C6H5
O
Br
+ HB+
O
Br
+
+ HB+
C6H5 H
Br
C6H5
B
B
O
O
O
H
O
H
O
Br
O
Br
C6H5
Br
C6H5
C6H5
O
C6H5
+ HB+
C 6H 5
O
Br
Br
H C6H5
3)3
+ Br
O
C6H5
–OC(CH
This reaction occurs
with both bases [LDA
and KOC(CH3)3].
+ Br
+ HOC(CH3)3
23.63
1st new C–C bond
O
H
H
OCH2CH3
LDA
THF
–78 oC
α
Remove
proton here.
O
H
[1]
A
O
OCH2CH3
[2]
H
OCH2CH3
[3]
Cl
Cl
B
O
OCH2CH3
LDA
THF
–78 oC
Cl
Cl
[4]
O
Cl
OCH2CH3
C
2nd new C–C bond
Cl
23.64 a. Since there are two C’s bonded to the α carbon, there are two possible intramolecular
alkylation reactions.
Path [1]
Br
Br
O
O
Path [1]
O
A
Path [2]
Path [2]
Br
O
Br
O
two different halo
ketones possible
667
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–23
b.
Form both bonds to the α carbon
during acetoacetic ester synthesis.
CO2Et
O
O
[1] NaOEt
Br
CO2Et
H3O+
NaOEt
O
Br
[2]
EtO2C
O
Δ
O
Br
23.65
O
COCH3
a.
O
Cl2
Cl
NHC(CH3)3
H2NC(CH3)3
H2O, HCl
COO–
COCH3 –OH, I
2
excess
b.
COOH
H3O+
+ CHI3
O
COCH3
OH
[1] LiAlH4
[2] H2O
[1] LDA, THF
c.
[2] CH3CH2CH2Br
O
Br2
[1] LDA, THF
d.
Li2CO3
CH3COOH
[2] CH3Br
Br
23.66
O
O
O
Br
Br2
a.
OCH3
NaOCH3
CH3COOH
O
OH
O
[1] LDA, THF
b.
[1] LiAlH4
[2] CH2=CHCH2Br
c.
[1] LDA, THF
[1] LDA, THF
[2] CH3CH2Br
O
Br
(from a.)
O
[2] CH3Br
O
d.
[2] H2O
O
O
Li2CO3
LiBr
DMF
O
O
COCH3
CH3CH2Br
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
[1] (CH3CH2)2CuLi
[2] H2O
O
LiBr
DMF
668
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–24
O
O
O
e.
[1] LDA, THF
[1] LDA, THF
[2] CH3CH2Br
[2] CH3CH2Br
O
Li2CO3
CH3COOH
O
Br
LiBr
DMF
O
[1] KOC(CH3)3
f.
O
O
Br2
O
Br
Br2
Li2CO3
LiBr
DMF
CH3COOH
[2] CH3CH2Br
(from e.)
23.67
O
O
O
Cl
AlCl3
O
Cl2
Br2
FeCl3
CH3CO2H
Br
Cl
Cl
H2NC(CH3)3
O
NHC(CH3)3
Cl
bupropion
23.68
NaCN
Br
Br
[1] LDA
[2] CH3I
(4 equiv each)
NC
(2 equiv)
CN
N
HN
NC
CN
Br2
hν
N
NaH
N
CH2Br
N
N
N
N
N
NC
NC
CN
CN
anastrozole
23.69
O
a.
OH
PBr3
Br
O
O
COOEt
[1] NaOEt
COOEt
H3O+
Δ
[2]
Br
mCPBA
O
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
669
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–25
O
O
b.
O
[1] NaOEt
COOEt
[2] HC CCH2Br
H 3 O+
HOCH2CH2OH
Δ
TsOH
O
COOEt
[1] NaH
[2] CH3I
O
H3 O +
O
H2
O
O
O
CH3
Lindlar catalyst
23.70
O
O
Br2
a.
O
CH3CH2NH2
CH3COOH
NH3 (excess)
Br
CH3CH2Br
NHCH2CH3
PBr3
CH3CH2OH
O
b.
O
Li2CO3
OH
NaBH4
LiBr
DMF
Br
CH3OH
(from a.)
Br2
c.
Br
FeBr3
O
(from b.)
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
O
[1] (C6H5)2CuLi
[2] H2O
C6H5
SOCl2
d.
CH2Br
O
(from b.)
Br2
hν
CH3
CH3OH
CH3Cl
AlCl3
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
[1] (C6H5CH2)2CuLi
[2] H2O
O
O
C6H5
NaBH4
CH3OH
OH
C6H5
670
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–26
23.71
O
O
O
CH3CH2I
LDA
THF
HO
–78 oC
A
O
CH3CH2O
most acidic H
To synthesize the desired product, a protecting group is needed:
O
O
O
LDA
TBDMS–Cl
imidazole
HO
THF
TBDMSO
A
TBDMSO
CH3CH2I
O
O
(CH3CH2CH2CH2)4N+F–
HO
B
TBDMSO
23.72
CH2(COOEt)2
NaOEt
OH
PBr3
Br
–
CH(COOEt)2
CH(COOEt)2
H3O+, Δ
Cl
Y
SOCl2
COOH
O
23.73
O
a.
O
Y =
O
[1] CH3Li
W
[2] CH3I
(CH3)2CH
Ha
O
Y
Hb
C
Hc Hd
CH2CH2CH3
He
C7H14O → one degree of unsaturation
IR peak at 1713 cm–1 → C=O
1
H NMR signals at (ppm)
He: triplet at 0.8 (3 H)
Ha: doublet at 0.9 (6 H)
Hd: sextet at 1.4 (2 H)
Hc: triplet at 1.9 (2 H)
Hb: septet at 2.1 (1 H)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
671
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–27
O
O
b.
CH3 Li
O
O
O
O
O
O
+ I–
Y
CH3
I
23.74 Removal of Ha with base does not generate an anion that can delocalize onto the carbonyl O
atom, whereas removal of Hb generates an enolate that is delocalized on O.
CO2CH3
CO2CH3
CO2CH3
Delocalization of this sort can't occur by
removal of Ha, making Ha less acidic.
H
H
H
B
O
H
O
O
Ha
Hb
B
H CO2CH3
CO2CH3
H
CO2CH3
H
O
CO2CH3
H
O
H
O
O
Removal of Hb gives an anion
that is resonance stabilized so
Hb is more acidic.
Mechanism:
CO2CH3
H
OCH3
CO2CH3
CO2CH3
O
O
CO2CH3
Br
O
O
+ CH3OH
CO2CH3
O
CO2CH3
CO2CH3
HO H
+
O
B
+ Br–
CO2CH3
O
–OH
H
O
+ HB+
672
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 23–28
23.75
O
OH
OH
OH
OH
H OSO3H
1,2-shift
OH
H
H2O
HSO4
HSO4
23.76
CH3 Li
O
OCH2CH3
H OH2
O
CH3
OCH2CH3
OH
H 3 O+
OH2
OCH2CH3
CH3
+ Li+
+ H 2O
H
OH
OCH2CH3
H OH2
OH
OCH2CH3
CH3
CH3
OCH2CH3
+ H2O
X
CH3
OCH2CH3
CH3
CH3
+ H2O
+ H2O
OH
OCH2CH3
CH3
OCH2CH3
H
+ H2 O
O H
+ H 2O
O
+ H2O
H3O+
HOCH2CH3
23.77
O
O
O
O
O
O
R X
e
H
e
H NH2
or
O
H NH2
OH
H
proton
transfer
OH
O
R
+ X–
+
NH2
e
673
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Substitution Reactions of Carbonyl Compounds at the α Carbon 23–29
23.78 a. In the presence of base an achiral enolate that can be protonated from both sides is formed.
CH3
OH
N
H
front
H
OH
OH
O
OH
H
O
RO
conjugated
H2O
RO
OH
O
H
O
back
achiral
A
(–)-hyoscyamine
RO
O
racemic mixture
optically inactive
b. The enolate A formed from (–)-hyoscyamine is conjugated with the benzene ring, making
it easier to form. The enolate B formed from (–)-littorine is not conjugated, so it is less
readily formed.
CH3
CH3
N
H
H OH
O
N
OH
H
OH
O
O
(–)-littorine
O
NOT conjugated
B
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
675
Carbonyl Condensation Reactions 24–1
Chapter 24 Carbonyl Condensation Reactions
Chapter Review
The four major carbonyl condensation reactions
Reaction type
Reaction
[1] Aldol reaction
(24.1)
2
O
RCH2
OH
–OH
C
H
H2O
[2] Claisen reaction
(24.5)
RCH2
C
OR'
[1] NaOR'
O
[2] H3O+
C
RCH2
C
or
R
H
H3O+
CHO
C
R
(E and Z)
α,β-unsaturated
carbonyl compound
β-hydroxy carbonyl
compound
O
2
CHCHO
H
aldehyde
(or ketone)
RCH2
–OH
RCH2 C
O
C
CH
ester
OR'
R
β-keto ester
O
O
[3] Michael reaction
(24.8)
–OR'
+
R
–
α,β-unsaturated
carbonyl compound
[4] Robinson
annulation (24.9)
R
1,5-dicarbonyl compound
carbonyl
compound
–OH
H2O
O
O
α,β-unsaturated
carbonyl
carbonyl compound compound
2-cyclohexenone
Useful variations
[1] Directed aldol reaction (24.3)
O
O
R'CH2
C
[1] LDA
R"
R" = H or alkyl
[2] RCHO
[3] H2O
HO
O
R C CH C
H R'
O
H2O
OH
+
O
O
or
R"
β-hydroxy carbonyl
compound
–
OH
or
H3O+
C
R
C
H
R"
C
R'
(E and Z)
α,β-unsaturated
carbonyl compound
676
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–2
[2] Intramolecular aldol reaction (24.4)
[a] With 1,4-dicarbonyl compounds:
O
O
NaOEt
O
[b] With 1,5-dicarbonyl compounds:
EtOH
O
O
NaOEt
O
EtOH
[3] Dieckmann reaction (24.7)
OEt
[a] With 1,6-diesters:
O
[1] NaOEt
O
O
O
C
[2] H3O+
OEt
OEt
[b] With 1,7-diesters:
OEt
O
O
O
[1] NaOEt
O
C
OEt
[2] H3O+
OEt
Practice Test on Chapter Review
1. a. Which compounds are possible Michael acceptors?
1.
3.
O
O
2.
O
4. Both compounds [1] and [2] are Michael acceptors.
OEt
5. Compounds [1], [2], and [3] are all Michael acceptors.
b. Which of the following compounds can be formed by an aldol reaction?
O
1.
CHO
HO
4. Both [1] and [2] can be formed by aldol reaction.
O
2.
3. HO
OH
5. Compounds [1], [2], and [3] can all be formed by aldol reaction.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
677
Carbonyl Condensation Reactions 24–3
c. Which compounds can be formed in a Robinson annulation?
O
O
1.
3.
CO2Et
2.
4. Compounds [1] and [2] can be formed by Robinson annulation.
O
5. Compounds [1], [2], and [3] can be formed by Robinson annulation.
d. What compounds can be used to form A by a condensation reaction?
O
O
O
O
CHO
CO2Et
1.
3.
and
A
O
and
O
2.
OEt
4. Compounds [1] and [2] can be used to form A.
5. Compounds [1], [2], and [3] can be used to form A.
2. Give the reagents required for each step.
O
O
O
(a)
O
(b)
H
HO
(c)
O
O
(e)
O
O
(d)
O
EtO
O
3. Draw the organic products formed in the following reactions.
O
O
O
a.
[1] LDA
c.
NaOEt
+
[2] CH3CH2CHO
[3] H2O
O
–
CHO
O
b.
+
HCO2Et
[1] NaOEt, EtOH
[2] H3
O+
d.
EtOH
O
+
OH, H2O
678
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–4
4. a. What organic starting materials are needed to synthesize D by a Robinson annulation reaction?
O
CO2CH3
D
b. What organic starting materials are needed to synthesize β-keto ester B by a Dieckmann reaction?
O
CO2Et
B
c. What starting materials are needed to synthesize A by an aldol reaction?
O
A
Answers to Practice Test
1.a. 4
b. 2
c. 1
d. 4
2.a. [1] O3; [2] (CH3)2S
b. CrO3, H2SO4, H2O
c. CH3CH2OH, H2SO4
d. [1] NaOEt, EtOH;
[2] H3O+
e. [1] LDA; [2] CH3I
4.
3.
O
OH
a.
O
CO2CH3
a.
O
OEt
b.
b.
O
CHO
CO2Et
O
O
c.
O
c.
O
d.
O
+
O
679
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–5
Chapter 24: Answers to Problems
24.1
O
CH2CHO
H
a.
OH
OH
c. CH3 C CH3
O
CH3 C CH2 C CH3
CH3
O
O
(CH3)3CCH2CHO
b.
HO
C(CH3)3
(CH3)3CCH2C
C CHO
H
O
HO
d.
H
24.2
O
O
O
CHO
a.
c.
b.
(CH3)3C
αH
no α H
no aldol reaction
d.
C
(CH3)3C
H
no α H
no aldol reaction
yes
C
CHO
e.
αH
CH3
αH
yes
yes
24.3
base
O
a.
O
O
OH
O
base
c.
HO
HO
CHO
CHO
base
b.
(E and Z isomers)
24.4
H OSO3H
O
C C C
H
H H
HO H
CH3
OH2 H
CH3 C
H
H
O
C C
H
CH3 C
H
H
+
+ HSO4
24.5
O
C C
H
O
HSO4
CH3CH CH C
H
H
+
H2O
H2SO4
Locate the α and β C’s to the carbonyl group, and break the molecule into two halves at this
bond. The α C and all of the atoms bonded to it belong to one carbonyl component. The β C
and all of the atoms bonded to it belong to the other carbonyl component.
β
β OH
α
CHO
a.
b. C6H5
c.
C6H5
OH
O
CHO
H
C6H5
O
β
O
α
H
CHO
CHO
α
O
C6H5
O
680
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–6
24.6
–OH
O
H
H H
C
H OH
O
H
C
O
C
H
H C
H
C H
O
O
+ H2O
OH
OH
H C
+
H C
–OH
C H
CHO
H
O
+ H2 O
C H
O
–OH
24.7
O
a.
CHO b. C6H5CHO
CHO
CH3CH2CH2CHO and CH2=O
or
and
OH
O
(E and Z isomers)
24.8
CO2Et
H
a.
CO2Et
CH2(CO2Et)2
O
H
c.
O
H
CH3COCH2CN
NC
COCH3
H
b.
COCH3
(E and Z isomers)
COCH3
CH2(COCH3)2
O
24.9
2
–
CHO
OH
CHO
a.
H2O
NaBH4
OH
CH3OH
OH
OH
CHO
b.
CHO
[1] (CH3)2CuLi
–OH
(E and Z mixture)
[2] H2O
CHO
NaBH4
c.
(from b.)
CH3OH
CH3CH2CH2CH C(CH2CH3)CH2OH
CHO
[1] CH3MgBr
d.
(from b.)
[2] H2O
CH3CH2CH2CH C(CH2CH3)CH(CH3)OH
681
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–7
24.10 Find the α and β C’s to the carbonyl group and break the bond between them.
O
OH
O
a.
b.
O
OH
O
c.
O
O
H
H
O
O
H
O
24.11
O
O
H
N
CH3O
CH3O
O
O
H2
N
CH3O
Pd-C
X
CH3O
N
CH3O
donepezil
CH3O
24.12 All enolates have a second resonance structure with a negative charge on O.
O
O
O
EtO
H
O
O
O
H
EtOH
H
EtOH
H H
H
EtO
O
O
+ OH
OH
H
H
OH
O
O
O
EtO H
EtOH
24.13
O
O
–
a.
CHO
O
O
OH
O
b.
H2O
[2] (CH3)2S
A
CHO
O
OH
H2O
24.14
[1] O3
–
–OH
O
H 2O
B
682
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–8
24.15 Join the α C of one ester to the carbonyl C of the other ester to form the β-keto ester.
O
a.
α
OCH3
O
O
O
b.
O
α
OCH2CH3
OCH3
O
α
α
OCH2CH3
24.16 In a crossed Claisen reaction between an ester and a ketone, the enolate is formed from the
ketone, and the product is a β-dicarbonyl compound.
O
a. CH3CH2CO2Et and HCO2Et
H
O
OEt
Only this compound
can form an enolate.
b.
O
and
CO2Et
H
HCO2Et
C
OEt
O
Only this compound
can form an enolate.
O
c.
CH3
C
O
and
CH3
C
CH3
O
O
OEt
The ketone
forms the enolate.
d.
O
O
O
C
C
and
OEt
O
The ketone
forms the enolate.
24.17 A β-dicarbonyl compound like avobenzone is prepared by a crossed Claisen reaction
between a ketone and an ester.
O
O
O
O
O
CH3O
CH3O
O
C(CH3)3
Break the molecule
into two components
at either dashed line.
avobenzone
C(CH3)3
or
O
O
CH3O
C(CH3)3
683
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–9
24.18
O
O
O
[1] NaOEt
a.
O
OEt
b.
[2] (EtO)2C=O
C6H5CH2
O
[1] NaOEt
C
C
[2] ClCO2Et
OEt
EtO
OEt
O
24.19
O
O
CO2Et
OEt
OEt
CH3
O
[1] NaOEt
[1] NaOEt
[2] CH3I
O
[2] (EtO)2C=O
A
EtO
B
EtO
H3O+
Δ
CO2H
ibuprofen
24.20
O
O
O
O
O
OEt
EtO
EtO
OEt
O
24.21 A Michael acceptor is an α,β-unsaturated carbonyl compound.
O
O
O
α,β-unsaturated
yes—Michael acceptor
O
d.
c.
b.
a.
O
not α,β-unsaturated
CH3O
not α,β-unsaturated
α,β-unsaturated
yes—Michael acceptor
24.22
O
a.
CH2 CHCO2Et
+
CH3
C
[1] NaOEt
CH2CO2Et
[2] H2O
CH3
O
O
C
C
COOEt
+ CH2(CO2Et)2
b.
[1] NaOEt
[2] H2O
O
O
EtO2C
O
O
c.
+
CH2
CH3
C
[1] NaOEt
CH2CN [2] H2O
O
CO2Et
CN
O
OEt
684
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–10
24.23
O
O
O
O
EtO2C
a.
CO2Et
b.
O
O
O
O
24.24 The Robinson annulation forms a six-membered ring and three new carbon–carbon bonds:
two σ bonds and one π bond.
new C–C bond
O
O
O
re-draw
+
a.
O
O–
CH3CH2
CH3CH2OH
O
O
O
O
new σ and π bonds
new C–C bond
O
O
COOEt
COOEt +
b.
re-draw
O
CH3CH2O–
CH3CH2OH
O
COOEt
O
new σ and π bonds
new C–C bond
O
c.
O
CH3CH2O–
CH3CH2OH
re-draw
+
O
O
O
new σ and π bonds
new C–C bond
EtO2C
O
d.
O
COOEt
EtOOC
re-draw
+
O
CH3CH2O–
CH3CH2OH
O
O
new σ and π bonds
24.25
a.
O
O
O
b.
c.
O
O
O
O
O
O
685
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–11
24.26
O
O
a. (CH CH ) C=O + CH =O
3
2 2
2
–OH
O
+ C6H5CHO
b.
H2O
or
–OH
H2O
(E and Z isomers)
O
OH
24.27
CHO
[1] O3
CHO
[2] (CH3)2S
CHO
NaOEt
EtOH
A
B
24.28 The product of an aldol reaction is a β-hydroxy carbonyl compound or an α,β-unsaturated
carbonyl compound. The α,β-unsaturated carbonyl compound is drawn as product unless
elimination of H2O cannot form a conjugated system.
OH
a. (CH3)2CHCHO only
–OH
(CH3)2CHCHC(CH3)2
H2O
CHO
–OH
c. C6H5CHO + CH3CH2CH2CHO
H2O
CHO
(E and Z isomers)
b. (CH3)2CHCHO + CH2=O
O
CH3
–OH
CH3 C CHO
H2O
d.
(CH3CH2)2C=O only
CH2OH
–OH
H 2O
24.29
HO
CHO
O
H
OH
CHO
H
O
OH
OH
CHO
CHO
24.30
O
O
a.
CH3
C
OH
[1] LDA
CH3 [2] CH3CH2CH2CHO
O
CH3CH2CH2 C CH2 C
CH3
H
[3] H2O
b. CH CH C OEt
3
2
OH
[1] LDA
[2]
O
O
CHO
O
C
CH
H
CH3
C
OEt
O
O
[3] H2O
24.31
O
O
CHO
a.
O
c.
O
b. OHC
CHO
CHO
O
686
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–12
24.32 Locate the α and β C’s to the carbonyl group, and break the molecule into two halves at this
bond. The α C and all of the atoms bonded to it belong to one carbonyl component. The β C
and all the atoms bonded to it belong to the other carbonyl component.
O
O
a.
OH
β
b.
β
O
β
d.
c.
C6H5
O
O
α C6H5
O
H
CH CHCN
β
α
O
e.
C6H5
α
α
CH3
O
O
O
C6H5
CH2=O
O
C6H5
α
β
CH3
CH3CN
CHO
C6H5
24.33
O
CHO
(CH3)2C=O
NaOEt
EtOH
CH3O
CH3O
X
24.34 Base removes the most acidic proton between the two C=O’s in B. This enolate reacts with
the aldehyde in A to form a product that loses H2O.
Form enolate here.
O
F
+
CO2Et
H
A
base
H OH
F
F
CO2Et
O
electrophilic C
– H2O
B
CO2Et
O
O
(E and Z isomers can form.)
24.35
O
O
O
O
c.
b.
a.
d.
HO
O
O
O
H
H
CH3
CH3
O
O
O
O
H
O
687
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–13
24.36 Ozonolysis cleaves the C=C, and base catalyzes an intramolecular aldol reaction.
O
O
NaOH
[1] O3
H2O
[2] (CH3)2S
O C
D
C10H14O
24.37 Aldol reactions proceed via resonance-stabilized enolates. K can form an enolate that allows
for delocalization of the negative charge on O. Delocalization is not possible in J, because a
double bond would be placed at a bridgehead carbon, which is geometrically impossible.
O
O
O
O
Bond angles
don't allow this.
from J
from K
24.38
O
a. C6H5CH2CH2CH2CO2Et
O
C6H5CH2CH2CH2
OEt
CH2CH2C6H5
O
b. (CH3)2CHCH2CH2CH2CO2Et
O
(CH3)2CHCH2CH2CH2
OEt
CH2CH2CH(CH3)2
CH3O
c. CH3O
O
O
CH2COOEt
OEt
OCH3
24.39
O
CH3CH2CH2CH2CO2Et + CH3CH2CO2Et
O
O
O
OEt
O
OEt
O
O
OEt
O
OEt
688
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–14
24.40
O
O
O
a. CH3CH2CH2CO2Et + C6H5CO2Et
d. CH3CH2CO2Et + (EtO)2C=O
OEt
CH2CH3
O
EtO
O
O
e.
O + HCO2Et
O
O
f.
OEt
c. EtO2CC(CH3)2CH2CH2CH2CO2Et
O
H
Cl
C
O
O
O
+
OEt
O
O
b. CH3CH2CH2CO2Et + (CH3)2C=O
O
O
EtO
OEt
24.41
O
O
O
CH3O
CH3O
CHO
c.
a.
O
OEt
or
H
CH3
O
CH3O
O
O
(EtO)2C=O
OEt
or
O
EtO
OEt
OEt
C6H5
b.
O
OEt
O
C6H5
O
O
CH3CHO
d. C6H5CH(COOEt)2
O
O
O
C6H5
OEt
EtO
OEt
24.42 Only esters with two H’s or three H’s on the α carbon form enolates that undergo Claisen
reaction to form resonance-stabilized enolates of the product β-keto ester. Thus, the enolate
forms on the CH2 α to one ester carbonyl, and cyclization yields a five-membered ring.
This is the only α carbon with 2 H's.
O
O
OCH3
CH3O
CH3O
O
NaOCH3
OCH3 [1] nucleophilic attack
CH3O
CH3OH
O
CH3O
O
O
CO2CH3
CH3O2C
CH3O–
[2] loss of
[3] deprotonation
O
B
O
CO2CH3
CH3O2C
H3
O+
CH3O2C
OCH3
CO2CH3
CH3O2C
H
O
acidic H between 2 C=O's
O
highly resonance-stabilized enolate
Formation of this enolate drives
the reaction to completion.
O
689
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–15
24.43
O
O
a.
+
C 6H 5
C6 H5
–OEt,
O
O
EtOH
C6 H5
C 6H 5
O
O
–
b.
OEt, EtOH
+ CH2(CN)2
CN
CN
24.44
O
O
O
a.
c.
b.
O
O
CO2Et
O
EtO2C
O
d.
CN
CO2Et
O
O
O
O
O
O
CO2Et
O
C6H5
EtO2C
CN
CO2Et
C6H5
24.45
O
O
H
a.
Michael
reaction
CH3 O
A
O
(E or Z isomer
can be used.)
O
O
b.
H OH
O
O
H
OH
O
O
H2O
OH
O
O
OH
H2O
H
OH
OH
O
690
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–16
24.46
O
O
−OH
a.
+
H2O
O
O
O
O
O
+
b.
−OH
re-draw
C6H5
+
O
O
H2O
C6H5
O
C6H5
O
O
re-draw
c.
−OH
+
+
H2O
O
O
O
O
O
−OH
re-draw
d.
+
O
H2O
O
O
24.47
O
O
c.
a.
O
O
O
O
O
O
O
O
d.
O
b.
O
O
O
24.48
CH3CH2CH2CHO
a.
–OH
CH3CH2CH2
H2O
b.
H
–OH
CH2
CH2=O, H2O
CHO
(E and Z)
CH2CH3
CHO
C
CH2CH3
g.
h.
CHO
c.
i.
[1] LDA
[2] CH3CHO; [3] H2O
CH3CH2CH2
d.
e.
j.
EtO2C
OH
[1] CH3Li
NaBH4
CH3OH
HOCH2CH2OH
TsOH
CH3NH2
mild acid
O
(CH3)2NH
mild acid
k.
CrO3
H
C
CH3CH2CH2CH2OH
CH3
N
H
CH3CH2CH=CHN(CH3)2
(E and Z)
CH3CH2CH2COOH
H2SO4
l.
O
CH3CH2CH2
CO2Et
[2] H2O
f.
CH3CH2CH2CH2OH
OH
H
CH2(CO2Et)2
NaOEt, EtOH
H2
Pd-C
O
Br2
H
CH3COOH
Br
691
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–17
m.
CH3CH2CH2
Ph3P=CH2
O
OH
n.
C CH2
NaCN, HCl
o.
CH3CH2CH2 C H
H
[1] LDA; [2] CH3I
H
CN
24.49
O
O
O
–OH
a.
e.
H2O
O
NaOEt, EtOH
f.
(CH3)2C=O
+
CHO
CH3
[2] CH3CH2CHO
O
C
C6H5
NaOCH3
C6H5
+
CH3OH
O
O
O
CO2Et
OH
H2O
O
C6H5
+
–OH
C6H5
H2O
g.
–
CH3
CHO
CN
O
d.
O
O
NaOEt, EtOH
c. NCCH2CO2Et
CH3
O
b.
O
OH
[1] LDA
[3] H2O
O
O
CH3
C
CH2CO2Et
[1] NaOEt, EtOH
CH2CO2Et
[2] H3O+
(E and Z)
h.
O
(E and Z) O
O
O
OEt
24.50
O
A
[1] NaOEt
O
CO2Et
[2](EtO)2CO
[3] H3O+
G
[1] LDA
[2] CH3CH2CHO
[3] H2O, –OH
H
H2, Pd-C
O
(1 equiv)
O
B
[1] NaOEt
[2] CH3CH2I
C
H3O+
Δ
CO2Et
O
HO
D
[1] CH3CH2MgBr
[2] H2O
E [1] LDA
[2] CH3I
O
I
[1] LDA
[2] HCO2Et
[3] H3O+
O
O
F H2SO4
O
H
major product
J [1] O3
O
O
K
–OH
H2O
O
[2] (CH3)2S
692
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–18
24.51
O
[1] O3
O
[2] (CH3)2S
O
NaOH
O
O
EtOH
H
O
B
O
H
O
O
OH
Form the enolate here to
generate a five-membered
ring in the product.
A
new C–C bond
The RCHO has the more
accessible carbonyl.
24.52
O
H
C
C
O
O
[1]
H
H
H H
C
C
+ H C H
H
O
C
C
C
H
HO
C
C
C
H H
H OCO2
O
O
HH
[3]
H H
H
+ HCO3–
CO32–
O
HH
[2]
HOCH2
H
C
C
H
HOCH2 CH2OH
+ CO32–
Repeat steps [1]–[3] with these
two H's and CO32–.
24.53 Enolate A is more substituted (and more stable) than either of the other two possible enolates and
attacks an aldehyde carbonyl group, which is sterically less hindered than a ketone carbonyl. The
resulting ring size (five-membered) is also quite stable. That is why 1-acetylcyclopentene is the
major product.
O
O
H
H OH
O
H
O
H H
–
OH
H O
–OH
OH
O
A
O
most stable enolate
less hindered carbonyl
O
H2 O
H
OH
O
H2 O
–
OH
1-acetylcyclopentene
major product
O
H H
–OH
O
–
H
O
H
O
H
O
B
The more hindered ketone
carbonyl makes nucleophilic
attack more difficult.
H2O
O
H
H OH
OH
H O
OH
OH
O
H
H
H2O
CHO
–OH
693
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–19
O
H
H
H H
–
H
O
H OH
O
O
–OH
OH
H
H
OH
H2O
H
O
O
O
O
OH
These two reacting functional groups
are farther away than the reacting
groups in the first two reactions,
making it harder for them to find each
other. Also, the product contains a
less stable seven-membered ring.
C
less stable enolate
H2O
O
–OH
24.54
Na+ –OCH3
H H
H
CH3O
CH3O
O
O
O
O
O
O
O
O
COOCH3
H OH
+ CH3OH
–OH
O
O
COOCH3
24.55 Removal of a proton from CH3NO2 forms an anion for which three resonance structures can
be drawn.
O
O
H CH2 N
O
CH2 N
O
CH2 N
O
O
CH2 N
O
O
–OH
O
C6H5
C
O
H
CH2 N
O
O
H OH
C6H5 C CH2
H
NO2
OH
OH
C6H5 C CH NO2
H H
C6H5 C CH NO2
H
–OH
+ H2O
C6H5CH CHNO2
+
–OH
694
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–20
24.56 All enolates have a second resonance structure with a negative charge on O.
O
O
O
O
H H
O
+ H2O
OH
H OH
H OH
O
O
O
O
H H
OH
H OH
HO
OH
O
O
HO
OH
H
O
O
O
H OH
24.57 Et3N reacts with phenylacetic acid to form a carboxylate anion that acts as a nucleophile to
displace Br, forming Y. Then an intramolecular crossed-Claisen reaction yields rofecoxib.
O
CO2H
CO2
Et3N
O
SN2
CH3SO2
H
Y
+ Et3NH
phenylacetic acid
O
X
Et3N
O
O
O
O
Et3N
H
O
O
proton
source
OH
CH3SO2
CH3SO2
O
CH3SO2
H A
+ Et3NH
O
O
O
O
+ –OH
OH
CH3SO2
+ Et3NH
CH3SO2
rofecoxib
O
695
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–21
24.58 Polymerization occurs by repeated Michael reactions.
B
H
B
H
[1]
B
[2]
O
O
O
O
O
O
O
O
O
O
O
O
Repeat step [2].
new C–C bonds
B
O
O
O
O
O
O
polytulipalin
24.59
O
O
H
O
H
O H
CH3
C
O
O
O
C
O
H
CH3
CH3
O
O
CH3COO–
+
C
H
O
O
O
O
CH3COO
CH2
H
O
H
CH3COO–
CH3COO–
OH
OH
O
H OCOCH3
H
+ CH3COOH
O
O
O
O
H
O
O
O
O
CH2
O
+ CH3COOH
O
O
+ –OH
coumarin
24.60
C6
CO2Et
a.
CO2Et
+ EtOH
H
ethyl 2,4-hexadienoate
CO2Et
OEt
O
EtO O
OEt
CO2Et
O
EtO
O
OEt
diethyl oxalate
O
CO2Et
EtO2C
O
OEt
CO2Et
+
OEt
696
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–22
b. The protons on C6 are more acidic than other sp3 hybridized C–H bonds because a highly
resonance-stabilized carbanion is formed when a proton is removed. One resonance
structure places a negative charge on the carbonyl O atom. This makes the protons on C6
similar in acidity to the α H’s to a carbonyl.
c. This is a crossed Claisen because it involves the enolate of a conjugated ester reacting
with the carbonyl group of a second ester.
24.61
O
O
–OH,
a.
H2 O
C6 H5
C6H5CHO
O
b.
–
O
OH, H2O
O
O
PCC
CH2OH
CHO
H2C=O
–OEt,
O
O
–OH,
c.
HCO2Et
CH3CH2OH
O
H 2O
O
O
O
OH
[1] LiAlH4
[1] LDA, THF
d.
[2] CH3CH2CH2CH2Br
[2] H2O
or
O
[1] LDA, THF
H2 (excess)
[2] CH3CH2CH2CHO
Pd-C
[3] H2O, –OH
(E and Z isomers)
O
e.
O
OH
–OH
H2 (excess)
H2 O
Pd-C
24.62
O
O
O
[1] Br2, CH3COOH
a.
O
[1] NaOEt
[2] Li2CO3, LiBr, DMF
O
H3O+
O
Δ
O
OEt
[2] H2O
COOEt
O
–OH,
b.
O
[1] (CH3)2CuLi
H2O
O
[2] H2O
O
O
697
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–23
24.63
O
OH
O
[1] LDA
a.
[2] CH3CH2CHO
[3] H2O
PCC
OH
O
O
O
O
O
[1] LDA
[1] NaOEt
[2] CH3CO2Et
[2] CH3Br
b.
OH
H2SO4
O
CrO3
OH
PBr3
H2SO4
H2O
CH3OH
OH
O
OH
O
[1] LDA
c.
CH3OH
C6H5
[2] C6H5CHO
SOCl2
[1] Br2, hν
O
HO
OH
C6H5
O
OH
H2SO4
[1] CH3MgBr
O
[1] O3
[2] H2O
–
CHO
OH, H2O
[2] (CH3)2S
Mg
PBr3
H2 (excess)
Pd-C
unstable
[2] –OH
AlCl3
CH3OH
C6H5
– H2O
PCC
CH3Cl
d.
O
major
product
CH3Br
24.64
OEt
a.
[2]
O
[1] NaOEt
[1] NaOEt
CO2Et
H3O+
[2]
O
CO2Et
Br
PBr3
O
OH
C6H5
[1] NaOEt
b.
CO2Et
O
[2]
C6H5
CO2Et
Br
O
(from a.)
CH3OH + SOCl2
[1] CH3Cl, AlCl3
[2] Br2, hν
Br
C6H5
[1] LiAlH4
OH
[2] H2O
OH
698
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–24
C6H5
OEt
c.
O
O
[1] NaOEt
C6H5
[2] H3O+
O
O
AlCl3
O
Cl
SOCl2
O
CrO3
OH
OH H2SO4
H2O
24.65
O
[1] O3
CHO
CrO3
[2] (CH3)2S
CHO
H2SO4
COOH
COOH
OH
H2SO4
CO2Et
[1] NaOEt
CO2Et
[2] H3O+
CO2Et
H2O
[1] NaOEt
O
O
H3
O+
CO2Et
Δ
24.66
O
O
CHO
O
a.
CH3O
CH3O
octinoxate
+
[2] CH3I
O
699
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–25
[1] NaH
b.
[2] CH3Cl
HO
Br
Br2
CH3O
[2] H2C=O
CH3O
SOCl2
CH2OH
[1] Mg
CH3O
PCC
(+ ortho isomer)
PCC
CH3OH
CH3OH
O
O
H2SO4
HO
HO
CHO
O
CH3O
CrO3
H2SO4, H2O
H2O
HO
CH2=O
[1] PBr3
BrMg
NaOR, ROH
HO
octinoxate
[2] Mg
H2O
O
Mg
BrMg
Br
PCC
PBr3
HO
HO
24.67
O
O
O
base
d.
a. EtO
O
O
O
O
O
EtO
b.
O
O
[1] NaOEt
c.
base
CH3
O
O
O
Two products
are possible.
O
O
O
O
O
O
OH
O
e.
O
O
OH
or
[2] CH3I
conjugated tautomers favored
24.68
O
a.
O
[1] HCO2Et
NaOEt, EtOH
[2] H3
O+
CHO
700
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–26
O
O
CHO
CHO
O
O
b.
NaOEt, EtOH
O
O
O
O
O
H
OEt
c.
OEt
H
O
H OEt O
O
+
O
O
+
O
HCO2Et
OEt
+
O
H OH2
d.
O
=
O
O
+ H
O
+OH
OH
H2O
H2O
+
OH
H OH2
OH
no α H
OH
O
OH
OH
+
H OH2
H2O
+O H
O
β-hydroxy ketone
+
OH2
H
+
H2O
H2O
H2 O
H3O+
O
O
O
H2O
This reaction is an acid-catalyzed aldol that proceeds by way of enols not enolates. The β-hydroxy
ketone initially formed cannot dehydrate to form an α,β-unsaturated carbonyl because there is no H
on the α carbon. Thus, dehydration occurs, but the resulting C=C is not conjugated with the C=O.
O
e.
H
CHO
This H is now more acidic because it is located between two carbonyl
groups. As a result, it is the most readily removed proton for the Michael
reaction in the next step.
701
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–27
24.69
O
O
O
O
O
Ar'
CH3O
OCH3
CH3O
OCH3
CH3O
H
N
O
F
OCH3
R2N
Ar
N
OCH2Ph
Ar
Ar'
OCH2Ph
O
O
Ar'
CH3O
CH3O
N
+ CH3O–
O
O
N
OCH3 Ar
F
24.70 Rearrangement generates a highly resonance-stabilized enolate between two carbonyl groups.
OEt
O
O
O OEt O
O
OEt
O
O
OEt
OEt
O
H
OCH2CH3
OCH2CH3
OCH2CH3
OCH2CH3
CH3CH2O
H OCH2CH3
OCH2CH3
O
O
CO2CH2CH3
(+ 2 resonance structures)
This product is a highly
resonance stabilized enolate.
This drives the reaction.
H3O+
O
H
CO2CH2CH3
H
O
OEt
O OEt
CO2CH2CH3
CO2CH2CH3
O
OCH2CH3
H OCH2CH3
702
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–28
24.71 All enolates have a second resonance structure with a negative charge on O.
O
O
O
O
B
H OH
[2]
[1]
+
O
O
H
OH
[3]
B
O
[4]
OH
[5]
+ OH
HB+
+ HB+
+ –OH
O
OH
O
O
HO
+
Repeat steps
[1]–[5] by
deprotonating the
indicated CH3.
O
O
H
isophorone
B
H
+ HB+
(+ 2 resonance
structures)
24.72 All enolates have a second resonance structure with a negative charge on O.
B
O
O
O
COOEt
H
H
O
CO2CH3
CO2CH3
CO2CH3
new bond
new bond
H2O
O
+ –OH
CO2CH3
703
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbonyl Condensation Reactions 24–29
24.73
a.
–
H CH2
OH
N
CH2
CH2
N
(+ other resonance structures)
one possible resonance structure
The negative charge is delocalized on the
electronegative N atom. This factor is what makes
the CH3 group bonded to the pyridine ring more
H OH acidic, and allows the condensation to occur.
O
C6H5
N
O
H
CH2
OH
C6H5 C CH2
N
N
C6H5 C CH
H
OH
N
C6H5 C CH
H H
N
H
H2O
OH
CH CH
N
A
OH
b. The condensation reaction can occur only if the CH3 group bonded to the pyridine ring has acidic
hydrogens that can be removed with –OH.
OH
N
CH2
N
H
CH2
OH
H CH2
3-methylpyridine
Since the negative charge is delocalized on the N, the CH3
contains acidic H's and reaction will occur.
CH2
(+ other resonance
structures)
2-methylpyridine
N
N
N
CH2
N
CH2
N
CH2
N
CH2
No resonance structure places the negative charge on the N,
so the CH3 is not acidic and condensation does not occur.
N
CH2
704
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 24–30
24.74 Since the reaction takes place in acid, enols are involved. After the initial condensation
reaction, the NH2 and C=O groups form an imine by an intramolecular reaction.
H OH2
O
OH
CO2Et
OH
CO2Et
CO2Et
enol nucleophile
H
H2O
Ar OH
CO2Et
H OH2
NH2 O
NH2 OH
NH2 OH
Ar
–H+
F
Ar
H OH2
H2O
Ar
H
Ar OH
Ar OH2
CO2Et
NH2 O
CO2Et
CO2Et
NH2 O
NH2 O
Ar
Ar
Ar
CO2Et
NH2 O
N
H H
H3O
CO2Et
CO2Et
proton
transfer
O
N
H
OH
H OH2
Ar
Ar
Ar
N
N
H
H3O
CO2Et
CO2Et
CO2Et
H2O
N
H
OH2
705
Carbonyl Condensation Reactions 24–31
24.75
B
B
H
O
CO2CH3
O
O
OCH3
O
O
O
OCH3
O
O
O
O
O
O
+ HB+
H
O
O
O
O
O
O
O
+ CH3O–
+ HB+
O
The enolate
opens the
epoxide ring.
O
O
O
O
HO
O
H+
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
707
Amines 25–1
Chapter 25 Amines
Chapter Review
General facts
• Amines are organic nitrogen compounds having the general structure RNH2, R2NH, or R3N, with
a lone pair of electrons on N (25.1).
• Amines are named using the suffix -amine (25.3).
• All amines have polar C–N bonds. Primary (1°) and 2° amines have polar N–H bonds and are
capable of intermolecular hydrogen bonding (25.4).
• The lone pair on N makes amines strong organic bases and nucleophiles (25.8).
Summary of spectroscopic absorptions (25.5)
Molecular ion Amines with an odd number of N atoms give an odd molecular ion.
Mass spectra
IR absorptions N–H
3300–3500 cm–1 (two peaks for RNH2, one peak for R2NH)
1
H NMR
absorptions
NH
CH–N
0.5–5 ppm (no splitting with adjacent protons)
2.3–3.0 ppm (deshielded Csp3–H)
13
C–N
30–50 ppm
C NMR
absorption
Comparing the basicity of amines and other compounds (25.10)
• Alkylamines (RNH2, R2NH, and R3N) are more basic than NH3 because of the electron-donating
R groups (25.10A).
• Alkylamines (RNH2) are more basic than arylamines (C6H5NH2), which have a delocalized lone
pair from the N atom (25.10B).
• Arylamines with electron-donor groups are more basic than arylamines with electronwithdrawing groups (25.10B).
• Alkylamines (RNH2) are more basic than amides (RCONH2), which have a delocalized lone pair
from the N atom (25.10C).
• Aromatic heterocycles with a localized electron pair on N are more basic than those with a
delocalized lone pair from the N atom (25.10D).
• Alkylamines with a lone pair in an sp3 hybrid orbital are more basic than those with a lone pair in
an sp2 hybrid orbital (25.10E).
Preparation of amines (25.7)
[1] Direct nucleophilic substitution with NH3 and amines (25.7A)
R X
+
NH3
excess
R NH2
+ NH4+ X–
1o amine
•
•
R'
R X + R' N R'
R'
+
R N R' X–
R'
ammonium salt
•
The mechanism is SN2.
The reaction works best for CH3X or
RCH2X.
The reaction works best to prepare
1o amines and ammonium salts.
708
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–2
[2] Gabriel synthesis (25.7A)
•
•
O
R X
+
CO2–
–OH
N
R NH2
H 2O
+
CO2–
1o amine
O
The mechanism is SN2.
The reaction works best for CH3X
or RCH2X.
Only 1o amines can be prepared.
•
[3] Reduction methods (25.7B)
[a] From nitro compounds
H2, Pd-C or
R NO2
[b] From nitriles
R NH2
Fe, HCl or
Sn, HCl
1o amine
[1] LiAlH4
[2] H2O
R C N
R CH2NH2
1o amine
O
[c] From amides
C
R
NR'2
[1] LiAlH4
[2] H2O
RCH2 N R'
R'
R' = H or alkyl
1o, 2o, and 3o amines
[4] Reductive amination (25.7C)
R
C O
•
R
NaBH3CN
R2"NH
+
R' C N R"
R'
H R"
•
R', R" = H or alkyl
1o, 2o, and 3o amines
Reductive amination adds one alkyl
group (from an aldehyde or ketone)
to a nitrogen nucleophile.
Primary (1°), 2°, and 3° amines can
be prepared.
Reactions of amines
[1] Reaction as a base (25.9)
R NH2
+
+
H A
R NH3
+
A
[2] Nucleophilic addition to aldehydes and ketones (25.11)
With 1o amines:
With 2o amines:
O
R
C
O
NR'
C
H
R = H or alkyl
R'NH2
R
C
C
imine
H
R
C
C
H
R = H or alkyl
NR'2
R'2NH
R
C
C
enamine
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
709
Amines 25–3
[3] Nucleophilic substitution with acid chlorides and anhydrides (25.11)
O
R
C
O
R'2NH
(2 equiv)
+
Z
R
Z = Cl or OCOR
R' = H or alkyl
C
NR'2
1o, 2o, and 3o amides
[4] Hofmann elimination (25.12)
C C
H NH2
[1] CH3I (excess)
•
C C
[2] Ag2O
[3] Δ
The less substituted alkene is the major
product.
alkene
[5] Reaction with nitrous acid (25.13)
With 1o amines:
R NH2
With 2o amines:
NaNO2
+
Cl–
R N N
NaNO2
R N H
HCl
HCl
R
alkyl diazonium salt
R N N O
R
N-nitrosamine
Reactions of diazonium salts
[1] Substitution reactions (25.14)
N2
Cl–
F
X
OH
phenol
With HBF4:
With CuX:
With H2O:
+
aryl chloride or
aryl bromide
aryl fluoride
X = Cl or Br
With NaI or KI:
With CuCN:
I
aryl iodide
With H3PO2:
CN
H
benzonitrile
benzene
[2] Coupling to form azo compounds (25.15)
N2+ Cl– +
Y
Y = NH2, NHR, NR2, OH
(a strong electrondonor group)
N N
azo compound
Y
+
HCl
710
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–4
Practice Test on Chapter Review
1. Give a systematic name for each of the following compounds.
CH3
a.
b.
NHCH2CH3
NHCH2CH3
2. (a) Which compound is the weakest base? (b) Which compound is the strongest base?
O
H
N
N
H
CH3
N
H
NH2
CH3
CH3CH2
A
B
C
D
3. (a) Which compound is the weakest base? (b) Which compound is the strongest base?
O
NH2
CH3
NH2
NH2
Cl
NH2
H 2N
A
B
C
D
4. Draw the organic products formed in each of the following reactions.
a. Cl
NH2
[1] NaNO2, HCl
CHO
d.
NaBH3CN
+
NH
[2] CuCN
O
NH2
[1] KOH
b.
NH
O
NH2
c.
[2] PhCH2CH2Br
[3] –OH, H2O
e.
[1] CH3I (excess)
[2] Ag2O
[3] heat
[1] NaNO2, HCl
[2] NaI
Cl
5. Draw the products formed when the given amine is treated with [1] CH3I (excess); [2] Ag2O; [3] ,
and indicate the major product. You need not consider any stereoisomers formed in the reaction.
N
H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
711
Amines 25–5
6. What organic starting materials are needed to synthesize B by reductive amination?
N
C(CH3)3
B
Answers to Practice Test
1.a. N-ethyl-2,4-dimethyl-3heptanamine
b. N-ethyl-3methylcyclohexanamine
4.
5.
a. Cl
CN
N
CO2
b.
2.a. B
b. C
+
Ph
CO2
3.a. C
b. B
NH2
I
c.
N
N
Cl
major
N
d.
6.
+
e.
O
H
N
C(CH3)3
Answers to Problems
25.1
Amines are classified as 1o, 2o, or 3o by the number of alkyl groups bonded to the nitrogen atom.
1o amine
2o amine
a. H N
2
N
H
C6H5
H
N
2o
NH2
b. CH3CH2O
amine
O
1o amine
25.2
3o amine
a.
CH3
b.
HO C CH2NH2
CH3
3o alcohol
CH3 N CH2CH2OH
CH3
1o amine
N CH3
1o alcohol
3o amine
712
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–6
The N atom of a quaternary ammonium salt is a stereogenic center when the N is surrounded
by four different groups. All stereogenic centers are circled.
25.3
OH
+
CH3
HO
CH3
+
a. CH3 N CH2CH2 N CH2CH3
CH3
H
N
b.
HO
H
N has 3 similar groups.
25.4
NHCH2CH3
a.
c.
CH3CH2CH(NH2)CH3
2-butanamine
or
sec-butylamine
e.
N(CH3)2
N,N-dimethylcyclohexanamine
N-ethyl-3-hexanamine
CH3
b.
(CH3CH2CH2CH2)2NH
d.
f.
dibutylamine
2-methyl-5-nonanamine
25.5
NHCH2CH2CH3
NH2
2-methyl-N-propylcyclopentanamine
An NH2 group named as a substituent is called an amino group.
a. 2,4-dimethyl-3-hexanamine
c. N-isopropyl-p-nitroaniline
e. N,N-dimethylethylamine
NHCH(CH3)2
g. N-methylaniline
NHCH3
N
O2N
NH2
b. N-methylpentylamine
d. N-methylpiperidine
f. 2-aminocyclohexanone
O
NHCH3
h. m-ethylaniline
NH2
NH2
N
CH2CH3
25.6
Primary (1°) and 2o amines have higher bp’s than similar compounds (like ethers) incapable
of hydrogen bonding, but lower bp’s than alcohols that have stronger intermolecular
hydrogen bonds. Tertiary amines (3°) have lower boiling points than 1o and 2o amines of
comparable molecular weight because they have no N–H bonds.
CH3
alkane
lowest
boiling point
O
CH3
ether
intermediate
boiling point
NH2
amine
N–H can hydrogen
bond.
highest boiling point
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
713
Amines 25–7
25.7
The NH signal occurs between 0.5 and 5.0 ppm. The protons on the carbon bonded to the
amine nitrogen are deshielded and typically absorb at 2.3–3.0 ppm. The NH protons are not
split.
molecular formula C6H15N
1H NMR absorptions (ppm):
0.9 (singlet, 1 H)
1.10 (triplet, 3 H)
1.15 (singlet, 9 H)
2.6 (quartet, 2 H)
25.8
NH
CH3 adjacent to CH2
(CH3)3C
CH2 adjacent to CH3
N
H
The atoms of 2-phenylethylamine are in bold.
O
a.
N
(CH3CH2)2N
CH3
b.
H
CH3O
O
H
N
H
LSD
lysergic acid diethyl amide
25.9
H
HO
N
CH3
codeine
SN2 reaction of an alkyl halide with NH3 or an amine forms an amine or an ammonium salt.
Cl
a.
NH3
NH2
excess
NH2
b.
CH3CH2Br
N(CH2CH3)3
Br–
excess
25.10 The Gabriel synthesis converts an alkyl halide into a 1o amine by a two-step process:
nucleophilic substitution followed by hydrolysis.
NH 2
a.
b.
Br
(CH3)2CHCH 2CH2NH2
c. CH3O
NH2
CH3O
Br
(CH3)2CHCH 2CH2Br
25.11 The Gabriel synthesis prepares 1° amines from alkyl halides. Since the reaction proceeds by an
SN2 mechanism, the halide must be CH3 or 1°, and X can’t be bonded to an sp2 hybridized C.
NH2
a.
aromatic
An SN2 does not occur
on an aryl halide.
cannot be made by
Gabriel synthesis
b.
NH2
can be made by Gabriel
synthesis
c.
N
H
2° amine
cannot be made by
Gabriel synthesis
NH2
d.
N on 3° C
An SN2 does not
occur on a 3° RX.
cannot be made by
Gabriel synthesis
714
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–8
25.12 Nitriles are reduced to 1o amines with LiAlH4. Nitro groups are reduced to 1o amines
using a variety of reducing agents. Primary (1°), 2o, and 3o amides are reduced to 1o, 2o,
and 3o amines respectively, using LiAlH4.
O
a.
CH3CHCH2NH2
CH3CHCH2NO2
CH3
CH3CHC N
CH3CHCNH2
CH3
CH3
CH3
O
b.
CH2NH2
CH2NO2
C N
C
NH2
c.
NH2
C
NO2
O
N
NH2
25.13 Primary (1°), 2o, and 3o amides are reduced to 1o, 2o, and 3o amines respectively, using
LiAlH4.
O
CONH2
CH2NH2
a.
NHCH3
c.
NHCH3
O
b.
N
N
25.14 Only amines with a CH2 or CH3 bonded to the N can be made by reduction of an amide.
NH2
a.
N bonded to benzene
cannot be made by reduction
of an amide
NH2
NH2
b.
c.
N bonded to a 3° C
cannot be made by
reduction of an amide
N bonded to CH2
can be made by
reduction of an amide
N
H
d.
N on 2° C on both sides
cannot be made by
reduction of an amide
25.15 Reductive amination is a two-step method that converts aldehydes and ketones into 1o, 2o,
and 3o amines. Reductive amination replaces a C=O by a C–H and C–N bond.
a.
CHO
NHCH3
CH3NH2
c.
NaBH3CN
b.
O
O
(CH3CH2)2NH
N(CH2CH3)2
NaBH3CN
NH2
NH3
NaBH3CN
d.
O
+
NH2
NHCH(CH3)2
NaBH3CN
25.16 Reductive amination occurs using the ketone in E and the amine in D.
CO2CH2CH3
O
OCH2CH3
O
E
O
H2N
+
N
H
O
N
OH
D
O
O
N
OH
enalapril
715
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–9
25.17
NH2
a.
O
NH3
OH
OH
b.
NHCH3
O
NH2CH3
or
OH
NH2
CH2=O
25.18
a.
NH2
Only amines that have a C bonded to a H and N atom can be made by
reductive amination; that is, an amine must have the following structural
feature:
H
C N
phentermine
In phentermine, the C bonded to N is not bonded to a H, so it cannot be
made by reductive amination.
b. systematic name: 2-methyl-1-phenyl-2-propanamine
25.19 The pKa of many protonated amines is 10–11, so the pKa of the starting acid must be less
than 10 for equilibrium to favor the products. Amines are thus readily protonated by strong
inorganic acids (e.g., HCl and H2SO4) and by carboxylic acids.
a. CH3CH2CH2CH2 NH2
CH3CH2CH2CH2 NH3 + Cl–
+ HCl
pKa ≈ 10
weaker acid
products favored
pKa = –7
b. C6H5COOH
+
c.
pKa = 10.7
weaker acid
products favored
pKa = 4.2
N
H H
pKa ≈ 10
pKa = 15.7
weaker acid
reactants favored
(CH3)2NH2 + C6H5COO–
(CH3)2NH
+ HO–
+ H2O
N
H
25.20 An amine can be separated from other organic compounds by converting it to a water-soluble
ammonium salt by an acid–base reaction. In each case, the extraction procedure would
employ the following steps:
• Dissolve the amine and either X or Y in CH2Cl2.
• Add a solution of 10% HCl. The amine will be protonated and dissolve in the aqueous
layer, while X or Y will remain in the organic layer as a neutral compound.
• Separate the layers.
a.
NH2
and
CH3
X
H Cl
+
NH3 Cl–
• soluble in H2O
• insoluble in CH2Cl2
CH3
X
• insoluble in H2O
• soluble in CH2Cl2
716
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–10
b.
(CH3CH2CH2CH2)3N
and
+
H Cl
(CH3CH2CH2CH2)2O
(CH3CH2CH2CH2)3NH Cl–
Y
(CH3CH2CH2CH2)2O
Y
• insoluble in H2O
• soluble in CH2Cl2
• soluble in H2O
• insoluble in CH2Cl2
25.21 Primary (1°), 2o, and 3o alkylamines are more basic than NH3 because of the electrondonating inductive effect of the R groups.
a. (CH3)2NH
and
NH3
2° alkylamine
CH3 groups are electron donating.
stronger base
b. CH3CH2NH2
1° alkylamine
stronger base
and
ClCH2CH2NH2
1° alkylamine
Cl is electron withdrawing.
weaker base
25.22 Arylamines are less basic than alkylamines because the electron pair on N is delocalized.
Electron-donor groups add electron density to the benzene ring making the arylamine more
basic than aniline. Electron-withdrawing groups remove electron density from the benzene
ring, making the arylamine less basic than aniline.
NH2
NH2
NH2
NH2
a.
NH2
NH2
b.
CH3OOC
O2N
CH3O
electronwithdrawing group
least basic
arylamine
intermediate
basicity
electrondonating group
most basic
electronwithdrawing group
least basic
arylamine
intermediate
basicity
alkylamine
most basic
25.23 Amides are much less basic than amines because the electron pair on N is highly delocalized.
CONH2
amide
least basic
NH2
NH2
arylamine
intermediate basicity
alkylamine
most basic
25.24
sp2 hybridized
more basic
This N is also sp2 hybridized
but the electron pair
CH3
occupies a p orpital, so it
N
N
can delocalize onto the
a. Electron pair on N
CH3 aromatic ring. Delocalization
occupies an sp2
hybrid orbital.
makes this N less basic.
DMAP
4-(N,N-dimethylamino)pyridine
b.
H
N
N
CH3
sp3 hybridized N
stronger base
nicotine
sp2 hybridized
N
higher percent s-character
weaker base
717
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–11
25.25
This electron pair is delocalized,
making it a weaker base.
Br
H
N
N
stronger base
sp3 hybridized N
25% s-character
HO
H
NH2
CH3O
a.
b.
N
sp2 hybridized N
33% s-character
N
N(CH3)2
c.
stronger base
sp3 hybridized N
25% s-character
sp2 hybridized N
33% s-character
stronger base
This compound is similar to
DMAP in Problem 25.24a.
25.26 Amines attack carbonyl groups to form products of nucleophilic addition or substitution.
CH3CH2CH2NH2
O
a.
O
NCH2CH2CH3
O
b. CH3 C O C CH3
COCl
c.
CH3CH2CH2NH2
CH3
O
O
O
C
C
C
NHCH2CH2CH3
CH3
O
CONHCH2CH2CH3
CH3CH2CH2NH2
(CH3CH2)2NH
O
N(CH2CH3)2
O
(CH3CH2)2NH
CH3
COCl
C
CH3
N(CH2CH3)2
CON(CH2CH3)2
(CH3CH2)2NH
25.27 [1] Convert the amine (aniline) into an amide (acetanilide).
[2] Carry out the Friedel–Crafts reaction.
[3] Hydrolyze the amide to generate the free amino group.
O
a.
NH2
CH3
C
O
C CH3
(CH3)3CCl (CH ) C
NH
3 3
AlCl3
Cl
NH2
CH3
C
C CH3
H3O+
NH
(CH3)3C
(+ ortho isomer)
O
b.
O
H
N
Cl
Cl
C
CH3
O
H
N
O
AlCl3
C
CH3
H3O+
O
CCH2CH3
NH2
O
O
(+ para isomer)
25.28
transition
state
δ+
N(CH3)3
H
OH
δ−
(no 3-D geometry shown here)
Ea
Energy
transition state:
starting
materials
H°
Reaction coordinate
products
NH2
718
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–12
25.29
[1] CH3I (excess)
a. CH3CH2CH2CH2 NH2
b.
[2] Ag2O
[3] Δ
[1] CH3I (excess)
(CH3)2CHNH2
[1] CH3I (excess)
NH2
c.
CH3CH2CH=CH2
[2] Ag2O
[3] Δ
CH3CH=CH2
[2] Ag2O
[3] Δ
25.30 In a Hofmann elimination, the base removes a proton from the less substituted, more
accessible β carbon atom, because of the bulky leaving group on the nearby α carbon.
[1] CH3I (excess)
a.
CH2CHCH3
CH=CHCH3
[2] Ag2O
[3] Δ
NH2
major product
[1] CH3I (excess)
b.
H2N
major product
+
β [1]
CH3I (excess)
β
c.
β
[2] Ag2O
[3] Δ
CH2CH=CH2
N
H
[2] Ag2O
[3] Δ
(3 β C's)
least substituted β carbon
CH3CH=CH(CH2)3N(CH3)2
+
CH2=CH(CH2)2CHN(CH3)2
+ CH3
CH2=CH(CH2)4N(CH3)2
major product, formed by
removal of a H from the least
substituted β C
25.31
K+ –OC(CH3)3
a.
Cl
Br
[1] CH3I (excess)
b.
K+ –OC(CH3)3
c.
[2] Ag2O
[3] Δ
NH2
[1] CH3I (excess)
d.
(E and Z)
[2] Ag2O
[3] Δ
NH2
25.32
N2+ Cl–
NH2
NaNO2
a.
CH3
b.
H
CH3CH2 N CH3
HCl
HCl
HCl
N
H
CH3
NaNO2
NaNO2
c.
CH3CH2 N CH3
NO
N
NO
NaNO2
d.
HCl
NH2
N2 Cl–
719
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–13
25.33
a.
NH2
b.
NH2
[1] NaNO2, HCl
c. CH3O
Br
[2] CuBr
[1] NaNO2, HCl
d.
OH
O2 N
CH3O
[2] HBF4
[1] CuCN
N2+ Cl–
[2] H2O
O2N
[1] NaNO2, HCl
NH2
CH2NH2
[2] LiAlH4
[3] H2O
Cl
F
Cl
25.34
NH2
NO2
a.
H2
H2SO4
Pd-C
NO2
b.
F
[1] NaNO2, HCl
HNO3
[2] HBF4
NO2
OH
NH2
HNO3
H2
[1] NaNO2, HCl
H2SO4
Pd-C
[2] H2O
(from a.)
NO2
OH
NH2
CH3
[1] NaNO2, HCl
c.
NH2
CH3Cl
[2] NaI
(+ para isomer)
AlCl3
I
I
(from a.)
NH2
NH2
Cl
Cl
Cl2 (excess)
d.
Cl
Cl
[1] NaNO2, HCl
[2] H3PO2
FeCl3
(from a.)
Cl
Cl
25.35
NH2
a.
OH
N N
NH2
c.
HO
OH
N N
HO
b.
HO
N N
OH
25.36 To determine what starting materials are needed to synthesize a particular azo compound,
always divide the molecule into two components: one has a benzene ring with a
diazonium ion, and one has a benzene ring with a very strong electron-donor group.
O2N
a. H2N
Cl
b. HO
N N
N N
O2N
H2N
Cl– N2
CH3
Cl
HO
Cl– N2
CH3
720
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–14
25.37
a.
N N
OH
b. O2N
NO2
N N
OH
alizarine yellow R
para red
COOH
+
+
N N
Cl–
O2N
NO2
N N
Cl–
OH
COOH
OH
25.38
O
CH3
O
O
O
O
O
–
O3S
N N
O
Dacron
methyl orange
O
To bind to fabric, methyl orange (an anion) needs to interact with positively charged
sites. Since Dacron is a neutral compound with no cationic sites on the chain, it does
not bind methyl orange well.
25.39
a.
4,6-dimethyl-1-heptanamine
N(CH2CH3)2
b.
NH2
N,N-diethylcycloheptanamine
25.40
or
N
H
A
weaker base
(delocalized electron
pair on N)
NH
B
stronger base
25.41
N
N
a, b.
NH
NH2 + Cl–
HCl
N
N
varenicline
most basic
only sp3 hybridized N
N
CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
721
Amines 25–15
25.42
a. CH3NHCH2CH2CH2CH3
N-methyl-1-butanamine
(N-methylbutylamine)
e. (C6H5)2NH
diphenylamine
h.
CH2CH3
N
H
2-ethylpyrrolidine
b.
N C(CH3)3
f.
NH2
1-octanamine
(octylamine)
CH3CH2CH2CH(NH2)CH(CH3)2
i.
CH2CH3
2-methyl-3-hexanamine
N-tert-butyl-N-ethylaniline
CH3
N
c.
g. O
CH2CH2CH3
N-methyl-N-propylcyclohexanamine
d.
NH2
j.
NH2
3-ethyl-2-methylcyclohexanamine
4-aminocyclohexanone
(CH3CH2CH2)3N
tripropylamine
25.43
e. N-methylpyrrole
a. cyclobutylamine
h. 3-methyl-2-hexanamine
NH2
NH2
N
b. N-isobutylcyclopentylamine
CH3
f. N-methylcyclopentylamine
N
i. 2-sec-butylpiperidine
NHCH3
H
c. tri-tert-butylamine
N
H
g. cis-2-aminocyclohexanol
N[C(CH3)3]3
NH2
d. N,N-diisopropylaniline
NH2
or
OH
N[CH(CH3)2]2
j. (2S)-2-heptanamine
H NH2
OH
25.44
NH2
1-butanamine
H
N
N-methyl-1-propanamine
NH2
2-butanamine
N
H
diethylamine
NH2
2-methyl-1-propanamine
N
H
N-methyl-2-propanamine
NH2
2-methyl-2-propanamine
N
N,N-dimethylethanamine
722
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–16
25.45 [* denotes a stereogenic center.]
a.
*
N
CH3
b.
CH2CH3
*
*
CH3CH2CHCH
2CH2CH2 N CH2CH2CH2CH3
CH2CH3
CH3
1 stereogenic center
2 stereoisomers
N
H
H CH3
CH3
CH3CH2
CH2CH2CH2
H CH3
H
CH3
CH3CH2
H CH3
N
Cl
CH2CH2CH2
N
Cl
CH3 CH2CH3
C
CH3CH2
CH2CH2CH2CH3
CH2CH2CH2
H CH3
CH3 CH2CH3
C
CH2CH3
Cl
CH3 CH2CH3
C
CH2CH3
N
CH3
2 stereogenic centers
4 stereoisomers
CH3CH2
CH2CH2CH2CH3
CH3 CH2CH3
C
CH2CH2CH2CH3
N
Cl
CH2CH2CH2
N
Cl
CH2CH2CH2CH3
25.46
a.
(CH3CH2)2NH
b. (CH3CH2)2NH
or
sp3 hybridized N
stronger base
N
or
2° alkylamine
Cl is electron withdrawing.
weaker base
2° alkylamine
stronger base
sp2 hybridized N
weaker base
(ClCH2CH2)2NH
25.47
a.
NH2
NH2
NH3
arylamine intermediate
least basic
basicity
alkylamine
most basic
O2N
b.
d.
N
H
N
delocalized
electron pair on N
least basic
sp2 hybridized N
intermediate
basicity
NH2
c.
N
H
sp3 hybridized N
most basic
NH2
Cl
CH3
electronwithdrawing group
least basic
intermediate
basicity
(C6H5)2NH
C6H5NH2
diarylamine
least basic
NH2
arylamine
intermediate
basicity
electrondonating group
most basic
NH2
alkylamine
most basic
25.48 The electron-withdrawing inductive effect of the phenyl group stabilizes benzylamine,
making its conjugate acid more acidic than the conjugate acid of cyclohexanamine. The
conjugate acid of aniline is more acidic than the conjugate acid of benzylamine, since loss of
a proton generates a resonance-stabilized amine, C6H5NH2.
NH3
pKa = 10.7
NH2
alkylamine
cyclohexanamine
CH2NH3
pKa intermediate
CH2NH2
electron-withdrawing inductive
effect of the sp2 hybridized C's
benzylamine
NH3
pKa = 4.6
NH2
resonance-stabilized
aromatic amine
aniline
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
723
Amines 25–17
25.49 The most basic N atom is protonated on treatment with acid.
O
O
O
O
NH
a.
N
O
CH2COOH
+ CH3CO2–
NH2
CH3CO2H
N
more
basic
O
CH2COOH
benazepril
a 3° alkylamine with an
sp3 hybridized N
most basic
O
O
Cl
b.
N
Cl
N
NH
CH3CO2H
N
NH
aripiprazole
O
Cl
O
Cl
N
H
CH3CO2–
25.50
Nc
O
C
a.
NHNH2
N
Nb < Na < Nc
Nb
Na
Na
Nc
N
NH2
b.
N
Nb H
Nb < Na < Nc
Order of basicity: Nb < Na < Nc
Nb – The electron pair on this N atom is delocalized
on the O atom; least basic.
Na – The electron pair on this N atom is not
delocalized, but is on an sp2 hybridized atom.
Nc – The electron pair on this N atom is on an sp3
hybridized N; most basic.
Order of basicity: Nb < Na < Nc
Nb – The electron pair on this N atom is delocalized
on the aromatic five-membered ring; least basic.
Na – The electron pair on this N atom is not
delocalized, but is on an sp2 hybridized atom.
Nc – The electron pair on this N atom is on an sp3
hybridized N; most basic.
25.51 The para isomer is the weaker base because the electron pair on its NH2 group can be
delocalized onto the NO2 group. In the meta isomer, no resonance structure places the
electron pair on the NO2 group, and fewer resonance structures can be drawn:
724
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–18
O2N
NH2
O2N
NH2
O2N
NH2
O2N
NH2
O2N
NH2
meta
NH2
NH2
O2 N
NH2
O
O2N
N
para
NH2
NH2
O2N
O2N
O
NH2
O
NH2
O
N
O
N
O
25.52
N
N
A
This two-carbon bridge makes it
difficult for the lone pair on N to
delocalize on the aromatic ring.
B
pKa of the conjugate acid = 5.2 pKa of the conjugate acid = 7.29 Resonance structures that place a double bond
between the N atom and the benzene ring are
stronger conjugate acid
weaker conjugate acid
destabilized. Since the electron pair is more
weaker base
stronger base
localized on N, compound B is more basic.
The electron pair of this arylamine
is delocalized on the benzene ring,
decreasing its basicity.
N
N
B
Geometry makes it difficult
to have a double bond here.
25.53
NH
B
N
pyrrole
pKa = 23
stronger acid
N
N
N
N
weaker conjugate base
The electron pair is delocalized, decreasing the
basicity. The N atom is sp2 hybridized.
NH
pyrrolidine
pKa = 44
weaker acid
B
N
stronger conjugate base
The electron pair is not
delocalized on the ring.
The N atom is sp3
hybridized.
25.54
a. C6H5CH2CH2CH2Br
b. C6H5CH2CH2Br
NH3
excess
NaCN
c. C6H5CH2CH2CH2NO2
C6H5CH2CH2CH2NH2
[1] LiAlH4
Pd-C
[1] LiAlH4
[2] H2O
C6H5CH2CH2CN
H2
d. C6H5CH2CH2CONH2
[2] H2O
C6H5CH2CH2CH2NH2
C6H5CH2CH2CH2NH2
C6H5CH2CH2CH2NH2
e. C6H5CH2CH2CHO
NH3
NaBH3CN
C6H5CH2CH2CH2NH2
725
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–19
25.55
a. (CH3CH2)2NH
NH2
b.
N(CH3)2
c.
H
N
d.
O
C
CH3
H
N
N(CH3)2
NH2
NHCH2CH3
O
or
O
O
CH3
N
C
H
O
25.56 In reductive amination, one alkyl group on N comes from the carbonyl compound. The
remainder of the molecule comes from NH3 or an amine.
NH2
a.
H
+ NH3
O
H
N
b.
H
C6H5
or
NH2
C6H5
O
(CH3CH2CH2)2NH
or
H
H
N
H
C
O
CH3CH2CH2
CH2CH3
O
d.
C6H5
O
O
c. (CH3CH2CH2)2N(CH2)2CH(CH3)2
H2N
H
N
H
NH2
H
25.57
a.
NH2
C6H5
O
b.
O
NaBH3CN
C6H5
c.
NH
O
CH2NH2
NH3
excess
CN
b.
[1] LiAlH4
CH2NH2
[2] H2O
CONH2
CH2NH2
[1] LiAlH4
c.
[2] H2O
CHO
d.
NH3
NaBH3CN
C6H5
CH2NH2
d.
N(CH3)2
25.58
Br
CHO
NH2
(CH3)2NH
NaBH3CN
a.
C6H5
NH3
NaBH3CN
CH2NH2
NaBH3CN
NH
726
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–20
CH3
CH2Br
e.
CH2NH2
Br2
NH3
hν
excess
COOH
CONH2
[1] SOCl2
f.
[2] NH3
NH2
g.
[2] H2O
CN
[1] NaNO2, HCl
[2] H2O
NO2
HNO3
CH2NH2
[1] LiAlH4
[2] CuCN
h.
CH2NH2
[1] LiAlH4
NH2
H2
Then as in (g).
Pd-C
H2SO4
25.59 Use the directions from Answer 25.20. Separation can be achieved because benzoic acid
reacts with aqueous base and aniline reacts with aqueous acid according to the following
equations:
COO–Na+
+ H2O
COOH
NaOH
benzoic acid
• soluble in CH2Cl2
• insoluble in H2O
(10% aqueous)
• soluble in H2O
• insoluble in CH2Cl2
+
NH2
NH3 Cl–
H Cl
aniline
• soluble in CH2Cl2
• insoluble in H2O
(10% aqueous)
• soluble in H2O
• insoluble in CH2Cl2
Toluene (C6H5CH3), on the other hand, is not protonated or deprotonated in aqueous solution, so it is
always soluble in CH2Cl2 and insoluble in H2O. The following flow chart illustrates the process.
CH3
COOH
NH2
CH2Cl2
10% NaOH
CH2Cl2
H2O
COO–Na+
CH3
NH2
10% HCl
H2O
CH2Cl2
+
NH3 Cl–
CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
727
Amines 25–21
25.60
H
N
a.
b.
N-ethylaniline
f.
H H
N
HCl
Δ
+ CH2=CH2
Cl–
O
CH3CH2COCl
CH3COOH
N
g.
H H
N
CH3COO–
O
HNO3
h. The product in (g)
c.
N(CH3)2
Ag2O
CH3I (excess)
N
N
H2SO4
N
(CH3)2C=O
O
O2N
NO2
CH3
N
d.
e.
i. The product in (g)
CH2O
[1] LiAlH4
N
[2] H2O
NaBH3CN
CH3I
CH3 CH3
N
(excess)
I–
O
H2
j. The product in (h)
O
N
N
Pd-C
NH2 H2N
25.61
NH2 p-methylaniline
CH3
f.
HCl
a.
CH3
CH3COCl
CH3
(CH3CO)2O
CH3
g.
h.
CH3I
excess
e.
CH3COOH
NH3 CH3COO–
CH3
NaNO2
N2+Cl–
CH3
HCl
C CH
3
NH
i.
d.
NH2
C CH3
NH
O
c.
AlCl3
AlCl3
CH3
NH3 Cl–
O
b.
CH3COCl
O
Step (b), then
CH3COCl
C CH3
NH
CH3
AlCl3
CH3
(CH3)2C=O
CH3
N(CH3)3 I–
N C(CH3)2
C O
CH3
j.
CH3CHO
NaBH3CN
25.62
a. CH3(CH2)6NH2
[1] CH3I (excess)
[2] Ag2O
[3] Δ
CH3(CH2)4CH=CH2
CH3
NHCH2CH3
728
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–22
[1] CH3I (excess)
b.
NH2
[2] Ag2O
[3] Δ
N
H
β1
β2
N
H
[2] Ag2O
[3] Δ
β3
N(CH3)2
+
β2
major product
+ (CH3)2CHN(CH3)2
+
CH3
(E + Z)
(CH3)2N
β3
(E + Z)
CH2=CHCH3 + CH3CH2N(CH3)2
d.
major product
(CH3)2N
CH2=CH2
[2] Ag2O
[3] Δ
β1
[1] CH3I (excess)
major product
[1] CH3I (excess)
c.
e.
(E + Z)
[1] CH3I (excess)
CH3
[2] Ag2O
[3] Δ
NH2
CH3
CH3
CH2
major product
CH3
CH3
CH3
25.63
N(CH3)2
O
HN(CH3)2
a.
mild acid
O
NH2CH2CH2CH3
b.
NCH2CH2CH3
mild acid
NH2
O
NH3
c.
NaBH3CN
NH2
O
d.
[1] NaBH4, CH3OH
[1] CH3I (excess)
[2] H2SO4
[2] Ag2O
[3] Δ
O
NH3
NaBH3CN
OH
O
CH3NH2
mCPBA
NHCH3
e.
(from d.)
Br2
f.
O
O
O
Br
CH3COOH
25.64
CH3
N
a.
one stereogenic center
benzphetamine
NH2CH2CH2CH2CH3
NHCH2CH2CH2CH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
729
Amines 25–23
b. Amides that can be reduced to benzphetamine:
CHO
N
N
and
O
c. Amines + carbonyl compounds that form benzphetamine by reductive amination:
NHCH3
O
or
H
O
or
β1
H
N
NH
CH2=O
CH3
α N
[1] CH3I
β2
[2] Ag2O
[3] Δ
d.
(CH3)2NCH2
major product
elimination across α, β1
(elimination across α, β2)
25.65
a.
CH2CH2Cl
NH3
g.
CH2CH2NH2
NH
NaNO2
N N O
HCl
excess
O
CO2
[1] KOH
b.
NH
O
c. Br
NH2 +
[2] (CH3)2CHCH2Cl
[3] –OH, H2O
NO2
Sn
NH + C6H5CHO
NaBH3CN
Br
NH2
O
+
i.
N
N
H
[1] LiAlH4
d.
CN
[2] H2O
CONHCH2CH3
N CH2C6H5
CO2
HCl
e.
h.
j. CH3CH2CH2 N CH(CH3)2
CH2NH2
[1] LiAlH4
H
CH2NHCH2CH3
[2] H2O
f. C6H5CH2CH2NH2 + (C6H5CO)2O
C6H5CH2CH2NHCOC6H5 + C6H5CH2CH2NH3+ C6H5COO–
[1] CH3I (excess)
[2] Ag2O
[3] Δ
CH3CH=CH2
(CH3)2NCH(CH3)2
CH3CH2CH2N(CH3)2
730
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–24
25.66 NH2 and H must be anti for the Hofmann elimination. Rotate around the C–C bond so the
NH2 and H are anti.
a.
C6H5
NH2
H
C6H5
NH2
CH3
b.
H
CH3
C6H5
CH2CH3
c.
CH3
(CH3)3C
C(CH3)3
rotate 120°
counterclockwise
three steps
(CH3)3C
NH2
CH2CH3
CH2CH3
C(CH3)3
H
rotate 120°
counterclockwise
C6H5
NH2
CH3
CH2CH3
H
C6H5
C(CH3)3
(CH3)3C
CH2CH3
NH2
CH3
H
three steps
three steps
(CH3)3C
(CH3)3C
25.67
O
F
KOH
F
NO2
F
DMSO
F
F
NO2
OH
Cl
F
KI, K2CO3
F
H2
NO2
Ni catalyst
F
F
NH2
O
O
O
O
A
B
C
not isolated
731
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–25
25.68
N2+ Cl–
Cl
HO
Cl
A
Br
a. H2O
Cl
h. C6H5NH2
d. CuBr
b. H3PO2
H
Cl
NC
Cl
F
Cl
N N
OH
e. CuCN
Cl
Cl
NH2
Cl
i. C6H5OH
c. CuCl
N N
Cl
f. HBF4
I
I
Cl
Cl
j. KI
g. NaI
25.69
R
CH3
C
CH3CH2CH2
[1] CH3I (excess)
H
NH2
A
H
[2] Ag2O
[3] Δ
H
B
O
[1] O3
H
C C
H
[2] CH3SCH3
CH2CH2CH3
C
O
H
H
C
CH2CH2CH3
25.70
a.
Br
Br
H H Na+
N
Br
OH
H
N
Br
Br
Na+
+ CH3CH2NH2
Br
N
H CH2CH3
+ H2O
Na+
OH
N
+ H2O + Na+
CH2CH3
H A
H
O
O
H
N
b.
NH2
O
H
N
proton
transfer
OH2
H
N
H3B–H
H
H
N
+ H2O
proton
source
+ A
H
N
+ BH3
732
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–26
25.71
OH
O
OH
HN
H N
H2C=O
H A
OH
OH
N
proton
transfer
A
OH
OH
OH
H
OH2
OH
H2C
N
N
N
N
A
H2O
H A
OH
H
OH
H
OH
H
N
N
N
25.72
NaNO2
Overall reaction:
HCl, H2O
NH2
The steps:
OH
+
+
OH
NaNO2 + HCl
NH2
N N O
Cl
H
+
+ HCl
+
N O
N N O
N N O
H
H
N N O H
H
+
H Cl
+
H Cl
–
N N OH2 + Cl
N N O H
H
Cl
+
O H
OH
H
+ HCl
H
+
B
A
O H
H
H OH
Cl
H
B
+ HCl
Cl
+
+N
A H OH
Cl
+
1,2-H shift
N N +
OH
+ HCl
N
+ H2O
733
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–27
25.73 A nitrosonium ion (+NO) is a weak electrophile so electrophilic aromatic substitution occurs
only with a strong electron-donor group that stabilizes the intermediate carbocation.
O N O
H Cl
NaCl
HO N O
H Cl
H O N O
H
Na
N(CH3)2
O N
H
N(CH3)2
O N
H
Cl
N(CH3)2
O N
H
O N
H
O N
H
N(CH3)2
O N
N(CH3)2
especially good
resonance structure
All atoms have octets
resonance-stabilized carbocation
N O
H2O
N O
N(CH3)2
N(CH3)2
H2O
OH
25.74
O
NH2
a.
NH3
[1] CH3I
NaBH3CN
[2] Ag2O
[3] Δ
O
(Hofmann elimination, less
substituted C=C favored)
OH
NaBH4
b.
H2SO4
(more substituted C=C favored)
CH3OH
(E + Z isomers formed)
25.75
HNO3
H2SO4
a.
NO2
CH3Cl
AlCl3
b.
CH3
Br2
FeBr3
ClCOCH3
H2
Br
HNO3
O 2N
H2SO4
NO2
CH3
Pd-C
H2
Pd-C
Br
NH 2
H2 N
CH3
Br
[1] NaNO2, HCl
[2] NaI
NHCOCH3
I
CH3
(+ ortho isomer)
O2N
H2N
H2
c.
[1] NaNO2, HCl
NC
Br2
[2] CuCN
Pd-C
NC
Br
FeBr3
(from a.)
d.
Br2
FeBr3
e. O2N
CH3
(from b.)
NO2
HNO3
Br
KMnO4
O 2N
H2SO4
Br
H2
Pd-C
(+ para isomer)
COOH
H2
Pd-C
H 2N
NH2
Br
COOH
[1] NaNO2, HCl
[2] H2O
[1] NaNO2, HCl
HO
[2] H2O
OH
Br
COOH
734
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–28
f.
NH2
[1] NaNO2, HCl
OH
C6H5N2+Cl–
N N
OH
[2] H2O
(from c.)
(from c.)
25.76
NH2
CN
[1] NaNO 2, HCl
a.
COOH
H 3 O+
[2] NaCN
[2] NH2CH3
Br
Br
Br2
CH3COCl
b.
NH2
CONHCH3
[1] SOCl2
Br
–
CH3Cl
NHCOCH3 FeBr3
OH
NHCOCH3 H2O
NHCOCH3 AlCl3
(+ ortho isomer)
NH2
[1] NaNO2, HCl
[2] H3PO2
CH3
Br
Br
c.
[1] LiAlH4
COOH
CH2OH
[2] H2O
Br2
Br
FeBr3
CH2OH
(3x)
(from a.)
Br
HO
H2N
HO
HO
[1] NaNO2, HCl
d.
C6H5N2+Cl–
CH3Cl
AlCl3
[2] H2O
CH3
N N
(from a., Step [1])
CH3
(+ ortho isomer)
25.77
CH3Cl
[1]
Br2
–OH
Br
[2]
Br
hν
AlCl3
NH3
OH
PCC
H2C=O
NH2
(excess)
CHO CH NH
3
2
NaBH3CN
(from [1])
N
H
[1] LiAlH4
[2] H2O
KMnO4
O
COOH [1] SOCl
2
[2] CH3NH2
[3]
NCH3
[1] LiAlH4
H
[2] H2O
(from [1])
[1] HNO3, H2SO4
[4]
N
H
NaBH3CN
[2] H2, Pd-C
[3] NaNO2, HCl
[4] CuCN
CN [1] DIBAL-H
[2] H2O
[1] LiAlH4
[2] H2O
NH2
Then route [1].
CHO
Then route [2].
N
H
735
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–29
Br
Br2
[5]
COOH
MgBr [1] CO
2
Mg
[2]
FeBr3
Then route [3].
H3O+
[1] H2C=O
[2] H2O
OH
Then route [2].
25.78
O
O
mCPBA
O
O
O
[1] LiAlH4
O
[2] H2O
O
PBr3
OH
O
Br
O
PCC
O
O
CH3NH2
NHCH3
O
CH3NH2
NaBH3CN
O
O
O
NHCH3
O
MDMA
MDMA
[part (b)]
[part (a)]
25.79
CH3O
CH3O
Br
CN [1] LiAlH
4
NaCN
a.
CH3O
[2] H2O
CH3O
CH3O
CH3O
Br
CH3O
HBr
b.
CH3O
ROOR
CH3O
Mg
CH3O
NH2
[1]
CH3O
CH3O
CH3O
O
[2] H2O
CH3O
CH3O
excess
MgBr
c.
CH3O
CH3O
CH3O
Br
mescaline
NH3
CH3O
CH3O
NH2
CH3O
CH3O
CH3O
CH3O
CH3O
OH
CH3O
PBr3
CH3O
Br
CH3O
CH3O
CH3O
NH3
excess
CH3O
NH2
CH3O
CH3O
736
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–30
25.80
NO2
HNO3
a.
H2SO4
NH2
H2
CN
[1] NaNO2, HCl
COOH
H3 O +
[2] CuCN
Pd-C
COOH
Cl2
FeCl3
Cl
NH2
NHCOCH3
CH3COCl
b.
NHCOCH3
HNO3
Cl2
H2SO4
FeCl3
(2X)
(from a.)
NHCOCH3
Cl
Cl
NH2
–OH
Cl
Cl
Cl
H2O
NO2
Cl
[1] NaNO2, HCl
[2] H3PO2
NO2
NO2
NO2
Cl
H2
Pd-C
Cl
Cl
Cl
[1] NaNO2, HCl
[2] H2O
NH2
OH
CH3Cl
c.
HNO3
CH3
AlCl3
CH3
H2SO4
H2
NO2
CH3
NH2
Pd-C
[1] NaNO2, HCl
[2] NaCN
(+ ortho isomer)
CH3
CH3CH2Cl
d.
HNO3
CH3CH2
AlCl3
H2SO4
CH2NH2
[1] LiAlH4
CH3
[2] H2O
H2
CH3CH2
NO2
CN
CH3CH2
Pd-C
NH2
[1] NaNO2, HCl
[2] CuCN
(+ ortho isomer)
CH3CH2
e. O2N
CH3
Br2
O2N
CH3
(from c.)
CH3
Br
H3O+
CH3CH2
[1] NaNO2, HCl
CN
HO
[2] H2O
CH3
Br
Br
Cl
NH2
Cl
H2 N
Pd-C
FeBr3
Cl
f.
H2
COOH
[1] NaNO2, HCl
N N
NH2
[2]
NH2
(from b.)
Cl
(from a.)
25.81
O
HNO3
a.
H2SO4
NO2
NO2 H
2
Cl2
FeCl3
NH2
O
H
N
O
O
Pd-C
Cl
Cl2
FeCl3
H
N
Cl
Cl
O
Cl
Cl
(+ isomer)
737
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–31
H2
b.
Pd-C H2N
O2N
NO2
[1] NaNO2, HCl
HNO3
[2] H2O
H2SO4 HO
HO
NH2
H2
Pd-C HO
O
(+ ortho isomer)
(from a.)
O
O
H
N
O
HO
O
O
ClCOCH2CH3
c.
Cl2
AlCl3
O
OH
NHCH3 NaBH4
Cl NH2CH3
NHCH3
CH3OH
H2O, HCl
25.82
CH3OH
OH
NH2
a.
[1] HNO3, H2SO4
[1] NaNO2, HCl
[2] H2, Pd-C
[2] H2O
OH
SOCl2
OH
CH3Cl
Br2
AlCl3
FeBr3
(2X)
OCH3
Br
Br
[1] NaH
Br
[2] CH3Cl
CH3
CH3
(+ ortho
isomer)
CH3CH2OH
Br
CH3
[1] Br2, hν
Br
[2] NH3
(excess)
NH2
CH3O
CrO3, H2SO4, H2O
Br
CH3COOH
NH2
SOCl2
H
N
CH3COCl
b.
O
H
N
Cl
O
H2O
O
NH2
O
AlCl3
(from a.)
–OH,
O
(+ ortho isomer)
CH3OH
SOCl2
c.
Cl2
FeCl3
[1] CH3Cl, AlCl3
Cl
[2] KMnO4
HOOC
Cl
[1] LiAlH4
[2] H2O
HO
Cl
PCC
O
Cl
H
(+ ortho isomer)
NH2
(from a.)
mild acid
N CH
Cl
CH3OH
SOCl2
d.
CH3Cl
AlCl3
CH3
CH3
HNO3
H2SO4
O2N
COOCH3
[1] KMnO4
[2] CH3OH, H+
O2N
COOCH3
H2
Pd-C
H2N
PCC
(+ ortho isomer)
O
(CH3)2CHNH
CO2CH3
NaBH3CN
OH
738
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–32
NH2
e. O2N
CH3
(from d.)
[1] H2, Pd-C
F
[1] KMnO4
CH3
[2] NaNO2, HCl
[3] HBF4
O
O
F
F
C
[2] SOCl2
Cl
HN
(from a.)
Probably a strong enough activator that the Friedel–Crafts reaction will still occur.
O
NHCOCH3
O
Make two parts:
AlCl3
Cl
CH3
C
NH
O
f.
I
I
CN
CN
NHCOCH3
NHCOCH3
NHCOCH3
[1] HNO3, H2SO4
[1] NaNO2, HCl
[2] H2, Pd-C
[2] CuCN
(from b.)
NH2
CN
(+ ortho isomer)
CH3
O2N
CH3
[1] H2, Pd-C
[2] NaNO2, HCl
(from d.)
O
[1] KMnO4
Cl
[2] SOCl2
I
I
[3] NaI
25.83
molecular weight = 87
C5H13N
two IR peaks = 1° amine
H
H
H
H
H
H
N
N
N
N
H
H
H
H
H
H
H
N
N
N
N
H
H
H
25.84
H2N
NHCH2CH3
CH2CH3
CH2NHCH3
Compound A: C8H11N
Compound B: C8H11N
IR absorption at 3400 cm–1→ 2° amine
IR absorption at 3310 cm–1 → 2° amine
1H NMR signals at (ppm):
1H NMR signals at (ppm):
1.3 (triplet, 3 H) CH3 adjacent to 2 H's
1.4 (singlet, 1 H) amine H
3.1 (quartet, 2 H) CH2 adjacent to 3 H's
2.4 (singlet, 3 H) CH3
3.8 (singlet, 2 H) CH2
3.6 (singlet, 1 H) amine H
6.8–7.2 (multiplet, 5 H) benzene ring
7.2 (multiplet, 5 H) benzene ring
Compound C: C8H11N
IR absorption at 3430 and
3350 cm–1 → 1° amine
1H NMR signals at (ppm):
1.3 (triplet, 3 H) CH3 near CH2
2.5 (quartet, 2 H) CH2 near CH3
3.6 (singlet, 2 H) amine H's
6.7 (doublet, 2 H) para disubstituted
7.0 (doublet, 2 H) benzene ring
739
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amines 25–33
25.85
C N
HO
HO
Compound D:
Molecular ion at m/z = 71: C3H5NO (possible formula)
IR absorption at 3600–3200 cm–1→ OH
2263 cm–1→ CN
Use integration values and the molecular formula to
determine the number of H's that give rise to each
signal.
1
H NMR signals at (ppm):
2.6 (triplet, 2 H) CH2 adjacent to 2 H's
3.2 (singlet, 1 H) OH
3.9 (triplet, 2 H) CH2 adjacent to 2 H's
NH2
Compound E:
Molecular ion at m/z = 75: C3H9NO (possible formula)
IR absorption at 3600–3200 cm–1→ OH
3636 cm–1 → N–H of amine
1H NMR signals at (ppm):
1.6 (quintet, 2 H) CH2 split by 2 CH2's
2.5 (singlet, 3 H) NH2 and OH
2.8 (triplet, 2 H) CH2 split by CH2
3.7 (triplet, 2 H) CH2 split by CH2
25.86 Guanidine is a strong base because its conjugate acid is stabilized by resonance. This
resonance delocalization makes guanidine easily donate its electron pair; thus it’s a strong
base.
NH
H2 N
C
HA
NH2
H2N
NH2
NH2
NH2
C
C
C
NH2
H2 N
NH2
H2 N
NH2
NH2
H2 N
C
NH2
pKa = 13.6
guanidine
25.87 The compound with the most available electron pair or the compound with the highest
electron density on an atom (N in this case) is the strongest base. Pyrrole is the weakest base
because its lone pair is delocalized on the five-membered ring to make it aromatic. Both
imidazole and thiazole contain sp2 hybridized N atoms with electron pairs that are localized
on N. Imidazole is a stronger base than thiazole, because its second N atom is more basic
than thiazole’s S atom, so it places more electron density on N by a resonance effect.
N
N
NH
imidazole
N
NH
N
pyrrole
least basic
N
S
thiazole This form contributes less
to the hybrid than the
equivalent resonance
structure in imidazole.
This N atom is
the strongest
base.
NH
S
N
S
thiazole
intermediate
basicity
NH
imidazole
most basic
25.88
CH3
N
[1] CH3I (excess)
CH3
N
CH3
I
[2] Ag2O
[2] Ag2O
[1] CH3I (excess)
[3] Δ
[3] Δ
N(CH3)2
I
N(CH3)3
C8H10
Y
740
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 25–34
25.89
One possibility:
O
O
O
CH3COCl
a.
O
O
Br2
O
AlCl3
H2NC(CH3)3
CH3CO2H
O
O
Br
O
O
(+ isomer)
NaBH4
CH3OH
HO
O
H3O+
HO
O
HN(CH2CH3)2
+
A
[1] NaNO2, HCl
H2
Pd-C
H2SO4 O N
2
N
H
OH
N(CH2CH3)2
HO
HNO3
O
N
H
OH
albuterol
b.
N
H
O
[2] H2O
H2N
[1] NaH
HO
O
Cl
[2]
Br2
O2N
COOH
HNO3
O
Mg
O
O
COCl
O2N
O
N(CH2CH3)2
O
H2
H2 N
N(CH2CH3)2
O
Pd-C
A
O
H3O+ CO2
O
SOCl2
O2N
Br
COOH
H2SO4
O
O
25.90 CH2=O reacts with the amine to form an intermediate imine, which undergoes an
intramolecular Diels–Alder reaction.
O
Ph
O
Ph
O
Ph
H
CH2=O
H2 N
H2O
X
OH
OH
H
H
one step
CH2
N
N
Y
C17H23NO
N
lupinine
+
N
epilupinine
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
741
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–1
Chapter 26 Carbon–Carbon Bond-Forming Reactions in Organic Synthesis
Chapter Review
Coupling reactions
[1] Coupling reactions of organocuprate reagents (26.1)
• R'X can be CH3X, RCH2X, 2o cyclic
+ RCu
R' X + R2CuLi
R' R
halides, vinyl halides, and aryl halides.
+ LiX
X = Cl, Br, I
• X may be Cl, Br, or I.
• With vinyl halides, coupling is
stereospecific.
[2] Suzuki reaction (26.2)
Y
R' X
+
Pd(PPh3)4
R B
NaOH
Y
X = Br, I
R' R
+
HO BY2
+
NaX
•
•
R'X is most often a vinyl halide
or aryl halide.
With vinyl halides, coupling is
stereospecific.
[3] Heck reaction (26.3)
Pd(OAc)2
R' X
Z
•
R'
Z
P(o-tolyl)3
X = Br or I
•
•
(CH3CH2)3N
(CH3CH2)3NH X–
•
R'X is a vinyl halide or aryl
halide.
Z = H, Ph, COOR, or CN
With vinyl halides, coupling is
stereospecific.
The reaction forms trans alkenes.
Cyclopropane synthesis
[1] Addition of dihalocarbenes to alkenes (26.4)
C
X X
C
CHX3
C
KOC(CH3)3
C
•
•
The reaction occurs with syn addition.
The position of substituents in the
alkene is retained in the cyclopropane.
•
•
The reaction occurs with syn addition.
The position of substituents in the
alkene is retained in the cyclopropane.
•
Metathesis works best when CH2=CH2,
a gas that escapes from the reaction
mixture, is formed as one product.
C
[2] Simmons–Smith reaction (26.5)
CH2I2
C C
Zn(Cu)
H H
C
C C
+ ZnI2
Metathesis (26.6)
2 RCH CH2
Grubbs
RCH CHR
catalyst
Grubbs
catalyst
diene
+ CH2 CH2
+ CH2 CH2
742
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–2
Practice Test on Chapter Review
1. a. Which functional groups react with lithium dialkyl cuprates?
1. epoxides
2. vinyl halides
3. acid chlorides
4. Compounds [1] and [2] both react with R2CuLi.
5. Compounds [1], [2], and [3] all react with R2CuLi.
b. Which of the following statements is (are) true for the Suzuki reaction?
1. Arylboranes can serve as one reactant.
2. The reaction is stereospecific.
3. The reaction occurs between a vinyl or aryl halide and an alkene in the presence of a
palladium catalyst.
4. Statements [1] and [2] are both true.
5. Statements [1], [2], and [3] are all true.
c. Which of the following compounds can react with CH2=CHCN in a Heck reaction?
1.
4. Both [1] and [2] can react.
Br
I
2.
5. Compounds [1], [2], and [3] can all react.
3. CH2 CH2
d. Which of the following compounds yields a pair of enantiomers on reaction with the Simmons–
Smith reagent?
1.
3.
4. Compounds [1] and [2] both yield a pair of enantiomers.
2.
5. Compounds [1], [2], and [3] all yield a pair of enantiomers.
e. Which of the following compounds can be made by a ring-closing metathesis reaction?
O
1.
3.
OH
4. Compounds [1] and [2] can both be prepared.
2.
5. Compounds [1], [2], and [3] can all be prepared.
HO
O
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
743
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–3
f. Which of the following compounds can be prepared from CH3–C≡C–H by a Suzuki reaction?
You may use other organic compounds or inorganic reagents.
1.
3.
4. Compounds [1] and [2] can both be prepared.
2.
5. Compounds [1], [2], and [3] can all be prepared.
2. Draw the product formed in each reaction. Indicate the stereochemistry around double bonds and
stereogenic centers when necessary.
a. HC CCH2CH2CH3
[1] catecholborane
[2]
e.
I , Pd(PPh3)4, NaOH
[1] 2 Li
Br
b.
[2] 0.5 CuI
[3] Ph
f.
Grubbs
cataylst
Br
[1] 2 Li
c.
CHBr3
KOC(CH3)3
g.
Br
[2] B(OCH3)3
[3]
Grubbs
cataylst
Br, Pd(PPh3)4, NaOH
OCH3
d.
CH3O
Pd(OAc)2
CO2CH3
I
P(o-tolyl)3
(CH3CH2)3N
3. What starting material is needed to synthesize each compound by ring-closure metathesis?
O
a.
b.
c.
HO
OH
OH
O
O
744
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–4
Answers to Practice Test
1.a. 5 2.
b. 4
a.
c. 4
d. 2
b.
e. 3
f. 4
3.
Br Br
e.
a.
Ph
O
b.
f.
c.
HO
OH
OCH3
d.
OH
g.
CH3O
O
c.
O
CO2CH3
Answers to Problems
26.1 A new C–C bond is formed in each coupling reaction.
a.
Cl
(CH2=CH)2CuLi
new C–C bond
b.
(CH3)2CuLi
Br
CH3
new C–C bond
c.
CH3O
CH3O
I
(CH3CH2CH2CH2)2CuLi
CH3O
new C–C bond
CH3O
Br
d.
2
new C–C bond
CuLi
26.2
I
OH
A=
(CH3CH2)2CuLi
OH
B=
several
steps
I
OH
(CH3)2CuLi
several
steps
OH
O
CO2CH3
C18 juvenile
hormone
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
745
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–5
26.3
Br
[1] Li
a. (CH3)2CHCH2CH2I
[(CH3)2CHCH2CH2]2CuLi
CH2CH2CH(CH3)2
[2] CuI
or
[1] Li
Br
CH2CH2CH(CH3)2
2
[1] Li
Cl
b.
(CH3)2CHCH2CH2I
CuLi
[2] CuI
Br
CuLi
[2] CuI
2
Cl
Cl
c.
[1] Li
CuLi
[2] CuI
2
26.4
I +
a.
b.
O
Pd(PPh3)4
O
NaOH
B
B(OCH3)2
Pd(PPh3)4
+
Br
NaOH
c.
[1] Li
Br
B(OCH3)2 + LiOCH3
[2] B(OCH3)3
O
O
B
C C H + H B
d.
O
O
26.5 The Suzuki reaction forms a new carbon–carbon bond between a vinyl halide and an arylborane.
O
CH3S
B(OH)2
+
O
O
O
Br
B
A
CH3S
26.6
O
O
H B
a. CH3CH2CH2 C C H
b.
c.
CH3O
I
Li
O
Li
Br2
I
B
H
O
Br
CH3O
(+ ortho isomer)
B(OCH3)3
Li
CH3O
B(OCH3)2
Li
B(OCH3)3
CH3O
Pd(PPh3)4, NaOH
I
Pd(PPh3)4
NaOH
B(OCH3)2
Br
Pd(PPh3)4
NaOH
CH3O
746
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–6
26.7
Br
Heck
a.
+
b.
I
reaction
OCH3
+
OCH3
Heck
reaction
Br
c.
+
CN
CN
Br
OCH3
d.
Heck
reaction
Heck
+
O
OCH3
reaction
O
26.8 Locate the double bond with the aryl, COOR, or CN substituent, and break the molecule into
two components at the end of the C=C not bonded to one of these substituents.
O
Br
O
OCH3
a.
OCH3
Br
b.
O
c.
O
CO2CH3
CO2CH3
Br
26.9 Add the carbene carbon from either side of the alkene.
H
CH3
a.
C C
H
H
Cl Cl
C
CHCl3
KOC(CH3)3
CH3 C
H
C
H
+
CH3
H
C
H
enantiomers
H
H
C
C
Cl Cl
CH3CH2
b.
CH2CH3
C C
H
H
Cl Cl
C
CHCl3
C
KOC(CH3)3 CH3CH2
H
C
+
CH2CH3
H
CH3CH2
H
CH3
CHCl3
KOC(CH3)3
CH3
CCl2
CH3
+
H
CCl2
H
enantiomers
C
C
identical
c.
C
Cl Cl
CH2CH3
H
747
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–7
26.10
CHCl3
a.
LiCu(CH3)2
c.
Cl
Cl
KOC(CH3)3
Br
Br
(from b.)
CHBr3
b.
Br
Br
KOC(CH3)3
26.11
CH2I2
a.
ZnI2
+
CH2I2
c.
+
Zn(Cu)
Zn(Cu)
ZnI2
H
CH2I2
b.
ZnI2
+
Zn(Cu)
H
26.12 The relative position of substituents in the reactant is retained in the product.
CH3CH2 C C
H
H H
C
H
CH2I2
CH2CH3
Zn(Cu)
CH3CH2 C
H
trans-3-hexene
C
CH3CH2
H
C
C
H
CH2CH3
C
H H
H
CH2CH3
two enantiomers of trans-1,2-diethylcyclopropane
26.13
Grubbs
catalyst
a.
Grubbs
catalyst
c.
(E and Z)
OCH3
Grubbs
catalyst
b.
(CH2=CH2 is also formed in each reaction.)
(E and Z)
OCH3
OCH3
26.14
+
cis-2-pentene
+
There are four products formed in this reaction including stereoisomers, and
therefore, it is not a practical method to synthesize 1,2-disubstituted alkenes.
26.15
O
a.
Ph
+
O
O
Ph
CO2CH3
b.
CO2CH3
O
CO2CH3
CO2CH3
748
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–8
26.16
O
O
RCM
O
H
O
OR
O
O
new C–C bond here
H
V
OR
26.17 Cleave the C=C bond in the product, and then bond each carbon of the original alkene to a
CH2 group using a double bond.
a.
O
O
c.
CO2CH3
CO2CH3
CHO
CHO
O
O
b.
CH3O
CH3O
OH
OH
26.18 Inversion of configuration occurs with the substitution of the methyl group for the tosylate.
O
SO2
CH3
(CH3)2CuLi
a.
CH3
(equatorial)
(CH3)2CuLi
b.
O SO2
CH3
CH3
(axial)
26.19
a.
O
O
RCM
b.
O
O
EtO2C CO2Et
EtO2C CO2Et
RCM
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
749
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–9
26.20
Cl
CH2CH2CH3
+
a.
(CH3CH2CH2)2CuLi
N
B(OCH3)2
Pd(PPh3)4
Br
+
b.
N
NaOH
O
O
Br
+
c.
N
d.
Cl
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
[1] Li
[2] CuI
[3]
Br
O
e.
Br
Pd(PPh3)4
B
+
O
f.
g.
NaOH
Br +
CH3O
Br
N
CO2CH3
Pd(OAc)2
CH3O
P(o-tolyl)3
(CH3CH2)3N
CO2CH3
[1] Li
[2] B(OCH3)3
Br + Pd catalyst
[3]
O
[1] H B
h. (CH3)3C C C H
O
[2] C6H5Br, Pd(PPh3)4, NaOH
26.21
Br
a.
c.
CO2CH3
Br
CO2CH3
Br
d.
b.
Br
CH3O
CH3O
750
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–10
26.22
Each coupling reaction uses Pd(PPh3)4
and NaOH to form the conjugated diene.
I
A
I
O
H
H B
O
C C H
B
B
O
O
H
ethynylcyclohexane
I
C
It is not possible to synthesize diene D using a Suzuki reaction with ethynylcyclohexane as
starting material. Hydroboration of ethynylcyclohexane adds the elements of H and B in a syn
fashion, affording a trans vinylborane. Since the Suzuki reaction is stereospecific, one of the
double bonds in the product must be trans.
D
26.23 Locate the styrene part of the molecule, and break the molecule into two components. The
second component in each reaction is styrene, C6H5CH=CH2.
a.
Br
b.
c.
Br
Br
styrene part
styrene part
styrene part
26.24
HBr
1-butene
Br
ROOR
+
CuLi
2
octane
[1] Li
[2] CuI
CuLi
2
26.25 Add the carbene carbon from either side of the alkene.
H
a.
Cl
CHCl3
c.
Cl
KOC(CH3)3
CHBr3
b.
Br
KOC(CH3)3
H
H
CH3
Br
CH3
Br
+
Br
H
H
H
CH2I2
H
d.
Zn(Cu)
CH2I2
H
+
Zn(Cu)
H
H
H
H
751
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–11
Cl Cl
C
CHCl3
e.
Cl Cl
C
C
H
CHCl3
f.
KOC(CH3)3
KOC(CH3)3
H
C
C
H
H
+
C
C
C
H
C
H
Cl Cl
26.26 Since the new three-membered ring has a stereogenic center on the C bonded to the phenyl
group, the phenyl group can be oriented in two different ways to afford two stereoisomers.
These products are diastereomers of each other.
CHI2
H
H
H
+
Zn(Cu)
H
H
H
26.27 High dilution conditions favor intramolecular metathesis.
H
OH
H
Grubbs
catalyst
N
a.
OH
N
OH
Grubbs
catalyst
c.
CO2CH3
OH
O
O
O
Grubbs
catalyst
O
b.
CO2CH3
26.28 Retrosynthetically break the double bond in the cyclic compound and add a new =CH2 at
each end to find the starting material.
O
a.
O
O
O
O
b.
O
CO2CH3
O
CO2CH3
c.
O
26.29 Alkene metathesis with two different alkenes is synthetically useful only when both alkenes
are symmetrically substituted; that is, the two groups on each end of the double bond are
identical to the two groups on the other end of the double bond.
752
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–12
a.
CH3CH2CH CH2
CH3CH2CH2CH CHCH2CH3 + CH3CH2CH CHCH2CH3 + CH3CH2CH2CH CHCH2CH2CH3
+
+ CH2 CH2 (Z + E)
CH3CH2CH2CH CH2
(Z + E)
b.
(Z + E)
This reaction is synthetically useful since it
yields only one product.
+
c.
+
+
+
+
+
+ CH3CH CHCH3 + CH3CH2CH CHCH2CH3
(Z + E)
(Z + E)
+
26.30
O
O
O
M
O
O
26.31 All double bonds can have either the E or Z configuration.
a.
c.
b.
26.32
Br Br
CHBr3
a.
KOC(CH3)3
Br
COOH
+
b.
CO2CH3
COOH
CH3O2C
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
Br
+
c.
NHCOCH3
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
NHCOCH3
CH3
CH3
d.
B(OCH3)2
O
e.
B
Br
+
O
CH3O
+
Br
Pd(PPh3)4
NaOH
Pd(PPh3)4
NaOH
CH3O
O
753
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–13
O
Grubbs
f.
O
catalyst
O
O
CH2I2
g.
Zn(Cu)
Br Br
h.
(CH3)2CuLi
Product in (a)
[1] Li
Cl
i.
[2] CuI
Br
[3]
O
[1]
j.
H B
H
O
CH3CH2 C C H
[2]
Br
H
Pd(PPh3)4
NaOH
26.33
O
Cl
Δ
Cl3C C
Cl
O Na+
C Na+ +
O C O
Cl
Cl
C
+
CO2
+
NaCl
Cl
26.34 This reaction follows the Simmons–Smith reaction mechanism illustrated in Mechanism 26.5.
Br
CHBr2
Zn(Cu)
[1]
CH
Zn
Br
+
[2]
ZnBr2
26.35
Ph2S
O
OCH3
Ph2S
a sulfur ylide
1,4-addition of the
sulfur ylide to the
β carbon
O
OCH3
OCH3
a resonance-stabilized
enolate
O
methyl trans-chrysanthemate
Ph2S
754
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–14
26.36
+ N2
A
N
N
CO2CH3
N
N
CO2CH3
CO2CH3
diazo compound
CO2CH3
B
26.37
I
PAr3
[1]
a.
I
Ar3P Pd
Pd(PAr3)2
oxidative addition
PAr3
[2]
+
Ar3P Pd
I
syn addition
PAr3
I
Ar3P Pd
PAr3
[3]
H
syn elimination
H
H
+
[4]
I
reductive
elimination
H
E
This H
is trans
to Pd.
The H cis to Pd
is eliminated.
Ar3P Pd
Pd(PAr3)2 + HI
b. This suggests that the stereochemistry in Step [3] must occur with syn elimination of H and
Pd to form E. Product F cannot form because the only H on the C bonded to the benzene
ring is trans to the Pd species, so it cannot be removed if elimination occurs in a syn fashion.
26.38
O
H B
Br NaNH2
O
C CH
Br
O
(– HBr)
B
O
(Z)-2-bromostyrene
Pd(PPh3)4
NaOH
A
26.39
Br2
FeBr3
O
O
H
H B
O
B
O
Br
Pd(PPh3)4, NaOH
755
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–15
26.40
HO
SOCl2
OH
HO
Cl
TBDMS–Cl
imidazole
NaH
HC CH
HC C
TBDMSO
Cl
O
B
OH
O
O
Bu4NF
BH
O
O
HC C
B
OTBDMS
OTBDMS
O
26.41
a. CH3O
CH3O
HNO3
NO2
H2SO4
H2
NH2
Pd-C
Br2
[1] NaH
OH
NaNO2
N2+Cl–
HCl
CH3O
CH3O
[2] CH3Br
PBr3
Synthesize these two components,
and then use a Heck reaction to
synthesize the product.
Br
H2O
OH
Br
CH3OH
CH3CH2OH
SOCl2
CH3CH2Cl
CH2CH3
AlCl3
CHCH3
hν
KOC(CH3)3
Br
Pd(OAc)2
Br
CH3O
Br2
CH3O
P(o-tolyl)3
(CH3CH2)3N
Br
CH3CH2OH
SOCl2
Br
CH3CH2Cl
b.
CH3O
AlCl3
(from a.)
Br
CO2Et
CH3O
CH3O
CO2Et
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
CO2Et
H2
Pd-C
CH3O
Br2
Br
CH3O
MgBr
Mg
[1] (2 equiv)
[2] H2O
FeBr3
OH
CH3O
756
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–16
26.42
Br Br
a.
OH
OH
Br Br
NBS
Br
NaOH
CHBr3
OH
hν
OH
KOC(CH3)3
Br
b.
Br
CH2I2
(CH3CH2CH2CH2)2CuLi
Zn(Cu)
(from a.)
Br
c.
CuLi
CH2I2
2
Br
Zn(Cu)
(from a.)
26.43
Br
Br2
a.
FeBr3
CH2I2
CH2=CH2
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
Zn(Cu)
styrene
CHCl3
b.
styrene
(from a.)
Cl
KOC(CH3)3
Br
Cl
CO2CH3
CO2CH3
c.
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
(from a.)
Br
d.
[1] Li
Br
CuLi
[2] CuI
(from a.)
2
For an alternate
synthesis of styrene,
see Problem 26.41.
757
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–17
26.44
[1] NaH
a.
C CH
HC CH
Cl
[2]
[1] NaH
H2
Lindlar
catalyst
[2]
Cl
CH2I2
Zn(Cu)
+ enantiomer
[1] NaH
b.
[1] NaH
C CH
HC CH
[2]
Na
Cl
[2]
NH3
Cl
CH2I2
Zn(Cu)
+ enantiomer
c.
O
CH2I2
CH2
Ph3P=CH2
Zn(Cu)
Br
d.
O
Br2
CH3CO2H
O
LiBr
O
Li2CO3, DMF
[1] (CH3)2CuLi
O
[2] H2O
Ph3P=CH2
[1] (CH3)2CuLi
CHCl3
[2] H2O
Cl
Cl
KOC(CH3)3
758
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–18
26.45
Either compound can be used to
a. CH3
OCH3
CH3
Br
Br
OCH3 synthesize the organoborane, so
two routes are possible.
Possibility [1]:
CH3Cl
AlCl3
Br2
Br2
CH3
[1] Li
CH3
FeBr3
HNO3
Br
FeBr3
Br
[2] B(OCH3)3
Br
H2
NO2
Br
CH3
B(OCH3)2
NH2
Pd-C
H2SO4
[1] NaNO2, HCl
Br
OH
[2] H2O
[1] NaH
[2] CH3I
Br
Pd(PPh3)4
CH3
B(OCH3)2
Br
OCH3
NaOH
CH3
OCH3
OCH3
Possibility [2]:
Br
OCH3
Pd(PPh3)4
[1] Li
(CH3O)2B
OCH3
[2] B(OCH3)3
CH3
Br
NaOH
(from Possibility [1])
(from Possibility [1])
CH3
HO
OCH3
HO
b.
Br
Br
The acidic OH makes it impossible to prepare an organolithium reagent
from this aryl halide, so this compound must be used as the aryl halide
that couples with the organoborane from bromobenzene.
O2N
O2N
HNO3
Br2
H2SO4
FeBr3
Br2
FeBr3
Br
H2N
Br
[1] Li
(CH3O)2B
[2] B(OCH3)3
HO
Pd(PPh3)4
Br
NaOH
[1] NaNO2, HCl
Br
Pd-C
HO
(CH3O)2B
HO
H2
Br
[2] H2O
759
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–19
Br
O
c.
O
Br
O
O
This can't be converted to an organoborane
reagent via an organolithium reagent.
three
B(OCH3)2
(as in 26.45b)
steps
Br2
FeBr3
HNO3
Br
[1] NaNO2, HCl
H2
NO2
Br
H2SO4
Pd-C
Br
NH2
[2] H2O
Br
OH
CH3CO
Cl
pyridine
B(OCH3)2
Br
Pd(PPh3)4
O
O
Br
O
NaOH
O
O
O
26.46
EtO2C CO2Et
EtO2C CO2Et
Grubbs
catalyst
a.
Synthesis of starting material:
[1] NaOEt
CH2(CO2Et)2
[2]
CH(COEt)2
EtO2C CO2Et
[1] NaOEt
[2]
Cl
Cl
SOCl2
OH
SOCl2
OH
OTBDMS
b.
OTBDMS
Grubbs
catalyst
Synthesis of starting material:
OH
PCC
H
OH
O
OH
PBr3
H2O
Br
Mg
MgBr
TBDMS–Cl
imidazole
OTBDMS
760
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–20
26.47
O
O
Grubbs
catalyst
a.
Synthesis of starting material:
O
SOCl2
CrO3
OH
H2SO4
H2O
OH
O
O
Cl
AlCl3
OH
O
NaBH4
[1] NaOEt
CH3OH
[2]
Cl
SOCl2
OH
b.
O
O
Grubbs
catalyst
O
O
Synthesis of starting material:
Br2
Br
MgBr
Mg
OH
FeBr3
OH
H2SO4
O
CH2 CH2
PCC
CHO
H2O
mCPBA
OH
TsOH
O
O
761
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–21
26.48
Cl
CH3
CH3O
N
Cl
CH3
CH3O
CO2CH3
N
[1] (CH3CH2CH2CH2)4N+ F—
CO2CH3 [2] PCC
Cl
CH3
CH3O
N
CO2CH3
OTBDMS
LiCu
OTBDMS
I
CHO
2
directed aldol
[1] CH3CHO + LDA → −CH2CHO
[2] H3O+
O
N
CH3O
O
O
Cl
O
Cl
O
N
CH3
CH3O
N
CO2CH3
several steps
O
CH3O
OH
N
H
O
CHO
A
maytansine
26.49
NO2
Br
NH2
Br H2
HNO3
a.
OH
Br [1] NaNO2, HCl
Pd-C
H2SO4
OCH3
Br [1] NaH
[2] H2O
OCH3
CuLi
Br [1] Li
[2] CH3I
[2] CuI
2
Br
OCH3
OCH3
OH
H2O
(2 enantiomers)
H2SO4
OH
b.
[1] OsO4
OTBDMS
TBDMS–Cl
HO
Br
Br
[2] NaHSO3, H2O
TBDMSO
imidazole
[1] Li
[2] CuI
Br
OTBDMS
Br
OTBDMS
TBDMSO
TBDMSO
CuLi
2
(CH3CH2CH2CH2)4N+F–
OH
HO
(2 enantiomers)
O
H
HB
c.
O
O
B
O
Br
Pd(PPh3)4, NaOH
762
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–22
Cl
Br
d.
Br
AlCl3
CO2CH3
CO2CH3
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
CO2CH3
Br
e.
CO2CH3
Pd(OAc)2
P(o-tolyl)3
(CH3CH2)3N
Br
CO2CH3
H
(+ enantiomer)
[1] Li
f.
H
Zn(Cu)
CH2I2
O
mCPBA
CuLi
[2] CuI
Br
2
(2 enantiomers)
26.50
O
a.
O
O
O
b.
O
O
O
O
Δ
Δ
2
[1] LiAlH4
CO2Et
EtO2C
CO2Et
OH
[2] H2O
CO2Et
OH
[1] NaH (2 equiv)
Cl (2 equiv)
[2]
O
O
26.51 There is more than one way to form Z by metathesis reactions. One possibility involves ring
opening of the bicyclic alkene followed by successive ring closures to generate the five- and
seven-membered rings.
O
O
O
RCM
H
O
O
H
O
RCM
O
H
O
Ru
H
Z
ring open
763
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–23
26.52
I
O
O
+
A
Pd(PPh3)4
B
O
N
H
OCH2CH3
N
NaOH
B
C
OCH2CH3
OCH2CH3
H3O+
H
D
C11H11NO
N
O
O
OCH2CH3
H OH2
H OH2
NH
OCH2CH3
NH
O
OCH2CH3
N H
+ H2O
O
N
+ H2O
O
O
C
H
H
H2O
N
N
D
OCH2CH3
O
H3O
N
O
O
+ H2O
resonance stabilized
+
H OCH2CH3
26.53
O
CH3O
O PPh3
PPh3
CH3O
H
O
X
CH3O
H
O
Y
+ Ph3P=O
O
Ph3P
CH3O
CH3O
O
Ph3P
O
+ PPh3
26.54 a. Reaction of a terminal alkene with the catalyst forms a metal–carbene that undergoes an
intramolecular reaction with the triple bond, generating a new metal–carbene. A second
intramolecular reaction forms the bicyclic product.
OSiEt3
OSiEt3
OSiEt3
OSiEt3
M
M
A
cyclization
two new C=C's
cyclization
B
764
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 26–24
b. Two products are possible because the cascade of reactions can begin at two different double bonds.
OSiEt3
OSiEt3
OSiEt3
M
OSiEt3
M
C
Begin here.
OSiEt3
OSiEt3
OSiEt3
M
M
C Begin here.
OSiEt3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
765
Pericyclic Reactions 27–1
Chapter 27 Pericyclic Reactions
Chapter Review
Electrocyclic reactions (27.3)
Woodward–Hoffmann rules for electrocyclic reactions
Number of π bonds
Even
Odd
Thermal reaction
Conrotatory
Disrotatory
Photochemical reaction
Disrotatory
Conrotatory
Examples
The stereochemistry of a thermal electrocyclic reaction is opposite to that of a photochemical
electrocyclic reaction.
CH3
• A thermal electrocyclic reaction with an even
Δ
number of π bonds occurs in a conrotatory
CH3
fashion.
CH3
+ enantiomer
CH3
CH3
hν
(2E,4E)-2,4-hexadiene
2 π bonds
CH3
• A photochemical electrocyclic reaction with
an even number of π bonds occurs in a
disrotatory fashion.
Cycloaddition reactions (27.4)
Woodward–Hoffmann rules for cycloaddition reactions
Number of π bonds
Even
Odd
Thermal reaction
Antarafacial
Suprafacial
Photochemical reaction
Suprafacial
Antarafacial
Examples
[1] A thermal [4 + 2] cycloaddition takes place in a suprafacial fashion with an odd number of
π bonds. An antarafacial photochemical [4 + 2] cycloaddition to form a six-membered ring cannot
occur, because of the geometrical constraints of forming a six-membered ring.
CH3
H
H
+
CH2
CH2
CH3
Δ
suprafacial
CH3
H
CH3
H
cis product only
766
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–2
[2] A photochemical [2 + 2] cycloaddition takes place in a suprafacial fashion with an even number
of π bonds. An antarafacial thermal [2 + 2] cycloaddition to form a four-membered ring cannot
occur, because of the geometrical constraints of forming a four-membered ring.
C6H5
C6H5
C
H
CH2
C
+
C6H5
hν
C6H5
suprafacial
H
cis product only
CH2
Sigmatropic rearrangements (27.5)
Woodward–Hoffmann rules for sigmatropic rearrangements
Number of electron pairs
Even
Odd
Thermal reaction
Antarafacial
Suprafacial
Photochemical reaction
Suprafacial
Antarafacial
Examples
[1] A Cope rearrangement is a thermal [3,3] sigmatropic rearrangement that converts a 1,5-diene
into an isomeric 1,5-diene.
Δ
isomeric 1,5-diene
1,5-diene
[2] An oxy-Cope rearrangement is a thermal [3,3] sigmatropic rearrangement that converts a
1,5-dien-3-ol into a δ, ε-unsaturated carbonyl compound, after tautomerization of an
intermediate enol.
O
HO
Δ
[3,3]
1,5-dien-3-ol
δ,ε-unsaturated carbonyl compound
[3] A Claisen rearrangement is a thermal [3,3] sigmatropic rearrangement that converts an
unsaturated ether into a γ,δ-unsaturated carbonyl compound.
Δ
O
unsaturated ether
O
γ,δ-unsaturated carbonyl compound
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
767
Pericyclic Reactions 27–3
Practice Test on Chapter Review
1. a. Which of the following pericyclic reactions is symmetry allowed and readily occurs?
1. a photochemical conrotatory electrocyclic ring closure of a conjugated triene
2. a disrotatory thermal electrocyclic ring opening of a substituted cyclohexadiene
3. a thermal [2 + 2] cycloaddition
4. Reactions [1] and [2] will both occur.
5. Reactions [1], [2], and [3] will all occur.
b. Which of the following reactions requires suprafacial stereochemistry to be symmetry allowed?
1. a photochemical [1,5] sigmatropic rearrangement
2. a thermal [8 + 2] cycloaddition
3. a photochemical [4 + 2] cycloaddition
4. Reactions [1] and [2] are both suprafacial.
5. Reactions [1], [2], and [3] are all suprafacial.
c. What product(s) are formed from the photochemical [2 + 2] cycloaddition of (3E)-3-hexene?
A
1.
2.
3.
4.
5.
B
C
A only
B only
C only
A and B
A, B, and C
d. What product(s) are formed from the photochemical electrocyclic ring opening of
cis-3,4-dimethylcyclobutene?
1.
2.
3.
4.
5.
(2E,4E)-2,4-hexadiene
(2E,4Z)-2,4-hexadiene
(3Z)-1,3,5-hexatriene
Compounds [1] and [2] are both formed.
Compounds [1], [2], and [3] are all formed.
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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–4
e. What product(s) are formed by the [3,3] sigmatropic rearrangement of 1,5-cyclodecadien-3-ol?
OH
A
1.
2.
3.
4.
5.
CHO
CHO
B
C
A only
B only
C only
A and B
A, B, and C
2. Consider the p orbitals of the terminal carbons of a conjugated polyene with like phases on the
same side of the molecule (as in A) or opposite sides of the molecule (as in B), and answer each
question.
A
B
a. Which drawing is consistent with the ground state HOMO of a conjugated triene?
b. Which drawing is consistent with the excited state LUMO of a conjugated diene?
c. Which drawing is consistent with the ground state LUMO for a conjugated tetraene?
3. What type of sigmatropic rearrangement is depicted in each reaction?
a.
b.
Answers to Practice Test
1. a. 4
b. 2
c. 4
d. 1
e. 3
2. a. A
b. B
c. A
3. a. [3,3]
b. [1,3]
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
769
Pericyclic Reactions 27–5
Answers to Problems
27.1
Use the following definitions:
• An electrocyclic ring closure is an intramolecular reaction that forms a cyclic product
containing one more σ bond and one fewer π bond than the reactant. An electrocyclic
ring opening is a reaction in which a σ bond of a cyclic reactant is cleaved to form a
conjugated product with one more π bond.
• A cycloaddition is a reaction between two compounds with π bonds that forms a cyclic
product with two new σ bonds.
• A sigmatropic rearrangement is a reaction in which a σ bond is broken in the reactant,
the π bonds rearrange, and a σ bond is formed in the product.
electrocyclic reaction
a.
2 π bonds
σ bond broken
3 π bonds
O
O
σ bond formed
+
b.
H2C CH2
cycloaddition
σ bond formed
σ bond formed
c.
4 π bonds
3 π bonds
d.
σ bond formed
1 π bond
σ bond broken
27.2
electrocyclic reaction
sigmatropic rearrangement
1 π bond
a. For a bonding molecular orbital, the number of bonding interactions is greater than the
number of nodes.
b. For an antibonding molecular orbital, the number of bonding interactions is less than the
number of nodes.
For 1,3-butadiene:
ψ1
ψ2
ψ 3*
ψ 4*
Bonding
3
2
1
0
Nodes
0
1
2
3
Type of MO
bonding MO
bonding MO
antibonding MO
antibonding MO
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–6
The molecular orbitals of all conjugated dienes look similar.
27.3
a.
c. Excited state
b. Ground state
ψ4∗
excited state LUMO
ψ3∗
excited state HOMO
ground state LUMO
Energy
770
ψ2
ground state HOMO
ψ1
a. There are 10 molecular orbitals from the 10 p orbitals of the five π bonds.
b. Five molecular orbitals are bonding and five molecular orbitals are antibonding.
c. The lowest energy molecular orbital (ψ1) has zero nodes.
27.4
ψ1
d. The highest energy molecular orbital (ψ10*) has nine nodes.
ψ10∗
To draw the product of an electrocyclic reaction, use curved arrows and begin at a π bond.
Move the π electrons to an adjacent carbon–carbon bond, and continue in a cyclic fashion.
27.5
a.
b.
Δ
c.
Δ
re-draw
Δ
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
771
Pericyclic Reactions 27–7
27.6
Thermal electrocyclic reactions occur in a disrotatory fashion for a conjugated polyene with
an odd number of π bonds, and in a conrotatory fashion for a conjugated polyene with an
even number of π bonds.
C6H5
C6H5
Δ
a.
disrotatory
C6H5
Δ
b.
conrotatory
C6H5
2 π bonds
3 π bonds
27.7
For an even number of π bonds, thermal electrocyclic reactions occur in a conrotatory fashion.
re-draw
Δ
re-draw
Δ
+ enantiomer
a.
b.
27.8
Photochemical electrocyclic reactions occur in a conrotatory fashion for a conjugated
polyene with an odd number of π bonds, and in a disrotatory fashion for a conjugated
polyene with an even number of π bonds.
C6H5
C6H5
hν
a.
conrotatory
C6H5
hν
b.
disrotatory
C6H5
+ enantiomer
27.9
[The (E,E) diene is favored
over the (Z,Z) diene.]
The photochemical electrocyclic reaction cleaves a six-membered ring to form a hexatriene.
hν
H
H
H
HO
HO
provitamin D3
7-dehydrocholesterol
27.10 Use the rules for electrocyclic reactions found in Answers 27.6 and 27.8.
Δ
a.
re-draw
disrotatory
3 π bonds
hν
conrotatory
+ enantiomer
772
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–8
Δ
re-draw
b.
+ enantiomer
disrotatory
3 π bonds
hν
conrotatory
27.11 Use the rules for electrocyclic reactions found in Answer 27.6. A reaction with three π bonds
and a disrotatory cyclization is thermal.
H
H
27.12 Count the number of π electrons in each reactant to classify the cycloaddition.
O
O
[2 + 2] cycloaddition
CH2
+
a.
CH2
2π
electrons
2π
electrons
(one possibility)
O
O
b.
[4 + 2] cycloaddition
CH2
CH2
4π
2π
electrons electrons
O
c.
O
[6 + 2] cycloaddition
CH2
CH2
6π
2π
electrons electrons
27.13 A thermal suprafacial addition is symmetry allowed in a [4 + 2] cycloaddition because like
phases interact.
LUMO of the diene
Like phases interact.
HOMO of the alkene
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
773
Pericyclic Reactions 27–9
27.14 A thermal [4 + 2] cycloaddition is suprafacial.
a.
CH2
re-draw
+ CH2 CH2
Δ
CH2
(E,E) diene
CN
b.
NC
re-draw
+
CN
Δ
CN
CN
+ enantiomer
CN
(E,Z) diene
27.15
CO2CH3
H O
CO2CH3
a.
O
Δ
H
endo
addition
H
O
X
O
CO2CH3
HH O
CO2CH3
b.
diene HOMO
H
O
H
H
O
dienophile LUMO
X
H
O
The dienophile is under the diene, by the rule of endo addition (Section 16.13).
The H's at the ring fusion are cis to each other, but trans to the CO2CH3 group.
27.16 A photochemical [2 + 2] cycloaddition is suprafacial.
C6 H5
a.
hν
+
C6 H5
C 6H 5
+ enantiomer
C 6H 5
O
O
H
H
b.
+
CN
CH2
CH2
+ enantiomer
CN
27.17 a. The photochemical [6 + 4] cycloaddition involves five π bonds (the total number of
π electrons divided by two) and is antarafacial.
b. A thermal [8 + 2] cycloaddition involves five π bonds and is suprafacial.
774
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–10
27.18 A photochemical [4 + 2] cycloaddition like the Diels–Alder reaction must proceed by an
antarafacial pathway. This would require either the 1,3-diene or the alkene component to
twist 180° in order for the like phases of the p orbitals to overlap. Such a rotation is not
possible in the formation of a six-membered ring.
excited state HOMO of the diene
180° twist required
LUMO of the alkene
27.19 Locate the σ bonds broken and formed, and count the number of atoms that connects them.
D 1
3
a.
2
1
D1
D
2
3
1
D
σ bond formed
σ bond broken
[1,3] sigmatropic rearrangement
1
2
3
1
3
1
2
3
b.
σ bond formed
1
2
2
3
σ bond broken
[3,3] sigmatropic rearrangement
27.20
D
2
a.
CD3
re-draw
1
3
4
D
D1
3
1
4
7
5
D
2
[1,7]
6
D
7
5
D1
6
b, c. The reaction involves four electron pairs (three π bonds and one σ bond), so it proceeds
by an antarafacial pathway under thermal conditions, and by a suprafacial pathway
under photochemical conditions.
27.21 Draw the products of each reaction.
CO2CH3
a.
re-draw
CH3O2C
[3,3]
CH3O2C
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
775
Pericyclic Reactions 27–11
[3,3]
b.
O
OH
[3,3]
tautomerize
c.
OH
27.22 Draw the product after protonation.
HO
O
[3,3]
base
protonation
O
O
27.23 Draw the product of Claisen rearrangement.
CHO
[3,3]
a.
[3,3]
c. O
OHC
O
[3,3]
b.
O
CHO
27.24
O
a.
O
O
O
RO
O
Z
O
O
O
O
O
RO
O
b.
O
O
O
O
O
O
O
O
O
+
metathesis
RO
RO
CH2 CH2
776
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–12
27.25 Predict the stereochemistry of each reaction using Table 27.4.
a. A [6 + 4] thermal cycloaddition involves five electron pairs, making the reaction
suprafacial.
b. A photochemical electrocyclic ring closure of 1,3,5,7,9-decapentaene involves five
electron pairs, making the reaction conrotatory.
c. A [4 + 4] photochemical cycloaddition involves four electron pairs, making the reaction
suprafacial.
d. A thermal [5,5] sigmatropic rearrangement involves five electron pairs, making the
reaction suprafacial.
27.26 Use the rules found in Answers 27.6 and 27.8.
A
Δ
Δ
(a)
disrotatory
(a)
disrotatory
3 π bonds
B
hν
hν
(b)
conrotatory
(b)
conrotatory
3 π bonds
27.27 Draw the product of [3,3] sigmatropic rearrangement of each compound.
OH
O
a.
[3,3]
tautomerization
OH
O
O
OH
b.
[3,3]
tautomerization
27.28 An electrocyclic reaction forms a product with one more or one fewer π bond than the
starting material. A cycloaddition forms a ring with two new σ bonds. A sigmatropic
rearrangement forms a product with the same number of π bonds, but the π bonds are
rearranged. Use Table 27.4 to determine the stereochemistry.
H
thermal electrocyclic ring closure
• 3 π bonds
• disrotatory
Δ
a.
H
3 π bonds
2 π bonds
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
777
Pericyclic Reactions 27–13
Δ
H
b.
hν
2 π bonds
2 π bonds
thermal [1,5] sigmatropic
rearrangement
• 3 electron pairs
(2 π + 1 σ)
• suprafacial
c.
3 π bonds
photochemical electrocyclic
ring opening
• 2 π bonds in diene formed
• disrotatory
2 new σ bonds
Δ
N(CH2CH3)2
N(CH2CH3)2
thermal [6 + 4] cycloaddition
• 5 π bonds
• suprafacial
27.29 Use the rules for thermal electrocyclic reactions found in Answer 27.6.
E
Δ
a.
disrotatory
Z
Δ
b.
re-draw
Z
conrotatory
2 π bonds
3 π bonds
27.30 Use the rules for photochemical electrocyclic reactions found in Answer 27.8.
E
a.
hν
conrotatory
Z
E
3 π bonds
Although conrotatory ring opening could also form, at least in theory, an all-(Z) triene, steric
hindrance during ring opening would cause the terminal CH3’s to crash into one another,
making this process unlikely.
hν
b.
re-draw
disrotatory
+
enantiomer
778
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–14
27.31 Use the rules found in Answer 27.8.
hν
a.
re-draw
disrotatory
4 π bonds
hν
b.
re-draw
4 π bonds
+ enantiomer
disrotatory
27.32 Use the rules found in Answers 27.6 and 27.8.
2E
a, b.
Δ
4Z
+ enantiomer
disrotatory
6Z
3 π bonds
hν
+ enantiomer
conrotatory
E
Δ
c.
disrotatory
E
3 π bonds
(one enantiomer)
E
hν
conrotatory
d.
E
3 π bonds
(one enantiomer)
27.33 The trans product is indicative of a disrotatory ring closure from the cyclic triene with the
given stereochemistry at the double bonds. A disrotatory ring closure with a polyene having
three π bonds must occur under thermal conditions.
E
Z
H
H
trans
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
779
Pericyclic Reactions 27–15
27.34 A disrotatory cyclization of a reactant with an even number of π bonds must occur under
photochemical conditions.
hν
H H
cis
M
2 π bonds
27.35 Use the rules found in Answers 27.6 and 27.8.
H
H
Δ
a, c.
hν
b, c.
disrotatory
conrotatory
H
N
3 π bonds
H
N
3 π bonds
cis
trans
+ enantiomer
27.36 Use the rules found in Answers 27.6 and 27.8.
H
E
Δ
conrotatory
H
Z
Q
2 π bonds
P
hν
disrotatory
Z
(A 10-membered ring cannot contain two E double bonds.)
Z
R
27.37 Since the reaction involves three π bonds in one reactant and two π bonds in the second
reactant, the reaction is a [6 + 4] cycloaddition. A suprafacial cycloaddition with five π
bonds must proceed under thermal conditions.
two new σ bonds formed
on the same side
[1]
+
O
O
O
O
C6H5
C6H5
[2]
+
C6H5
C6H5
780
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–16
27.38 The Diels–Alder reaction is a thermal, suprafacial [4 + 2] cycloaddition.
O
a.
+
H
Δ
O
O
O
O
H
O
+
b.
H
Δ
O
O
O
H
O
O
O
+ enantiomer
27.39 A photochemical [2 + 2] cycloaddition is suprafacial.
hν
a. 2
b.
hν
2
27.40 A thermal [4 + 2] cycloaddition is suprafacial.
CO2CH3
a.
CO2CH3
Δ
CO2CH3
C
c.
C
CO2CH3
CO2CH3
CO2CH3
CO2CH3
b.
CO2CH3
Δ
CO2CH3
Δ
CH3O2C
CO2CH3
CO2CH3
27.41 1,3-Butadiene can react with itself in a symmetry-allowed thermal [4 + 2] cycloaddition to
form 4-vinylcyclohexene.
[4 + 2]
Δ
4-vinylcyclohexene
1,5-Cyclooctadiene would have to be formed from 1,3-butadiene by a [4 + 4] cycloaddition,
which is not allowed under thermal conditions.
Δ
[4 + 4]
1,5-cyclooctadiene
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
781
Pericyclic Reactions 27–17
27.42 A series of three [2 + 2] cycloadditions with E alkenes forms X.
CO2CH2CH3
CO2CH2CH3
[2 + 2]
CO2CH2CH3
X
CO2CH2CH3
27.43 Re-draw the reactant and product to more clearly show the relative location of the bonds
broken and formed.
2
1
3
2
1
3
a.
re-draw
1
[3,3]
3
2
1
D
5
5
4
b. D
re-draw
3
1
2
[1,5]
H
3
2
O
SC6H5
re-draw
1
2
1
3
c.
CD2H
4
D
D
3
2
2
3
1
O
S2 1
[2,3]
2
1
S
O
1
C6H5
C6H5
27.44 Draw the products of each reaction.
H
[3,3]
a.
[3,3]
c.
O
H
O
[3,3]
b.
O
tautomerize
OH
OH
27.45 a. Two [1,5] sigmatropic rearrangements occur.
H1
[1,5]
1
2
3
3
2
1
5
4
5-methyl1,3-cyclopentadiene
H
[1,5]
H1
4
5
1-methyl1,3-cyclopentadiene
H H
2-methyl1,3-cyclopentadiene
782
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–18
b.
H
1
2
3
[1,3]
H H
5-methyl
isomer
2-methyl
isomer
A [1,3] sigmatropic rearrangement requires photochemical conditions not thermal conditions, so
5-methyl-1,3-cyclopentadiene cannot rearrange directly to its 2-methyl isomer by a [1,3] shift.
27.46
3
4
3
1
2
2
O
4
4
1
3
5
4
2
[5,5]
5
1
O
OH
H
5
3
1
2
tautomerization
5
27.47
O
OH
+
O
Cl
O
LDA
O
pyridine
O
O
O
B
A
re-draw
OH
The conversion of B to C is a
[3,3] sigmatropic rearrangement
of an intermediate enolate.
O
H3O+
CO2H
re-draw
O
O
O
[3,3]
C
27.48 Use the definitions found in Answer 27.1.
Δ
an electrocyclic ring opening
[1]
2 π bonds
3
4
3
1
5
3 π bonds
2
7
Δ
H [2]
1
4
6
6
3 π bonds
2
1
5
7
3 π bonds
Δ
an electrocyclic ring closure
[3]
3 π bonds
a [1,7] sigmatropic rearrangement
H
2 π bonds
O
Δ
783
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Pericyclic Reactions 27–19
27.49
CO2CH3
a.
CH3O2C
+
CO2CH3
D
hν
CH2 CH2
disrotatory
D
D
H
[4 + 2]
c.
Δ
D
H
D
hν
d.
CO2CH3
D
Δ
b.
+ enantiomer
[2 + 2]
D
conrotatory
D
27.50
O
O
Δ
a.
[3,3]
Claisen
HO
O
[1] KH
b.
[2] H3O+
anionic oxy-Cope
O
18-crown-6
O
1
2
c.
d.
O 1
Δ
2
3
3
Claisen
hν
[2 + 2]
C7H8
O
OH
tautomerization
784
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–20
27.51 The mechanism consists of sequential [3,3] sigmatropic rearrangements, followed by
tautomerization.
O
O
O
OH
H+ source
[3,3]
[3,3]
H
H
OH
OH
H
B
27.52
O
H+
B
O
H
HO
HB+
O
O
O
O
O
[3,3]
O
H+
O
O
O
allyl vinyl ether
H
O
H
B
+
HB
OH
H
27.53
Z
H
Δ
H
conrotatory electrocyclic
ring opening forming
4 π bonds
H
Δ
E
disrotatory ring
closure involving only
3 of the 4 π bonds
H
785
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Pericyclic Reactions 27–21
27.54
SCH3
SCH3
Δ
OCH3
OCH3
electrocyclic ring opening
forming two π bonds
O
CH3S
H
Δ
CH3S
[4 + 2]
cycloaddition
suprafacial
CH3O
O
CH3O
O
O
O
H
O
27.55 The mechanism consists of a [4 + 2] cycloaddition, followed by intramolecular imine formation.
H H
NH2
O
NH2
O
H
N
O
N
H HO
[4 + 2]
O
O
H
O
proton
transfer
H
O
H
H
H2O
H
N
O
H
O
H
O
H3O
27.56
This bridged bicyclic system is cleaved.
HO
H
=
H
H
OH
OH
H
H
[3,3]
This bond forms a new
bridged ring system.
N
H2O
N
H
H
786
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–22
27.57
O
O
O
B
O H
O
OCH2CH3
O
OCH2CH3
O
+ HB+
H OH
O
O
O
O
O
O
O
O
O
CH3CH2O
H
B
O
H
CH3CH2OH
B
O
O
O
H
H2O
O
O
OH
H OH
CO2
O
H
HO–
787
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Pericyclic Reactions 27–23
27.58 Conrotatory cyclization of Y using four π bonds forms X. Disrotatory ring closure of X can
occur in two ways—on the top face or bottom face of the eight-membered ring to form
diastereomers.
Δ
C6H5
C6H5
4 π bonds
conrotatory
CO2CH3
CO2CH3
+ enantiomer
X
Y
disrotatory
H
C6H5
diene
=
H
H
H
H
C6H5
CO2CH3
C6H5
CO2CH3
CO2CH3
endiandric acid E
methyl ester
dienophile
endiandric acid D
methyl ester
Only this stereoisomer has the
side chain close to the sixmembered ring for Diels–Alder.
[4 + 2]
C6H5
H
H
H
H
H
Endiandric acid A has a free COOH
group instead of the CO2CH3 group.
methyl ester of
endiandric acid A
H
CO2CH3
27.59
O
O
a.
O
O
O
O
C6H5
dienophile
O
diene dienophile
O
b.
or
O
O
diene
C6H5
or
C6H5
diene
O
C6H5
dienophile
O
O
dienophile
O
diene
O
O
788
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 27–24
O2CCH3
O2CCH3
O2CCH3
O
C4H9
O
O
[4 + 2]
c.
C4H9
O
[3,3]
cycloaddition
O2CCH3
C4H9
O2CCH3
sigmatropic
rearrangement
O2CCH3
O
C4H9
O
C4H9
C4H9
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
789
Carbohydrates 28–1
Chapter 28 Carbohydrates
Chapter Review
Important terms
A monosaccharide containing an aldehyde (28.2)
• Aldose
A monosaccharide containing a ketone (28.2)
• Ketose
A monosaccharide with the O bonded to the stereogenic center farthest from the
• D-Sugar
carbonyl group drawn on the right in the Fischer projection (28.2C)
Two diastereomers that differ in configuration around one stereogenic center
• Epimers
only (28.3)
Monosaccharides that differ in configuration at only the hemiacetal OH
• Anomers
group (28.6)
• Glycoside An acetal derived from a monosaccharide hemiacetal (28.7)
Acyclic, Haworth, and 3-D representations for D-glucose (28.6)
CHO
CH2OH
O
H
H
OH
H
Haworth
projection
H
H
HO
OH
HO
H
OH
CH2OH
O
H
H
OH
OH
H
H
H
OH
H
OH
H
HO
H
CH2OH
α anomer
OH
H
HO
HO
3-D representation
H
HO
HO
H C OH
OH
OH
OH
H
HO C H
H
H
H
H C OH
O
OH
β anomer
acyclic form
CHO
H
OH
OH
H
H C OH
O
H
OH
H
CH2OH
Reactions of monosaccharides involving the hemiacetal
[1] Glycoside formation (28.7A)
OH
OH
O
HO
HO
ROH
HCl
OH
α-D-glucose
OH
OH
•
OH
O
HO
HO
O
+ HO
HO
OR
OH
OR
•
β glycoside
α glycoside
Only the hemiacetal OH
reacts.
A mixture of α and β
glycosides forms.
[2] Glycoside hydrolysis (28.7B)
OH
OH
HO
HO
O
OH
H 3O +
HO
HO
OH
O
OH
OR
HO
+
HO
OH
OH
OH
β anomer
α anomer
+
O
ROH
•
A mixture of α and β anomers
forms.
790
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–2
Reactions of monosaccharides at the OH groups
[1] Ether formation (28.8)
OR
OH
O
HO
HO
OH
OH
RX
•
•
O
RO
RO
Ag2O
All OH groups react.
The stereochemistry at all stereogenic
centers is retained.
OR
OR
[2] Ester formation (28.8)
OH
HO
HO
O
Ac2O or
OH
AcCl
OH
OAc
O
AcO
AcO
•
•
OAc
All OH groups react.
The stereochemistry at all stereogenic
centers is retained.
AcO
pyridine
Reactions of monosaccharides at the carbonyl group
[1] Oxidation of aldoses (28.9B)
CHO
COOH
COOH
•
[O]
or
•
CH2OH
CH2OH
aldose
aldonic acid
Aldonic acids are formed using:
• Ag2O, NH4OH
• Cu2+
• Br2, H2O
Aldaric acids are formed with HNO3, H2O.
COOH
aldaric acid
[2] Reduction of aldoses to alditols (28.9A)
CHO
CH2OH
NaBH4
CH3OH
CH2OH
CH2OH
aldose
alditol
[3] Wohl degradation (28.10A)
This C–C bond
is cleaved.
•
CHO
H
OH
CHO
•
[1] NH2OH
[2] Ac2O, NaOAc
[3] NaOCH3
CH2OH
•
CH2OH
The C1–C2 bond is cleaved to shorten
an aldose chain by one carbon.
The stereochemistry at all other
stereogenic centers is retained.
Two epimers at C2 form the same
product.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
791
Carbohydrates 28–3
[4] Kiliani–Fischer synthesis (28.10B)
CHO
CHO
H
CHO
OH
HO
[1] NaCN, HCl
H
•
+
[2] H2, Pd-BaSO4
•
[3] H3O+
CH2OH
CH2OH
One carbon is added to the aldehyde
end of an aldose.
Two epimers at C2 are formed.
CH2OH
Other reactions
[1] Hydrolysis of disaccharides (28.12)
O
H3O+
This bond is cleaved.
O
O
+
OH
O
O
OH
OH
A mixture of anomers is formed.
[2] Formation of N-glycosides (28.14B)
CH2OH
O
H
H
OH
H
RNH2
H
mild
OH
OH
H+
CH2OH
O
H
H
OH
H
+
H
NHR
OH
CH2OH
O
H
H
OH
NHR
H
H
OH
•
Two anomers are formed.
Practice Test on Chapter Review
1. a. How are the following two representations related to each other?
HO
HO
CH2OH
O
H
OH
HO
A
H
CH2OH
O
H
OH H
OH
H
OH
H
OH
B
1. A and B are anomers of each other.
2. A and B are epimers of each other.
3. A and B are diastereomers of each other.
4. Statements [1] and [2] are both true.
5. Statements [1], [2], and [3] are all true.
b. Which of the following statements is (are) true about monosaccharide C?
1. C is a D-sugar.
OH
2. The β anomer is drawn.
HOO
HO
3. C is an aldohexose.
OH
4. Statements [1] and [2] are both true.
HO
C
5. Statements [1], [2], and [3] are all true.
792
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–4
c. Which of the following are different representations for monosaccharide D?
CHO
H
OH
H
OH
HO
HO
H
OH
O
1.
H
CH2OH
OH
2.
HO
HO
OH
OH
O
H
H
HO
OH
OH
H
H
OH
O
3.
H
OH
CH2OH
HO
OH
OH
4. Both [1] and [2] are representations for D.
D
5. Compounds [1], [2], and [3] all represent D.
d. Which aldoses give an optically active compound upon reaction with NaBH4 in CH3OH?
CHO
H
1. HO
HO
CHO
OH
H
H
2. HO
H
H
CH2OH
CHO
OH
H
3.
OH
H
OH
H
OH
H
CH2OH
4. Both [1] and [2] give an optically active product.
OH
CH2OH
5. Compounds [1], [2], and [3] all give optically active
products.
2. Answer each question about monosaccharide D as True (T) or False (F).
OH
HO
O
HO
HO
OH
D
D is a D-sugar.
D is drawn as an α anomer.
D is an aldohexose.
Reduction of D with NaBH4 in CH3OH forms
an optically inactive alditol.
e. Oxidation of D with Br2, H2O forms an
optically active aldonic acid.
f. Oxidation of D with HNO3 forms
an optically active aldaric acid.
g. C2 has the R configuration.
h. Treatment of D with CH3OH, HCl
forms two products.
i. Treatment of D with Ag2O, and
CH3I (excess) forms two products.
j. An epimer of D at C3 has an axial
OH group.
a.
b.
c.
d.
3. Answer the following questions about the three monosaccharides (A–C) drawn below.
CHO
H
OH
HO
HO
H
a.
b.
c.
d.
CHO
HO
H
H
H
OH
H
H
OH
OH
H
OH
CH2OH
CH2OH
A
B
CH2OH
HO
O
H
H
OH
H
H
H
OH
OH
C
Draw the α anomer of A in a Haworth projection.
Draw the β anomer of B in a three-dimensional representation using a chair conformation.
Convert C into the acyclic form of the monosaccharide using a Fischer projection.
What two aldoses yield A in a Wohl degradation?
793
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–5
4. Draw the product of each reaction with the starting material D-xylose.
a.
b.
c.
d.
e.
CHO
H
HO
H
OH
H
OH
CH2OH
CH3OH, HCl
NaBH4, CH3OH
Br2, H2O
[1] NaCN, HCl; [2] H2, Pd-BaSO4; [3] H3O+
Ac2O, pyridine
D-xylose
Answers to Practice Test
1. a. 5
b. 5
c. 4
d. 1
3.
4.
CH2OH
HO
a.
CH2OH
O
H
OH
OH
H
b. HO
c.
OH
HO
H
OCH3
OH
CH2OH
b.
HO
H
HO
CHO
H
H
H
OH
OH
H
+ β anomer
OH
OH
HO
O
H
c. HO
OH
H
CHO
CHO
OH
HO
H
OH
H
HO
H
H
OH
CH2OH
OH
H
OH
CH2OH
H
H
H
OH
COOH
H
H
HO
OH
CH2OH
CH2OH
d.
H
O
OH
H
H
H
2. a. T
b. F
c. T
d. F
e. T
f. T
g. F
h. T
i. F
j. T
a.
H
+
HO
OH
HO
H
HO
H
H
CHO
H
d.
H
HO
H
OH
CHO
H
H
OH
OH
H
OH
H
+ HO
OH
H
CH2OH
CH2OH
CH2OAc
O
OAc
e.
H
H
H
OH
CH2OH
H
H
+ β anomer
OAc
OAc
Answers to Problems
28.1
A ketose is a monosaccharide containing a ketone. An aldose is a monosaccharide
containing an aldehyde. A monosaccharide is called: a triose if it has three C’s; a tetrose if
it has four C’s; a pentose if it has five C’s; a hexose if it has six C’s, and so forth.
794
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–6
CHO
CH2OH
C O
a. a ketotetrose
b. an aldopentose
H C OH
H C OH
H C OH
H C OH
CH2OH
CH2OH
Rotate and re-draw each molecule to place the horizontal bonds in front of the plane and the
vertical bonds behind the plane. Then use a cross to represent the stereogenic center in a
Fischer projection formula.
28.2
COOH
COOH
a. CH3 C OH
=
CH3
CHO
HO
CH3
OHC
OH
HO
H
C
CH3
CHO
CH2CH3
re-draw
H
C
HOCH2 H
CHO
re-draw
C
C
c.
CH2CH2OH
CH2CH2OH
CHO
= HO
CH3
CHO
re-draw
d. H C CHO
HO
C
CH3
CH2OH
CH2CH3
CH2CH3
OH
H
H
CHO
CH2OH = H
CHO
= HO
CH3
CH3
H
H
For each molecule:
[1] Convert the Fischer projection formula to a representation with wedges and dashes.
[2] Assign priorities (Section 5.6).
[3] Determine R or S in the usual manner. Reverse the answer if priority group [4] is
oriented forward (on a wedge).
28.3
CH2NH2
a. Cl
[1]
H
[1]
Cl C H
CH2NH2
[2]
H
CH2NH2
CH2NH2
[3]
1 Cl C CH2Br 2
1
H
H
2
[2]
CHO
CHO
[3]
1 Cl C H 4
1
[1]
3
[2]
Cl C H
CH2OH
d. Cl
CH2Br
H
[1]
COOH
Cl C CH2Br
H
[2]
2
1 Cl C H
CH2OH
CH2OH
3
3
COOH
1 Cl C CH2Br 2
H
4
S configuration
CHO
[3]
3
COOH
H forward
3
1 Cl C H 4
CH2OH
Cl C H
CH2NH2
CH2NH2
CHO
S configuration
2
2
CHO
Cl C CH2Br 2
4
CH2NH2
CHO
Cl
CHO
3
3
H
CHO
b. Cl
CH2NH2
Cl C CH2Br
CH2Br
H
c.
c. an aldotetrose
H C OH
CH2OH
b.
CHO
H C OH
H forward
S configuration
3
COOH
[3]
1
Cl C CH2Br 2
H
S configuration
795
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–7
28.4
OHC
HO H H OH
HO
HO
rotate
CHO
H
OH
H
H
HO
HO H HO H
OHC
H
OH
Convert
staggered to
eclipsed.
HO
OH
H
OH
H
OH
H
H
OH
CHO
re-draw
CHO
HO
H
HO
H
HO
H
HO
H
HO
H
H
=
OH
HO
H
CH2OH
H
OH
CH2OH
OH
28.5
R
CHO
S
H C OH
R
HO C H
R
H C OH
H C OH
CH2OH
D-glucose
A D sugar has the OH group on the stereogenic center farthest from the carbonyl on the right.
An L sugar has the OH group on the stereogenic center farthest from the carbonyl on the left.
28.6
CHO
a.
CHO
H
OH
HO
H
OH
H
HO
H
HO
CH2OH
CHO
H
HO
OH
HO
H
C
OH group on the left: L sugar
b. A and B are diastereomers.
A and C are enantiomers.
B and C are diastereomers.
28.7
There are 32 aldoheptoses; 16 are D sugars.
C2 R
CHO
CHO
CHO
CHO
C3 R H
H
OH
H
OH
H
OH
H
OH
OH
H
OH
H
OH
H
OH
H
OH
HO
H
HO
H
H
OH
H
OH
HO
H
H
OH
HO
H
OH
H
OH
H
OH
H
CH2OH
CH2OH
CH2OH
OH
CH2OH
B
A
H
H
CH2OH
OH group on the left: L sugar
H
H
OH
CH2OH
OH group on the right: D sugar
796
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–8
28.8
Epimers are two diastereomers that differ in the configuration around only one stereogenic
center.
epimers
CHO
28.9
a.
b.
c.
d.
e.
f.
CHO
H
OH
HO
H
OH
H
CHO
H
and
OH
CH2OH
CH2OH
D-erythrose
D-threose
H
HO
OH
H
CH2OH
L-threose
D-allose and L-allose: enantiomers
D-altrose and D-gulose: diastereomers but not epimers
D-galactose and D-talose: epimers
D-mannose and D-fructose: constitutional isomers
D-fructose and D-sorbose: diastereomers but not epimers
L-sorbose and L-tagatose: epimers
28.10
a.
HO
H
H
CH2OH
CH2OH
CH2OH
C O
C O
C O
H
H
OH
HO
OH
HO
CH2OH
b.
OH
HO
H
H
HO
H
H
H
CH2OH
D-fructose
CH2OH
C O
c.
HO
OH
OH
HO
CH2OH
H
CH2OH
D-tagatose
L-fructose
H
H
L-sorbose
enantiomers
28.11
CH2OH
S
CH2OH
S
C O
C O
H
HO
H
OH
HO
H
OH
H
HO
H
H
OH
CH2OH
CH2OH
D-tagatose
D-fructose
28.12 Step [1]: Place the O atom in the upper right corner of a hexagon, and add the CH2OH group
on the first carbon counterclockwise from the O atom.
Step [2]: Place the anomeric carbon on the first carbon clockwise from the O atom.
Step [3]: Add the substituents at the three remaining stereogenic centers, clockwise around
the ring.
a. Draw the α anomer of:
CHO
[1]
CH2OH
O
H
OH
H
OH
D sugar, CH2OH
H
OH
is drawn up.
OH
CH2OH
H
OH
H
H
[2]
CH2OH
O
H
farthest away C,
OH on right = D sugar
α anomer
OH is down
for a D sugar.
[3]
H
CH2OH
O
H
H
H
HO
OH
H
OH
OH
First three substituents are
on the right so they are
drawn down.
797
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–9
b. Draw the α anomer of:
CHO
HO
H
HO
H
H
[2]
O
CH2OH
is drawn down.
H
farthest away C,
OH on left
= L sugar
CH2OH
O
H
c. Draw the β anomer of:
[1]
CHO
[2]
CH2OH
O
H
OH
H
OH
H
[3]
HO
OH
HO
OH
CH2OH
β anomer
OH is up for
a D sugar.
D sugar, CH2OH
is drawn up.
farthest away C,
OH on right = D sugar
H
[3]
H
H
H
O OH
CH2OH
OH OH
H
H
H
H
The α anomer
The first two substituents are
has the OH and
on the left so they are
CH2OH trans. In
drawn up. The third is on
an L sugar, the
the right, drawn down.
OH must be drawn up.
L sugar, CH2OH
CH2OH
HO
OH
H
OH
HO
O
CH2OH
[1]
H
H
H
CH2OH
O OH
H
OH
H
H
OH
H
The first substituent is
on the left so it is
drawn up. The other two are on
the right, drawn down.
28.13 To convert each Haworth projection into its acyclic form:
[1] Draw the C skeleton with the CHO on the top and the CH2OH on the bottom.
[2] Draw in the OH group farthest from the C=O.
A CH2OH group drawn up means a D sugar; a CH2OH group drawn down means an
L sugar.
[3] Add the three other stereogenic centers, counterclockwise around the ring.
“Up” groups go on the left, and “down” groups go on the right.
"up" group
on left
CH2OH
CH2OH is up =
D sugar
O
HO
H
H
a.
OH
[1]
CHO
CHO
[2]
H
H
H
OH
OH
CH2OH
"down" groups
on right
H
b.
HO
"down" groups
on right
OH
H
OH
OH
H
CH2OH
H
OH
CH2OH
OH on right =
D sugar
CH2OH is down =
L sugar
O
H
CH2OH
OH
H
OH
OH
H
HO
H
H
CHO
[3]
[1]
CHO
CHO
[2]
HO
H
"up" group
on left
CHO
[3]
HO
CH2OH
H
CH2OH
OH on left =
L sugar
H
H
OH
H
OH
HO
H
CH2OH
798
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–10
28.14 To convert a Haworth projection into a 3-D representation with a chair cyclohexane:
[1] Draw the pyranose ring as a chair with the O as an “up” atom.
[2] Add the substituents around the ring.
O is an "up" atom.
CH2OH
HO
O
HO
H
H
a.
OH
O
[1]
OH
H
[2]
H
H
OH
H
H
OH
HO
OH
H
O is an "up" atom.
O
H
b.
H OH
H
H
[1]
[2] HO
O
H
H
OH
HO
OH
H
HO
H
H
OH
O
H
CH2OH
OH
H
O
H
With so many axial groups,
this is not the more stable
conformation of this sugar.
OH
OH
28.15 Cyclization always forms a new stereogenic center at the anomeric carbon, so two different
anomers are possible.
Two anomers of D-erythrose:
CHO
H
H
OH
H
OH
H
OH
OH
OH
H
CH2OH
H
O
H
H
OH
O
H
H
OH
OH
H
D-erythrose
H
28.16
HO
HO
CH3CH2OH HO
OH
HO
HCl
HO O
a. HO
HO
H
OH
OH
CH3CH2OH
H
OH
OH
OH
CH2OH O
OH
CH3CH2OH
HO
H
HCl
H
CH2OH
H
OH
β-D-fructose
OH
H
OH
O
OH
O
+
H
HCl
α-D-gulose
c.
OCH2CH3
HO
OCH2CH3
O
b.
HO O
+ HO
H
β-D-mannose
OH
HO
HO O
OH
HO
OCH2CH3
OCH2CH3
OH HO H
CH2OH O
OCH2CH3
+
HO
H
H
CH2OH
H
OH
CH2OH O
CH2OH
HO
H
H
OCH2CH3
H
OH
799
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–11
28.17
H OH2
H
+
HOCH2
O
H
H
OH
H
OH
H
+
HOCH2
OCH2CH3
OCH2CH3
O
H
H
OH
H
OH
H
HOCH2
O
H
H
OH
+ H2O
H
+
H
+ CH3CH2OH
OH
resonance-stabilized
carbocation
+
HOCH2
O
H
H
H
OH
OH
H
H
H2O
above
HOCH2
O
+
H
H
OH
OH
+
HOCH2
O H
O
H
H
OH
H
OH
H
H
H2O
HOCH2
OH
O
+ H3O+
H
H
OH
H
OH
H
H
planar carbocation
HOCH2
H2O
below
H
H
HO
OH
HOCH2
H
O
H
H
O
H
+
H
O H
H
+ H3O+
H
OH
HO
OH
H2O
28.18
a. All circled O atoms are part of a glycoside.
OH
HO
OH
OH
b. Hydrolysis of rebaudioside A breaks each bond
indicated with a dashed line and forms four
molecules of glucose and the aglycon drawn.
OH
OH
O
O
O
HO
O
OH
OH
O
O
HO
OH O
HO
O
OH
+
OH
HO
OH
HO
+
OH
HO
OH
OH
OH
HO
H
OH
H
O
O
HO
O
rebaudioside A
Trade name: Truvia
HO
OH
+
OH
O
H
+
H
O
OH
HO
OH
HO
HO
OH
O
OH
aglycon
HO
Both anomers of glucose
are formed, but only the β
anomer is drawn.
800
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–12
28.19
OH
CH3O
OH
Ag2O
O
a.
OH
HO
NaH
OH
HO
c.
C6H5CH2O
OH
C6H5CH2O
O
H
O
pyridine
OAc
AcO
AcO
OH
H
C6H5COO
O
+
OH
C6 H5
C
Cl
pyridine
OOCC6H5
O
C6H5COO
OH
H
OOCC6H5
C6H5COO
O
product in (c)
+ α anomer + HOCH2C6H5
OAc
Ac2O
OH
O
f.
OH
C6H5CH2O
OAc
HO
H
OCH2C6H5
H3O+
H
OH
H
e.
OCH2C6H5
C6H5CH2O
OH
HO
OH
C6H5CH2O
C6H5CH2O
O
d.
O
C6H5CH2Cl
OH
H
OCH2C6H5
O
OCH2C6H5
C6H5CH2O
H
OCH2C6H5
C6H5CH2O
OH
C6H5CH2O
OCH3
CH3O
O
b.
O
CH3O
CH3I
OH
H
OH
OCH3
+
C6H5
C
C6H5CH2O
Cl
H
OCH2C6H5
O
pyridine
C6H5CH2O
+ α anomer
OCOC6H5
C6H5CH2O
H
28.20
CH2OH
CH2OH
CH2OH
O
H
OH
HO
H
H
HO
H
NaBH4
HO
H
HO
HO
H
CH3OH
HO
H
HO
H
OH
H
CH2OH
OH
CH2OH
CH2OH
D-tagatose
H
H
OH
D-galactitol
D-talitol
28.21 Carbohydrates containing a hemiacetal are in equilibrium with an acyclic aldehyde, making
them reducing sugars. Glycosides are acetals, so they are not in equilibrium with any acyclic
aldehyde, making them nonreducing sugars.
acetal
hemiacetal
HO
a.
CH2OH
O
H
H
OH
OH
b.
H
H
H
OH
reducing sugar
CH2OH O
H
H
OH
hemiacetal
HO
HO
H
H
OCH2CH3
OH
nonreducing sugar
O
c.
O
HO
O
HO
OH
OH
HO
lactose
reducing sugar
OH
801
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–13
28.22
CHO
a. HO
COOH
H
H
OH
H
OH
HO
Ag2O
NH4OH
H
OH
H
OH
H
OH
CH2OH
HO
Br2
H2O
OH
H
HO
HNO3
H
H
H2O
OH
H
OH
CH2OH
COOH
COOH
H
H
COOH
H
OH
CHO
b.
c. HO
H
CH2OH
HO
CHO
H
H
H
OH
OH
H
CH2OH
OH
CH2OH
28.23 Molecules with a plane of symmetry are optically inactive.
CHO
CHO
a.
COOH
H
OH
H
OH
H
OH
H
CH2OH
D-erythrose
HO
HO
H
HO
OH
H
H
HO
OH
HO
H
HO
COOH
H
H
CH2OH
OH
OH
H
H
H
CH2OH
OH
COOH
optically inactive
D-galactose
COOH
H
COOH
OH
HO
CHO
H
c.
optically inactive
HO
b.
H
H
H
OH
COOH
D-lyxose
optically active
28.24
CHO
HO
H
HO
H
OH
OH
H
OH
H
H
CHO
H
OH
or
HO
H
CH2OH
D-idose
CHO
H
H
HO
OH
H
OH
H
OH
CH2OH
CH2OH
D-gulose
D-xylose
28.25
CHO
HO
a.
H
H
OH
CH2OH
D-threose
CHO
HO
H
H
HO
H
HO
H
OH
CH2OH
CHO
CHO
H
OH
H
OH
CH2OH
b.
H
OH
H
OH
H
OH
CH2OH
D-ribose
CHO
HO
CHO
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
CH2OH
OH
CH2OH
802
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–14
CHO
CHO
H
c.
HO
OH
HO
H
HO
H
H
H
OH
H
OH
OH
H
OH
HO
H
HO
HO
H
HO
H
CH2OH
CHO
H
OH
H
H
CH2OH
D-galactose
H
OH
CH2OH
28.26
Possible optically inactive D-aldaric acids:
CHO
COOH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
COOH
H
plane of
symmetry
HO
H
H
H
COOH
A'
CHO
OH
HO
OH
H
COOH
OH
H
OH
CH2OH
A"
This OH is on the right for a D sugar.
There are two possible structures for the D-aldopentose (A' and A''), and the Wohl degradation
determines which structure corresponds to A.
H
OH
H
OH
H
OH
CH2OH
This is A.
Product of Wohl degradation:
CHO
CHO
CHO
Wohl
COOH
H
OH
H
OH
[O]
CH2OH
COOH
H
OH
HO
H
OH
H
COOH
A'
H
CHO
[O]
HO
OH
H
COOH
Wohl
OH
CH2OH
optically active
optically inactive
H
H
OH
HO
H
H
OH
CH2OH
A"
B
Since this compound has no plane of symmetry, its
precursor is B, and thus A'' = A.
28.27
CHO
HO
H
HO
H
H
OH
H
OH
CH2OH
CH2OH
HO
H
HO
H
CHO
H
HO
OH
rotate
H
H
180°
HO
OH
CHO
identical
H
H
OH
H
OH
CH2OH
D-idose
803
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–15
28.28
Optically inactive alditols formed from NaBH4 reduction of a D-aldohexose.
CHO
CH2OH
CH2OH
H
OH
H
OH
H
H
OH
H
OH
HO
H
H
OH
H
OH
HO
H
H
OH
H
OH
H
NaBH4
CH2OH
CH2OH
H
NaBH4
OH
OH
H
OH
H
H
H
HO
H
H
OH
CH2OH
optically inactive
A''
This is A.
This is B.
CHO
H
OH
HO
CH2OH
optically inactive
A'
CHO
OH
CHO
K–F
COOH
H
OH
OH
H
OH
H
CH2OH
H
OH
HO
H
OH
H
OH
HO
H
OH
H
OH
H
[O]
CH2OH
A'
COOH
COOH
[O]
OH
HO
H
HO
H
H
COOH
optically inactive
B'
CHO
CHO
optically active
H
K–F
OH
OH
HO
H
HO
H
H
OH
CH2OH
CH2OH
B''
A''
Two D-aldohexoses (A' and A'') give optically inactive alditols on reduction. A'' is formed from B''
by Kiliani–Fischer synthesis. Since B'' affords an optically active aldaric acid on oxidation, B'' is B
and A'' is A. The alternate possibility (A') is formed from an aldopentose B' that gives an optically
inactive aldaric acid on oxidation.
28.29
OH
HO
HO
O
OH
O
HO
OH
OH
H3O+
OH
O
HO
HO
O
+
OH
OH
HO OH
α-D-glucose
HO
HO
O
OH
The same products are formed
on hydrolysis of the α and β
anomers of maltose.
HO
β-D-glucose
α anomer
28.30
β glycoside bond
OH
O
HO
HO
4
1 O
OH
O
HO
OH
cellobiose
OH
OH
Two possible anomers here. β OH is drawn.
804
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–16
28.31
a.
4
OH
HO
OH
HO
4
O
O HO
4
O
O HO
OH
HO
O
O
O
HO
b.
O HO
1
1
dextran
1
HO
HO
O
1
6
O
HO
HO
1
HO O
6
O
HO
HO
HO
28.32
O
HO
OHO
NHAc
O
HO
NHAc
HO
NHAc
OH
O
OH
O
OH
O
OH
O
O
NHAc
chitin—a polysaccharide composed of NAG units
OH
O
OH
O
HO
O
NH2
HO
OH
O
O
HO
NH2
OH
O
OHO
NH2
chitosan
O
NH2
28.33
OH
a.
CH2OH
O
H
H
OH
H
OH
H
H
OH
CH3NH2
mild H+
CH2OH
O
H
H
OH
H
OH
b.
C6H5NH2
OH
OH
+
OH
CH2OH
O
H
H
OH
OH
H
mild
OH
OH
NHC6H5 +
H+
OH
+ H2O
H
H
O
HO
NHCH3
H
OH
O
HO
OH
NHCH3
H
OH
H
OH
H
O
HO
H
OH
OH
+ H2O
NHC6H5
805
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–17
28.34
OH
OH
OH
O
HO
HO
O
HO
HO
OH
OH
+
OH2
OH
OH
H OH2
OH
O
HO
HO
HO
HO
+
O+
+
H2 O
OH
+ H2O
CH3CH2NH2
above
OH
OH
H
H2O
NHCH2CH3
O
HO
HO
O
HO
HO
+
NHCH2CH3 +
OH
HO
OH
H3O+
O
HO
HO
+
OH
OH
OH
HO
HO
below
HO
CH3CH2NH2
O
HO
HO
O
+
OH
+
NHCH2CH3
H
H3O+
NHCH2CH3
H2O
28.35
O
O
NH
a.
N
HO CH2
O
N
HO CH2
O
N
N
NH2
O
H
H
OH
OH
NH
b.
OH
28.36
a. Two purine bases (A and G) are both bicyclic bases. Therefore they are too big to
hydrogen bond to each other on the inside of the DNA double helix.
b. Hydrogen bonding between guanine and cytosine has three hydrogen bonds, whereas
between guanine and thymine there are only two. This makes hydrogen bonding between
guanine and cytosine more favorable.
H
H
N
N
H
H
O
N
H
guanine
G
H
cytosine
C
H
O
NH
N
N
N
O
N
O
N
N
H
N
N
H
N
H
guanine
G
O
H
CH3
N
N
H
thymine
T
806
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–18
28.37
OHC
HO H H OH
a. HO
H
rotate
CHO
H
OH
HO
HO
H
H OH H OH
CHO
OHC
H
OH
H
OH
OH
H
H
OH
H
OH
Convert
staggered to
eclipsed.
HO
H
HO
re-draw
CHO
OH
H
H
H
=
OH
H
OH
HO
H
H
OH
OH
H
CH2OH
OH
CH2OH
OH
OHC
HO H H OH
b. HO
H
rotate
CHO
OH
H
HO
HO
H
HO H H OH
CHO
OHC
H
OH
H
OH
OH
H
OH
H
H
OH
Convert
staggered to
eclipsed.
HO
H
re-draw
CHO
OH
HO
H
HO
H
H
H
OH
HO
=
H
HO
OH
H
H
CH2OH
OH
CH2OH
OH
28.38
CHO
HO
CHO
OH
O
=
H
OH
OH
H
OH
H
OH
H
OH
OH
O
HO
OH
OH
OH
CH2OH
A
β anomer
=
H
OH
H
OH
H
OH
H
B
α anomer
D-ribose
OH
CH2OH
D-allose
28.39 Label the compounds with R or S and then classify.
CHO
H
H
OH
CH2CH3
c.
R
identical
A
R
d.
C
CH3CH2
H OH
CHO
CHO
OH
H
CHO
b.
a. CH3CH2 C OH
R
identical
C
HO
CHO
S
enantiomer
CH2CH3
H
S
enantiomer
28.40 Use the directions from Answer 28.2 to draw each Fischer projection.
Cl
COOH
a.
CH3 C Br
COOH
=
CH3 S
Br
b. CH3 C Cl
H
H
CH3O OCH2CH3
CH3
CH3CH2
re-draw CH3
CH2CH3
S
C Br = H
Cl
OCH3
C
H
re-draw
CH3CH2
H
Br
CH3
Br
CH2CH3 = CH3
OCH2CH3
=
Cl C H
Cl
S
CH2CH3
OCH3
CH2CH3 f. H
OCH2CH3
R
C
Br
Cl
Cl = CH3CH2 S
S
Cl
C
H
Br
S
Br
S
Cl
H
H
CH2CH3
Cl
S
re-draw Br C H
Cl C
R
H
H
Cl
Br
CH3
H C Br
e.
C
Br
CH3
CH3
re-draw
C
Br
C
H
H
c.
d.
Br
Cl
=
Br
Cl
S
R
H
H
Br
807
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–19
CHO
CH3
H
C
g. H
CH3
Br
H C Br = H S
Br C H
Br S
re-draw
C
CH3
Br
CH3
Br
h. HO
CHO
H
HO
H C OH
HO H
CH3
CH3
CHO
S
=
S
HO C H
HO
R
HO H H OH re-draw HO C H
H
H
H
OH
CH2OH
CH2OH
28.41 Epimers are two diastereomers that differ in the configuration around only one stereogenic
center.
CHO
H
CHO
OH
HO
H
H
H
HO
OH
HO
OH
H
H
CH2OH
C4
CH2OH
D-xylose
L-arabinose
28.42
CHO
HO
a.
CHO
H
H
H
H
OH
HO
OH
HO
CH2OH
CHO
OH
HO
b. HO
H
H
H
C3
H
H
c. HO
OH
CH2OH
D-arabinose
CHO
H
enantiomer
O
d. HO
H
H
CH2OH
OH
OH
OH
OH
OH
CH2OH
epimer
diastereomer
(but not epimer)
constitutional
isomer
28.43
CHO
HO
CHO
H
HO
CHO
H
H
CH2OH
C O
OH
CH2OH
H
OH
H
OH
HO
H
H
OH
H
OH
H
OH
HO
H
H
OH
H
OH
HO
OH
H
OH
H
H
CH2OH
CH2OH
CH2OH
A
B
C
a. A and B epimers
b. A and C diastereomers
CH2OH
O
H
H
H
OH
OH
H
OH
OH
D
HO
OH
HO
F
H
E
c. B and C enantiomers
d. A and D constitutional isomers
e. E and F diastereomers
28.44
OH
H
H
OH
OH
H
a. anomers, epimers, diastereomers, reducing sugars
b.
CHO
H
OH
OH
B
H
HO
OH
H
A
H
H
OH
OH
O
O
H
H
O
OH
HO
OH
H
OH
CH2OH
D-xylose
This is the acyclic form of
both A and B.
808
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–20
28.45 Use the directions from Answer 28.12.
a. β-D-talopyranose
CH2OH
O
H
[1]
CHO
HO
H
HO
H
HO
H
H
OH
CH2OH
CH2OH
O
H
[2]
OH
[3]
OH
H
H
β anomer
OH is up for
a D sugar.
D sugar, CH2OH
is drawn up.
farthest away C,
OH on right = D sugar
CH2OH
O OH
H
OH OH
H
H
H
D-talose
b. β-D-mannopyranose
CH2OH
O
H
[1]
CHO
HO
CH2OH
O
H
[2]
OH
H
H
HO
H
H
OH
CH2OH
β anomer
OH is up for
a D sugar.
D sugar, CH2OH
is drawn up.
OH
H
CH2OH
O OH
H
OH OH
H
OH
H
H
H
[3]
farthest away C,
OH on right = D sugar
D-mannose
c. α-D-galactopyranose
CH2OH
O
H
[1]
CHO
H
CH2OH
O
H
[2]
H
OH
OH
HO
H
HO
H
H
OH
CH2OH
α anomer
OH is down.
D sugar, CH2OH
is drawn up.
CH2OH
O
H
[3] OH H
OH
H
H
H
OH
OH
farthest away C,
OH on right = D sugar
D-galactose
d. α-D-ribofuranose
[1]
[2]
H
CHO
H
OH
H
OH
H
CH2OH
O
OH
CH2OH
CH2OH
O
H
H
OH
[3]
CH2OH
H O
H
OH
α anomer
OH is down.
D sugar, CH2OH
is drawn up.
H
H
OH
OH
farthest away C,
OH on right = D sugar
D-ribose
e. α-D-tagatofuranose
CH2OH
C O
HO
H
HO
H
H
OH
CH2OH
D-tagatose
[1]
CH2OH
O
H
[2]
CH2OH
O
CH2OH
H
OH
[3]
CH2OH
OH O
H
OH
H
D sugar, CH2OH
is drawn up.
farthest away C,
OH on right = D sugar
α anomer
OH is down.
CH2OH
OH
H
809
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–21
28.46
CHO
a. HO
H
HO
H
HO
H
HO
OH
α anomer OH
H
D sugar
HOOH OH
D sugar
HOOH
HO
O
O
HO
OH
β anomer
CH2OH
farthest away C,
OH on right = D sugar
CHO
b.
H
OH
H
OH
HO
O
OH
OH OHOH
OH OH
α anomer
OH
D sugar
OH
HO
O
H
H
D sugar
OH
HO
β anomer
CH2OH
farthest away C,
OH on right = D sugar
c.
CHO
HO
D sugar
HOOH OH
H
H
O
OH
HO
H
H
OH
OH
D sugar
HOOH
HO
O
OH
OH
OH
α anomer
β anomer
CH2OH
farthest away C,
OH on right = D sugar
28.47 Use the directions from Answer 28.13.
"up" group
on left
a.
OH
CH2OH
CH2OH is up =
D sugar
O
H
OH
OH
[3]
H
OH
HO
H
"down" group
on right
CH2OH
OH
H
CH2OH
[1]
CHO
CHO
[2]
H
"down" group
on right
HO
HO
H
"up" group
on left
H
OH
CHO
[3]
H
OH
OH
H
CH2OH
HO
CH2OH
OH
H
H
OH
OH on right =
D sugar
CH2OH is down =
L sugar
O
CHO
H
HO
"up" group
on left
b.
CHO
[2]
H
H
OH
CHO
H
H
"up" group
on left
[1]
CH2OH
H
CH2OH
OH on left =
L sugar
HO
H
OH
H
H
CH2OH
810
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–22
OH
c.
H
HO O
HO
CH2OH is up =
D sugar
O
[1]
OH
CH2OH
H
H
OH =
CHO
[2]
CHO
[3]
HO
OH
H
OH
OH
CHO
OH
H
H
CH2OH
OH
H
H
OH
H
OH
H
CH2OH
OH
CH2OH
OH on right =
D sugar
CHO
d.
OH
H
OH
O
H
OH
HO
OH
OH
H
OH
CHO
OH
H
OH
H
O
f. HO
=
HO
HO
CH2OH
O
OH
e. HOCH2
OH
O
H
H
H
OH
H
H
OH
H
OH
H
OH
CH2OH
28.48
CHO
HO
H
H
a.
H
OH
OH
H
CH2OH
b.
CH2OH
OH CH2OH
H
H O HO
H O HO
H
OH
H
β anomer
D-arabinose
H
H
OH
OH
H
O OH
OH
OH
H
OH
H
α anomer
H
H
H
O H
OH
OH
H
OH
OH
two anomers in the pyranose form
28.49
Two anomers of D-idose, as well as two conformations of each anomer:
α anomer
axial
O
HO
OH
OH OH
OH
O
OH
OH
HO
OH
4 axial substituents
β anomer
OH
equatorial CH2OH group
OH OH
OH
O
4 equatorial OH groups
More stable conformation for the α anomer—the
CH2OH is axial, but all other groups are equatorial.
equatorial CH2OH group
OH
OH
3 axial substituents
OH
O
HO
axial
OH
OH
H
OH
CH2OH
OH
C O
CH2OH
OH
OH
H
= HO
OH
OH
CH2OH
D sugar
H
axial
OH
HO
3 equatorial OH groups
The more stable conformation for the β anomer—the
CH2OH is axial, as is the anomeric OH, but three
other OH groups are equatorial.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
811
Carbohydrates 28–23
28.50
CH3O
HO
CH3O
HO
OH
f. C6H5COCl, pyridine
OH
g. The product in (a), then H3O+
O
OH
D-gulose
HO
C6H5CH2O
c. C6H5CH2Cl, Ag2O
C6H5COO
OCH3
OOCC6H5
O
O OOCC H
6 5
COC6H5
OCH3 OCH
3
O
+ β anomer
+ β anomer
HO
HO
OH
CH3O
b. CH3OH, HCl
O
C6H5COO
OCH3
O
a. CH3I, Ag2O
OCH3
OCH2C6H5
O
CH3O
CH3O
h. The product in (b), then Ac2O, pyridine
OH
OAc
OAc
O
+ β anomer
O
OCH2C6H5
CH2C6H5
C6H5CH2O
HO
OAc
O
CH3O
HO
OH
OCH3
O
+ β anomer
d. C6H5CH2OH, HCl
OAc
OCH2C6H5
CH3O
OAc
CH3O
+ β anomer
OCH2C6H5
j. The product in (d), then CH3I, Ag2O
O
e. Ac2O, pyridine
OAc
OCH3
i. The product in (g), then C6H5CH2Cl, Ag2O
OH
CH3O
OCH3
O
OAc
OAc
OAc
CH3O
CH3O
+ β anomer
OCH2C6H5
28.51
OH
OH
O
a. CH3OH, HCl
HO
CHO
HO
OH
COOH
d. Br2, H2O
+ β anomer
OCH3
H
H
OH
H
OH
H
OH
b. (CH3)2CHOH, HCl
HO
OH
OH
O
OH
c. NaBH4, CH3OH
HO
OH
H
OH
OH
CH2OH
+ β anomer
OCH(CH3)2
COOH
e. HNO3, H2O
HO
H
H
OH
CH2OH
H
OH
H
H
H
OH
H
OH
H
H
H
H
CH2OH
D-altrose
HO
OH
CH2OH
OH
COOH
812
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–24
CHO
f. [1] NH2OH
[2] (CH3CO)2O, NaOCOCH3
[3] NaOCH3
H
OH
H
OH
H
OH
OCH3
OCH3
O
h. CH3I, Ag2O
CH3O
OCH3 OCH3
CH2OH
CHO
HO
H
H
OH
H
H
H
HO
+
AcO
OH
H
OH
OH
H
OH
OH
H
OH
+ β anomer
OAc
OAc
H
CH2OH
OAc
OAc
O
i. Ac2O, pyridine
CHO
g. [1] NaCN, HCl
[2] H2, Pd-BaSO4 HO
H
[3] H3O+
+ β anomer
j. C6H5CH2NH2, mild H+ OH
OH
O
HO
+ β anomer
CH2OH
NHCH2C6H5
OH
28.52
OH
H
H
OH
H
HO
HO
O
H
H
OH
H3O+
H
O
HO
H
H
N
H
O
H
H
HO
H3O+
OH
O
solanine
HO
OH
H
H
HO
HO
OH
OH
O
HO
+
HO
OH
HO
HO
H
H
H
monosaccharide
(both anomers)
monosaccharide
(both anomers)
CHO
HO
OH
H
OH
H
CH2OH
D-glucose
OH
OH
CHO
OH
O
OH
28.53
H
H
aglycon
monosaccharide
(both anomers)
OH
N
H
+
O
HO
O
O
H
OH
H
HO
aglycon
H
OH
O
OH
HO
+
OH
OH
monosaccharide
(both anomers)
H
HO
OH
O
HO
HO
H
salicin
OH
OH
H
O
CHO
HO
H
HO
OH
H
OH
H
CH2OH
D-arabinose
H
H
OH
OH
CH2OH
D-mannose
813
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–25
28.54
CHO
CHO
HO
a. H
OH
H
OH
H
H
H
OH
H
OH
H
OH
CH2OH
CH2OH
CHO
b.
HO
H
H
HO
H
HO
H
HO
HO
H
HO
H
HO
OH
H
OH
CH2OH
HO
H
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
OH
CHO
CHO
H
c.
H
CH2OH
OH
OH
OH
H
H
CH2OH
OH
CH2OH
H
H
H
CH2OH
CHO
CHO
OH
H
OH
CH2OH
CHO
CHO
OH
CH2OH
28.55
OH
OH
CH3OH
O
a. HO
OH
HO
HO
HO
HCl
OH
+ α anomer
H
H
HO
H
OH CH3OH
H
OH
H
OH
H
OH
CH2OH
H
H
H
H
H
Ag2O
OCH3
H
OCH3
CH2OCH3
COOH
OH
HO
OCH3
CH3O
CH2OH
CHO
c.
H
H
Br2
HO
OH
H2O
H
OH
H
CH2OH
OCH3
OCH2CH3
+ α anomer
H
CH3I
H
O
CH3CH2O
CH3CH2O
CH2OCH3
OH
NaBH4
H
Ag2O
OH
OCH2CH3
+ α anomer
OH
HO
OCH3
CH2OH
CHO
b.
CH3CH2I
O
COOH
OH
H
Ac2O
H
OH
OAc
AcO
pyridine
H
H
OH
OAc
H
OAc
CH2OH
CH2OAc
28.56 Molecules with a plane of symmetry are optically inactive.
CHO
H
OH
H
NaBH4
OH
H
CHO
CH2OH
CH3OH
OH
H
H
H
CH2OH
OH
H
OH
HO
OH
H
CH2OH
OH
NaBH4
H
OH
CH3OH
H
HO
H
CH2OH
CH2OH
OH
H
OH
CH2OH
D-xylose
D-ribose
28.57
OH
OH
O
a. HO
OCH3
H3
O+
HO
OH
OH
OH
OH
O
O
HO
OH
OH
OH
OH
+ CH3OH
OH
814
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–26
b.
OCH3
HO
O
HO
H3O+
HOCH2
HO
OCH2CH3
OH
c.
OCH3
HO
O
OH
HOCH2
H3O+
OH
HO
OH
OH
NHCH2CH3
O
OCH3
HO
O
OH
+ CH3CH2OH
OH
HOCH2
OH
O
OH
OH
O
+ NH2CH2CH3
OH
OH
OH
28.58
OH
OH
O
HO
HO
HO
HO
OH
OH
OH
O
+
HO
HO
OH2
OH
OH
OH
H OH
H OH2
OH
OH
O
HO
HO
O
HO
HO
resonance-stabilized carbocation
H OH
+
OH
O
HO
HO
OH
OH
O
O
HO
HO
H OH2
OH
OH
OH +
O H
H
H OH
H OH
28.59
H
O
C6H5
C
H
OH
C6H5
C
H
C6H5
CH2OH
HO
HO
O
H
OH
B (any base)
OH
H
O
HO
HO
C6H5
O
OCH3
OCH3
O
HO
HO
OH
O
+ HB+
OCH3
OH
OH
OH2
C6H5
O
H O
HO
OH
B
C6H5
O
OCH3
O
O
HO
+ H2O
O
OH
+ HB+
OCH3
C6H5
O
OH
(any base)
C6H5
O
HO
HO
OCH3
O
HO
HO
O
OCH3
OH
815
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–27
28.60
OH
OH
O
HO
HO
OH
OH
OH
HO
H O
HO
O
O
HO
OH
O
HO
HO
OCH3
HO
HO
O
+
OH
OCH3
OH
OH
OH
O
A
OH
+
+ H2O
H OH2
O+
HO
HO
+
OH
O
HO
HO
OCH3
B HO
OH
OH
O
HO
HO
HO
HO
OCH3
OH
O
HO
HO
B
HO
HO
OCH3
H
OH
H OH2
+
O+
HO
HO
OH
A
+ H2O
+
OH
O
+ CH3OH
OH
H2O
above
OH
H
O
HO
HO
H2O
+
OH
HO
OH
O
HO
HO
OH
HO
O
HO
HO
+ H3O+
+
A OH
OH
below
OH
O
HO
HO
H2O
HO
O
HO
HO
+
HO
OH
H
OH
H2O
28.61
COOH
H
OH
H
OH
H
OH
H
H
=
CH2OH
OH O
H
H
H
OH
OH
OH
H A
H
OH
OH
OH
CH2OH
+
OH OH
H
H
H
OH
OH
HO
H
O
OH
HO
OH
A
=
H
H
H
OH
OH
OH
H
O
H
O
H
OH
OH
+
H A
OH
H
H
H
OH
OH
OH
OH
OH
O
OH
OH
OH
H
H
H A
OH
A
OH
O
A
O+
OH
OH
+
CH2OH
H
OH H
OH
O
H
H
H
H
O H
A
+
O
H
H
OH2
H
OH
OH
OH
OH
+ H2O
OH
OH
+
A
816
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–28
28.62
OH
CHO
H
OH
HO
H
H
H
C O
C O
C OH
C OH
HO
OH
H
H
H
H
H2O
OH
H
CH2OH
HO
OH
H
OH
H
CH2OH
Protonation of this
enolate can occur
from two directions.
D-glucose
H
C OH
H OH
C OH
H
HO
OH
H
OH
H
H
OH
+
OH
OH
CH2OH
CH2OH
enediol
A
Protonation on O
forms an enediol.
H2O
two protonation
products
CHO
HO
H
HO
H
H
OH
H
OH
CH2OH
H OH
H
H
C OH
C OH
C O H
HO
OH
H
H
HO
OH
H
OH
H
H
CH2OH
enediol
A
Deprotonation of the OH at C2
of the enediol forms a new
enolate that goes on to
form the ketohexose.
H C OH
C O
HO
OH
H
OH
H
CH2OH
H
C OH
C O
H
H
C O
H
HO
OH
H
OH
H
CH2OH
H
OH
+
OH
CH2OH
+ H2O
28.63
Two D-aldopentoses (A' and A'') yield optically active aldaric acids when oxidized.
Optically active D-aldaric acids:
CHO
HO
H
H
OH
H
OH
CH2OH
A'
COOH
COOH
[O]
H
HO
H
H
OH
HO
H
H
OH
H
HO
COOH
optically active
OH
COOH
optically active
CHO
[O]
HO
HO
H
H
H
OH
CH2OH
A"
D-lyxose
OH
817
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–29
CHO
HO
CHO
H
H
OH
H
OH
Wohl
H
OH
H
OH
[O]
H
OH
HO
H
OH
H
CH2OH
CH2OH
COOH
COOH
A'
[O] HO
H
OH
H
HO
Wohl
H
H
HO
OH
COOH
COOH
optically
inactive
CHO
CHO
H
H
CH2OH
OH
CH2OH
no plane of
symmetry
optically active
A"
Only A" undergoes Wohl degradation to an aldotetrose that is oxidized to an optically active aldaric
acid, so A'' is the structure of the D-aldopentose in question.
28.64
CHO
HO
CH2OH
H
HO
CH2OH
HO
H
HO
H
HO
H
OH
H
OH
HO
H
OH
H
OH
H
CH2OH
CHO
H
CH2OH
OH
H
H
OH
CH2OH
CH2OH
identical
(by rotating 180°)
D-arabinose
H
D-lyxose
28.65
Only two D-aldopentoses (A' and A'') yield optically inactive aldaric acids (B' and B'').
CHO
COOH
H
OH
H
OH
H
OH
[O]
COOH
H
OH
H
H
OH
HO
H
OH
H
CHO
OH
H
H
[O]
OH
CH2OH
COOH
A'
B'
B''
optically inactive
optically inactive
HO
H
COOH
OH
H
OH
CH2OH
A"
Product of Kiliani–Fischer synthesis:
CHO
CHO
H
OH
H
OH
H
OH
K–F
CH2OH
CHO
OH
HO
H
OH
H
OH
H
H
OH
H
OH
HO
OH
H
OH
H
H
CH2OH
A'
D'
[O]
COOH
plane of
symmetry
H
CHO
H
HO
CHO
H
H
OH
OH
H
OH
H
OH
HO
H
H
OH
CH2OH
CH2OH
CH2OH
C'
C"
D"
[O]
[O]
[O]
COOH
COOH
COOH
H
OH
HO
H
OH
H
OH
H
H
H
OH
H
OH
HO
H
OH
H
OH
H
HO
H
H
OH
OH
H
OH
H
OH
HO
H
H
OH
COOH
COOH
COOH
COOH
optically
inactive
optically
active
optically
active
optically
active
CHO
H
K–F HO
H
OH
H
OH
CH2OH
A"
818
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–30
Only A' fits the criteria. Kiliani–Fischer synthesis of A' forms C' and D', which are oxidized to one
optically active and one optically inactive aldaric acid. A similar procedure with A'' forms two
optically active aldaric acids. Thus, the structures of A–D correspond to the structures of A'–D'.
28.66
Only two D-aldopentoses (A' and A'') are reduced to optically active alditols.
CHO
HO
CH2OH
H
H
OH
H
OH
HO
[H]
CH2OH
H
HO
H
H
OH
HO
H
H
OH
H
CH2OH
[H]
H
H
CH2OH
optically
active
H
HO
OH
CH2OH
A'
CHO
HO
OH
CH2OH
A"
optically
active
Product of Kiliani–Fischer synthesis:
CHO
CHO
HO
H
H
H
OH K–F
H
OH
HO
H
H
CH2OH
CHO
H
HO
H
H
H
HO
H
HO
H
HO
H
OH
HO
H
HO
H
OH
H
OH
H
OH
H
CH2OH
C'
C"
[O]
[O]
COOH
HO
OH
CH2OH
B'
H
CHO
HO
CH2OH
A'
CHO
OH
H
COOH
HO
H
H
H
HO
H
HO
H
HO
H
HO
OH
HO
H
OH
H
OH
H
COOH
COOH
optically
active
OH
OH
CH2OH
A"
OH
H
H
H
COOH
optically
active
OH
COOH
H
H
H
H
[O]
HO
OH
H
K–F HO
B''
OH
H
HO
CH2OH
[O]
COOH
CHO
OH
OH
COOH
optically
active
optically
inactive
Only A'' fits the criteria. Kiliani–Fischer synthesis of A'' forms B'' and C'', which are oxidized to
one optically inactive and one optically active diacid. A similar procedure with A' forms two
optically active diacids. Thus, the structures of A–C correspond to A''–C''.
28.67
D-gulose
CHO
CH2OH
CHO
H
OH
H
OH
H
H
OH
H
OH
HO
HO
H
H
OH
CH2OH
A
HO
H
H
OH
CH2OH
B
OH
H
H
OH
CH2OH
C
CHO
CH2OH
H
OH
HO
H
H
OH
CH2OH
D
HO
H
H
OH
CH2OH
E
COOH
HO
CH2OH
H
H
OH
OH
H
OH
COOH
HO
H
F
H
H
OH
CHO
G
819
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–31
28.68 A disaccharide formed from two galactose units in a 1→4-β-glycosidic linkage:
β glycoside bond
OH
O
HO
HO
OH
O
1 O
OH 4
HO
OH
OH
28.69 A disaccharide formed from two mannose units in a 1→4-α-glycosidic linkage:
α glycoside bond
HO
HO
O
HO
HO
HO
1
4
O
OH
O
OH
HO
28.70
OH
a.
OH
O
OH
O
HO
OH
HO
OH
OH
OCH3 OCH
3
O
CH3I
O
Ag2O
CH3O
OCH3
OCH3
(Both anomers of B and C are
formed, but only one is drawn.)
O
CH3O
OCH3
OH
O
CH3I
b.
HO
OH
O
HO
HO
HO
H3O+
OCH3
O
CH3O
OCH3
+
O
OH
D
CH3O
OCH3
OH
E
OCH3
O
OH
OCH3
+ CH3OH
C
OCH3 OCH
3
O
H3O+
OCH3
O
HO
CH3O
OH
OCH3
O
CH3O
OCH3
O
CH3O
CH3O
Ag2O
OCH3
OCH3
A
B
OH
OCH3
O
HO
CH3O
CH3O
+
OCH3
O
OH
F + CH3OH
(Both anomers of E and F are
formed, but only one is drawn.)
OCH3
28.71
OH
a.
β glycoside bond
OH
O
1
HO
OH
b.
c.
4
O
OH
OH
OH
O
O
OH
HO
OH
1 4-β-glycoside bond
reducing sugar
(hemiacetal)
HO
HO
1
O
HO
HO
reducing sugar
(hemiacetal)
α glycoside bond
6
OH
O
1 6-α−glycoside bond
OH
820
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–32
28.72
a, b.
OH
OH
CH2OH
O
HO
O CH2
OH
1 6-α-glycoside bond
HO
O
OH
OH
stachyose
1 2-α-glycoside bond
OH
HOCH2 O
OH O
c.
OH
CH2OH
OH
O
CH2OH
O
OH
O
OH
O
OH OH
HO
OH
OH
HOCH2
O
HO OH
OH
OH
O
O
CH2OH
OH
OH
OH
HO
OH
OH
OH
O
2
HO
CH2OH
O
O
=
O CH2
OH
O
HO
O
OH OH
H3O+
OH
1 6-α-glycoside bond
OH
OH
CH2OH
OH
OH
OH
identical
Two anomers of each monosaccharide are formed,
but only one anomer is drawn.
d. Stachyose is not a reducing sugar since it contains no hemiacetal.
e.
CH3I
Ag2O
OCH3
CH3O
OCH3
O
CH3O
O
CH3O
O
CH3O
O
OCH3
CH3O
CH3O
f.
product
in (e)
H3O+
OCH3
O
O
O
O OCH3
OCH3 CH3
CH2OCH3
O
CH3O
OH
OCH3
OCH3
OCH3
OCH3
CH2OH
O
CH2OH
O
OH
OCH3
OCH3
OH
OCH3
CH3O
OCH3
Two anomers of each monosaccharide are formed.
CH3OCH2 CH3O OH
O
CH3O
CH2OCH3
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
821
Carbohydrates 28–33
28.73
OH
O
HO
HO
Isomaltose must be composed of two glucose units
in an α-glycosidic linkage. Since it is a reducing sugar
it contains a hemiacetal. The free OH groups in the
hydrolysis products show where the two
monosaccharides are joined.
isomaltose + α anomer
OH
O
O
HO
HO
OH
the hemiacetal
OH
[1] CH3I, Ag2O
[2] H3O+
OCH3
O
CH3O
CH3O
OH
6
1
CH3O
O
CH3O
CH3O
OH
CH3O
(Both anomers are present.)
OH
28.74
OH
O
HO
HO
CH3O
CH3O
CH3I
trehalose
OH
O
Ag2O
OH
OCH3
O
CH3O
H3O+
O
O
OCH3
O
HO
OH
CH3O
CH3O
CH3O
CH3O
(both anomers)
OCH3
OH
OCH3
O
OCH3
Trehalose must be composed of D-glucose units only, joined in an α-glycosidic linkage. Since
trehalose is nonreducing it contains no hemiacetal. Since there is only one product formed after
methylation and hydrolysis, the two anomeric C’s must be joined.
28.75
HO
a. HO
O
H2N
CH2OH
1
O
HO
HO
6
α
c.
O
H
O
HO
H
H
H
OH
NH2
O
NHCH2C6H5
O
CH3
b.
HO
HO
OH
OH
O
4
1
d.
OH
O
HO
OH
mannose
glucose
N
HO CH2
O
NH
OH
OH
O
H
OH
O
OH
822
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–34
28.76
OHC
H OH HO H
HO
rotate
a. OHC
OH
H
H
OH
H
OH
OH
H
H
OH
H
H
HO
H OH HO H
CHO
OHC
OH
H
HO
re-draw
H
H
OH
H
OH
HO
H
CH3
b. OH on left in Fischer
L-monosaccharide
OH
O
Ring
OH
OH
OH
flip.
more stable chair
OH
O
HO
OH
OH
The α anomer has the CH3 on
C5 and the anomeric OH trans.
c. Fucose is unusual because it is an L-monosaccharide and it contains a CH3 group rather
than a CH2OH group on its terminal carbon.
28.77
CH2OH
OH
a.
O
HO
H
OH
OH
O
OH O
H
OH
OH
OH
HO
OH
=
HO
H
H
OH
H
H
OH
CH2OH
D-fructose
H
HO
O
b. HO
OH
CHO
OH
H
OH
OH
OH
H
CHO
OH
=
H
OH
H
OH
H
OH
CH2OH
HO
D-ribose
CHO
HO
H
c.
HO
H
OH
O
OH
HO
H
OH H
OH CHO =
H
H
OH
OH
HO
H
H
OH
H
OH
OH
CH2OH
D-glucose
CHO
OH
HO
d. HO
O
OH
OH
HO
H
H
HO
HO
H
OH
OH CHO =
OH
H
H
HO
HO
H
OH
H
H
CH2OH
L-glucose
823
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Carbohydrates 28–35
28.78
Ignoring stereochemistry along the way:
A
HO
OH
OH H
H A
O
O
[1]
OH
A
OH
A
O
O
OH
H
HO
HO
OH
HO
C
[5]
O
CH3
OH
O
CH3
O
OH
C
CH3
[8]
OH
[4]
OH
OH
HO
H A
[7]
HO
HO
OH
O
[6]
OH2
HO
OH
HO
[3]
HO
OH
OH
OH
OH
O
OH
[2]
OH
OH
OH
H
OH
OH
HO
H A
HO
OH
+ HA
CH3
– H2O
O
O
O
HO
HO
[9]
OH
O
O
O
H
OH
HO
A
[10]
OH
OH
O
O
HO
OH
C
OH
CH3
OH
+ H2O
[11]
O
O
O
O
[14]
OH
O
O
– H2O
HO
O
[13]
OH
O
O
O
H2O
HO
O
OH
O
O
OH
HO
HO
O
HO
O
H A
A
[15]
[12]
HO
O
H
A
A
O
H
O
O
O
HO
O
O
[16]
O
O
O
O
HO
O
+ HA
=
O
O
H
O
HO
H
O
CH3
824
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 28–36
28.79 The hydrolysis data suggest that the trisaccharide has D-galactose on one end and D-fructose
on the other. D-Galactose must be joined to its adjacent sugar by a β-glycosidic linkage.
D-Fructose must be joined to its adjacent sugar by an α-glycosidic linkage.
HO
CH3O
OH
O
HO
OH
galactose
β O
HO
HO
HO
O
HO
α
O
CH3I
Ag2O
CH3O
OCH3
O
O
CH3O CH3O
CH3O
O
CH3O
O
OH
OH
glucose
HO
CH3O
O
OCH3
(Hydrolysis cleaves all acetals,
indicated with arrows.)
OCH3
O
CH3O
O
CH3O
CH3O
H3O+
fructose
2 anomers
CH3O
CH3O
CH3O
OH
OH
CH3O
CH3O
CH3O
O
O
CH3O
OH
OCH3
OH
OH
2,3,4,6-tetra-O-methyl2,3,4-tri-O-methyl-D-glucose
1,3,6-tri-O-methyl-D-fructose
D-galactose
(Both anomers of each compound are formed.)
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
825
Amino Acids and Proteins 29–1
Chapter 29 Amino Acids and Proteins
Chapter Review
Synthesis of amino acids (29.2)
[1] From α-halo carboxylic acids by SN2 reaction
NH3
R CHCOOH
R CHCOO–NH4+
(large excess)
SN2
Br
+
NH4+ Br–
•
Alkylation works best with
unhindered alkyl halides—that is,
with CH3X and RCH2X.
NH2
[2] By alkylation of diethyl acetamidomalonate
O
H
H
C N C COOEt
CH3
COOEt
R
[1] NaOEt
H2N C COOH
[2] RX
H
[3] H3O+, Δ
[3] Strecker synthesis
O
R
C
NH4Cl
H
NaCN
NH2
NH2
H3O+
R C CN
R C COOH
H
H
α-amino nitrile
Preparation of optically active amino acids
[1] Resolution of enantiomers by forming diastereomers (29.3A)
• Convert a racemic mixture of amino acids into a racemic mixture of N-acetyl amino acids
[(S)- and (R)-CH3CONHCH(R)COOH].
• React the enantiomers with a chiral amine to form a mixture of diastereomers.
• Separate the diastereomers.
• Regenerate the amino acids by protonation of the carboxylate salt and hydrolysis of the
N-acetyl group.
[2] Kinetic resolution using enzymes (29.3B)
H2N
C
COOH
(CH3CO)2O
AcNH
H R
C
COOH
acylase
H R
H2N
C
COOH
H R
(S)-isomer
(S)-isomer
separate
H2N
C
COOH
R H
(CH3CO)2O
AcNH
C
COOH
R H
acylase
AcNH
C
COOH
R H
(R)-isomer
enantiomers
enantiomers
NO REACTION
826
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–2
[3] By enantioselective hydrogenation (29.4)
R
NHAc
H
AcNH
H2
C C
Rh*
COOH
C
H2O, –OH
COOH
H2 N
H CH2R
C
COOH
H CH2R
S enantiomer
S amino acid
Rh* = chiral Rh hydrogenation catalyst
Summary of methods used for peptide sequencing (29.6)
• Complete hydrolysis of all amide bonds in a peptide gives the identity and amount of the
individual amino acids.
• Edman degradation identifies the N-terminal amino acid. Repeated Edman degradations can
be used to sequence a peptide from the N-terminal end.
• Cleavage with carboxypeptidase identifies the C-terminal amino acid.
• Partial hydrolysis of a peptide forms smaller fragments that can be sequenced. Amino acid
sequences common to smaller fragments can be used to determine the sequence of the
complete peptide.
• Selective cleavage of a peptide occurs with trypsin and chymotrypsin to identify the location
of specific amino acids (Table 29.2).
Adding and removing protecting groups for amino acids (29.7)
[1] Protection of an amino group as a Boc derivative
R
H2N
H
C
R
[(CH3)3COCO]2O
(CH3CH2)3N
CO2H
Boc N
H
H
C
CO2H
[2] Deprotection of a Boc-protected amino acid
R
H
C
Boc N
H
CO 2H
CF3CO2H
or
HCl or HBr
R
H2N
H
C
CO2H
[3] Protection of an amino group as an Fmoc derivative
H2N
C
C
R H
O
R H
OH
+
CH2O
C
O
Fmoc Cl
Na2CO3
Cl
H2O
Fmoc N
H
C
C
O
OH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
827
Amino Acids and Proteins 29–3
[4] Deprotection of an Fmoc-protected amino acid
R H
C
Fmoc N
H
R H
C
OH
H2N
C
O
C
OH
O
N
H
[5] Protection of a carboxy group as an ester
O
O
H2N
C
C
OH
CH3OH, H+ H2N
C
C
OCH3
H R
H R
O
O
H2N
C
C
C6H5CH2OH,
OH
H+
H2N
C
C
OCH2C6H5
H R
H R
methyl ester
benzyl ester
[6] Deprotection of an ester group
O
H2N
C
C
O
O
–OH
OCH3
H2N
C
H2O
H R
C
OH
H R
H2N
C
C
O
H2O, –OH
OCH2C6H5
H2N
or
H2, Pd-C
H R
C
H R
benzyl ester
methyl ester
Synthesis of dipeptides (29.7)
[1] Amide formation with DCC
R H
Boc N
H
C
OH +
C
O
O
H2N
C
H R
C
R H
OCH2C6H5
DCC
Boc N
H
C
C
O
H
N
O
C
C
OCH2C6H5
H R
[2] Four steps are needed to synthesize a dipeptide:
a. Protect the amino group of one amino acid using a Boc or Fmoc group.
b. Protect the carboxy group of the second amino acid using an ester.
c. Form the amide bond with DCC.
d. Remove both protecting groups in one or two reactions.
Summary of the Merrifield method of peptide synthesis (29.8)
[1] Attach an Fmoc-protected amino acid to a polymer derived from polystyrene.
[2] Remove the Fmoc protecting group.
[3] Form the amide bond with a second Fmoc-protected amino acid using DCC.
[4] Repeat steps [2] and [3].
[5] Remove the protecting group and detach the peptide from the polymer.
C
OH
828
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–4
Practice Test on Chapter Review
1. a Which statement is true about the peptide Ala–Gly–Tyr–Phe?
1. The N-terminal amino acid is Ala.
2. The N-terminal amino acid is Phe.
3. The peptide contains four peptide bonds.
4. Statements [1] and [3] are true.
5. Statements [2] and [3] are true.
b. Which of the following peptides is hydrolyzed by trypsin?
1. Glu–Ser–Gly–Arg
2. Arg–Gln–Trp–Asp
3. Glu–Val–Leu–Lys
4. Peptides [1] and [2] are hydrolyzed.
5. Peptides [1], [2], and [3] are all hydrolyzed.
c. In which types of protein structure is hydrogen bonding observed?
1. α-helix
2. β-pleated sheet
3. 3° structure
4. Hydrogen bonding is present in [1] and [2].
5. Hydrogen bonding is present in [1], [2], and [3].
2. Answer the following questions about peptides.
O
O
H 2N
H
C
C
CH3
Ala
H 2N
OH
C
C
O
OH
H 2N
H CH(CH3)2
Val
C
C
OH
H CH2OH
Ser
a. Draw the structure of the following tripeptide: Val–Ser–Ala.
b. Give the three-letter abbreviation for the N-terminal amino acid.
c. Give the three-letter abbreviation for the C-terminal amino acid.
3. Answer the following questions about the amino acid leucine (2-amino-4-methylpentanoic acid),
which has pKa’s of 2.33 and 9.74 for its ionizable functional groups.
a. Draw a Fischer projection for L-leucine and label the stereogenic center as R or S.
b. What is the pI of leucine?
c. Draw the structure of the predominant form of leucine at its isoelectric point.
d. Draw the structure of the predominant form of leucine at pH 10.
e. Is leucine an acidic, basic, or neutral amino acid?
4. What product is formed when the amino acid phenylalanine is treated with each reagent?
a. PhCH2OH, H+
b. Ac2O, pyridine
c. PhCOCl, pyridine
d. (Boc)2O
e. C6H5N=C=S
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
829
Amino Acids and Proteins 29–5
5. Draw the amino acids and peptide fragments formed when the octapeptide
Tyr–Gly–Ala–Lys–Val–Ser–Phe–Met is treated with each reagent or enzyme:
a. chymotrypsin
b. trypsin
c. carboxypeptidase
d. C6H5N=C=S
Answers to Practice Test
1. a. 1
b. 2
c. 5
2. a
4.
O
Val
OCH2Ph
a.
H CH2OH
O
H
H2N
C
C
C
N
N
C
C
OH
C
H
O
H CH(CH3)2
H CH3
O
H2N H
O
OH
b.
HN H
Ala
Ser
5.a. Tyr, Gly–Ala–Lys–Val–Ser–Phe, Met
b. Tyr–Gly–Ala–Lys, Val–Ser–Phe– Met
c. Tyr–Gly–Ala–Lys–Val–Ser–Phe, Met
d. Tyr, Gly–Ala–Lys–Val–Ser–Phe–Met
O
b. Val
c. Ala
O
3.
a. H2N S
OH
c.
COOH
HN H
H
O
Ph
CH2CH(CH3)2
O
b. 6.04
O
c.
OH
d.
O
HN H
NH3
Boc
O
O
C6H5
d.
O
e.
NH2
CH2
S
N
H
e. neutral
Answers to Problems
29.1
H CH3 O
S
S C
CH3
OH
H2N H
L-isoleucine
R
O
H CH3 O
S C
R C
H
H2N H
OH
S
H NH2
CH3
OH
H O
R
R C
H NH2
OH
830
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–6
29.2
a.
b.
NH3+
(CH3)2CH
C
COO–
c.
NH3+
COO–
+
HOOCCH2CH2 C COO–
N
H
H
NH3+
d.
C COO–
(CH3)2CHCH2
H
H H
29.3
In an amino acid, the electron-withdrawing carboxy group destabilizes the ammonium ion
(–NH3+), making it more readily donate a proton; that is, it makes it a stronger acid. Also,
the electron-withdrawing carboxy group removes electron density from the amino group
(–NH2) of the conjugate base, making it a weaker base than a 1o amine, which has no
electron-withdrawing group.
29.4
The most direct way to synthesize an α-amino acid is by SN2 reaction of an α-halo
carboxylic acid with a large excess of NH3.
a.
NH3
Br CH COOH
H
b. Br CH COOH
CH CH3
c. Br CH COOH
H2N CH COO– NH4+
large excess
H
NH3
CH2
glycine
NH3
large excess
H2N CH COO– NH4+
CH2
H2N CH COO– NH4+
large excess
CH CH3
CH2
phenylalanine
CH2
CH3
CH3
isoleucine
29.5
a.
CH3
CH3I
H2N C COOH
alanine
c.
CH3CH2CH(CH3)Br
H
b.
CH(CH3)CH2CH3
H2N C COOH
isoleucine
H
CH2CH(CH3)2
(CH3)2CHCH2Cl
H2N C COOH
leucine
H
29.6
O
H
H
C N C COOEt
CH3
COOEt
[1] NaOEt
[2] CH2=O
[3] H3O+, Δ
CH2OH
H2N C COOH
H
serine
29.7
O
a. H2N CHCOOH
CH CH3
CH3
valine
(CH3)2CH
C
O
b. H2N CHCOOH
H
CH2
(CH3)2CHCH2
C
O
c. H2N CHCOOH
H
CH2
CH CH3
CH3
leucine
phenylalanine
C6H5CH2
C
H
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
831
Amino Acids and Proteins 29–7
29.8
a. BrCH2COOH
NH3
large excess
H
b. CH3CONH C COOEt
COOEt
NH2CH2COO– NH4+
H
[1] NaOEt
[2] (CH3)2CHCl
CH(CH3)CH2CH3
H
[1] NaOEt
d. CH3CONH C COOEt
CH(CH3)2
H2N CHCOOH
[2] H3O+
H
H2N C COOH
[3] H3O+, Δ
[1] NH4Cl, NaCN
c. CH3CH2CH(CH3)CHO
H2N C COOH
[2] BrCH2CO2Et
COOEt
CH2CO2H
[3] H3O+, Δ
A chiral amine must be used to resolve a racemic mixture of amino acids.
29.9
N
CH3 H
a. C6H5CH2CH2NH2
achiral
c.
b.
d.
C
N
CH3CH2
achiral
H
NH2
N
chiral
(can be used)
H
O
H
HO
chiral
(can be used)
29.10
NH2
+
Ac2O
AcNH
COOH
C
C
R
H2 N
proton transfer
C6H5
C
(R isomer only)
CH3 H
+
COO–
H3N
C
C6H5
CH3 H
H CH2CH(CH3)2
enantiomers
(CH3)2CHCH2 H
S
AcNH
COOH
C
+
H CH2CH(CH3)2
Step [1]:
React both enantiomers with the
R isomer of the chiral amine.
enantiomers
R
S
AcNH
COOH
C
(CH3)2CHCH2 H
H CH2CH(CH3)2
To begin:
Convert the amino acids into N-acetyl
amino acids (two enantiomers).
NH2
COOH
C
S
R
AcNH
COO–
C
(CH3)2CHCH2 H
+
H3 N
C
C6H5
CH3 H
diastereomers
R
R
These salts have the same configuration around one stereogenic center,
but the opposite configuration about the other stereogenic center.
separate
Step [2]:
Separate the diastereomers.
AcNH
C
COO–
H CH2CH(CH3)2
S
+
H3N
C
CH3 H
R
C6H5
AcNH
C
(CH3)2CHCH2 H
R
COO–
+
H3N
C
CH3 H
R
C6H5
832
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–8
H2O, –OH
Step [3]:
Regenerate the amino acid
by hydrolysis of the amide.
NH2
C
H2O, –OH
COOH
NH2
H CH2CH(CH3)2
COOH
C
H2N
+
(CH3)2CHCH2 H
(S)-leucine
C
C6H5
CH3 H
(R)-leucine
The chiral amine is
also regenerated.
The amino acids are now separated.
29.11
COOH
H2N
[1] (CH3CO)2O
H2N C H
[2] acylase
CH2CH(CH3)2
C
H
COOH
N
C
C
O
(CH3)2CHCH2 H
COOH
CH3
+
H CH2CH(CH3)2
(S)-leucine
(mixture of enantiomers)
N-acetyl-(R)-leucine
29.12
H
a. H2N CHCOOH
CH3
(CH3)2CH
NHAc
b. H2N CHCOOH
C C
H
COOH
CH2
H2NCOCH2
NHAc
C C
H
c. H2N CHCOOH
CH2
COOH
CH CH3
CH2
CH3
CONH2
NHAc
C C
COOH
H
29.13 Draw the peptide by joining adjacent COOH and NH2 groups in amide bonds.
amide
O
a.
H2N CH C OH
(CH3)2CH H
O
H2N CH C OH
CH CH3
CH2
CH3
CH2
Val
COOH
H2N
C
N-terminal
O
O
H2N CH C OH H2N CH C OH
H
CH2
Gly
C
C
C-terminal
OH
H CH2CH2COOH
amide
O
H2N CH C OH
O
O
Val–Glu
Glu
b.
C
H
N
NH
N
CH2
Leu
His
H2N
CH CH3
CH3
(CH3)2CHCH2
H
O
H H
C
C
O
N-terminal
H
N
C
C
H CH2
N
H
amide
NH
Gly–His–Leu
N
C
C
O
OH
C-terminal
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
833
Amino Acids and Proteins 29–9
O
O
O
O
c. H2N CH C OH H2N CH C OH H2N CH C OH H2N CH C OH
CH2
CH3
CH OH
CH OH
CH2SCH3
M
A
CH3
CH3
T
T
N-terminal
CH2SCH3 amide
CH3
CH2 H
O
CHOH
H
C
C
C
H2N
H
N
C
O
C
CH3
H
N
H
amide
C
O
H
N
O
C
H
OH
C-terminal
CHOH
CH3
M–A–T–T
29.14
N
HN
CONH2
a.
H2N
C
O
CH2 H
C
C
N
H
C
O
H CH2
O
H
N
C
C
H2N
b.
OH
H CH(CH3)2
C
Arg–Asn–Val
R–N–V
NH
C
N
H
C
O
H
N
O
C
C
OH
H CH2
CH2
CH2
CH2
CONH2
CH2
C
HN
CH2 H
C
H CH2
CH2
CH2
O
Lys–His–Gln
K–H–Q
NH2
NH2
29.15 There are six different tripeptides that can be formed from three amino acids (A, B, C):
A–B–C, A–C–B, B–A–C, B–C–A, C–A–B, and C–B–A.
29.16
O
H2 N
N
H
H
N
O
O
N
H
O
H
N
OH
O
leu-enkephalin
HO
29.17
O
HO
HS
O
a. H2N
H
N
N
COOH
H
O
glutathione
O
N
O
H
H
O
H2N
NH2
O
S
OH
COOH
N
S
H
N
N
COOH
H
O
O
OH
834
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–10
b.
O
HS
H2N
α COOH
H
O
N
N
COOH
H2N
H
The peptide bond beween glutamic acid and its adjacent
OH amino acid (cysteine) is formed from the COOH in the R
group of glutamic acid, not the α COOH.
O
This comes from the amino acid glutamic acid.
O
α
OH
This carboxy group is used to form the amide bond in the peptide, not the α
COOH, as is usual. That's what makes glutathione's structure unusual.
COOH
glutamic acid
29.18
O
C6H5
a.
S
O
C6H5
N
CH3
N
H
b.
S
from Ala
N
N
H
from Val
29.19 Determine the sequence of the octapeptide as in Sample Problem 29.2. Look for overlapping
sequences in the fragments.
common amino acids
Answer:
Ala–Leu–Tyr
Tyr–Leu–Val–Cys
Ala–Leu–Tyr–Leu–Val–Cys–Gly–Glu
Val–Cys–Gly–Glu
29.20 Trypsin cleaves peptides at amide bonds with a carbonyl group from Arg and Lys.
Chymotrypsin cleaves at amide bonds with a carbonyl group from Phe, Tyr, and Trp.
a. [1] Gly–Ala–Phe–Leu–Lys + Ala
[2] Phe–Tyr–Gly–Cys–Arg + Ser
[3] Thr–Pro–Lys + Glu–His–Gly–Phe–Cys–Trp–Val–Val–Phe
b. [1] Gly–Ala–Phe + Leu–Lys–Ala
[2] Phe + Tyr + Gly–Cys–Arg–Ser
[3] Thr–Pro–Lys–Glu–His–Gly–Phe + Cys–Trp + Val–Val–Phe
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
835
Amino Acids and Proteins 29–11
29.21
Edman degradation gives N-terminal
amino acid:
Carboxypeptidase identifies the C-terminal
amino acid:
Leu–___–___–___–___–___–___
Leu–___–___–___–___–___–Glu
Partial hydrolysis
common amino acids
Ala–Ser–Arg
Gly–Ala–Ser
Leu–Gly–Ala–Ser–Arg–___–Glu
or
Gly–Ala–Ser–Arg
Leu–___–Gly–Ala–Ser–Arg–Glu
Cleavage by trypsin is after Arg and yields a
dipeptide; therefore, this must be the peptide:
Leu–Gly–Ala–Ser–Arg–Phe–Glu
29.22
a.
O
(CH3)2CHCH2 H
H2 N
C
Leu
OH
C
O
(CH3)2CHCH2 H
Boc N
H
C
(CH3)2CHCH2 H
[(CH3)3COCO]2O
C
OH
Boc N
C
(CH3CH2)3N
H
O
O
C
OH + H2N
OCH2C6H5
C
C
O
H CH(CH3)2
H2N
C
O
C
C6H5CH2OH, H+
OH
H CH(CH3)2
H2N
C
(CH3)2CHCH2 H
Boc N
H
C
C
O
O
H
N
C
C
OCH2C6H5
H CH(CH3)2
HBr, CH3COOH
(CH3)2CHCH2 H
H2N
C
C
O
H
N
OCH2C6H5
H CH(CH3)2
Val
new amide bond
DCC
C
O
C
C
OH
H CH(CH3)2
Leu–Val
836
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–12
O
CH3 H
b.
C
H2N
Ala
C
CH3 H
[(CH3)3COCO]2O
OH
Boc N
H
(CH3CH2)3N
O
C
H2N
C
OH
C6H5CH2OH, H+
H2N
H CH(CH3)CH2CH3
O
A
C
OH
O
C
C
OCH2C6H5
CH(CH3)CH2CH3
H
Ile
C
B
new amide bond
A
+
CH3 H
DCC
B
Boc N
H
C
C
O
O
H2N
C
C
H H
H
N
C
O
H2
C
Pd-C
OCH2C6H5
CH3 H
Boc N
H
C
OH
H2N
C
C
C
C
OH
H CH(CH3)CH2CH3
C
O
C6H5CH2OH, H+
C
O
H CH(CH3)CH2CH3
O
H
N
OCH2C6H5
H H
Gly
new amide bond
O
C
+ H2N
C
H H
C
CH3 H
OCH2C6H5
DCC
Boc N
H
C
C
O
O
H
N
C
H
C
N
H
C
OCH2C6H5
HBr
CH3COOH
O
CH(CH3)CH2CH3
CH3 H
Ala–Ile–Gly
H2N
C
C
O
H
N
H
O
C
C
N
H
C
OH
O
CH(CH3)CH2CH3
837
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–13
CH3 H
c.
H2N
C
CH3 H
C
OH [(CH3)3COCO]2O
(CH3CH2)3N
Ala O
C
Boc N
H
O
OH
C
H2N
O
C
C
H H
A
O
C6H5CH2OH, H+
OH
+
A
DCC
B
Boc N
H
CH3 H
H2N
C
C
O
C
C
H2
OCH2C6H5
Boc N
H
Pd-C
C
C
H H
C
OH
H2N
+
H2N
C
C
O
C
OH
H H
C
C
C
OCH2C6H5
new amide bond
OCH2C6H5
DCC
CH3 H
BOC N
H
C
C
O
H
N
O
C
C
CH3 H
N
H
H H
C
C
H2
OCH2C6H5
Pd-C
O
CH3 H
Boc N
H
C
B
DCC
CH3 H
CH3 H
O
O
H
H
C
N
C
N
C
C
C
C
C
C
N
OCH2C6H5
Boc N
H
H
H H
O
O
H H
O
H
N
C
CH3 H
C
C
N
H
C
OH
O
H H
D
HBr
CH3COOH
CH3 H
H2N
C
O
new amide bond
+
C
O
CH3 H
D
B
O
H
N
O
OCH2C6H5
CH3 H
C6H5CH2OH, H+
Ala O
C
C
CH3 H
O
H
N
C
C
H H
Gly
new amide bond
CH3 H
H2N
C
C
O
H
N
O
C
C
H H
CH3 H
N
H
C
C
O
Ala–Gly–Ala–Gly
H
N
O
C
H H
C
OH
838
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–14
29.23 The dipeptide depicted in the 3-D model has alanine as the N-terminal amino acid and
cysteine as the C-terminal amino acid.
O
H2N
H
Boc N
[(CH3)2COCO]2O
OH
O
OH
H2N
OH
(CH3CH2)3N
H CH3
HSCH2 H
O
A
O
H
Boc N
OCH2C6H5
H2N
H CH3
H
Boc N
DCC
HSCH2
H
O
H CH3
O
A
O
B
Cys
HSCH2 H
OH
OCH2C6H5
H2 N
H+
H CH3
Ala
HSCH2 H
C6H5CH2OH
OCH2C6H5
N
H
O
B
HBr, CH3COOH
HSCH2
H
O
H2N
H CH3
N
H
Ala–Cys
29.24
All Fmoc-protected amino acids are made by the following general reaction:
R
H2N
H
C
C
O
R
OH
Fmoc–Cl
Na2CO3, H2O
Fmoc N
H
H
C
C
O
OH
OH
O
839
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–15
H H
The steps:
Fmoc N
H
C
OH
C
[1] base
[2] Cl CH2 POLYMER
O
H H
C
Fmoc N
H
C
O CH2 POLYMER
O
[1]
[2] DCC
N
H
H
Fmoc N
O
C
C
OH
H CH(CH3)CH2CH3
(CH3)2CHCH2 H
C
Fmoc N
H
C
H
N
O
C
C
O
H
H
H
C
N
H
[1]
C
O CH2 POLYMER
O
CH(CH3)CH2CH3
N
H
[2] DCC
H
Fmoc N
Fmoc N
H
N
H
CH3 H
[2] DCC
C
OH
Fmoc N
C
H
O
C
C
C
C
N
H
C
O CH2 POLYMER
O
H CH(CH )CH CH
3
2
3
(CH3)2CHCH2 H
[1]
H H
O
OH
C
O
[1]
(CH3)2CHCH2
O H H
O
H
H
H2N
C
C
C
C
OH
N
C
N
N
C
C
C
H
H
O
O
H CH3
H CH(CH )CH CH
N
(CH3)2CHCH2
H
H H
O
O
H
H
[2] HF
H
Fmoc N
C
C
C
C
O CH2 POLYMER
N
C
N
N
C
C
C
H
H
O
O
H CH3
H CH(CH3)CH2CH3
3
2
3
Ala–Leu–Ile–Gly
+ F CH2 POLYMER
29.25 In a parallel β-pleated sheet, the strands run in the same direction from the N- to C-terminal
amino acid. In an antiparallel β-pleated sheet, the strands run in the opposite direction.
H
H
O CH3 H
N
C
C
N
H CH3 H
H
C
C
O
H
N
O CH3 H
C
C
N
H CH3 H
C
C
OH
O
H CH3 O
H H CH3 O
H
HO
C
C
N
C
C
N
C
N
C
C
N
C
H
O
H CH H O
H CH H
3
H
H
N
N
O CH3 H H
O CH3 H
C
C
N
C
C
OH
C
N
C
C
N
C
H
H
O H CH3
H CH3
O
parallel
H
C
O
C
H CH3
3
CH3 H
H
C
N
N
H
C
O
O CH3 H
C
C
H CH3
antiparallel
N
H
C
C
O
OH
840
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–16
29.26
a. Ser and Tyr
b. Val and Leu
H2N CH COOH
H2N CH COOH
CH2
CH2
H2N CH COOH
OH
side chains with
OH groups
c. 2 Phe residues
H2N CH COOH
CH CH3
CH2
CH3
CH CH3
H2N CH COOH
H2N CH COOH
CH2
CH2
CH3
side chains with only
C–C and C–H bonds
OH
van der Waals forces
van der Waals forces
hydrogen bonding
29.27 a. The R group for glycine is a hydrogen. The R groups must be small to allow the β-pleated
sheets to stack on top of each other. With large R groups, steric hindrance prevents
stacking.
b. Silk fibers are water insoluble because most of the polar functional groups are in the
interior of the stacked sheets. The β-pleated sheets are stacked one on top of another, so
few polar functional groups are available for hydrogen bonding to water.
29.28
H CH2C6H5
H2N
C
c.
H CH2C6H5
a. CH3OH, H+
H2N
C
C
H3N
COOCH3
H CH2C6H5
O
b. CH3COCl, pyridine
CH3
C
N
H
COOH
H CH2C6H5
d. NaOH (1 equiv)
COOH
phenylalanine
H CH2C6H5
HCl (1 equiv)
C
C
O
H2 N
C
COO
OH
O
C6H5
e. C6H5N=C=S
N
CH2C6H5
N
H
S
29.29
a. N-terminal amino acid: alanine
C-terminal amino acid: serine
b. A–Q–C–S
c. Amide bonds are bold.
C-terminal
HSCH2
O
CH3 H
H2N
C
C
O
H
N
H
C
C
H
C
N
H
C
O
H
N
C
C
OH
O
H CH2OH
CH2CH2CONH2
N-terminal
Ala–Gln–Cys–Ser
A–Q–C–S
841
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–17
29.30 The dipeptide is composed of phenylalanine and leucine.
O
C6H5CH2 H
H2N
C
C
C6H5CH2 H
[(CH3)3COCO]2O
OH
(CH3CH2)3N
Boc N
H
O
Phe
C
C
OH
C
+ B
DCC
Boc N
H
C
H
N
C
O
O
C
C6H5CH2OH, H+
OH
O
C
C
OCH2C6H5
B
C6H5CH2 H
HBr
OCH2C6H5
C
H CH2CH(CH3)2
Leu
O
C
H2N
H CH2CH(CH3)2
A
C6H5CH2 H
A
H2N
CH3COOH
H2N
H CH2CH(CH3)2
C
C
O
H
N
O
C
C
OH
H CH2CH(CH3)2
Phe–Leu
29.31
R COOH
a.
S
H2 N C H
COOH
CH3 C SH
COOH
b.
H C NH2
H2 N C H
CH3 C SH
CH3
CH3 C S
CH3
(R)-penicillamine
CH3
(S)-penicillamine
COOH
H2 N C H
S C CH3
CH3
29.32 Amino acids are insoluble in diethyl ether because amino acids are highly polar; they exist as
salts in their neutral form. Diethyl ether is weakly polar, so amino acids are not soluble in it.
N-Acetyl amino acids are soluble because they are polar but not salts.
NH3+
C
R
H
NHCOCH3
R
H
COO–
C
COOH
N-acetyl amino acid
ether soluble
amino acid, a salt
H2O soluble and ether insoluble
29.33 The electron pair on the N atom not part of a double bond is delocalized on the fivemembered ring, making it less basic.
H2N CH COOH
CH2
HA
H2N CH COOH
CH2
+
NH
N
N
When this N is protonated...
H2N CH COOH
CH2
NH
N
sp3 hybridized N atom
NH2
...the ring is no longer aromatic.
H2N CH COOH
HA
preferred
path
When this N is protonated...
H2N CH COOH
CH2
CH2
+
NH
NH
+N
H
N
H
...the ring is still aromatic.
6 π electrons
842
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–18
29.34
H2N CH COOH
H2N CH COOH
CH2
CH2
The ring structure on tryptophan is aromatic since
each atom contains a p orbital. Protonation of the N
atom would disrupt the aromaticity, making this a
less favorable reaction.
H2N
HN
+
no p orbital on N
This electron pair is delocalized on the bicyclic ring
system (giving it 10 π electrons), making it less
available for donation, and thus less basic.
29.35 At its isoelectric point, each amino acid is neutral.
a.
+
COO–
H3N C H
b.
COO–
+
H3N C H
CH3
alanine
CH2CH2SCH3
methionine
c.
+
COO–
H3N C H
CH2COOH
aspartic acid
d.
COO–
H2N C H
+
CH2CH2CH2CH2NH3
lysine
29.36
a. [1] glutamic acid: use the pKa’s 2.10 + 4.07
[2] lysine: use the pKa’s 8.95 + 10.53
[3] arginine: use the pKa’s 9.04 + 12.48
b. In general, the pI of an acidic amino acid is lower than that of a neutral amino acid.
c. In general, the pI of a basic amino acid is higher than that of a neutral amino acid.
29.37
a. threonine
pI = 5.06
(+1) charge at pH = 1
H3N CH COOH
b. methionine
pI = 5.74
(+1) charge at pH = 1
H3N CH COOH
c. aspartic acid
pI = 2.98
(+1) charge at pH = 1
H3N CH COOH
d. arginine
pI = 5.41
(+2) charge at pH = 1
H3N CH COOH
CH OH
CH2
CH2
CH2
CH3
CH2
COOH
CH2
S
CH2
CH3
NH
C NH2
NH2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
843
Amino Acids and Proteins 29–19
29.38
a. valine
pI = 6.00
(–1) charge
at pH = 11
b. proline
pI = 6.30
(–1) charge
at pH = 11
c. glutamic acid
pI = 3.08
(–2) charge
at pH = 11
COO–
H2N CH COO–
H2N CH COO–
CH2
CH2
CH2
CH2
COO–
CH2
H2N CH COO–
CH CH3
HN
CH3
d. lysine
pI = 9.74
(–1) charge
at pH = 11
CH2
NH2
29.39 The terminal NH2 and COOH groups are ionizable functional groups, so they can gain or lose
protons in aqueous solution.
O
a.
CH3 H
H2N CH C OH
CH3
H2N
Ala
b. At pH = 1
C
C
O
CH3 H
H3N
C
C
O
H
N
O CH3
C
C
H CH3
N
H
H
N
O CH3
C
C
H CH3
N
H
H
C
C
OH
A–A–A
O
H
C
C
OH
O
c. The pKa of the COOH of the tripeptide is higher than the pKa of the COOH group of
alanine, making it less acidic. This occurs because the COOH group in the tripeptide is
farther away from the –NH3+ group. The positively charged –NH3+ group stabilizes the
negatively charged carboxylate anion of alanine more than the carboxylate anion of the
tripeptide because it is so much closer in alanine. The opposite effect is observed with the
ionization of the –NH3+ group. In alanine, the –NH3+ is closer to the COO– group, so it is
more difficult to lose a proton, resulting in a higher pKa. In the tripeptide, the –NH3+ is
farther away from the COO–, so it is less affected by its presence.
844
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–20
29.40
H2N
H CH2CH(CH3)2
H2N
C
C
g. C6H5COCl, pyridine
C6H5
COOCH3
b. CH3COCl, pyridine
C
CH3
N
H
C
C
C
H2N
N
H
C
C
O
Boc N
H
C
COOH
O
i. The product in (d), then
NH2CH2COOCH3 + DCC
H CH2CH(CH3)2
H
C
N
COOCH3
AcNH
C
C
COOCH2C6H5
O
H CH2CH(CH3)2
d. Ac2O, pyridine
C
AcNH
e.
C
H3N
H CH2CH(CH3)2
H
C
N
COOCH
Boc N
C
C 3
H
O H H
H2N
C
H CH2CH(CH3)2
k. Fmoc–Cl, Na2CO3, H2O
Fmoc N
H
O
COOH
H CH2CH(CH3)2
f. NaOH (1 equiv)
C6H5
l. C6H5N=C=S
COO
N
S
Br
b. CH3CONHCH(COOEt)2
NH3
(CH3)2CHCH2CHCOO– NH4+ + NH4+Br–
excess
NH3
NH2
[1] NaOEt
[2] O
C O
CH3
HO
CH2 CHCOOH
CH2Br
[3] H3O+, Δ
O
NH2
O
c.
[1] NH4Cl, NaCN
H
N
H
[2] H3O+
HOOC
O
NH2
O
[1] NH4Cl, NaCN
d.
CH3O
CHO
e. CH3CONHCH(COOEt)2
[2] H3O+
COOH
HO
[1] NaOEt
[2] ClCH2CH2CH2CH2NHAc
[3] H3O+, Δ
NH2
CH2CH2CH2CH2NH2
H2N C COOH
H
C
COOH
CH2CH(CH3)2
N
H
29.41
a. (CH3)2CHCH2CHCOOH
H H
j. The product in (h), then NH2CH2COOCH3 + DCC
COOH
H CH2CH(CH3)2
HCl (1 equiv)
OH
H CH2CH(CH3)2
h. [(CH3)3COCO]2O
(CH3CH2)3N
OH
H CH2CH(CH3)2
c. C6H5CH2OH, H+
C
H CH2CH(CH3)2
O
COOH
leucine
H CH2CH(CH3)2
O
H CH2CH(CH3)2
+
a. CH3OH, H
845
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–21
29.42
a. Asn
b. His
O
H2N CH C OH
c. Trp
O
H2N CH C OH
CH2Br
CH2
C O
C O
NH2
NH2
CH2Br
CH2
O
CH2Br
H2N CH C OH
CH2
NH
NH
HN
N
N
HN
29.43
OH
O
H
H
C N C COOEt
CH3
CHCH3
[1] NaOEt
H2N C COOH
[2] CH3CHO
[3] H3O+, Δ
COOEt
H
threonine
29.44
Br2
a. CH3CHO
CH3
C
+
NH3
H3NCH2COO–
excess
glycine
NH2
[1] NH4Cl, NaCN
H
BrCH2COOH
H2SO4, H2O
CH3COOH
O
b.
CrO3
BrCH2CHO
CH3 C COOH
[2] H3O+
H
alanine
29.45
O
N–K+
CH2(COOEt)2
Br2
Br CH(COOEt)2
CH3COOH
O
O
N
CH(COOEt)2
A
B
O
[1] NaOEt
[2] ClCH2CH2SCH3
O
H
H2N C COOH
CH2CH2SCH3
D
[1] NaOH, H2O
COOEt
N
[2] H3O+, Δ
C COOEt
CH2CH2SCH3
O
C
846
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–22
29.46
O
OEt
H
H
C N C COOEt
CH3
COOEt
[1]
O
NaOEt
CH2 CHCOOEt
H
EtOH
C N C COOEt +
CH3
O
[2]
COOEt
H
C N C COOEt
CH3
COOEt
CH2–CHCOOEt
H OH2
[3]
H2N CHCOOH
H3O+
CH2
Δ
CH2COOH
O
COOEt
H
C N C COOEt
CH3
CH2CH2COOEt
glutamic acid
+ H O
2
29.47
CN
N
CHO
[1] DIBAL-H
[2] H2O
CO2Et
Ph3P=CHCO2Et
N
N
CO2Et
H2
Pd-C
(+ cis isomer)
N
B
A
C
KOH
H2O
O
H2N
O
NH
N
OCH3
O
E
OCH3
Boc NH H
CO2H
Boc NH H
DCC
N
D
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
847
Amino Acids and Proteins 29–23
29.48
CH3O
O
H
CH3O
Cl
O
H
N
Cl
enantiomers
N
S
S
R isomer
S isomer
proton
transfer
O
HO3S
CH3O
O
H
CH3O
Cl
O
H
NH
Cl
NH
S
diastereomers
S
O
O3S
O
O3S
separate
CH3O
O
H
CH3O
Cl
O
H
NH
Cl
NH
S
S
O
O3S
O
O3S
–OH
–OH
CH3O
O
H
CH3O
Cl
N
S
O
H
N
S
R isomer
S isomer
Cl
HO3S
O
The chiral sulfonic
acid is
regenerated.
848
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–24
29.49
H CH3
C
NH2
S
COOH
To begin:
Convert the amino acids into amino
acid esters (two enantiomers).
H
C
H2N
enantiomers
COOH
R
CH3OH, H+
H CH3
C
H2N
S
COOCH3
CH3
H2N
H
C
enantiomers
COOCH3
R
H OH
Step [1]:
React both enantiomers with the
R isomer of mandelic acid.
proton transfer
C
COO–
H3N
C
S
COOCH3
COOH
CH3
H OH
H CH3
R
(R)-mandelic acid
C
C6H5
H OH
C6H5
CH3
C
C6H5
R
COO– H3N
H
C
R
COOCH3 diastereomers
These salts have the same configuration around one stereogenic center,
but the opposite configuration about the other stereogenic center.
separate
Step [2]:
Separate the diastereomers.
H OH
C6H5
Step [3]:
Regenerate the amino acids
by hydrolysis of the esters.
C
R
H CH3
–
COO
H3N
C
S
COOCH3
C6H5
C
COO–
H2N
C
S
COOH
H3N
R
H2O, –OH
H CH3
CH3
H OH
H
C
R
COOCH3
H2O, –OH
CH3
H2N
H
C
R
COOH
+
H OH
C6H5
C
COOH
The chiral acid is
regenerated.
849
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–25
29.50
NH2
COOH
C
NH2
+
R
S
AcNH
enantiomers
C6H5CH2 H
H CH2C6H5
To begin:
Convert the amino acids into N-acetyl
amino acids (two enantiomers).
COOH
C
Ac2O
COOH
C
AcNH
+
COOH
C
enantiomers
C6H5CH2 H
H CH2C6H5
R
S
N
proton transfer
Step [1]:
React both enantiomers with the
R isomer of the chiral amine.
CH3O
CH3O
N
brucine
H
O
O
H
H
N
AcNH
C
COO–
H CH2C6H5
S
AcNH
CH3O
CH3O
N
COO–
C
C6H5CH2 H
R
H
O
N
CH3O
CH3O
N
H
O
O
O
diastereomers
separate
Step [2]:
Separate the diastereomers.
H
AcNH
C
N
COO–
CH3O
H CH2C6H5
S
H
AcNH
COO–
C
N
CH3O
C6H5CH2 H
CH3O
N
O
Step [3]:
Regenerate the amino acid
by hydrolysis of the amide.
R
H
CH3O
H2O, –OH
C
COOH
H CH2C6H5
(S)-phenylalanine
H
O
O
NH2
N
O
H2O, –OH
NH2
C
N
COOH
C6H5CH2 H
(R)-phenylalanine
The amino acids are now separated.
+
CH3O
CH3O
N
O
H
O
The chiral amine is
also regenerated.
850
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–26
29.51
NH2
Ac2O
a. (CH3)2CH CH COOH
racemic mixture
H2N
acylase
C
AcNH
COOH
H CH(CH3)2
H CH(CH3)2
CH3CONH
NHCOCH3
b.
COOH
H2
–OH
chiral
Rh catalyst
H2O
H2N
COOH
C
+
COOH
C
CH2CH2CH2CH2NH2
H
(S)-isomer
COOH
c.
NHAc
OH
chiral
Rh catalyst
N
H
H2N
–
H2
H2O
C
COOH
(S)-isomer
CH2
H
N
H
29.52
CH3O
COOH
O
CH3
N
H
O
H2
CH3O
chiral
Rh catalyst
CH3
COOH
CH3
H N
H
O
HO
strong
acid
COOH
H NH2
HO
O
O
CH3
A
L-dopa
O
29.53
a.
C6H5CH2 H
H2N
C
C
O
H
N
O
C
C
H
c.
H2NCH2CH2CH2CH2 H
OH
H2N
CH3
H H
H2N
C
C
O
H
N
C
O
C
C
OH
H H
Lys–Gly
d.
O
C
H
N
O
Phe–Ala
b.
C
C
NH2C(NH)NHCH2CH2CH2 H
OH
H2N
C
C
O
H CH2
H
N
O
C
C
OH
CH2
H
R–H
CH2CONH2
NH
Gly–Gln
N
29.54 Amide bonds are bold lines (not wedges).
C-terminal
(CH3)2CH
O
H
O
H
H
C
N
C
OH
C
C
N
C
C
C
N
C
H2N
H
H CH2C6H4OH
O
O
H
CH2CH2CH2NHC(NH)NH2
HO2CCH2 H
N-terminal
Asp–Arg–Val–Tyr
D–R–V–Y
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
851
Amino Acids and Proteins 29–27
29.55 Name a peptide from the N-terminal to the C-terminal end.
C-terminal
CH2COOH
H H
a.
H2N
C
C
H
N
O
C
O
CH2 H
C
C
N
H
CH
H
2
COOH
C
b.
OH
HOOC
H
N
C
H CH2
O
H H
C
O
C
O
N
H
C
N-terminal
C
NH2
H CH3
CH2
CH2
Gly–Asp–Glu
G–D–E
NH
Ala–Gly–Arg
A–G–R
C
HN
NH2
29.56 A peptide C–N bond is stronger than an ester C–O bond because the C–N bond has more
double bond character due to resonance. Since N is more basic than O, an amide C–N bond
is more stabilized by delocalization of the lone pair on N.
O
R
C
A
O
R
NH2
C
B
+
NH2
Structure B contributes greatly to the resonance hybrid and
this shortens and strengthens the C–N bond.
29.57
O CH3 H
H2 N
C
C
N
C
H CH3 H
C
O
OH
O
s-trans
H 2N
C
C
N
H
H CH3 C
OH
C
s-cis HCH3
O
29.58
a.
b.
c.
d.
A–P–F + L–K–W + S–G–R–G
A–P–F–L–K + W–S–G–R + G
A–P–F–L–K–W–S–G–R + G
A + P–F–L–K–W–S–G–R–G
29.59
common amino acids
a.
Answer:
Gly–Ala
Ala–His
Gly–Ala–His–Tyr
His–Tyr
common amino acids
b.
Answer:
Lys–His
His–Gly–Glu
Gly–Glu–Phe
Lys–His–Gly–Glu–Phe
852
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–28
29.60
common amino acids
Arg–Arg–Val
Answer:
Val–Tyr
Tyr–Ile–His
Arg–Arg–Val–Tyr–Ile–His–Pro–Phe
Ile–His–Pro–Phe
29.61 Gly is the N-terminal amino acid (from Edman degradation), and Leu is the C-terminal
amino acid (from treatment with carboxypeptidase). Partial hydrolysis gives the rest of the
sequence.
common amino acids
Answer:
Gly–Ala–Phe–His
Gly–Gly–Ala–Phe–His–Ile–His–Leu
Phe–His–Ile
Ile–His–Leu
29.62 Edman degradation data give the N-terminal amino acid for the octapeptide and all smaller
peptides.
A
B
A: Glu–Arg–Val–Tyr
B: Ile–Leu–His–Phe
C: Glu–Arg
D: Val–Tyr
C
D
Glu–Arg–Val–Tyr–Ile–Leu–His–Phe
octapeptide
cleavage with
trypsin
cleavage with
chymotrypsin
carboxypeptidase
Glu–Arg–Val–Tyr–Ile–Leu–His + Phe
29.63 A and B can react to form an amide, or two molecules of B can form an amide.
H H
Boc N
H
C
H
O
OH + H2N
C
O
C
C
H
DCC
OH
C
Boc N
H
O
H CH(CH3)2
A
C
H
N
C
29.64
O
C
C
O
CH3OH, H+
OH
H2 N
H CH(CH3)2
O
b.
H2 N
C
C
C
C
OCH3
H CH(CH3)2
O
C6H5CH2OH, H+
OH
H2 N
H CH2CH(CH3)2
c. NH2CH2COOH
[(CH3)3COCO]2O
(CH3CH2)3N
H
Boc N
C
C
OCH2C6H5
H CH2CH(CH3)2
O
C
H H
C
OH
+
OH
H CH(CH3)2
B
a. H2N
(CH3)2CH
O
H2 N
H
C
C
O
H
N
O
C
OH
H CH(CH3)2
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
853
Amino Acids and Proteins 29–29
d. product in (b) + product in (c)
H
O
H
Boc N
DCC
C
C
N
H
CH2CH(CH3)2
C
C
O
H H
e. (CH3)3CO
C
H
O
N
C
O
C
H
H2
OCH2C6H5
(CH3)3CO
Pd-C
H CH(CH3)2
O
N
C
OCH2C6H5
O
C
C
OH +
C6H5CH3
H CH(CH3)2
O
HBr
f. starting material in (e)
H2N
CH3COOH
C
C
OH
H CH(CH3)2
O
CF3COOH
g. product in (e)
H2N
C
C
OH
H CH(CH3)2
O
C
h.
O
Na2CO3
+ Fmoc–Cl
OH
C
H2O
H2N H
OH
HN H
Fmoc
29.65
O
H H
a.
H2N
C
Gly
C
H H
[(CH3)3COCO]2O
OH
(CH3CH2)3N
Boc N
H
O
DCC
Boc N
H
C
C
O
H
N
C
OH
O
A
H H
A + B
C
H2N
H
C
C
O
OH
C
H CH3
H2N
CH3
B
H H
OCH2C6H5
HBr
CH3COOH
C
C
H CH3
Ala
O
C
C6H5CH2OH, H+
H2N
C
C
O
H
N
O
C
C
H CH3
Gly–Ala
OH
OCH2C6H5
854
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–30
CH3
CH3CH2CH
b.
C
H2N
CH3
H
Ile
OH
C
(CH3CH2)3N
O
C6H5CH2
H2N
C
C
C
C
OH
O
OH
C6H5CH2OH, H+
H CH3
O
C
Phe
C
O
C6H5CH2OH, H
C
OCH2C6H5
B
Ala
H
N
C
O
H
O
C
C
CH3CH2CH H
H2
Pd-C
OCH2C6H5
C
Boc N
H
CH3
C6H5CH2 H
OH
C
H CH3
C
C
+
H2N
C
C
O
DCC
Boc N
H
C
H
N
O
C
C
C
OH
H CH3
C6H5CH2
H
O
CH3CH2CH H
C
OCH2C6H5
O
H
N
O
CH3
H
H2N
CH3
DCC
Boc N
H
H2N
A
CH3CH2CH H
+ B
C
Boc N
H
CH3
A
O
CH3CH2CH H
[(CH3)3COCO]2O
C
C
N
H
C
C
OCH2C6H5
O
H CH3
HBr
CH3COOH
CH3
C6H5CH2
H
O
CH3CH2CH H
H2N
C
C
O
H
N
C
C
H CH3
Ile–Ala–Phe
N
H
C
C
O
OH
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
855
Amino Acids and Proteins 29–31
29.66 Make all the Fmoc derivatives as described in Problem 29.24.
C6H5CH2 H
a.
Fmoc N
H
C
C
OH
C6H5CH2 H
[1] base
[2] Cl CH2 POLYMER
C
Fmoc N
H
O
C
O CH2 POLYMER
O
[1]
(Fmoc-Phe)
N
H
H
Fmoc N
[2] DCC
O
C
C
OH
H CH2C6H5
C6H5CH2
H
O
(CH3)2CHCH2 H
Fmoc N
H
C
C
O
H
N
N
H
C
O CH2 POLYMER
N
C
[2] DCC +
H
(CH3)2CHCH2 H
H CH2C6H5 O
C
C6H5CH2
H
O
[1]
C
Fmoc N
H
[1]
N
H
C
C
H
Fmoc N
(Fmoc-Phe)
C
C
O CH2 POLYMER
N
C
H
H CH2C6H5 O
C
OH
O
(Fmoc-Leu)
[2] DCC CH3 H
+
Fmoc N
H
C
C
O
OH
(Fmoc-Ala)
(CH3)2CHCH2
C6H5CH2
O
H
H
O
H
H
Fmoc N
C
C
N
C
C
O CH2 POLYMER
C
N
C
C
N
C
H
H
H CH3
O H
CH2C6H5 O
[1]
N
H
[2] HF
(CH3)2CHCH2
C6H5CH2
H
O
H
O
H
H2N
C
C
N
C
C
OH
C
N
C
C
N
C
H
H
H CH3
O
H CH2C6H5 O
Ala–Leu–Phe–Phe
+
F CH2 POLYMER
856
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–32
CH3
CH3
CH3CH2CH H
b.
Fmoc N
H
C
C
OH
CH3CH2CH H
[1] base
[2] Cl CH2 POLYMER
C
Fmoc N
H
O
C
O
(Fmoc-Ile)
O CH2 POLYMER
[1]
N
H
H
Fmoc N
[2] DCC
CH3CH2CH
H
O
H
C
N
C
C
O CH2 POLYMER
Fmoc N
C
C
N
C
H
H
O
O
H CH3
H H
[1]
N
H
H
Fmoc N
C
N
H
C
C
N
H
C
H CH3
H H
Fmoc N
H
[1]
C
C
OH
(Fmoc-Ala)
CH3CH2CH
H
O
+
[2] DCC
C
H CH3
CH3
CH3
O
C
O CH2 POLYMER
O
OH
O
(Fmoc-Gly)
C6H5CH2 H
[2] DCC
C
OH
+ Fmoc N
C
H
(Fmoc-Phe)
O
CH3
CH3CH2CH
H H
O
H
O
H
H
Fmoc N
C
C
N
C
C
O CH2 POLYMER
C
N
C
C
N
C
H
H
H
O H
O
CH2C6H5
CH3
CH3
[1]
H H
O
N
H
CH3CH2CH
H
O
H
C
N
C
C
OH
N
C
C
N
C
H
H
O
H CH2C6H5 O H CH3
H2N
[2] HF
C
C
Phe–Gly–Ala–Ile
+
F CH2 POLYMER
29.67
a. A p-nitrophenyl ester activates the carboxy group of the first amino acid to amide formation by
converting the OH group into a good leaving group, the p-nitrophenoxide group, which is highly
resonance stabilized. In this case the electron-withdrawing NO2 group further stabilizes the
leaving group.
O
NO2
O
NO2
O
NO2
O
NO2
O
O
N+
O
p-nitrophenoxide
O
O
N+
O
O
O
N
The negative charge is delocalized
on the O atom of the NO2 group.
b. The p-methoxyphenyl ester contains an electron-donating OCH3 group, making CH3OC6H4O– a
poorer leaving group than NO2C6H4O–, so this ester does not activate the amino acid to amide
formation as much.
O
857
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–33
29.68
R H
a. H2N
C
C
= RNH
2
OH
O
O
O
CH2O
C
O
O
O
O
N
CH2O
O
C
O N
CH2O
HNR
O
RNH2
CO3
O
O
H OCO2–
O
O
+
HO N
CH2O
O N
O
H
2–
C
HNR
C
O
CH2O
NHR
C
O N
HNR H
O
N-hydroxysuccinimide
O
Fmoc-protected amino acid
+ CO32–
N
H
H
b.
CH2 O
O
R H
C
C
N
H
C
OH
O
CH2 O
O
R H
C
C
N
H
R H
C
OH
+ CO2 +
DMF
HN
O
C
C
OH
O
+
+
Fmoc-protected amino acid
N
H2
proton
transfer
R H
H2N
C
C
O
OH
+
N
H
29.69 Reaction of the OH groups of the Wang resin with the COOH group of the Fmoc-protected
amino acids would form esters by Fischer esterification. After the peptide has been
synthesized, the esters can be hydrolyzed with aqueous acid or base, but the conditions
cannot be too harsh to break the amide bond or cause epimerization.
858
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–34
H
Fmoc N CHCOOH
H2SO4
R
O
O
O
O
Fischer
esterification
Wang resin
OH
OH
O
new ester
C O
new ester
O
C O
CH R
CHR
NHFmoc
NHFmoc
29.70 Amino acids commonly found in the interior of a globular protein have nonpolar or weakly
polar side chains: isoleucine and phenylalanine. Amino acids commonly found on the
surface have COOH, NH2, and other groups that can hydrogen bond to water: aspartic acid,
lysine, arginine, and glutamic acid.
29.71 The proline residues on collagen are hydroxylated to increase hydrogen bonding interactions.
O
O
[O]
N
N
The new OH group allows more hydrogen bonding interactions
between the chains of the triple helix, thus stabilizing it.
OH
29.72
O
C
(CH3)2CHCH2
O
C
(CH3)2CHCH2
Br
Br2
H
CH3COOH
(CH3)2CHCHCHO
[1] NH4Cl, NaCN
H [2] H3
O +,
Δ
Br
CrO3
H2SO4, H2O
(CH3)2CHCHCOOH
NH2
(CH3)2CHCH2
NH3+
NH3
excess
(CH3)2CHCHCOO–
valine
(Racemic valine and leucine are formed as
products, but the synthesis of the tripeptide is
drawn with one enantiomer only.)
C COOH
H
leucine
O
H CH(CH3)2
H2N
C
C
[(CH3)3COCO]2O
OH
(CH3CH2)3N
O
valine
H CH(CH3)2
C
C
O
Boc N
H
CH(CH3)2
C
A
C6H5CH2OH, H+
H2N
H
OCH2C6H5
C
C
O
H2N
OH
H
C
C
O
OH
C6H5CH2OH, H+
CH2CH(CH3)2
leucine
H2N
C
C
OCH2C6H5
H CH2CH(CH3)2
B
859
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Amino Acids and Proteins 29–35
DCC
A + B
H CH(CH3)2 O
H
C
N
C
C
C
Boc N
OCH2C6H5
H
O
CH
CH(CH
H
2
3)2
H CH(CH3)2 O
H
C
N
C
C
C
Boc N
OH
H
O
H CH2CH(CH3)2
DCC
C
H CH(CH3)2
H
C
N
C
C
Boc–NH
H2
Pd-C
O
O
C
OH
H CH2CH(CH3)2
H CH(CH3)2 O
H CH(CH3)2
H
C
N
C
C
C
C
Boc N
N
COOCH2C6H5
H
H
O
H CH2CH(CH3)2
HBr
CH3COOH
H CH(CH3)2
H
C
N
C
C
H2N
O
O
H CH(CH3)2
C
N
COOH
H
CH
CH(CH
H
2
3)2
C
Val–Leu–Val
29.73 Perhaps using a chiral amine R*NH2 (or related chiral nitrogen-containing compound) to
make a chiral imine, will now favor formation of one of the amino nitriles in the Strecker
synthesis. Hydrolysis of the CN group and removal of R* would then form the amino acid.
O
R
NR*
NH2R*
H
R
H
chiral imine
chiral amine
CN
NR*
NR*
R C H Hydrolyze R C H
CN
nitrile
amino nitrile
Perhaps a large excess of one
stereoisomer will be formed.
COOH
NH2
Remove
R*
R C H
COOH
amino acid
enantiomerically enriched (?)
29.74 This reaction is similar to the reaction of penicillin with the glycopeptide transpeptidase
enzyme discussed in Section 22.14. Serine has a nucleophilic OH, which can open the
strained β-lactone to form a covalently bound, inactive enzyme.
860
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 29–36
HO
Enzyme
NHCHO
from serine residue
O
O
O
O
orlistat
Three operations occur: nucleophilic
addition; loss of the alkoxide leaving group;
proton transfer.
NHCHO
O
O
OH
O
O
Enzyme
29.75
H OH2
S
C6H5
OH
S
R
C6H5
N
H
N
H
H2O
O
N
H O OH
R
N
H
C6H5
N
HO OH
proton
S
R
N
H
S
transfer C6H5
R
N
H
N
H
thiazolinone
H2O
OH2
OH
N
R
N
S
H
O H
C6H5
H2O
N
C6H5
R
N
H
S
O
C6H5
S
N
R
N
H
N-phenylthiohydantoin
H3O
H
proton C6H5 N
transfer
S
OH
OH
R
N
H
HO
OH
S
C6H5
N
H
N
H
R
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
861
Lipids 30–1
Chapter 30 Lipids
Chapter Review
Hydrolyzable lipids
[1] Waxes (30.2)—Esters formed from a long-chain alcohol and a long-chain carboxylic acid
O
R
R, R' = long chains of C's
C
OR'
[2] Triacylglycerols (30.3)—Triesters of glycerol with three fatty acids
O
O
R
O
R, R', R" = alkyl groups with 11–19 C's
O
R'
O
R"
O
[3] Phospholipids (30.4)
[a] Phosphatidylethanolamine [b] Phosphatidylcholine
(cephalin)
(lecithin)
O
O
[c] Sphingomyelin
O
R
O
(CH2)12CH3
R
O
O
NH
O
O
O
HO
O
R'
+
O P O CH2CH2NH3
O–
O
O
R'
+
O P O CH2CH2N(CH3)3
O–
R, R' = long carbon chain
R, R' = long carbon chain
O
R
+
P O CH2CH2NR'3
–
O
R = long carbon chain
R' = H or CH3
Nonhydrolyzable lipids
[1] Fat-soluble vitamins (30.5)—Vitamins A, D, E, and K
[2] Eicosanoids (30.6)—Compounds containing 20 carbons derived from arachidonic acid. There
are four types: prostaglandins, thromboxanes, prostacyclins, and leukotrienes.
[3] Terpenes (30.7)—Lipids composed of repeating five-carbon units called isoprene units
Isoprene unit
C
Types of terpenes
[1] monoterpene
10 C’s
[4] sesterterpene
25 C’s
[2] sesquiterpene
[3] diterpene
15 C’s
20 C’s
[5] triterpene
[6] tetraterpene
30 C’s
40 C’s
C C
C
C
862
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–2
[4] Steroids (30.8)—Tetracyclic lipids composed of three six-membered and one five-membered ring
18
1
2
A
3
CH3
12
19
17
11
CH3
10
D
9
8
B
14
16
15
7
5
4
13
C
6
Practice Test on Chapter Review
1. a. Which of the following compounds contains an ester?
1. a wax
2. a cephalin
3. a sphingomyelin
4. Both [1] and [2] contain esters.
5. Compounds [1], [2], and [3] all contain esters.
b. Which of the following compounds is an eicosanoid?
1. a leukotriene
2. a thromboxane
3. a prostaglandin
4. Both [1] and [2] are eicosanoids.
5. Compounds [1], [2], and [3] are all eicosanoids.
c. Which of the following statements is (are) true about A?
(CH2)12CH3
HO
NHCO(CH2)14CH3
O
O
+
P O CH2CH2N(CH3)3
O–
A
1.
2.
3.
4.
5.
A is a lecithin.
A is a phosphoacylglycerol.
A has two tetrahedral stereogenic centers.
Both [1] and [2] are true.
Statements [1], [2], and [3] are all true.
2. Label each compound as a (1) hydrolyzable or (2) nonhydrolyzable lipid.
a. a triacylglycerol
e. oleic acid
b. vitamin A
f. PGE1
c. cholesterol
g. a wax
d. a lecithin
h. a phosphoacylglycerol
863
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Lipids 30–3
3. For each compound: How many isoprene units does the compound contain? Classify the
compound as a monoterpene, sesquiterpenoid, etc.
OH
O
O
O
O
A
B
C
4. Answer True (T) or False (F) for each statement.
a. Eicosanoids are biosynthesized from dimethylallyl diphosphate and isopentenyl diphosphate.
b. Prostaglandins are biosynthesized from arachidonic acid.
c. Fats have lower melting points than oils, and are generally solids at room temperature.
d. A cephalin is one type of phosphoacylglycerol.
e. Sphingomyelins possess a polar head and two nonpolar tails.
f. A sesterterpene contains six isoprene units.
g. The typical steroid skeleton has four six-membered rings joined with trans ring fusions.
h. Waxes are high molecular weight lipids formed from a fatty acid and a long-chain amine.
5. Which terms describe each compound: (a) hydrolyzable lipid; (b) triacylglycerol;
(c) phosphoacylglycerol; (d) phospholipid; (e) nonhydrolyzable lipid; (f) prostaglandin;
(g) terpenoid? More than one term may apply to a compound.
O
O
O
(CH2)12CH3
O
COOH
O
OH
HO
O
B
OH
(CH2)12CH3
+
O P O CH2CH2NH3
A
O–
C
Answers to Practice Test
1. a. 4
b. 5
c. 3
2. a. 1
b. 2
c. 2
d. 1
e. 2
f. 2
g. 1
h. 1
3. A: 2, monoterpenoid
B: 4, diterpenoid
C: 3, sesquiterpenoid
4. a. F
b. T
c. F
d. T
e. T
f. F
g. F
h. F
5. A: e, f
B: e, g
C: a, c, d
864
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–4
Answers to Problems
Join the carboxylic acid and alcohol together to form an ester.
30.1
O
O
Eicosapentaenoic acid has 20 C’s and 5 C=C’s. Since an increasing number of double bonds
decreases the melting point, eicosapentaenoic acid should have a melting point lower than
arachidonic acid; that is, < –49 °C.
30.2
30.3
O
O
HO
O
a.
O
O
OH
H2O
O
HO
OH
H+
O
HO
OH
O
A
O
b.
O
H2 (excess)
c.
O
Pd-C
H2 (1 equiv)
O
O
Pd-C
O
O
B
two possible products:
O
O
O
O
O
O
O
O
or
O
O
O
O
C
A
2 double bonds
lowest melting point
<
C
1 double bond
intermediate melting point
<
C
B
0 double bonds
highest melting point
30.4
O
CH3(CH2)9CH2
C
OH
lauric acid
Lauric acid is a saturated fatty acid but has only 12 C’s. The carbon
chain is much shorter than palmitic acid (16 C’s) and stearic acid
(18 C’s), making coconut oil a liquid at room temperature.
865
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Lipids 30–5
30.5
O
O
O
O
O
O
O
O
O
O
O
30.6
O
A lecithin is a type of phosphoacylglycerol. Two of the hydroxy groups of glycerol are
esterified with fatty acids. The third OH group is part of a phosphodiester, which is also
bonded to another low molecular weight alcohol.
O
O
O
one possibility:
R
O
O
O
palmitic acid
(CH2)14CH3
O
O
O
R'
O
+
O P O CH2CH2N(CH3)3
(CH2)7CH=CH(CH2)7CH3
oleic acid
+
O P O CH2CH2N(CH3)3
O–
O–
general structure
of a lecithin
30.7
Soaps and phosphoacylglycerols have hydrophilic and hydrophobic components. Both
compounds have an ionic “head” that is attracted to polar solvents like H2O. This head is
small in size compared to the hydrophobic region, which consists of one or two long
hydrocarbon chains. These nonpolar chains consist of only C–C and C–H bonds and exhibit
only van der Waals forces.
O
R
O
O
O
O–
Na+ – O
The R groups are
hydrophobic and
nonpolar.
O
R
nonpolar, hydrophobic region
polar head
R'
O P O CH2CH2N(CH3)3
30.8
soap
phosphoacylglycerol
O
+
polar head
Phospholipids have a polar (ionic) head and two nonpolar tails. These two regions, which
exhibit very different forces of attraction, allow the phospholipids to form a bilayer with a
central hydrophobic region that serves as a barrier to agents crossing a cell membrane, while
still possessing an ionic head to interact with the aqueous environment inside and outside the
cell. Two different regions are needed in the molecule. Triacylglycerols have three polar,
uncharged ester groups, but they are not nearly as polar as phospholipids. They do not have
an ionic head with nonpolar tails and so they do not form bilayers. They are largely nonpolar
C–C and C–H bonds so they are not attracted to an aqueous medium, making them H2O
insoluble.
866
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–6
30.9
Fat-soluble vitamins are hydrophobic and therefore are readily stored in the fatty tissues of
the body. Water-soluble vitamins, on the other hand, are readily excreted in the urine, so
large concentrations cannot build up in the body.
30.10
O
O
COOCH3
OH
HO
COOCH3
HO
+
HO
R
S
misoprostol
diastereomers
Only one tetrahedral stereogenic center is different in these two compounds.
30.11 Isoprene units are shown in bold.
OH
a.
OH
geraniol
OH
b.
grandisol
c.
camphor
d.
vitamin A
O
30.12
HO
manoalide
O
O
HO
O
30.13
four isoprene units = diterpene
biformene
867
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Lipids 30–7
30.14
OPP
OPP
farnesyl diphosphate
isopentenyl diphosphate
resonance-stabilized carbocation
–
OPP
OPP
+
+ H B+
OPP
H
B
30.15
single bond
free rotation
OPP
resonance-stabilized
carbocation
B
H
+
1,2-H shift
+
H
+ –OPP
H
α-terpinene
resonance-stabilized
carbocation
30.16
S
a.
methyl group at C13
O
carbonyl at C17
b.
O
R
H
methyl group at C10
H
O
double bond
between C4 and C5
carbonyl at C3
O
H
868
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–8
30.17
H
H
H
H
H
H
H
H
H
HO
H
HO
cholesterol
enantiomer
equatorial OH
H
H
H
H
different here
H
H
different here
H
H
H
HO
H
HO
diastereomer
diastereomer
30.18
H
H
H
H
H
H
HO
H
H
H
H
HO
A
O
H
O
H
B
All four rings are in the same plane. The bulky CH3 groups (arrows) are located above the
plane. Epoxide A is favored, because it results from epoxidation below the plane, on the
opposite side from the CH3 groups that shield the top of the molecule somewhat to attack by
reagents. In B, the epoxide ring is above the plane on the same side as the CH3 groups.
Formation of B would require epoxidation of the planar C=C from the less accessible, more
sterically hindered side of the double bond. This path is thus disfavored.
30.19
a.
b.
α-pinene
humulene
30.20
CH3
a.
CH3
b.
H
H
trans
H
H
cis
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
869
Lipids 30–9
30.21
O
O
H
H
b.
a.
H
HO
H
H
OH
H
H
H
30.22
H
H
H
O
CH3(CH2)16
C
H
H
O
30.23 Each compound has one tetrahedral stereogenic center (circled), so there are two
stereoisomers (two enantiomers) possible. All C=C’s have the Z configuration.
O
O
O
O
(CH2)14CH3
O
O
O
(CH2)7CH=CH(CH2)7CH3
(CH2)6(CH2CH=CH)2(CH2)4CH3
O
O
(CH2)7CH=CH(CH2)7CH3
O
O
O
O
O
(CH2)14CH3
O
O
(CH2)14CH3
(CH2)6(CH2CH=CH)2(CH2)4CH3
O
(CH2)6(CH2CH=CH)2(CH2)4CH3
(CH2)7CH=CH(CH2)7CH3
O
30.24
O
O
(CH2)7CH=CH(CH2)7CH3
O
O
O
O
(CH2)14CH3
(CH2)7CH=CH(CH2)7CH3
There is no stereogenic center in this triacylglycerol since these R
groups are identical, making this triacylglycerol optically inactive.
870
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–10
30.25
O
O
O
O
O
O
O
O
O
CHO
CH3(CH2)4CHO = O
O
M
O
CH2(CHO)2 = P
N
O
[1] O3
[2] CH3SCH3
H2, Pd-C
O
O
The C=C's are assumed to be Z,
since that is the naturally occurring
configuration.
O
O
O
O
L
30.26 When R" = CH2CH2NH3+, the compound is called a phosphatidylethanolamine or cephalin.
O
a.
O
(CH2)12CH3
(CH2)16CH3
O
b.
HO
O
NH
O
O
(CH2)16CH3
(CH2)14CH3
O
+
O P O CH2CH2NH3
O
O–
+
P
O CH2CH2NH3
O–
cephalin
sphingomyelin
30.27
a. HO
O
b.
O
c.
O
O
COOH
O
HO
OH
PGF2α
C15
S
[1] (R)-CBS reagent O
[2] H2O
O
A
O
X
[1] Zn(BH4)2
[2] H2O
O
O
O
O
O
O
HO H
R
O
OH
Use a chiral reducing agent to
add hydride from one side only
to form a single diastereomer.
B and C
diastereomers
O
O
H OH
S
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
871
Lipids 30–11
30.28
OH
g.
e.
a.
neral
CHO
COOH
patchouli alcohol
O
O
b.
f.
O
dextropimaric acid
h.
O
carvone
periplanone B
HO
β-amyrin
c.
lycopene
d.
β-carotene
30.29 A monoterpene contains 10 carbons and two isoprene units; a sesquiterpene contains 15
carbons and three isoprene units, etc. See Table 30.5.
OH
a.
e.
g.
monoterpenoid CHO
diterpenoid
sesquiterpenoid
O
O
b.
COOH
O
monoterpenoid
f.
O
h.
sesquiterpenoid
HO
triterpenoid
c.
tetraterpene
d.
tetraterpene
872
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–12
30.30
tail
tail head
tail
head tail
lycopene
tail
head
tail
head tail
head
tail head
head
head tail
tail
squalene
tail
head
tail
head tail
tail
30.31
B
+
H
+
=
–OPP
OPP
+
α-pinene
resonance-stabilized
carbocation
30.32 The unusual feature in the cyclization that forms flexibilene is that a 2° carbocation rather
than a 3° carbocation is generated. Cyclization at the other end of the C=C would have given
a 3° carbocation and formed a 14-membered ring. In addition, the 2° carbocation does not
rearrange to form a 3° carbocation.
OPP
OPP
OPP
isopentenyl diphosphate
farnesyl diphosphate
OPP
OPP
H
HB
B
=
2° carbocation
B
H
OPP
OPP
flexibilene
HB
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
873
Lipids 30–13
30.33
a.
OH
+ H2O
+ H2O
OH2
H OH2
H
H2O
X
[1] O3
[2] Zn, H2O
b.
O
O
–OH,
H2O
CHO
O
O
O
16,17-dehydroprogesterone
30.34
H
H
a.
HO
HO
c.
OH
H
OH
H
equatorial OH
H
HO
HO
H
H
H
H
H
HO
H
d.
b.
HO
H
H
H
H
axial OH
H
equatorial OH
axial OH
30.35
H
H
OH
a.
HO
HO
(eq)
H
H
H
OH (ax)
Axial reacts
faster.
H
b.
HO
HO
OH
H
(eq)
H
OH
(ax)
30.36
O
OH
a, b.
H
R
O
H
S
S
O
c.
H
CH3(CH2)5COCl
H
H
methenolone
pyridine
H
O
H
H
Primobolan
Axial reacts
faster.
874
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–14
30.37
CH3 groups make this face more sterically hindered.
a.
H
b.
H
H
O
H
H2, Pd-C
=
H
H
O
H
O
H
H
H H
H
H
H
H2 added from below
The bottom face is more accessible, so the H2 is
added from this face to form an equatorial OH.
30.38
H
H
H
H
HO
H
cholesterol
H
H
H
H
a. CH3COCl
H
d. oleic acid, H+
H
O
O
H
H
H
O
H
b. H2, Pd-C
CH3(CH2)7CH=CH(CH2)7
O
H
H
H
H
H
HO
c.
H
H
H
e. [1] BH3, THF
H
[2] H2O2, –OH
H
PCC
H
HO
H
H
H
OH
H
H
H
O
30.39
H
1,2-shift
OPP
+ –OPP
resonancestabilized
carbocation
1,2-shift
B
=
875
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Lipids 30–15
30.40 Re-draw the starting material in a conformation that suggests the structure of the product.
=
OH
OH
OH2
H OH2
H2O
H2O
H2
O
H2O
O H
H
H2O
OH
H OH2
O H
O H
H
O
H OH2
876
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 30–16
30.41
(any base)
B
OPP
H
=
farnesyl diphosphate
OPP
+ –OPP
resonance-stabilized
carbocation
CH3
CH3
CH3
H A (any acid)
1,2-H shift
H
CH3
CH3
A
+ A
1,2-CH3 shift
H
B
+ HB
CH3
CH3
epi-aristolochene
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
877
Synthetic Polymers 31–1
Chapter 31 Synthetic Polymers
Chapter Review
Chain-growth polymers—Addition polymers
[1] Chain-growth polymers with alkene starting materials (31.2)
•
General reaction:
Z
initiator
Z
•
Z
Z
Z
Mechanism—three possibilities, depending on the identity of Z:
Type
Identity of Z
Initiator
Comments
[1] radical
polymerization
Z stabilizes a radical.
Z = R, Ph, Cl, etc.
A source of
Termination occurs by
radicals (ROOR) radical coupling or
disproportionation. Chain
branching occurs.
[2] cationic
polymerization
Z stabilizes a carbocation.
Z = R, Ph, OR, etc.
H–A or a Lewis
acid (BF3 +
H2O)
[3] anionic
polymerization
Z stabilizes a carbanion.
An
Z = Ph, COOR, COR, CN, organolithium
etc.
reagent (R–Li)
Termination occurs by
loss of a proton.
Termination occurs only
when an acid or other
electrophile is added.
[2] Chain-growth polymers with epoxide starting materials (31.3)
O
–OH
R
R
O
O
R
•
•
The mechanism is SN2.
Ring opening occurs at the less substituted
carbon of the epoxide.
878
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 31–2
Examples of step-growth polymers—Condensation polymers (31.6)
Polyamides
Polyesters
H
O
O
O
N
O
O
nylon 6
polyethylene terephthalate
O
H
O
N
O
N
H
O
Polyurethanes
O
H
N
N
O
copolymer of
glycolic and lactic acids
Kevlar
H
O
O
Polycarbonates
O
O
O
a polyurethane
O
C
O
Lexan
Structure and properties
•
•
•
•
•
•
Polymers prepared from monomers having the general structure CH2=CHZ can be isotactic,
syndiotactic, or atactic, depending on the identity of Z and the method of preparation (31.4).
Ziegler–Natta catalysts form polymers without significant branching. Polymers can be
isotactic, syndiotactic, or atactic, depending on the catalyst. Polymers prepared from 1,3-dienes
have the E or Z configuration, depending on the monomer (31.4, 31.5).
Most polymers contain ordered crystalline regions and less ordered amorphous regions (31.7).
The greater the crystallinity, the harder the polymer.
Elastomers are polymers that stretch and can return to their original shape (31.5).
Thermoplastics are polymers that can be molded, shaped, and cooled such that the new form is
preserved (31.7).
Thermosetting polymers are composed of complex networks of covalent bonds, so they cannot
be melted to form a liquid phase (31.7).
Practice Test on Chapter Review
1. a. Which of the following statements is (are) true about chain-growth polymers?
1. The reaction mechanism involves initiation, propagation, and termination.
2. The reaction may occur with anionic, cationic, or radical intermediates.
3. Epoxides can serve as monomers.
4. Statements [1] and [2] are both true.
5. Statements [1], [2], and [3] are all true.
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
879
Synthetic Polymers 31–3
b. Which of the following alkenes is likely to undergo anionic polymerization?
1. CH2=CHCO2CH3
2. CH2=CHOCH3
3. CH2=CHCH2CO2CH3
4. Both [1] and [2] will react.
5. Compounds [1], [2], and [3] will all react.
c. Which of the following compounds can serve as an initiator in cationic polymerization?
1. butyllithium
2. (CH3)3COOC(CH3)3
3. BF3
4. Both [1] and [2] can serve as initiators.
5. Compounds [1], [2], and [3] can all serve as initiators.
d. Which of the following statements is (are) true about step-growth polymers?
1. A small molecule such as H2O or HCl is extruded during synthesis.
2. Polycarbonates are an example of a step-growth polymer.
3. Step-growth polymers are also called addition polymers.
4. Statements [1] and [2] are both true.
5. Statements [1], [2], and [3] are all true.
2. Label each statement as True (T) or False (F).
a. A polyester is the most easily recycled polymer.
b. Natural rubber is a polymer of repeating isoprene units in which all double bonds have the
E configuration.
c. A syndiotactic polymer has all Z groups bonded to the polymer chain on the same side.
d. A polyether can be formed by anionic polymerization of an epoxide.
e. An epoxy resin is a chain-growth polymer.
f. A branched polymer is more amorphous, giving it a higher Tm.
g. Polystyrene is a thermoplastic that can be melted and molded in shapes that are retained
when the polymer is cooled.
h. A polyurethane is a condensation polymer.
i. Ziegler–Natta catalysts are used to form highly branched chain-growth polymers.
j. Using a feedstock from a renewable source is one method of green polymer synthesis.
3. What monomer(s) are needed to synthesize each polymer?
O
O
a.
O
O
O
O
O
d.
O
O
b.
O
O
O
O
O
c.
CO2CH3
CO2CH3
e.
F Cl F Cl F Cl
O
880
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 31–4
Answers to Practice Test
1. a. 5
2. a. T
b. 1
b. F
c. 3
c. F
d. 4
d. T
e. F
f. F
g. T
h. T
i. F
j. T
3.
O
a.
O
b.
and HO
Cl
OH
Cl
CO2CH3 and CH2 CH2
c.
O
d.
OH
HO
Cl
e.
F
Answers to Problems
Place brackets around the repeating unit that creates the polymer.
31.1
Cl
Cl
Cl
Cl
Cl
poly(vinyl chloride)
O
H
O
N
O
n
H
N
N
H
nylon 6,6
O
H
N
N
H
N
H
O
O
n
Draw each polymer formed by chain-growth polymerization.
31.2
Cl
Cl Cl Cl Cl Cl Cl
OCH3 OCH3 OCH3
c.
a.
Cl
OCH3
CH3 CH3 CH3
b.
O
OCH3
d.
OCH3
OCH3 OCH3
CO2CH3
O
O
O C O C O C
881
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Synthetic Polymers 31–5
31.3
Draw each polymer formed by radical polymerization.
CH3 CH3 CH3
O
a.
31.4
CO2CH3
O
CN
O
b.
O C O C O C
CN
CN
CN
Use Mechanism 31.1 as a model of radical polymerization.
Initiation:
(CH3)3CO
[1]
OC(CH3)3
Propagation:
[3]
O
+
O
O
O
O
(CH3)3CO
[2]
O
2 (CH3)3CO
(CH3)3CO
Repeat Step [3] over and over.
(CH3)3CO
O
O
O
O
O
O
Termination:
O
O
O
O
O
O
O
O
[4]
O
OAc OAc O
O
O
(Ac = CH3CO–)
31.5
Radical polymerization forms a long chain of polystyrene with phenyl groups bonded to every
other carbon. To form branches on this polystyrene chain, a radical on a second polymer chain
abstracts a H atom. Abstraction of Ha forms a resonance-stabilized radical A'. The 2° radical
B' (without added resonance stabilization) is formed by abstraction of Hb. Abstraction of Ha is
favored, therefore, and this radical goes on to form products with 4° C’s (A).
Ph
3° C
Ph
Hb
RO
Ph
Ph
abstraction of Hb
Ha
Ph
B'
Ph
Ph
Ph
Ph
Ph
2° radical
no resonance stabilization
Ph
Ph
abstraction of Ha
A'
Ph
Ph
Ph
resonance-stabilized
benzylic radical
Ph
Ph
B
Ph
Ph
Ph
4° C
Ph
A
Ph
Ph
Ph
favored
882
Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition
Chapter 31–6
31.6
Cationic polymerization proceeds via a carbocation intermediate. Substrates that form more
stable 3° carbocations react more readily in these polymerization reactions than substrates
that form less stable 1° carbocations. CH2=C(CH3)2 will form a more substituted carbocation
than CH2=CH2.
3° carbocation
31.7
Cationic polymerization occurs with alkene monomers having substituents that can stabilize
carbocations, such as alkyl groups and other electron-donor groups. Anionic polymerization
occurs with alkene monomers having substituents that can stabilize a negative charge, such
as COR, COOR, or CN.
a. CH2=C(CH3)COOCH3
b. CH2=CHCH3
electron-withdrawing group
anionic polymerization
31.8
1° carbocation
c. CH2=