Course
Study Guide/Solutions Manual
to accompany:
Organic Chemistry, 3rd Edition
University of Pretoria
Department of Chemistry
http://create.mcgraw-hill.com
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ISBN-10: 1121180612
ISBN-13: 9781121180611
Contents
1. Structure and Bonding 1
2. Acids and Bases 33
3. Introduction to Organic Molecules and Functional Groups 57
4. Alkanes 75
5. Stereochemistry 111
6. Understanding Organic Reactions 139
7. Alkyl Halides and Nucleophilic Substitution 159
8. Alkyl Halides and Elimination Reactions 193
9. Alcohols, Ethers, and Epoxides 223
10. Alkenes 257
11. Alkynes 287
12. Oxidation and Reduction 309
13. Mass Spectrometry and Infrared Spectroscopy 337
14. Nuclear Magnetic Resonance Spectroscopy 351
15. Radical Reactions 373
16. Conjugation, Resonance, and Dienes 397
17. Benzene and Aromatic Compounds 421
18. Electrophilic and Aromatic Substitution 443
19. Carboxylic Acids and the Acidity of the O-H Bond 479
20. Introduction to Carbonyl Chemistry 501
21. Aldehydes and Ketones — Nucleophilic Addition 535
22. Carboxylic Acids and Their Derivatives — Nucleophilic Acyl Substitution 567
23. Substitution Reactions of Carbonyl Compounds at the a Carbon 603
24. Carbonyl Condensation Reactions 631
25. Amines 659
26. Carbon-Carbon Bonding-Forming Reactions in Organic Synthesis 693
27. Carbohydrates 715
28. Amino Acids and Proteins 751
29. Lipids 785
30. Synthetic Polymers 801
iii
Credits
1. Structure and Bonding: Chapter 1 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
by Smith 1
2. Acids and Bases: Chapter 2 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 33
3. Introduction to Organic Molecules and Functional Groups: Chapter 3 from Study Guide/Solutions Manual to
accompany Organic Chemistry, Third Edition by Smith 57
4. Alkanes: Chapter 4 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 75
5. Stereochemistry: Chapter 5 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 111
6. Understanding Organic Reactions: Chapter 6 from Study Guide/Solutions Manual to accompany Organic Chemistry,
Third Edition by Smith 139
7. Alkyl Halides and Nucleophilic Substitution: Chapter 7 from Study Guide/Solutions Manual to accompany Organic
Chemistry, Third Edition by Smith 159
8. Alkyl Halides and Elimination Reactions: Chapter 8 from Study Guide/Solutions Manual to accompany Organic
Chemistry, Third Edition by Smith 193
9. Alcohols, Ethers, and Epoxides: Chapter 9 from Study Guide/Solutions Manual to accompany Organic Chemistry,
Third Edition by Smith 223
10. Alkenes: Chapter 10 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 257
11. Alkynes: Chapter 11 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 287
12. Oxidation and Reduction: Chapter 12 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third
Edition by Smith 309
13. Mass Spectrometry and Infrared Spectroscopy: Chapter 13 from Study Guide/Solutions Manual to accompany
Organic Chemistry, Third Edition by Smith 337
14. Nuclear Magnetic Resonance Spectroscopy: Chapter 14 from Study Guide/Solutions Manual to accompany Organic
Chemistry, Third Edition by Smith 351
15. Radical Reactions: Chapter 15 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 373
16. Conjugation, Resonance, and Dienes: Chapter 16 from Study Guide/Solutions Manual to accompany Organic
Chemistry, Third Edition by Smith 397
17. Benzene and Aromatic Compounds: Chapter 17 from Study Guide/Solutions Manual to accompany Organic
Chemistry, Third Edition by Smith 421
18. Electrophilic and Aromatic Substitution: Chapter 18 from Study Guide/Solutions Manual to accompany Organic
Chemistry, Third Edition by Smith 443
19. Carboxylic Acids and the Acidity of the O-H Bond: Chapter 19 from Study Guide/Solutions Manual to accompany
Organic Chemistry, Third Edition by Smith 479
iv
20. Introduction to Carbonyl Chemistry: Chapter 20 from Study Guide/Solutions Manual to accompany Organic Chemistry,
Third Edition by Smith 501
21. Aldehydes and Ketones — Nucleophilic Addition: Chapter 21 from Study Guide/Solutions Manual to accompany
Organic Chemistry, Third Edition by Smith 535
22. Carboxylic Acids and Their Derivatives — Nucleophilic Acyl Substitution: Chapter 22 from Study
Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 567
23. Substitution Reactions of Carbonyl Compounds at the a Carbon: Chapter 23 from Study Guide/Solutions Manual to
accompany Organic Chemistry, Third Edition by Smith 603
24. Carbonyl Condensation Reactions: Chapter 24 from Study Guide/Solutions Manual to accompany Organic Chemistry,
Third Edition by Smith 631
25. Amines: Chapter 25 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 659
26. Carbon-Carbon Bonding-Forming Reactions in Organic Synthesis: Chapter 26 from Study Guide/Solutions Manual
to accompany Organic Chemistry, Third Edition by Smith 693
27. Carbohydrates: Chapter 27 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 715
28. Amino Acids and Proteins: Chapter 28 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third
Edition by Smith 751
29. Lipids: Chapter 29 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by Smith 785
30. Synthetic Polymers: Chapter 30 from Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition by
Smith 801
v
Smith: Study Guide/
Solutions Manual to
accompany Organic
Chemistry, Third Edition
1. Structure and Bonding
© The McGraw−Hill
Companies, 2011
Text
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
Structure and Bonding 1–1
C
Chhaapptteerr 11:: SSttrruuccttuurree aanndd B
Boonnddiinngg
IIm
mppoorrttaanntt ffaaccttss
• 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 of an atom and the
number of electrons it “owns” (Section 1.3C). See Sample Problem 1.4 for a stepwise example.
Definition:
Examples:
formal charge
number of
valence electrons
C
C
• C shares 8 electrons.
• C "owns" 4 electrons.
• FC = 0
•
=
–
number of electrons
an atom "owns"
C
• Each C shares 6 electrons.
• Each C "owns" 3 electrons.
• FC = +1
• 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 1.5).
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 and
electron deficient regions (Section 1.11).
1
2
Smith: Study Guide/
1. Structure and Bonding
Text
Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition
Solutions Manual to
accompany Organic
Chemistry, Third Edition
© The McGraw−Hill
Companies, 2011
Chapter 1–2
TThhee iim
mppoorrttaannccee ooff LLeew
wiiss ssttrruuccttuurreess ((SSeeccttiioonnss 11..33,, 11..44))
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.
[linear, trigonal planar, or tetrahedral] (Section 1.6)
Geometry
Hybridization
Lewis structure
[sp, sp2, or sp3] (Section 1.8)
Types of bonds
[single, double, or triple] (Sections 1.3, 1.9)
R
Reessoonnaannccee ((SSeeccttiioonn 11..55))
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 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
O
delocalized 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
G
Geeoom
meettrryy aanndd hhyybbrriiddiizzaattiioonn
The number of groups around an atom determines both its geometry (Section 1.6) and hybridization
(Section 1.8).
Number of
groups
2
3
4
Geometry
Bond angle (o)
Hybridization
Examples
linear
trigonal planar
tetrahedral
180
120
109.5
sp
sp2
sp3
BeH2, HCCH
BF3, CH2=CH2
CH4, NH3, H2O
Smith: Study Guide/
Solutions Manual to
accompany Organic
Chemistry, Third Edition
1. Structure and Bonding
© The McGraw−Hill
Companies, 2011
Text
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
Structure and Bonding 1–3
D
Drraaw
wiinngg oorrggaanniicc m
moolleeccuulleess ((SSeeccttiioonn 11..77))
• Shorthand methods are used to abbreviate the structure of organic molecules.
CH3 H
=
CH3 C
C
C
H
H
CH3
skeletal structure
•
CH3
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
C
H
H
H
H
H
H
HH
C
C
C
H
H
H
H
H
HH
Each drawing has two solid lines, one wedge, and one dashed line.
B
Boonndd lleennggtthh
• Bond length decreases across a row and increases down a column of the periodic table
(Section 1.6A).
C H
>
N H
>
O H
H F
H Cl
<
<
H Br
Increasing bond length
Increasing bond length
•
Bond length decreases as the number of electrons between two nuclei increases (Section 1.10A).
CH3 CH3
<
CH2 CH2 < H C C H
Increasing bond length
•
Bond length increases as the percent s-character decreases (Section 1.10B).
Csp H
Csp2 H
Csp3 H
Increasing bond length
•
Bond length and bond strength are inversely related. Shorter bonds are stronger bonds (Section
1.10).
longest C–C bond
weakest bond
C C
C C
Increasing bond strength
C C
shortest C–C bond
strongest bond
3
4
Smith: Study Guide/
1. Structure and Bonding
Text
Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition
Solutions Manual to
accompany Organic
Chemistry, Third Edition
© The McGraw−Hill
Companies, 2011
Chapter 1–4
•
Sigma () bonds are generally stronger than bonds (Section 1.9).
C C
1 strong m bond
C C
1 stronger m bond
1 weaker / bond
C C
1 stronger m bond
2 weaker / bonds
EElleeccttrroonneeggaattiivviittyy aanndd ppoollaarriittyy ((SSeeccttiioonnss 11..1111,, 11..1122))
• Electronegativity increases 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.
D
Drraaw
wiinngg LLeew
wiiss ssttrruuccttuurreess:: A
A sshhoorrttccuutt
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 were no bonds between atoms and every atom
has a filled shell of valence electrons; i.e., hydrogen gets two electrons, and second-row elements
get eight.
[3] Subtract the number obtained in Step [2] from the sum obtained in Step [1]. This difference
tells how many electrons must be shared to give every H two electrons and every second-row
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.
Example: Draw all valid Lewis structures for CH3NCO using the shortcut procedure.
[1] Arrange the atoms.
H
• In this case the arrangement of atoms is implied by the way the structure is
H C N C O
drawn.
H
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Solutions Manual to
accompany Organic
Chemistry, Third Edition
1. Structure and Bonding
© The McGraw−Hill
Companies, 2011
Text
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
Structure and Bonding 1–5
[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; i.e., 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
H C N C O
H C N C O
H
H
H
H
H C N C O
H
H C N C O
H
Bonds to H's added.
H C N C O
H
H
H
H
H C N C O
H C N C O
H
All bonds drawn in.
Three arrangements possible.
•
Calculate the formal charge on each atom.
H
H
H C N C O
H
H C N C O
H
+1
•
H
Electron pairs drawn in.
Every atom has an octet.
–1
H
H C N C O
H
–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.
5
6
Smith: Study Guide/
1. Structure and Bonding
Text
Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition
Solutions Manual to
accompany Organic
Chemistry, Third Edition
© The McGraw−Hill
Companies, 2011
Chapter 1–6
C
Chhaapptteerr 11:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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 The atomic number is the number of protons. The total number of electrons in the neutral atom
is equal to the number of protons. The number of valence electrons is equal to the group number
for second-row elements. The group number is located above each column in the periodic table.
a. atomic number
[1] 31P 15
15
1.3
c. valence e–
5
d. group number
5A
[2] 19
F
9
9
9
7
7A
[3] 12H
1
1
1
1A
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
1.4
b. total number of e–
15
b. Li+ Br
ionic
H H
c. H C C H
H H
All C–H and C–C
bonds are covalent.
d. Na+ N H
H
ionic
Both N–H bonds
are covalent.
a. Ionic bonding is observed in NaF since Na is in group 1A and has only one valence electron,
and F is in group 7A and has seven valence electrons. When F gains one electron from Na,
they form an ionic bond.
b. Covalent bonding is observed in CFCl3 since carbon is a nonmetal in the middle of the
periodic table and does not readily transfer electrons.
1.5 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
Smith: Study Guide/
Solutions Manual to
accompany Organic
Chemistry, Third Edition
1. Structure and Bonding
© The McGraw−Hill
Companies, 2011
Text
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
Structure and Bonding 1–7
1.6 [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]
[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
H H
H C C H
H H
b.
[1]
H
H C N H
H H
c.
[1]
[2] Count valence e.
1C x 4 e =4
3H x 1 e =3
negative charge = 1
total e
=8
H
H C
H
d.
[1]
[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.
[3]
H
H
H C
H C
H
H
[The –1 charge on C is
6 used.
explained in Section 1.3C.]
C needs 2 more
electrons for an octet.
e
H
H
[2] Count valence e.
[3]
= 4
1C x 4 e H C Cl
H C Cl
= 3
3H x 1 e H
H
1Cl x 7 e–
= 7
8 e used.
Complete octet.
= 14
total e
Cl needs 6 more
electrons for an octet.
H
H C Cl
H
1.7 Follow the directions from Answer 1.6.
a. HCN
b. H2CO
H C N
H C O
H
c. HOCH2CO2H
Count valence e.
1C x 4 e = 4
1H x 1 e = 1
1N x 5 e = 5
total e = 10
Count valence e.
1C x 4 e = 4
2H x 1 e = 2
1O x 6 e = 6
= 12
total e
H O
Count valence e.
H O C C O H 2C x 4 e = 8
H
4H x 1 e = 4
3O x 6 e = 18
total e = 30
H C N
4
e
used.
H C N
Complete N and C octets.
H C O
H C O
H
H
6 e used.
Complete O and C octets.
H O
H O C C O H
H
16 e used.
H O
H O C C O H
H
Complete octets.
7
8
Smith: Study Guide/
1. Structure and Bonding
Text
Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition
Solutions Manual to
accompany Organic
Chemistry, Third Edition
© The McGraw−Hill
Companies, 2011
Chapter 1–8
1.8 Formal charge (FC) = number of valence electrons – [number of unshared electrons +
1/2 (number of shared electrons)]
H
a.
6 [2 + 1/2(6)] = +1
5 [0 + 1/2(8)] = +1
+
c.
CH3 N C
b.
H N H
O O O
H
4 [0 + 1/2(8)] = 0
5 [0 + 1/2(8)] = +1
4 [2 + 1/2(6)] = 1
6 [4 + 1/2(4)] = 0
6 [6 + 1/2(2)] = 1
1.9
H
a. CH3O
[1] H C O
H
[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
H
H
[2] Count valence e.
[3] H C O
1C x 4 e = 4
H
3H x 1 e = 3
8 e used.
1O x 6 e = 6
total e
= 13
Add 1 for () charge = 14
H C O
[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
H
[4] H C O
H
H
Assign charge.
[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
used.
10
e
total e = 13
Add 1 for () charge = 14
[4]
H H
Assign charge.
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
C Cl
Cl
H
H C
Cl
H
Complete octet and
assign charge.
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
= 26
total e
H H H
H C C C O H
H H H
H O H
H C C C H
H H H
H H
H
H C C O C H
H H
H
H C N H
1.10
a. C2H4Cl2 (two isomers)
H H
H C N H
H
Smith: Study Guide/
Solutions Manual to
accompany Organic
Chemistry, Third Edition
1. Structure and Bonding
© The McGraw−Hill
Companies, 2011
Text
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
Structure and Bonding 1–9
c. C3H6 (two isomers)
H
Count valence e.
3C x 4 e =12
6H x 1 e = 6
total e = 18
H H
C
C
C H
H
H
H
H
H C C C
H H H
1.11 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
3 lone pairs
O
a.
N C O
and
C N O
b.
O
HO C O
different arrangement of atoms = isomers
and
HO C O
same arrangement of atoms =
resonance structures
1.12 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
O
c.
same arrangement of atoms =
resonance structures
(C2H5O2)–
C2H4O2
O
CH3 C OH
A
C
D
different arrangement of atoms = isomers
O
CH3 C OH
H C CH2OH
A
B
A
b.
CH3 C OH
O
O
O
d. CH3 C OH
H
H C CH2OH
B
D
different arrangement of atoms = isomers
different molecular formulas = neither
1.13 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
The net charge is the same
in both resonance structures.
b.
CH3 C C CH2
CH3 C C CH2
H H
H H
The net charge is the same
in both resonance structures.
9
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1.14 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
C CH3
H
CH2 C
H
C CH3
H
b.
O C O
O C O
O
O
H
One electron pair moves:
one curved arrow.
Two electron pairs move:
two curved arrows.
1.15 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
c.
H C C Cl
H
H C C Cl
H H
H H
1.16 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
CH3 C N CH3
b.
CH2
C
CH2
CH2
H
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:
H H
C
CH2
C
CH2
H
1.17 Draw a second resonance structure for nitrous acid.
+
H O N O
H O N O
major contributor
fewer charges
minor contributor
H O
–
N
O
hybrid
1.18 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
H
H
Cl Cl
C
C
C
Cl
Cl
Cl
Cl
H
H
Cl
Cl
C
H
HH
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Structure and Bonding 1–11
1.19 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
b.
4 groups = tetrahedral
4 groups = tetrahedral
4 groups = tetrahedral
CH3 C N
d.
CH3 O CH3
2 groups = linear
2 groups = linear
4 groups = tetrahedral (or bent molecular shape)
1.20 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°
1.21 To predict the geometry around an atom, use the rules in Answer 1.19.
3 groups
H
H trigonal planar H
C
HO C C
C C
C C C C C C
H
H
H
H
H
4 groups
2 groups
tetrahedral
linear
(or bent molecular shape)
4 groups
tetrahedral (or bent molecular shape)
H
4 groups
H H O H H tetrahedral
C C C C C CH3
H H H H H
3 groups
trigonal planar
enanthotoxin
11
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Chapter 1–12
1.22 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
(CH3)2CHCHO
H C H
H
H
H C H
H H H H H
a. H C C C C C C H
H
H C H
HH C H
H C H
b. H C
H H H H H
H C H
c. H C
C H
H
CH3(CH2)4CH(CH3)2
C
H C H
H H
O H H
C C C H
HH C H
H C H
H
H
O
d. H C C
C H
H
H C H
H H H
H
H
H C H
double bond
needed to
give C an octet
H
1.23 Simplify each condensed structure using parentheses.
CH2CH3
a.
CH3CH2CH2CH2CH2Cl
CH2OH
b. CH3CH2CH2 C CH2CH3
H
CH3(CH2)4Cl
CH3
CH3
c. HOCH2 C CH2CH2CH2 C CH2 C CH3
H
CH3(CH2)2CH(CH2CH3)2
CH3
CH3
(HOCH2)2CH(CH2)3C(CH3)2CH2C(CH3)3
1.24 Draw the Lewis structure of lactic acid.
H
H O
H C C
CH3CH(OH)CO2H
O
C O H
H H
1.25 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)
3 H's
avobenzone
1.26 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
a.
H C C
CH3
CH3
H C C H
H H
b.
H
H
O
H C
C H
H
H C
C
H
H
C
H H
CH3 H
c.
CH3
C
C
H
C
H H
C
Cl
C
C
H H H
H
HH
d.
CH3
N
C
C
N
CH3 H H
CH3
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Structure and Bonding 1–13
1.27 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
b.
a.
positive charge
no lone pairs
no H's needed
O
H
d.
c.
positive charge
no lone pairs
one H needed
negative charge
one lone pair
one H needed
H
negative charge
one lone pair
one H needed
1.28 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
=
d.
HO
O
= HO(CH2)2CH=CHCO2CH(CH3)2
H2N
1.29 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 , and all double bonds contain one and one bond.
Each H uses a
1s orbital.
H H H
H C C C H
All single bonds are
bonds.
Each C–C bond is Csp3–Csp3.
Each C–H bond is Csp3–H1s.
H H H
Total of 10 bonds.
Each C has 4 groups and is
sp3 hybridized.
1.30 [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
Be has 2 bonds.
H C Be
H
H
4 groups
sp3
2 groups
sp
[3] All C–H bonds: Csp3–H1s
C–Be bond: Csp3–Besp
Be–H bond: Besp–H1s
13
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CH3
b. [1] CH3 B CH3
[2] Count groups around each atom:
[3] All C–H bonds: Csp3–H1s
C–B bonds: Csp3–Bsp2
CH3
CH3 B CH3
B forms 3 bonds.
4 groups
sp3
H
3 groups
sp2
H
[2] Count groups around each atom:
c. [1] H C O C H
H
H
H
[3] All C–H bonds: Csp3–H1s
C–O bonds: Csp3–Osp3
H
H C O C H
4 groups
sp3
H
H
4 groups
sp3
1.31 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.
b.
CH3 C CH
3 groups 3 groups
sp2
sp2
2 groups
sp
4 groups
sp3
c. CH2 C CH2
N CH3
3 groups
sp2
2 groups
sp
1.32 All single bonds are . Multiple bonds contain one bond, and all others are bonds.
All CH bonds are bonds.
O
a.
CH3
C
H
bond
one bond,
one bond
bond
b. CH3 C N
one + two bonds
bond bond
c.
H
O
C
bond
one bond, one bond
O CH3
bond
1.33 Bond length and bond strength are inversely related: longer bonds are weaker bonds. Single
bonds are weaker and longer than double bonds, which are weaker and longer than triple bonds.
bond 1:
single bond
a.
bond 3:
double bond
C C
increasing bond strength: 1 < 3 < 2
increasing bond length: 2 < 3 < 1
bond 2:
triple bond
b.
bond 1:
single bond
CH3
H
N
N
bond 2:
double bond
C N
bond 3:
triple bond
increasing bond strength: 1 < 2 < 3
increasing bond length: 3 < 2 < 1
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Structure and Bonding 1–15
1.34 Bond length and bond strength are inversely related: longer bonds are weaker bonds. 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.
CH3
a.
CH3 C C H
or
H
Csp–H1s
CH2 N H
H
CH3 N H
or
Nsp2–H1s
33% s-character
shorter bond
Csp2–H1s
33% s-character
50% s-character
shorter bond
Nsp3–H1s
25% s-character
H
H
b.
c.
C CH2
C O
H C OH
or
H
H
2
Csp –H1s
33% s-character
shorter bond
Csp3–H1s
25% s-character
1.35 Electronegativity increases 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 electropositive
most electronegative
most electronegative
a.
Se < S < O
Na < P < Cl
b.
increasing
electronegativity
most electropositive
most electropositive
most electronegative
most electronegative
c.
increasing
electronegativity
S < Cl < F
increasing
electronegativity
d.
P<N<O
increasing
electronegativity
1.36 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.
+ a. H F
b.
+
B C
c.
+
C Li
d.
+
C Cl
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1.37 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.
Electronegative
atom pulls e density.
c.
Br
+C
a.
H
net dipole
C
F
H
H
+
F
no resulting dipole =
nonpolar molecule
F F
Four polar bonds cancel.
All C–H bonds have no dipole.
one polar bond
net dipole = polar molecule
+
H
b.
Br C Br
H
re-draw
C
Br
resulting dipole =
polar molecule
d.
H H
Br
resulting dipole =
polar molecule
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 Cl
+
C
H
H
Cl
e.
+
C
H
no resulting dipole =
nonpolar molecule
+C C +
H
Cl Two polar bonds are
equal and opposite
and cancel.
1.38
The C–O and O–H bonds are polar.
+
H O
H
C C
+ H
O C
H O
H H
C C C
C C
C
+
H N H+ O H The two circled C's are
hybridized. H
H
H
+
All the C–H bonds are nonpolar.
All H's bonded to O and N bear a partial positive charge (+).
sp3
H
H H
C C
O
H O C
C C C
H N H O H
C C
H
O
C
H
H
one possibility
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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 1.3.
H
a. Na I
+
–
b. Br Cl
c.
H
e. Na+
d. H C N H
H Cl
O C H
H H
covalent
ionic
ionic
all covalent bonds
covalent
H
All other bonds are covalent.
1.41 Formal charge (FC) = number of valence electrons – [number of unshared electrons +
1/2 (number of shared electrons)]. C is in group 4A.
H H
a.
CH2 CH
b.
c.
H C H
d.
H 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
H C C
H H
4 [0 + 1/2(8)] = 0
4 [0 + 1/2(6)] = +1
1.42 Formal charge (FC) = number of valence electrons – [number of unshared electrons +
1/2 (number of shared electrons)]. N is in group 5A and O is in group 6A.
a.
CH3 N CH3
5 [4 + 1/2(4)] = 1
c.
e. CH3 O
CH3 N N
5 [0 + 1/2(8)] = +1
6 [5 + 1/2(2)] = 0
5 [2 + 1/2(6)] = 0
6 [2 + 1/2(6)] = +1
5 [2 + 1/2(6)] = 0
5 [0 + 1/2(8)] = +1
OH
b.
d.
N N N
CH3 C CH3
f.
CH3 N O
6 [4 + 1/2(4)] = 0
5 [4 + 1/2(4)] = 1
5 [4 + 1/2(4)] = 1
17
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Chapter 1–18
1.43 Follow the steps in Answer 1.6 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
H
H C N O
valence e
2C x 4 e =
3H x 1 e =
1N x 5 e =
1O x 6 e =
total e =
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
1H x 1 e
= 1
3O x 6 e
= 18
or H O C O
1 for () charge = 1
total e
= 24
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
1H x 1 e
= 1
2O x 6 e
= 12
1 for () charge = 1
= 18
total e
or
H C N N
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
O
O
e
H
b. CH3NO2
c. CH3CNO
d. HCO2
H C N N
H 2–
or H C C N O
H
H C C N
e
valence
2C x 4 e
= 8
2H x 1 e
= 2
1N x 5 e
= 5
1 for () charge = 1
total e
= 16
or H C C N
H
H
1.44 Follow the steps in Answer 1.6 to draw Lewis structures.
a. N2
[1] N N
b. (CH3OH2)+ [1]
H
H C O H
H H
c. (CH3CH2)
[1]
H
H C C H
H H
d. HNNH
[1] H N N H
[2] Count valence e.
2N x 5 e = 10
total e = 10
[3]
[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
N N
N N
2
e
used.
H
Complete N octets.
[4]
H C O H
H H
12
e
H H
Add charge
and lone pair.
used.
[4]
H
H
H C O H
H
H C C H
H C C H
H H
H H
12
e
used.
6 e used.
Add charge
and lone pair.
H N N H
Complete N octets.
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[1]
e. H6BN
[2] Count valence e.
1B x 3 e = 3
6H x 1 e = 6
1N x 5 e = 5
total e = 14
H H
H B N H
H H
[3]
[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 1.6 to draw Lewis structures.
a. (CH3CH2)2O
b. CH2CHCN
[2] Count valence e.
1O x 6 e = 6
H C C O C C H
10H x 1 e = 10
H H
H H
4C x 4 e = 16
= 32
total e
[1] H H
[2] Count valence e.
1N x 5 e = 5
C C C N
3H x 1 e = 3
H
3C x 4 e = 12
total e = 20
[1]
[3] H H
H H
H H
H H
H H
H H
[4]
H H
C C C N
C C C N
H
H
12 e used.
[3]
[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
[3] H O
Add lone pairs
and bonds.
[4]
H O H
H O C C C O H
H
H O H
H O C C C O H
H
H
22 e used.
[4]
O H
H C C O C C H
H
Cl H H
H Cl H
H C C C H
H C C C H
H O
H H H
H H H
H
24 e used.
O H
O H
H C C H
H C C H
H
H
H
H
O
C C
H H
H
H C C C N H
H C C N C H
H H H H
H H H H
H
H
H
Add lone pairs
and bonds.
c. Four isomers of molecular formula C3H9N
b. Three isomers of molecular formula C2H4O
O H
H C C O C C H
H
H H H
H
Add lone pairs
and bonds.
1.46 Isomers must have a different arrangement of atoms.
a. Two isomers of molecular formula C3H7Cl
H H
Add lone pairs.
[3] H H
O H
H H
H C C O C C H
28 e used.
H O H
[1] H O
[4]
H H
H C C O C C 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 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
19
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Chapter 1–20
1.47
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
H C C C H
HO
H
H C O
H O H
C
H
C C H
H
H
HO 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
1.48 Use the definition of isomers and resonance structures in Answer 1.11.
O
O
O
O
O
CH2
a.
b.
c.
d.
5 membered
ring
C6 H8 O
A
C6H8O
isomers
C6H10O
different molecular formula
neither isomers nor
resonance structures
C6 H8 O
C6H8O
same arrangement
different arrangement
of atoms
of atoms
resonance structures
isomers
1.49 Use the definitions of isomers and resonance structures in Answer 1.11.
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
1.50 Use the definitions of isomers and resonance structures in Answer 1.11.
CH3
a.
CH3 O CH2CH3
and CH3 C OH
c.
and
H
one O–H bond
two C–O bonds
Different arrangement of atoms.
Both have molecular formula C3H8O = isomers
b.
and
CH3 C C CH3
H H
ring
double bond
Different arrangement of atoms.
Both have molecular formula C4H8 = isomers
Same arrangement of atoms.
Both have molecular formula (C4H7).
Different arrangement of electrons = resonance structures
d.
CH3CH2CH3
and
CH3CH2CH2
molecular formula C3H8 molecular formula (C3H7)
different molecular formulas = neither
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Structure and Bonding 1–21
1.51 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
CH2
One electron pair moves = one arrow
O
b.
O
H C NH2
CH2
c.
Four electron pairs move = four arrows
H C NH2
Two electron pairs move = two arrows
1.52 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
CH3 N N
O
b.
c.
CH3
CH3
NH2
O
CH3 C CH CH2
NH2
d.
CH3 C CH CH2
1.53 Use the rules in Answer 1.15.
O
O
O
a. CH3 C O
O
c.
CH3 C O
Two electron pairs move = two arrows
Two electron pairs move = two arrows
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
1.54 To draw the resonance hybrid, use the rules in Answer 1.16.
Charge is on both O's.
Double bond can
be in 2 locations.
O
a.
CH3 C
O Charge is on both atoms.
O
c.
Double bond can be in 2 locations.
+
partial double bond character
b.
+
+
CH2 NH2
Charge is on both atoms.
Charge is on both atoms.
OH
d.
H C H
+
C–O bond has
partial double bond character.
21
22
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1.55 For the compounds where the arrangement of atoms is not given, first draw a Lewis structure.
Then use the rules in Answer 1.15.
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
= 24
total e
Count valence e.
3N x 5 e = 15
() charge = 1
total e
= 16
c. N3
d.
CH2 CH CH
CH CH2
O
O
O N
O
O N
O
O N
O
O
2–
2–
H O H O H
H
O O O
N N N
H C C C C C H
H
O O O
N N N
N N N
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
H
H
H
e.
f.
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 1.16.
H
H
H
H
H
CH2
H
CH2
H
CH2
H
H
H
H
H
H
H
H
H
H
H
H
resonance hybrid
H
H
+
+
CH2
+
+
H
H
H
H
CH2
H
H
H
CH2
H
H
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Structure and Bonding 1–23
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.
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
CH3
CH3 C N NH2
CH3
CH3 C N NH2
3 bonds for this N
2 charges
2
3 bonds for this N
no charges
contributes the most = 3
2 bonds for this N
2 charges
contributes the least = 1
1.58
This C would have 5 bonds.
H H
H
H
invalid
a.
c. CH3CH2 C N
b.
CH3CH2 C N
O
d.
H H
[Note: The pentavalent C's in (a) and (d) bear a (–1) formal charge.]
1.59 Use the rules in Answer 1.20.
3 groups = 120°
H
Cl
C
H
H H
This C would have 5 bonds.
invalid
a. CH3Cl
O
c. 120°
H
H
H
4 groups = 109.5°
C N
H
CH3
4 groups = 109.5°
4 groups = ~109.5°
H
b. H N O H
H
4 groups = ~109.5°
d.
HC C C OH
H
both C's surrounded by
2 groups = 180°
Cl
H
H
e.
3 groups = 120°
4 groups = 109.5°
H
4 groups = ~109.5°
H 120°
All C atoms have
3 groups = 120°.
24
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1.60 To predict the geometry around an atom, use the rules in Answer 1.19.
O
CH3CH2CH2CH3
a.
c.
3 groups
(3 atoms)
trigonal planar
4 groups
(4 atoms)
tetrahedral
b.
d.
(CH3)2N
e.
3 groups
(3 atoms)
trigonal planar
f.
BF4
(CH3)3N
4 groups
(3 atoms, 1 lone pair)
tetrahedral
(trigonal pyramidal molecular shape)
4 groups
(4 atoms)
tetrahedral
4 groups
(2 atoms, 2 lone pairs)
tetrahedral
(bent molecular shape)
CH3 C OH
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
H
C
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.
H
CH3
O
H H
*
*
H
H
H H H H
H
H
H
H
OH
CH3
N *
a.
HO *
O
H H H H
H
H CH3
b.
H
H HH
*
H
N
* H
H
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
*
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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
C
C
C
H H
H
H
C
H
H H O
C H H
H H
H
N
CH3
C
C
C
C
N
C
C
CH3
H H
HH
H H C H
H H OH
c.
H
C
ethambutol
(drug used to treat tuberculosis)
H
H
H
H
d.
H
C
C
O
C
C
H
H H CH3 O H
H
C
C
C C
H
H
C
C
C C
H
C
C
H
H
CH H
C
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
C
f.
N
CH3(CH2)2C(CH3)2CH(CH3)CH(CH3)CH(Br)CH3
H
H
b. CH3CH(Cl)CH(OH)CH3
Cl
H
e.
OH
N
Br
H H
H
H CH3
C
C C
C C
CH2
H
H H
CH3
c. (CH3)3C(CH2)5CH3
C C
limonene
(oil of lemon)
1.65 For Lewis structures, all atoms including H’s and all lone pairs must be drawn in.
a. CH3CH2COOH
H H
O
H C C
H
C O H
H H
H C
H
H C
O
C
H
b. CH3CONHCH3
H
c. CH3COCH2Br
O
H
C N
H
C H
H
e. (CH3)3CCHO
CH3 O
H
C Br
CH3 C
CH3
H
d. (CH3)3COH
CH3
CH3 C O H
CH3
C H
f. CH3COCl
H
H C
H
O
C Cl
g. CH3COCH2CO2H
H O H O
H C C C C O H
H
H
h. HO2CCH(OH)CO2H
O H O
H O C C C O H
O H
25
26
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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
d.
H
H
H
H
H
H
H
CH3
H
H
H
H
1.67
H
CH3
a.
CH
NCH3
b.
CH3 C NH
H
H
c. O
CH3
H H H H
H
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
e. CH2CH3
d.
c.
This C has 5 bonds.
b. (CH3)3CHCH2CH2CH3
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.
This C has 5 bonds.
It should be CH3CH2–.
1.69 To determine the hybridization around the labeled atoms, use the procedure in Answer 1.31.
a. CH3CH2
4 groups
(3 atoms, 1 lone pair)
sp3, tetrahedral
b.
4 groups
(4 atoms)
sp3, tetrahedral
(Each C has 2 H's.)
c. (CH3)3O+
e. CH3 C C H
4 groups
(3 atoms, 1 lone pair)
sp3, tetrahedral
d.
CH2Cl
4 groups
(4 atoms)
sp3, tetrahedral
2 groups
(2 atoms)
sp, linear
f. CH2=NOCH3
3 groups
(2 atoms, 1 lone pair)
sp2, trigonal planar
g. CH3CH=C=CH2
3 groups
(3 atoms)
sp2
trigonal planar
2 groups
(2 atoms)
sp, linear
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Structure and Bonding 1–27
1.70 To determine what orbitals are involved in bonding, use the procedure in Answer 1.29.
O
Csp2–Csp3
H H
: Csp2–Osp2
: Cp–Op
Csp3–H1s
a.
c.
Csp3–Csp3
Csp–Csp2
: Csp2–Csp2
: Cp–Cp
H
Csp2–H1s
b.
H C C C N CH3
d.
: Csp3–Nsp2
H
Csp2–Csp3
Csp–H1s : Csp–Csp
: Cp–Cp
: Cp–Cp
1.71 To determine what orbitals are involved in bonding, use the procedure in Answer 1.29.
Csp3–Osp3
Osp3–H1s
H
a.
: Csp2–Osp2
: Cp–Op
O
Csp3–Osp3
CO2H
O
O
O
b.
HO
Csp2–Csp3
OH
CH3
Csp2–Csp3
O
Csp3–Csp3
HO
Csp3–Csp3
H
Csp2–H1s
zingerone
(responsible for the pungent
taste of ginger)
citric acid
(responsible for the tartness
of citrus fruits)
1.72
bonds
: Csp–Csp2
: Cp–Cp
sp sp2
sp2
bond
bonds
H
CH2 C O
ketene
sp2
C
1s
: Csp–Osp2
: Cp–Op
H
C
sp2
O
H
sp
sp2
bond
H
C
C
sp2
[For clarity, only the large bonding lobes of the hybrid orbitals are drawn.]
1.73
CH2 CH
sp2
sp
: Csp2–Csp
: Cp–Cp
CH2 CH
sp2
sp2
: Csp2–Csp2
: Cp–Cp
C
27
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1.74 To determine relative bond length, use the rules in Answer 1.34.
H
H
a.
C C H
C C H
H H
H C C CH2 C C C H
H H H
H
double bond
middle
triple bond
shortest
b.
C CH3
single bond
longest
Csp3–H1s
Csp–H1s
Csp2–H1s
lowest
highest
middle
% s-character
% s-character
% s-character
longest
shortest
middle
1.75 To determine relative bond length, use the rules in Answer 1.34.
a.
HO
NH2
b.
longer
(N is to the left of O in
the second row.)
Cl
Br
c.
single bond
longer
CH3
N
NH2
CH3
longer
(Br is below Cl
in group 7A.)
1.76
c. shortest, strongest C–C bond
b. and d. bond (1)
longest,
weakest C–C
bond (2)
single 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.
1.77 Remember shorter bonds are stronger bonds. A bond formed from two sp2 hybridized C’s is
stronger than a bond formed from two sp3 hybridized C’s because the sp2 hybridized C orbitals
have a higher percent s-character.
1.78 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
1.79
Cl
33% s-character
higher percent s-character
stronger bond
chloroethane (ethyl chloride)
CH3 CH2 Cl
25% s-character
Csp3
a. No, a compound with only one polar bond must be polar. The single bond dipole is not
cancelled by another bond dipole, so the molecule as a whole remains polar.
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b. Yes, a compound with multiple polar bonds can be nonpolar since the dipoles can cancel each
other out, making a nonpolar molecule.
Cl
C
Cl
Cl no net dipole
Cl
c. No, a compound cannot be polar if it contains only nonpolar bonds. There must be differences
in electronegativity to make a compound polar.
H H
nonpolar
1.80 Dipoles result from unequal sharing of electrons in covalent bonds. More electronegative atoms
“pull” electron density towards them, making a dipole.
a.
+
Br Cl
+
b.
+
NH2 OH
c.
d.
CH3 NH2
Li +
1.81 Use the directions from Answer 1.37.
Br
c. CBr4
a. CHBr3
Br
H
C
Br
Br
C
net dipole
Br
b. CH3CH2OCH2CH3
Br
d.
e.
CH3
net dipole
net dipole
1.82
1 and 2 bonds
polar bond
(essentially) nonpolar
tetrahedral
sp3 hybridized
CH3 C N
linear
sp hybridized
net dipole
Cl
CH3
C O
O
Br
Br no net dipole
Br
linear
sp hybridized
The lone pair is in an
sp hybrid orbital.
All C–H bonds are nonpolar bonds.
All H's use a 1s orbital in bonding.
f.
no net dipole
Cl
29
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1.83
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
[Only the larger bonding lobe
of each orbital is drawn.]
H
H
H
p orbitals on C's overlap
sp2 hybrid orbitals on C
bonds
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.]
1.84
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
CH3
H C O
3 groups
sp2
trigonal planar
4 groups
HO
sp3
tetrahedral
(trigonal pyramidal molecular shape)
H
H
H O H H H
b.
HO
S
C C N
NH2
H
H
N
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 +.
CH3
CH3
H C O
H O
skeletal structure:
NH2 H
NH2 H
N
N
c.
O
HO
S
N
O
d.
HO
O
O
S
N
O
6 bonds
O
OH
OH
e. 33% s-character = sp2 hybridized
H
H
These C–H bonds have 33% s-character.
HO
H
H
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Structure and Bonding 1–31
1.85
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
1.86 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
C C H
H H
N O
shortest longest C–C bond
C–N bond
O
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 (CN); increasing the number of electrons between
atoms decreases bond length.
H
f.
O
H
N CH2CH3
H O
H
C N
H
N CH2CH3
CH2CH3
H O
H
N O
O
O
H
O
H O
H
N CH2CH3
H O
H
C N
O
H O
N O
H
g.
H
H O
C N
H
CH2CH3
O
H O
N CH2CH3
CH2CH3
H O
H
N O
N O
O
O
C N
CH2CH3
31
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Chapter 1–32
1.87
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.
1.88 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.
33% s-character
higher percent s-character
pulls more electron density
more electronegative
+
sp2
Csp3
C
25% s-character
1.89
Isomers of C4H8:
H
H
C
C
H
CH2CH3
CH3
H
C
C
H
CH3
CH3
H
C
H
C
H
H
C
C
CH3
CH3
CH3
These two compounds are different
because of restricted rotation around
the C=C (Section 8.2B).
1.90 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
1.91
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
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Acids and Bases 2–1
C
Chhaapptteerr 22:: A
Acciiddss aanndd B
Baasseess
wiiss aacciiddss aanndd bbaasseess
A
A ccoom
mppaarriissoonn ooff B
Brrøønnsstteedd––LLoow
wrryy aanndd LLeew
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 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 bond
Examples
HCl, H2SO4, H2O,
CH3COOH, TsOH
–
OH, –OCH3, H–, –NH2,
CH2=CH2
BF3, AlCl3, HCl,
CH3COOH, H2O
–
OH, –OCH3, H–, –NH2,
CH2=CH2
A
Acciidd––bbaassee rreeaaccttiioonnss
[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.
IIm
mppoorrttaanntt ffaaccttss
• Definition: pKa = log Ka. The lower the pKa, the stronger the acid (2.3).
NH3
pKa = 38
versus
H 2O
pKa = 15.7
lower pKa = stronger acid
33
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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 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—
These bases
HCCH
25
HCC—
can deprotonate
H–H
35
H
—
CH3COO–H.
higher pKa
than
CH3COO–H
FFaaccttoorrss tthhaatt ddeetteerrm
miinnee aacciiddiittyy ((22..55))
[1] Element effects (2.5A)
The acidity of HA increases both 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
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Acids and Bases 2–3
[2] Inductive effects (2.5B)
The acidity of HA increases with the presence of electronwithdrawing groups in A.
CH3CH2OH
weaker acid
CF3CH2OH
CH3CH2O
No additional electronegative
atoms stabilize the conjugate base.
–
F
stronger acid
F
–
C
+
H
C
O
F– H
CF3 withdraws electron density,
stabilizing the conjugate base.
[3] Resonance effects (2.5C)
The acidity of HA 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 HA 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
35
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Chapter 2–4
C
Chhaapptteerr 22:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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 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— bonds
no lone pairs
or bonds
not a base
O
c.
CH3 C
CH3CH2CH2CH3
CH3CH2OH
OCH3
not a base—no lone pairs
or bonds
acid—contains H atoms
base—lone pairs on O
acid—contains H atoms
base—lone pairs on O's, 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.
Br
HBr
SO42–
HSO4
CH3O
CH3OH
(CH3)2C=OH
2.3 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
2.4 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
+
acid
Cl
H2O
base
CH3
C
acid
H3O
conjugate base conjugate acid
O
b.
+
O
CH2
H
+
OCH3
CH3
base
C
CH2
+
CH3OH
conjugate base conjugate acid
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Acids and Bases 2–5
2.5 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.
O
a.
O
+
Cl3C C
O CH3
O H
H
+
acid
+
H C C
()1 charge on each side
H2
base
+
CH3 NH2
CH3 NH3
H Cl
base
d.
()1 charge on each side
base
H C C H
c.
HO CH3
O
acid
b.
+
Cl3C C
+
net neutral on each side
Cl
acid
CH3CH2 O H
+
CH3CH2 O H +
H OSO3H
base
OSO3H
net neutral on each side
H
acid
2.6 Draw the products in each reaction as in Answer 2.5.
a. CH3OH
HCl
b. (CH3CH2)2O
CH3OH2 + Cl–
HCl
(CH3CH2)2OH + Cl–
c.
(CH3)3N
d.
NH
HCl
(CH3)3NH + Cl–
HCl
–
NH2 + Cl
2.7 The smaller the pKa, the stronger the acid. The larger Ka, the stronger the acid.
OH
a.
CH3CH2CH3
pKa = 50
or
CH3CH2OH
CH3
or
b.
pKa = 16
Ka = 10–10
smaller pKa
stronger acid
Ka = 10–41
larger Ka
stronger acid
2.8 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
Ka = 10–21
Ka = 5.2 x 10–5
pKa = 21
pKa = 4.3
b. pKa = 7
pKa = 11
pKa = 3.2
Ka = 10–7 Ka = 10–11 Ka = 6.3 x 10–4
37
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Chapter 2–6
2.9
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.
Increasing acidity
Increasing acidity
b.
H2O, NH3, CH4
pKa = 15.7 38 50
conjugate bases:
OH, NH
2,
CH
HCCH, CH2=CH2, CH4
pKa =
CH
3
(CH3)3CCO2H
pivalic acid
pKa = 5.0
c. weaker acid = stronger conjugate base
a. smaller pKa = larger Ka
b. smaller pKa = stronger acid
d. stronger acid = equilibrium farther to the right
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.
For NH3, pKa is 38.
estimated pKa = 38
2.12
2,
Use the definitions in Answer 2.9 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 farther to
the right.
HCO2H
formic acid
pKa = 3.8
2.11
50
Increasing basicity
Increasing basicity
2.10
44
CCH, CH=CH
conjugate bases:
3
25
c. BrCH2COO H
O H
For CH3CH2OH,
pKa is 16.
estimated pKa = 16
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
b.
CH4
+
acid
pKa = 50
weaker acid
c.
CH3COOH
acid
pKa = 4.8
CH2 CH
conjugate base
OH
CH3
base
conjugate base
+
+
H
base
CH3CH2O
base
+
H2
conjugate acid
pKa = 35
H2O
conjugate acid
pKa = 15.7
CH3COO
conjugate base
+
Equilibrium favors
the starting materials.
Equilibrium favors
the starting materials.
CH3CH2OH
conjugate acid
pKa = 16
weaker acid
Equilibrium favors
the products.
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Acids and Bases 2–7
d.
Cl
+
base
2.13
CH3CH2OH
acid
pKa = 16
weaker acid
Acid
HCl
HCCH
H2
Acid
H2
NH3
CH2=CH2
CH4
pKa
–7
25
35
Conjugate base
Cl
not strong enough
HCC
strong enough
H
pKa
35
38
44
50
Conjugate base
H
NH
2
CH2=CH
CH3
An acid can be deprotonated by the conjugate base of any acid with a higher pKa.
CH3CN
pKa = 25
Any base having a conjugate
acid with a pKa higher than
25 can deprotonate this acid.
Base
NaH
Na2CO3
NaOH
NaNH2
NaHCO3
Conjugate acid
H2
HCO3–
H2O
NH3
H2CO3
pKa
35
10.2
15.7
38
6.4
Only NaH and NaNH2
are strong enough to
deprotonate acetonitrile.
The acidity of H–Z increases across a row and down a column of the periodic table.
b. HBr, HCl
a. NH3, H2O
O is farther to the right in
the periodic table.
stronger acid
2.16
conjugate base
Equilibrium favors
the starting materials.
An acid can be deprotonated by the conjugate base of any acid with a higher pKa.
HCCH
pKa = 25
All of these acids have a higher pKa
than HCCH, and a conjugate
base that can deprotonate HCCH.
2.15
CH3CH2O
conjugate acid
pKa = 7
CH3COOH
pKa = 4.8
Any base having a conjugate
acid with a pKa higher than
4.8 can deprotonate this acid.
2.14
+
HCl
Br is farther down
the periodic table.
stronger acid
c. H2S, HBr
Br is farther across and down
the periodic table.
stronger acid
Compare the most acidic protons in each compound to determine the stronger acid.
a.
CH3CH2CH2NH2
or
(CH3)3N
b. CH3CH2OCH3
N–H bond
N is farther to the right in
the periodic table.
stronger acid
C–H bond
C–H bond
or CH3CH2CH2OH
O–H bond
O is farther to the right in
the periodic table.
stronger acid
39
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Chapter 2–8
2.17
Look at the element bonded to the acidic H and decide its acidity based on the periodic trends.
Farther right and down the periodic table is more acidic.
most acidic
a. CH3CH2CH2CH2OH
most acidic
b. HOCH2CH2CH2NH2
Molecule contains C–H
and O–H bonds.
O is farther right; therefore,
O–H hydrogen is the most acidic.
2.18
most acidic
Molecule contains C–H,
N–H, and O–H bonds.
O is farthest right; therefore,
O–H hydrogen is the 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.
The acidity of HA increases across the periodic table. Pseudoephedrine contains C–H, N–H, and
O–H bonds. The O–H bond is most acidic.
H O
H
N
pseudoephedrine
2.19
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
2.20
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 -hydroxy acid
The extra OH group contains an electronegative
O, which stabilizes the conjugate base.
stronger acid
2.21
CH3COOH
CH3CO2H
acetic acid
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
CH3
acetone
pKa = 19.2
O
O
C
C
2 resonance structures
more stable conjugate base
CH3
CH2
CH3
CH2
Acetone is more acidic.
One resonance structure places the (–) charge on the
more electronegative O atom. This is especially good.
base
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Acids and Bases 2–9
base
CH3CH2CH3
CH3CH2CH2
propane
pKa = 50
2.22
only one Lewis structure
less stable conjugate base
(Any C–H bond in the starting
material can be removed.)
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 C C N
H
H
base
The negative charge is stabilized by
delocalization on the C and N atoms.
H C C N
H C C N
H
acetonitrile
(one Lewis structure)
2.23
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
b.
CH3CH2CH2CH2–H
sp hybridized C
50% s-character
base
more acidic
sp3 hybridized C
25% s-character
base
H
or
sp2 hybridized C
33% s-character
more acidic
H
sp3 hybridized C
25% s-character
base
CH3CH2 C C
or
H
CH3CH2CH2CH2
or
stronger
conjugate base
2.24
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.
41
42
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Chapter 2–10
2.25
Look at the element bonded to the acidic H and decide its acidity based on the periodic trends.
Farther right and down the periodic table is more acidic.
most acidic
O
most acidic
OH
b.
a.
O
THC
tetrahydrocannabinol
The molecule contains C–H
and O–H bonds.
O is farther right; therefore,
O–H hydrogen is the most acidic.
2.26
N
H
most acidic
The molecule contains C–H,
N–H, and O–H bonds.
O is farthest right; therefore,
O–H hydrogen is the most
acidic.
The molecule contains C–H
and O–H bonds.
O is farther right; therefore,
O–H hydrogen is the most acidic.
(CH3)2CHO Na+
d. (CH3)2CHOH + H–OCOCH3
base
acid
+
conjugate base
(CH3)2CHOH2
H2
conjugate acid
HSO4–
+
conjugate acid
conjugate base
(CH3)2CHO Li+
c. (CH3)2CHOH + Li+ –N[CH(CH3)2]2
acid
base
+
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
NH2
absorption here
in the neutral form
Lewis bases are electron pair donors: they contain a lone pair or a bond.
a. NH3
yes - has
lone pair
2.29
propranolol
OH
Draw the products of proton transfer from the acid to the base.
b. (CH3)2CHOH + H–OSO3H
base
acid
2.28
O
ketoprofen
a. (CH3)2CHOH + Na+ H
acid
base
2.27
c.
COOH
b. CH3CH2CH3
no - no lone pair
or / bond
c. H
d. H C C H
yes - has
lone pair
yes - has
2 / bonds
Lewis acids are electron pair acceptors. Most Lewis acids contain a proton or an unfilled
valence shell of electrons.
a. BBr3
yes
unfilled valence shell
on B
b. CH3CH2OH
yes
contains a proton
c. (CH3)3C+
d. Br
yes
no
unfilled valence shell
no proton
on C
no unfilled valence shell
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Acids and Bases 2–11
2.30
Label the Lewis acid and Lewis base and then draw the curved arrows.
new bond
F
a.
BF3
+ CH3 O CH3
F
Lewis base
Lewis acid
unfilled valence shell lone pairs
on B
on O
2.31
CH3
b.
F B O
CH3
(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
2.32
+
BBr3
Cl Al Cl
CH3CH2 O CH2CH3
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
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
Cl
Cl Al Cl
c. CH3CH2 N CH2CH3
+
AlCl3
CH2CH3
Lewis base
nucleophile
lone pair
on N
CH3CH2 N CH2CH3
CH2CH3
Lewis acid
electrophile
unfilled valence shell
on Al
CH3
44
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Chapter 2–12
2.33
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
H2 O
+
H
+
H2O
Lewis acid
Lewis base
contains a bond contains a proton
2.34
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
2.35
e. CH3OCH3
H2CO3
f. CH3COO
CH3 O CH3
H
CH3COOH
To draw the conjugate base of a Brønsted–Lowry acid, remove a proton from the acid.
H
a. HCN
CO32
H
c. (CH3)2NH2
H
d. HC CH
CN
H
b. HCO3
2.36
NH3
H
(CH3)2NH
HC C
H
e. CH3CH2COOH
H
f. CH3SO3H
CH3CH2COO
CH3SO3
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.
2.37
N CH3
+
+ H
N
H OSO3H
+
CH3
HSO4
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
K+
+ H2O
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Acids and Bases 2–13
O
O
+ K+ –OH
b.
+ H2O
O K+
O H
d. CH3
2.38
+ K+ –OH
C C H
c.
+ K+ –OH
O H
+
C C K+
H2O
O K+
CH3
+
H2O
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
acid
CH3CH2 C
+
O
base
HO CH3
conjugate acid
conjugate base
c.
CH3CH2 C C H
+
acid
d.
(CH3CH2)3N
+
base
e.
CH3CH2 O H
CH3CH2 C C
+
conjugate base
conjugate acid
H Cl
(CH3CH2)3NH
acid
conjugate acid
+
base
f.
H
base
H Br
acid
+
CH3C C
H OH
base
+
H2
Cl
conjugate base
+
CH3CH2 O H
Br
H
conjugate acid
CH3C CH
+
conjugate base
HO
conjugate acid conjugate base
acid
2.39 Label the acid and base in the starting materials and then draw the products of proton transfer
from acid to base.
O
C
a.
O
O
H
C
O
+
acid
NH3
+
base
O
b.
F3C C
acid
c.
NH4
O
+
O H
CH3 C C H
acid
Na+ HCO3
base
+
Na+ NH2
base
+
F3C C
O
H2CO3
Na
CH3 C C
Na
+ NH3
45
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Chapter 2–14
d.
K+
+
CH3CH2 O H
OH
CH3CH2 O
base
acid
O
e.
CH3 N H
+ CH3
H
S O H
CH3
f.
CH3
acid
CH3
CH3
C
CH3
+
HSO4
acid
Label the acid and base in the starting materials and then draw the products of proton transfer
from acid to base.
a.
+
CH2CH(CH3)NH2
H Cl
base
b.
CH2CH(CH3)NH3
Cl
CH2CH(CH3)NH
H2
acid
CH2CH(CH3)N H
+
Na+ H
H
acid
2.41
Na
base
H CH
3
COO– Na+
COO–H
+
CH3O
acid
Na+
OH
base
H CH2CH2NH2CH3
CF3 + H–Cl
O
b.
+ H2O
CH3O
H CH2CH2NHCH3
O
CF3
+ Cl
acid
base
2.42
+
Draw the products of proton transfer from acid to base.
H CH
3
a.
SO3
O H
H OSO3H
base
2.40
CH3
CH3
O
C
+
CH3 N H
O
base
H2O
+
K
Draw the products of proton transfer from acid to base.
CF3
CF3
CH2CH(CH3)NHCH2CH3
+ H OCOCH3
base
CH2C(CH3)2NH2
base
CH2CH(CH3)NHCH2CH3
H
acid
+ H OCOCH3
acid
CH2C(CH3)2NH3 +
OCOCH3
+
OCOCH3
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Acids and Bases 2–15
2.43
To convert pKa to Ka, take the antilog of (–) the pKa.
a. H2S
pKa = 7.0
Ka = 10–7
2.44
b. ClCH2COOH
pKa = 2.8
Ka = 1.6 x 10–3
To convert from Ka to pKa, take (–) the log of the Ka; pKa = –log Ka.
+
CH2NH3
a.
+
NH3
b.
Ka = 4.7 x 10–10
pKa = 9.3
2.45
c. HCN
pKa = 9.1
Ka = 7.9 x 10–10
Ka = 2.3 x 10–5
pKa = 4.6
Acid
CH3CH2OH
pKa
Conjugate base
CH3CH2O
HCCH
16
25
H2
NH3
CH2=CH2
35
38
44
HCC
H
NH
2
CH2=CH
CH4
50
CH3
b. NH3
Acid
pKa = 38
Any base with a conjugate CH2=CH2
CH4
acid having a pKa higher than
38 can deprotonate it.
Strong
enough to
deprotonate
H2O.
pKa Conjugate base
CH2=CH Strong enough to
44
50
CH3
deprotonate NH3.
c. CH4
pKa = 50
There is no base with a
conjugate acid having a pKa
higher than 50 in the
table.
An acid can be deprotonated by the conjugate base of any acid with a higher pKa.
CH3CH2CH2CCH
pKa = 25
Any base having a conjugate
acid with a pKa higher than
25 can deprotonate this acid.
2.47
Ka = 5.9 x 10–1
pKa = 0.23
An acid can be deprotonated by the conjugate base of any acid with a higher pKa.
a. H2O
pKa = 15.7
Any base with a conjugate
acid having a pKa higher than
15.7 can deprotonate it.
2.46
c. CF3COOH
–
Base
H2O
NaOH
NaNH2
NH3
NaH
CH3Li
Conjugate acid
H3O+
H2O
NH3
NH4+
H2
CH4
pKa
–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.
a. HCOOH
pKa = 3.8
stronger acid
deprotonated
b. H2S
pKa = 7.0
stronger acid
deprotonated
c.
CH3
pKa = 41
weaker acid
d. CH3NH2
pKa = 40
weaker acid
These acids are too weak to be
deprotonated by OH.
47
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Chapter 2–16
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
pKa = ~5
c.
(CH3)3COH +
HOSO3H
+
(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
starting material favored
pKa = 50
CH3NH2 + HOSO3H
pKa = –9
2.49
products favored
Li+ + CH3CH3
pKa = 25
f.
HSO4
starting material favored
O Na+
O H
d.
+
HCl
pKa = –7
+
CH3NH3 + HSO4
products favored
products favored
pKa = 10.7
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
b. Acidity increases down a column:
HF < HCl < HBr
c. increasing acidity: OH < H2O < H3O +
d. increasing acidity: NH3 < H2O < H2S
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
f. increasing acidity: H2O < H2S < HCl
Compare HCl and SH bonds first:
acidity increases across a row.
H–Cl is more acidic.
Compare OH and SH bonds:
acidity increases down a column.
SH is more acidic.
g. CH3CH2CH3, ClCH2CH2OH, CH3CH2OH
O–H bond and
O–H bond
electron-withdrawing Cl
strongest acid
increasing acidity: CH3CH2CH3 < CH3CH2OH < ClCH2CH2OH
only C–H bonds
weakest acid
h. HC CCH2CH3
CH3CH2CH2CH3
CH3C CCH3
H H
sp C–H
strongest acid
sp3 C–H
all
weakest acid
sp2 C–H
increasing acidity: CH3CH2CH2CH3 < CH3CH=CHCH3 < HC CCH2CH3
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Acids and Bases 2–17
2.50
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.
CH CH2 <
CH2CH3 <
Draw the conjugate acid.
Increasing acidity of conjugate acids:
CH4 < H2O < HBr
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 scharacter of the carbon atoms in each molecule. Nearby C’s with a higher percent s-character can
help to stabilize the conjugate base.
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
CH3CH2CH2
CH2 C
CH2
one Lewis structure
weakest acid
CH3
O
CH2 C
CH2
two resonance structures
negative charge delocalized
on two carbons
CH3
C
O
CH2
CH3
C
CH2
two resonance structures
negative charge delocalized
on one O and one C
strongest acid
49
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Chapter 2–18
2.54
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
NH2
conjugate acid
2.55
OH
most basic site
A
NH2
O
conjugate base
most acidic proton
Remove the most acidic proton to form the conjugate base. Protonate the most basic electron
pair to form the conjugate acid.
only O–H bond
most acidic proton
N
most basic electron pair
O
OCH3
Increasing basicity:
COOH
O
O
N
ibuprofen
O
cocaine
conjugate base:
conjugate acid:
H N
O
OCH3
COO
O
O
2.56
A lower pKa means a stronger acid. The pKa is low for the C–H bond in CH3NO2 due to
resonance stabilization of the conjugate base.
H
H
O
B
H C N
H
O
H C N
O
H
H
O
H C N
O
O
O
The negative charge is delocalized on the
electronegative O atom. This stabilizes
the conjugate base.
2.57
Compare the isomers.
dimethyl ether
CH3
O
CH3
All H's are on C.
O
H C N
CH3CH2OH
ethanol
One O–H bond
O–H bonds are more acidic
than C–H bonds.
more acidic
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Acids and Bases 2–19
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
H
B
N
H
O
O
H
O
O
OH
O
H
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.
N
OH
B
less acidic proton
no resonance stabilization
formed from the weaker acid
Draw the conjugate base to determine the most acidic hydrogen.
O
O
O
O
O
H H
O
H
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
N
OH
N
N
O
H
H
H
N
O
N
OH
N
2.59
O
H
more acidic proton
O
H
O
N
OH
H
N
H
resonance-stabilized conjugate base
Look at the element bonded to the acidic H and decide its acidity based on the periodic trends.
Farther right across a row and down a column of the periodic table is more acidic.
F
a.
OH
b.
CO2H
most acidic
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.
c.
O
O
O
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.
52
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Chapter 2–20
2.61
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
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.
OH
H OH
c
a
b
O
conjugate base
by loss of (c):
O
O
O
H OH
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
no resonance stabilization, but
negative charge on O, an electronegative atom
OH
O
conjugate base
by loss of (a):
2.62
OH
OH
OH
OH
This conjugate base has two
resonance structures, but one
places a negative charge on C.
Lewis bases are electron pair donors: they contain a lone pair or a bond. Brønsted–Lowry
bases are proton acceptors: to accept a proton they need a lone pair or a bond. This means
Lewis bases are also Brønsted–Lowry bases.
lone pairs on O
both
O
a.
b.
2.63
O
H
C
CH3 Cl
lone pairs on Cl
both
d.
c.
H
neither = no lone pairs
or bond
bonds
both
A 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.
a. H3O+
both contains a H
b.
Cl3C+
Lewis acid unfilled valence
shell on C
c. BCl3
d. BF4
Lewis acid unfilled valence
shell on B
neither no H or unfilled
valence shell
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Acids and Bases 2–21
2.64
Label the Lewis acid and Lewis base and then draw the products.
Cl
a.
+ BCl3
Cl
Lewis base
O
c.
Cl B Cl
CH3
Cl
Lewis acid
new bond
b.
CH3
CH3
C
C
CH3
CH3
H OSO3H
+
CH3
Cl
OH
Lewis base
C C H
CH3
+
HSO4
new bond
BF3
CH3CH2OH + BF3
nucleophile
electrophile
OH2
CH3CH2 O H
+ H2O
d.
electrophile
b.
CH3
C O
CH3
c.
CH3 S CH3
Br
electrophile
+
e. Br Br
CH3
+ BF3
nucleophile
nucleophile
AlCl3
CH3SCH3 + AlCl3
nucleophile
nucleophile
FeBr3
Br Br Fe Br
Br
electrophile
C O BF3
CH3
electrophile
Draw the product of each reaction.
CH2CH3
CH2CH3
+ H2O
a. CH3CH2 C
CH2CH3
CH2CH3
b. CH3CH2 C
CH3CH2 C
+ NH3
CH2CH3
CH2CH3
+ (CH3)2NH
CH3CH2 CH3
e. CH3CH2 C
OH
CH2CH3
CH2CH3
CH2CH3
CH3CH2 C NH3
CH2CH3
CH3CH2 CH3
+ CH3OH
CH2CH3
c. CH3CH2 C
CH2CH3
d. CH3CH2 C
CH3CH2 C OH2
CH2CH3
CH3CH2 C
CH2CH3
CH3CH2 C
O CH3
CH2CH3
2.67
nucleophile
OH
H Br
proton transfer
OH2
Br
+ Br
electrophile
NHCH3
CH2CH3
CH3CH2 CH3
+ (CH3)2O
CH2CH3
a.
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.
2.66
CH3 C Cl
Lewis acid
Lewis base
2.65
O
+ OH
CH3
CH3
Lewis acid
C
+ H2 O
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Chapter 2–22
Br
b.
proton transfer
H Br
+ Br
electrophile
nucleophile
H
H
Br
+ Br
c.
H
Br
+ Br
proton transfer
+ HBr
H
H
nucleophile electrophile
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.
2.68
a.
CH2
+
C(CH3)2
b. OH + (CH3)3C+
H2O + CH2=C(CH3)2
(CH3)3COH
H
OH
Answer each question about esmolol.
2.69
most acidic
second most acidic
Esmolol contains C–H, N–H, and O–H bonds.
H
Since acidity increases across a row of the
N
periodic table, the OH bond is most acidic,
followed by the NH bond.
OH
a.
O
CH3O
esmolol
O
O H
b.
O
Na+ O
H
N
O
Na+ H
CH3O
+ H2
CH3O
O
O
O H
c.
O
OH H H
H
N
O
H Cl
CH3O
O
OH
d, e, f.
O
*
CH3O
*
O
*
*
*
H
N
*
N
+ Cl–
CH3O
O
*
H
N
All sp2 C are indicated with an arrow.
The N is the only trigonal pyramidal atom.
The + C's are indicated with a (*).
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Acids and Bases 2–23
2.70
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
CH3
O
CH3
C
NH2
C
H
O
NH2
CH3
C
H
resonance stabilization of the + charge
O is more readily protonated
because the product is
resonance stabilized.
NH2
O
protonate N
acetamide
CH3
C
no other resonance structure
NH3
2.71
O
O
HO
O
OH
pKa = 2.86
HO
O
O
+
O
O
O
O
O
O
O
This group destabilizes the
second negative charge.
O
HO
O
+
+
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.
2.72
O
pKa = 5.70
O
The COOH group of glycine gives up a proton to the basic NH2 group to form the zwitterion.
O
a.
acts as a base
NH2CH2 C
proton transfer
acts as an acid
OH
glycine
b.
O
O
+ Cl NH3CH2 C
O
OH
most basic site
O
c.
H
NH2
CH2 C
O
most acidic site
Na+
OH
O
zwitterion form
H Cl
NH3CH2 C
O
NH3CH2 C
O
NH2CH2 C
O
+ Na+ + H2O
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Chapter 2–24
2.73
Use curved arrows to show how the reaction occurs.
O
[1]
O
H
O
O
H
H
OH
O
O
[2]
+
H O H
OH
H
Protonate the negative charge on this carbon to form the product.
2.74
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
OH
OH
O
OH
O
HO
O
HO
O
O
OH
The most delocalized anion
with 3 resonance structures.
Removal of either
of these H's does not
give a resonancestabilized anion.
HO
OH
O
HO
O
OH
loss of H+ HO
OH
O
HO
O
O
HO
O
HO
only 2 resonance structures
This proton is less acidic since its conjugate base is less
resonance stabilized.
O
O
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57
Introduction to Organic Molecules and Functional Groups 3–1
C
Moolleeccuulleess aanndd FFuunnccttiioonnaall G
Grroouuppss
Chhaapptteerr 33:: IInnttrroodduuccttiioonn ttoo O
Orrggaanniicc M
Increasing strength
TTyyppeess ooff iinntteerrm
moolleeccuullaarr ffoorrcceess ((33..33))
Type of force
van der Waals
(VDW)
Cause
Due to the interaction of temporary dipoles
• Larger surface area, stronger forces
• Larger, more polarizable atoms, stronger
forces
Examples
All organic compounds
dipole–dipole
Due to the interaction of permanent dipoles
(DD)
hydrogen bonding Due to the electrostatic interaction of a H atom
(HB or H-bonding) in an O–H, N–H, or H–F bond with another N,
O, or F atom.
H2O
ion–ion
NaCl, LiF
Due to the interaction of two ions
(CH3)2C=O, H2O
PPhhyyssiiccaall pprrooppeerrttiieess
Property
Boiling
point
(3.4A)
•
Observation
For compounds of comparable molecular weight, the stronger the forces the higher
the bp.
CH3CH2CH2CH2CH3
CH3CH2CH2CHO
VDW, DD
MW = 72
bp = 76 oC
VDW
MW = 72
bp = 36 oC
CH3CH2CH2CH2OH
VDW, DD, HB
MW = 74
bp = 118 oC
Increasing strength of intermolecular forces
Increasing boiling point
3.1 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.
CH3I
CH3F
bp = 78
oC
bp = 42 oC
Increasing polarizability
Increasing boiling point
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Chapter 3–2
Melting
point
(3.4B)
•
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
oC
(CH3)4C
mp = 17 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
R
Reeaaccttiivviittyy ((33..88))
• Nucleophiles react with electrophiles.
• Electronegative heteroatoms create electrophilic carbon atoms, which tend to react with
nucleophiles.
• Lone pairs and bonds are nucleophilic sites that tend to react with electrophiles.
CH3CH2 Cl
+
electrophilic site
OH
+
CH3
CH3 O CH3
CH3 N CH3
basic and nucleophilic site
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Molecules and Functional
Groups
Introduction to Organic Molecules and Functional Groups 3–3
Chapter 3: A
Annssw
weerrss ttoo PPrroobblleem
mss
3.1
CH3CH2 OH
CH3CH2 O H
H OSO3H
Na+ H
+ HSO4
CH3CH2 OH2
H2SO4
CH3CH3
CH3CH2 O Na+ + H2
no reaction
NaH
CH3CH3
no reaction
3.2 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
shikimic acid
alcohols
(hydroxy groups)
ester
NH2
oseltamivir
amide
amine
3.3 One possible structure for each functional group:
O
a. aldehyde = R
b. ketone =
R
C
H
CH3CH2CH2
C
c. carboxylic acid =
H
O
O
O
C
C
C
CH3
R
d. ester =
CH2CH3
R
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
C
CH3
ketone
ketone
CH3
alkene
O
O
O
CH3CH2
O
C
H
C
C
H H
C
H
alkene
H
R
C
CH3CH2CH2
OH
O
O
R
CH3CH2
C
O
CH3
C
OH
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Chapter 3–4
3.5 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.
only nonpolar C–C
and C–H bonds
VDW only
e. CH3CH2CH2COOH
(CH3CH2)3N
c.
• 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.
f.
CH2 CHCl
• VDW forces
• polar C–Cl bond - DD
• VDW forces
• 2 polar C–O bonds
and a net dipole - DD
• no H on O so
no H-bonding
CH3 C C CH3
only nonpolar C–H and
C–C bonds
VDW only
3.6 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
c. CH3(CH2)4CH3 or CH3(CH2)5CH3
longer molecule, more surface area
higher boiling point
only VDW
VDW and DD
polar, stronger intermolecular forces
higher boiling point
b. CH3CH2COOH or CH3COOCH3
d.
CH2 CHCl
or
CH2 CHI
I is more polarizable.
higher boiling point
no H-bonding
VDW, DD, and H-bonding
stronger intermolecular forces
higher boiling point
3.7 Increasing intermolecular forces: van der Waals < dipole–dipole < hydrogen bonding
O
CH3CH2
C
O
NH2
N–H bonds allow for hydrogen bonding.
stronger intermolecular forces
higher boiling point
H
C
N
CH3
CH3
no hydrogen bonding
weaker intermolecular forces
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Introduction to Organic Molecules and Functional Groups 3–5
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 In the more ordered solid phase, molecules are much closer together than in the less ordered
liquid phase. The shape of a molecule determines how closely it can pack in the solid phase so
symmetry is important. In the liquid phase, molecules are already farther apart, so symmetry is
less important and thus it doesn’t affect boiling point.
3.11 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
3.12 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
O
COOH
a.
c.
b.
HO
O
norethindrone
arachidonic acid
OH
benzo[a]pyrene derivative
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Chapter 3–6
3.13 Like dissolves like.
• To be soluble in water, a molecule must be ionic, or have a polar functional group capable of Hbonding 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
vitamin B3
(niacin)
soluble in water due to
two polar functional groups
and only 6 C's in the molecule
nonpolar
O
vitamin K1
(phylloquinone)
soluble in organic solvents
two polar C–O bonds but the
compound has > 10 C's
water insoluble
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
ionic salt
b. CH3(CH2)14CO2–Na+
c. CH3(CH2)12COOH
long chain
no salt
This is a soap because it contains
both a long chain and a carboxylic
acid 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).
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.
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Introduction to Organic Molecules and Functional Groups 3–7
3.16
amide
ester
ester
ester
ether
ether
ester
H
O
ester
O
O
O
O
O
O
O
O
HN
O
O
O
ether
N
O
O
H
N
amide
ether
H
O
O
O
O
N
O
O
ester amide
O
O
O
O
O
O
O
N
O
O
N
H
O
ester
ester
ester
ester
amide
amide
valinomycin
nonactin
3.17 Electronegative heteroatoms like N, O, or X make a carbon atom an electrophile.
A lone pair on a heteroatom makes it basic and nucleophilic.
Bonds create nucleophilic sites and are more easily broken than bonds.
nucleophilic
a.
electrophilic
b. H O H
Br
C bonded to Br
electrophilic
nucleophilic
nucleophilic
c.
d.
nucleophilic
N
CH3
electrophilic
3.18 Electrophiles and nucleophiles react with each other.
O
a.
CH3CH2 Br
electrophile
b.
+
YES
c.
+
Br
nucleophile
CH3
C
Cl
+
electrophile
nucleophile
CH3 C C CH3
nucleophile
OH
NO
d.
CH3 C C CH3
nucleophile
OCH3
YES
nucleophile
+ Br
electrophile
YES
H
amide
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Chapter 3–8
3.19 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
O
N
carboxylic acid
sulfide
CH3
CH3
COOH
penicillin G
ibuprofen
carboxylic acid
amine
amine
NH2
alkene
amine
N
CO2H
b.
S
O
CH2N(CH3)2
ester
amide
H
N
O
d.
alkene
carboxylic acid
pregabalin
f.
OH
H
O
alkyne
alkene
alkenes
O
O
alcohol
ester
ketone
pyrethrin I
histrionicotoxin
3.20
CH3
OH
CH2OH
CH3 CH CH2 CH3
CH3CH2CH2CH2 OH
alcohol
alcohol
CH3 C OH
CH3 CH CH3
CH3
alcohol
alcohol
CH3
CH3CH2 O CH3CH2
CH3 O CH CH3
ether
ether
CH3 O CH2CH2CH3
ether
3.21 A cyclic ester is called a lactone. A cyclic amide is called a lactam.
O
a.
N CH3
O
b.
amine
c.
ether
d.
O
NH
O
ester
lactone
amide
lactam
3.22 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
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Introduction to Organic Molecules and Functional Groups 3–9
3.23 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)
3.24 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
d. increasing intermolecular forces:
CH3Cl < CH3Br < CH3I
VDW
dipole–dipole
H-bonding
CH3Cl < CH3OH < NaCl
Increasing polarizability
stronger intermolecular forces
VDW
VDW
ionic
dipole–dipole dipole–dipole
H-bonding
3.25
O
CH3
H O
C CH3
C
O H
O
hydrogen bonding between
two acetic acid molecules
3.26 A = VDW forces; B = H-bonding; C = ion–ion interactions; D = H-bonding; E = H-bonding;
F = VDW forces.
3.27 Use the principles from Answer 3.6.
a.
I
CH3(CH2)4
CH3(CH2)5
I
CH3(CH2)6
I
Increasing size, increasing surface area, increasing boiling point
b.
CH3CH2CH2CH3 < (CH3)3N < CH3CH2CH2NH2
VDW
VDW
dipole–dipole
VDW
dipole–dipole
H-bonding
Increasing boiling point
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Chapter 3–10
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
VDW
dipole–dipole
H-bonding
Increasing boiling point
3.28
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.
3.29
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
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Introduction to Organic Molecules and Functional Groups 3–11
3.30
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
3.31
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.
3.32
91 oC
symmetrical
higher mp
25 oC
most spherical
highest 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
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Chapter 3–12
3.33 Increasing polarity = increasing water solubility.
Neither compound is very H2O soluble.
a.
CH3CH2CH2CH3 <
(CH3)3CH
<
CH3OCH2CH3 < CH3CH2CH2OH
VDW
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.)
b.
Br
polar
no H-bonding
OH
O
polar
H-bonding to H2O,
not itself
VDW
DD
H-bonding
polar and
H-bonding
More opportunities
for H-bonding with its
O atom and its H on O.
3.34 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 need only 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
3.35 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
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 like itself, and this increases its boiling point.
VDW forces
dipole–dipole forces
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Introduction to Organic Molecules and Functional Groups 3–13
3.36
Use the solubility rule from Answer 3.13.
OH
CCl3
a.
Cl
C
Cl
H
O
CH3
b.
O
COOH
2 polar functional groups
but > 10 C's
not water soluble
O
HO
d.
N
many polar bonds with N and O
atoms
many opportunities for H-bonding
water soluble
OH
OH
O
HO
O
HO
O
CH3
caffeine
HO
OCH3
aspartame
many polar bonds with N and O atoms
many opportunities for H-bonding
water soluble
OH
N
N
N
H
CH3O
CH3
N
H2N
mestranol
DDT
no N or O
not water soluble
O
e.
c.
OH
f.
OH
sucrose
many polar bonds with O
11 O's and 12 C's
many opportunities for H-bonding with H2O
water soluble
carotatoxin
1 polar functional group
but > 10 C's
not water soluble
3.37
(CH3)2CHCH(CH3)2
B
6 C's
Branching makes less
surface area, weaker VDW.
lowest bp
CH3(CH2)4CH3
C
6 C's
no branching
CH3(CH2)5CH3
D
7 C's
CH3(CH2)6CH3
A
8 C's
highest bp
C, D, and A are all long chain hydrocarbons,
but the size increases from C to D to A, increasing
the VDW forces and increasing bp.
3.38
a.
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
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
HOCH2
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.
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Chapter 3–14
3.39
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
C(CH3)3
avobenzone
two ketones
one ether
3.40
O
OCH3
dioxybenzone
two hydroxy groups
one ketone
one ether
more water soluble
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.
O
H-bond
H
no H-bonding
H
O
O
O
O
Cl
poly(ethylene glycol)
PEG
water soluble
3.41
OH
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.
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71
Introduction to Organic Molecules and Functional Groups 3–15
3.42
Compare the intermolecular forces of crack and cocaine hydrochloride. Stronger intermolecular
forces increase both the boiling point and the water solubility.
CH3
ionic bond
H
N
CH3
N
Cl
COOCH3
COOCH3
O
O
O C
O C
H
H
cocaine hydrochloride
a salt
cocaine (crack)
neutral organic molecule
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 passage and lungs.
3.43
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/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 O H
H O H
CH2CH2O H
N CH2CH2O
b.
O
H
O
H
H
H O H
H O H
nonpolar
interacts with
organic material
polar
interacts with water
by H-bonding
at O and H atoms
H
O
H
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Chapter 3–16
3.44
An emulsifying agent is one that dissolves a compound in a solvent in which it is not normally
soluble. In this case the phospholipids can dissolve the oil in its nonpolar tails and bring it into
solution in the aqueous vinegar solution. Or, the nonpolar tails dissolve in the oil, and the polar
head brings the water-soluble compounds into solution. In any case, the phospholipids make a
uniform medium, mayonnaise, from two insoluble layers.
vinegar
oil
aqueous
organic
hydrophilic
hydrophobic
These two ingredients will not mix. The emulsifying agent (egg yolk) has phospholipids that
have both hydrophobic and hydrophilic portions, making the mayonnaise uniform.
3.45
O
a.
OH
O
O
O
H
N
HCl
O
N
H
OH
O
O
N
O
H
N
OH HH
O
O
N
H
N
H
five functional groups that have
many opportunities for H-bonding
water soluble
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.
3.46
Use the rules from Answer 3.17.
nucleophilic
a.
+
I nucleophilic
nucleophilic
+
c.
+
O
+
+
e. CH3OH
electrophilic
electrophilic
electrophilic
O
b.
CH2
Cl–
N
H
O
b.
OH H H
O
d.
f.
nucleophilic
All the C=C's are
nucleophilic.
CH3
nucleophilic
C
+ Cl
electrophilic
(All lone pairs on O and
Cl are nucleophilic.)
N
Cl–
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Introduction to Organic Molecules and Functional Groups 3–17
3.47
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
3.48
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
3.49
amine
can H-bond
NH2
HO
hydroxy group
can H-bond
OH
HO
O
OH
O
O
O
O
HO
O
O
HN
HO
O
most
acidic
proton HO
N
H
O
Cl
H
N
O
O
C
Cl
N
H
H2N
H
N
O
O
OH
OH
OH
O
vancomycin
N
H
amide
can H-bond
NHCH3
a. 7 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 Hbonding are ROH, RCOOH, amides,
and amines.
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Chapter 3–18
3.50
O
OH
=
C
CHO
O
OH
H
=
O
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 Hbonding, 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
a. melting point
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 + +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
H
COOH
loss of 1 proton
loss of 1 proton
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
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
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.
3.51
HOOC
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.
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Alkanes 4–1
C
Chhaapptteerr 44:: A
Allkkaanneess
G
Geenneerraall ffaaccttss aabboouutt aallkkaanneess ((44..11––44..33))
• 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 CC and CH bonds and no functional group so they undergo few
reactions.
• Alkanes are named with the suffix -ane.
C
Cllaassssiiffyyiinngg C
C’’ss aanndd H
H’’ss ((44..11A
A))
• 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
2o H
CH3
3o H
2o H
N
Naam
meess ooff aallkkyyll ggrroouuppss ((44..44A
A))
CH3
=
methyl
CH3CH2
=
isopropyl
CH3CH2CHCH3
=
sec-butyl
=
propyl
(CH3)2CH
=
butyl
ethyl
CH3CH2CH2
CH3CH2CH2CH2
(CH3)2CHCH2
=
isobutyl
=
(CH3)3C
tert-butyl
3o H
CH3CH2 C CH3
C
1o H
H
=
75
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Chapter 4–2
C
Coonnffoorrm
maattiioonnss iinn aaccyycclliicc aallkkaanneess ((44..99,, 44..1100))
• 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
gauche
HH
H
CH3
H
H
H
H
H
H
H
H
H
•
•
•
H
H
H
H
H
CH3
H
CH3
H
•
H
CH3
•
Dihedral angle of • Dihedral angle of 2
2 CH3’s = 180o
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
TTyyppeess ooff ssttrraaiinn
• 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).
TTw
woo ttyyppeess ooff iissoom
meerrss
[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
C
Coonnffoorrm
maattiioonnss iinn ccyycclloohheexxaannee ((44..1122,, 44..1133))
• 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. Ring-flipping
converts axial to equatorial H’s, and vice versa.
An axial H flips equatorial.
Hax
Heq
Ring-flip.
Heq
Hax
An equatorial H flips axial.
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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
CH3
H
H
trans isomer
CH3
H
cis isomer
O
Oxxiiddaattiioonn––rreedduuccttiioonn rreeaaccttiioonnss ((44..1144))
• Oxidation results in an increase in the number of CZ bonds or a decrease in the number of
CH 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 CZ bonds or an increase in the number of
CH bonds.
H
H
C C
H
H H
H C C H
H
ethylene
H H
ethane
Increase in C–H bonds = reduction
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Chapter 4–4
C
Chhaapptteerr 44:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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 Isopentane has 4 C’s in a row with a 1 C branch.
H
H
a. CH3CH2 C CH3
b.
CH3
H C CH3
CH3 C CH3
H
c. CH3CH2CH(CH3)2
re-draw
H
isopentane
d.
isopentane
H C
H
C H
e.
f. CH3CH(CH3)CH2CH3
CH3 C H CH3
H
re-draw
isopentane
5 C's in a row
pentane
isopentane
isopentane
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
1° H's
CH3
H H
H
3° H
All others are 2° H's.
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Alkanes 4–5
4.4 Use the definition of 1°, 2°, 3°, or 4° carbon atoms from Answer 4.3.
1° C
4° C
O
3° C
All other tetrahedral C's are 2° C's.
HO
3° C
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
CH3CH2 C CH3
H
H
CH3 C
CH3
H
C CH3
CH3 CH3
4.6
Molecular formula C8H18 with one CH3 substituent:
CH3
CH3
CH3CH2CH2CH2CH2 C CH3
CH3
CH3CH2CH2CH2 C CH2CH3
H
CH3CH2CH2 C CH2CH2CH3
H
H
4.7 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
4.8 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
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Chapter 4–6
4.9 Use the steps from Answer 4.5 to draw the constitutional isomers.
Five constitutional isomers of molecular formula C5H10 having one ring:
[2]
[1]
[3]
CH3
CH3
CH3
CH3
CH3
CH2CH3
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.
4.10
4-tert-butyl
[1]
CH3
a. CH3CH2CH2
[2]
CH3
C CH3
C CH2CH2CH2CH3
CH3
4
8 carbons = octane
H
[1]
H C CH3
H C CH2 CHCH3
CH3
CH3
b.
[3] 4-tert-butyl-4-methyloctane
CH3
CH3 C CH3
CH3CH2CH2 C CH2CH2CH2CH3
CH3 6 7 8
1 2 3
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
4.11 Use the steps in Answer 4.10 to name each alkane.
a. CH3CH2CH(CH3)CH2CH3
[1]
re-draw
[2]
1
2
3
4
5
CH3CH2 CH CH2CH3
CH3CH2 CH CH2CH3
CH3
CH3
5 carbons = pentane
[3] 3-methylpentane
3-methyl
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Alkanes 4–7
b. (CH3)3CCH2CH(CH2CH3)2
re-draw
[1]
[2] 2
CH3
1
CH3
CH3 C CH2 CH CH2CH3
CH3
[3] 4-ethyl-2,2-dimethylhexane
4
5
6
CH3 C CH2 CH CH2CH3
CH2CH3
CH3
CH2CH3
4-ethyl
6 carbons = hexane
2,2-dimethyl
c. CH3(CH2)3CH(CH2CH2CH3)CH(CH3)2
re-draw
[1]
4-isopropyl [3] 4-isopropyloctane
[2]
CH3
CH3CH2CH2CH2 CH CH CH3
CH3
8 7 6 5 4
CH3CH2CH2CH2 CH CH CH3
CH2CH2CH3
CH2CH2CH3
3
8 carbons = octane
2
1
2,2,4,4-tetramethyl
[2]
d. [1]
[3] 2,2,4,4-tetramethylpentane
1 2 3 4 5
5 carbons = pentane
2-methyl
[2]
[1]
1
[3] 3-ethyl-2,5-dimethylheptane
34 5 6 7
e.
2
or
3-ethyl 5-methyl
longest chain = 7 carbons = heptane
Number so there are more substituents.
Pick the upper option.
2-methyl
f.
[2] 1
[1]
3
5
[3] 5-sec-butyl-3-ethyl-2,7-dimethyldecane
5-sec-butyl
2
10 carbons = decane
3-ethyl
6
8 9 10
7-methyl
4.12 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
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Chapter 4–8
b. 3,3-dimethylpentane
[1]
5 carbon alkane
methyl groups on C3
[2]
[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
[3]
methyl groups on C3 and C5
CH3
CH3
CH3
CH3
CH3CH2 CH CH2 C
CH3
CH2CH2CH3
CH3
d. 3-ethyl-4-methylhexane
[1]
[2]
[3]
ethyl group on C3
6 carbon alkane
CH2CH3
CH3CH2 CH CH CH2CH3
CH3
CH2CH3
C
C C C C C
C C C C C C
CH3
methyl group on C4
e. 3-ethyl-5-isobutylnonane
[1]
[2]
[3]
isobutyl group on C5
9 carbon alkane
C C C C C C C C C
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
4.13 Use the steps in Answer 4.10 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
H H H H
C H
H
H H H CH3 H
H C C C C
5
H H H H
1
[3] 2-methylpentane
1
[3] 3-methylpentane
C 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
5 carbons = pentane
H C C C
5
H H H
C C H
H H
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Alkanes 4–9
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
1
H H
H
H
H C C
C
C H
[3] 2,3-dimethylbutane
H CH3 CH3 H
4 carbons = butane
2,3-dimethyl
4.14 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.)
1
2
CH3
[2] 3 C C C CH
3
[1]
a.
C
4
6 carbons in ring =
cyclohexane
C
1,1-dimethyl
[3] 1,1-dimethylcyclohexane
C 6
5
Number so the
substituents are at C1.
1,2,3-trimethyl
[1]
[2]
3
4C
b.
CH3
C 2
C CH3
[3] 1,2,3-trimethylcyclopentane
5C C
CH3
5 carbons in ring =
cyclopentane
1
Number so the first substituent
is at C1, second at C2.
1
2
3 C
C
[2] C
[1]
c.
C
CH3 4 C
5
6 carbons in ring =
cyclohexane
C
[3] 1-butyl-4-methylcyclohexane
6
1-butyl
4-methyl
Number so the earlier alphabetical
substituent is at C1, butyl before methyl.
83
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Chapter 4–10
1
[1]
[2]
[3] 1-sec-butyl-2-isopropylcyclohexane
C
5C
d.
4C
C
C
C
3
6 carbons in ring =
cyclohexane
1-sec-butyl
6
2
2-isopropyl
Number so the earlier alphabetical
substituent is at C1, butyl before isopropyl.
[1]
[2]
e.
1
3
5
C
C
C
C
C
[3] 1-cyclopropylpentane
4
2
1-cyclopropyl
longest chain =
5 carbons =
pentane
Number so the
cyclopropyl is at C1.
5
[1]
6
4
3
[2]
[3] 3-butyl-1,1-dimethylcyclohexane
2
f.
3-butyl
6 carbons in ring =
cyclohexane
1,1-dimethyl
1
Number so the two
methyls are at C1.
4.15 To draw the structures, use the steps in Answer 4.12.
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] CH3
C C 3
CH3 2
CH3
b. 1,1,2-trimethylcyclopropane
[1] 3 carbon cycloalkane
c. 4-ethyl-1,2-dimethylcyclohexane
[1] 6 carbon cycloalkane
[2]
C
C
C
[1] 5 carbon cycloalkane
C
[3]
CH3
C
C C
3
CH3CH2 4 C 2 CH3
C
C
CH3
[2]
4C
CH
CH3
CH3
[3] CH CH
3
2
C
C
ethyl 5 C 1 CH3
6
on C4
2 CH3's
C
d. 1-sec-butyl-3-isopropylcyclopentane
C
3 CH3's
3
C C
C
2C
C C CH3
1 CH
3
C
C
CH3
[2]
CH3
C3
isopropyl
[3]
C 2
C
C
1
5
CH CH3
CH3 CH2
sec-butyl
CH3
CH3
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Alkanes 4–11
e. 1,1,2,3,4-pentamethylcycloheptane
[1] 7 carbon cycloalkane
[2]
5 CH3's
CH3
[3]
CH3
CH3
CH3 3 C 4
C
C5
C
C
C
C
C
C C
2
CH3 C 1 C 6
C C7
CH3
CH3
CH3
CH3
CH3
4.16 To name the cycloalkanes, use the steps from Answer 4.14.
[1]
5 carbons in ring =
cyclopentane
[1]
[2]
C C
[3] methylcyclobutane
C C
CH3
CH3
4 carbons in ring =
cyclobutane
[1]
methyl
[2]
CH3
CH3
CH3
3 carbons in ring =
cyclopropane
[1]
[2]
CH3
3 carbons in ring =
cyclopropane
CH3
C
C C
[3] ethylcyclopropane
CH2CH3
ethyl
3 carbons in ring =
cyclopropane
CH3
[3] 1,2-dimethylcyclopropane
1,2-dimethyl
CH2CH3
[1]
C
C C
[2]
CH3
[3] 1,1-dimethylcyclopropane
CH3
1,1-dimethyl
4.17 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).
CH3(CH2)6CH3
8 C's
linear
largest number of C's
no branching
highest bp
CH3(CH2)5CH3
7 C's
linear
CH3CH2CH2CH2CH(CH3)2
(CH3)3CCH(CH3)2
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
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Chapter 4–12
4.18 To draw a Newman projection, visualize the carbons as one in front and one in back of each
other. The CC bond is not drawn. There is only one staggered and one eclipsed conformation.
Br
rotation here
H H
H C C Br
C in front
H
H
H H
60o
H
H
H Br
H
H
H
H
H
1
staggered
2
eclipsed
C behind
4.19 Staggered conformations are more stable than eclipsed conformations.
H
H
H CH3 HH
H
H
HH
CH3
H
H
HH
H
H CH3
H
H
H
H
CH3
H
eclipsed
energy maximum
H H
H C C CH3
Energy
rotation here
H
H
CH3 CH3
H
H
H
H
H
60o
0o
CH3
H
H
H H
120o
H
H
H
H
H
H
180o
240o
300o
staggered
energy minimum
360o = 0o
Dihedral angle
4.20
4.0 kJ/mol
H
H
H CH3
To calculate H,CH3 destabilization:
H,H eclipsing
4.0 kJ/mol of destabilization
H
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
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Alkanes 4–13
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.
4.21
rotation here
H
H
CH3
CH3
CH3 C CH2CH3
CH3
CH3
60o
H
CH3
H
H
H
1
staggered
most stable
H
CH3
H
CH3
CH3
H
2
eclipsed
3
staggered
most stable
60o
CH3
CH3
CH3
60o
H
H
60o
CH3
CH3
CH3
H
H
H
6
eclipsed
least stable
H
CH3
CH3
60o
H
CH3
CH3
CH3
60o
H
H
H
5
staggered
4
eclipsed
least stable
4.22 To determine the most and least stable conformations, use the rules from Answer 4.21.
Cl
1,2-dichloroethane
H
H
ClCH2 CH2Cl
60o
H
H
H
H
H
H Cl
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
Cl
6
eclipsed
60o
H
H
H
Cl
60o
H
H
H H
Cl
Cl
Cl
5
staggered, gauche
4
eclipsed
87
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Chapter 4–14
highest energy
Cl groups eclipsed
least stable
4
3
5
1
most stable
180o
Eclipsed forms are
higher in energy.
6
Energy
2
Staggered forms
are lower in energy.
1
most stable
Cl groups anti
60o 120o 180o
0o
120o 60o
Dihedral angle between 2 Cl's
4.23 Add the energy increase for each eclipsing interaction to determine the destabilization.
HH
H CH
CH3
H
3
HCH3
4.24
H CH
3
b.
a.
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
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 Br
H H
HO
H
equatorial
H
H
Cl H
OH
equatorial
CH3
axial up
H
axial down
Br
HO
Cl
OH
Up carbons are dark circles.
Down carbons are clear circles.
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Alkanes 4–15
4.25
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; i.e., 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
Cl
Draw in the H
b.
and label the C
as up or down.
Draw second conformation.
Up carbons switch
to down carbons.
H
Cl
axial
eq
Axial bond is up =
up carbon
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
4.26 Larger axial substituents create unfavorable diaxial interactions, whereas equatorial groups have
more room and are favored.
H H
H
H
C C H
CH2CH3
H H
larger substituent
more important to be equatorial
equatorial CH2CH3
H
C
H
H
H
H
C C H
more compact substituent
less important to be equatorial
H
C
C
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.
The H's and CH3 of the sp3 hybridized
C have severe 1,3-diaxial interactions
with the two other axial H's.
H
C CH
equatorial C CH
CH3
H
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.
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Chapter 4–16
4.27 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
trans-1-ethyl-2-methylcyclopentane
b.
or
CH3
CH3
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
4.28 Cis and trans isomers are stereoisomers.
cis-1,3-diethylcyclobutane
a. trans-1,3-diethylcyclobutane
b. cis-1,2-diethylcyclobutane
trans = opposite sides of the ring
one group on a wedge,
one group on a dash
cis = same side of the ring
both groups on wedges or
both on dashes
constitutional isomer
different arrangement of atoms
4.29 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.
down bond
(up) (equatorial)
H
H
(up)
Br
Cl
H
HO
up bond
(axial)
OH
OH
down bond
(equatorial)
both groups down =
cis isomer
Cl
Cl
b.
up bond
H
(down) (equatorial)
H
H
c. H
(down)
H
H
Br
Br
down bond
H
(equatorial)
one group up, one down =
trans isomer
Cl
one group up, one down =
trans isomer
Br
4.30
CH3
a.
trans: CH3
CH3
c.
CH3
groups on same side
cis isomer
CH3
cis:
H
CH3
b.
H
CH3
groups on opposite sides
trans isomer
(one possibility)
CH3
CH3
H
H
two chair conformations for the cis isomer
Same stability since they both have
one equatorial, one axial CH3 group.
H
CH3
H
H
CH3
H
CH3
both groups equatorial
more stable
two chair conformations for the trans isomer
d. The trans isomer is more stable because
it can have both methyl groups in the more roomy
equatorial position.
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Alkanes 4–17
4.31
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
4.32 Oxidation results in an increase in the number of CZ bonds, or a decrease in the number of CH
bonds.
Reduction results in a decrease in the number of CZ bonds, or an increase in the number of CH
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
4.33 The products of a combustion reaction of a hydrocarbon are always the same: CO2 and H2O.
flame
a.
CH3CH2CH3
b.
+
+
5 O2
9 O2
flame
3 CO2 +
6 CO2 +
4 H2O + heat
6 H2O + heat
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Chapter 4–18
4.34 Lipids contain many nonpolar C–C and C–H bonds and few polar functional groups.
NH2
HOOC
a. CH3(CH2)7CH=CH(CH2)7COOH
b.
oleic acid
H
N
O
CH3O
O
aspartame
only one polar functional group
18 carbons
a lipid
many polar functional groups
only 14 carbons
not a lipid
4.35 “Like dissolves like.” Beeswax is a lipid, and therefore, 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
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Alkanes 4–19
4.36 Use the rules from Answer 4.3.
1°
4°
3°
CH3 CH3
2°
a.
[1]
2°
[2]
4°
[2]
CH3
CH3
3°
1°
CH2
CH2
C
b. [1] CH3
CH
C
CH2
CH3
CH2
CH
All CH3's have 1° H's.
All CH2's have 2° H's.
All CH's have 3° H's.
2°
1°
CH3
CH2
CH2
CH
CH3
CH3
4.37
One possibility:
CH3
a. CH3 C CH3
b.
d. (CH3)3CH
c. CH3CH2CH3
CH3
4.38 Use the rules from Answer 4.3.
1°
O
O
O
O
2°
O
4°
4°
O
OH
OH
C(CH3)3
2° 2°
2°
3°
4° 1°
4.39
a. Five constitutional isomers of molecular formula C4H8:
CH2
CH3CH CHCH3
CH2 CHCH2CH3
CH3
CH3
C
CH3
b. Nine constitutional isomers of molecular formula C7H16:
H
H
CH3 C CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH3
CH3CH2 C CH2CH2CH3
CH3
CH3
CH3
CH3 C CH2CH2CH3
CH3CH2 C CH2CH3
CH3
CH3
H
CH3CH2 C CH2CH3
CH2CH3
H
CH3 C
CH3
C CH3
CH3 CH3
CH3
H
CH3 C
H
C CH2CH3
CH3 CH3
H
CH3 C
H
CH2 C CH3
CH3
CH3
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Chapter 4–20
c. Twelve constitutional isomers of molecular formula C6H12 containing one ring:
4.40 Use the steps in Answers 4.10 and 4.14 to name the alkanes.
a. CH3CH2CHCH2CHCH2CH2CH3
[1]
CH2CH3
CH3
1 2 3 4 5 6 7 8
[2] CH3CH2CHCH2CHCH2CH2CH3
CH3
8 carbons = octane
CH2CH3
3-methyl
7-methyl
7
b. CH3CH2CCH2CH2CHCHCH2CH2CH3
[1]
5-ethyl
CH3
CH2CH3
CH3
2 3 CH2CH3
[2] 1
[3] 5-ethyl-3-methyloctane
[3] 3,3,6-triethyl-7-methyldecane
CH3CH2CCH2CH2CHCHCH2CH2CH3
CH2CH3 CH2CH3
CH2CH3 CH2CH3
10 carbons = decane
3,3,6-triethyl
c. CH3CH2CH2C(CH3)2C(CH3)2CH2CH3
[1]
[2]
re-draw
CH3 CH3
CH3CH2CH2 C
[3] 3,3,4,4-tetramethylheptane
CH3 CH3 3
CH3CH2CH2 C
C CH2CH3
CH3 CH32
4
CH3 CH3
C CH2CH3
1
3,3,4,4-tetramethyl
7 carbons = heptane
d. CH3CH2C(CH2CH3)2CH(CH3)CH(CH2CH2CH3)2
3,3-diethyl
re-draw
[1]
CH3CH2 H
CH3CH2 C
C
4
[2]
CH3CH2 H
H
CH3CH2 C
C CH2CH2CH3
1
CH3CH2 CH3 CH2CH2CH3
CH3CH2 H
CH3CH2 C
C CH2CH2CH3
CH3CH2 CH3
7 carbons = heptane
C CH2CH2CH3
4-methyl
5-propyl
3,3-diethyl
e. (CH3CH2)3CCH(CH3)CH2CH2CH3
re-draw
[3] 3,3-diethyl-4-methyl-5-propyloctane
CH3CH2 CH3 CH2CH2CH3
8 carbons = octane
[1]
C
5
H
CH3CH2 H
[2]
CH3CH2 C
1
4
[3] 3,3-diethyl-4-methylheptane
C CH2CH2CH3
CH3CH2 CH3
6
4-methyl
7
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Alkanes 4–21
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
3,4-dimethyl
CH3 CH3 CH2CH2CH3
[3] 3,4-dimethyl-5-propylnonane
C CH2CH2CH2CH3
5-propyl
9 carbons = nonane
g. (CH3CH2CH2)4C
re-draw
[1]
4
4,4-dipropyl
[2]
CH2CH2CH3
CH2CH2CH3
CH3CH2CH2 C CH2CH2CH3
[3] 4,4-dipropylheptane
CH3CH2CH2 C CH2CH2CH3
CH2CH2CH3
1
2
3
CH2CH2CH3
7 carbons = heptane
1
h.
[1]
[2]
5
3
7
6-isopropyl
3-methyl
10 carbons = decane
i.
[3] 6-isopropyl-3-methyldecane
6
[2]
[1]
10 carbons = decane
8
6
4-isopropyl
4
[3] 8-ethyl-4-isopropyl-2,6-dimethyldecane
1
8-ethyl
j.
2,6-dimethyl
4-isopropyl
4
[3]
[2]
[1]
4-isopropyloctane
1
8 carbons = octane
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
3
1
2
1-sec-butyl
2-isopropyl
1-sec-butyl-2-isopropylcyclopentane
95
96
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Chapter 4–22
5
6
1
n.
4
1-isobutyl-3-isopropylcyclohexane
3
2
1-isobutyl
3-isopropyl
4.41
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
2
CH3
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
H
CH3CH2 C CH2 C CH2CH3
C CH2CH2CH3
1
CH3 CH3
4
3,4-dimethyl
3
CH3
CH3
5
3,5-dimethyl
3,5-dimethylheptane
3,4-dimethylheptane
4.42 Use the steps in Answer 4.12 to draw the structures.
a. 3-ethyl-2-methylhexane
[1]
6 C chain
C C
[2]
C C C C C C
C C C C
H
CH3 CH2CH3
methyl
on C2
[3] CH3 C
H
C CH2CH2CH3
CH3 CH2CH3
ethyl on C3
b. sec-butylcyclopentane
[1]
5 C ring
[2]
isopropyl on C4
c. 4-isopropyl-2,4,5-trimethylheptane
[1]
7 C chain
[2]
CH3
CH
C C C C
CH3
C C C C C C C
CH3
C C C
CH3 CH3
methyls on C2, C4, and C5
d. cyclobutylcycloheptane
[1] 7 C cycloalkane
[2]
CH3
[3]
CH3
CH
CH3 CH CH2 C
CH3
CHCH2CH3
CH3 CH3
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Alkanes 4–23
e. 3-ethyl-1,1-dimethylcyclohexane
[1]
6 C cycloalkane
[2]
[3]
CH3CH2
C
C
C
C
C
ethyl on C3 C
C
2 ethyl groups
8 C cycloalkane
C
C
C
C
C
C C
g. 6-isopropyl-2,3-dimethylnonane
[1]
9 C alkane
C
[2]
C
C
C C
C
isopropyl
methyl
CH3
CH3
C
C
C
C
C
C
C
C
C
8 C alkane
C
C
C
C
C
C
C
C
C
C
CH3
[2] CH3 CH3
C
C
C
C
C
[1]
C
C
CH3 CH3
C
C
C
C
CH2CH3
[3]
ethyl on C1
[2]
CH2CH3
or
j. trans-1-tert-butyl-4-ethylcyclohexane
6 C ring
C
C
methyl
CH3
[1]
[3]
CH3
CH3
i. cis-1-ethyl-3-methylcyclopentane
5 C ring
C
CH
5 methyl groups
h. 2,2,6,6,7-pentamethyloctane
[1]
C
[3]
C C
C
C
2 methyl
groups on
C1
CH3
C
C C
C
C
CH3
C
C C
[2]
C
C
C
f. 4-butyl-1,1-diethylcyclooctane
[1]
C
methyl on C3
C(CH3)3
[2]
CH3CH2
CH3
97
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Chapter 4–24
4.43 Draw the compounds.
a. 2,2-dimethyl-4-ethylheptane
CH3
e. 1-ethyl-2,6-dimethylcycloheptane
CH2CH3
alphabetized incorrectly
ethyl before methyl
CH3
CH2CH3
2
CH3
CH3 C CH2 C CH2CH2CH3
Numbered incorrectly.
Re-number so methyls
are at C1 and C4.
H
1
4
CH3
4-ethyl-2,2-dimethylheptane
2-ethyl-1,4-dimethylcycloheptane
b. 5-ethyl-2-methylhexane
f. 5,5,6-trimethyloctane
Longest chain was not
chosen = heptane
3
Numbered incorrectly.
Re-number so methyls
are at C3 and C4.
4
2,5-dimethylheptane
3,4,4-trimethyloctane
g. 3-butyl-2,2-dimethylhexane
H
c. 2-methyl-2-isopropylheptane
CH3C CH3
Longest chain was not
chosen = octane
CH3 C CH2CH2CH2CH2CH3
CH3
2,3,3-trimethyloctane
d. 1,5-dimethylcyclohexane
1
4-tert-butyloctane
CH3
1
Numbered incorrectly.
Re-number so methyls
are at C1 and C3.
4
Longest chain not
chosen = octane
h. 1,3-dimethylbutane
H
Longest chain not
chosen = pentane
3
CH3
CH3 CCH2
CH3
1,3-dimethylcyclohexane
CH2
CH3
2-methylpentane
4.44
CH3
a.
H
CH3
CH2CH2CH3
CH3
H
CH2CH2CH3
b.
CH3
H
H
CH2CH3
re-draw
4
4-isopropylheptane
CH3
CH2CH3
re-draw
3
3-ethyl-3-methylpentane
c.
CH3CH2
CH2CH2CH3
CH3CH2CH2
H
CH2CH3
re-draw
4
5
4,4-diethyl-5-methyloctane
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Alkanes 4–25
4.45 Use the rules from Answer 4.17.
a.
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)3CC(CH3)3
branching = lower surface area
lower boiling point
more spherical, better packing =
higher melting point
CH3(CH2)6CH3
no branching = higher surface area
higher boiling point
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.
CH3
H
or
H
H
H
H
CH3
H
1 gauche CH3,CH3
= 3.8 kJ/mol
of destabilization
CH3
higher energy
2 gauche CH3,CH3
3.8 kJ/mol x 2 = 7.6 kJ/mol
of destabilization
Energy difference =
7.6 kJ/mol – 3.8 kJ/mol =
CH3
H
CH3
b.
H
H
or
CH3
H
CH3
CH3
H
CH3
CH3
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
Energy difference =
3.8 kJ/mol
H CH3
18 kJ/mol – 7.6 kJ/mol = 10.4 kJ/mol
99
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Chapter 4–26
4.48 Use the rules from Answer 4.21 to determine the most and least stable conformations.
a. CH3 CH2CH2CH2CH3
b. CH3CH2CH2 CH2CH2CH3
H CH CH CH
2
2
3
H
H
CH2CH2CH3
H
H
HH
H
H
H
H
eclipsed
least stable
staggered
most stable
CH3CH2
CH2CH3
CH2CH3
H
H
H
CH2CH3
staggered
ethyl groups anti
most stable
All staggered conformations are equal in energy.
All eclipsed conformations are equal in energy.
H
H
HH
eclipsed
ethyl groups eclipsed
least stable
4.49
b.
a.
c.
Br
Cl
H
Cl
Cl
H
H
H
H
Cl
H
H
H
CH3
H
H
CH3
H
4.50
CH3
a.
C
H
H
H
C
H
Br
C
b.
H
c.
H
Br
Cl H
H
Cl
CH3
H
C
Br
Cl
CH3
Cl H
Cl
H
H
Br
Cl
H
Br
Br
CH3
H
CH2CH3
Cl
Cl
CH2CH3
4.51
CH3CH2
H
H
60°
[1]
H
CH3CH2 CH2CH2CH3
H
CH2CH3
H
H
H
CH3
H
60°
H
H
CH3
H
1
60°
H
H
H
CH3
H
6
CH2CH3
CH3
3
2
CH3CH2
H
60°
60°
H
H
H
60°
H
H
H
H
H
CH3CH2
CH3
5
CH3
CH2CH3
4
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Alkanes 4–27
least stable
4
Energy
2
Eclipsed forms are
higher in energy.
6
Staggered forms
are lower in energy.
5
3
1
most stable
1
most stable
60o
120o
180o
60o
0o
120o
180o
Dihedral angle between two alkyl groups
most stable
CH3CH2
H
[2] CH3CH2 CHCH2CH3
H
H
CH3
H
CH2CH3
H
H
60°
H
H
H
60°
CH3
CH3
CH3
CH3
CH3
1
CH2CH3
CH3
3
2
60°
CH3CH2
60°
H
CH3
H
H
CH3
60°
CH3
H
60°
H
CH3CH2
CH3
H
H
CH3
CH2CH3
CH3
5
6
H
H
least stable
4
least stable
4
Energy
2
3
120o
Staggered forms
are lower in energy.
5
1
most stable
180o
Eclipsed forms are
higher in energy.
6
1
most stable
60o
0o
60o
120o
180o
Dihedral angle
(between CH3CH2 in back and CH3 in front)
101
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Chapter 4–28
4.52 Two types of strain:
4.1 Torsional strain is due to eclipsed groups on adjacent carbon atoms.
4.2 Steric strain is due to overlapping electron clouds of large groups (ex: gauche interactions).
H
CH3
H
CH3
H
a.
H
H
H CH3
b.
CH3
H
c.
H
CH3
CH3
two sites
three bulky methyl groups close =
steric strain
HH
CH2CH3
CH3CH2
two bulky ethyl groups close =
steric strain
eclipsed conformation =
torsional strain
eclipsed conformation =
torsional strain
4.53 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
CH3
H
CH3
CH3
H
H
H
b. CH3 C(CH3)3
H
most stable
H CH3
H
H
CH3
CH3
H
H
least stable
Destabilization energy =
CH3
H
H CH3
H
CH3
H
most stable
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
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
4.54
Cl
H
H
H
H
H
most stable
HH
H Cl
H
H
least stable
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.
15.0 kJ/mol – 8.0 kJ/mol = 7.0 kJ/mol
destabilization due to H,Cl eclipsing
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Alkanes 4–29
4.55 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.
4.56
axial
H
OH axial
a.
[1]
c. HO
d.
down HO
HO
eq
eq HO
H
H
eq
Br
a.
[2]
H
H
H
eq
H
H
ax
ax
OH
c.
down
up
CH
eq Br 3
H eq
H eq
CH3eq
CH3
d.
OH
HO
ax
CH3 up
H
b.
OH eq
HO
OH ax
H
ax
Br
d.
both up =
cis
axial
a.
eq HO
ax
Br
c.
H
axial
H eq
eq
OH
Br
b.
H eq
CH3 eq
H ax
OH
H
one up, one down =
trans
up
axial
[3]
ax
H up
OH
b.
HO
ax
ax
H
OH
OH eq
eq HO
eq H
H eq
H
H
one up, one down =
trans
axial
H
ax
4.57
ax
ax
H
H
CH3
CH3
eq
eq
both groups equatorial
more stable
eq H
H eq
CH3
CH3
ax
ax
OH
ax
104
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Chapter 4–30
4.58
Axial/equatorial substituent location
Disubstituted cyclohexane
a. 1,2-cis disubstituted
b. 1,2-trans disubstituted
c. 1,3-cis disubstituted
d. 1,3-trans disubstituted
e. 1,4-cis disubstituted
f. 1,4-trans disubstituted
Conformation 1
Axial/equatorial
Axial/axial
Axial/axial
Axial/equatorial
Axial/equatorial
Axial/axial
Conformation 2
Equatorial/axial
Equatorial/equatorial
Equatorial/equatorial
Equatorial/axial
Equatorial/axial
Equatorial/equatorial
4.59 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
eq
eq
larger group equatorial
more stable
c. trans isomer
c. trans isomer
ax
eq
larger group equatorial
more stable
c. trans isomer
ax
ax
eq
eq
eq
eq
eq
ax
eq
ax
larger group equatorial
more stable
ax
both groups equatorial
more stable
d.
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
4.60 Compare the isomers by drawing them in chair conformations. Equatorial substituents are more
stable. See the definitions in Problem 4.59.
a.
cis
CH2CH3 ax
ax
CH2CH3 ax
CH2CH3
H
CH2CH3 eq
trans
H
CH2CH3 eq
H
H
CH3CH2
H
eq
CH2CH3
H
ax
b. 1-ethyl-3-isopropylcyclohexane
ax
(CH3)2CH
cis
(CH3)2CH
H
CH2CH3
eq
H
both groups equatorial
most stable of all conformations
trans isomer
ax
H
trans
CH2CH3
H
H
CH2CH3
H
(CH3)2CH
H
(CH3)2CH
eq
ax
both groups equatorial
most stable of all conformations
cis isomer
The cis isomer is more stable than the trans isomer because
its more stable conformation has two groups equatorial.
4.61
Only the more stable conformation is drawn.
CH3
CH3
CH3
CH3
or
CH3
CH3
re-draw to see
axial and equatorial
CH3
CH3
CH3
CH3
CH3
CH3
more stable
substituents on C1, C3, C5 =
all equatorial
4.62
OH
re-draw to see
axial and equatorial
OH
CH3
CH3
HO
all equatorial
menthol
HO
eq
CH2CH3
The trans isomer is more stable than the cis isomer because
its more stable conformation has two groups equatorial.
eq
H
H
eq
ax
CH2CH3
H
106
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Chapter 4–32
4.63
a. HO
OH
O
OH
HO
HO
or
HO
HO
HO
O
OH
O
b. HO
HO
or
HO
O
OH
OH
HO
most stable
All groups are equatorial.
HO
OH
HO
OH
OH
4.64
OH
OH OH
a.
O
OH
HO
OH
OH
OH
O
HO
c.
O
OH
OH
OH
more stable
More groups are equatorial.
O
HO
OH
HO
HO
constitutional isomer
OH OH
OH
O
OH
OH OH
b.
OH
HO
HO
HO
OH
galactose
O
d.
OH
glucose
All groups are equatorial.
more stable
OH
OH
stereoisomer
4.65
CH2CH3
a.
and
d.
same molecular formula C4H8
different connectivity
constitutional isomers
b.
same molecular formula C10H20
different connectivity
constitutional isomers
CH3
H
H
CH3
CH3
H
H
=
CH2CH3
1 down, 1 up =
1 down, 1 up =
trans
trans
same arrangement in three dimensions
identical
and
f.
H
CH3
molecular formula: C6H12
different molecular formulas
not isomers
CH3
and
CH3
molecular formula: C6H10
different arrangement in three dimensions
stereoisomers
c.
and
e.
CH3
CH2CH3
CH2CH3
and
H
CH2CH3
and
CH3CH2
CH2CH3
CH3CH2
1 down, 1 up =
both down =
cis
trans
different arrangement in three dimensions
stereoisomers
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Alkanes 4–33
ax
ax
g.
eq
eq
H
CH3CH2
CH2CH3
up
CH3
H
up
up
both up = cis
h.
H
CH3
and
up
and
H
3,4-dimethylhexane
both up = cis
2,4-dimethylhexane
same molecular formula C8H18
different IUPAC names
constitutional isomers
same arrangement in three dimensions
identical
4.66
CH3
a.
CH
H
CH3
CH3
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
CH3 CH2CH3
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
4.67
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
107
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Chapter 4–34
4.68
Three constitutional isomers of C7H14:
1,1-dimethylcyclopentane
1,2-dimethylcyclopentane
1,3-dimethylcyclopentane
or
or
trans
cis
cis
trans
4.69 Use the definitions from Answer 4.32 to classify the reactions.
O
a.
=
CH3CHO
CH3
C
CH3CH2OH
H
d.
CH2 CH2
Decrease in the number of C–O
bonds. Reduction
H C C H
Decrease in the number of C–H
bonds. Oxidation
CH3
b.
Increase in the number of C–O
bonds. Oxidation
c.
CH2 CH2
CH2Br
e.
O
Increase in the number of C–Z
bonds. Oxidation
HOCH2CH2OH
f.
CH3CH2OH
Two new C–O
bonds. Oxidation
CH2 CH2
Loss of one C–O
bond and one C–H
bond. Neither
4.70 Use the rule from Answer 4.33.
flame
a. CH3CH2CH2CH2CH(CH3)2
flame
7 CO2 + 8 H2O + heat
4 CO2 + 5 H2O + heat
b.
11 O2
(13/2) O2
4.71
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
b.
OH
H
1 C–H bond
phenol
[2] loss of 1 C–O bond,
loss of 1 C–H bond
neither
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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Alkanes 4–35
4.72 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
O HO
c.
d. HO
estradiol
O
HO
many polar functional groups
not a lipid
OH
O
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
4.73
O
OH
O
COH
OH
CNHCH2CH2SO3Na+
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.
4.74 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.
4.75 The amide in the four-membered ring has 90° bond angles giving it angle strain, and therefore
making it more reactive.
amide
H
N
penicillin G
S
O
N
CH3
CH3
O
COOH
strained amide
more reactive
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Chapter 4–36
4.76
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
H
longer bond
H
H
4.77
H
H
H
decalin
H
trans-decalin
cis-decalin
H
H
H
H
H
H
trans
The trans isomer is more stable since
the carbon groups at the ring junction are
both in the favorable equatorial position.
1,3-diaxial interaction
cis
This bond is axial, creating
unfavorable 1,3-diaxial interactions.
4.78
4
2
1
5
3
1
1
(1-methylbutyl)cyclopentane
3
4
(1,1-dimethylpropyl)cyclopentane
4
3
2
1
2
3
3
pentylcyclopentane
1
2
2
(2-methylbutyl)cyclopentane
(2,2-dimethylpropyl)cyclopentane
3
2
2
1
(1-ethylpropyl)cyclopentane
2
3
1
(1,2-dimethylpropyl)cyclopentane
1
3
4
(3-methylbutyl)cyclopentane
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111
Stereochemistry 5–1
C
Chhaapptteerr 55:: SStteerreeoocchheem
miissttrryy
wiitthh tthhee ssaam
mee m
moolleeccuullaarr ffoorrm
muullaa ((55..22,, 55..1111))..
IIssoom
meerrss aarree ddiiffffeerreenntt ccoom
mppoouunnddss w
[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
CH2CH3
C
CH3CH2
C
H
Br
CH3
C
H
Br
H
Br
A
CH3
C
CH2CH3
C
H
Br
Br
C
H
Br
C
enantiomers
CH3
C
H
Br
H
B
CH3CH2
C
H
Br
D
enantiomers
A and B are diastereomers of C and D.
A
Assssiiggnniinngg pprriioorriittyy ((55..66))
• 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, consider a multiply bonded atom as an
equivalent number of singly bonded atoms.
highest atomic number =
highest priority
1
3
Br
CH2CH2CH3
4 H C CH2I
*
3
Cl
2
I is NOT bonded directly
to the stereogenic center.
4 CH3 C CH2CH2CH2CH2CH3
*
1
OH
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
bonded to 3 O's.
112
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Chapter 5–2
SSoom
mee bbaassiicc pprriinncciipplleess
• 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.
mppoouunndd ttoo rroottaattee ppllaannee--ppoollaarriizzeedd lliigghhtt ((55..1122))..
O
Oppttiiccaall aaccttiivviittyy iiss tthhee aabbiilliittyy ooff aa ccoom
• 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.
TThhee pprreeffiixxeess RR aanndd SS ccoom
mppaarreedd w
wiitthh dd aanndd ll
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, 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.
meerrss ccoom
mppaarreedd ((55..1122))
TThhee pphhyyssiiccaall pprrooppeerrttiieess ooff iissoom
Type of isomer
Physical properties
Constitutional isomers
Different
Enantiomers
Identical except the direction of rotation of polarized light
Diastereomers
Different
Racemic mixture
Possibly different from either enantiomer
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113
Stereochemistry 5–3
EEqquuaattiioonnss
• Specific rotation (5.12C):
specific =
rotation
•
[]
=
l x c
= observed rotation (o)
l = length of sample tube (dm)
c = concentration (g/mL)
Enantiomeric excess (5.12D):
ee
=
% of one enantiomer – % of other enantiomer
=
[] mixture
x 100%
[] pure enantiomer
dm = decimeter
1 dm = 10 cm
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Chapter 5–4
C
Chhaapptteerr 55:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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
O
b.
and
c.
and
different connectivity of atoms
constitutional isomers
OH
d.
four-membered ring three-membered ring
and
trans isomer
different connectivity of atoms
constitutional isomers
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).
CH3
a.
C
Cl
CH3
CH3 C
Br
CH3
Br
c.
Cl
CH3
O
CH3
identical
CH3
Br
H
C
Cl
Cl
Br
H
stereogenic center
nonsuperimposable mirror images
chiral molecules
CH3
achiral molecule
CH3
C
O
identical
achiral molecule
b.
CH3
H Br
d.
Br H
C
F
C
CH2CH3
CH3CH2
stereogenic center
nonsuperimposable mirror images
chiral molecules
F
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Stereochemistry 5–5
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
H
H
CH3
c.
H
CH3
C
CH3
d.
H
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 CC bond so that the Br atoms are eclipsed.
rotate
CH3 here H
C
Br
Br
Br
H
C
C
CH3
H
H
Br
C
CH3
CH3
C2 C3
plane of symmetry
5.6 To locate a stereogenic center, omit:
All C’s with 2 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
a.
CH3CH2 C CH2CH3
d.
Cl
bonded to 2 identical
ethyl groups
0 stereogenic centers
CH3CH2CH2OH
0 stereogenic centers
CH3
e. (CH3)2CHCH2CH2 C CH2CH3
H
b.
(CH3)3CH
This C is bonded to
4 different groups.
1 stereogenic center
0 stereogenic centers
c.
H
CH3 C CH=CH2
OH
This C is bonded to
4 different groups.
1 stereogenic center
f.
H
CH3CH2 C CH2CH2CH3
CH3
This C is bonded to
4 different groups.
1 stereogenic center
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Chapter 5–6
5.7 Use the directions from Answer 5.6 to locate the stereogenic centers.
Br
H
a.
c.
CH3CH2CH2 C CH3
d.
OH
Br
stereogenic center
both C's bonded to 4
different groups
2 stereogenic centers
H
b. (CH3)2CHCH2 C COOH
3 C's bonded to 4
different groups
3 stereogenic centers
NH2
stereogenic center
5.8 Use the directions from Answer 5.6 to locate the stereogenic centers.
O
OH
H2N
CH3O
O
O
CH3O
NH2
H
N
4 C's bonded to 4
different groups
4 stereogenic centers
aliskiren
5.9 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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Stereochemistry 5–7
5.10 Draw the chiral molecule with only C and H atoms.
CH2CH3
CH2CH3
CH3 C CH2CH2CH3
or
CH3 C CH(CH3)2
H
H
5.11 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.
Cl
b.
Cl
c.
Each labeled C
is bonded to:
d.
H, Cl, CH2, CHCl:
2 stereogenic centers
O
4 C's bonded to 4
different groups:
4 stereogenic centers
NH2
O
O
e.
O
N
H
CO2H
C bonded to
H, 2 different O's and 1 C:
1 stereogenic center
no stereogenic centers
CO2H
5.12
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.
5.13 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
c. –H, –D
higher mass
higher priority
e. –CH2CH2Cl, –CH2CH(CH3)2
higher priority
H
b. –I, –Br
higher priority
d. –CH2Br, –CH2CH2Br
higher priority
f. –CH2OH, –CHO
=
C O
H
=
C O
O C
2 H's, 1 O
2 O's, 1 H
2 C–O bonds
C bonded to 2 O's has
higher priority.
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Chapter 5–8
5.14 Rank by decreasing priority. Lower atomic number = lower priority.
Highest priority = 1, Lowest priority = 4
priority
priority
2
c. –CH2CH3 C bonded to 2 H's + 1 C
a. –COOH C = second lowest
3
atomic number
3
–CH3
C bonded to 3 H's
H = lowest atomic number 4
–H
–NH2
4
H = lowest
atomic number
1
–CH(CH3)2 C bonded to 1 H + 2 C's
decreasing priority: –CH(CH3)2, –CH2CH3, –CH3, –H
–H
N = second highest
2
atomic number
O = highest atomic number 1
–OH
decreasing priority: –OH, –NH2, –COOH, –H
b. –H
–CH3
priority
4
H = lowest
atomic number
C bonded to 3 H's
d. –CH=CH2 C bonded to 1 H + 2 C's
3
1
Cl = highest
atomic number
–CH2Cl C bonded to 2 H's + 1 Cl 2
decreasing priority: –Cl, –CH2Cl, –CH3, –H
–Cl
priority
2
–CH3
C bonded to 3 H's
3
–CCH
C bonded to 3 C's
1
4
H = lowest
atomic number
decreasing priority: –CCH, –CH=CH2, –CH3, –H
–H
5.15 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, the
stereogenic center is labeled R; if the circle is counterclockwise, 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
1
ClCH2
CH3
counterclockwise
S isomer
CH2Br
rotate
C
OH
CH3
HO
H 3
3
C
d.
CH2Cl
1
counterclockwise
S isomer
clockwise
R isomer
lowest priority group
now back
5.16
CH3O
O
2
Cl
CH3OOC
N
1
S
H Cl
N
Cl H
3
3
2
COOCH3
N
1
S
S
clopidogrel
clockwise
R isomer
counterclockwise
S isomer
Plavix
5.17 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
2
1
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Stereochemistry 5–9
5.18
a. CH3CH2CH(Cl)CH(OH)CH2CH3
b. CH3CH(Br)CH2CH(Cl)CH3
2 stereogenic centers = 4 possible stereoisomers
CH3CH2
CH2CH3
C
H
Cl
C
CH3CH2
H
HO
CH2CH3
C
CH2CH3
C
H
OH
A
H
Cl
CH3CH2
H Cl H Br
H
Cl
CH3CH2
HO
H
Br H Cl H
A
B
H Cl Br H
H Br Cl H
C
D
CH2CH3
C
H
C
C
B
C
OH
2 stereogenic centers = 4 possible stereoisomers
C
H
Cl
D
5.19
a. CH3CH(OH)CH(OH)CH3
b. CH3CH(OH)CH(Cl)CH3
2 stereogenic centers = 4 possible stereoisomers
CH3
CH3
C
HO
CH3
C
H
H
OH
A
CH3
H
HO
CH3
C
C
B
C
H
OH
H
HO
CH3
C
H
CH3
C
H
HO
CH3
C
C
OH
HO
H
CH3
H
OH
H
HO
C
C
CH3
CH3
H
Cl
H
Cl
C
A
CH3
C
identical
C is a meso compound.
A and B are enantiomers.
Pairs of diastereomers: A and C, B and C.
5.20
2 stereogenic centers = 4 possible stereoisomers
C
C
H
B
CH3
CH3
H
Cl
H
Cl
C
C
CH3
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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Chapter 5–10
5.21 Use the definition in Answer 5.20 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
5.22 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
One designation remains the same,
the other changes.
Its diastereomers are: R,R and S,S
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.
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
5.24 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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Stereochemistry 5–11
5.25 Meso compounds generally have a plane of symmetry. They cannot have just one stereogenic
center.
Cl
a.
b.
c.
OH
plane of symmetry
meso compound
no plane of symmetry
not a meso compound
no plane of symmetry
not a meso compound
5.26
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
A
Cl
Cl
identical
A
CH3
CH3
CH3
CH3
Cl
Cl
trans
B
identical
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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Chapter 5–12
5.27 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
c.
H
CH3
d.
C
CH3
H
NH2
(S)-alanine
[] = +8.5
mp = 297 oC
OH
HO
same molecular formula,
opposite configuration at one
stereogenic center
enantiomers
COOH
and
OH
1,3-isomer
1,4- isomer
constitutional isomers
and
5.28
OH
HO
same molecular formula
same R,S designation:
identical
b.
HO
and
OH
HO
trans
cis
Both 1,3 isomers,
cis and trans:
diastereomers
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
5.29
[] =
l xc
= observed rotation
l = length of tube (dm)
c = concentration (g/mL)
[] =
10°
= +100 = specific rotation
1 dm x (1 g/10 mL)
5.30 Enantiomeric excess = ee = % of one enantiomer % of other enantiomer.
a. 95% 5% = 90% ee
5.31
b. 85% 15% = 70% ee
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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Stereochemistry 5–13
5.32
a.
[] mixture
ee =
+10
x 100% = 42% ee
+24
x 100%
[] pure enantiomer
b.
[] solution
x 100% = 80% ee
+24
[] solution = +19.2
5.33
a.
[] mixture
x 100% = 60% ee
+3.8
b. % one enantiomer – % other enantiomer = ee
80% – 20% = 60% ee
[] mixture = +2.3
80% dextrorotatory (+) enantiomer
20% levorotatory (–) enantiomer
5.34 • Enantiomers have the same physical properties (mp, bp, solubility), and rotate the plane of
polarized light to an equal but opposite extent.
• Diastereomers have different physical properties.
• A racemic mixture is optically inactive.
cis isomer
trans isomers
CH3
CH3
A
CH3
enantiomers
CH3
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).
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Chapter 5–14
5.35 Use the definitions from Answer 5.2.
CH3
O
a.
c.
and
H
CH3
and
O
b.
both up
cis
Both compounds are
1,2-dimethylcyclohexane.
one cis, one trans = stereoisomers
O
and
H
one up, one down
trans
same molecular formula C4H8O
different connectivity
constitutional isomers
and
d.
O
same molecular formula C7H14
different connectivity
constitutional isomers
C5H8O
C5H10O
different molecular formulas
not isomers
5.36 Use the definitions from Answer 5.3.
CH3
a.
C
CH3
OH
CH3
CH2OH
H
HOCH2
H
O
c.
C
O
e. OHC
OH
OH
CH3
HO
OH
threose
identical
achiral
identical
achiral
COOH
b.
C
HSCH2
chiral
COOH
H
NH2
H
H2N
d.
C
CH2SH
H
Br
Br
H
cysteine
chiral
identical
achiral
5.37
CHO
C
CH3
CH3
a.
OH
H
A
R isomer
HO
C
CH3
b.
H
CHO
S
enantiomer
C
OHC
HO H
c.
H
OH
R
identical
CH3
C
CHO
S
enantiomer
CHO
OH
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Stereochemistry 5–15
5.38 A plane of symmetry cuts the molecule into two identical halves.
H
CH3CH2
a.
C
Cl
b.
C
H
C
COOH
H
HO
C
H
H
CH3
H and OH
are aligned.
The plane of symmetry
bisects the molecule.
plane of symmetry
CH2CH3
C
Cl
d.
C
C
HOOC
CH2CH3
CH3CH2
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
5.39 Use the directions from Answer 5.6 to locate the stereogenic centers.
a. CH3CH2CH2CH2CH2CH3
All C's have 2 or more H's.
0 stereogenic centers
g.
O
bonded to 4 different groups
1 stereogenic center
H
b.
CH3CH2O C CH2CH3
CH3
h.
1 stereogenic center
All C's have 2 or more H's, or
are sp2 hybridized.
0 stereogenic centers
c. (CH3)2CHCH(OH)CH(CH3)2
0 stereogenic centers
H
H
d. (CH3)2CHCH2 C CH2 C
H
C
CH3
Cl
CH2CH3
CH3 CH3
i.
3 stereogenic centers
Each indicated C bonded to
4 different groups =
2 stereogenic centers
H
e.
CH3 C CH2CH3
D
OH
bonded to 4 different groups
1 stereogenic center
HO
j.
OH
OH
OH
O
HO
OH
OH
f.
OH
OH
OH
Each indicated C bonded to 4 different groups =
6 stereogenic centers
Each indicated C bonded to 4 different groups =
5 stereogenic centers
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Chapter 5–16
5.40 Stereogenic centers are circled.
Eight constitutional isomers:
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
5.41
a.
H2N H
NH2
H NH2
amphetamine
O
O
CH3
COOH
b.
H
COOH
H CH3
O
HOOC
ketoprofen
5.42 Draw a molecule to fit each description.
a.
OH
O
c.
b.
O
alcohol
ketone
cyclic ether
5.43 Assign priority based on the rules in Answer 5.13.
a. OH, NH2
c. CH(CH3)2, CH2OH
higher atomic number
higher priority
b. CD3, CH3
D higher mass than H
higher priority
e. CHO, COOH
C bonded to O
higher priority
d. –CH2Cl, –CH2CH2CH2Br
C has 3 bonds to O.
higher priority
f. CH2NH2, NHCH3
higher atomic number
higher priority
C bonded to Cl
higher priority
5.44 Assign priority based on the rules in Answer 5.13.
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. –CCH > –CH=CH2 > –CH(CH3)2 > –CH2CH3
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Stereochemistry 5–17
5.45 Use the rules in Answer 5.15 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
3 H
Cl
C
d.
CH2CH3
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
HOOC
C
f.
CH3
SH
H
HO
NH2
H
C
CH3
C
H
g.
H
S
S
CH3
HO
H
S, R
R, R
Cl
h.
Cl
S
5.46
a. (3R)-3-methylhexane
c. (3R,5S,6R)-5-ethyl-3,6-dimethylnonane
CH3 H
R
CH3 H
S
CH3 H
b. (4R,5S)-4,5-diethyloctane
4R
R
H CH2CH3
d. (3S,6S)-6-isopropyl-3-methyldecane
H CH2CH3
H CH(CH3)2
S
S
CH3CH2 H
5S
H CH3
5.47
5
a.
6
H
3
6R
9
1
b.
4
2
7
4R
5
6R
3
5S
3R
1
10
c.
9
7
1
S
(3S)-3-methylhexane
(4R,6R)-4-ethyl-6-methyldecane
(3R,5S,6R)-5-isobutyl-3,6-dimethylnonane
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Chapter 5–18
5.48 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
5.49
S
a.
OH
NH Cl
b.
H NH2
HO
CH3CH2O2C
S CH3
COOH
ketamine
L-dopa
S
enalapril
O
N
N
H
c.
O
CO2H
S
5.50
CO2CH3
H
H
N
methylphenidate
H
N
H
CO2CH3
H
H
R,R
S,S
5.51
a. 1R,2S C1
OH
d.
NHCH3
OH
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
S
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Stereochemistry 5–19
5.53
O
a. CH3CH(OH)CH(OH)CH2CH3
c.
b. CH3CH2CH2CH(CH3)2
OH
OH
HO
OH
HO
0 stereogenic centers
4 stereogenic centers
24 = 16 possible stereoisomers
2 stereogenic centers
22 = 4 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
Br
Br
C
B
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
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Chapter 5–20
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
NH2
H
I H
I H
H2N
OH
d.
C
C
H
HO
OH
H
diastereomer
I H
H2N
H2N
or
HO
enantiomer
CH2OH
diastereomer
HO
CH3
or
C
H
OH
enantiomer
c.
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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Stereochemistry 5–21
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
OHC
H
HO
C2 C3
OH
H
C
R
C
R
OHC
a.
CH2OH
HO
D-erythrose
C
S
H
HOCH2
CH2OH
C
R
b.
HO
H
OH
2S,3R
diastereomer
2R,3R
H
C
R
CHO
OHC
C
R
H
C
HO S
H
OH
2R,3R
identical
OHC
H
OH
c.
C
S
d.
CH2OH
H
OH
C
H R
HO
2S,3S
enantiomer
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
H
CHO
CH2OH
H
OH
HO
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
H
H
OH
CHO
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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Chapter 5–22
5.61
R
R
trans
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
CH3
and
a.
g.
C
and
C
H
Br
enantiomers
H
2S,3S
CH3
b.
CH3
CHO
C
H
HO
OH
OH
OH
1,4-cis
1,4-trans
diastereomers
i.
C
H
HO
H
OH
and
6 H's
12 H's
different molecular formulas
not isomers
2R,3R
one different configuration
diastereomers
C
d.
CH3
H
j.
BrCH2
and
and
Cl
k.
and H
Cl
Br on end
HO CH3
Cl
Cl
Br
H
and
C
H
enantiomers
H
1,3-trans
diastereomers
1,3-cis
mirror images
not superimposable
enantiomers
C
C
CH3
CH3
and
I
HOCH2
BrCH2
CH2OH
CH3
enantiomers
different molecular formulas
not isomers
f.
H
HO
H
H
C
2R,3S
e.
CH3
2S,3S
CH3
and
H
C
and
HO
OHC
C
C
H
h.
same molecular formula
different connectivity
constitutional isomers
c.
H
Br
H
Br
Br
identical
and
CH3
CH3
CH3
l.
Br
I
and
C
H
CH2Br
CH3
CH2OH
C
Br H
Br in middle
different connectivity
constitutional isomers
H
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Stereochemistry 5–23
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.
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
[] mixture
x 100%
[] 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%
[] mixture
x 100%
e. 60% =
–165
120
x 100% = 73% ee
165
c. [] = +165
d. 80% 20% = 60% ee
[] mixture = 99
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Chapter 5–24
5.65
OH
HO
OH
O
HO
O
O
OH
O
HO
HCl, H2O
OH
amygdalin
(laetrile)
COOH
only one of the
products formed
CN
mandelic acid
OH
a. The 11 stereogenic centers are circled. Maximum number of stereoisomers = 211 = 2048
b. Enantiomers of mandelic acid:
HO H
H OH
HOOC
COOH
R
c. 60% 40% = 20% ee
20% = [] mixture/154 x 100%
[] mixture = 31
+50
d. ee =
+154
S
x 100% = 32% ee
[] 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 7 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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Stereochemistry 5–25
5.67
c. diastereomer
a. Each stereogenic center is circled.
O
H S
N
N
N S
H
O
N
OH
CONH2 O
saquinavir
Trade name: Invirase
N
S
S
O
H
N
HS
R
H
N
H
O
N
OH
CONH2 O
H
NH
H
NH
(CH3)3C
(CH3)3C
d. constitutional isomer
b. enantiomer
O
H
N
N
N
H
O
N
OH
CONH2 O
N
H
H
N
H
O
O
H
NH
(CH3)3C
new amine
NH
O
H
N
H
new aldehyde
N
OH
O
H
NH
(CH3)3C
5.68
HO
a.
HO
O
HO
O
OH
re-draw
more stable ring
all groups equatorial
salicin
b. diastereomer
O
HO
HO
O
re-draw
OH
O
HO
HO
OH
All other groups on the ring are equatorial.
HO
OH
O
OH
axial
HO
c. enantiomer
HO
O
OH
OH
OH
HO
OH
O
HO
HO
O
HO
O
OH
OH
re-draw
HO
HO
O
O
OH
HO
OH
In the enantiomer, all groups are still
equatorial, but all down bonds are up bonds
and all up bonds are down bonds.
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Chapter 5–26
5.69 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
CH3
C C C
CH3
H
H
CH3
HC C C C CH C CH CH CH CH CHCH2CO2H
C C C
B
allene
H
no plane of symmetry
mycomycin
re-draw
chiral
A
CH CH CH CHCH2CO2H
HC C C C
These two substituents are at 90o to these two substituents.
Allene A contains two planes of symmetry,
making it achiral.
C C C
H
H
The substituents on each end of the allene in mycomycin
are different. Therefore, mycomycin is chiral.
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.
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Stereochemistry 5–27
5.71
racemic mixture of
2-phenylpropanoic acid
salts formed by proton transfer
COO
COOH
C
H NH2
H
CH3
S
C
+
(R)-sec-butylamine
diastereomers
COO
COOH
H NH2
C
+
R
R
S
enantiomers
H
CH3
H
CH3
+
H NH3
(R)-sec-butylamine
H
CH3
+
H NH3
C
R
R
These salts are diastereomers, and they are now
separable by physical methods since they have
different physical properties.
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Reactions
139
Understanding Organic Reactions 6–1
C
Reeaaccttiioonnss
Chhaapptteerr 66:: U
Unnddeerrssttaannddiinngg O
Orrggaanniicc R
W
Wrriittiinngg oorrggaanniicc rreeaaccttiioonnss ((66..11))
• 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.
TTyyppeess ooff rreeaaccttiioonnss ((66..22))
[1] Substitution
C Z
+ Y
C Y
+
Z
Z = H or a heteroatom
Y replaces Z
[2] Elimination
C
C
X
Y
+ reagent
Two bonds are broken.
[3] Addition
C C
+
+
C C
X Y
bond
X Y
C C
X Y
This bond is broken.
IIm
mppoorrttaanntt ttrreennddss
Values compared
Bond dissociation
energy and bond
strength
Two bonds are formed.
Trend
The higher the bond dissociation energy, the stronger the bond
(6.4).
Increasing size of the halogen
CH3 F
Ho = 456 kJ/mol
CH3 Cl
CH3 Br
351 kJ/mol
293 kJ/mol
Increasing bond strength
CH3
I
234 kJ/mol
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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
Go > 0
Go reactants
Keq < 1
more stable reactants
Free energy
Free energy
Equilibrium always favors the species lower in energy.
Go reactants
Go < 0
Go products
Equilibrium favors the starting materials.
Keq > 1
more stable products
Equilibrium favors the products.
R
Reeaaccttiivvee iinntteerrm
meeddiiaatteess ((66..33))
• 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
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Understanding Organic Reactions 6–3
EEnneerrggyy ddiiaaggrraam
mss ((66..77,, 66..88))
Energy
transition state
Ea
reactants
Ea determines the rate.
Ho is the difference in bonding energy
between the reactants and products.
Ho
products
Reaction coordinate
C
Coonnddiittiioonnss ffaavvoorriinngg pprroodduucctt ffoorrm
maattiioonn ((66..55,, 66..66))
Variable Value
Meaning
Keq
Keq > 1
More product than starting material is present at equilibrium.
Go
Go < 0
The energy of the products is lower than the energy of the reactants.
Ho
Ho < 0
Bonds in the products are stronger than bonds in the reactants.
So
So > 0
The product is more disordered than the reactant.
EEqquuaattiioonnss ((66..55,, 66..66))
Go = 2.303RT log Keq
Keq depends on the energy difference
between reactants and products.
R = 8.314 J/(K•mol), the gas constant
T = Kelvin temperature (K)
Go
free energy
change
=
Ho
TSo
change in
bonding energy
T = Kelvin temperature (K)
FFaaccttoorrss aaffffeeccttiinngg rreeaaccttiioonn rraattee ((66..99))
Factor
Effect
energy of activation
higher Ea slower reaction
concentration
higher concentration faster reaction
temperature
higher temperature faster reaction
change in
disorder
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Chapter 6–4
C
Chhaapptteerr 66:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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 bond is formed.
[3] In an addition reaction, elements are added to the starting material.
OH
O
Br
c.
a.
CH3
C
Br replaces OH =
substitution reaction
b.
CH3
CH3
C
CH2Cl
Cl replaces H =
substitution reaction
H
O
O
d.
CH3CH CHCH3
CH3CH2CH(OH)CH3
OH
bond formed
elements lost
(H + OH)
elimination reaction
addition of 2 H's
addition reaction
6.2 Alkenes undergo addition reactions.
HCl
CH2 CH2
CH3CH2Cl
Br2
CH2 CH2
BrCH2CH2Br
addition of 1 H and 1 Cl
addition reaction
addition of 2 Br's
addition reaction
6.3 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, Br.
Electrons go to the more
electronegative atom, O.
Electrons go to the more
electronegative atom, C.
CH3
CH3 C
OH
Br
CH3CH2
CH3
carbocation
Li
carbanion
carbocation
6.4 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
+
N N
c.
CH3 C
CH3
+
Br
CH3 C Br
CH3
b.
CH3
+
CH3
CH3 CH3
d.
HO OH
CH3
2 HO
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Understanding Organic Reactions 6–5
6.5 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
CH3 SH
or
S is larger than O.
longer,
higher bond weaker bond
dissociation energy
c. (CH3)2C O
CH3 OCH3
or
higher bond
dissociation energy
single bond
fewer electrons
6.6 To determine Ho 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 Ho for the reaction.
A positive Ho means the reaction is endothermic. A negative Ho means the reaction is
exothermic.
a. CH3CH2 Br + H2O
CH3CH2 OH
+ HBr
[2] Bonds formed
[1] Bonds broken
Ho (kJ/mol)
Ho (kJ/mol)
CH3CH2 Br
+ 285
CH3CH2 OH
389
H OH
+ 498
H Br
368
+ 783 kJ/mol
Total
757 kJ/mol
Total
[3] Overall Ho =
sum in Step [1]
+
sum in Step [2]
+ 783 kJ/mol
757
kJ/mol
ANSWER: + 26 kJ/mol
endothermic
b. CH4 + Cl2
CH3Cl + HCl
[1] Bonds broken
[2] Bonds formed
Ho (kJ/mol)
Ho (kJ/mol)
CH3 H
+ 435
CH3 Cl
351
Cl Cl
+ 242
H Cl
431
Total
+ 677 kJ/mol
Total
782 kJ/mol
[3] Overall Ho =
sum in Step [1]
+
sum in Step [2]
+ 677 kJ/mol
782 kJ/mol
ANSWER: 105 kJ/mol
exothermic
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Chapter 6–6
6.7 Use the directions from Answer 6.6. 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
Ho (kJ/mol)
CH3 H
O O
Ho (kJ/mol)
OC O 535 x 2 = 1070
+ 435 x 4 = + 1740
+ 497 x 2 = + 994
sum in Step [1]
+
sum in Step [2]
HO H 498 x 4 = 1992
+ 2734 kJ/mol
Total
[3] Overall Ho =
[2] Bonds formed
[1] Bonds broken
+ 2734 kJ/mol
3062 kJ/mol
Total
3062 kJ/mol
ANSWER:
b. 2 CH3CH3 + 7 O2
4 CO2 + 6 H2O
[3] Overall Ho =
[2] Bonds formed
[1] Bonds broken
Ho (kJ/mol)
CH3CH2 H + 410 x 12 = + 4920
O O
+ 497 x 7 = + 3479
C C
+ 368 x 2 =
Total
328 kJ/mol
+736
Ho (kJ/mol)
OC O
535 x 8 = 4280
sum in Step [1]
+
sum in Step [2]
HO H 498 x 12 = 5976
Total
10256 kJ/mol
+ 9135 kJ/mol
+ 9135 kJ/mol
10256 kJ/mol
ANSWER: 1121 kJ/mol
6.8 Use the following relationships to answer the questions:
Keq = 1 then G° = 0; Keq > 1 then G° < 0; 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 = 102
is larger than Keq = 105, so G° is lower.
6.9 Use the relationships from Answer 6.8.
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.
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Understanding Organic Reactions 6–7
6.10 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.
6.11
H
H
OCH3
Keq = 2.7
OCH3
a. The Keq is > 1 and therefore the product (the conformation on the right) is favored at
equilibrium.
b. The G° for this process must be negative since the product is favored.
c. G° is somewhere between 0 and 6 kJ/mol.
6.12 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.
6.13
a. False.
b. True.
c. False.
d. True.
e. False.
The reaction is endothermic.
This assumes that G° is approximately equal to H°.
Keq < 1
The starting material is favored at equilibrium.
6.14
a. True.
b. False. G° for the reaction is negative.
c. True.
d. False. The bonds in the product are stronger than the bonds in the starting material.
e. True.
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Chapter 6–8
6.15
Energy
transition
state
Ea
product
H°
starting
material
Reaction coordinate
6.16 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
CH3
b. CH3O H
H2O
CH3 C
CH3
CH3
+
transition state: CH3 C
OH
CH3O
transition state:
+
CH3O
H2O
H
OH
OH2
CH3
6.17
transition
state 2: E
Energy
transition
state 1: D
B
Ea(A–B)
Ea(A–C)
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 lower-energy
product.
d. See labels.
e. See labels.
f. See labels.
C
Reaction coordinate
6.18
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 since there are two energy barriers.
See labels.
See labels.
One reactive intermediate is formed (see label).
The first step is rate determining since its transition
state is at higher energy.
f. The overall reaction is endothermic since the energy
of the products is higher than the energy of the
reactants.
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Reactions
147
Understanding Organic Reactions 6–9
6.19
Energy
Ea(B–C)
B
Ea(A–B)
relative energies: C < A < B
B
C is rate-determining.
A
C
Reaction coordinate
6.20 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 since 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.
Energy
6.21 The Ea of an endothermic reaction is at least as large as its H° because the Ea essentially
“includes” the H° in its total. The Ea measures the difference between the energy of the starting
material and the energy of the transition state, and in an endothermic reaction, the energy of the
products is somewhere in between these two values.
Ea
H° = (+) value for an endothermic reaction
Reaction coordinate
6.22
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.
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Chapter 6–10
6.23 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.
6.24 The rate equation is determined by the rate-determining step.
a.
CH3CH2 Br
b.
(CH3)3C
+
CH2 CH2
OH
(CH3)3C +
Br
slow
+
OH
fast
Br
one step
rate = k[CH3CH2Br][–OH]
+ H2O + Br
(CH3)2C
CH2
+ H2O
two steps
The slow step determines the rate equation.
rate = k[(CH3)3CBr]
6.25 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.
a.
H2O
CH2 CH2
H2 adds to the starting material.
I– not used up = catalyst.
CH3CH2OH
b.
I–
CH3Cl
CH3OH
–OH
H2SO4
OH
H2SO4 is not used up = catalyst.
substitutes for
O
H2
c.
Cl.
Pt
Pt not used up = catalyst.
6.26 Use the directions from Answer 6.1.
O
HO OH
O
H OH
c.
a.
addition of 2 H's
addition reaction
bond formed
elements lost
(H + OH)
elimination reaction
b.
H
H
Cl replaces H =
substitution reaction
H
Cl
O
d.
C
OH
O
Cl
H replaces Cl =
substitution reaction
C
H
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6.27
H
a. homolysis of
c. heterolysis of CH3 MgBr
b. heterolysis of CH3 O H
CH3 C H
H
H
CH3 C
H
CH3 O
CH3
H
H
MgBr
carbanion
radical
6.28 Use the rules in Answer 6.4 to draw the arrows.
Br
a.
Br
O
b.
CH3
Cl
CH3
Br
d.
+
Cl
e.
+
+ Br
Br Br
O
CH3 C CH3
c.
+
+
Cl
C
CH3
f.
CH3 Cl
CH3CH2 Br
+
H
CH3
C
CH3
CH3CH2OH +
OH
CH3
C H +
H
+ H2O
C C
OH
CH3
H
Br
H
6.29
a.
I
+
OH +
OH
H
OH
I
H C
c.
H
O
H
C
O
b. CH3 C CH2CH2CH3
OCH2CH3
CH3
C
d.
CH2CH2CH3
H
H
C H
H
Br
+
H
H
+ H2O +
C C
Br
H
H
C
Cl
H
H
+ HCl
OCH2CH3
6.30 Draw the curved arrows to identify the product X.
O
+
H Br
[1]
O H
+
Br
Br
[2]
OH
A
B
X
6.31 Follow the curved arrows to identify the intermediate Y.
CO2H [1]
O
CO2H
O
O
C
[2]
O
COOH
O
O
D
Y
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Chapter 6–12
6.32 Use the rules from Answer 6.5.
a.
I
Br CCl3
CCl3
largest halogen intermediate
weakest bond bond strength
b. H2N NH2
Cl CCl3
smallest halogen
strongest bond
single bond
weakest bond
HN NH
N N
double bond
intermediate
bond strength
triple bond
strongest bond
6.33 Use the directions from Answer 6.6.
a. CH3CH2 H
+ Br2
CH3CH2 Br
+ HBr
[2] Bonds formed
[1] Bonds broken
Ho (kJ/mol)
[3] Overall Ho =
Ho (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
[2] Bonds formed
Ho (kJ/mol)
CH3 H
c.
ANSWER: 51 kJ/mol
Ho (kJ/mol)
H OH
+ 435 kJ/mol
CH3 OH + HBr
CH3 Br
498 kJ/mol
+ 435 kJ/mol
498 kJ/mol
ANSWER: 63 kJ/mol
+ H2O
[2] Bonds formed
[1] Bonds broken
[3] Overall Ho =
Ho (kJ/mol)
[3] Overall Ho =
Ho (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
Total
ANSWER: 34 kJ/mol
H + CH3Br
[1] Bonds broken
[2] Bonds formed
Ho (kJ/mol)
CH3 H
791 kJ/mol
+ 435 kJ/mol
Ho (kJ/mol)
CH3 Br
293 kJ/mol
[3] Overall Ho =
+ 435 kJ/mol
293 kJ/mol
ANSWER: + 142 kJ/mol
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Understanding Organic Reactions 6–13
6.34
propane
propene
CH3 CH2CH3
CH3
CH2CH3
Ho = 356 kJ/mol
This bond is formed from two sp3
hybridized C's.
CH3 CH CH2
CH3
CH CH2
Ho = 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.
6.35
H
H
CH2 CH C H
H
CH2 CH C H
CH2 CH
C H
hybrid:
H
CH2 CH C H
H
6.36 The more stable radical is formed by a reaction with a smaller H°.
H
CH3 CH2
C H
CH3 CH2
C H
H
Ho = 410 kJ/mol = less stable radical
H
This C–H bond is stronger.
A
H
CH3 C
CH3
CH3 C
H
CH3
Ho = 397 kJ/mol = more stable radical
H
This C–H bond is weaker.
B
Since 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.
6.37 Use the bond dissociation energy for the CC bond in ethane as an estimate of the bond
strength in ethylene. Then you can estimate the bond strength as well.
CH3 CH3
o
H = 368 kJ/mol
CH2 CH2
635 – 368 = 267 kJ/mol = bond
o
H = 635 kJ/mol
6.38 Use the rules from Answer 6.10.
a. Keq = 0.5. Keq is less than one so the starting material is favored.
b. Go = 100 kJ/mol. Go is less than 0 so the product is favored.
c. Ho = 8.0 kJ/mol. Ho is positive, so the starting material is favored.
d. Keq = 16. Keq is greater than one so the product is favored.
e. Go = 2.0 kJ/mol. Go is greater than zero so the starting material is favored.
f. Ho = 200 kJ/mol. Ho is positive so the starting material is favored.
g. So = 8 J/(K•mol). So is greater than zero so the product is more disordered and favored.
h. So = 8 J/(K•mol). So is less than zero so the starting material is more disordered and
favored.
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6.39
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.
6.40
R
axial
R
equatorial
H
R
Keq
–CH3
–CH2CH3
–CH(CH3)2
–C(CH3)3
H
18
23
38
4000
a. The equatorial conformation is always present in the larger amount at equilibrium since 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 since this group has the highest Keq.
c. The cyclohexane with the –CH3 group will have the greatest amount of axial conformation at
equilibrium since this group has the lowest Keq.
d. The cyclohexane with the –C(CH3)3 group will have the most negative G° since it has the
largest Keq.
e. The larger the R group, the more favored the equatorial conformation.
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
H
severe 1,3-diaxial interactions with
the CH3 group and the axial H's
C
CH3
CH3
less severe 1,3-diaxial interactions
6.41 Calculate Keq, and then find the percentage of axial and equatorial conformations present at
equilibrium.
F (axial)
a.
H
fluorocyclohexane
1 part
F (equatorial)
H
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
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6.42 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
S° is positive.
products favored
decreased number of molecules
S° is negative.
starting material favored
increased number of molecules
H2O
S° is positive.
products favored
+
a.
b.
CH3
c.
(CH3)2C(OH)2
+
CH3CH3
CH3
(CH3)2C=O
d. CH3COOCH3
+
H2O
+
CH3COOH
+
CH3OH
no change in the number of molecules
neither favored
6.43 Use the directions in Answer 6.16 to draw the transition state. Nonbonded electron pairs are drawn
in at reacting sites.
Br
+
a.
OH
+
c.
Br
O
+
+ Br
transition
state:
+
BF3
transition
state:
Cl
F B Cl
d.
C
CH3
F
F
Cl
C H
+ H2O
transition
state:
H
CH3
CH3 +
C
CH3
H
+
C C
H
F
F B
CH3
H
CH3
NH3
O H NH2
transition
state:
F
b.
+
NH2
H3O
+
H
+
C H OH2
H
6.44
A
B
Reaction coordinate
• one step A B
• exothermic since
B lower than A
• low Ea
(small energy barrier)
H°
B
A
Reaction coordinate
• one step A B
• endothermic since
B higher than A
• high Ea
(large energy barrier)
A
B
EaA–B
Energy
Ea
Energy
Ea
H°
d.
c.
b.
Energy
Energy
a.
EaB–C
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.
Ea = 16 kJ/mol
A
H° = 80 kJ/mol
B
Reaction coordinate
• one step A B
• exothermic since
B lower than A
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Chapter 6–16
6.45
a.
CH3 H
b.
Cl + CH4
CH3 + HCl
Cl
CH3 + HCl
[1] Bonds broken
Ho (kJ/mol)
431 kJ/mol
H Cl
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.
c.
Ea = 16 kJ/mol
Energy
+ 435 kJ/mol
Ho (kJ/mol)
+ 435 kJ/mol
CH3 H
[3] Overall Ho =
[2] Bonds formed
16 kJ/mol
Energy
H°
= 4 kJ/mol
12 kJ/mol
4 kJ/mol
Reaction coordinate
Reaction coordinate
6.46
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
A
E
G
Reaction coordinate
6.47
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, H° is negative, and the products are lower in energy than
the starting materials.
transition
state
O
CH3
C O
CH3
H
O C CH3
CH3
Energy
transition state:
Ea
starting
materials
Reaction coordinate
H°
products
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6.48
H
H
H
H
H Cl
+H
[1]
H
H
+ Cl
Cl
[2]
H
a. Step [1] breaks one bond and the HCl bond, and one CH bond is formed. The H° for this
step should be positive since more bonds are broken than formed.
b. Step [2] forms one bond. The H° for this step should be negative since one bond is formed
and none is broken.
c. Step [1] is rate-determining since it is more difficult.
d. Transition state for Step [1]:
Transition state for Step [2]:
H
H
H
H
+
Cl
+ H
Cl H
e.
Energy
Ea2
H°1 is positive.
Ea1
H°2 is negative.
H°overall is negative.
Reaction coordinate
6.49
B
C
Energy
Ea
Ea
(CH3)3C +
A
+ H 2O
+ I
Ea
(CH3)3C
+ HI
OH (CH3)3C
+
OH2
H°3
H°2
H°1 = 0
+ I
H°overall
(CH3)3C
Reaction coordinate
a. The reaction has three steps, since there are three energy barriers.
b. See above.
I
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Chapter 6–18
c. Transition state A (see graph for
location):
+
(CH3)3C
OH
H
Transition state B:
(CH3)3C
+
I
Transition state C:
+ I
OH2
+
(CH3)3C
d. Step [2] is rate-determining since this step has the highest energy transition state.
6.50 Ea, concentration, catalysts, rate constant, and temperature affect reaction rate so (c), (d), (e), (g),
and (h) affect rate.
6.51
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.
6.52
O
acetyl
C
Cl
chloride CH3
CH3O
[1]
slow
O
CH3O
O
[2]
CH3 C Cl
fast
CH3
C
+ Cl–
OCH3
methyl acetate
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).
6.53
a.
b.
c.
d.
e.
f.
True: Increasing temperature increases reaction rate.
True: If a reaction is fast, 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.
6.54
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 since it
is slow and therefore rate-determining: Rate = k[(CH3)3CI]
c. Possibility [1] is second order; possibility [2] is first order.
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Understanding Organic Reactions 6–19
d. These rate equations can be used to show which mechanism is plausible by changing the
concentration of –OH. If this affects the rate, possibility [1] is reasonable. If it does not affect
the rate, possibility [2] is reasonable.
CH3
e.
CH3 C
CH2
I
Energy
–
H
–
OH
Ea
H°
A = (CH3)3CI + –OH
B = (CH3)2C=CH2 + I– + H2O
B
A
Reaction coordinate
transition state
[1]
f.
transition state
[2]
Energy
B
Ea[1]
H°[1]
(CH3)3C+
+ I–
+ –OH
CH3
[1] CH3 C +CH3
I –
Ea[2]
H°[2]
H°overall
(CH3)3CI
(CH3)2C=CH2
+ H2O
CH3
[2] CH3 C CH2
+
H
OH
–
Reaction coordinate
6.55 The difference in both the acidity and the bond dissociation energy of CH3CH3 versus HCCH 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.
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Chapter 6–20
6.56
Ha Hb
a.
Hb
Hb
Hb
C C CH3
C C CH3
C C CH3
C C CH3
H H
H H
H H
H H
Hb
b.
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 the 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.
6.57 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.
6.58
O
CH3
C
O
+ CH3CH2OH
OH
CH3
C
OCH2CH3
+ H2O
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.
6.59
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
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Nucleophilic Substitution
159
Alkyl Halides and Nucleophilic Substitution 7–1
C
Chhaapptteerr 77:: A
Allkkyyll H
Haalliiddeess aanndd N
Nuucclleeoopphhiilliicc SSuubbssttiittuuttiioonn
G
Geenneerraall ffaaccttss aabboouutt aallkkyyll hhaalliiddeess
• 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).
TThhee cceennttrraall tthheem
mee ((77..66))
• 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 CNu bond
comes from the nucleophile.
•
•
One bond is broken and one bond is formed.
There are two possible mechanisms: SN1 and SN2.
SSNNN11 aanndd SSNNN22 m
meecchhaanniissm
mss ccoom
mppaarreedd
SN2 mechanism
One step (7.11B)
•
[2] Alkyl halide
•
Order of reactivity: CH3X >
RCH2X > R2CHX > R3CX
(7.11D)
•
Order of reactivity: R3CX >
R2CHX > RCH2X > CH3X
(7.13D)
[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
•
•
•
favored by polar protic solvents
(7.17D)
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Chapter 7–2
Increasing rate of the SN1 reaction
H
H
H
R
H C Br
R C Br
R C Br
R C Br
H
H
R
R
1o
2o
3o
both
SN1 and SN2
SN1
methyl
SN2
Increasing rate of an SN2 reaction
IIm
mppoorrttaanntt ttrreennddss
• The best leaving group is the weakest base. Leaving group ability increases 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
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Alkyl Halides and Nucleophilic Substitution 7–3
IIm
mppoorrttaanntt pprriinncciipplleess
•
Principle
Electron-donating groups (such as R
groups) stabilize a positive charge (7.14A).
•
Example
3 Carbocations (R3C+) are more stable
than 2o carbocations (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.
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Chapter 7–4
C
Chhaapptteerr 77:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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.
CH3CH2CH2CH2CH2 Br
F
b.
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.
F
This F is bonded to a C which is
not bonded to any other C's. Therefore,
it cannot be classified as 1°, 2°, or 3°. HO
S
O
OCOC2H5
H
F
O
F
H
bonded to C bonded to 3 C's
3° alkyl halide
bonded to C bonded to 2 C's
2° alkyl halide
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
[3] 3-chloro-2-methylpentane
2 Cl
3-chloro
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Alkyl Halides and Nucleophilic Substitution 7–5
2-bromo
b.
Br
[1]
[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.
F
[1]
F
[2]
[3] 2-fluoro-5,5-dimethylheptane
5,5-dimethyl
1 2 3 4
7
7 carbon alkane = heptane
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.
a. 3-chloro-2-methylhexane
[1]
6 carbon alkane
[2]
methyl at C2
1 2 3 4 5 6
Cl
chloro at C3
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
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Chapter 7–6
e. propyl chloride
[1] 3 carbon alkyl group
[2] chloride on end
CH3CH2CH2
CH3CH2CH2 Cl
f. sec-butyl bromide
[1] 4 carbon alkyl group
[2] bromide
CH3 CHCH2CH3
CH3 CHCH2CH3
Br
7.6 Boiling points of alkyl halides increase as the size (and polarizability) of X increases. Remember:
stronger intermolecular forces = higher boiling point.
a.
smallest halogen
least polarizable
lowest boiling point
b.
CH3(CH2)4CH3
CH3CH2CH2I
middle size halogen
intermediate
boiling point
largest halogen
most polarizable
highest boiling point
CH3(CH2)5Br
weakest forces
nonpolar
lowest boiling point
7.7
CH3CH3CH2Cl
CH3CH2CH2F
VDW, DD forces
intermediate
boiling point
CH3(CH2)5OH
OH is capable of hydrogen bonding.
strongest forces
highest boiling point
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.
Cl
b.
Cl
sp3 C–Cl bond
intermediate
boiling point
lowest boiling point
Br
larger halogen, sp3 C–Br bond
highest boiling point
7.8 Since more polar molecules are more water soluble, look for polarity differences between
methoxychlor and DDT.
methoxychlor
DDT
H
CH3O
C
H
OCH3
CCl3
2 methoxy groups
more polar
The O atoms can hydrogen bond to H2O.
more biodegradable
Cl
C
Cl
CCl3
2 chloro groups
less polar
readily soluble in an organic medium
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Alkyl Halides and Nucleophilic Substitution 7–7
7.9 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 bonds.
[3] Substitute the nucleophile for the leaving group on the electrophilic carbon.
+
a.
Br
leaving group
+ Br
OCH2CH3
OCH2CH3
nonbonded e pairs
nucleophile
Cl
OH
b.
+
leaving group
nonbonded e pairs
nucleophile
+
I
c.
leaving group
+ Na+Cl–
Na+ OH
N3
+ I
N3
nonbonded e pairs
nucleophile
Br
CN
+
d.
leaving group
+ Na+Br–
Na+ CN
nonbonded e pairs
nucleophile
7.10 Use the steps from Answer 7.9 and then draw the proton transfer reaction.
a.
+
Br
N(CH2CH3)3
substitution
+
N(CH2CH3)3 + Br
nucleophile
leaving group
+
b. (CH3)3C Cl
H2O
substitution
(CH3)3C
proton
+ Cl
(CH3)3C
transfer
H
nucleophile
leaving group
O H
7.11 Draw the structure of CPC using the steps from Answer 7.9.
N
nucleophile
+
Cl
substitution
+
N
leaving group
+
CPC
Cl
O H
+ HCl
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7.12 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–
c. H2O, H2S
neutral compound
less basic
better leaving group
farther down a column
of the periodic table
less basic
better leaving group
farther down a column
of the periodic table
less basic
better leaving group
7.13 Good leaving groups include Cl–, Br–, I–, H2O.
a.
CH3CH2CH2 Br
c. CH3CH2CH2 OH2
b. CH3CH2CH2OH
Br is a good
leaving group.
No good leaving group.
is too strong a base.
H2O is a good
leaving group.
OH
d. CH3CH3
No good leaving group.
H is too strong a base.
7.14 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
+
CN
nucleophile
pKa (HCN) = 9.1
+
CH3CH2 Br
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.
7.15 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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Alkyl Halides and Nucleophilic Substitution 7–9
7.16 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. CH3 , HO
c. CH3NH2, CH3OH
d.
Across a row of the periodic
A negatively charged
Across a row of the periodic
nucleophile is stronger table, nucleophilicity decreases table, nucleophilicity decreases
than its conjugate acid. with species of the same charge. with species of the same charge.
stronger nucleophile
stronger nucleophile
stronger nucleophile
C
CH3
O
CH3CH2O
same attacking
atom (O)
stronger base
stronger
nucleophile
7.17 Polar protic solvents are capable of H-bonding, and therefore must contain a H bonded to an
electronegative O or N. Polar aprotic solvents are incapable of H-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
7.18 • 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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7.19 The stronger base is the stronger nucleophile except in polar protic solvents when nucleophilicity
increases down a column. For other rules, see Answers 7.16 and 7.18.
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
OH
weakest nucleophile
strongest nucleophile
weaker base than
intermediate nucleophile
7.20 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
(CH3)2CHCH2CH2
C CH
CCH replaces Br.
HCC is needed.
OCH2CH3 replaces Br.
CH3CH2O is needed.
7.21 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).
7.22 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 b.
Br
+
SH
SH
+
Br
SH
Br
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7.23 All SN2 reactions have one step.
Cl CH3CH2CH2
Energy
CH3O Ea
CH3CH2CH2 Cl
H°
+ OCH3
CH3CH2CH2 OCH3 + Cl
Reaction coordinate
7.24 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.
D CH CH
2
3
C
+
Br
CH3CH2O
OCH2CH3
H
b.
C
+
I
CN
CN
H
7.25 Increasing the number of R groups increases crowding of the transition state and decreases the rate
of an SN2 reaction.
Cl
or
a. CH3CH2 Cl
1° alkyl halide
CH3 Cl
b.
methyl halide
faster reaction
or
2° alkyl halide
c.
Cl
Br
2° alkyl halide
faster reaction
1° alkyl halide
faster reaction
or
Br
3° alkyl halide
7.26
These three methyl groups make the alkyl halide sterically hindered.
This slows the rate of an SN2 reaction even though it is a 1° alkyl halide.
CH3
CH3 C CH2Br
CH3
7.27
+
H
N
N
H
CH3 SR2
H
N
A
+ SR2
N H
loss of
CH3
a proton
H
N
N
CH3
nicotine
7.28 In a first-order reaction, the rate changes with any change in [RX]. The rate is independent of
any change in [nucleophile].
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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Chapter 7–12
7.29 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
C
Br
(CH3)2CH
CH2CH3
CH3
H
CH3CH2
Cl
CH2CH3
+
HBr
OH
enantiomers
CH3COO
CH3
(CH3)2CH
CH2CH3
nucleophile
b.
C
+
OH
H
CH3
CH3CH2
H
OOCCH3
+
OOCCH3
CH3CH2
+
Cl–
CH3
diastereomers
leaving group
7.30 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
7.31 For carbocations: Increasing number of R groups = Increasing stability.
+
+
CH3CH2CH2CH2
+
CH3CHCH2CH3
CH3 C CH3
CH3
1° carbocation
least stable
2° carbocation
intermediate
stability
3° carbocation
most stable
7.32 For carbocations: Increasing number of R groups = Increasing stability.
+
a.
+
+
(CH3)2CHCH2CH2
(CH3)2CHCHCH3
1° carbocation
least stable
2° carbocation
intermediate
stability
CH2
+
(CH3)2CCH2CH3
3° carbocation
most stable
+
b.
+
1° carbocation
least stable
2° carbocation
intermediate
stability
3° carbocation
most stable
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7.33 The rate of an SN1 reaction increases with increasing alkyl substitution.
a.
or
(CH3)3CBr
(CH3)3CCH2Br
CH3
1° alkyl halide
3° alkyl halide
faster SN1 reaction slower SN1 reaction
b.
Br
or
2° alkyl halide
slower SN1 reaction
Br
Br
c.
or
3° alkyl halide
faster SN1 reaction
Br
2° alkyl halide
slower SN1 reaction
3° alkyl halide
faster SN1 reaction
7.34 • 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
SN 2
2° alkyl halide
SN1 and SN2
Br
c.
3° alkyl halide
SN1
7.35 • 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
2° alkyl halide
Both SN1 and SN2
are possible.
OCH3
7.36 First decide whether the reaction will proceed via an SN1 or SN2 mechanism. Then draw the
products with stereochemistry.
+
a.
H Br
2° alkyl halide
SN1 and SN2
H2O
Weak nucleophile
favors SN1.
+ HBr
+
H OH
HO H
enantiomers
SN1 = racemization at the
stereogenic C
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Cl
HC C
C C H
+
b.
+
Cl
SN2 = inversion at the stereogenic C
D H
H D
1° alkyl halide
SN2 only
7.37 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
d. CH3CH2CH2OH
or
weaker base
better leaving group
7.38
+
OH2
weaker base
better leaving group
(CH3)3CI
or
(CH3)3C
CH3CH2CH2 OCOCH3
weaker base
better leaving group
• 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.
b. CH3CN
polar aprotic solvent
no O–H or N–H bond
favors SN2
a. CH3CH2OH
polar protic solvent
contains an O–H bond
favors SN1
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
7.39 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.
a.
(CH3)3CBr
3o RX
+
H2O
H2O
or
H2O
(CH3)3COH
+
HBr
– SN1 reaction (CH3)2C=O
+ CH3OH
b.
Cl
CH3OH
+
or
DMSO
HCl
Br
+
OH
1o RX – SN2 reaction
d.
+
H Cl
CH3O
H2O
OH
+
Br
or
CH3OH
+
or
2o RX strong nucleophile
SN2 reaction
DMF [HCON(CH3)2]
Polar aprotic solvent
increases the rate of an
SN2 reaction.
DMF
HMPA
CH3OH
Polar protic solvent
increases the rate of an
SN1 reaction.
OCH3
3o RX – SN1 reaction
c.
Polar protic solvent
increases the rate of an
SN1 reaction.
H OCH3
Cl
HMPA [(CH3)2N]3P=O
Polar aprotic solvent
increases the rate of an
SN2 reaction.
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Alkyl Halides and Nucleophilic Substitution 7–15
7.40 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
+
Br
Strong nucleophile
favors SN2.
2° alkyl halide
SN1 or SN2
I
c.
N3
N3
3° alkyl halide
SN1
OCH3
CH3OH
+
SN2 reaction = inversion at the stereogenic center
The leaving group was "up."
The nucleophile attacks from below.
+
SN1 reaction
HI
Weak nucleophile
favors SN1.
d.
+
+
H2O
Weak nucleophile
favors SN1.
Cl
3° alkyl halide
SN1
OH
+ HCl
SN1 reaction
forms two enantiomers.
HO
7.41 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
7.42
Na+ CN
a.
CN
carbon
framework
CN
nucleophile
b. (CH3)3CCH2CH2
carbon
framework
Cl
SH
(CH3)3CCH2CH2
nucleophile
Cl
Na+ SH
(CH3)3CCH2CH2
SH
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Chapter 7–16
OH
Cl
OH
Na+ OH
c.
nucleophile
carbon
framework
Na+ C
d.
CH3CH2 C C H
carbon
framework
CH
CH3CH2 C C H
CH3CH2 Cl
nucleophile
7.43
Cl CH2CH3
CH3O
CH3OCH2CH3
Cl CH3
CH3CH2O
CH3OCH2CH3
7.44 Use the directions from Answer 7.4 to name the compounds.
[1]
a.
[2]
CH3
CH3 C CH2CH2 F
[3] 1-fluoro-3,3-dimethylbutane
1
CH3 C CH2CH2 F
3 CH3
CH3
1-fluoro
3,3-dimethyl
4 carbon alkane = butane
b.[1]
CH3
[2]
3-ethyl
3
I
2
2-methyl
[2]
3-ethyl-1-iodo-2-methylhexane
1-iodo
6 carbon alkane = hexane
c. [1] (CH3)3CCH2Br
[3]
1 I
CH3
[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
I
5 carbon cycloalkane =
cyclopentane
f.
[1]
Cl
1-bromo
[3] cis-1-bromo-3-iodocyclopentane
3-iodo
1
Cl
[2]
trans-1,2-dichloro
Cl
6 carbon cycloalkane =
cyclohexane
[3] 6-bromo-2-chloro-6-methyloctane
Cl
6-bromo 2-chloro
Br
e. [1]
1
Cl
2
[3] trans-1,2-dichlorocyclohexane
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Alkyl Halides and Nucleophilic Substitution 7–17
g. [1] (CH3)3CCH2CH(Cl)CH2Cl
CH3
CH3
[2]
CH3
CH3 C CH2 C CH2Cl
CH3
H
[3] 1,2-dichloro-4,4-dimethylpentane
CH3 C CH2 C CH2Cl
H
Cl
Cl
4,4-dimethyl
1,2-dichloro
5 carbon alkane = pentane
h.
[1]
[2]
4
I H
2 1
[3] (2R)-2-iodo-4,4-dimethylhexane
I H
6 carbon alkane = hexane
(Indicate the R/S
designation also)
4,4-dimethyl
2
3
(2R)-2-iodo
I H4
Clockwise
R
1
7.45 To work backwards to a structure, use the directions in Answer 7.5.
e. 1-bromo-4-ethyl-3-fluorooctane
a. isopropyl bromide
Br
CH3 CHCH3
Bromine on middle C
makes it an isopropyl group.
b. 3-bromo-4-ethylheptane
3
4-ethyl
4
3-bromo
Br
c. 1,1-dichloro-2-methylcyclohexane
1 Cl
1,1-dichloro
Cl
2
2-methyl
d. trans-1-chloro-3-iodocyclobutane
1-chloro
3 Cl
1
I
3-iodo
4-ethyl
Br
1
1-bromo
3
F
4
3-fluoro
f. (3S)-3-iodo-2-methylnonane
2-methyl
4
3
1
I H
3S
3-iodo
g. (1R,2R)-trans-1-bromo-2-chlorocyclohexane
1R
1-bromo
Br
1
Cl
2-chloro
2R
h. (5R)-4,4,5-trichloro-3,3-dimethyldecane
5R
3,3-dimethyl
Cl H
1
4
5
3
Cl Cl
4,4,5-trichloro
176
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7.46
H H
CH3
a.
CH3 C CH2CH2F
CH3
g.
d.
Br
1° halide
I
h.
I H
e.
2° halide
1° halide
2° halide
I
c.
1° halide
2° halide
2° halide
2° halide
Br
b.
C C Cl
Cl H
Cl
3° halide
(CH3)2CCH2
(CH3)3CCH2Br
Cl
1° halide
f.
Cl
Both are
2° halides.
7.47
1-chloro
3-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
2
*
Cl
2-chloropentane
[* denotes stereogenic center]
3
Cl H 4
1
Clockwise
"4" in back =
R
2
3
4
1 Cl H
Clockwise
"4" in front =
S
Two stereoisomers
3-methyl
3
2
3
*
2-chloro
Cl
2-chloro-3-methylbutane
[* denotes stereogenic center]
4H
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
Cl
Cl
1-chloro-2-methylbutane
[* denotes stereogenic center]
3
H
2
1
Clockwise
"4" in front =
S
3
1-chloro
1-chloro-3-methylbutane
2-chloro
2-methyl
2
3-methyl
Cl
4
H
Cl
2
1
Clockwise
"4" in back =
R
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7.48 Use the directions from Answer 7.6.
Br
a. (CH3)3CBr
or
CH3CH2CH2CH2Br
b.
larger surface area =
stronger intermolecular forces =
higher boiling point
I
c.
Br
or
or
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
CH3CH2CH2CH2OH + HBr
CH3CH2CH2CH2NH2 + HBr
CH3CH2CH2CH2I + Na+ Br
CH3CH2CH2CH2N3 + Na+ Br
7.50 Use the steps from Answer 7.9 and then draw the proton transfer reaction, when necessary.
O
a.
+
Cl
CH3
C
CH3
nucleophile
leaving group
I
b.
leaving group
+
I
c.
leaving group
d.
Cl
+
Cl
CN
+ NaI
O
O
Na+ –CN
nucleophile
H2O
OH + HI
nucleophile
+ CH3CH2OH
leaving group
+
O
nucleophile
OCH2CH3
+ HCl
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Br
OCH3
+ Na+ –OCH3
e.
nucleophile
leaving group
Cl
f.
CH3
+
leaving group
+ NaBr
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
e.
This has only C–C
and C–H bonds.
No good leaving group.
OH2
d.
Cl good leaving group
weak base
bad leaving group
NH2 is a strong base.
f.
good leaving group
H2O is a weak base.
CH3CH2NH2
CH3CH2CH2
I
I good leaving group
weak base
7.52 Use the rules from Answer 7.12.
a. increasing leaving group ability: NH2 < OH < F
c. increasing leaving group ability: Cl < Br < I
least basic
best leaving
group
most basic
worst leaving
group
most basic
worst leaving
group
b. increasing leaving group ability: NH2 < OH < H2O
most basic
worst leaving
group
least basic
best leaving
group
d. increasing leaving group ability: NH3 < H2O < H2S
most basic
worst leaving
group
least basic
best leaving
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 (CH3OH) = 15.5
+
NH2
Reaction will not occur.
stronger base
pKa (NH3) = 38
CH3CH2OCH3
+ I
weaker base
pKa (HI) = –10
Reaction will occur.
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Alkyl Halides and Nucleophilic Substitution 7–21
c.
OH
F
+ F
weaker base
pKa (HF) = 3.2
CN
d.
+
Reaction will not occur.
OH
stronger base
pKa (H2O) = 15.7
I
+ I
+
CN
Reaction will not occur.
stronger base
pKa (HCN) = 9.1
weaker base
pKa (HI) = –10
7.54
Br
A
SCH3 replaces Br.
SCH is needed.
3
SCH3
a.
b.
OCH(CH3)2 OCH(CH3)2 replaces Br.
OCH(CH )
3 2
needed.
C
c.
C
CH3
N(CH3)3
d.
Br
C CCH3 replaces Br.
C CCH3 is needed.
N(CH3)3 replaces Br.
N(CH3)3 is needed.
7.55 Use the directions in Answer 7.16.
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.
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
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
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
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7.56 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.
e.
no O–H or N–H bond
aprotic
contains O–H bond
protic
b.
CH2Cl2
c.
(CH3)2CHOH
d.
CH3NO2
no O–H or N–H bond
aprotic
f. HCONH2
contains an N–H bond
protic
NH3
contains N–H bond
protic
no O–H or N–H bond
aprotic
N(CH3)3
7.57
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
The amide N is less nucleophilic
since the electron pair is
delocalized by resonance.
7.58
1° alkyl halide
SN2 reaction
Br
a. Mechanism:
+
CN
CN
Energy
b. Energy diagram:
c. Transition state:
Ea
Br
+
+ Br
acetone
CN
Br H°
CN
+ Br
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).
CH3
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Alkyl Halides and Nucleophilic Substitution 7–23
7.59 Use the directions for Answer 7.25.
Br
Br
Br
a.
2° alkyl halide
intermediate
reactivity
3° alkyl halide
least reactive
1° alkyl halide
most reactive
Br
Br
b.
Br
3° alkyl halide
least reactive
c.
2° alkyl halide
intermediate
reactivity
1° alkyl halide
most reactive
Br
Br
Br
2° alkyl halide
intermediate
reactivity
vinyl halide
least reactive
1° alkyl halide
most reactive
7.60
better leaving group
a. CH3CH2Br
CH3CH2Cl
+
OH
+
OH
faster reaction
stronger nucleophile
b.
Br
+
OH
Br
+
H2O
faster reaction
stronger nucleophile
c.
d.
Cl
+
NaOH
Cl
+
NaOCOCH3
I
+
OCH3
+
OCH3
I
faster reaction
CH3OH
DMSO
faster reaction
polar aprotic
solvent
less steric hindrance
e.
Br
+
OCH2CH3
Br
+
OCH2CH3
faster reaction
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7.61 All SN2 reactions proceed with backside attack of the nucleophile. When nucleophilic attack
occurs at a stereogenic center, inversion of configuration occurs.
CH3
CH3
a.
C
H
+
Cl
* D
I
b.
+
Cl
*
c.
C
OCH3
+
H
OH
OH
*
*
+ I
OCH2CH3
*
OCH2CH3
H
d.
inversion of configuration
+ Cl
D
* OCH3
Br
+
CN
CN
* *
+ Cl
No bond to the stereogenic center is broken,
since the leaving group is not bonded to
the stereogenic center.
H
inversion of configuration
+ Br
[* denotes a stereogenic center]
7.62 Follow the definitions from Answer 7.30.
a.
CH3CH2CHCH2CH3
c.
2° carbocation
e.
(CH3)2CHCH2CH2
1° carbocation
2° carbocation
CH2CH3
b.
d.
f.
3° carbocation
CH2
3° carbocation
1° carbocation
7.63 For carbocations: Increasing number of R groups = Increasing stability.
a.
CH2
b.
1° carbocation
least stable
2° carbocation
intermediate
stablity
3° carbocation
most stable
1° carbocation
least stable
2° carbocation
intermediate
stablity
7.64
Cl H
H H
Cl C C
H C C
Cl H
H H
+
3 Cl groups –
electron-withdrawing
destabilizing
least stable
methyl group
without added Cl's
more stable
H
H
H C O C
H
H
H
H
H C O C
H
resonance stabilized
most stable
H
3° carbocation
most stable
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Alkyl Halides and Nucleophilic Substitution 7–25
7.65
Step [1]
CH3
a. Mechanism:
SN1 only
CH3 C CH2CH3
I
+ H2O
CH3
CH3
A
CH3
Step [2]
C CH2CH3 + H2O
B
CH3 C CH2CH3
OH2
+ I
C
Energy
b. Energy diagram:
Ea1
B
H°1
Ea2
H°2
H°overall = 0
C CH3
A CH3
CH3 C CH2CH3
CH3 C CH2CH3
OH2
I
Reaction coordinate
c. Transition states:
CH3
CH3
+
CH3 +
C CH2CH3
CH3
OH2
+
C CH2CH3
I –
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.)
7.66 The rate of an SN1 reaction increases with increasing alkyl substitution.
Br
a.
Br
Br
1° alkyl halide
least reactive
3° alkyl halide
most reactive
2° alkyl halide
intermediate
reactivity
b.
Br
Br
Br
1° alkyl halide
least reactive
Br
2° alkyl halide
intermediate
reactivity
3° alkyl halide
most reactive
Br
Br
c.
aryl halide
least reactive
2° alkyl halide
intermediate
reactivity
3° alkyl halide
most reactive
184
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Chapter 7–26
7.67 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.
Br
Cl
c.
better leaving group
faster reaction
Br + CH OH
3
Cl
2° halide
faster SN1 reaction
+ H2O
3° halide
d.
faster SN1 reaction
I
1° halide
slower SN1 reaction
I
CH3OH
+
aryl halide
slower reaction
+ H O
2
+ CH3CH2OH
CH3CH2OH
+ CH3CH2OH
DMSO
polar protic solvent
faster reaction
polar aprotic solvent
slower reaction
7.68
CH3CH2
a.
Br
CH3
C
CH3
b.
+ H2O
HO
Cl
CH3
+ CH3OH
CH3CH2
CH3
+
C
CH3
CH3
C
CH3CH2
CH3CH2
CH2CH3
C
OH
CH3
C
+ HCl
OCH3
Br
OCH2CH3
c.
+
+ CH3CH2OH
Br
d.
+ HBr
CH3CH2O
+ HBr
OH
+
H2 O
+
OH + HBr
7.69 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.
CH3OH
CH3CH2CH2CH CHCH2OCH3
CH3CH2CH2CH CHCH2Br
CH3CH2CH2CHCH CH2
HBr
OCH3
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
Br
CH3CH2CH2CH CH CH2
CH3CH2CH2CH CH CH2
H OCH3
CH3OH
Br
CH3CH2CH2CHCH CH2
OCH3
HBr
HBr
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185
Alkyl Halides and Nucleophilic Substitution 7–27
7.70
H
H
Br
a.
+
CN + Br
CN
acetone
1° alkyl halide
SN2 only
b.
+
OCH3
DMSO
+
Br H
strong nucleophile
polar aprotic solvent
Both favor SN2.
2° alkyl halide
SN1 and SN2
Br
c.
Br
H OCH3
reaction at a stereogenic center
inversion of configuration
OCH3
+ CH3OH
+ HBr
CH2CH2CH3
CH2CH2CH3
3° alkyl halide
SN1 only
CH2CH3
d.
C
H
CH2CH3
C
+ CH3COOH
I
CH3
H
CH2CH3
C
CH3COO
CH3
OOCCH3
CH3
H
2° alkyl halide Weak nucleophile
favors SN1.
SN1 and SN2
e.
+
OCH2CH3
Br
2° alkyl halide
SN1 and SN2
f.
OCH2CH3
DMF
+ Br
HI
reaction at a stereogenic center
racemization of product
reaction at a stereogenic center
inversion of configuration
strong nucleophile
polar aprotic solvent
Both favor SN2.
OCH2CH3
OCH2CH3
Cl
+ CH3CH2OH
+
+ HCl
two products – diastereomers
Nucleophile attacks
from above and below.
Weak nucleophile
favors SN1.
2° alkyl halide
SN1 and SN2
7.71
H
CN (axial)
CN
Br
a. (CH3)3C
(eq) acetone
H
(CH3)3C
H
inversion (equatorial to axial)
H
polar aprotic solvent
SN2 reaction Large tert-butyl group in
more roomy equatorial
position.
Br (axial)
H
b. (CH3)3C
CN
acetone
H
(CH3)3C
H
H
polar aprotic solvent
SN2 reaction
inversion (axial to equatorial)
CN (eq)
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Chapter 7–28
7.72
Br
[2]
C
[1] Na+H
(CH3)2NCH2CH2O H
OCH2CH2N(CH3)2
H
Na+
(CH3)2NCH2CH2O
C
+ NaBr
+ H2
H
diphenhydramine
7.73 First decide whether the reaction will proceed via an SN1 or SN2 mechanism (Answer 7.40), and
then draw the mechanism.
H
Br
OCH2CH3
Br
OCH2CH3
H
CH3CH2OH
can attack from
above or below
3° alkyl halide
SN1 only
+
Br
OCH2CH3
OCH2CH3
+ HBr
+
7.74
nucleophile
O
O
C O H
O
C O
C
OH
+ Br
O
Br
Br
C7H10O2
leaving group
7.75
Br
H H
N
CH3
Br–
N
Br
H
N
CO32–
CH3
N H
CH3
N
CO32–
N
+ HCO3–
Br–
N
CH3
N
nicotine
+ NaHCO3
+ NaBr
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Alkyl Halides and Nucleophilic Substitution 7–29
7.76
a. Hexane is nonpolar and therefore few nucleophiles will dissolve in it.
b. (CH3)3CO– is a stronger base than CH3CH2O–:
The three electron-donating CH3 groups add electron density to the negative charge of the
conjugate base, destabilizing it and making it a stronger base.
c. By the Hammond postulate, the SN1 reaction is faster with RX that form more stable carbocations.
(CH3)3C
(CH3)2C
CF3
3° Carbocation is stabilized by
three electron-donor CH3 groups.
Although this carbocation is also 3°,
the three electron-withdrawing F atoms
destabilize the positive charge. Since
the carbocation is less stable, the
reaction to form it is slower.
d. The identity of the nucleophile does not affect the rate of SN1 reactions since the nucleophile does
not appear in the rate-determining step.
Polar aprotic solvent
favors SN2 reaction.
e.
2° alkyl halide
SN1 or SN2
H
Br
H
Br
acetone
Br
Br
S
(2R)-2-bromobutane
optically active
Strong nucleophile
favors SN2 reaction.
This compound reacts with Br until a 50:50 mixture results, making the mixture optically inactive.
Then either compound can react with Br and the mixture remains optically inactive.
7.77
CH3NH2
Cl
N
Cl
+
CH3NH2
N
Cl
+
N
Cl
N CH3
+
H
N H
CH3
CH3NH3
N
Cl
N
H
+
CH3NH3
CH3
N
N
H
CH3
CH3NH2
+
Cl
188
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Chapter 7–30
7.78 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
OCH3
configuration
retained
Br
optically active
Reaction does not occur at
the stereogenic center.
(5R)-2-bromo-2,5-dimethylnonane
7.79
The nucleophile has replaced the leaving group.
Missing reagent:
a.
O
I
O
C CH
Cl
b.
The nucleophile has replaced the leaving group.
Missing reagent:
C CH
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.
7.80 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.
carbon
framework
functional
group
Cl
Na
SH
SH
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189
Alkyl Halides and Nucleophilic Substitution 7–31
Na
O
b.
O
Cl
O
functional
carbon
group
framework
Na
c.
CH3CH2CN
CN
CH3CH2Cl
CH3CH2CN
functional
group
carbon
framework
O
Cl
d.
functional
group
carbon
framework
Na
O
O
2o halide
O Na
or
O
Cl
O
1o halide
This path is preferred.
The strong nucleophile favors
an SN2 reaction so an unhindered
1o alkyl halide reacts faster.
carbon
functional framework
group
e.
CH3CH2 OCOCH3
carbon
framework
Na
CH3CH2Cl
functional
group
OCOCH3
CH3CH2OCOCH3
7.81 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
b.
A
CH3
CH3(CH2)17N(CH3)2
CH2Cl
CH2
N
(CH2)17CH3
CH3
A
C
7.82
I
B
very crowded 3° halide
O Na
D
OCH3
Na+ OCH3
C
E
CH3I
OCH3
A
E
unhindered methyl halide
preferred method
The strong nucleophile favors SN2
reaction so the alkyl halide should
be unhindered for a faster reaction.
Cl–
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Chapter 7–32
7.83
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
7.84
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
+ Na+ Br–
[1] Na+ –OCH2CH3
CH(CO2CH2CH3)2
[2] C6H5CH2–Br
(Chapter 23)
+ Na+ Br–
+ NH3
C6H5CH2 CH(CO2CH2CH3)2
+ Na+ Br–
+ CH3CH2OH
7.85
quinuclidine
The three alkyl groups are "tied back"
in a ring, making the electron pair
more available.
N
triethylamine
CH3CH2
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
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Alkyl Halides and Nucleophilic Substitution 7–33
7.86
O
O
H
O
[1]
H
+
H2
O
O
CH3 Br
CH3
[2]
minor product
O
+ NaBr
O
CH3 Br
CH3
[2]
major product
7.87
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
SN2
(CH3)3C
O
O
base
Br
(CH3)3C
3° alkyl halide
harder reaction
intramolecular
SN2
(CH3)3C
192
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Chapter 7–34
7.88
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 bonded to sp3 C
Resonance-stabilized carbocation forms.
best for SN1
Cl
Cl
Cl
Cl
Cl
+
(1 equiv)
O
J
Cl
O
re-draw
Cl
Cl
O
Cl
Cl
+
CH3OH
O
O CH3
Cl
H
Cl
Cl
+
O
K
OCH3
HCl
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193
Alkyl Halides and Elimination Reactions 8–1
C
miinnaattiioonn R
Reeaaccttiioonnss
Chhaapptteerr 88:: A
Allkkyyll H
Haalliiddeess aanndd EElliim
miinnaattiioonn
A
A ccoom
mppaarriissoonn bbeettw
weeeenn nnuucclleeoopphhiilliicc ssuubbssttiittuuttiioonn aanndd --eelliim
Nucleophilic substitutionA nucleophile attacks a carbon atom (7.6).
Nu
H
H Nu
+
C C
C C
X
substitution
product
X
good
leaving group
-EliminationA base attacks a proton (8.1).
B
H
C C
C C
+ H B+ +
X
elimination
product
•
•
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.
X
good
leaving group
•
•
Differences
In substitution, a nucleophile attacks a
single carbon atom.
In elimination, a Brønsted–Lowry base
removes a proton to form a bond, and
two carbons are involved in the reaction.
TThhee iim
mppoorrttaannccee ooff tthhee bbaassee iinn EE22 aanndd EE11 rreeaaccttiioonnss ((88..99))
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
+
H2O
+ Br
CH3
same product
different mechanism
CH3 C CH3
Br
H2O
weak base
E1
CH3
C CH2
CH3
+
H3O
+
+ Br
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Chapter 8–2
EE11 aanndd EE22 m
meecchhaanniissm
mss ccoom
mppaarreedd
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)
SSuum
mm
maarryy cchhaarrtt oonn tthhee ffoouurr m
meecchhaanniissm
mss:: SSNNN11,, SSNNN22,, EE11,, oorr EE22
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
ZZaaiittsseevv rruullee
• -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.
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Alkyl Halides and Elimination Reactions 8–3
C
Chhaapptteerr 88:: A
Annssw
weerrss ttoo PPrroobblleem
mss
8.1 • The carbon bonded to the leaving group is the carbon. Any carbon bonded to it is a carbon.
• To draw the products of an elimination reaction: Remove the leaving group from the carbon and a H from the carbon and form a bond.
a.
b.
CH3CH2CH2CH2CH2 Cl
1
2
K+ OC(CH3)3
1
CH3CH2CH2CH CH2
(CH3CH2)2C
CH2
CH3CH=C(CH3)CH2CH3
Cl
1
c.
K+ OC(CH3)3
2
Br
K+ OC(CH3)3
1
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 8–4
8.4 Two definitions:
• Constitutional isomers differ in the connectivity of the atoms.
• Stereoisomers differ only in the 3-D arrangement of atoms in space.
and
c.
C C
H
and
d.
trans
H
C C
and
H
H
CH3
cis
trans
different arrangement of atoms in space
stereoisomers
different connectivity of atoms
constitutional isomers
b.
CH3CH2
CH3
CH3CH2
a.
trans
CH3CH2
CH3
identical
H
CH3CH2
C C
and
CH3
C C
H
H
H
different connectivity of atoms
constitutional isomers
8.5 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
CH3CH2
trisubstituted
more stable
H
H
CH3
or
c.
C C
or
H
H
CH3
disubstituted
more stable
disubstituted
CH2CH3
trans
more stable
cis
8.6 Use the rules from Answer 8.5 to explain the energy differences.
cis-2-butene
trans-2-butene
cis-2,2,5,5-tetramethyl-3-hexene
less steric interaction between
smaller CH3 groups
smaller energy difference
trans-2,2,5,5-tetramethyl-3-hexene
more steric interaction between larger
tert-butyl groups in the cis isomer
larger difference in stability
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.
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Alkyl Halides and Elimination Reactions 8–5
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 carbon.
[2] A bond forms.
[3] The leaving group comes off.
OCH2CH3
CH3CH2 H
CH3CH2 C
CHCH3
transition state:
(CH3CH2)2C=CHCH3 +
Br
HOCH2CH3
+
CH3CH2 C
new bond
carbon
CH3CH2 H
Br
OCH2CH3
CHCH3
Br
8.9 For E2 elimination to occur there must be at least one hydrogen on a carbon.
carbon
CH3 H
CH3 C
C H
no H's on carbon
inert to E2 elimination
CH3 Br
8.10 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
8.11 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
2° alkyl halide
intermediate reactivity
1° alkyl halide
least reactive
3° alkyl halide
most reactive
CH3
Cl
Cl
b.
Cl
1° alkyl halide
least reactive
(CH3)2C(Br)CH2CH2CH3
CH3
2° alkyl halide
intermediate reactivity
3° alkyl halide
most reactive
8.12 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 8–6
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.)
8.13 The Zaitsev rule states: In a -elimination reaction, the major product has the more substituted
double bond.
CH3 H
a.
CH3 C
C
H
Br
CH2CH3
(CH3)2C
loss of H and Br
CH3
b.
CHCH2CH3
+
(CH3)2CHCH
trisubstituted
major product
Br
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
c.
CH3
monosubstituted
minor product
disubstituted
major product
CH3
Cl
d.
CH3CH2CH2CH2CH=CHCH3
loss of H and Cl
trisubstituted
CH3
CH3 ONLY product
loss of H and Cl
CH3
CH3
8.14 An E1 mechanism has two steps:
[1] The leaving group comes off, creating a carbocation.
[2] A base pulls off a proton from a carbon, and a bond forms.
CH3
CH3
C
CH2CH3
[1]
+ CH3OH
CH3
CH3
[2]
+
(CH3)2C=CHCH3 + CH3OH2 + Cl
H
Cl
transition state [1]:
C CH CH3 CH3OH + Cl
transition state [2]:
CH3
CH3 +C
CH2CH3
Cl CH3
CH3 C
+
CH CH3
H
OCH3
+
H
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8.15 The Zaitsev rule states: In a -elimination reaction, the major product has the more substituted
double bond.
2
CH3
CH3
1
C CHCH3
CH3CH2
CH3CH2 C CH2CH3 + H2O
a.
CH3CH2
Cl 1
trisubstituted
major product
1
CH2CH2CH3
CH3
b.
3
I
2
CH2CH2CH3
+
CH3
CH3OH
+
C CH2
CH3CH2
disubstituted
CH2CH2CH3
+
tetrasubstituted
major product
CH2
CH2CH2CH3
CH3
+
disubstituted
trisubstituted
8.16 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.)
8.17 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 carbon and create a new bond.
CH3
CH3
Br
a.
+ H2O
leaving
group nucleophile
and base
SN1 product
CH3
b. CH3 C CH2CH2CH3
nucleophile
and base
CH3
E1 products
CH3
+ CH3CH2OH
Cl
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 8–8
8.18 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
H
HO
C
CH3
CH3
C
CH3
H
Br
C
base
H
C
CH3
H
H and Br are on opposite sides =
anti periplanar
8.19 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.
8.20 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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H
CH3
H
H
H
H
1
H
H
CH(CH3)2
OH
CH3
Cl
2
CH(CH3)2
H
H
CH3
H
H
H
[loss of H(2) + 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
1
2
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(1) + Cl]
re-draw
two different axial H's
CH(CH3)2
CH(CH3)2
B
only one axial H
on a carbon
CH(CH3)2
H
H
H
H
CH(CH3)2
[loss of H(1) + Cl]
=
CH3
disubstituted
only product
8.21 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 8–10
8.22 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
Cl
c.
OCH3
+
strong negatively
charged base
E2
I
b.
+
CH3OH
weak neutral
base
E1
H2O
d. CH3CH2Br
weak neutral
base
E1
+
OC(CH3)3
strong negatively
charged base
E2
8.23 Draw the alkynes that result from removal of two equivalents of HX.
Br
Cl Cl
a.
C C CH2CH3
NH2
C C CH2CH3
c. CH3 C CH2CH3
NH2
CH3C CCH3
Br
H H
+ HC CCH2CH3
Br
KOC(CH3)3
b. CH3CH2CH2CHCl2
CH3CH2C
DMSO
CH
NH2
d.
C
Br
8.24
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
C
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Alkyl Halides and Elimination Reactions 8–11
8.25
weak base
SN1 and E1
3° halide
no SN2
CH3
CH3
Br
CH3OH
CH3
overall
reaction
The steps:
CH3
SN1
CH3
CH3OH
+ Br
CH3
H
CH3
CH3
Br
+
+
CH3
CH3
or
E1
OCH3
O CH3
H
CH3
Br
CH3
HBr
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Chapter 8–12
8.26
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.
8.27 To give only one product in an elimination reaction, the starting alkyl halide must have only one
type of carbon with H’s.
a.
CH2 CHCH2CH2CH3
CH2
CH2CH2CH2CH3
Cl
b. (CH3)2CHCH CH2
Cl
C(CH3)3
e.
CH2Cl
c.
Two carbons
are identical.
CH2Cl
CH2
CH3
CH3
(CH3)2CHCH2
Cl
d.
C(CH3)3
Two carbons
are identical.
8.28 To have stereoisomers, the two groups on each end of the double bond must be different from each
other.
farnesene
geranial
CHO
b.
a.
two methyl groups
no stereoisomers
2 H's — no
two different groups stereoisomers
at each end
can have
stereoisomers
two methyl groups
no stereoisomers
two different groups
at each end
can have
stereoisomers
8.29 Use the definitions in Answer 8.4.
CH3
a.
CH2
and
and
c.
trans trans
different connectivity
constitutional isomers
trans trans
identical
H
CH3CH2
b.
CH3
and
C C
CH3
CH3CH2
CH2CH3
stereoisomers
CH2CH3
C C
CH3
CH3
C
d.
CH3
CH3
C
and
CH3
CH3
stereoisomers
H
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Alkyl Halides and Elimination Reactions 8–13
8.30 There are three different isomers. Cis and trans isomers are diastereomers.
H
Cl
H
C C
H
Br
H
C C
Br
Cl
H
C C
H
Cl
B
A
constitutional isomer
of B and C
Br
C
diastereomers
8.31
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
PGF2
HO
CH CHCH(OH)(CH2)4CH3
sp3 stereogenic center
Double bond can be cis or trans.
8.32 Use the rules from Answer 8.5 to rank the alkenes.
CH3CH2
H
monosubstituted
b.
CH3
CH3CH2
H
C C
C C
a. CH2 CHCH2CH2CH3
H
H
least stable
disubstituted
cis
intermediate
stability
CH2 CHCH(CH3)2
CH2 C(CH3)CH2CH3
monosubstituted
least stable
disubstituted
intermediate
stability
CH3
disubstituted
trans
most stable
(CH3)2C
CHCH3
trisubstituted
most stable
8.33 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
CH3
CH3
+ H2
C C
H
CH3CH2CH2CH3
Ho = 127 kJ/mol
1-butene
H
cis-2-butene
CH3CH2CH2CH3
Ho = 120 kJ/mol
Energy
CH2 CHCH2CH3
cis-2-butene
larger H° for
1-butene
higher in energy
butane
smaller H° for cis-2-butene
lower in energy, more stable
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Chapter 8–14
8.34
Cl
a.
1
(CH3)3CO
(loss of 2 H)
major product
disubstituted
(loss of 1 H)
monosubstituted
DBU
b.
O
O
only product
Cl
O
O
3
I
c.
CH3CH=CHCH2CH2CH(CH3)2
2
CH3
1
OH
CH3CH2C(CH3)=C(CH3)CH2CH2CH3
2
CH3CH2CH(CH3)C(CH3) CHCH2CH3
(loss of 1 H)
major product
tetrasubstituted
(loss of 2 H)
trisubstituted
CH2
(loss of 3 H)
disubstituted
d.
Cl
OC(CH3)3
only product
2
e.
1
2
f.
OH
I
Br
CH3CH2CH2CH2CH=CHCH3
(loss of 1 H)
major product
disubstituted
(loss of 2 H)
monosubstituted
OH
1
(loss of 2 H)
major product
trisubstituted
(loss of 1 H)
disubstituted
8.35 To give only one alkene as the product of elimination, the alkyl halide must have either:
• only one carbon with a hydrogen atom
• all identical 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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Cl
Cl
c.
8.36 Draw the products of the E2 reaction and compare the number of C’s bonded to the C=C.
1
A
2
2
1 major product
trisubstituted
Br
Br
1
(CH3)2CHCH=CHCH3
disubstituted
2
B
(CH3)2CHCH=CHCH3
1
major product
disubstituted
2
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.
8.37
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)
8.38
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.
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Chapter 8–16
8.39
H
a.
KOH
CH3CH2CH2 C
NaOCH2CH3
b.
Cl
Br
trans isomer more stable
major product
trans isomer more stable
major product
8.40
CH3
a.
Cl
CH3
CH3
CH3
CH2
CH3
CH3
CH3
trisubstituted
disubstituted
tetrasubstituted
major product
Br
b.
trisubstituted
This isomer is more stable —
large groups farther away.
major product
disubstituted
trisubstituted
Cl
c.
disubstituted
trisubstituted
major product
8.41 Use the rules from Answer 8.22.
a.
OCH3
CH3CH CHCH3
strong base
E2
Br
2° halide
CH3OH
b.
weak base
E1
Br
2° halide
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
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8.42 The order of reactivity is the same for both E2 and E1: 1° < 2° < 3°
a.
Br
Br
Br
1° halide
2° halide
3° halide
Increasing reactivity in E1 and E2
CH3
b.
Br
CH3
Cl
CH3
Cl
3° halide
2° halide
3° halide +
better leaving group
Increasing reactivity in E1 and E2
8.43
CH3
a.
Cl
Cl
OH
3° halide – faster reaction
CH3
Cl
H2O
b.
CH3 Cl
c. (CH3)3CCl
H2O
strong base – E2
H2O
(CH3)3CCl
OH
OH
OH
DMSO
3° halide – faster reaction
polar aprotic solvent
faster reaction
8.44
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
bromocyclodecane
cis-cyclodecene
8.45 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
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Chapter 8–18
8.46 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
CH3
E2
CH2CH3
CH3
CH2CH3
CH3
Br
b.
Br
H
CH3
C6H5
CH3
CH3
rotate
CH2CH3
C6H5
E2
CH3
CH2CH3
CH3
CH3
CH2CH3
C6H5
CH2CH3
H
C6H5
c.
C6H5
C6H5
CH3
H
CH3
Br
H
rotate
CH2CH3
CH3
CH2CH3
E2
Br
CH3
CH3
CH3
8.47
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
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Alkyl Halides and Elimination Reactions 8–19
Cl
b.
H
two chair
conformations
H
(CH3)2CH
CH(CH3)2
H
H
H
c.
1
re-draw
CH3
CH3
CH(CH3)2
CH(CH3)2
D
D
= D
D
H
(loss of 2 H)
(loss of 2 H)
D
Cl
H
Cl
D
H
H
(CH3)2CH
CH3
(loss of 1 H)
major product
trisubstituted
re-draw
D
H
H
H
(CH3)2CH
CH3
H
1 H
two axial H's
B
B
Choose this conformation.
axial Cl
H
H
H
A
H
Cl
Cl
(CH3)2CH
CH3
2
Cl
axial
H
H
Cl
H
CH3
(CH3)2CH
CH3
CH3
H
2
D
H
H
=
D
H H
D
enantiomers
OH
D
This conformation reacts.
axial Cl
D
D =
H
D
H
(loss of 1 H)
D
Cl
d.
=
D
H
H
1
H
H
D
(loss of 2 D)
H
Cl
D
2
This conformation reacts.
axial Cl
H
D
Cl
H
D
=
D
H
D
H
enantiomers
OH
D
H
H
H
D
(loss of 1 D)
=
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Chapter 8–20
8.48
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
H
CH2CH3
C C
CH3
CH3
C C
CH3
CH3
CH3
CH2CH3 CH3CH2
identical
–HCl
H
H
CH3
C C
Cl
D
C
–HCl
H
CH3
C C
B
A
2-chloro-3-methylpentane
H
H
H
CH3
CH3
–HCl
CH3
H
C C
C C
CH3
CH2CH3 CH3CH2
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.
8.49 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 tertbutyl and Br groups are axial, a conformation that is not energetically favorable.
Br
Br
H
=
This conformation must react,
but it contains two axial groups.
cis
Br
Br
H
=
trans
preferred conformation
8.50
CH2CHCl2
a.
C CH
NaNH2
(2 equiv)
CH3
b.
CH3CH2 C
CHCH2Br
CH3 Br
Cl
c.
CH3 C CH2CH3
Cl
NaNH2
(2 equiv)
NaNH2
(excess)
CH3
CH3CH2 C
C CH
CH3
HC C CH2CH3
CH3 C C CH3
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H H
d.
NaNH2
C C
C C
(2 equiv)
Cl Cl
8.51
H H
Br
a. CH3C CCH3
CH3 C CH2CH3
or
CH3 C C CH3
Br Br
Br
CH3
CH3 Br
b. CH3 C C CH
CH3 C
CH3
CH3
CH3 Br
C CH3
or
CH3 C
CH3 Br
CH CH2Br
or
C C
CH2CHBr2
CH3
CH3
Br H
c.
CH3 C
Br Br
or
C C
C C
Br H
H H
8.52
H H
CH3 C C CH3
CH3 C C CH3
CH3 CH=C=CH2
CH2=CH CH=CH2
C
Br Br
2,3-dibromobutane
sp sp
A
sp
B
8.53 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 have increased reaction rates with better leaving groups.
Both SN2 and E2 have increased rates when changing from CH3OH (protic solvent) to (CH3)2SO
(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.
8.54
Br
a.
OC(CH3)3
sterically
1° halide hindered base
SN2 or E2
OCH2CH3
I
b.
1° halide
SN2 or E2
Cl
c. CH3 C CH3
Cl
dihalide
strong
nucleophile
NH2
(2 equiv)
strong base
HC C CH3
E2
OCH2CH3
SN2
214
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Chapter 8–22
Br
d.
sterically
hindered
base
1° halide
SN2 or E2
CH2CH3
2° halide
SN1, SN2, E1, E2
CH2CH3
E2
sterically
hindered
base
Br
major product
OCH2CH3
Br
CH3CH2OH
CH2CH3
(CH3)2CH
SN1 product
2 NaNH2
CHCH2Br
CHCH3
CH2CH3
CH2CH3
weak base
3° halide
no SN2
g.
CH2CH3
OC(CH3)3
e.
f.
E2
DBU
(CH3)2CH
E1 products
C CH
dihalide Br
Cl Cl
h.
KOC(CH3)3
dihalide
CH3
OCH2CH3
I
CH3CH2OH
i.
2° halide
SN1, SN2, E1, E2
j.
(2 equiv)
DMSO
CH3
CH3 C C CH
CH3CH CHCH3
SN1 product
weak base
E1 product
H2O
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
8.55 [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 hydrogen in the alkyl halide.
a.
CH3Cl
[1] NaOCOCH3
[2] NaOCH3
[3] KOC(CH3)3
Cl
CH3OCOCH3
CH3OCH3
CH3OC(CH3)3
b.
[1] NaOCOCH3
[2] NaOCH3
[3] KOC(CH3)3
OCOCH3
OCH3
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Alkyl Halides and Elimination Reactions 8–23
c.
[1] NaOCOCH3
Cl
Cl
d.
OCOCH3
[2] NaOCH3
[1] NaOCOCH3
OCOCH3
[2] NaOCH3
OCH3
+
E2
SN2
[3] KOC(CH3)3
[3] KOC(CH3)3
8.56
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:
C=
(CH3)3C
OCH3
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
(CH3)3C
H
OCH3
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.
216
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Chapter 8–24
8.57
Cl
H
H
a.
strong base
SN2 and E2
2° halide
SN1, SN2, E1, E2
Cl
OH
OH
H
SN2 product
major E2 product
inversion at
stereogenic center
HO
H2O
b.
H
H
weak base
SN1 and E1
2° halide
SN1, SN2, E1, E2
c.
Cl
C6H5
3° halide
no SN2
minor E2 product
OH
SN1 products
major E1 product
CH3
CH3
minor E2 product
minor E1 product
CH3
CH3
CH3OH
OCH3
C6H5
weak base
SN1 and E1
minor E1 product
CH3
CH3
CH3
CH3
C6H5
OCH3
SN1 products
C6H5
E1 product
NaOH
d.
strong base
SN2 and E2
Cl
OH
SN2 product
2° halide
SN1, SN2, E1, E2
CH3
Br
e.
3° halide
no SN2
weak base
good nucleophile
SN1
f.
D
2° halide
SN1, SN2, E1, E2
achiral
SN1 product
OH
KOH
strong base
SN2 and E2
minor E2 product
CH3
OOCCH3
CH3COO
Br
major E2 product
(trans diaxial elimination of D, Br)
D
SN2 product
inversion at
stereogenic center
E2 product
8.58
a.
b.
CH3OH
Cl
weak base
SN1 and E1
KOH
Cl
strong base
E2
OCH3
OCH3
SN1
SN1
E1
E1
E1
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Alkyl Halides and Elimination Reactions 8–25
8.59
CH3
CH3
Br
a.
OC(CH3)3
OC(CH3)3
CH3
strong
bulky base
E2
3° halide
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.
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.
c.
CH3
Cl
3° halide
minor product only
OH
strong base
E2
I
d.
Cl
2° halide
good nucleophile,
weak base
SN2 favored
CH3
More substituted
alkene is favored.
minor product only
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.
218
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Chapter 8–26
8.60
3° halide, weak base:
SN1 and E1
CH3CH2OH
a.
+
overall
reaction
Cl
+
+ HCl
OCH2CH3
The steps:
+ HCl
SN1
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.
OCH2CH3
CH3CH2OH
H
Cl
or
E1 H
+ HCl
Cl
or
E1
+ HCl
H
Cl
CH3
Cl
b.
3° halide
strong base
E2
CH3
OH
CH2
+ H2O + Cl
+
overall
reaction
CH3
CH3
Cl
Each product:
one step
H
OH
or
+ H2O + Cl
CH2 H
OH
CH2
Cl
one step
8.61 Draw the products of each reaction with the 1° alkyl halide.
Cl
a.
strong
nucleophile
SN2
H
Cl
b.
H
NaOCH2CH3
OCH2CH3
H
KCN
strong
nucleophile
SN2
Cl
c.
H
CN
H
DBU
sterically
hindered base
E2
H
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Alkyl Halides and Elimination Reactions 8–27
8.62
H
H
H
Br
H
H2O
H
O H
OH
H2O
Br
above
H
H
H
+
HBr
+
HBr
H
Br–
H
H
H
O H
re-draw
Br
below
H
OH
H
H
H
+
H
H
HBr
H
Br
8.63
good nucleophile
O
CH3
C
O
CH3 CHCH3
O
C
CH3
CH3COO is a good nucleophile and a weak base
and so it favors substitution by SN2.
O
CH3 CHCH3
(only)
Br
CH3CH2O
strong base
CH3 CHCH3 + CH3CH CH2 The strong base gives both SN2 and
E2 products, but since the 2° RX is
OCH2CH3
20%
80%
somewhat hindered to substitution,
the E2 product is favored.
220
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Chapter 8–28
8.64
Cl
OCH3
CH3OH
+
+
+
+
HCl
OCH3
3° halide
weak base
SN1 and E1
Cl
H
CH3OH
Cl
OCH3
OCH3
Cl–
HCl
Cl
H
or
HCl
Cl–
CH3OH
HCl
OCH3
Cl
or
OCH3
H
HCl
Cl
H
8.65 E2 elimination needs a leaving group and a hydrogen in the trans diaxial position.
Two different
conformations:
Cl
Cl
Cl
Cl
Cl
Cl
This conformation has Cl's
axial, but no H's axial.
Cl
Cl
Cl
Cl
Cl
Cl
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.
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Alkyl Halides and Elimination Reactions 8–29
8.66
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.
8.67
leaving group
H
N
C6H5O
O
S
N
C6H5O
DBN
N
O
B
H
H H
H H
O
overall reaction
S
N
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
8.68
a.
H
D
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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Epoxides
223
Alcohols, Ethers, and Epoxides 9–1
C
Chhaapptteerr 99:: A
Allccoohhoollss,, EEtthheerrss,, aanndd EEppooxxiiddeess
G
Geenneerraall ffaaccttss aabboouutt R
RO
OH
H,, R
RO
OR
R,, aanndd eeppooxxiiddeess
• All three compounds contain an O atom that is sp3 hybridized and tetrahedral (9.2).
CH3
O
CH3
H
109o
an alcohol
•
60o
O
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
=
O
=
H
C
H H
H
H
H
O
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,
O
O
C C
Nu
OH
H OH
C C
[1]
C C
[2]
Nu
+ OH
Nu
Nu
A
A nneew
w rreeaaccttiioonn ooff ccaarrbbooccaattiioonnss ((99..99))
• 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).
C C
1,2-shift
R
(or H)
C C
R
(or H)
PPrreeppaarraattiioonn ooff aallccoohhoollss,, eetthheerrss,, aanndd eeppooxxiiddeess ((99..66))
[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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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.
R
Reeaaccttiioonnss ooff aallccoohhoollss
[1] Dehydration to form alkenes
[a] Using strong acid (9.8, 9.9)
C C
H OH
H2SO4
or
TsOH
•
C C
+
•
•
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.
H2O
•
[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
•
•
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[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
pyridine
O
R O S
O
CH3
O
R OTs
• The C–O bond is not broken so the
configuration at a stereogenic center is
retained.
R
Reeaaccttiioonnss ooff aallkkyyll ttoossyyllaatteess
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
+
R
Reeaaccttiioonnss ooff eetthheerrss
Only one reaction is useful: Cleavage with strong acids (9.14)
R O R'
+
H X
R X
+
+
R' X
(2 equiv)
(X = Br or I)
H2O
•
•
With 2o and 3o R groups, the mechanism is
SN1.
With CH3 and 1o R groups the mechanism is
SN2.
R
Reeaaccttiioonnss ooff eeppooxxiiddeess
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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Chapter 9–4
C
Chhaapptteerr 99:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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
CH3
O
1° alcohol
symmetrical ether
2° alcohol
OH
CH3
OH
3° alcohol
O
unsymmetrical ether
O
unsymmetrical ether
1° alcohol
9.2 Use the definition in Answer 9.1 to classify each OH group in cortisol.
O
OH
HO
2° alcohol
1° alcohol
OH
H
H
3° alcohol
H
O
cortisol
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.
[1]
OH
OH
[2]
3
9 carbons = nonanol
5-ethyl
[3] 5-ethyl-6-methyl-3-nonanol
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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
3
OH
OH
d. trans-1,2-cyclohexanediol
or
OH
OH
OH
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
butyl
methyl
butyl methyl ether
b. common name:
OCH3
CH3 O CH2CH2CH2CH3
substituent:
methoxy
1-methoxybutane
IUPAC name:
OCH3
substituent –
methoxy
methyl
larger group – 6 C's
cyclohexane
cyclohexyl
cyclohexyl methyl ether
c. common name:
CH3CH2CH2 O CH2CH2CH3
propyl
propyl
dipropyl ether
larger group – 4 C's
butane
methoxycyclohexane
IUPAC name:
CH3CH2CH2 O CH2CH2CH3
propoxy
propane
1-propoxypropane
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Chapter 9–6
9.6 Name each ether using the rules from Answer 9.5.
1 CH3
a.
b.
CH3CH2 O CH2CH3
CH3 C
O CH3
CH3
ethoxy
ethane
2-methoxy
2-methyl
propane
2-methoxy-2-methylpropane
1-ethoxyethane
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.
H
Three possibilities:
[1] methyloxirane
[2] 1,2-epoxypropane
[3] propene oxide
O
a.
CH3
c.
O
H
Three possibilities:
[1] cis-2-methyl-3-propyloxirane
[2] cis-2,3-epoxyhexane
[3] cis-2-hexene oxide
1
CH3
b.
O
1-methyl
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
a.
CH3
OH
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
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9.9 Draw dimethyl ether and ethanol and analyze their intermolecular forces to explain the observed
trend.
dimethyl ether
CH3
O
ethanol
CH3CH2OH
CH3
VDW
DD
no HB
much lower bp
VDW
DD
HB
Two molecules of CH3CH2OH
can hydrogen bond to each other.
stronger forces =
much higher bp
Both molecules contain an O atom
and can hydrogen bond with water. They
have fewer than 5 C's and are
therefore water soluble.
H
H
O
H
CH3
O
O
H
CH3
CH3CH2
O
H
9.10 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.
9.11 Compare the number of carbons and functional groups to determine if each compound is soluble in
water, an organic solvent, or both.
CH3
O
O
N
CH3
O
O
H
O
Br–
H
O
H
O
S
CO2CH3
eplerenone
24 C's and 6 O's (from four functional groups)
soluble in organic solvents
probably borderline water solubility
HO
S
tiotropium bromide
19 C's, 4 O's
a salt – water soluble
Because it has many C's, it is also
probably soluble in organic solvent.
9.12 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
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Chapter 9–8
9.13 To synthesize an ether using a Williamson ether synthesis:
[1] First find the two possible alkoxides and alkyl halides needed for nucleophilic substitution.
[2] Classify the alkyl halides as 1°, 2°, or 3°. The favored path has the less hindered halide.
Two possibilities:
Two possibilities:
CH3
CH3 O
a.
CH3 O
b.
CH3
CH3CH2 O C CH3
CH3CH2 O C CH3
H
H
CH3
CH3O
+ Br
CH3CH2 O
CH3Br + O
+
CH3CH2
H
methyl
halide
2° halide
CH3
Br C CH3
2° halide
Br +
1° halide
O C CH3
H
less hindered RX
preferred
less hindered RX
preferred
9.14 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
9.15 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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CH3
OH
c.
CH2
CH3
TsOH
+
+ H2O
trisubstituted
disubstituted
major product minor product
9.16 The rate of dehydration increases as the number of R groups increases.
(CH3)2CHCH2CH2CH2OH
a.
1° alcohol
slowest reaction
(CH3)2CHCH2CH(OH)CH3
(CH3)2C(OH)CH2CH2CH3
2° alcohol
intermediate
reactivity
3° alcohol
fastest reaction
CH3
OH
OH
b.
HO
1° alcohol
slowest reaction
CH3
3° alcohol
fastest reaction
2° alcohol
intermediate
reactivity
9.17 There are three steps in the E1 mechanism for dehydration of alcohols and three transition states.
transition state [1]:
transition state [2]:
CH3
CH3
+
CH3 C CH3
+ OH
H
CH3 C
transition state [3]:
CH3
CH3
+OH
+
C CH2
CH3
H
2
OSO3H
OSO3H
9.18
transition state [1]:
transition state [2]:
H
H
CH3CH2 C H
+ OH
CH3 C
H
H
OSO3H
OSO3H
CH2
OH2
+
9.19
H H
HSO4
H
+
This alkene is also formed in addition
to Y from the rearranged carbocation.
+
rearranged 3° carbocation
H2SO4
The initially formed 2° carbocation gives two alkenes:
or
HSO4
H
H
+
H
H
+ CH
2
H
HSO4
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Chapter 9–10
9.20
CH3 H
a. CH3 C
H
C CH2CH3
CH3 H
rearrangement
+
CH3 C
+
1,2-H shift
3° carbocation
more stable
H
rearrangement
1,2-H shift
+
c.
+
rearrangement
1,2-methyl shift
H
2° carbocation
b.
+
C CH2CH3
2° carbocation
3° carbocation
more stable
+
3° carbocation
more stable
2° carbocation
9.21
CH3 H
CH3 C
C CH3
H
CH3 H
H2O
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
9.22
CH3
a. CH3 C CH2CH3
HCl
CH3
CH3 C CH2CH3
c.
OH
HBr
Br
+ H2O
Cl
OH
HI
b.
+ H2O
I
OH
+ H2O
9.23 • 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
CH3CH2CH2
I H
D
HI
C
CH3CH2CH2
D
1° alcohol so inversion of configuration
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CH3
b.
Br
HBr
3° alcohol, so Br attacks from above and below.
The product is achiral.
CH3
OH
achiral starting material
HO
achiral product
Cl
HCl
c.
Cl
3° alcohol = racemization
9.24
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)
9.25 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
9.26 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
9.27 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
b.
3° alcohol, SN1 =
racemization
I
SN2 =
inversion
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Chapter 9–12
9.28 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.
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
9.29
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
9.30
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.)
9.31
HO H
S
TsCl
TsO H
NaOH
H OH
SN2
pyridine
retention
S
enantiomers
inversion
R
One inversion from starting material
to product.
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9.32 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.
H
a.
CH3 C OH
CH3
H
SOCl2
pyridine
CH3 C OH
d.
CH3
H
b.
H
CH3 C Cl
CH3
TsCl
H
CH3 C OH
CH3
CH3
H
c.
CH3
H
e.
CH3 C OTs
pyridine
CH3 C Br
CH3
H
CH3 C OH
H
HBr
[1] PBr3
[2] NaCN
CH3 C CN
CH3
H
H2SO4
CH3 C OH
CH2 CHCH3
CH3 C OH
f.
CH3
CH3
POCl3
CH2 CHCH3
pyridine
9.33
HBr
a. CH3CH2 O CH2CH3
CH3
b.
2 CH3CH2 Br
HBr
CH3 C O CH2CH3
c.
+ H2O
O CH3
HBr
Br
CH3
CH3 C Br
H
+ CH3Br
+
+ H2O
CH3CH2Br + H2O
H
9.34 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.
HBr
OCH3
anisole
OH
+
CH3Br
phenol
Br
bromobenzene
NOT formed
SN1:
SN2:
O CH3
CH3OH
H
highly unstable
carbocation
CH3
O
H
Br
9.35 Compare epoxides and cyclopropane. For a compound to be reactive towards nucleophiles, it must
be electrophilic.
O
+
+
epoxide
cyclopropane
O is electronegative and pulls electron density
away from C's. This makes them electrophilic
and reactive with nucleophiles.
Cyclopropane has all C's and H's,
so all nonpolar bonds. There
are no electrophilic C's so it will not
react with nucleophiles.
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Chapter 9–14
9.36 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
CH3
OH
[1] CH3CH2O
[2] H2O
O
C C
b. CH3
OCH2CH3
H
Attack here:
less substituted C
backside attack
[1] H
H
H
HO
C C
H [2] H O
2
CH3
H
H
C C
C
CH
Attack here:
less substituted C
backside attack
9.37 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
H
CH3
CH3
C
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
trans-2,3-dimethyloxirane
HO
HO
+
C
H
CH3
CH3
H
C
H
CH3
C
OH
identical
Rotate around the C–C bond to
see the plane of symmetry.
HO
OH
C
CH3
H
C
H
CH3
meso compound
[1]
O
–OH
[2] H2O
H
CH3
C
C
CH3
H
HO
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9.38 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
c.
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
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
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Chapter 9–16
9.39 Draw the structure of each alcohol, using the definitions in Answer 9.1.
OH
OH
a.
OH
1°
b.
OH
2°
3°
enol
9.40 Use the directions from Answer 9.3.
a.
(CH3)2CHCH2CH2CH2OH
[1]
[2]
H
CH3 C CH2CH2CH2OH
H
[3]
4-methyl-1-pentanol
[3]
4-ethyl-6-methyl-3-heptanol
[3]
4-ethyl-5-methyl-3-octanol
[3]
cis-1,4-cyclohexanediol
[3]
3,3-dimethylcyclohexanol
[3]
(2R,3R)-2,3-butanediol
[3]
5-methyl-2,3,4-heptanetriol
CH3 C CH2CH2CH2OH
CH3
CH3
5 carbons = pentanol
b.
1
4-methyl
(CH3)2CHCH2CH(CH2CH3)CH(OH)CH2CH3
[1]
H
[2]
H OH
CH3 C CH2 C CH
CH3
CH2CH3
H
CH3 C CH2 C CH
CH2CH3
CH3
7 carbons = heptanol
c.
[1]
CH2CH3
CH2CH3
6-methyl
4-ethyl
5-methyl
[2]
OH
OH
8 carbons = octanol
d.
[1]
3
H OH
3
4
HO
OH
4-ethyl
[2]
1
HO
OH
cis
cyclohexanediol
e.
[1]
[2]
HO
HO
6 carbons = cyclohexanol
f.
[1]
HO H
3,3-dimethyl
[2]
HO H
2
HO H
3
HO H
2R,3R
4 carbons = butanediol
g.
OH
[1]
OH
OH
7 carbons = heptanetriol
[2]
4
OH
OH
5-methyl
2
OH
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Alcohols, Ethers, and Epoxides 9–17
h.
[1]
[2]
HO
CH(CH3)2
[3]
1
CH(CH3)2
HO
trans-3isopropylcyclopentanol
5 carbons = cyclopentanol
3-isopropyl
9.41 Use the rules from Answers 9.5 and 9.7.
a.
d.
O
O
1,2-epoxy-2-methylhexane
or 2-butyl-2-methyloxirane
or 2-methylhexene oxide
dicyclohexyl ether
4,4-dimethyl
b.
CH2CH3
e.
OCH2CH2CH3
longest chain =
heptane
substituent =
3-propoxy
4,4-dimethyl-3-propoxyheptane
c.
O
epoxy
2
5 carbons =
cyclopentane
1,2-epoxy-1-ethylcyclopentane
or 1-ethylcyclopentene oxide
CH3
O
f.
ethyl isobutyl ether
or 1-ethoxy-2-methylpropane
CH3
CH3 C O C CH3
CH3
CH3
tert-butyl tert-butyl
di-tert-butyl ether
9.42 Use the directions from Answer 9.4.
a. 4-ethyl-3-heptanol
3
d. 6-sec-butyl-7,7-diethyl-4-decanol
4
OH
6
OH
4
1
OH
b. trans-2-methylcyclohexanol
OH
or
CH3
HO
1
e. 3-chloro-1,2-propanediol
HO
CH3
OH
3
c. 2,3,3-trimethyl-2-butanol
f. diisobutyl ether
2
2
O
7
3
Cl
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Chapter 9–18
1
g. 1,2-epoxy-1,3,3-trimethylcyclohexane
i. (2R,3S)-3-isopropyl-2-hexanol
3
O
2
1
OH
3
j. (2S)-2-ethoxy-1,1-dimethylcyclopentane
1
h. 1-ethoxy-3-ethylheptane
O 1
3
2
O
9.43
Eight constitutional isomers of molecular formula C5H12O containing an OH group:
OH 1-pentanol
OH
1 OH 3-methyl-1-butanol
1° alcohol
3
1° alcohol
1 OH 2-methyl-1-butanol
1° alcohol
2-pentanol
OH
2 1
2° alcohol
OH
OH
3-pentanol
2° alcohol
2,2-dimethyl-1-propanol
1° alcohol
2
3-methyl-2-butanol
2° alcohol
OH
2-methyl-2-butanol
3° alcohol
2
9.44 Use the boiling point rules from Answer 9.8.
a.
CH3CH2OCH3
ether
no hydrogen bonding
lowest bp
b.
(CH3)2CHOH
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
9.45 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
–108 °C
lowest melting point
OH
–90 °C
intermediate melting point
OH
26 °C
highest melting point
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Alcohols, Ethers, and Epoxides 9–19
9.46 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
OH
HO
OH
1-butanol
118 °C
1,2-propanediol
187 °C
1,3-propanediol
215 °C
9.47
a. CH3CH2CH2OH
b. CH3CH2CH2OH
c. CH3CH2CH2OH
d. CH3CH2CH2OH
e. CH3CH2CH2OH
f. CH3CH2CH2OH
g. CH3CH2CH2OH
h. CH3CH2CH2OH
i.
CH3CH2CH2OH
H2SO4
CH3CH CH2
NaH
+ H2O
CH3CH2CH2O Na+
HCl
CH3CH2CH2Cl
ZnCl2
HBr
+ H2
+ H2O
CH3CH2CH2Br + H2O
SOCl2
CH3CH2CH2Cl
pyridine
PBr3
CH3CH2CH2Br
TsCl
CH3CH2CH2OTs
pyridine
[1] NaH
[2] CH3CH2Br
CH3CH2CH2O Na+
[1] TsCl
CH3CH2CH2OTs
[2] NaSH
CH3CH2CH2OCH2CH3
CH3CH2CH2SH
9.48
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
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Chapter 9–20
9.49 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
CHCH3
TsOH
TsOH
trisubstituted disubstituted
major product
d.
TsOH
CH3CH2CH2CH2OH
CH3CH2CH CH2
OH
TsOH
e.
tetrasubstituted
major product
disubstituted
Two products formed
by carbocation rearrangement
9.50 The more stable alkene is the major product.
OH
H2SO4
trans and disubstituted
major product
monosubstituted
cis and disubstituted
9.51 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
SN2
NaOCH2CH3
SN2
NaOH
SN2
K+
CH3CH2CH2CH2 SCH3 + HOTs
CH3CH2CH2CH2 OCH2CH3
CH3CH2CH2CH2 OH
+ Na+
+ Na+
–OTs
–OTs
–
OC(CH3)3
E2
CH3CH2CH CH2 + (CH3)3COH + K+ –OTs
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Alcohols, Ethers, and Epoxides 9–21
9.52
HBr
a.
b.
+
HO H
OH
H D
Br H
HCl
ZnCl2
H Br
Cl
1° Alcohol will undergo SN2.
inversion
H D
SOCl2
c.
pyridine
HO H
SOCl2 always implies SN2.
inversion
H Cl
TsCl
pyridine
d.
2° Alcohol will undergo SN1.
racemization
KI
HO H
TsO H
SN2
inversion
H
I
Configuration is maintained.
C–O bond is not broken.
9.53
NaH
(a)
(b)
(c)
(2R)-2-hexanol
B=
O H
TsCl
pyridine
HO H
CH3I
A=
PBr3
CH3O H
CH3O
C=
TsO H
E=
D=
H OCH3
CH3O
F=
H Br
CH3O H
Routes (a) and (c) given identical products, labeled B and F.
9.54 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 CO bond to form a carbocation.
[3] Remove a hydrogen to form the bond.
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Chapter 9–22
OH
a.
CH3 H OSO3H
CH3
CH3
+
overall
reaction
The steps:
CH2
+
+ H2O
H
O H
CH3
H
H
+ HSO4
CH3
+ H2SO4
CH3
+ H2O
H
HSO4
2° carbocation
and
+ HSO4
H
CH3
1,2-H shift
CH2
CH2
H
3° carbocation
and
2° carbocation
H
+ HSO4
CH3
CH3
CH3
b.
CH3
OH
H OSO3H
overall
reaction
+ H2SO4
+ H2SO4
CH3
+ H2O
CH3
The steps:
CH3
CH3
CH3
+ HSO4
CH3 + H O
2
1,2-CH3 shift
2° carbocation
CH3
+ H2SO4
CH3
H
O H
H
CH3
+ HSO4
3° carbocation
CH3
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Alcohols, Ethers, and Epoxides 9–23
9.55
a.
Br
OH
Br
C
B
A
(E1 with acid)
OH
(E2 with base)
D
(E2 with base)
(E1 with acid)
3,3-dimethylcyclopentene
major
product
b. The best starting material to form 3,3-dimethylcyclopentene would be C since the alkene can be
formed as a single product by an E2 elimination with base.
9.56 With POCl3 (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has a 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.
O
OH
Cl
P
Cl
H
Cl
O
POCl2
N
POCl2
N
O
H
+
N
H
V
+
W
Cl
+ –OPOCl2
+
N
H
OH
H OSO3H
+
OH2
+
+ H2O
V
+ HSO4–
2° carbocation
1,2–CH3 shift
+
3° carbocation
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Chapter 9–24
HSO4–
H
H
+
+
+
HSO4–
HSO4–
H
3° carbocation
3° carbocation
X
3° carbocation
Y
+ H2SO4
+
Z
+
H2SO4
H2SO4
9.57 To draw the mechanism:
[1] Protonate the oxygen to make a good leaving group.
[2] Break the CO bond to form a carbocation.
[3] Look for possible rearrangements to make a more stable carbocation.
[4] Remove a hydrogen to form the bond.
Dark and light circles are meant to show where the carbons in the starting material appear in the product.
OH
+ H2O
O H
H
H OSO3H
H
2o carbocation
+ HSO4–
+ HSO4
–
3o carbocation
9.58
HO
H
HBr
a.
Br
H
H
S
HO
H
b.
Br
R
HO
H
HCl
c.
Cl
H
SOCl2
pyridine
PBr3 follows
SN2 = inversion.
H
H
Cl
2° alcohol
SN1 = racemization
R
S
HO
d.
2° alcohol
SN1 = racemization
R
H
PBr3
Br
H
R
Cl
SOCl2 follows
SN2 = inversion.
+ H2SO4
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9.59
OH
3-methyl-2-butanol
[1] HBr
H
1,2-H shift
H
Br
Br
[2] –H2O
2° carbocation
2-methyl-1-propanol
HBr
HO
3° carbocation
H2O
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
9.60 Conversion of a 1° alcohol into a 1° alkyl chloride occurs by an SN2 mechanism. SN2 mechanisms
occur more readily in polar aprotic solvents by making the nucleophile stronger. No added ZnCl2
is necessary.
HCl
R OH
R Cl
HMPA
polar aprotic solvent
This makes Cl< a better nucleophile.
9.61
H Cl
A
Cl
OH
H2O
OH2
two resonance structures
for the carbocation
B
Cl
Cl
Cl
Cl
C
9.62
CH3CH2CH2CH2OH
H2SO4, NaBr
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
H
Br
CH3CH2CH2CH2Br + H2O
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Chapter 9–26
Formation of CH3CH2CH=CH2:
CH3CH2CH CH2 O H
H
CH3CH2CH CH2
+ H2O
H2SO4
H
HSO4–
Ether formation (from the protonated alcohol):
CH3CH2CH2CH2
CH3CH2CH2CH2
OH2
CH3CH2CH2CH2
O CH2CH2CH2CH3 + H2O
H
HSO4
O H
CH3CH2CH2CH2 O CH2CH2CH2CH3
H2SO4
9.63
H OSO3H
OH
O
OH2
+
O
+ HSO4–
O
+ HSO4
–
H
1,2-shift
O +
O
+ H2SO4
+ H2O
H
O
+
9.64
HSO4–
OH
H OSO3H
OH2
OH
O
OH
+ HSO4–
H
O
OH
+ H2O
+ H2SO4
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Alcohols, Ethers, and Epoxides 9–27
9.65
a.
c.
O
Br
O
CH3CH2OCH2CH2CH3
O
2° halide
CH3CH2O + BrCH2CH2CH3
O
Br
1° halide
1° halide
less hindered RX
preferred path
OCH2CH2CH3
b.
CH3CH2OCH2CH2CH3
CH3CH2Br + OCH2CH2CH3
1° halide
Neither path preferred.
OCH2CH2CH3
Br
O
OCH2CH2CH3
+ BrCH2CH2CH3
2° halide
1° halide
less hindered RX
preferred path
9.66 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
CH3
CH3
O C CH3
Br C CH3
Br
O C CH3
CH3
CH3
aryl halide
unreactive in SN2
O
CH3
3° alkyl halide
too sterically
hindered for SN2
9.67
a.
b.
(CH3)3COCH2CH2CH3
O
HBr
(CH3)3CBr + BrCH2CH2CH3
(2 equiv)
HBr
(2 equiv)
2
Br
H2O
H2O
c.
OCH3
HBr
(2 equiv)
Br
CH3Br
H2O
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Chapter 9–28
9.68
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
CH3
CH3
O
Cl
Na+ H
H
O
Cl
CH3
O
I
CH3
H
I
H
b.
H
I
I
+ H2 + NaCl
+ H2 + Na+
O
9.69
OC(CH3)3
O C(CH3)3
H–OCOCF3
CH3
OH
H
CH2
+ CF3CO2–
C
OH
CH3
+
CH3
C CH
2
H
CH3
CF3CO2–
+ CF3CO2H
9.70
O
a. H
C
O
C
H
H
H
HBr
d. H C
BrCH2CH2OH
H
O
b. H
C
C
H
H
H
O
c. H
C
C
H
H2O
e. H
H2SO4
[2] H2O
[1] HC C–
H
H
C
H
HOCH2CH2OH
H
H
[2] H2O
O
CH3CH2OCH2CH2OH
C
H
CH2 CH2OH
[1] OH
C
f. H C
HC C
[2] H2O
O
HOCH2CH2OH
[1] CH3CH2O
H
H
C
[1] CH3S
H
H
CH3SCH2CH2OH
[2] H2O
9.71
O
a. CH
3
CH3
H
H
CH3
b.
O
CH3CH2OH
H2SO4
[1] CH3CH2O Na+
[2] H2O
O
CH3 OH
CH3 C
CH3CH2O
C H
H
CH3
OH
OCH2CH3
OH
HBr
c.
O
d.
Br
OH
[1] NaCN
[2] H2O
CN
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9.72
CH3
H
Cl
C
a.
CH3
C
H O
Cl
H
C
H
CH3
H
CH3 C
C
O
H
CH3
O
C4H8O
Na+ H
H
Cl
CH3
C
b.
H O
H
Cl
CH3
H
C
C
CH3
CH3
H C
C
O
H
CH3
OH
C
c.
CH3
C
H
H
CH3
H
Cl
rotate
C
CH3
CH3
H
C
CH3
H
O
C4H8O
Na+ H
Cl
C
The 2 CH3 groups are anti in the
starting material, making them
trans in the product.
CH3
H
CH3
H
Cl
C
C
CH3
O H
The 2 CH3 groups are gauche in the
starting material, making them
cis in the product.
C
H
O
backside
attack
CH3
H C
C
H
CH3
O
C4H8O
Na+ H
9.73
O
O
CH3CH2O
O
CH3CH2O
+
Cl
CH3CH2OCH2
Cl
9.74
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]
HO
pyridine
Cl
OH
TsCl
(1S,3S)-3-isobutylcyclopentanol
[6]
HO
e. constitutional isomer with an ether
O
butoxycyclopentane
pyridine
TsO
CH3
Cl–
252
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Chapter 9–30
9.75
KOC(CH3)3
OTs
a.
b.
OH
*
Bulky base favors E2.
HBr
H CH3
H CH3
O
c.
Keep the stereochemistry at the stereogenic
center [*] the same here since no bond is
broken to it.
Br
*
CH3CH2
H
[1] CN
CH3CH2 C C H
CH2CH3
H
[2] H2O
C
+
C
H
CH3CH2
CH2CH3
H
NC
KCN
(CH3)3C
H
SN2
inversion
Br
CH3
O
O
CH3CO2
OTs
pyridine
H D
CN
OH
H D
Br
Br
OH
CH3CO2
H D
HBr
OCH3
OH
[1] NaOCH3
[2] H2O
OH
OH
OCH3
O
NaH
OCH2CH3
CH3CH2I
i.
CH2CH3
CH3
j.
CH3CH2 C O CH3
CH3
CH2CH3
CH2CH3
HI
(2 equiv)
CH3
CH3CH2 C
I
+
I
CH3
+ H2O
CH3
9.76
a.
CH2CH3
SN2
inversion
TsCl
OH
h.
C
CH3
f.
g.
C
CN
(CH3)3C
PBr3
OH
e.
H
H
OTs
d.
HO
OH
OH
b.
OH
c.
OH
HBr
or
PBr3
Br
HCl, ZnCl2
or
SOCl2, pyridine
Cl
[1] Na+ H
O
[2] CH3CH2Cl
O
identical
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Epoxides
253
Alcohols, Ethers, and Epoxides 9–31
[1] TsCl, pyridine
d.
OH
OTs
[2] N3
N3
Make OH a good leaving
group (use TsCl); then add N3.
9.77
a.
OH
HCl
or
SOCl2, pyridine
b.
OH
H2SO4
c.
OH [1] Na+ H
d.
OH
Cl
[2] CH3Cl
O
OCH3
CN
OTs [2] –CN
[1] TsCl, pyridine
Make OH a good
leaving group (use TsCl);
then add CN.
9.78
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
(h) =
OH
+ enantiomer
O
Cl
OH,
H2O
OH
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Chapter 9–32
9.79
OH
O
O
Cl
O
O
O
NaH
(CH3)2CHNH2
N
H H
O
proton
transfer
O
OH
N
H
propranolol
9.80
a. All 2° OH groups on stereogenic centers are circled (40 stereogenic centers).
OH
OH
O
O
HO
OH
O
OH
O
OH
NH2
OH
OH
OH
HO
OH
OH
OH
OH
OH
OH
O
HO
O
N
N
H
H
HO
OH
OH
O
OH
OH
O
OH
OH
OH
H
OH
HO
HO
O
OH
O
O
HO
OH
O
palytoxin
OH
OH
HO
OH
H
OH
OH
OH
OH
OH
OH
OH
This carbocation is resonance stabilized,
so loss of H2O to form it is easier than
loss of H2O from a 2° alcohol, where the
carbocation is not resonance stabilized.
R
b.
R C OH
OR'
hemiacetal
H+
R
R C OH2
OR'
R
R
R C
R C
OR'
OR'
resonance-stabilized cation
H2O
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Alcohols, Ethers, and Epoxides 9–33
9.81 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(CH
H
2CH3)2
Li+
O
+
H OH
O
+
HN(CH2CH3)2
LiOH
OH
9.82 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
OH2
H
1o alcohol
no 1o carbocation
at this step
H
H
1,2-shift
CH3CH2 C CH2
+
H2O
H
+ H2O
2o carbocation
H
CH3CH2 C CH2
H
CH3CH2 C CH2
H2O
H
+
CH3
H
+
H
+
C C
CH3
H
H
CH3CH2CH=CH2
H3O
H
C C
CH3 CH C CH3
=
CH3
H3O
CH3
H2O
9.83
OH OH
CH3 C
C CH3
OH
OH2 OH
H OSO3H
CH3 C
CH3 CH3
CH3 C
C CH3
pinacol
C CH3
CH3 C
shift
CH3 CH3
CH3 CH3
CH3
1,2-CH3
CH3
OH
C
CH3
+ H2O
+ HSO4
CH3
H2SO4 +
CH3 C
CH3
O
CH3 O H
C
CH3 C
CH3
CH3
HSO4
C
CH3
pinacolone
9.84
O
O H
O
C
I
O
H
H
O
Na+ OH
C
I
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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257
Alkenes 10–1
C
Chhaapptteerr 1100:: A
Allkkeenneess
G
Geenneerraall ffaaccttss aabboouutt aallkkeenneess
• Alkenes contain a carbon–carbon double bond consisting of a stronger bond and a weaker 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
H
CH3
C C
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 bond is electron rich and much weaker than a bond, alkenes undergo addition reactions
with electrophiles (10.8).
SStteerreeoocchheem
miissttrryy ooff aallkkeennee aaddddiittiioonn rreeaaccttiioonnss ((1100..88))
A reagent XY adds to a double bond in one of three different ways:
• Syn addition—X and Y add from the same side.
H BH2
C C
•
BH2
C C
•
Syn addition occurs in hydroboration.
Anti addition—X and Y add from opposite sides.
C C
•
H
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.
C C
H X
or
H2O, H+
H
X(OH)
C C
•
H
and
C C
X(OH)
Syn and anti addition occur in
hydrohalogenation and
hydration.
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Chapter 10–2
A
Addddiittiioonn rreeaaccttiioonnss ooff aallkkeenneess
[1] Hydrohalogenation—Addition of HX (X = Cl, Br, I) (10.9–10.11)
RCH CH2 + H X
R CH CH2
X
H
alkyl halide
•
•
•
•
•
The mechanism has two steps.
Carbocations are formed as intermediates.
Carbocation rearrangements are possible.
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)
RCH CH2 + H OH
H2SO4
R CH CH2
OH H
alcohol
RCH CH2 + H OR
H2SO4
R CH CH2
OR H
ether
For both reactions:
• The mechanism has three steps.
• Carbocations are formed as intermediates.
• Carbocation rearrangements are possible.
• Markovnikov’s rule is followed. H bonds
to the less substituted C to form the more
stable carbocation.
• Syn and anti addition occur.
[3] Halogenation—Addition of X2 (X = Cl or Br) (10.13–10.14)
RCH CH2
+
X X
R CH CH2
X
X
vicinal dihalide
•
•
•
•
The mechanism has two steps.
Bridged halonium ions are formed as
intermediates.
No rearrangements occur.
Anti addition occurs.
[4] Halohydrin formation—Addition of OH and X (X = Cl, Br) (10.15)
RCH CH2 + X X
H2O
R CH CH2
OH X
halohydrin
•
•
•
•
•
•
The mechanism has three steps.
Bridged halonium ions are formed as
intermediates.
No rearrangements occur.
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)
[1] BH3 or 9-BBN
• Hydroboration has a one-step mechanism.
RCH CH2
R CH CH2
• No rearrangements occur.
[2] H2O2, HO
H OH
• OH bonds to the less substituted C.
alcohol
• Syn addition of H2O results.
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259
Alkenes 10–3
C
Chhaapptteerr 1100:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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.
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 bonds
c. a compound with 2 rings
b. a compound that has 1 ring and 1 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.
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Chapter 10–4
3-methyl
1
a. [1] CH2 CHCH(CH3)CH2CH3
CH3
CH
CHCHCH
2
[2]
2CH3
5 C chain with double bond
pentene
[3] 3-methyl-1-pentene
1-pentene
3
CH3CH2
b. [1] (CH3CH2)2C CHCH2CH2CH3
C CHCH2CH2CH3
[2]
CH3CH2
7 C chain with double bond
heptene
[3] 3-ethyl-3-heptene
3-heptene
3-ethyl
2-ethyl
c. [1]
[2]
4-methyl
[3] 2-ethyl-4-methyl-1-pentene
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
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Alkenes 10–5
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.
higher priority
Cl
CH3
a.
higher priority
C C
higher priority
Two higher priority groups are
on opposite sides: E isomer.
OCH3
Br
H
H
H
c.
higher priority
higher priority
C6H5
O
H
higher priority
CH3CH2
b.
CH2CH3
higher priority
kavain
C C
H
higher priority
In both double bonds, the two higher priority groups are on opposite
sides: E isomers.
CH3
Two higher priority groups are
on the same side: Z isomer.
10.7 To name an alkene: First follow the rules from Answer 10.4. Then, when necessary, assign an E
or Z prefix based on priority, as in 10.6.
2-hexene
Br
CH2CH2CH3
CH3
[2]
C C
H
higher priority
CH2CH2CH3
CH3
a. [1]
[3] (2E)-3-bromo-2-hexene
C C
E alkene
6 C chain with double bond
hexene
Br
H
higher priority
3-bromo
3-decene
CH3CH2
b. [1]
CH2CH2CH2CH2C(CH3)3
CH3CH2
[2]
C C
H
higher priority
CH2CH3
10 C chain with double bond
decene
CH3
C C
Z alkene
H
CH3
CH2CH2CH2CH2CCH3
9,9-dimethyl
higher priority
9,9-dimethyl
CH2CH3
4-ethyl
[3] (3Z)-4-ethyl-9,9-dimethyl-3-decene
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Chapter 10–6
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
E
(2E,4E)-2,4-hexadiene
Z
Z
(2E,4Z)-2,4-hexadiene
Z
(2Z,4Z)-2,4-hexadiene
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
CH3CH2
C C
CH3
H
CH3CH2
C C
CH3
H
All dipoles cancel.
smallest surface area
no net dipole
lowest bp
CH2CH3
C C
CH2CH3
Two dipoles cancel.
no net dipole
trans isomer
intermediate bp
H
H
Two dipoles reinforce.
net dipole
cis isomer
highest bp
10.11 Recall from Section 5.13B that the odor of a molecule is determined more by shape than by
functional groups. That is why the R and S isomers of limonene smell so differently.
H
(R)-limonene
H
(S)-limonene
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Alkenes 10–7
10.12 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
10.13
Br
a.
H2SO4
OH
CH2=CHCH2CH2CH2CH3
NaOCH2CH3
b.
+
CH3CH=CHCH2CH2CH3
10.14 To draw the products of an addition reaction:
[1] Locate the two bonds that will be broken in the reaction. Always break the bond.
[2] Draw the product by forming two new bonds.
HCl
H
CH3
two new bonds
a.
c.
Cl
b.
H
HCl
CH3
CH3
CH3
CH3CH2CH2CH CHCH2CH2CH3
two new bonds
Cl
H H
HCl
CH3CH2CH2 C C CH2CH2CH3
H Cl
two new bonds
10.15 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 CX bond.
transition state
step [1]:
H
+
H
H
H Cl
[1]
transition state
step [2]:
H
H
H
H
[2]
Cl
Cl
+ H
Cl
+
Cl
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Chapter 10–8
10.16 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.
10.17 To determine which alkene will react faster, draw the carbocation that forms in the ratedetermining step. The more stable, more substituted carbocation, the lower the Ea to form it and
the faster the reaction.
CH3
CH3
H
C C
CH3
CH3
C
H X
H
C C
H
CH3
H
CH2
H
CH3CHCH2
H X
3° carbocation
faster reaction
H
H
2° carbocation
slower reaction
10.18 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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Alkenes 10–9
10.19 Addition of HX to alkenes involves 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
CH3 C C CH2CH3
CH2CH3
CH3 C C CH2CH3
H Br
H CH3
CH3 C C CH(CH3)2
CH3 C C CH(CH3)2
CH(CH3)2
(cis or trans)
CH3 C C CH2CH3
H
H H
H
H
H H
Br H
2° carbocation
2° carbocation
Rearrangement would not further
stabilize either carbocation.
no rearrangement
C C
c.
H H
CH3 C C CH2CH3
H
(cis or trans)
CH3
H H
1,2-H shift
H CH3 CH3
CH3 C C
H
H
2° carbocation
Rearrangement would not further
stabilize the carbocation.
C CH3
H H
2° carbocation
rearrangement
3° carbocation
more stable
H H
H CH3 CH3
CH3 C C CH(CH3)2
CH3 C C
Br H
H H
C CH3
Br
10.20 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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Chapter 10–10
10.21 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
10.22
CH3
CH2CH2CH3
CH3
or
CH2 C
a. CH3 C CH2CH2CH3
OH
C CHCH2CH3
CH3
CH3
H+ would add here to form
a 3° carbocation.
CH3
OH
b.
CH3
CH2
or
H+
H+ would add here to form
a 3° carbocation.
would add here to form
a 3° carbocation.
or
c.
H+
OH
H+ would add here to form
a 3° carbocation.
CH3CH CHCH3
would add here to form
a 2° carbocation.
(cis or trans)
10.23
H
H2O
OH
H2SO4
1-pentene
HO
H
enantiomers
10.24 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
transition state [1]:
X
X X
Step [1]
C C
transition state [2]:
X
X +
+
X
C C
X
Step [2]
C C
C
+
C
C
X
X
X
C
10.25 Halogenation of an alkene adds two elements of X in an anti fashion.
a.
Br2
Br
Br
Cl2
CH3
Cl
CH3
Cl
Cl
Cl
b.
Br
Br
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Alkenes 10–11
10.26 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.
Cl
Cl
Cl
CH3
Cl
CH3
Cl2
a.
c.
Br2
CH3
CH3
CH3
Br
enantiomers
H
b.
C
Br
Br2
Br
H
C
C
H
C
H
Br
achiral meso compound
10.27 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
H
Attack of Br– can occur
from the left or right:
C
CH3
C C
CH3
H
Br
Br Br
below
CH3
CH3
H
Br
H
above
Br
H
H
C
H
enantiomers
C
C
CH3
Br
CH3
right
left
left
right
+ Br
CH3
H
Br
C
C
H
CH3
CH3
H
Br
C
C
+ Br
CH3
H
C
Br
Br
C
H
CH3
H
CH3
CH3
C
H
C
Br
+ Br
H
CH3
CH3
C
H
C
Br
H
CH3
+ Br
Br
H
C
CH3
H
Br
diastereomers
C
Br
CH3
Br
H
C
CH3
H
C
CH3
Br
CH3
H
C
Br
All four compounds are identical—an achiral meso compound.
Br
C
H
CH3
Br
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Chapter 10–12
10.28 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.
Br
NBS
a.
DMSO, H2O
HO
CH3
Cl2
b.
HO
H2O
Br
CH3
OH
CH3
OH
Cl
Cl
Cl bonds to the carbon
with more H's to begin with.
10.29
H
a. (CH3)2S
H
H CH2CH3
H B N CH2CH3
CH3
H B S
b. (CH3CH2)3N
CH3
H CH2CH2CH2CH3
H B P CH2CH2CH2CH3
c. (CH3CH2CH2CH2)3P
H CH2CH3
H CH2CH2CH2CH3
10.30 In hydroboration the boron atom is the electrophile and becomes bonded to the carbon atom that
had more H’s to begin with.
CH3
a.
C CH2
CH3
BH3
CH3 C CH2
CH3
BH3
CH2BH2
H BH2
C with more H's.
B will add here.
C with more H's.
B will add here.
BH3
b.
CH2
c.
BH2
H
C with more H's.
B will add here.
10.31 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
BH3
H2O2, –OH
CH3CH2CH CH2
BH2
H
OH
CH2CH3
H
CH2CH3
H2O2, –OH
H
CH2CH3
H
CH2CH3
H
BH2
BH2
OH
OH
CH3
CH3
H
CH3
BH2
BH2
H2O2,
CH3
CH3
H
OH
CH3
H
–OH
CH3
CH3
H
OH
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Alkenes 10–13
10.32 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
10.33
CH2
a.
CH2
H2O
H2SO4
OH
Hydration places the OH
on the more substituted carbon.
CH3
[1] BH3
CH2OH
[2] H2O2, HO
Hydroboration–oxidation places the OH
on the less substituted carbon.
OH
Hydration places the OH
on the more substituted carbon.
H2O
H2SO4
b.
OH
[1] BH3
[2] H2O2, HO
OH
H2O
H2SO4
c.
[1] BH3
[2] H2O2, HO
Hydroboration–oxidation places the OH
on the less substituted carbon.
Hydration places the OH
on the more substituted carbon.
Hydroboration–oxidation places the OH
on the less substituted carbon.
HO
10.34 There are always two steps in this kind of question:
[1] Identify the functional group and decide what types of reactions it undergoes (for
example, 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
a.
CH2
HBr
CH3
c.
Br
alkene:
addition reactions
H2SO4
OH
alcohol:
substitution and elimination
nucleophile and base
Cl
b.
CH2CH2CH3
OCH3
NaOCH3
2° alkyl halide:
substitution and elimination
CH3CH2CH=CHCH3
(cis and trans)
CHCH2CH3
CH2CH2CH3
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Chapter 10–14
10.35 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
OH
alkyl halide
Can undergo
substitution and elimination.
K+ –OC(CH3)3
halohydrin:
Can form from an
alkene with Cl2 and H2O.
Cl2
Cl
Br
H2O
OH
OH is added to
the more substituted C.
CH3
?
OH
b.
OH
alcohol
Can undergo
substitution and elimination.
alcohol
Can be formed by
substitution and addition.
CH3
OH
CH3 [1] BH
3
H2SO4
[2] H2O2, HO–
(major
product)
CH3
OH
OH is added to
the less substituted C.
10.36 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
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Alkenes 10–15
10.37 First determine the number of degrees of unsaturation in the compound. Then decide which
combinations of rings and 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 bonds
3 bonds + 1 ring
2 bonds + 2 rings
1 bond + 3 rings
4 rings
10.38 The statement is incorrect because in 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
higher
priority
Cl
higher
priority
higher
priority
N(CH2CH3)2
O
O
enclomiphene
E isomer
N(CH2CH3)2
zuclomiphene
Z isomer
10.39 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
d.
CH3
CH3
4-methyl
CH3
C C
H
CH3
C C
CH2CH(CH3)2
CH3
higher
3-methyl
priority
5-methyl
CH2CHCH3
H
e.
1-ethyl
5-isopropyl
CH2CH(CH3)2
Higher priority groups
are on opposite sides =
E alkene.
6 C chain with a double bond =
hexene
6 C ring with a double bond =
cyclohexene
C C
H
2-hexene
CH3 (2E)-3,5-dimethyl-2-hexene
CH3
higher
priority
1-ethyl-5-isopropylcyclohexene
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Chapter 10–16
1-sec-butyl-2-methylcyclopentene
2-methyl
f.
1-sec-butyl
5 C ring with a double bond =
cyclopentene
E double bond (higher priority groups on
opposite sides with bold bonds)
4-isopropyl
g.
(4E)-4-isopropyl-4-hepten-3-ol
3-ol
OH
OH
7 C chain with a double bond =
heptene
4-heptene
2-cyclohexene 5-sec-butyl-2-cyclohexenol
1-ol
h.
OH
OH
5-sec-butyl
6 C ring with a double bond =
cyclohexene
10.40 Use the directions from Answer 10.8.
2
4-ethyl
a. (3E)-4-ethyl-3-heptene
7 carbons
e. (2Z)-3-isopropyl-2-heptene
3-isopropyl
7 carbons
3
Higher priority groups on
opposite sides = E.
3,3-dimethyl
b. 3,3-dimethylcyclopentene
5 carbon ring
1
Higher priority groups on
the same side = Z.
1
f. cis-3,4-dimethylcyclopentene
or
3
4
5 carbon ring
3,4-dimethyl
c. cis-4-octene
g. trans-2-heptene
8 carbons
4
Higher priority groups on
the same side = cis.
7 carbons
2 Higher priority groups on
opposite sides = trans.
4-propyl
2
d. 4-vinylcyclopentene
5 carbon ring
4
1-isopropyl
1
h. 1-isopropyl-4-propylcyclohexene
6 carbon ring
10.41
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
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Alkenes 10–17
10.42
CH3
CH3
H
CH3
a.
c.
(1Z,4S)-1,4-dimethylcyclodecene
diastereomer
(1E,4R)-1,4-dimethylcyclodecene
CH3
CH3
CH3
H
b.
(1Z,4R)-1,4-dimethylcyclodecene
diastereomer
(1E,4S)-1,4-dimethylcyclodecene
enantiomer
10.43 Name the alkene from which the epoxide can be derived and add the word oxide.
1-ethyl
H
derived
a.
O
O
c.
from
H
derived
from
H
6 carbon ring
cyclohexene
1-ethylcyclohexene
1-ethylcyclohexene oxide
(3E)- 3-heptene oxide
or
trans-3-heptene oxide
H
7 carbon chain
heptene
(3E)- 3-heptene
or
trans-3-heptene
2-methyl
O
b.
derived
derived
d.
C(CH3)3
C(CH3)3
from
from
2-methyl-2-hexene oxide
O
6 carbon chain
hexene
2-methyl-2-hexene
5 carbon ring
cyclopentene
4-tert-butylcyclopentene oxide 4-tert-butylcyclopentene
10.44
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.
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Chapter 10–18
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
1
5
As written this methyl
is at C2. Re-number
to put it at C1.
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.
1-cyclohexen-4-ol
new name:
5-isobutyl-1-methylcyclohexene
OH
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.
6
OH
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
10.45
COOH
a. and b.
COOH CH3 H
CH3O
H
E
Z
HOOC
E
S
E
E
bongkrekic acid
Z
R
Z
c. Since there are 7 double bonds
and 2 tetrahedral stereogenic
centers, 29 = 512 possible stereoisomers.
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Alkenes 10–19
10.46
H
2-methyl-1-pentene (2E)-4-methyl-2-pentene
4-methyl-1-pentene
(2E)-3-methyl-2-pentene
(3R)-3-methyl-1-pentene
2-methyl-2-pentene (2Z)-4-methyl-2-pentene
2-ethyl-1-butene
(2Z)-3-methyl-2-pentene
(3S)-3-methyl-1-pentene
H
10.47
O
stearic acid
OH
hIghest melting point
no double bonds
OH
intermediate melting point
one E double bond
O
elaidic acid
O
OH
oleic acid
lowest melting point
one Z double bond
10.48
O
a.
O
OH
all trans double bonds
higher melting point
OH
O
eleostearic acid
b.
OH
all cis double bonds
lower melting point
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Chapter 10–20
10.49
O
a.
O
O
O
O
O
O
*
O
O
O
O
O
A
A has one tetrahedral stereogenic center,
labeled with an asterisk [*].
B
O
O
O
b.
O
O
O
R
S
O
H
O
H
O
O
O
O
10.50
transition state [1]:
Br
[1]
+
Br H
C C
H
H
[2]
Br
H C C H
H
Br H
H
Br
H
H Br
C C H
Energy
Br2 H
CH2 CH2
A
transition state [2]:
+
Br
A
H
CH2 CH2
BrCH2CH2Br
+ Br2
H
H
C C
Br
H
Reaction coordinate
10.51 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
Ho (kJ/mol)
C–C bond
+ 267
I
+ 297
H
Total
+ 564 kJ/mol
Ho (kJ/mol)
CH2ICH2 H
410
I
222
C
Total
[3] Overall Ho =
sum in Step [1]
+
sum in Step [2]
+ 564 kJ/mol
632
kJ/mol
– 68
kJ/mol
632 kJ/mol
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Alkenes 10–21
CH2 CH2 + HCl
CH3CH2 Cl
[1] Bonds broken
[3] Overall Ho =
[2] Bonds formed
C–C 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
Ho (kJ/mol)
Total
Ho (kJ/mol)
749
kJ/mol
– 51 kJ/mol
Compare the H°:
Addition of HI: –68 kJ/mol more negative H°, larger Keq.
Addition of HCl: –51 kJ/mol
10.52
H
HBr
a.
Br
Br2, H2O
f.
Br
H
HI
b.
NBS
g.
aqueous DMSO
I
H
H2O
H2SO4
CH3CH2OH
d.
H2SO4
Cl2
e.
[2] H2O2,
i.
OH
Br
Br
+
OH
OH
HO–
OH
OH
H
OH
H
[1] BH3
h.
c.
Br
+
H
[1] 9-BBN
–
[2] H2O2, HO
OCH2CH3
OH
Cl
+
Cl
Cl
Cl
10.53
CH3
a.
C CH2
CH3
CH3
b.
C CH2
HBr
c.
C CH2
CH3
CH3
d.
C CH2
CH3
HI
e.
CH3
f.
CH3 C CH3
C CH2
C CH2
H2SO4
CH3CH2OH
H2SO4
Cl2
CH3
CH3 C CH3
OH
CH3
h.
CH3
CH3
CH3
CH3 C CH2Cl
Cl
NBS
aqueous DMSO
[1] BH3
[2] H2O2, HO–
CH3
CH3 C CH3
OCH2CH3
C CH2
i.
C CH2
CH3
CH3 C CH2Br
OH
CH3
g.
CH3
H2O
CH3
Br2, H2O
CH3
CH3
I
CH3
C CH2
CH3 C CH3
Br
CH3
CH3
CH3
CH3
[1] 9-BBN
[2] H2O2, HO–
CH3
CH3 C CH2Br
OH
CH3
CH3 C CH2OH
H
CH3
CH3 C CH2OH
H
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Chapter 10–22
10.54
Br
Br2
Br
Halogenation
a.
OH
Br2
b.
Halohydrin formation: Br adds to the C
that had more H's to begin with.
Br
H2O
OCH3
Br2
Same as halohydrin formation, except
CH3OH in place of H2O.
Br
c.
CH3OH
10.55
Br
a.
Br
d.
CH3CH2CH2CH=CHCH3
Br
Cl
b.
e.
CH2 or
CH3
Cl
Cl
CH3
CH3
c.
CH3
C CH3
C CH2
CH3CH2
f.
(CH3CH2)3CBr
Br
C CHCH3
CH3CH2
10.56 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
acid-catalyzed addition
OH
b.
hydroboration–oxidation
e.
OH
OH
d.
hydroboration–oxidation
or
OH
Both methods would give
product mixtures.
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Alkenes 10–23
10.57
a.
Cl
(CH3CH2)2C=CHCH2CH3
CH3CH2
H
CH3CH2
CH3CH2 H
HCl
C C
Cl2
d.
CH3CH2 C
Cl
CH2CH3
Cl
C H
CH2CH3
e.
CH2
b.
CH2Br
OH
Br adds here
to less substituted C.
(CH3CH2)2C=CH2
CH3CH2
f.
CH3CH2 C CH3
H2SO4
CH3CH2
[1] 9-BBN
CH3CH2
H2O
C CH2
CH2
OH
H
[1] BH3
(CH3)2C=CHCH3
CH2OH
[2] H2O2, HO
OH adds here
to less substituted C.
H adds here
to less substituted C.
c.
Br2
H2O
H adds here
to less substituted C.
(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
Br2
h.
CHCH3
Br Br
10.58
CH3CH2
H
or
C C
CH3CH2CH2
CH3
CH2CH2CH3
or
CH3CH C
H
Cl
CH3CH2 C CH2CH2CH3
C CHCH2CH3
CH3
CH3
CH3CH2
10.59
H
a.
CH3
H
C C
C C
=
H
CH3
C
H
CH3
b.
CH3
CH3CH2
H
CH3CH2
CH3
CH3
C C
H
CH3CH2
CH3 CH3CH2
=
C C
c.
CH3
=
C C
H
CH3
C C
CH3
H
Cl2
HO
H2O
C
CH3
CH3CH2
NBS
DMSO, H2O
H
C
Br
H
CH3
H
C
C
CH3
H
b.
CH2
H2O
H2SO4
(CH3)3C
CH3
(CH3)3C
OH
I H
H
I
C
HO
CH3
H
C
CH3CH2
CH3
CH3
CH3
H
HO
OH
HI
Cl
CH3
CH3CH2
C
10.60
a. (CH3)3C
C
Br
Cl
HO
C
CH3
Br
CH3CH2
CH3
C
CH3
H
Br
Br
H
Br2
C
Br
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Chapter 10–24
CH3
Cl
Cl2
c.
CH3
CH3
CH3
Cl
d.
Cl
CH3
CH3
Cl
[1] BH3
CH3
only anti addition
CH3
H
[2] H2O2, HO
CH3
CH3
CH3
H
OH
Br CH2CH3
CH3
only syn addition
OH
Br CH2CH3
HBr
e.
Cl
Cl2
f.
OH
only anti addition
+
H2O
OH
Cl
NBS
g.
Br
DMSO, H2O
h.
H2O
CH3CH=CHCH2CH3
Br
HO CH3
HO CH3
H OH
HO H
H2SO4
OH
10.61
Cl
CH3
CH2 C
CH3
Br2
H
NaCl
H
C
CH2 C CH3
C
Br
Cl– acts as the nucleophile.
Cl
CH3
CH3
Br
+ Br
Br acts as the electrophile and is therefore added to
the C with more H's to begin with.
CH3
+
Br Br
10.62 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
Br2
a.
H
Br
Br
H
H
Br
cis-4-octene
Br
(4R,5R)-4,5-dibromooctane
(4S,5S)-4,5-dibromooctane
enantiomers
H
Br2
b.
H
trans-4-octene
H
Br
H
H
Br
(4R,5S)-4,5-dibromooctane
meso compound
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Alkenes 10–25
10.63
CH3CH2
CH3CH2
CH2CH3
C C
H
H
H
H Cl
CH3CH2
CH3CH2
CH2CH3
H
C C H
H
H
Cl–
CH3CH2
H
C
Cl
Cl
H
CH2CH3
By protonation of the alkene, the cis and
trans isomers produce identical
carbocation intermediates.
Cl–
CH3CH2
CH2CH2CH3
CH2CH3
H Cl
C C H
H
H
C C
C
CH3CH2
C
CH2CH2CH3
H
CH3CH2
CH2CH2CH3
Cl
C
Cl
CH2CH2CH3
H
Both cis- and trans-3-hexene give the same
racemic mixture of products, so the
reaction is not stereospecific.
10.64 The alkene that forms the more stable carbocation reacts faster, according to the Hammond
postulate.
H Br
a.
CH CH
CH CH
CH CH
H
H
CH CH
CH CH
CH CH
H
H
H
This 2° carbocation is resonance stabilized, making it more
stable, so the starting alkene reacts faster with HBr.
CH3 CH CH CH3
H Br
b.
CH3 CH CH CH3
H
2° carbocation
no resonance stabilization
CH3
Markovnikov
CH3
addition
CH2 C
CH3
CH2OCH3
H Br
C(CH3)2
faster
H
3° carbocation
H Br
CH2 C
CH2
Markovnikov
addition
CH3
CH2
H
C
CH2–OCH3
3° carbocation
This carbocation is still 3°, but the nearby
electronegative O atom withdraws electron density
from the carbocation, destabilizing it. Thus, the
reaction to form this carbocation occurs more slowly.
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Chapter 10–26
10.65
Cl
H
H
H Cl
a.
Cl
H
H
H
H
O
H
HO
H
C
O
O
O C
H
H
CH3
O C
CH3
H
H
H
H
H
H Br
b.
Cl
H
and
H
Br
Br
C H
CH3
C
CH3
2o carbocation
+ HCl
H
H
1,2-H shift
CH3
3o carbocation
+ Br–
10.66
a.
H
CH3
CH3
CH3
1,2-shift
H2O
CH3
CH3
O H
–
CH3 HSO4
CH3
OH
CH3
H2SO4
HSO4–
H OSO3H
2o carbocation
H2O
3o
+ H2SO4
carbocation
H OSO3H
b.
OH
OH
H2SO4
+ H2SO4
O
HSO4–
CH3
H
–OSO
3H
O
CH3
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Alkenes 10–27
10.67
H
H OH2
H
+ H3O+
O H
H2O
OH
H2O
H
H
O H
1,2-shift
H2O
H2O
H
H
OH
+ H3O+
H2O
O H
1,2-shift
H2O
OH
H
H
+ H3O+
10.68 The isomerization reaction occurs by protonation and deprotonation.
H
CH2
OSO3H
C
CH3
CH3
CH3
C H
CH3
CH3
C
CH3
C H
CH3
CH3
HSO4–
2,3-dimethyl-1-butene
CH3
C C
CH3
CH3
2,3-dimethyl-2-butene
10.69
H
H
C C
H
H Br
H
H
CH3 C
C C
H
H
C C
H
H
CH3CH CHCH2
Br
CH3CH=CH–CH2Br
H
Br
Br
CH3CHCH CH2
Since two resonance structures can be drawn
for the intermediate carbocation, two different
products result from attack by Br–.
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Chapter 10–28
10.70
Br
HBr
A
OCH3
OCH3
B
H Br
COOCH3
C
COOCH3
D
H Br
Br
O
OCH3
OCH3
H
Br
HBr
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
+ O
H
This carbocation is destabilized
by the + 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.
10.71
Br Br
H
Br
OH
OH
O
Br
O
Br
+ HBr
+ Br
+ Br
10.72
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
OH
c.
H
H2SO4
[1] BH3
[2] H2O2,
d.
CH3 CH CH2I
K+ –OC(CH3)3
OH
HO–
+ H2
CH3
(CH3)2C
CH2
HCl
CH3 C Cl
CH3
e.
CH3
Cl
Br
Br2
K+ –OC(CH3)3
H2O
f.
CH3CH CH2
O–
NaH
Br2
Br
OH
Br
CH3CH CH2
NaNH2
(2 equiv)
CH3C CH
CH3I
OCH3
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Alkenes 10–29
10.73
Cl
Cl2
a.
H2O
+ H2
Br
CN
NaCN
Br
OH
O
O
NaH
NaOH
c.
O
O
OH
HBr
b.
Cl
Na+ H–
CH3CH2CH2I
(from b.)
[1] –SH
d.
O
[2] H2O
OH
+ enantiomer
SH
(from a.)
e.
O
[1] –CCH
OH
[2] H2O
C
+ enantiomer
(from a.)
CH
10.74
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.)
10.75
OH
Br
OH
POCl3
Br2
pyridine
H2O
A
B
major product
H2SO4
OH
Br2
NaH
O
H2O
O
NaH
Br
C
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Chapter 10–30
10.76
TsO H
TsO–
H
+ TsOH
+ TsO–
A
isocomene
10.77
Na+ HCO3–
O
H O
O
O
O
O
O
O
I
I
I
OCH3
HO
OCH3
OCH3
B
HO
HO
Na+
I
OCH3
HO
I–
C
H2CO3
10.78
TsO H
OH
OH2
nerol
+ TsOH
H2O
O H
H
H
OH
OH
-terpineol
OSO2Cl
OH
OH
OH
OTs
H OSO2Cl
HSO3Cl
-cyclogeraniol
nerol
OSO2Cl
10.79
O
H OSO3H
O H
OH
+
HSO4–
H2O
H O
H
HSO4–
OH
HO
OH
+ H2SO4
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Alkynes 11–1
C
Chhaapptteerr 1111:: A
Allkkyynneess
G
Geenneerraall ffaaccttss aabboouutt aallkkyynneess
• Alkynes contain a carbon–carbon triple bond consisting of a strong bond and two weak 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 bonds make an alkyne electron rich, alkynes undergo addition reactions with
electrophiles (11.6).
A
Addddiittiioonn rreeaaccttiioonnss ooff aallkkyynneess
[1] Hydrohalogenation—Addition of HX (X = Cl, Br, 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.
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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] BH3
OH
C
C
H
H H
enol
The unstable enol, first formed after
oxidation, rearranges to a carbonyl
group.
aldehyde
R
Reeaaccttiioonnss iinnvvoollvviinngg aacceettyylliiddee aanniioonnss
[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.
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289
Alkynes 11–3
C
Chhaapptteerr 1111:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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 CH2CH2CH3
CH3 C C CH2CH3
HC C CH CH3
CH3
terminal alkyne
internal alkyne
terminal alkyne
11.2
Csp–Csp3
(c)
Csp2–Csp
(b)
OH
(a)
Csp3–Csp2
Increasing bond strength: a < c < b
O
santalbic acid
11.3 Like alkenes, the larger the number of alkyl groups bonded to the sp hybridized C, the more stable
the alkyne. This makes internal alkynes more stable than terminal alkynes.
11.4 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.
H C C CH2CCH2CH2CH3
CH2CH3 CH3
c.
CH2=CHCH2CHC
CH2CH2CH3
CH3
4,4-dipropyl-1-heptyne
b. CH3C CCClCH2CH3
CCCH2CH2CH3
(Number to give the lower
number to the first site of
unsaturation.)
4-ethyl-7,7-dimethyl-1-decen-5-yne
d.
1
5 (The longest chain must contain both functional groups.)
CH3
3
4-chloro-4-methyl-2-hexyne
3-isopropyl-1,5-octadiyne
11.5 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.
OH
a. trans-2-ethynylcyclopentanol
5 C ring with OH at C1
OH
OH on C1
C CH
ethynyl at C2
or
C CH
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Chapter 11–4
tert-butyl at C4
b. 4-tert-butyl-5-decyne
10 C chain with
a triple bond
triple bond at C5
c. 3-methylcyclononyne
9 C ring with a
triple bond at C1
3-methyl
triple bond at C1
11.6 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.
11.7 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
BrCH CHCH2CH2CH2CH3
Na+ –NH2
HC CCH2CH2CH2CH3
not isolated
b. CH2=CCl(CH2)3CH3
c. CH2=CH(CH2)3CH3
Na+ –NH2
Cl2
HC CCH2CH2CH2CH3
CH2CHCH2CH2CH2CH3
Cl Cl
Na+ –NH2
HC CCH2CH2CH2CH3
(2 equiv)
11.8 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.
c. CH2=CH– [pKa (CH2=CH2) = 44]
pKa > 25 = Can deprotonate acetylene.
d. (CH3)3CO– {pKa [(CH3)3COH] = 18}
pKa < 25 = Cannot deprotonate acetylene.
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Alkynes 11–5
11.9 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.
Br
2 HBr
a. CH3CH2CH2CH2 C C H
CH3CH2CH2CH2 C CH3
Br
b. CH3 C C CH2CH3
Br
2 HBr
Br
CH3 CH2 C CH2CH3
CH3 C CH2 CH2CH3
Br
Br
Br
c.
C CH
2 HBr
C CH3
Br
11.10
a.
+
Cl
c.
Cl
+
+
CH3
b. CH3 O CH2
O
NH
NH
CH2
11.11 Addition of one equivalent of X2 to alkynes forms trans dihalides.
Addition of two equivalents of X2 to alkynes forms tetrahalides.
CH3CH2 C C CH2CH3
2 Br2
Br Br
CH3CH2 C C CH2CH3
Br Br
CH3CH2 C C CH2CH3
Cl
Cl2
CH2CH3
C C
CH3CH2
trans dihalide
Cl
11.12
Cl2
CH3 C C CH3
Cl
CH3
C C
CH3
Cl
The two Cl atoms are electron withdrawing,
making the bond less electron rich and
therefore less reactive with an electrophile.
11.13 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
O
new C–H bond
OH
H H
OH
H
O
H
H
new C–H bond
O
b.
H
c.
new C–H bond
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Chapter 11–6
11.14 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
11.15
OH
OH
OH
a.
OH
b.
enol tautomers
constitutional isomers,
but not tautomers
11.16 Reaction with H2O, H2SO4, and HgSO4 adds the oxygen to the more substituted carbon.
Reaction with [1] BH3, [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] BH3
[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
Forms a ketone. H2O is added with the
O atom on the more substituted carbon.
C
H2SO4, HgSO4
CH3
H H
C
C O
H
[1] BH3
[2] H2O2, HO
Forms an aldehyde. H2O is added with the
O atom on the less substituted carbon.
11.17
a. H C C H
[1] NaH
[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
CH3
+
C CH
+ NaCl
1° alkyl halide
substitution product
3° alkyl halide
elimination product
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Alkynes 11–7
11.18
a.
(CH3)2CHCH2C
CH3
CH3 C CH2
H
CH
CH3
CH3 C CH2Cl
H
C C H
terminal alkyne
only one possibility
C C H
1° RX
b. CH3C CCH2CH2CH2CH3
CH3
C C
[1]
[1]
CH2CH2CH2CH3
[2]
C C CH2CH2CH2CH3
CH3Cl
[2] CH3 C C
Cl CH2CH2CH2CH3
internal alkyne
two possibilities
1° RX
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
11.19
Na+ H
HC C H
HC C
Na+ H
H C C CH2CH3
CH3CH2–Br
C C CH2CH3 + H2
H
+ H2
+ Na+Br–
+ Na+
(CH3)2CHCH2C
Br
H
(CH3)2CHCH2CH2
–
C C CH2CH3 + Br
11.20
CH3
CH3
CH3
CH3 C C C C CH3
CH3
CH3 C C C
CH3
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
11.21
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
b.
O
[1]
C C H
[2] H2O
C
OH
CH
+
C
OH
CH
enantiomers
Backside attack of the nucleophile
(–CCH) at either C since both
ends are equally substituted.
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Chapter 11–8
11.22
a. CH3CH2CH2Br
CH3CH2C C–Na+
b. (CH3)2CHCH2CH2Cl
c. (CH3CH2)3CCl
CH3CH2C C–Na+
CH3CH2C C–Na+
d. BrCH2CH2CH2CH2OH
e.
f.
CH3CH2CH2C CCH2CH3
(CH3CH2)2C
CH3CH2C C–Na+
O
CH3CH2C C–Na+
O
CH3CH2C C–Na+
(CH3)2CHCH2CH2C
CCH2CH3
CHCH3
CH3CH2C CH
BrCH2CH2CH2CH2O–Na+
CH3CH2C CH
H2O
CH3CH2C CCH2CH2OH
H2O
CH3CH2C CCH2CHCH3
OH
11.23 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
HC C
2 C's
CH3CH2C C
CH3CH2–Br
+ CH3CH2Br
CH3CH2C C H
Na+ H
HC C
+ CH3CH2Br
CH3CH2–Br
CH3CH2C C
CH3CH2C CCH2CH3
11.24
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, CC functional group
(can form an acetylide
anion by reaction with NaH)
O
Retrosynthetic
analysis:
CH3CH2CH2 C
Forward direction: H C C H
CH3CH2C C H
H C C H
H
Na+ H
C C H
CH3CH2–Br
CH3CH2C C H
[1] BH3
[2] H2O2,
HO–
O
CH3CH2CH2 C
H
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Alkynes 11–9
11.25
a., b.
N
O
O
O
N
O
c.
d. * = sp hybridized C
e. 3 < 1 < 2
H
HN
(1)
(2)
*
* *
O
*
C*
(3)
HO
erlotinib
most acidic
C–H proton
phomallenic acid C
most acidic
proton
shortest C–C single bond
Csp–Csp2
11.26 Use the rules from Answer 11.4 to name the alkynes.
3-hexyne
1-hexyne
2-hexyne
3-methyl-1-pentyne
4-methyl-1-pentyne
3,3-dimethyl-1-butyne
4-methyl-2-pentyne
11.27 Use the rules from Answer 11.4 to name the alkynes.
5
3
a. CH3CH2CH(CH3)C CCH2CH3
CH3
5-methyl-3-heptyne
3-heptyne
5-methyl
f.
CH3CHC CCHCH3
CH3
5
2
CH3
7-methyl
CH3CH2CHC CCHCHCH2CH3
CH2CH3
3,6-diethyl
HC C CH(CH2CH3)CH2CH2CH3
1-hexyne
3-ethyl
6-methyl
h.
3,6-diethyl-7-methyl-4-nonyne
d.
2,5-octadiyne
2
5-ethyl
6
4-ethyl
4-nonyne
CH3CH2
CCH3
g.
2,5-dimethyl
c.
CH2CH2CH3
CH3CH2C CCH2C
2,5-dimethyl-3-hexyne
CH3
3-methyl
3-ethyl-3-methyl-1-hexyne
CH3CH2 C C CH
3-ethyl
3-hexyne
b.
e.
3-ethyl-1-hexyne
(2E)-4,5-diethyl-2-decen-6-yne
1
ethynyl
1-ethynyl-6-methylcyclohexene
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Chapter 11–10
11.28 Use the directions from Answer 11.5 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
2-pentyne
Z
1
11.29 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
CH3
• C=O
• C=C
• one more CH bond
• OH on C=C
keto–enol tautomers
b.
and
OH is not bonded
to the C=C.
NOT keto–enol tautomers
H
and
O
• C=O
• one more CH bond
• C=C
• OH on C=C
keto–enol tautomers
O
OH
O
OH
c.
d.
OH
and
NOT keto–enol tautomers
OH is not bonded
to the C=C.
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Alkynes 11–11
11.30 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.
O
OH
E/Z isomers
possible
a.
O
b. CH3CH2CHO =
O
OH
CH2CH3
CH2CH3
c.
OH
+
CH2CH3
OH
H
H
E/Z isomers
possible
11.31 Use the directions from Answer 11.13 to draw each keto form.
a.
OH
O
OH
c.
OH
O
O
b.
H
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
O H
A
H2O H
O
OH + H3O
H
B
H
+ H2O
+ H2 O
11.34
O H
HO
O
O
H OH
HO
O
H OH
11.35
NHCH3
X
H O H
enamine 2
NHCH3
H
+ H2O
NCH3
NCH3
H
+ H2O
Y
imine
+ H3O
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Chapter 11–12
11.36 The equilibrium always favors the formation of the weaker acid and the weaker base.
a. HC C
+ CH3OH
c.
HC CH + CH3O
pKa = 15.5
pKa = 25
weaker acid
Equilibrium favors products.
b. CH3C CH
pKa = 25
+
CH3
+
CH3C C
Na+Br–
HC CH +
pKa = 25
weaker acid
Equilibrium favors
starting materials.
d. CH3CH2C C
CH4
HC C– Na
+ HBr
pKa = –9
CH3CH2C CH + CH3COO
+ CH3COOH
pKa = 25
pKa = 4.8
pKa = 50
weaker acid
Equilibrium favors products.
weaker acid
Equilibrium favors products.
11.37
HC CCH2CH2CH2CH3
a.
e.
CH3 C CH2CH2CH2CH3
(2 equiv)
Cl
Br
HBr
b.
O
Cl
HCl
Br
g.
H
[2] H2O2, HO–
NaH
f.
CH3 CCH2CH2CH2CH3
(2 equiv)
[1] BH3
C
CH2CH2CH2CH2CH3
Na+ C CCH2CH2CH2CH3
[1] –NH2
CH3CH2C CCH2CH2CH2CH3
[2] CH3CH2Br
Cl Cl
Cl2
c.
HC CCH2CH2CH2CH3
Cl Cl
(2 equiv)
H2O
d.
h.
H
[2]
OH
O
C C
H2SO4, HgSO4 H
[1] –NH2
CH3
CH2CH2CH2CH3
C
O
HOCH2CH2 C CCH2CH2CH2CH3
[3] H2O
CH2CH2CH2CH3
11.38
Br Br
HBr
a.
c.
(2 equiv)
O
[1] BH3
d.
(2 equiv)
[2] H2O2, HO–
Br Br
11.39
O
a.
(CH3CH2)3C
C CH
H2O
H2SO4, HgSO4
b.
(CH3CH2)3C
C CH
c.
(CH3CH2)3C
C CH
d.
(CH3CH2)3C
C CH
[1] BH3
[2] H2O2, HO–
HCl
(2 equiv)
[1] NaH
[2] CH3CH2Br
O
H2SO4
Br Br
Br2
b.
H2O
(CH3CH2)3C
(CH3CH2)3C
C
CH3
CH2CHO
(CH3CH2)3CCCl2CH3
(CH3CH2)3C
C CCH2CH3
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Alkynes 11–13
11.40 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'
HX
R C C R'
H X
R C C R'
R C C R'
H H
H H
H H
sp2 hybridized carbocation
more stable
faster reaction
sp hybridized carbocation
less stable
slower reaction
11.41
O
OH
C
a.
OH
O
b.
C
CH3
CH3
CH3
O
C
c.
CH
C
CH3 C CH
CH2
OH
C
CH3
d.
CH3CH2
O
CH
C
CH2
C C
CH2CH3
OH
11.42 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
CH2CH3
HO
O
CH3
C
H
CH3 C C CH3
C C
CH2CH3
CH3
2-butanone
CH3
11.43
a.
CH2CHO
C CH
b.
(CH3)2CHC
CCH(CH3)2
O
11.44 Equilibrium favors the weaker acid. CH3CH2CH2CH2–Li+ is a strong enough base to remove the
proton of an alkyne because its conjugate acid, CH3CH2CH2CH3, is weaker than a terminal alkyne.
RC C H
CH3CH2CH2CH2–Li+
RC C–Li+
pKa = 25
CH3CH2CH2CH3
pKa = 50
much weaker acid
11.45
a.
C CH
2 HBr
Br Br
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Chapter 11–14
b.
Cl
2 Cl2
(CH3)3CC
(CH3)3CC
CH
CHCl2
Cl
c.
CH
Cl Cl
[1] Cl2
CH
[2] NaNH2
C C
C C
(2 equiv)
H H
[1] BH3
C CH
d.
C
[2] H2O2, HO
e.
O
Cl
HCl
C CH
C
CH2
(1 equiv)
HC C + D2O
f.
g.
HC CD
C C CH3
h.
+ DO–
O
H2O
C
CH2
H2SO4
CH3CH2C C + CH3CH2CH2–OTs
CH3
O
C
CH2CH3
CH3
CH3CH2C CCH2CH2CH3
CH3
[1] HC C
+ OTs
CH3
[2] HO–H
O
O
i.
H
OH
C CH
C CH
H
j.
CH3C C–H
[1] Na+ H–
[2]
CH3C
Br
+
C–
CH3C C–H
2° halide
E2
CH2–I
k. CH3CH2C C–H
[1]
Na+ –NH2
C C–H [1] Na+ H–
l.
CH3CH2C C
C C
[2] O
[2]
CH3CH2C C
CH2
C CCH2CH2O– [3] HO–H
C C CH2CH2OH
11.46
CH2CH2Br
KOC(CH3)3
CH=CH2
Br2
H H
C C H
Br Br
A
B
KOC(CH3)3
C CH
DMSO
(2 equiv)
C
D
C CCH3
CH3I
NaNH2
C C
E
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Alkynes 11–15
11.47
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
11.48
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
O
H
CH3
[1] HC C
H
CH3 [2] H2O
HO
C
CH3
C
H
CH3
C
OH
H
C
O
C CH
d.
CH3 H
H
CH3
[1] HC C
H
CH3
[2] H2O
HO
H
C
H
CH3
CH3
CH3
11.49
TsCl
OH pyridine
H D
HC C
OTs
D H
SN2
B
A
retention
PBr3
SN 2
D H
Br
A
inversion
inversion
HC C
SN2
H D
C
inversion
OH
H
C
C
C
C
CH
OH
CH3
HC
HC
enantiomers
H D
H
Configuration is retained.
CH3 H
H D
C
C
CH
inversion
Cl HC C
b.
c.
CH3
H
C
H
CH3
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Chapter 11–16
11.50
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.
OH
OR
H C C
[2] H2O
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
11.51
Br H
CH3CH2 C C
CH3CH2
H
Br H
H
CH3–C
H
H
C C
Br
NH2
Br
1-butyne
H2N
H
C CH3
CH3
CH3 C C CH3
C C
CH3
Br
CH3CH2 C C H
NH2
H
Br
major product
more substituted alkyne
2-butyne
(E and Z isomers possible)
NH2
Reaction by-products:
H
Br H
CH3CH2 C C
NH2 H
H
Br H
NH2
C H
2 NH3 + 2 Br–
CH3CH C CH2
CH3 C C
Br
H
1,2-butadiene
11.52 Draw two diagrams to show and bonds.
bond
sp2 sp
sp3
+
CH2 C CH3
vinyl cation
sp3
sp2 sp
sp3
H
H
H
H
C
H
C
H
sp2
C
H
vacant p orbital
for the carbocation
C
C
H
C
H
H
sp3
All H's use 1s orbitals.
All bonds above are bonds.
The positive charge in a vinyl carbocation resides on a carbon that is sp hybridized, while in (CH3)2CH+,
the positive charge is located on an sp2 hybridized carbon. The higher percent s-character on carbon
destabilizes the positive charge in the vinyl cation. Moreover, the positively charged carbocation is now
bonded to an sp2 hybridized carbon, which donates electrons less readily than an sp3 hybridized carbon.
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Alkynes 11–17
11.53 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 HCC+, the positively charged C uses two p
orbitals to form two bonds. If the bond is formed using an sp hybrid orbital, the second hybrid
orbital would have to remain vacant, a highly unstable situation. See also Problem 11.52.
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.
more stable
less stable
11.54
CH3
H Cl
CH3
+
C
Cl– attack on the opposite side
to the H yields the Z isomer.
C C
Cl
CH3C CCH3
Cl
C CH3
H
CH3
CH3
CH3
C C
H
H
Cl
Cl
Cl– attack on the same side as
the H yields the E isomer.
11.55
a.
C
C
C
C
C
C
C
C
H
H
C O
H
H
CH3CH2 Li+
C H
O
OH
H OH
+ CH3CH3
+
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
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Chapter 11–18
11.56
H
C C OCH3
CH3CH2CH2
H OH2
H
CH3CH2CH2C C OCH3
H OH2
H2O
H
OH
C C
OCH3
CH3CH2CH2
OH
C C
OCH3
CH3CH2CH2
H
C C OCH3
CH3CH2CH2
H
H2O
O
CH3CH2CH2CH2 C
OCH3
O H
H2O
CH3CH2CH2CH2 C
OCH3
OH
CH3CH2CH2CH2 C
OCH3
H2O
H OH2
11.57
a.
H
H
CH3 C C CH2
CH3 C C CH2
NH2
CH3 C C CH2
CH3 C C CH
H
H
2-butyne
NH3
NH2
H
H2O
NH2
NH2
CH3CH2 C C
NH2
NH3
CH3 C C CH
CH3CH2 C C H
acetylide anion
CH3 C C CH
H
1-butyne
NH3
H
b. A more stable internal alkyne can be isomerized to a less stable terminal alkyne under these reaction
conditions because when CH3CH2CCH is first formed, it contains an sp hybridized C–H bond,
which is more acidic than any proton in CH3–CC–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.
11.58
Cl
a.
2 –NH2
Cl
Cl2
b.
KOC(CH3)3
Cl
(2 equiv)
DMSO
Cl
c.
KOC(CH3)3
Cl2
POCl3
OH
Cl
pyridine
(2 equiv)
DMSO
Cl
11.59
a. C6H5CH2CH2Br
KOC(CH3)3
C6H5CH=CH2
Br2
C6H5CHBrCH2Br
NaNH2
excess
C6H5C CH
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Alkynes 11–19
KOC(CH3)3
b. C6H5CHBrCH3
H2SO4
c. C6H5CH2CH2OH
Br2
C6H5CH=CH2
Br2
C6H5CH=CH2
NaNH2
C6H5CHBrCH2Br
excess
NaNH2
C6H5CHBrCH2Br
excess
C6H5C CH
C6H5C CH
11.60 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
11.61
a.
HC C–H
b.
HC C–H
Na+ H–
(CH3)2CHCH2–Cl
HC C
Na+ H–
(CH3)2CHCH2C
NaH
CH3CH2CH2–Cl
HC C
CH
CH3CH2CH2C CH
CH3CH2CH2C C
CH3CH2CH2–Cl
[1] BH3
c. CH3CH2CH2C CH
(from b.)
d.
[2] H2O2, HO–
H2SO4
HgSO4
(from b.)
e.
O
H2O
CH3CH2CH2C CH
CH3CH2CH2
2 HCl
CH3CH2CH2C CH
CH3CH2CH2C CCH2CH2CH3
CH3CH2CH2CH2CHO
C
CH3
CH3CH2CH2CCl2CH3
(from b.)
f.
CH3CH2CH2C CCH2CH2CH3
O
H2O
H2SO4, HgSO4
CH3CH2CH2
C
CH2CH2CH2CH3
(from b.)
11.62
Cl2
a. CH3CH2CH CH2
b. CH3CH2C CH
(from a.)
c. CH3CH2C CH
(from a.)
d. CH3CH2CH CH2
HBr
(2 equiv)
Cl2
(2 equiv)
Br2
Cl
Cl
CH3CH2CH CH2
CH3CH2CBr2CH3
CH3CH2CCl2CHCl2
CH3CH2CHBrCH2Br
2 –NH2
CH3CH2C CH
306
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Chapter 11–20
e.
NaH
CH3CH2C CH
CH3CH2CH2 Br
CH3CH2C C
CH3CH2C C CH2CH2CH3
(from a.)
[1]
f.
CH3CH2C C
O
CH3CH2C C CH2CH2OH
[2] H2O
(from e.)
O
OH
[1]
g. CH3CH2C C
[2] H2O
(from e.)
C CCH2CH3
(+ enantiomer)
11.63
NaH
CH3CH2CH2CH2CH2CH2Br
a.
HC CH
b.
CH3CH2CH2CH2CH2CH2C CH
HC C
NaH
CH3CH2CH2CH2CH2CH2C CH
CH3CH2CH2CH2CH2CH2C C
CH3CH2Br
CH3CH2CH2CH2CH2CH2C CCH2CH3
(from a.)
c.
CH3CH2CH2CH2CH2CH2C C
[2] H2O
(from b.)
d.
O
[1]
CH3CH2CH2CH2CH2CH2C CCH2CH2OH
CH3CH2CH2CH2CH2CH2C CCH2CH2OH
[1] NaH
CH3CH2CH2CH2CH2CH2C CCH2CH2OCH2CH3
[2] CH3CH2Br
(from c.)
11.64
K+ –OC(CH3)3
CH2 CH2
Br
Br
Br2
NaNH2
NaH
HC CH
HC C
Br (2 equiv)
Br
H2O
Br
C C
NaH
C C
HC C
H2SO4, HgSO4
O
11.65
a.
H2SO4
2 –NH2
Br2
OH
Br
CH3 C CH
NaH
CH3 C C
Br
CH3 C CCH2CH2CH3
SOCl2
OH
NaH
Cl2
b.
(from a.)
[1] CH3 C C
Cl
H2O
OH
O
(from a.)
[2] H2O
Cl
OH
CH3C CCH2CHCH3
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Alkynes 11–21
11.66
a.
H2SO4
OH
CH2 CH2
Br
Br2
NaNH2
(2 equiv)
Br
NaH
HC CH
HC C
Br
PBr3
OH
[1] NaH
C C
HC C
[2] O
HO
[3] H2O
OH
b.
H2SO4
Br2
CH2 CH2
C C
Br
H2O
NaH
O
OH
C C
NaH
HO
C C
CH3CH2Br
O
CH3CH2 O
(from a.)
(from a.)
11.67
Br Br
Br
CH3 C C H
CH3
Br
CH3
H
HO
C C
CH3 C C H
H O
H2O
CH3
Br
C C
C C H
H
H O
H
Br–
Br
H
+ H2O
H OH2
H2O
CH3
O
CH3
C
Br
C C H
H O
CH2Br
H3O+
H
H2O
11.68
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
+ CH3OH2
O
Y
308
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Chapter 11–22
11.69
CH3
C
C
O
H
C
O
H
CH3
CH3
OH
OH2 C
C
C
C
O
+
H
C
CH3
H2O
C
CH3
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
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309
Oxidation and Reduction 12–1
C
Chhaapptteerr 1122:: O
Oxxiiddaattiioonn aanndd R
Reedduuccttiioonn
SSuum
mm
maarryy:: TTeerrm
mss tthhaatt ddeessccrriibbee rreeaaccttiioonn sseelleeccttiivviittyy
• A regioselective reaction forms predominately or exclusively one constitutional isomer
(Section 8.5).
CH3
I
CH3
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.
D
Deeffiinniittiioonnss ooff ooxxiiddaattiioonn aanndd rreedduuccttiioonn
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]
R
Reedduuccttiioonn rreeaaccttiioonnss
[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.
310
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Chapter 12–2
[2] Reduction of alkynes
H H
2 H2
R C C R
R C C R
Pd-C
[a]
•
Two equivalents of H2 are added and four
new C–H bonds are formed (12.5A).
•
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.
H H
alkane
[b]
R C C R
R
R
H2
C C
Lindlar
catalyst
H
•
H
cis alkene
[c]
R C C R
R
Na
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] H2O
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
O
Oxxiiddaattiioonn rreeaaccttiioonnss
[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
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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 and bonds of the alkene
are cleaved to form two carbonyl
groups.
1,2-diol
[d] Oxidative cleavage (12.10)
R'
R
R
C O
[2] Zn, H2O or
CH3SCH3
H
R'
R
[1] O3
C C
O C
+
R
H
ketone
aldehyde
[2] Oxidative cleavage of alkynes (12.11)
[a]
R
[1] O3
R C C R'
internal alkyne
R'
C O
[2] H2O
+
•
O C
HO
OH
The bond and both bonds of the
alkyne are cleaved.
carboxylic acids
[b]
[1] O3
R C C H
terminal alkyne
[2] H2O
R
C O
+
CO2
HO
[3] Oxidation of alcohols (12.12, 12.13)
H
[a]
[b]
PCC
R C OH
or
HCrO4–
H
A-26
1o alcohol Amberlyst
resin
H
R C OH
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.
C O
H
aldehyde
R
CrO3
C O
H2SO4, H2O
1o alcohol
HO
carboxylic acid
H
[c]
R
PCC or CrO3
or
R
HCrO4–
2o 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
H
R
O
C C
CH2OH
H
with (–)-DET
or
H
CH2OH
R C C
H
O
with (+)-DET
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Chapter 12–4
C
Chhaapptteerr 1122:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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.
CH3
d.
C
O
CH3
CH3
CH2 CH2
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 bond. To draw the alkane product, add a H to each C of the double
bond.
CH3
CH2CH(CH3)2
CH3 CH2CH(CH3)2
C C
a.
CH3
CH3 C
H
H
c.
C H
H
b.
12.3 Draw the alkenes that form each alkane when hydrogenated.
a.
or
H2
Pd-C
or
c.
or
or
or
or
or
H2
Pd-C
or
b.
H2
Pd-C
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, decreasing the heat of
hydrogenation.
CH3CH2
H
H
cis alkane
less stable
larger heat of hydrogenation
H
H
or
b.
C C
or
C C
a.
CH3CH2
CH2CH3
CH2CH3
trans alkane
trisubstituted
disubstituted
less stable
larger heat of hydrogenation
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Oxidation and Reduction 12–5
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 Increasing alkyl substitution on the C=C decreases the rate of hydrogenation.
With one equivalent of
H2, only the more reactive
of the double bonds will
be reduced.
disubstituted
more reactive
CH3
trisubstituted
CH2
H2, Pd-C
CH3
(1 equiv)
C
CH3
CH3
C H
CH3
limonene
12.7
Two enantiomers are formed in equal amounts:
new stereogenic center
H
a.
CH3 C CH2CH2CH3
C C
H
b.
CH3
CH2CH3
CH2CH3
CH2CH3
CH2CH2CH3
CH2
H
CH3
CH3
H
=
C
CH3
+ CH3
CH2CH3
+
CH2CH2CH3
H
CH3CH2CH2 C CH
3
H
H
CH3
diastereomers
c.
C(CH3)3
C(CH3)3
C(CH3)3
+
diastereomers
12.8
A
Molecular formula
before hydrogenation
C10H12
Molecular formula
after hydrogenation
C10H16
Number
of rings
3
Number of
bonds
2
B
C
C4H8
C6H8
C4H10
C6H12
0
1
1
2
Compound
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Chapter 12–6
12.9
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
1 equiv
A
CH2OCO(CH2)16CH3
CH2OCO(CH2)16CH3
CHOCO(CH2)6CH2CH=CH(CH2)7CH3 CHOCO(CH2)9CH2CH=CH(CH2)4CH3
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
12.10 Hydrogenation of HCCCH2CH2CH3 and CH3CCCH2CH3 yields the same compound. The heat
of hydrogenation is larger for HCCCH2CH2CH3 than for CH3CCCH2CH3 because internal
alkynes are more stable (lower in energy) than terminal alkynes.
12.11
O
CH2 C
CH3
O
C CH2CH3
H2
H H
C
CH2CH3
C
C
Lindlar catalyst
A
CH3
cis-jasmone
H
(perfume component
isolated from jasmine flowers)
H
12.12 To draw the products of catalytic hydrogenation remember:
• H2 (excess), Pd will reduce alkenes and alkynes to alkanes.
• H2 (excess), Lindlar catalyst will reduce only alkynes to cis alkenes.
a.
b.
CH2=CHCH2CH2 C C CH3
CH2=CHCH2CH2 C C CH3
H2 (excess)
Pd-C
H2 (excess)
Lindlar
catalyst
CH3CH2CH2CH2CH2CH2CH3
CH2 CH CH2 CH2
CH3
C C
H
H
12.13 Use the directions from Answer 12.12.
a. CH3OCH2CH2C CCH2CH(CH3)2
H2 (excess)
CH3OCH2CH2CH2CH2CH2CH(CH3)2
Pd-C
b. CH3OCH2CH2C CCH2CH(CH3)2
H2 (1 equiv)
CH3OCH2CH2
C C
Lindlar catalyst
c.
CH3OCH2CH2C CCH2CH(CH3)2
H2 (excess)
Lindlar catalyst
H
CH3OCH2CH2
H
CH2CH(CH3)2
C C
H
H
CH3OCH2CH2
d. CH3OCH2CH2C CCH2CH(CH3)2
CH2CH(CH3)2
Na, NH3
H
C C
H
CH2CH(CH3)2
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Oxidation and Reduction 12–7
12.14
D2
CH3CD2CD2CH2CH2CH3
Pd-C
CH3
D2
CH3 C C CH2CH2CH3
C
Lindlar
catalyst
CH2CH2CH3
C
D
D
CH3
Na
C
ND3
D
C
D
CH2CH2CH3
12.15
H2
H
R
Lindlar
catalyst
A
B
H
12.16 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
12.17 To draw the product, add an O atom across the 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
12.18 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.
CH3
H
H
H
CH3CH2
b.
H
mCPBA
CH3CH2
H
H
CH3
c.
H
CH2CH3
C C
H
O
C C
mCPBA
CH3
H
H C C H enantiomers
O
O
C C
CH3CH2
CH2CH3
H C C H
CH2CH3
O
H
identical
CH3
CH3
O
O
H
H
enantiomers
O
CH2
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Chapter 12–8
12.19 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-2Butene 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
CH3
[1] RCO3H
H
[2] H2O, HO
C
CH3
C
CH3
H
OH
H
cis-2-butene
CH3 C C
H
CH3
H
OH
C
C
HO
H
CH3
enantiomers
HO
H
[1] RCO3H
CH3
[2] H2O, HO
H
C
CH3
CH3
H
CH3
C
OH
H
OH
C
C
H
CH3
HO
identical
meso compound
trans-2-butene
12.20 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
C
H
H
[1] OsO4
HO
H
CH3
[2] NaHSO3, H2O
CH3
C
C
HO
OH
C
trans-2-butene
CH3
CH3
H
C
H
CH3
H
CH3
H
OH
identical
meso compound
cis-2-butene
CH3 C C
H
CH3
H
H
C
CH3
C
HO
OH
enantiomers
12.21 To draw the oxidative cleavage products:
• Locate all the bonds in the molecule.
• Replace all C=C’s with two C=O’s.
Replace this bond with two C=O's.
[1] O3
a. (CH3)2C
CHCH2CH2CH2CH3
[2] Zn, H2O
[1] O3
[2] Zn, H2O
+
O CHCH2CH2CH2CH3
aldehyde
ketone
+
O
One ketone and
one aldehyde
are formed.
Two aldehydes are formed.
O
[2] Zn, H2O
c.
O
H
[1] O3
b.
(CH3)2C
H
O
O
H
A dicarbonyl compound is formed.
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Oxidation and Reduction 12–9
12.22 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
CH3
CH3
b.
CHCH3
+
C
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.
O
CH3
C
CH3CHO
Join these two C's.
12.23
O
a.
[1] O3
O
O
[2] CH3SCH3
H
H
H
O
[1] O3
b.
[2] CH3SCH3 H
c.
[2] CH3SCH3
O O
O
O
O
O
H
O
H
H
H
a.
[2] H2O
O
O
[1] O3
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
H
O
12.24 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.
CH3CH2 C C CH2CH2CH3
H
O
O
[2] CH3SCH3
H
H
H
[1] O3
d.
H
O
[1] O3
CH3
O
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12.25
a.
c.
CH3CH2CH(CH3)CO2H
b.
CO2 + CH3(CH2)8CO2H
CH3CH2CH C C CHCH2CH3
CH3(CH2)8 C CH
CH3
CH3CH2CO2H, HO2CCH2CO2H, CH3CO2H
CH3CH2 C C CH2 C C CH3
CH3
d.
HO2C(CH2)14CO2H
12.26 For 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
PCC
OH
a.
OH
H
c.
CrO3
OH
H2SO4, H2O
O
OH
OH
O
O
PCC
b.
CrO3
d.
H2SO4, H2O
12.27 Upon treatment with HCrO4–– Amberlyst A-26 resin:
• 1° Alcohols are oxidized to aldehydes.
• 2° Alcohols are oxidized to ketones.
H
OH
O
HCrO4–
a.
c.
Amberlyst A-26 resin
b. HO
OH
H
HCrO4–
OH
OH
O
HCrO4–
O
Amberlyst A-26 resin
O
O
Amberlyst A-26 resin
12.28
NaOCl
a.
OH
b.
O
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.
+ NaCl + H2O
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.
12.29
O
HO
OH
ethylene glycol
O
H
H
O
OH
HO
O
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Oxidation and Reduction 12–11
12.30 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.
re-draw
OH
(CH3)3C
H O
OOH
OH
Ti[OCH(CH3)2]4
(–)-DET
()-DET adds O
above the plane.
12.31 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.
geraniol
OH
This alkene is not part of an
allylic alcohol and will not be epoxidized.
320
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Chapter 12–12
12.32 Use the rules from Answer 12.1.
OH
a.
d.
CH2CH3 reduction
C CH
O
reduction
O
b.
c.
CH3CH2Br
e.
OH oxidation
OH
CH2 CH2
OH
HO
f.
neither
1 C–H and 1 C–Br
bond are removed.
ClCH2CH2Cl oxidation
(2 new C–Cl bonds)
CH2 CH2
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, decreasing the heat of hydrogenation.
2-methyl-2-butene
trisubstituted
smallest Ho = –112 kJ/mol
2-methyl-1-butene
disubstituted
intermediate Ho = –119 kJ/mol
3-methyl-1-butene
monosubstituted
largest Ho = –127 kJ/mol
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 bond
b. Compound B: molecular formula C10H16: hydrogenated to C10H18.
3 degrees of unsaturation, 1 is hydrogenated.
2 rings and 1 bond
c. Compound C: molecular formula C8H8: hydrogenated to C8H16.
5 degrees of unsaturation, 4 are hydrogenated.
1 ring and 4 bonds
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Oxidation and Reduction 12–13
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
+
H
C
H
O
[1] O3
O
[2] Zn, H2O
B
+ O
identical
c.
a. trisubstituted
intermediate heat of hydrogenation
b. intermediate reaction rate
O
[1] O3
[2] Zn, H2O
O
+
H
C
12.37 Work backwards to find the alkene that will be hydrogenated to form 3-methylpentane.
2 possible enantiomers:
or
H
R isomer
3-methylpentane
H
S isomer
12.38
O
O
H2 (excess)
OH
a.
OH
Pd-C
stearidonic acid
stearic acid
O
b.
H2 (1 equiv)
O
OH
OH
Pd-C
O
O
OH
O
OH
c.
trans
one possibility
OH
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Chapter 12–14
O
O
d.
OH
O
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
12.39
H2
a.
H2
b.
Lindlar catalyst
Na
NH3
c.
[1] LiAlH4
h.
Pd-C
no reaction
no reaction
(CH3)3COOH
j.
[1] CH3CO3H
OH
k.
O
OH
O
syn addition
H2O,
[2] H2O
OH
OH
KMnO4
g.
OH
[1] LiAlH4
l.
OH
[2] NaHSO3, H2O
no reaction
mCPBA
OH
anti addition
[1] OsO4 + NMO
f.
H
Ti[OCH(CH3)2]4
()-DET
OH
[2] H2O, HO
O
H
[2] CH3SCH3
O
e.
O
[1] O3
i.
CH3CO3H
d.
no reaction
[2] H2O
syn addition
HO
OH
12.40
a. CH3CH2CH2 C C CH2CH2CH3
H2 (excess)
Pd-C
CH3CH2CH2
H2
b. CH3CH2CH2 C C CH2CH2CH3
C C
Lindlar catalyst
c. CH3CH2CH2 C C CH2CH2CH3
d. CH3CH2CH2 C C CH2CH2CH3
CH2CH2CH3
Na
H
H
H
CH2CH2CH3
[1] O3
C C
CH3CH2CH2
H
O
[2] H2O
CH3CH2CH2
NH3
cis alkene
C
trans alkene
O
+
OH
CH3CH2CH2
C
OH
identical
12.41
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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b.
CH3CH2
H
C
C
CH3
O
mCPBA
CH3CH2
CH3
CH2OH
H
CH2OH
CH3CH2
CH3
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
CH3CH2
CrO3
CH2OH
C
H
[1] PBr3
C
CH2OH
CH3CH2
CH3
COOH
CH3CH2
C
H2
HCrO4–
H
C
C
CH3
O
OH
H2SO4, H2O
O
d.
OH
PCC
OH
CF3CO3H
H
OH
O
O
H
e.
f.
OH
[1] OsO4
[2] NaHSO3, H2O
OH
OH
H
OH
OH
OH
OH
OH
OH
[1] HCO3H
[2] H2O, HO–
OH
OH
OH
OH
OH
OH
H O
OH
g.
OH
(CH3)3COOH
re-draw
Ti[OCH(CH3)2]4
(+)-DET
h.
OH
KMnO4
H2O, HO
OH
OH
OH
OH
OH
OH
CH3
CH3CH2
Amberlyst A-26
CH2OH
OH
Na2Cr2O7
C
CH3
Pd-C
c.
H
C
12.42
OH
CH2Br
CH3CH2
H
C C
CH3
b.
C
CH3
CHO
H2SO4, H2O
OH
H
[2] LiAlH4
[3] H2O
h.
a.
O
CH3
CH3CH2
CH3CH2
C
CH3
C
CH3
H
C
g.
H
C
CH2OH
d. CH3CH2
CH3
PCC
C
CH3
C
CH3CH2
H
CH2OH
H
O
(CH3)3COOH
Ti[OCH(CH3)2]4
()-DET
H
CH3
CH3CH2
OH
H
C
CHO
CH2OH
H
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Chapter 12–16
12.43
OH
O
PCC
a.
OH
c.
CrO3
OH
H2SO4, H2O
O
O
PCC
b. CH3CH2CH2CH2OH
CH3CH2CH2
C
H
12.44
OH
a.
b.
[1] SOCl2
CH2
[3] H2O
[1] OsO4
OH
[2] NaHSO3, H2O
CH2OH
O
[1] mCPBA
c.
[2] LiAlH4
Cl
pyridine
CH2
[2] LiAlH4
OH
[3] H2O
CH3
H2
d.
Lindlar
catalyst
12.45
a. H2, Pd-C
d. [1] LiAlH4
b. mCPBA
O
c. KMnO4
H2O, HO–
or [1] OsO4
[2] NaHSO3, H2O
OH
OH f. PBr3
h. [1] LiAlH4
[2] H2O
or HBr
[2] H2O
g. CrO3,
or PCC or HCrO4–,
H2SO4, H2O
Amberlyst
A-26 resin
O
e. H2O (–OH)
OH
OH
Br
OH
(+ enantiomer)
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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Oxidation and Reduction 12–17
b.
C6H5
[1]
H
C C
H
C6H5
C6H5
[2]
C6H5
[1] CH3CO3H
H
[2] –OH, H2O
rotate
C
H
C6H5
syn addition
HO
OH
C
[2] NaHSO3, H2O
C C
H
HO
[1] OsO4
H
HO
H
C
C6H5
H
C6H5
C6H5
C
H
C6H5
C6H5
C
H
C
+ enantiomer
OH
+ enantiomer
OH
OH
CH3CH2CH2
c. [1]
CH2CH2CH3
C C
H
HO
[1] OsO4
OH
C
CH3CH2CH2
[2] NaHSO3, H2O
H
CH3CH2CH2
H
C
H
CH2CH2CH3
syn addition
CH3CH2CH2
[2]
H
H
H
[1] CH3CO3H
C C
CH2CH2CH3
[2] –OH, H2O
H
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
H
CH2CH2CH3
OH
OH
CH3CH2CH2
H
rotate
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Chapter 12–18
12.48
Cl
H OH
[1]
C
mCPBA
O
O
O
O
O
[2]
Br
OH
Br
Br
OH
H
O
OH
H
Na+ H
Br
[3]
Cl
C
OH
OH
O
Br
Na+ + H2 +
O
Br–
O
+
HO
12.49 Use the directions from Answer 12.21.
a. (CH3CH2)2C CHCH2CH3
[1] O3
[2] CH3SCH3
(CH3CH2)2C
O CHCH2CH3
CH3
[1] O3
b.
+
O
O
[2] Zn, H2O
+ O C
CH3
O
C CH
c.
C
[1] O3
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
O
(CH3)2C
CH2
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
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Oxidation and Reduction 12–19
12.51 Use the directions from Answer 12.22.
Join these two C's.
O
[1] O3
a. C10H18
H
[2] CH3SCH3
O
2 degrees of unsaturation
one ring + one bond
O
[1] O3
b. C10H16
[2] CH3SCH3
3 degrees of unsaturation
O
two rings + one bond
12.52
Join these two C's.
a. CH3CH2CH2CH2COOH and
Join these two C's.
COOH
and
CH3CH2CH2CH2C CH c.
CO2
C CCH3
CH3COOH
Join these two C's.
CH3CH2C CCH2CH2CH3
b. CH3CH2COOH and CH3CH2CH2COOH
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
4 equiv
(from portion B)
1 equiv
(from portion C)
COOH
b.
c.
linolenic acid
[1] O3
[2] Zn, H2O
[1] O3
[2] Zn, H2O
O
zingiberene
H
O
+
+
H
COOH
O
O
2 equiv
12.54
A
a.
C8H12
O
[1] O3
[2] CH3SCH3
H
+
+
H
H
O
O
O
O
H
H
H
O
H
H
O
O
328
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Chapter 12–20
[1] NaNH2
H2 (excess)
b.
C6H10
[2] CH3I
Pd-C
B
C
C7H12
12.55
The hydrogenation reaction tells you that both
oximene and myrcene have 3 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
H
C
(2 equiv)
C
CH2CH2
CHO
12.56
a.
H2 (2 equiv)
H2
Pd-C
Lindlar catalyst
C7H12
A
A
oxidative
cleavage
B
Na, NH3
O
OH
+ other product(s)
C
b. A does not react with NaH because it is not a terminal alkyne.
12.57
OH
H2SO4
H2
Pd-C
decalin
C10H18O
A
C10H16
B
C10H16
C
ozonolysis
O
O
C10H16O2
E
CHO
D
O
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Oxidation and Reduction 12–21
12.58 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.
DHA
CO2H
A
B
B
B
B
C
B
[1] O3
[2] Zn, H2O
+
CH3CH2CHO
OHCCH2CHO
+
OHCCH2CH2CO2H
5 equiv
(from portion B)
(from portion A)
(from portion C)
12.59 The stereogenic center (labeled with *) in both structures can be R or S.
*
O
O
H
O
H
H
H2
ozonolysis
O
O
H
Pd-C
or
H
*
butylcyclopheptane
O
possible structures for
dictyopterene D'
H
12.60
OH
OH
a.
re-draw
(CH3)3COOH
Ti[OC(CH3)2]4
()-DET
b.
H C C
(CH3)3C
CH2OH
H
H
(CH3)3C
(CH3)3COOH
C
Ti[OC(CH3)2]4
(+)-DET
C
H
O
OH
H
CH2OH
H
O
12.61
OH
re-draw
OH
(CH3)3COOH
Ti[OC(CH3)2]4
()-DET
O
CH2OH
H
major product
87%
CH2OH
+
O
H
minor product
13%
enantiomeric excess =
% one enantiomer – % second enantiomer
ee = 87% – 13% = 74%
330
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Chapter 12–22
12.62
O
H
a.
Replace this O
to make an alkene.
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
12.63 Use retrosynthetic analysis to devise a synthesis of each hydrocarbon from acetylene.
a.
CH3CH2CH CH2
CH3CH2CH CH
NaH
HC CH
CH3CH2Cl
C CH
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
CH3Cl
C CH
NaH
CH3 C CH
CH3 C C
CH3Cl
CH3
H2
CH3 C C CH3
C C
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.
CH3
NaH
Cl
C CH
HC C
NaH
CH3 C C
CH3Cl
CH3 C C CH3
Na
CH3
NH3
H
HC CCH2CH(CH3)2
NaH
C C
C CH
Cl
C C
HC CH
H2
(2 equiv)
Pd-C
H
C C
CH3
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Oxidation and Reduction 12–23
12.64
Cl
NaH
HC CH
HC C
NaH
Cl
H2
Lindlar catalyst
12.65
Br
–NH
2
Br2
Na
NH3
(2 equiv)
cis
trans
Br
12.66
CH3
O
a.
CH3 C
H
C
HC CH
C
CH3
H
NaH
CH3
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
mCPBA
CH3
H
HC CH
CH3 C C CH3
CH3
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.
C
H
H
C
H
CH3
H
H
CH3
H
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
OH
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
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Chapter 12–24
HO
d.
OH
C
H
CH3
H
C
H
CH3
CH3 C C CH3
C C
CH3
CH3
HC C CH3
HC CH
H
(+ enantiomer)
H
CH3
HO
KMnO4
C C
CH3
H
H2O, HO
(from b.)
OH
(+ enantiomer)
C
C
H
CH3
H
CH3
12.67
[1] BH3
a. C6H5CH CH2
PCC
C6H5CH2 CH2OH
[2] H2O, –OH
C6H5CH2 CHO
HO
H2O
b. C6H5CH CH2
H2SO4
[1] BH3
c. C6H5CH CH2
[2] H2O,
O
PCC
C6H5CH CH3
C5H5
CrO3
C6H5CH2 CH2OH
–OH
C
CH3
C6H5CH2 COOH
H2SO4, H2O
HC CH
NaH
O
mCPBA
d. C6H5CH CH2
C6H5
C C H
H H
HO
[1] C CH
C6H5CH CH2C CH
[2] H2O
12.68
Br
Br2
Br
NaH
2 NaNH2
1-pentene
CH3Cl
Na
(2E)-2-hexene
NH3
12.69
a.
b.
CH3CH2CH CH2
CH3CH2CH2CH2OH
[1] 9-BBN or BH3
[2] H2O2,
HO–
POCl3
pyridine
CrO3
CH3CH2CH2CH2OH
CH3CH2CH CH2
mCPBA
H2SO4, H2O
CH3CH2CH2COOH
HO
O
[1] LiAlH4
CH3CH2CH CH3
CH3CH2CH CH2
[2] H2O
H2O, H2SO4
CrO3
H2SO4, H2O
CH3CH2
C
O
CH3
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Oxidation and Reduction 12–25
c.
OH
[1] OsO4
[2] NaHSO3, H2O
OH
[2] CH3I (2 equiv)
O
mCPBA
d.
OCH3
OH
H2O
HC C
O
OCH3
[1] NaH (2 equiv)
C CH
NaH
O
PCC
C CH
C CH
HC CH
12.70
NaH
a. HC CH
CH3CH2Br
HC C
NaH
HC C CH2CH3
C C CH2CH3
O
H2
Lindlar catalyst
OH
OH
b.
HO CH2CH2 C C CH2CH3
Na
c.
OH
O CH2CH2 C C CH2CH3
OH
NH3
(from a.)
H2O
PBr3
Br
(from b.)
PCC
d.
CHO
OH
(from b.)
12.71
H2
a.
[2] H2O2, HO–
Lindlar catalyst
[1] LiAlH4
mCPBA
b.
OH
[1] 9-BBN or BH3
C CH
O
[2] H2O
OH
(from a.)
c.
C CH
HO
NaH
C C
O
H
mCPBA
d.
(from c.)
H
CH3
(+ enantiomer)
CH3Cl
C C CH3
Na
NH3
KMnO4
H2O, HO–
OH
334
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Chapter 12–26
12.72
HC CH
NaH
Br
C CH
NaH
CH3CH2 C CH
Br
CH3CH2 C C
Na, NH3
Cl
H
Cl2
anti
addition
Cl
H
(3R,4S)-3,4-dichlorohexane
12.73
a.
OH
HC CH
K+–OC(CH3)3
PBr3
NaH
CH2 CH2
Br
Br
C CH
Br2
2 NaNH2
Br
NaH
CH3CH2 C CH
HC CH
Br
Br
CH3CH2 C C
Na, NH3
H2
b.
(from a.)
H
Lindlar catalyst
HO
[1] OsO4
c.
(from b.)
O
mCPBA
[2] NaHSO3, H2O
H
C
CH3CH2
H
C
H
CH2CH3
OH
12.74
OCH2CH3
H OCH2CH3
H
OCH3
Li
OCH3
OCH3
Li
OCH3
Li
H
OCH2CH3
OCH3
OCH3
H OCH2CH3
Li
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Oxidation and Reduction 12–27
12.75
The favored conformation
for both molecules places
the tert-butyl group equatorial.
OH
(CH3)3C
OH
H
H
= (CH ) C
3 3
A
A
OH
(CH3)3C
=
(CH3)3C
OH
B
This OH is axial and
will react faster because
the OH group is more
hindered.
B
This OH is equatorial
and will react more slowly
because the OH group
is less hindered.
12.76
O
H O
Cl
O
O
O C
H
O
Cl
O
O
O
O
O
C
O
mCPBA
OSiR3
OSiR3
R = alkyl group
O
C
O
H
Cl
O
O
O
H
O
O
O
O
OH
O
C
Cl
O
O
O
OH
OSiR3
C
O
O
H O
O
O
Cl
OSiR3
OSiR3
O
O
C
O
Cl
HO
C
Cl
C
Cl
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Chapter 12–28
12.77
O
mCPBA
"down" bond
a.
O comes in from below.
X
"up" bond
OH
Br2
b.
Br
H2O
Br
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.
NaH
O
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Infrared Spectroscopy
337
Mass Spectrometry and Infrared Spectroscopy 13–1
C
Chhaapptteerr 1133:: M
Maassss SSppeeccttrroom
meettrryy aanndd IInnffrraarreedd SSppeeccttrroossccooppyy
M
Maassss ssppeeccttrroom
meettrryy ((M
MSS))
•
•
•
•
•
•
•
•
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).
EElleeccttrroom
maaggnneettiicc rraaddiiaattiioonn
• The wavelength and frequency of electromagnetic radiation are inversely related by the following
equations: = c/ or = c/ (13.5).
• The energy of a photon is proportional to its frequency; the higher the frequency the higher the
energy: E = h (13.5).
IInnffrraarreedd ssppeeccttrroossccooppyy ((IIR
R,, 1133..66 aanndd 1133..77))
• Infrared spectroscopy identifies functional groups.
• IR absorptions are reported in wavenumbers:
wavenumber = ~ = 1/
•
•
•
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 ~
•
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 ~
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Chapter 13–2
C
Chhaapptteerr 1133:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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). Do the
same for Br, using 79Br and 81Br.
a. C4H935Cl = 92
C4H937Cl = 94
Two peaks in 3:1 ratio at m/z 92 and 94
d. C4H11N = 73
One peak at m/z 73
b. C3H7F = 62
One peak at m/z 62
e. C4H4N2 = 80
One peak at m/z 80
c. C6H1179Br = 162
C6H1181Br = 164
Two peaks in a 1:1 ratio at m/z 162 and 164
13.4
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
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Mass Spectrometry and Infrared Spectroscopy 13–3
13.5
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.
13.6
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
– H2O
CH3 C CH2 CH3
CH2=CHCH2CH3
+
CH3CH=CHCH3
m/z = 56
H
m/z = 56
13.7
O
a.
C
O C
CH3CH2 C O
CH3CH2
(from cleavage of bond [2])
[1]
b.
13.8
(from cleavage of bond [1])
[2]
CH3CH2CH2CH2CH2CH2OH
+ CH2OH
c. CH3CH2CH2CHO
CH3CH2CH2C O
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
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Chapter 13–4
13.9
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.
13.10 Wavelength and frequency are inversely proportional. The higher frequency light will have a
shorter wavelength.
a. Light having a of 102 nm has a higher than light with a of 104 nm.
b. Light having a of 100 nm has a higher than light with a of 100 m.
c. Blue light has a higher than red light.
13.11 The energy of a photon is proportional to its frequency, and inversely proportional to its
wavelength.
a. Light having a of 108 Hz is of higher energy than light having a of 104 Hz.
b. Light having a of 10 nm is of higher energy than light having a of 1000 nm.
c. Blue light is of higher energy than red light.
13.12 The larger the energy difference between two states, the higher the frequency of radiation needed
for absorption. The 400 kJ/mol transition requires a higher of radiation than a 20 kJ/mol
transition.
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 a of 10 m is higher in energy than IR light having a of 20 m.
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
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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.
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.
O
e.
a.
no additional peaks
b.
OH
Csp2–H at 3150–3000 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).
OH
O–H bond at 3600–3200 cm–1
c.
O
Csp2–H at 3150–3000 cm–1
C=C bond at 1650 cm–1
d.
f.
CH3O
HO
N
H
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
O
C=O bond at ~1700 cm–1
13.19 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.
13.20
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 = CC 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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Chapter 13–6
13.21
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
Cl
d.
b.
molecular formula: C10H16
molecular ion (m/z): 136
molecular formula: C8H17Br
molecular ions (m/z): 192, 194
molecular formula: C5H11Cl
molecular ions (m/z): 106, 108
13.22
O
CH2CH2CH3
C9H12
molecular weight = 120
C
OCH2CH3
CH2CH3
C9H10O
molecular weight = 134
C8H10O
molecular weight = 122
13.23 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
13.24 Likely molecular formula, C8H16 (one degree of unsaturation—one ring or one bond).
Four structures with m/z = 112
13.25
CH3
CH3 CH CO2CH3
Cl
B
C4H7O2Cl
molecular weight: 122, 124
should show 2 peaks for the
molecular ion with a 3:1 ratio
Mass spectrum [1]
C
OCH3
C8H10O
molecular weight: 122
Mass spectrum [2]
CH3CH2CH2Br
A
C3H7Br
molecular weight: 122, 124
should show 2 peaks for the
molecular ion with a 1:1 ratio
Mass spectrum [3]
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Mass Spectrometry and Infrared Spectroscopy 13–7
13.26
H2
or
or
or
or
Pd-C
Possible structures
C7H12
(exact mass 96.0940)
13.27
[1] [2]
a.
OH
OH
OH
(from cleavage of bond [1])
[1]
O
[2]
(from cleavage of bond [2])
O
O
b.
(from cleavage of bond [1])
(from cleavage of bond [2])
[1] [2]
c.
O
O
O
(from cleavage of bond [1])
(from cleavage of bond [2])
13.28
[1]
a.
– H2O
C6H5 CH CH2
H H
Cleave bond [1].
H H
m/z = 104
H
e–
C6H5 C C OH
m/z = 122
(resonance-stabilized
carbocation)
C6H5 C
H
m/z = 91
e–
CH2 C
Cleave bond [1].
[3]
CH3 H H
b.
CH2 C
C C
H
m/z = 68
– H2O
[1]
CH3 H H
CH2 C C C OH
[2]
H H
m/z = 86
H H
C C OH
H H
m/z = 71
e–
Cleave bond [2].
e–
Cleave bond [3].
CH2 C CH3
m/z = 41
CH2 OH
m/z = 31
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Chapter 13–8
13.29
Cleave bond [1].
[1]
[2]
CH3
+
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
13.30
Ketone A
Ketone B
O
O
m/z = 128
m/z = 128
cleavage
cleavage
O
CH2CH3
+
O
+
CH3
+
+
m/z = 99
m/z = 113
This is ketone A since cleavage gives a fragment
with m/z of 99.
This is ketone B since cleavage gives a fragment
with m/z of 113.
13.31 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
13.32 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.
13.33 Molecules with an odd number of N's have an odd number of H's, making the molecular ion odd
as well.
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Mass Spectrometry and Infrared Spectroscopy 13–9
13.34 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.
13.35 Cleavage of a 1° alcohol (RCH2OH) forms an alkyl radical R• and a resonance-stabilized
carbocation with m/z = 31.
CH2OH
CH2 OH
m/z = 31
resonance-stabilized carbocation
13.36 An ether fragments by cleavage because the resulting carbocation is resonance stabilized.
[1]
(CH3)2CH
[2]
CH2
O
CH2 CH3
Cleave
bond [1].
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
13.37
a. (CH3)2C O or (CH3)2CH OH
stronger bond
higher ~
absorption
b. (CH3)2C NCH3
or (CH3)2CH NCH3
stronger bond
higher ~
absorption
c.
H or
H
stronger bond
higher ~
absorption
13.38 Locate the functional groups in each compound. Use Table 13.2 to determine what IR
absorptions each would have.
a.
Csp3–H at 2850–3000 cm–1
d.
O
b.
C CH
Csp–H at 3300 cm–1
Csp3–H at 2850–3000 cm–1
C–C triple bond at 2250 cm–1
e.
OH
O
OH
O–H at 3200–3600 cm–1
Csp3–H at 2850–3000 cm–1
f.
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
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
c.
Csp3–H at 2850–3000 cm–1
C=O at 1700 cm–1
OH
CH2
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Chapter 13–10
13.39
a.
and
C=C bond
1650 cm–1
Csp2–H at 3150–3000 cm–1
HC CCH2CH2CH3
CC bond
2250 cm–1
Csp–H at 3300 cm–1
O
b. CH CH
3
2
OCH3
OCH3
d.
O
and
CH3(CH2)5
no C=O bond
C
OCH3
C=O bond
~1700 cm–1
O
C
and
OH
CH3
C
e.
OCH3
O–H bond
no O–H bond
> 3000 cm–1
[See note on OH in Answer 13.38f.]
and
CH3C CCH3
CH3CH2C CH
Csp–H bond
3300 cm–1
CC bond at ~2250 cm–1
no CC absorption
due to symmetry
O
c.
CH3CH2
C
and
CH3
f.
CH3CH CHCH2OH
HC CCH2N(CH2CH3)2
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
and
CH3(CH2)5C N
Csp–H bond
3300 cm–1
13.40 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
13.41 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.
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Mass Spectrometry and Infrared Spectroscopy 13–11
13.42 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)
Csp2–H at ~3150 cm–1
Phenyl peaks at 1600 and 1500 cm–1
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
13.43 In addition to Csp3–H at ~3000–2850 cm–1:
O
CH3
C
O
CH3
~1700 cm–1
CH3CH2
C
H
CH2OH
C
H
~1700 cm–1
H
OH
C
CH3 O CH=CH2
H
(OH) 3200–3600 cm–1 (OH) 3200–3600 cm–1 (C=C)1650 cm–1
(C=C)1650 cm–1
(Csp2–H) 3150–3000 cm–1
(Csp2–H) 3150–3000 cm–1
No enols (such as CH3CH=CHOH) are drawn
since these compounds are not stable.
13.44
a. Compound with a molecular ion at m/z = 72
IR absorption at 1725 cm–1 = C=O bond
Molecular formula: C4H8O
O
O
O
CH3
No additional peaks above 1500 cm–1
c. Compound with a molecular ion at m/z = 74
IR absorption at 3600–3200 cm–1 = O–H bond
Molecular formula: C4H10O
OH
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 = CN bond
CH3CH2C N
OH
OH
OH
or
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Chapter 13–12
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 = CC bond
Two possible enantiomers: CH3
*
HC CCHCH2CH3
C
CH3
stereogenic center
CH3CH2
CH3
or
C CH
C
HC C
H
H
CH2CH3
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
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
13.48
H
O
Br
O
H–Br
H
+ Br
H Br
H
CH3
CH3
HBr
O
C CH2
CH3
CH3
H
C
H O
CH2
H
Br
Br
H2O
13.49
O
fragments:
J
C6H12O
m/z = 100
IR absorption at 2962 cm–1 = Csp3–H bonds
1718 cm–1 = C=O bond
O
O
cleavage product
cleavage product
m/z = 43
m/z = 85
The fragment at m/z = 57 could be due to (C4H9)+ or (C3H5O)+.
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Mass Spectrometry and Infrared Spectroscopy 13–13
13.50
O
C
NH2
CHO
L
C 7 H6 O
m/z = 106
K
C7H9N
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
2820 cm–1 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.
13.51
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.
13.52 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
Br
N
N H
H
+
Na+ Br– + H2
W
+ Na+ + H2
Na+ H
13.53 The ,-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
+
three resonance structures for 2-cyclohexenone
O
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Chapter 13–14
13.54
PCC
a. and b.
H
PCC
NaOCOCH3
O
O
O Cr OH
OH
H
A
molecular ion at 154
C10H18O
IR at 1730 cm–1 (C=O)
O
+
H B
PCC
+
O Cr OH
O
H
OH
OH
O
+
B
+
OH
H
+
+H B
B
a. and c.
+
O
H
O
HO
isopulegone
+
+
H B
+ Cr4+
B
citronellol
O
isopulegone
+ Cr4+ + H B+
O
+
H2O
O
O
H OH
B
OH
B
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Resonance Spectroscopy
351
Nuclear Magnetic Resonance Spectroscopy 14–1
C
Chhaapptteerr 1144:: N
Nuucclleeaarr M
Maaggnneettiicc R
Reessoonnaannccee SSppeeccttrroossccooppyy
11
1H
HN
NM
MR
R ssppeeccttrroossccooppyy
[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 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.
1133
CN
NM
MR
R ssppeeccttrroossccooppyy ((1144..1111))
13C
[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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Chapter 14–2
C
Chhaapptteerr 1144:: A
Annssw
weerrss ttoo PPrroobblleem
mss
14.1
Use the formula = [observed chemical shift (Hz)/ of the NMR (MHz)] to calculate the
chemical shifts.
a. CH3 protons:
= [1715 Hz] / [500 MHz]
= 3.43 ppm
14.2
Calculate the chemical shifts as in Answer 14.1.
a. one signal:
= [1017 Hz] / [300 MHz]
= 3.39 ppm
14.3
b. The positive direction of the scale is downfield
from TMS. The CH3 protons absorb upfield
from the OH proton.
OH proton:
= [1830 Hz] / [500 MHz]
= 3.66 ppm
second signal:
= [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
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. CH3CH2OCH2CH3
2 kinds of H's
2 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
Each C is a different distance from the Cl. This
makes each C different, and each set of H's
different. There are 8 different kinds of protons.
CH3CH2CH2CH2CH2CH2CH2CH2Cl
14.5
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
Hb
H
CH3
H
2 NMR signals
14.6
Ha
CH3
b.
Hc
H
CH3
Ha
Ha
H
Ha
H
c.
Hb
H
H
H
Hd
4 NMR signals
Hb
Hb
H
CH3
H
Hc
Hb
CH3
Hc
3 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.
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Resonance Spectroscopy
Nuclear Magnetic Resonance Spectroscopy 14–3
CH3CH2CH2CH2CH2CH3
a.
c.
replacement
of H with X
CH3
C
CH3CH2CH2CH2
Pick one configuration at the
existing stereogenic center.
CH3
C
H
X
CH3CH2CH2CH2
CH3
X
H
H
C
HO
enantiomers =
enantiotopic H's
b.
CH3CH(OH)CH2CH2CH3
X
replacement
of H with X
CH3
C
C
CH2CH3
H
X
HO
H
C
CH2CH3
H
diastereomers =
diastereotopic H's
CH3CH2CH2CH2CH3
replacement
of H with X
no stereogenic center
CH3CH2CHCH2CH3
neither enantiotopic nor diastereotopic
X
14.7
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.
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
H H He
Ha
c.
b.
Hc
H H
Hd
Cl C C O CH3
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
14.8
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.
14.9
O
a. ClCH2CH2CH2Br
Ha Hb Hc
3 types of protons:
Hb < Hc < Ha
b. CH3OCH2OC(CH3)3
Ha Hb
Hc
3 types of protons:
Hc < Ha < Hb
c. CH3
C
CH2CH3
Ha
Hb Hc
3 types of protons:
Hc < Ha < Hb
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Chapter 14–4
14.10
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
C
Ha
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
14.11 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
14.12 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
14.13
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
14.14 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.
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Nuclear Magnetic Resonance Spectroscopy 14–5
a.
O
O
C
C
CH3CH2
c.
Cl
Ha Hb
Ha: 3 peaks - triplet
Hb: 4 peaks - quartet
Ha
CH2CH2Br
e.
d.
CH3 C Br
H
Ha
H
Hb
Cl
Br
Ha
Ha: 2 peaks - doublet
Hb: 4 peaks - quartet
Ha
Ha: 2 peaks - doublet
Hb: 2 peaks - doublet
f.
C C
Br
H
C C
CH3
Hb Hc
Ha: 1 peak - singlet
Hb: 3 peaks - triplet
Hc: 3 peaks - triplet
Hb
H
b.
CH3
CH3CH2
ClCH2CH(OCH3)2
Hb
H
Ha Hb
Ha: 2 peaks - doublet
Hb: 3 peaks - triplet
Ha: 2 peaks - doublet
Hb: 2 peaks - doublet
14.15 Identical protons do not split each other.
Ha
Ha
H Cl
H Cl
Cl C C Cl
Br C C Cl
Cl H
Br H
Ha
Ha: 1 singlet
All protons are equivalent.
Hb
Ha: 2 peaks - doublet
Hb: 2 peaks - doublet
14.16 Use the directions from Answer 14.14.
Ha
Ha
Ha O
a.
O
Hb
Hb
c. CH3 C H
Ha: quartet
Hb: triplet
2 NMR signals
Hb
Ha: doublet
Hb: quartet
2 NMR signals
O
b. CH3
C
OCH2CH2OCH3 Ha and Hd are both singlets.
Ha
Hb Hc
Hd
Hb: triplet
Hc: triplet
4 NMR signals
Ha: triplet
Hb: doublet
Hc
Hc: singlet
3 NMR signals
d. Cl2CHCH2CO2CH3
Ha Hb
14.17 CH3CH2Cl
3 units
2 units
3
chemical shift (ppm)
1
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.
356
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Chapter 14–6
14.18
Cl
a.
(CH3)2CHCO2CH3
b.
CH3CH2CH2CH2CH3
c.
CH2Br
Ha
C C
d.
H
Hb
Ha H
Ha Hb Hc
Ha: split by 1 H
Ha: split by 2 H's
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
14.19
CH3CH2
O
b.
a. CH3OCH2CH3
Ha
CH3CH2
H b Hc
Ha: singlet at ~3 ppm
Hb: quartet at ~3.5 ppm
Hc: triplet at ~1 ppm
C
c. CH3OCH2CH2CH2OCH3
OCH(CH3)2
H a Hb
Hc Hd
Ha: triplet at ~1 ppm
Hb: quartet at ~2 ppm
Hc: septet at ~3.5 ppm
Hd: doublet at ~1 ppm
Ha
d.
Hb Hc Hb Ha
Ha
CH2CH3
Ha
C C
H
Ha: singlet at ~3 ppm
Hb: triplet at ~3.5 ppm
Hc: quintet at ~1.5 ppm
Hb
H
Hc
Ha: triplet at ~1 ppm
Hb: multiplet (8 peaks) at ~2.5 ppm
Hc: triplet at ~5 ppm
14.20
Hb
Splitting diagram for Hb
Hb
Cl
Jab = 13.1 Hz
Cl
Ha
Hc
Hc
1 trans Ha proton splits Hb into
1 + 1 = 2 peaks
a doublet
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
14.21
Cl
H
Hb
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
singlet Cl
H
doublet
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
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Nuclear Magnetic Resonance Spectroscopy 14–7
14.22 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
7 peaks
3 NMR signals
12 peaks (maximum)
6 peaks (more likely,
resulting from peak overlap)
4 NMR signals
14.23
H
Hc
C CH3
Hd
5 H's on
benzene ring
OH
Hb
A
Ha
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.
14.24 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 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
1.2
26
26 units/3 units per H = 9 H's
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
The singlet (9 H) is likely from a tert-butyl group:
The CH3 and CH2 groups split each other: CH3
CH3
C CH3
CH3
CH2
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Chapter 14–8
• Join the pieces together.
O
C
CH3CH2O
O
CH3
C CH3
or
CH3CH2
C
CH3
O C CH3
CH3
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.
14.25
•
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)
•
Calculate degrees of unsaturation
2n + 2 = 2(3) + 2 = 8
8 – 8 = 0 degrees of unsaturation
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
Put information together:
CH3
HO C CH3
H
14.26
a.
Absorption [A]:
Absorption [B]:
Absorption [C]:
Absorption [D]:
singlet at ~3.8 ppm
multiplet at ~3.6 ppm
triplet at ~2.9 ppm
singlet at ~1.9 ppm
b.
H
N
H
CH3
N
CH3O
split by 2 adjacent
nonequivalent H's
into a triplet
O
H H HH
CH3O–
CH2N
CH2 adjacent to five-membered ring
CH3C=O
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Nuclear Magnetic Resonance Spectroscopy 14–9
14.27 Identify each compound from the 1H NMR data.
singlet
at 2.5
a.
CH2 CHCOCH3
O
HCl
H H
CH3
b.
triplet
at 3.6
Cl
(CH3)2C
O
base
O
OH
H2O
H H
singlet
A
B
singlet
at 2.2
triplet
at 3.05
singlet
at 1.3
singlet
at 3.8
14.28 Each different kind of carbon atom will give a different 13C NMR signal.
O
b.
a. CH3CH2CH2CH3
CH3CH2
Ca Cb Cb Ca
C
c. CH3CH2CH2 O CH2CH2CH3
OCH3
Ca Cb Cc
Cc Cb Ca
same groups on both sides of O
3 kinds of C's
3 13C NMR signals
Each C is different.
4 kinds of C's
4 13C NMR signals
2 kinds of C's
2 13C NMR signals
CH3CH2
d.
H
C C
H
H
Each C is different.
4 kinds of C's
4 13C NMR signals
14.29
Hb Ha
Ha Hb Ha
Hd
a.
Ha
H H H
Hc
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
b.
2 1H NMR signals
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 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
c.
H H Cl
Hc
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
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.
14.30 Electronegative elements shift absorptions downfield. The carbons of alkenes and benzene rings,
and carbonyl carbons are also shifted downfield.
O
a.
CH3CH2OCH2CH3
The CH2 group is closer
to the electronegative O
and will be farther downfield.
b.
BrCH2CHBr2
The C of the CHBr2 group has two
bonds to electronegative Br atoms
and will be farther downfield.
c.
H
C
OCH3
The carbonyl carbon is
highly deshielded and
will be farther downfield.
d. CH3CH CH2
The CH2 group is part of
a double bond and will
be farther downfield.
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Chapter 14–10
14.31
a. In order of lowest to highest chemical shift:
Ca C b C c Cd
CH3CHCH2CH3
OH
C d < C a < C c < Cb
b. In order of lowest to highest chemical shift:
(CH3CH2)2C=O
Ca Cb
Cc
C a < C b < Cc
14.32
• 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.
14.33
O
OH
• molecular formula C4H8O
2n + 2 = 2(4) + 2 = 10
10 – 8 = 2/2 = 1 degree of unsaturation
•
13C
NMR signal at > 160 ppm due to
C=O
• molecular formula C4H8O
2n + 2 = 2(4) + 2 = 10
10 – 8 = 2/2 = 1 degree of unsaturation
• all 13C NMR signals at < 160 ppm
NO C=O
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Resonance Spectroscopy
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Nuclear Magnetic Resonance Spectroscopy 14–11
14.34 Use the directions from Answer 14.3.
a.
e.
(CH3)3CH
O
2 kinds of H's
b.
h.
f.
CH3CH2
H
Hb
Br
Hc
H H
g.
Hd
CH3 C C CH3
Ha
HO H
Hf
He
Hc
6 kinds of H's
CH3CH2CH2OCH2CH2CH3
3 kinds of H's
C C
H
i.
3 kinds of H's
CH2CH3
H
Hc
H
H
C C
c. CH3CH2OCH2CH2CH2CH2CH3
7 kinds of H's
Hb
CH3
4 kinds of H's
(CH3)3CC(CH3)3
Ha CH3
Hd
CH2CH3
C C
Hb
5 kinds of H's
1 kind of H
d.
CH3
Ha
Hd
j.
4 kinds of H's
Ha
H
Hb
H
Hc
CH3
O
H
Hd
4 kinds of H's
14.35
CH3
H
H
Hb
CH3
H
a.
H
Hc
Ha
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
H H
c
Ha
CH3
Hd
H
Ha CH3
d.
Hc
H
CH3
c.
H
Hb
3 kinds of protons
He
Hc
4 kinds of protons
Hb
Hc
H
Hb
4 kinds of protons
Hd CH3
Ha
Hc
7 kinds of protons
14.36
equivalent
O
O
a.
CH3
O
b.
CH3
CHO
c.
d.
N
N
OH
H
N
N
CH3
caffeine
4 NMR signals
OCH3
OH
vanillin
6 NMR signals
thymol
7 NMR signals
CH3O
N
H
HO
capsaicin
15 NMR signals
equivalent
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Chapter 14–12
14.37
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
(in ppm) = [observed chemical shift (Hz)] / of the NMR (MHz)]
14.38
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
14.39 Use the directions from Answer 14.8.
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
14.40 Use the directions from Answer 14.12.
[total number of integration units] / [total number of protons]
[13 + 33 + 73] / 10 = ~12 units per proton
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.
14.41
Hb Hb
Hb
Ha
CH3
a. CH3CO2C(CH3)3 and CH3CO2CH3
c.
Ha
Hb
Ha
Hb
CH3
CH3
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
Hb
CH3
Ha
Hb Hb
Hb
CH3
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
14.42 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
Ha
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Nuclear Magnetic Resonance Spectroscopy 14–13
14.43
CH3
Ha
CH3CH2CH2CH2CH2CH3
CH3CHCH2CH2CH3
Ha Hb Hc Hc Hb Ha
Ha Hb Hc Hd He
3 signals:
Ha: split by 2 Hb protons - triplet
Hc: split by 2 Hb protons - triplet
Hb: split by 3 Ha + 2 Hc protons - 12 peaks (maximum)
Since Hb is located in a flexible alkyl chain, peak overlap
occurs, so that only 3 + 2 + 1 = 6 peaks will likely be
observed.
Hc
5 signals:
Ha: split by 1 Hb proton - doublet
Hb: split by 6 Ha + 2 Hc protons - 21 peaks (maximum)
Hc: split by 1 Hb + 2 Hd protons - 6 peaks (maximum)
Hd: split by 2 Hc + 3 He protons - 12 peaks (maximum)
He: split by 2 Hd protons - triplet
Since Hb, Hc, and Hd are located in a flexible alkyl chain,
it is likely that peak overlap occurs, so that the following is
observed:
Hb (6 + 2 + 1 = 9 peaks), Hc (1 + 2 + 1 = 4 peaks), and
Hd (2 + 3 + 1 = 6 peaks).
Ha Ha
CH3
Hc
CH3CH2CHCH2CH3
CH3
Ha Hb
Hb Ha
Hd
4 signals:
CH3 CH3
Ha: split by 2 Hb protons - triplet
CH
CH CHCH3
3
Hb: split by 3 Ha + 1 Hc protons - 8 peaks (maximum)
Hc: split by 4 Hb + 3 Hd protons - 20 peaks (maximum)
Ha Hb Hb Ha
Hd: split by 1 Hc proton - doublet
Since Hb and Hc are located in a flexible alkyl chain, it
2 signals:
is likely that peak overlap occurs, so that the following
Ha: split by 1 Hb proton - doublet
is observed: Hb (3 + 1 + 1 = 5 peaks) and Hc (4 + 3 +
H
b: split by 6 Ha protons - septet
1 = 8 peaks).
CH3CH2 C CH3
Ha Hb
CH3
Hc
Hc
3 signals:
Ha: split by 2 Hb protons - triplet
Hb: split by 3 Ha protons - quartet
Hc: no splitting - singlet
14.44
O
a. CH3CH(OCH3)2
e. (CH3)2CH
CH3 protons split by 1 H = doublet
CH proton split by 3 H's = quartet
O
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
d. CH3OCH2CHCl2
CH2 protons split by 1 H = doublet
CH proton split by 2 H's = triplet
H a Hb
C
O
OCH2CH3
CH3CH2CH2
Hc H d
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
f.
h.
HOCH2CH2CH2OH
g. CH3CH2CH2CH2OH
Ha Hb Hc Hd
OH
Ha H b H c
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.
Ha Hb
Ha protons split by 2 CH2 groups =
quintet
Hb protons split by 2 H's = triplet
C
O
i.
CH3CH2
Ha
C
H
Hb
Ha: split by CH3 group + Hb
= 8 peaks (maximum)
Hb: 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).
364
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Chapter 14–14
H
CH3
j.
k.
C C
CH3CH2
H
l.
C C
Br
Hb
Ha
H
CH3
Ha
Hb
H
Hc
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
CH3
H
Ha
C C
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
H
Hb
H
Br
H
Br
C C
Ha
C C
CO2CH3
Hb H
Ha: split by 1 H = doublet
Hb: split by 1 H = doublet
Ha and Hb are geminal.
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.
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
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Nuclear Magnetic Resonance Spectroscopy 14–15
14.48 Only two compounds in Problem 14.42 give one signal in their 13C NMR spectrum:
CH3CH3
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 +, 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
d.
a. HC(CH3)3
2 signals
CH2
g.
O
5 signals
7 signals
e. CH3CH2
b.
CH2CH3
h.
O
C C
5 signals
H
4 signals
H
3 signals
c. CH3OCH(CH3)2
f.
i.
OH
3 signals
7 signals
3 signals
14.51
O
a. CH3CH2
Ca Cb
O
OH
C
OH
Cc
Ca < Cb < Cc
b. CH3CH2CHCH2CH3
c.
C
Ca
Ca Cb Cc
Ca < Cb < Cc
d. CH2 CHCH2CH2CH2Br
CH2CH3
Ca
Cb Cc
Cc < Cb < Ca
Cb Cc
Cb < Cc < Ca
14.52
19 ppm
62 ppm
a. CH3CH2CH2CH2OH
14 ppm 35 ppm
16 ppm
205 ppm
b. (CH3)2CHCHO
41 ppm
143 ppm
23 ppm
c. CH2=CHCH(OH)CH3
113 ppm
69 ppm
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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
Cb
CO2CH3
Cb Ca
3 signals
3 different C's
3 signals
H
Cc
C C
Cc
Cd
CO2CH3
C C
H
O
Cc
CO2CH3
Cb Ca
4 signals
O
Ce O
C b Cc
Ca
5 signals
14.54 Use the directions from Answer 14.24.
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
Br
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. 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
CH3CH2
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
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
H
CH3 C O C CH3
CH3
CH3
f. C3H6O: 1 degree of unsaturation
IR peak at 1730 cm–1: C=O
O
NMR: triplet at 1.11 ppm
C
multiplet at 2.46 ppm
CH3CH2
H
triplet at 9.79 ppm
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Nuclear Magnetic Resonance Spectroscopy 14–17
14.55
Two isomers of C9H10O: 5 degrees of unsaturation (benzene ring likely)
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)
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)
O
CH2
C
O
C
CH3
CH2CH3
14.56
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
Hb
C
H
O C O
CH3
O
C
CH3
Hb
Ha
14.57
Hb
O
[1] LiC
[2] H2O
CH
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)
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Chapter 14–18
14.58
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
C
Hb
O
OCH2CH3
CH3
Ha Hc
Hc
C
CH2OCH3
Ha
Hb
14.59
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)
Ha
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
CH3
CH2CH2NH2
H
Hb
Hc
C NH2
Hc
Hb
Ha
14.60
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
O
C
OCH2CH3
downfield due to the O atom
b. C9H12:
4 degrees of unsaturation
IR absorption at 2850–3150 cm–1:
C–H bonds
NMR data:
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
C CH3
H
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14.61
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):
CH
0.9 (triplet, 3 H), split by 2 H's
3 C CH3
CH2CH3
1.2 (singlet, 6 H), due to 2 CH3 groups
1.5 (quartet, 2 H), split by 3 H's
1.6 (singlet, 1 H), due to the OH proton
14.62
Compound L has a molecular ion at 90: molecular formula C4H10O2
0 degrees of unsaturation
total integration units/# H's
(25 + 46 + 7)/10 = ~8 units per H
IR absorptions at 2992 and 2941 cm–1: Csp3–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
14.63
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
Hb
Hb
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Chapter 14–20
14.64
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
O
C CH2CH2CH3
Hc Hb Ha
O
14.65
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
14.66
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
CH3
C
Hc Ha
= Q
CH(CH3)2
MW = 86
14.67
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
Ha
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)
Hc
O
CH3CH2
C
OCH2CHCH3
CH3
Hb Hd
He
Ha
Ha
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Nuclear Magnetic Resonance Spectroscopy 14–21
14.68
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
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.
14.69 Both A and B have the same molecular ion—since they are isomers—and show a C=O peak in
their IR spectra. 1H NMR spectroscopy is the best way to distinguish the two compounds.
O
CH3O
C
O
or
C(CH3)3
CH3
C
OC(CH3)3
B
A
Both A and B have two singlets in a 3:1 ratio in their 1H NMR
spectra. But A has a peak at ~3 ppm due to the deshielded
CH3 group bonded to the O atom. B has no proton that is so
deshielded. Both of its singlets are in the 1–2.5 ppm region.
14.70
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 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: CC bond
13
C NMR: 4 signals: 4 kinds of C's
CH3
HC C C CH3
CH3
14.71
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.
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Chapter 14–22
14.72
Ho
Ho
Ho
Ho
Ho
Hi
Hi
Ho
Hi
Hi
Ho
Hi
Hi
Ho
Ho
Ho
Ho
Ho
18-Annulene has 18 electrons that create an induced magnetic field
similar to the 6 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 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 protons and the
absorption is therefore very far upfield, even higher than TMS
(–1.8 ppm).
14.73
Ca
stereogenic center
CH3 H
CH3 C
H
C CH3 Replace a CH3 group
with X.
OH
Cb
H
HO C CH3
X C CH3
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.
14.74
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.
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Radical Reactions 15–1
C
Chhaapptteerr 1155:: R
Raaddiiccaall R
Reeaaccttiioonnss
G
Geenneerraall ffeeaattuurreess ooff rraaddiiccaallss
• 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
o
1
R2CH
R3C
o
3o
2
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
R
Raaddiiccaall rreeaaccttiioonnss
[1] Halogenation of alkanes (15.4)
R H
X2
h or 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
h 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
h, , 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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Chapter 15–2
C
Chhaapptteerr 1155:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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
CH3 CH2
H Cl
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 generating an
alkyl halide.
Cl2
a.
Cl
b. CH3CH2CH2CH2CH2CH3
c. (CH3)3CH
H
Cl2
ClCH2CH2CH2CH2CH2CH3
Cl
H
CH3 C CH3
H
CH3CH2 C CH2CH2CH3
Cl
Cl
Cl2
CH3 C CH2CH2CH2CH3
CH3 C CH2Cl
CH3
CH3
15.5
A
Cl2
Cl
Cl2
B
Cl
Cl
Cl
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Radical Reactions 15–3
15.6
Br
Initiation:
Propagation:
CH3
h
or Br
Br
H Br
CH3 Br
Br
+
Br
CH3
Br Br
CH3
Termination:
+
Br
+
H
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
H° = –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
H° = +67 kJ/mol
H Br
H
I
1 bond formed
–297 kJ/mol
H° = +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
376
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Chapter 15–4
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
Br
Br
Br2
b.
Br2
Br
d.
15.11
Cl2
Cl
Cl
h
Cl
Cl
Cl
This is the desired product,
1-chloro-1-methylcyclohexane,
but many other products are formed.
15.12 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.
15.13
CH3
Br2
a. CH3 C H
CH3
CH3
CH3
CH3 C Br
c.
CH3
H2O
C CH2
CH3
CH3 C OH
H2SO4
CH3
CH3
(from b.)
CH3
b.
(CH3)3CO–K+
CH3 C Br
CH3
CH3
C CH2
CH3
d.
CH3
(from a.)
Cl2
C CH2
CH2Cl
CH3 C Cl
CCl4
CH3
CH3
(from b.)
15.14
Br2
a.
Br K+ –OC(CH )
3 3
h
mCPBA
b.
(from a.)
Br2
Br
+ enantiomer
Br
O
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Radical Reactions 15–5
15.15 Since the reaction does not occur at the stereogenic center, leave it as is.
C1
C4
H Br
CH3
H Br
Cl2
C
C
h
CH2CH3
H Br
C
CH2CH3
ClCH2
(2R)-2-bromobutane
CH3
CH2CH2Cl
S
R
15.16
Cl
Cl
Cl2
a. CH3CH2CH2CH2CH3
CH3CH2CH2CH2CH2Cl
Cl
b.
Cl2
CH3
H
c.
H Cl
Cl2
d.
H Cl
CH3
CH3
C
CH3
Cl H
CH(CH2CH3)2
H Cl
H Cl
h
(Consider attack at
C2 and C3 only.)
CH3
H
Cl H
CH2CH3
Cl Cl
CH3
ClCH2CH2CH(CH2CH3)2
CH3CH2 C CH2CH3
h
CH2CH3
C
Cl
CH3
Cl
Cl2
CH3CH2 C CH2CH3
Cl
CH3
CH3
H
CH3 CH3CH2CH2
CH3CH2CH2
Cl
Cl
CH2Cl
C
CH3CH2CHCH2CH3
Cl
C
CH(CH2CH3)2
H Cl
15.17
O N
Chain propagation:
O N O
+
O3
+
O
O N O + O2
O N
+
O2
The radical is re-formed.
15.18 Draw the resonance structure by moving the 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 • is used on any atom that has an unpaired electron in any resonance structure.
a.
CH3 CH CH CH2
CH3 CH CH CH2
hybrid: CH3 CH CH
c.
CH2
hybrid:
b.
d.
hybrid:
hybrid:
378
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Chapter 15–6
15.19 Reaction of an alkene with NBS or Br2 + h yields allylic substitution products.
Br
NBS
a.
c. CH2 CH CH3
h
Br2
CH2CHCH3
Br Br
NBS
b. CH2 CH CH3
CH2 CH CH2Br
h
15.20
c. CH2 C(CH2CH3)2
CH3 CH CH CH2Br
h
+
h
b.
+
BrCH2C(CH2CH3) CHCH3
NBS
h
CH3
CH3
Br
CH3
CH3
+
Br
15.21
CH3
CH3
CH3
NBS
CH3
Br
CH3
Br
h
Br
Br
15.22 Reaction of an alkene with NBS + h yields allylic substitution products.
a.
one possible product: high yield
Br
b.
CH3CH2CH CHCH2Br
CH2 C
CH2CH3
CH3CH(Br)CH CH2
CH3
CH3
CHBrCH3
NBS
NBS
a. CH3 CH CH CH3
c.
Br
Cannot be made in high yield by allylic halogenation.
Any alkene starting material would yield a mixture of allylic halides.
15.23
OOH
second resonance structure
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Radical Reactions 15–7
15.24 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
O
C
H
C
OH
OH
linoleic acid
This allylic C–H bond is most readily cleaved.
O
O
C
C
OH
OH
hydroperoxide products:
HO
O
O
O
C
C
OH
O
O
C
HO
OH
OH
(E/Z isomers are possible.)
OH
O
15.25
R
O H
(CH3)3C
O
C(CH3)3
(CH3)3C
C(CH3)3
BHT
H
H
H
H
CH3
O
(CH3)3C
O
C(CH3)3
H
CH3
O
(CH3)3C
H
C(CH3)3
H
CH3
H
(CH3)3C
H
CH3
O
C(CH3)3
(CH3)3C
H
H
CH3
CH2 CHCH2CH2CH2CH3
HBr
CH3CHCH2CH2CH2CH3
Br
CH2 CHCH2CH2CH2CH3
HBr
CH2BrCH2CH2CH2CH2CH3
ROOR
b.
HBr
Br
HBr
ROOR
H
CH3
15.26
a.
C(CH3)3
Br
380
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Chapter 15–8
HBr
c. CH3CH CHCH2CH2CH3
CH3CH CH2CH2CH2CH3
or HBr, ROOR
CH3CH2 CHCH2CH2CH3
Br
Br
15.27 In addition of HBr under radical conditions:
• Br• adds first to form the more stable radical.
• Then H• is added to the carbon radical.
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
15.28
Br
HBr
ROOR
a.
c.
Br2
Br
Br
Br
HBr
b.
15.29
Transition state 1:
H
CH3 C
Energy
Transition state 2:
CH2
Br
H
CH3 C CH2
H Br
Br
CH3 CH CH2
+
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
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Radical Reactions 15–9
15.30
H
a.
H
CH2 C
H
H
CH2 C CH2 C CH2
C6H5
C6H5
b.
C
C6H5
CH2 CH OCOCH3
C6H5
O
O
O
COCH3 COCH3 COCH3
polystyrene
poly(vinyl acetate)
15.31
Initiation:
RO
OR
Cl
[1]
+ CH2 C
2 RO
Cl
[2]
ROCH2 C
H
H
Cl
Cl
Propagation:
ROCH2 C
Cl
[3]
ROCH2 C
CH2 C
H
Cl
Cl
H
H
CH2 C
Cl
CH2 C
H
H
H
Repeat Step [3] over and over.
Termination:
carbon radical
new C–C bond
Cl Cl
[4]
C CH2
CH2 C C CH2
[one possibility]
H H
15.32
a. increasing bond strength: 2 < 3 < 1
b. and c.
CH3
CH2 C CHCH3
CH3
H CH2 C CHCH3
H H
H
1° radical
least stable
2° radical
intermediate stability
CH3
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
382
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Chapter 15–10
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
15.35
Hb
Hd
H H
CH2 CHCHCHC(CH3)CH2 H
Ha
H
Hc
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
Increasing ease of abstraction:
Hc < Hd < Ha < Hb
15.36
Cl
a.
Cl
b.
(CH3)3CCH2CH2CH2CH3
(CH3)3CCH2CH2CH2CH2Cl
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.
CH3
Br
CH3
Br
b.
(CH3)3CCH2CH(CH3)2
(CH3)3CCH2C(CH3)2
Br
d.
(CH3)3CCH2CH3
(CH3)3CCHCH3
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Radical Reactions 15–11
15.38 Draw all of the alkane isomers of C6H14 and their products on chlorination. Then determine
which letter corresponds to which alkane.
Cl
*
*
h
A
Cl
Cl
Cl2
Cl
Cl
Cl2
Cl
*
Cl
*
* *
h
Cl
Cl
B
Cl2
*
h
Cl
Cl
C
Cl
Cl2
Cl2
*
Cl
h
D
Cl
Cl
h
*
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 a H.
Br
a.
Cl
c.
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
These two compounds can be formed
in high yield from an alkane.
three different
C–H bonds
Br
Br on 2° carbon
The product with Br on 3° carbon
will form predominantly.
These two compounds cannot be
formed in high yield from an alkane.
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Chapter 15–12
15.41 Chlorination with two equivalents of Cl2 yields a variety of products.
Cl
Cl2 (2 equiv)
Cl
Cl
Cl
Cl
Cl
Cl
Cl
The desired product is only one of four products formed.
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
CH3
CH2Br
CH3
weaker bond
H
o
C
stronger bond H = 356 kJ/mol
o
As
usual,
more
product is formed
H = 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.
Cl
Y
Br2
b.
Br
Y
K+–OC(CH3)3
Br2
c.
Y
Br
15.44 Draw the resonance structures by moving the bonds and the radical.
CH2
a.
b.
Cl
Cl
CH2
c.
Z
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Radical Reactions 15–13
15.45 Reaction of an alkene with NBS + h yields allylic substitution products.
Br
NBS
a.
h
Br
NBS
b. CH3CH2CH CHCH2CH3
c. (CH3)2C CHCH3
h
NBS
(CH3)2C
h
Br
CH3CH2CH CHCHCH3
CHCH2Br
(CH3)2C(Br)CH
Br
CH3CH2CHCH CHCH3
CH2 C(CH3)CH(Br)CH3
BrCH2C(CH3) CHCH3
CH2
Br
NBS
d.
h
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
h
*
Br
Br
5-bromo-1-methylcyclopentene
15.47
Br
NBS
h
Br
X
Br
Br
Br
Br
15.48
Cl
+ Cl2
a.
h
Cl
Cl
Cl
CH3
+ Br2
b.
CH3
CH3
CH3
Br
CH3
CH3
(major product)
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Chapter 15–14
Br
c.
+ Br2
Br
Br
h
(minor product)
(two major products)
Br
h
+ NBS
CH2
d.
CH2
Br
Br
Br2
HBr
e.
Br
h.
HBr
ROOR
f.
Br
Br
Br
i.
g.
h
HBr
CH2
ROOR
CH=CHCH2Br
NBS
Br
15.49
Br
Br
HBr
a.
Br
Br2
b.
NBS
c.
h
Br
+
enantiomer
15.50
KOC(CH3)3
Br2
h
O
Br
Br
A
ozonolysis
cyclohexanone
B
C
D
15.51
CH3
a.
CH3
b.
Br2
h
Cl2
h
CH3
Br
CH2Cl
CH3
CH3
CH3
Cl
Cl
CH3
CH3
Cl
Cl
Cl
CH3
Cl
CH3
Cl
O
acetone
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Radical Reactions 15–15
Cl
Cl2
c.
Cl
Cl
Cl
h
Cl
Cl
Cl
Cl
CH3
d.
Br2
h
CH3
CH3
CH3
Br2
e.
H
CH3
CH3
Br
Br
CH3
CH3
+
CH3 Br
C
CH3CH2
Br CH3
CCl3
CCl3
f.
CH3
Cl2
F
H
C
ClCH2CH2
F
H
CCl3
+
C
F
Cl
CH3CH2
CCl3
+
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
Cl
Cl
Cl
a.
h
Cl
Cl
Cl
Cl
Cl
A
B
C
D
E Cl
(2R)-2-chloropentane
Cl
Cl Cl
G
F
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
h
Cl
1
2
3
Cl
4
Cl
Cl
Cl
5
6
9
Cl
7
Cl
10
11
8
Cl
Cl
388
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Chapter 15–16
a. There would be 10 fractions, since 4 and 5 (two enantiomers) would be in one fraction.
b. All fractions except the one that contains 4 and 5 would be optically active.
15.54
Br
NBS
CH2
Br
CH2
CH2
CH2
CH2
h
A
Br
CH2Br
Br
CH2Br
15.55
CH3
a.
CH3
h
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:
(CH3)3C
Propagation:
h
or Br
H
(CH3)3C
Termination:
(one possibility)
Br
+
total bonds formed = –640 kJ/mol
Br
Br
+
H–Br bond formed
–368 kJ/mol
+
Br
Br Br
(CH3)3C
c. H° = (bonds broken) – (bonds formed)
= (+381 kJ/mol) + (–368 kJ/mol)
= +13 kJ/mol
H° = (bonds broken) – (bonds formed)
= (+192 kJ/mol) + (–272 kJ/mol)
= –80 kJ/mol
H Br
(CH3)3C
Br
H° = –67 kJ/mol
Br
Br
Br Br
d. and e.
Transition state 1:
Transition state 2:
+13
kJ/mol
Energy
Transition state 1:
(CH3)3C
H
CH3
(CH3)3C
CH3
+ Br
C CH
3
H
Br
–80
kJ/mol
(CH3)3C
Reaction coordinate
Br
Br
Transition state 2:
CH3
CH3 C CH3
Br
Br
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Radical Reactions 15–17
15.56
O
Initiation:
O
N
Br
N
+
Br
h
O
NBS
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
Ho = +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.
Ho = –83 kJ/mol
I
–234 kJ/mol
15.58 Calculate H° for each of these steps, and use these values to explain why this alternate
mechanism is unlikely.
[1]
+ Cl
C–H bond broken
+435 kJ/mol
C–Cl bond formed
–351 kJ/mol
+ Cl2
Cl–Cl bond broken
+242 kJ/mol
H–Cl bond formed H° = –189 kJ/mol
–431 kJ/mol
CH4
CH3Cl + H
H
[2]
HCl +
H° = +84 kJ/mol
The H° for Step [1] is very endothermic,
making this mechanism unlikely.
Cl
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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Br
3,3-dimethyl-1-butene
HBr
Br
HBr
peroxide
The 2o radical does
NOT rearrange.
Br
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
Cl
Initiation:
CH3 H
Propagation:
One possibility for
termination:
h
or Cl
+
Cl
+
Cl
CH3
Cl Cl
CH3
CH3
Cl
CH3
H Cl
CH3 Cl
Cl
CH3 CH3
CH3 CH2 H
CH3CH2
+
Cl
Cl Cl
CH3CH2
CH3CH2 Cl
H Cl
Cl
15.61
H
Step 1:
CH3 CH CH2
+
Cl
CH3
C CH2
H1° = –72 kJ/mol
Cl
1 bond broken
+267 kJ/mol
H
Step 2:
1 bond formed
–339 kJ/mol
H
CH3 C CH2
+ H–Cl
Cl
CH3 C CH2
H Cl
1 bond broken
+431 kJ/mol
1 bond formed
–397 kJ/mol
H2° = +34 kJ/mol
This step of propagation is endothermic. It
prohibits chain propagation from occurring over
and over.
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Radical Reactions 15–19
15.62
Cl2
a.
Cl
mCPBA
f.
O
(from b.)
Cl
b.
K+ –OC(CH3)3
g.
(from a.)
Br
Br2
Br
(from c.)
Br
Br
NBS
c.
Br
ROOR
h.
(from b.)
(from b.)
OH
i.
H2SO4
OH
mCPBA
OH
O
(from h.)
Cl
Cl2
e.
OH
(from c.)
H2O
d.
OH
Br
OH
j.
[1] NaCN
O
Cl
(from b.)
[2] H2O
(from f.)
CN
15.63
Br2
a.
Br
H2O
K+–OC(CH3)3
OH
H2SO4
h
major
product
mCPBA
b.
Cl2
c.
O
Cl
+ enantiomer
Cl
(from a.)
(from a.)
15.64
a. CH3CH2CH3
b.
Br
Br2
h
CH3CHCH3
K+ –OC(CH3)3
CH3CH CH2
Br Br
Br2
CH3 C C H
Br
H H
K+ –OC(CH3)3
HBr
K+ –OC(CH3)3
CH3C CH
(2 equiv)
DMSO
Br
ROOR
c. HC CH
NaH
HC C
CH3CH2Br
H2
HBr
Lindlar catalyst
ROOR
Br
15.65
a. CH3 CH3
b.
HC CH
(from a.)
Br2
h
NaH
CH3 CH2Br
HC C
K+ –OC(CH3)3
CH3 CH2Br
(from a.)
CH2 CH2
HC CCH2CH3
Br2
BrCH2 CH2Br
2 NaNH2
HC CH
392
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Chapter 15–20
O
[1] HC
mCPBA
c. CH2 CH2
C
HC CCH2CH2OH
(from b.)
(from a.)
[2] H2O
d. HC CCH2CH3
NaH
C CCH2CH3
CH3 CH2Br
CH3CH2 C C CH2CH3
(from a.)
(from b.)
Na
NH3
H2O
e. CH3CH2 C C CH2CH3
H2SO4
HgSO4
(from d.)
O
15.66
Cl
Cl2
K+ –OC(CH3)3
[1] O3
CHO
OHC
h
[2] (CH3)2S
15.67
O2 abstracts a H here.
O O
COOH
HH
COOH
COOH
O O
C5H11
C5H11
C5H11
arachidonic acid
+ HOO
R H
OOH
COOH
C5H11
O O
COOH
+
COOH
R H
C5H11
C5H11
another molecule of
arachidonic acid
5-HPETE
This process is repeated.
15.68
[1]
O
H
O O
O O
[2]
H
O
O
O O
O
[3]
+ HOO
O
O OH
+
O
Then, repeat
Steps [2] and [3].
15.69
HOO
a.
OOH
(cis and trans)
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Radical Reactions 15–21
O2 abstracts a H here.
O O
b.
O O
[1]
H H
+ HOO
[2a]
[3a]
R H
O O
O O
[2b]
OOH
[3b]
HOO
+ R
R H
O O
O O
[RH = 1-hexene]
+ R
Repeat Steps [2] and [3] again and again.
15.70
(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
O
OCH3
OCH3
15.71 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
15.72 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
394
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Chapter 15–22
15.73
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
15.74
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
[4]
C CH2
H
CN
CH2 C
H
H
H
Repeat Step [3] over and over.
Termination:
carbon radical
new C–C bond
CN CN
CH2 C C CH2
H
[one possibility]
H H
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
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Radical Reactions 15–23
15.76
singlet singlet
at 2.23 at 4.04
Cl2
Cl
Cl
doublet
at 1.69
Cl
Cl
h
Cl
Cl Cl
multiplet Cl B
at 4.34
A
C
doublet at 5.85
15.77
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
triplet
Cl2
Cl
h
Cl
quintet
minor product
15.78
a. CH3CH3
Cl2
equivalent H's
singlet
doublet quartet
ClCH2CH2Cl
CH3CHCl2
h
Cl
b. Initiation:
Propagation:
X
h
Cl
Y
Cl
or +
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
ClCH2CH2Cl
X
Cl
+
Cl
H Cl
CH2 CH2Cl
CH2 CH2Cl
Termination:
Cl
CH3CHCl2
Y
CH3 CH Cl
Formation of X:
H Cl
CH3 CH Cl
Cl Cl
Cl
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Chapter 15–24
15.79
RO
OR
RO
O
O
CH3
C
OR
O
[1]
H
OR
CH3
[2]
C
O
H
C
O
CH3 [3]
+
H OR
CH3
O
C
(Repeat Steps [2]
and [3].)
15.80
Initiation:
R3SnH
R3Sn
+ Z
Propagation:
+ HZ
CH2
Br
+ R3SnBr
R3Sn
R3SnH
CH2
R3SnH
CH2
CH2
R3SnH
+
+
R3Sn
+
R3Sn
R3Sn
15.81
CO2H
O
O
COOH
O
O
O
COOH
O
O
A
O
COOH
O
O
COOH
O
O
OOH
+ R
O O
R H
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Resonance, and Dienes
397
Conjugation, Resonance, and Dienes 16–1
C
Diieenneess
Chhaapptteerr 1166:: C
Coonnjjuuggaattiioonn,, R
Reessoonnaannccee,, aanndd D
C
Coonnjjuuggaattiioonn aanndd ddeellooccaalliizzaattiioonn ooff eelleeccttrroonn ddeennssiittyy
• 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 any system X=Y–Z:, Z is sp2 hybridized to allow the lone pair to occupy a p orbital, making the
system conjugated (16.5).
FFoouurr ccoom
mm
moonn eexxaam
mpplleess ooff rreessoonnaannccee ((1166..33))
[1] The three atom “allyl” system:
X
Y Z
*
X Y Z
*
* = +, –, •, or ••
[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
+
Electronegativity of Y > X
X Y
R
Ruulleess oonn eevvaalluuaattiinngg tthhee rreellaattiivvee ““ssttaabbiilliittyy”” ooff rreessoonnaannccee ssttrruuccttuurreess ((1166..44))
[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
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Chapter 16–2
TThhee uunnuussuuaall pprrooppeerrttiieess ooff ccoonnjjuuggaatteedd ddiieenneess
[1] The C–C bond joining the two double bonds is unusually short (16.8).
[2] Conjugated dienes are more stable than similar isolated dienes. Ho 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 bonds
increases, the absorption shifts to longer wavelength (16.15).
R
Reeaaccttiioonnss ooff ccoonnjjuuggaatteedd ddiieenneess
[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
•
•
•
•
•
•
•
Z
The three new bonds
are labeled in bold.
dienophile
The reaction forms two and one 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).
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Resonance, and Dienes
399
Conjugation, Resonance, and Dienes 16–3
C
Chhaapptteerr 1166:: A
Annssw
weerrss ttoo PPrroobblleem
mss
16.1 Isolated dienes have two double bonds separated by two or more bonds.
Conjugated dienes have two double bonds separated by only one bond.
a.
b.
One bond separates
two double bonds =
conjugated diene
c.
Two bonds separate
two double bonds =
isolated diene
d.
One bond separates
two double bonds =
conjugated diene
Four bonds separate
two double bonds =
isolated diene
16.2
isolated isolated
isolated
O
OOH
C
OH
5-HPETE
conjugated
16.3
a.
Conjugation occurs when there are overlapping p orbitals on three or more adjacent atoms.
Double bonds separated by 2 bonds are not conjugated.
CH2 CH CH CH CH CH2
All of the carbon atoms are sp2
hybridized. Each bond is
separated by only one bond.
conjugated
b.
The two bonds are
separated by three bonds.
NOT conjugated
c.
e.
O
The two bonds are
separated by only one bond.
conjugated
d.
+
This carbon is not sp2 hybridized.
NOT conjugated
+
Three adjacent carbon atoms are sp2 hybridized
and have an unhybridized p orbital.
conjugated
16.4 Two resonance structures differ only in the placement of electrons. All bonds stay in the same
place. Nonbonded electrons and bonds can be moved. To draw the hybrid:
• Use a dashed line between atoms that have a bond in one resonance structure and not the
other.
• Use a symbol for atoms with a charge or radical in one structure but not the other.
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Chapter 16–4
resonance hybrid:
a.
+
+
+
The + charge is delocalized
on two carbons.
+
resonance hybrid:
b.
+
+
+
The + charge is delocalized
on two carbons.
+
The + charge is delocalized
on two carbons.
+
resonance hybrid:
c.
+
+
+
Each different kind of carbon atom will give a different 13C signal. When a carbocation is
delocalized as in structure B, carbons become equivalent.
16.5
+
The two + carbons are identical.
CH3
+
B
4 different kinds of C
4 13C NMR signals
A
5 different kinds of C
5 13C NMR signals
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.
16.6
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
16.7
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
C
CH3
H
Move the charge
and the double bond.
d.
Move the charge
and the double bond.
CH3CH2
C
+
C
H
CH3
c. CH3 CH Cl
Move the lone pair.
+
CH3 CH Cl
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Conjugation, Resonance, and Dienes 16–5
16.8 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
a. CH3
one more bond
+
NH2
+
C NH2
CH3 C
CH3
+
+
NH2
+
CH3 C
c.
hybrid CH3
least stable
most stable
one more bond
All atoms have an octet.
better resonance structure
intermediate stability
CH3
b.
CH3
C
CH3
NH
least stable
C
O
O
+
one more bond
better resonance structure
intermediate stability
least stable
+
most stable
O
O
O
O
NH
CH3
negative charge on the
more electronegative atom
better resonance structure
intermediate stability
C
d.
NH
hybrid
most stable
N
least stable
–
+
N
one more bond
better resonance structure
N
most stable
intermediate stability
16.9
OCH3
OCH3
OCH3
OCH3
largest contribution
more bonds and all
atoms have octets
16.10
O
a.
CH3
C
O
OH
A
no charges
All atoms have an octet.
4 bonds (C–C + C–O)
most stable
lowest energy
resonance structure
CH3
C
O O
OH
CH3
C
b.
OH
+
+ OH
C
CH3
D
B
C
2 charges
2 charges
C does not have an octet. All atoms have an octet.
3 bonds (C–C + C–O)
4 bonds (C–C + C–O)
least stable
intermediate stability
highest energy
intermediate energy
resonance hybrid
lower in energy
than any single
resonance form
c. In order of increasing energy: D < A < C < B
16.11 Remember that in any allyl system, there must be p orbitals to delocalize the lone pair.
CH2
O
a.
O
b.
CH3 C
c.
O
sp2 hybridized
trigonal planar geometry
sp2 hybridized
trigonal planar geometry
sp2 hybridized
trigonal planar geometry
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Chapter 16–6
16.12 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.
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
16.13
s-cis
conjugated
OH
conjugated
s-trans
E E
conjugated
isolated
Z
Z
HO
Z
O
isolated
OH
Z
isolated
isolated
16.14 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
16.15 Two equivalent resonance structures delocalize the bond and the negative charge.
O
CH3 C
O
CH3
CH3 C
O
O hybrid:
O
C
O These bond lengths are equal
because they are identical.
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16.16 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
one bond = conjugated diene
smaller heat of hydrogenation
Double bonds separated by
two bonds = isolated diene
larger heat of hydrogenation
or
b.
Double bonds separated by
two bonds = isolated diene
larger heat of hydrogenation
Double bonds separated by
one bond = conjugated diene
smaller heat of hydrogenation
16.17 Isolated dienes are higher in energy than conjugated dienes. Compare the location of the double
bonds in the compounds below.
*
*
*
3 conjugated double bonds
most stable
*
2 conjugated double bonds
intermediate stability
0 conjugated double bonds
least stable
*
16.18 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.
404
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Chapter 16–8
16.19 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
16.20 Label the products as 1,2- or 1,4-products. The 1,2-product is the kinetic product, and the 1,4product, 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
16.21 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.
re-draw
COOH
COOH
dienophile
diene
CH3
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
O
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Conjugation, Resonance, and Dienes 16–9
16.22 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
16.23
(2Z,4Z)-2,4-hexadiene
(2E,4E)-2,4-hexadiene
sterically unhindered
more reactive
sterically hindered
Steric interactions between the two CH3 groups make it difficult
for the diene to adopt the needed s-cis conformation.
16.24 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
16.25 A cis dienophile forms a cis-substituted cyclohexene.
A trans dienophile forms a trans-substituted cyclohexene.
CH3OOC
a.
COOCH3
+
H
COOCH3
C C
+
H
COOCH3
cis dienophile
CH3OOC
b.
COOCH3
+
COOCH3
COOCH3
COOCH3
trans-substituted products
O
+
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
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Chapter 16–10
16.26 The endo product (with the substituents under the plane of the new six-membered ring) is the
preferred product.
+
a.
CH2 CHCOOCH3
COOCH3
COOCH3
+
b.
endo substituent
COOCH3 both groups endo
COOCH3
COOCH3
16.27 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
c.
Cl
Cl
H
H
H
H
CH3OOC
O
O
Cl
O
Cl
O
O
O
O
O
O
+
16.28
CH3O
CH3O
+
CH3O
CH3O
NC
O
O
NC
H
(+ enantiomer)
H O
A
O
16.29 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
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Conjugation, Resonance, and Dienes 16–11
16.30 Sunscreens contain conjugated systems to absorb UV radiation from sunlight. Look for
conjugated systems in the compounds below.
O
a.
O
b.
CH3O
c.
O
CH3O
OH
conjugated system
could be a sunscreen
O
not a conjugated system
conjugated system
could be a sunscreen
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Chapter 16–12
16.31 Use the definition from Answer 16.1.
CH2=CHC N
3 bonds with only
1 bond between
conjugated
1 bond with
no adjacent sp2 hybridized atoms
NOT conjugated
2 multiple bonds with only
1 bond between
conjugated
CH2OCH3
CH2
3 bonds with 2 or
more bonds between
NOT conjugated
This C is sp2.
1 bond with
an adjacent sp2 hybridized atom
1 bond with
no adjacent sp2 hybridized atoms
NOT conjugated
The lone pair occupies a p orbital,
so there are p orbitals on
three adjacent atoms.
conjugated
16.32 Use the definition from Answer 16.1.
conjugated
a.
conjugated
conjugated
isolated
conjugated
isolated
c.
b.
isolated
isolated
conjugated
isolated
conjugated
16.33 Although 2,3-di-tert-butyl-1,3-butadiene has four adjacent p orbitals, the bulky tert-butyl groups
prevent the diene from adopting the s-cis conformation needed for the Diels–Alder reaction.
Thus, this diene does not undergo a characteristic reaction of conjugated dienes.
16.34
a. (CH3)2C
CH CH2
(CH3)2C CH CH2
e.
CH3O
CH CH CH2
CH3O
b.
c.
d.
CH CH CH2
CH3O CH CH CH2
CH2
CH2
f.
O
O
N(CH3)2
O
N(CH3)2
g.
O
h.
O
O
O
O
O
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Conjugation, Resonance, and Dienes 16–13
16.35
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
16.36
resonance hybrid:
Five resonance structures delocalize the negative charge
on five C's making them all equivalent.
All of the carbons are identical in the anion.
16.37 The N atom of C6H5CH2NH2 is surrounded by four groups and is sp3 hybridized. Although the N
atom of C6H5NH2 is surrounded by four groups, it is also bonded to a benzene ring. To be
conjugated with the benzene ring, the N atom must be sp2 hybridized and its lone pairs must
occupy a p orbital. In this way the lone pair can be delocalized, as shown in one resonance
structure.
NH2
sp2
conjugated
NH2
[+ three other
resonance structures]
NH2
sp3
The N atom is not conjugated
with the benzene ring.
410
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Chapter 16–14
16.38
CH2
a.
CH CH2
H
CH3CH2CH2 H
more acidic
CH2 CH CH2
CH2 CH
less acidic
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
+
1-butene CH3 CH2CH=CH2
CH3
CH3
+
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.
Two unstable radicals form.
16.39 Use the directions from Answer 16.12.
a. (3Z)-1,3-pentadiene in the s-trans conformation
c. (2E,4E,6E)-2,4,6-octatriene
d. (2E,4E)-3-methyl-2,4-hexadiene in the s-cis conformation
double bonds on opposite sides
s-trans
b. (2E,4Z)-1-bromo-3-methyl-2,4-hexadiene
Br
s-cis
16.40
E
Z
C
1,4-pentadiene
(3E)-1,3-pentadiene
(3Z)-1,3-pentadiene
2-methyl-1,3-butadiene
C
C
3-methyl-1,2-butadiene
2,3-pentadiene
16.41
2E,4E
2E,4Z
2Z,4E
2Z,4Z
1,2-pentadiene
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16.42
and
a.
and
c.
Z
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
16.43 Use the directions from Answer 16.16 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
16.44 Conjugated dienes react with HX to form 1,2- and 1,4-products.
Br
a.
HBr
major product, formed by addition of HBr to
the more substituted C=C
(1 equiv)
isolated diene
b.
Br
HBr
Br
(E and Z isomers can form.)
(1 equiv)
1,2-product
1,4-product
Br
Br
c.
Br
HBr
(1 equiv)
Br
1,2-product
1,4-product
1,2-product
(E and Z isomers)
1,4-product
412
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Chapter 16–16
16.45
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
2o
H Br
CH CH2CH3
1,2H shift
Br
Br
CHCH2CH3
C
This 2o carbocation
is also benzylic, making it
resonance stabilized, as above.
carbocation
16.46 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.
RO OR
RO
OR
HOR +
H Br
Br
Use each resonance structure to react with HBr.
Br
Br
H Br
Br
Br
Br
H Br
Br
Br
Br
16.47
C2
CH2
a. and b.
HCl
X
H adds here at C1.
CH3
Cl
Y
Cl added at C2.
1,2-product
kinetic product
CH3
C4
Cl
Z
Cl added at C4.
1,4-product
thermodynamic
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.
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Conjugation, Resonance, and Dienes 16–17
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
16.48 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
1,4-product
Cl
(+ three more resonance structures that delocalize
the positive charge onto the benzene ring)
16.49 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.
16.50 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.
414
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Chapter 16–18
16.51 Use the directions from Answer 16.21.
a.
re-draw
dienophile
diene
COOCH3
b.
Cl
diene
COOCH3
Cl
Cl
trans-substituted products
trans dienophile
COOCH3
c.
COOCH3
COOCH3
COOCH3
Cl
Cl
Cl
diene
cis dienophile
cis-substituted products
O
H O
d.
H
diene
dienophile
endo ring
e.
O
re-draw
O
H
endo substituent
dienophile
diene
O
O
f.
H
H
+
H
O
H
Both C=C's of the
dienophile react so two
Diels–Alder reactions
take place.
or
excess
H
O
diene
O
O
H
H
O
H
dienophile
16.52 Use the directions from Answer 16.27.
CH3
a.
CH3
COOCH3
COOCH3
COOCH3
COOCH3
COOCH3
b.
+
CH3
O
c.
CH3
COOCH3
CH3
CH3
O
H
H
COOCH3
COOCH3
O +
COOCH3
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Conjugation, Resonance, and Dienes 16–19
+
d.
identical
Cl
Cl
O
e.
O
O C
CH3
Cl
CH3
O
O
O
f.
O
+
O C
C
CH3
O
+
O
O
O
O
O
O
16.53
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
16.54
COOCH3
a.
diene
COOCH3
+ HC C COOCH3
dienophile
CO2CH3
+ CH3O2C C C CO2CH3
b.
diene
dienophile
CO2CH3
CO2CH3
CO2CH3
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Chapter 16–20
16.55
OCH3
OCH3
+
CHO
CHO
OCH3
+
CHO
C
OCH3
H
1,2-disubstituted product
major
O
C
OCH3
1,3-disubstituted product
minor
The major product is formed when the circled
carbons with a + and – react.
H
O
resonance hybrids:
+OCH3
+OCH3
O
+
+ H
+H
+
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.
O For the 1,3-product, there are no partial
charges of opposite sign on reacting
carbons. This arrangement is less attractive.
16.56
COOH
+
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.
16.57
H
C
+
Cl
Cl
Cl
Cl
X
Cl
Cl
Y Cl
mCPBA
Cl
Cl
1 equiv
Cl
Cl
Cl
Cl
Cl
C
H
Cl
Cl
Cl
Cl
Z
aldrin
O
dieldrin
This double bond is more electron rich,
so it is epoxidized more readily.
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Resonance, and Dienes
417
Conjugation, Resonance, and Dienes 16–21
16.58
In each problem, the synthesis must
begin with the preparation of
cyclopentadiene from dicyclopentadiene.
2
dicyclopentadiene
a.
HO
[1] OsO4
COOCH3
cyclopentadiene
HO
[2] NaHSO3, H2O
COOCH3
COOCH3
O
O
b.
O
mCPBA
O
O
O
c.
O
O
O
O
H2 (excess)
CHO
[1] NaH
[2] CH3CH2CH2CH2Br
Pd-C
CHO
OH
O
16.59
O
re-draw
O
a.
diene
O
dienophile
COOCH3
re-draw
b.
COOCH3
COOCH3
dienophile
diene
16.60 A transannular Diels–Alder reaction forms a tricyclic product from a monocyclic starting
material.
O
O
O
O
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Chapter 16–22
16.61
CH3CH CHCH2OH
CH3CH CHCH2 OH2
+ Br
CH3CH CHCH2
two resonance
structures
H Br
+ H2O
CH3CH CHCH2Br
+ Br
CH3CHCH CH2
CH3CHCH CH2
Br
+ Br
16.62
a.
HCl
(CH3)2C=CHCH2CH2CH=CH2
(CH3)2C(Cl)CH2CH2CH2CH=CH2
isolated diene
(CH3)2C=CHCH2CH2CHClCH3
major product
minor product
I
b.
HI
I
conjugated diene
1,4-product
1,2-product
O
O
H
c.
O
+
O
H
O
O
O
O
diene
dienophile
COOCH3
+
d.
diene
COOH
+
COOH
COOH
diene
cis dienophile
HBr
f.
COOCH3
+
trans dienophile
+
e.
COOCH3
COOH
COOH
COOH
Br
Br
conjugated diene
1,2-product
1,4-product
16.63 The mechanism is E1, with formation of a resonance-stabilized carbocation.
OH
OH2
H
H A
A
H2O
A
H A
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Conjugation, Resonance, and Dienes 16–23
16.64
Cl
2
a.
1
loss of H (from 1 C ) + Cl
conjugated
more stable
major product
loss of H (from 2 C) + Cl
more substituted
b. Dehydrohalogentaion generally forms the more stable product. In this reaction, loss of H from the 1
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
isoprene
H
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
Each H is a doublet
of doublets in the
4.9–5.2 ppm region.
OH
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 system
makes it a sunscreen.
420
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Chapter 16–24
16.69
O
+
H
H
O
O
H
O3 cleaves
the C=C.
O
[1] O3
H O
O
O
[2] (CH3)2S
O
H
H
O
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.
Endo product formed.
16.70
O
O
O
CH3
CH3
+
O
+
H
O
Diels–Alder
reaction
O
H
COOCH3
CH3O2C
COOCH3
A
O
loss of CO2
O C O
B
16.71
diene
1st
Diels–Alder
reaction
C
CO2CH3
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
16.72 Retro Diels–Alder reaction forms a conjugated diene. Intramolecular Diels–Alder reaction then
forms N.
CO2CH3
CO2CH3
CO2CH3
NOCH3
HN
NOCH3
NOCH3
retro
Diels–Alder
HN
M
intramolecular
Diels–Alder
reaction
HN
N
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Compounds
421
Benzene and Aromatic Compounds 17–1
C
Chhaapptteerr 1177:: B
Beennzzeennee aanndd A
Arroom
maattiicc C
Coom
mppoouunnddss
maattiicc ccoom
mppoouunnddss ((1177..77))
C
Coom
mppaarriinngg aarroom
maattiicc,, aannttiiaarroom
maattiicc,, aanndd nnoonnaarroom
•
Aromatic compound
•
A cyclic, planar, completely conjugated compound that
contains 4n + 2 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 electrons.
•
•
Antiaromatic
compound
•
A cyclic, planar, completely conjugated compound that
contains 4n 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 electrons.
•
•
A compound that is not
aromatic
•
A compound that lacks one (or more) of the requirements to
be aromatic or antiaromatic.
maattiicc ccoom
mppoouunnddss
PPrrooppeerrttiieess ooff aarroom
• Every carbon has a p orbital to delocalize electron density (17.2).
• They are unusually stable. Ho 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).
EExxaam
mpplleess ooff aarroom
maattiicc ccoom
mppoouunnddss w
wiitthh 66 eelleeccttrroonnss ((1177..88))
benzene
N
N
H
pyridine
pyrrole
+
cyclopentadienyl
anion
tropylium
cation
maattiicc ((1177..88))
EExxaam
mpplleess ooff ccoom
mppoouunnddss tthhaatt aarree nnoott aarroom
not cyclic
not planar
not completely
conjugated
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Chapter 17–2
C
Chhaapptteerr 1177:: A
Annssw
weerrss ttoo PPrroobblleem
mss
17.1
Move the electrons in the bonds to draw all major resonance structures.
N(CH3)2
O
N(CH3)2
O
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.
b.
H
Csp2–Csp2
Csp2–H1s
Csp2–Csp2
Cp–Cp
Csp2–Csp2
Cp–Cp
shortest of all
the indicated bonds
in (a) and (b)
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
3
isopropylbenzene
b.
I 4
2
c.
PhCH(CH3)2
1 CH2CH3
Two groups are 1,3 = meta.
butyl group
phenol
m-butylphenol
CH3
1
2 Br
ethyl
2
3
iodo Two groups are 1,4 = para.
p-ethyliodobenzene
2-bromo
d.
Cl 5
5-chloro
toluene (CH3 group must be at the "1" position,
if the molecule is named as a toluene derivative.)
2-bromo-5-chlorotoluene
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Compounds
423
Benzene and Aromatic Compounds 17–3
17.4
Work backwards to draw the structures from the names.
a. isobutylbenzene
c. cis-1,2-diphenylcyclohexane
e. 4-chloro-1,2-diethylbenzene
isobutyl
group
Cl
b. o-dichlorobenzene
f. 3-tert-butyl-2-ethyltoluene
d. m-bromoaniline
Br
Cl
Cl
NH2
aniline
17.5
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
1,2,3-trichlorobenzene
Cl
1,2,4-trichlorobenzene
1,3,5-trichlorobenzene
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
a.
Cb
CH2CH3
Cd
Ca
Cc
Cb Cc
c. Cd
b.
Cl
Cb
4 types of C's in the benzene ring
6 signals
17.8
Cc Cb
CH3
Cd
Cc Cb
C C
Ca b c
4 signals
All C's are different.
7 signals
Each of the three isomeric trichlorobenzenes exhibits a different number of 13C NMR signals.
Cd
Cc
Cc
Ca
Cl
Ca Cl
Cl
Cb
4 signals
Cc Cb
Cd
Cl
Ca
Cl
Cf
Cl
Ce
6 signals
Cl
Cb
Ca
Cb
Ca
Cl
Ca
Cl
Cb
2 signals
424
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Chapter 17–4
17.9
The less stable compound has a larger heat of hydrogenation.
CH3
CH2
A
B
no benzene ring, less stable
larger H°
benzene ring, more stable
smaller H°
17.10 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
a.
H
H
H
not aromatic
alkene H's ~ 4.5–6 ppm
H
H
c.
b.
H
H
aromatic ring
H's ~ 6.5–8 ppm
aromatic ring
H's ~ 6.5–8 ppm
17.11 To be aromatic, a ring must have 4n + 2 electrons.
16 e
4n
4(4) = 16
antiaromatic
20 e
4n
4(5) = 20
antiaromatic
22 e
4n + 2
4(5) + 2 = 22
aromatic
17.12 Annulenes have alternating double and single bonds. An odd number of carbon atoms in the ring
would mean there would be two adjacent single bonds. Therefore an annulene having an odd
number of carbon atoms cannot exist.
17.13
17.14 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.
O
Count one lone pair from O.
4n + 2 = 4(1) + 2 = 6
aromatic
+
O
no lone pair from O
4n + 2 = 4(1) + 2 = 6
aromatic
c.
d.
O
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
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Compounds
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Benzene and Aromatic Compounds 17–5
17.15
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.
17.16
a. The five-membered ring is aromatic
because it has 6 electrons, two from
each bond and two from the N atom
that is not part of a double bond.
F
N
2
N
N
F
CF3
N
CF3
N
N
N
sp2 hybridized N
lone pair in sp2 orbital
sp2 hybridized N
lone pair in sp2 orbital
NH2 O
F
N
F
sp3 hybridized N
lone pair in sp3 orbital
2
NH2 O
sp2 hybridized N
lone pair in p orbital
F
2
F
sp2 hybridized N
lone pair in p orbital
b. and c.
sitagliptin
17.17
17.18 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.
B
B
H H
1,3,5-cycloheptatriene
pKa = 39
H
8 Electrons make this
conjugate base
especially unstable
(antiaromatic).
Since the conjugate base is unstable,
the pKa of 1,3,5-cycloheptatriene is high.
H H
cyclopentadiene
H
6 electrons
aromatic conjugate base
very stable anion
Since the conjugate base is very stable, the pKa
of cyclopentadiene is much lower.
426
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Chapter 17–6
17.19 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.
17.20
17.21 To be aromatic, the ions must have 4n + 2 electrons. Ions in (b) and (c) do not have the right
number of electrons to be aromatic.
2 electrons
4(0) + 2 = 2
aromatic
a.
d.
10 electrons
4(2) + 2 = 10
aromatic
17.22
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, and not
delocalized in the ring. Although A has 24 e– total, only
18 e– are delocalized around the ring.
2 antibonding MOs
1 bonding MO
+
17.23 In using the inscribed polygon method, always draw the vertex pointing down.
2 electrons
All bonding MOs are filled.
aromatic
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Benzene and Aromatic Compounds 17–7
17.24 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 electrons
Not all bonding MOs are filled.
not aromatic
5 electrons
Not all bonding MOs are filled.
not aromatic
17.25 C60 would exhibit only one 13C NMR signal because all the carbons are identical.
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. 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
[1]
Br
Br
Br
[2]
Br
Br
Br
Br
Br
A
B
C
D
17.27
propylbenzene
1,2,3-trimethylbenzene
isopropylbenzene
o-ethyltoluene
p-ethyltoluene
m-ethyltoluene
1,2,4-trimethylbenzene
1,3,5-trimethylbenzene
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Chapter 17–8
17.28
C8H10:
Br
Br
Br
C8H9Br:
Br
Br
Br
Br
Br
Br
3 isomers
2 isomers
3 isomers
1 isomer
17.29 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
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Benzene and Aromatic Compounds 17–9
17.30
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
17.31
a. constitutional isomers of molecular formula C8H9Cl, and b. 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
17.32 Count the electrons in the bonds. Each bond holds two electrons.
a.
b.
10 electrons
7 electrons
c.
d.
10 electrons
e.
14 electrons
12 electrons
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Chapter 17–10
17.33 To be aromatic, the compounds must be cyclic, planar, completely conjugated, and have 4n + 2 electrons.
Circled C's are not sp2.
not completely conjugated
not aromatic
a.
14 electrons in outer ring
aromatic
b.
4 benzene rings
joined together
aromatic
c.
Circled C is not sp2.
not completely conjugated
not aromatic
d.
12 electrons
does not have 4n + 2
electrons
not aromatic
e.
12 electrons
does not have 4n + 2
electrons
not aromatic
f.
17.34 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 electrons
counting a lone pair from S
4(1) + 2 = 6
aromatic
not aromatic
not aromatic
N
N
b.
H
N
f.
h.
10 electrons
4(2) + 2 = 10
aromatic
6 electrons,
counting a lone pair from O
4(1) + 2 = 6
aromatic
17.35
Circled C's are
not sp2.
not aromatic
a.
c.
Circled C is
not sp2.
not aromatic
d.
4 electrons
4(1) = 4
antiaromatic
O
b.
10 electrons
in 10-membered ring
4(2) + 2 = 10
aromatic
Count
these 2e–.
N
N
N
6 electrons
counting a lone pair from O
4(1) + 2 = 6
aromatic
N
6 electrons
counting the lone
pair from N
4(1) + 2 = 6
aromatic
O
d.
O
g.
e.
10 electrons
4(2) + 2 = 10
aromatic
These lone
pairs are on
doubly
bonded N
atoms, so they
can't be
counted.
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Benzene and Aromatic Compounds 17–11
17.36
6 electrons
in this ring
6 electrons
in this ring
+
A
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 6 electrons, making it aromatic. The molecule possesses a dipole such that the sevenmembered ring is electron deficient and the five-membered ring is electron rich.
17.37 Each compound is completely conjugated. A compound with 4n + 2 electrons is especially
stable, while a compound with 4n electrons is especially unstable.
pentalene
azulene
8 electrons
4(2) = 8
antiaromatic
unstable
heptalene
10 electrons
4(2) + 2 = 10
aromatic
very stable
12 electrons
6(2) = 12
antiaromatic
unstable
17.38 Benzene has C–C bonds of equal length, intermediate between a C–C double and single bond.
Cyclooctatetraene is not planar and not aromatic so its double bonds are localized.
cyclooctatetraene
a
c
b
6 electrons: aromatic
all bonds of equal length
intermediate
d
not aromatic
longer single bond
localized double bond: shorter
17.39
N
N
N
N
H
purine
d<a=b<c
sp2 hybridized but
with lone pair in p orbital
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 electrons
d. Purine is cyclic, planar, completely conjugated, and
has 10 electrons [4(2) + 2] so it is aromatic.
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17.40
a. 16 total electrons
b. 14 electrons delocalized in the ring. [Note: Two of the electrons in the
triple bond are localized between two C's, perpendicular to the 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 electrons delocalized in
the ring is 14. 4(3) + 2 = 14, so the ring is aromatic.
C
17.41 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.
+
6 electrons
4 electrons
2o carbocation
aromatic
antiaromatic
very stable intermediate
very unstable intermediate
Increasing stability
17.42
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
17.43 -Pyrone reacts like benzene because it is aromatic. A second resonance structure can be drawn
showing how the ring has six electrons. Thus, -pyrone undergoes reactions characteristic of
aromatic compounds—that is, substitution rather than addition.
O
O
-pyrone
O
+
O
6 electrons
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Compounds
433
Benzene and Aromatic Compounds 17–13
17.44
a.
cyclopropenyl radical
b.
N
N
N
H
H
H
and
pyrrole
N
N
N
H
H
H
c.
phenanthrene
17.45
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.
17.46
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.
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Chapter 17–14
17.47 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
6 electrons, aromatic
more stable conjugate base
Its acid is more acidic.
or
b.
or
more acidic
6 electrons, aromatic
more stable conjugate base
Its acid is more acidic.
antiaromatic
highly destabilized
conjugate base
17.48
+ NaNH2
+ NH3
Na+
indene
The conjugate base of indene has 10 electrons making it aromatic
and very stable. Therefore, indene is more acidic than many hydrocarbons.
17.49
Ha
Ha
Hb
Hc
– Hb
CH3
CH3
A
B
Hb is most acidic because its
conjugate base is aromatic (6 electrons).
– Hc
CH3
CH3
Hd
Hd
Hc is least acidic because its
conjugate base has 8 electrons,
making it antiaromatic.
17.50
conjugate base
pyrrole
N
N
H
conjugate base
cyclopentadiene
more acidic
H H
H
Both pyrrole and the conjugate base of pyrrole have 6 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.
Cyclopentadiene is not aromatic, but the conjugate base has 6 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.
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Benzene and Aromatic Compounds 17–15
17.51
H H
+ H+
a.
N H
pyrrole
or
+ H+
H H
C2
H
N
N
H H
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.
N H
A
N
H
H
B
b. H H
Both C and its conjugate base pyridine are
aromatic. Since C has six 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
Loss of a proton from A (which is not aromatic)
gives two electrons to N, and forms pyrrole,
which has six 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
N H
17.52
a.
3 antibonding MOs
2K
2 nonbonding MOs
3 bonding MOs
cyclooctatetraene
cyclooctatetraene and its 8 electrons
dianion of
cyclooctatetraene
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 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.
17.53
4 antibonding MOs
4 antibonding MOs
+
cyclononatetraenyl
cation
8 electrons
antiaromatic
5 bonding MOs
4 antibonding MOs
5 bonding MOs
cyclononatetraenyl
radical
9 electrons
not aromatic
5 bonding MOs
cyclononatetraenyl
anion
10 electrons
aromatic
All bonding MOs are filled.
436
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17.54 The number of different types of C’s = the number of signals.
3
4
1
2 CH3
CH3
1
a.
2
3
1
1
c.
b.
1
3
5
2
4
CH2CH3
5 different C's
all unique
9 different C's
2
d.
1
3
2
1
2
2
3 different C's
3
4
3
2
3 4 3
2
1
1
4 different C's
17.55 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]
17.56
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 2 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
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Benzene and Aromatic Compounds 17–17
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
17.57
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
O
CH3
C
OCH3
It would be hard to distinguish
these two compounds with
the given data.
17.58
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
OH
H
CH3 CH3
1H
septet at ~3.2 ppm
6 equivalent H's
doublet at ~1.2 ppm
IR absorptions:
3500–3200 cm–1 (O–H)
3150–2850 cm–1 (C–H bonds)
1621 and 1585 cm–1 (benzene ring)
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Chapter 17–18
X
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.
Y
Z
basic structure of
thymol
17.59
The induced magnetic field by the circulating electrons
opposes the applied field in this vicinity, shifting the absorption upfield to
a lower chemical shift than other sp3 C–H protons.
In this case the protons absorb upfield from TMS, an unusual
phenomenon for C–H protons.
Induced magnetic field
H H
2.84 ppm
H
–0.6 ppm
(upfield from TMS)
H
Induced magnetic field
B0
The induced magnetic field by the circulating electrons
reinforces the applied field in this vicinity, shifting the absorption
downfield to a somewhat higher chemical shift.
17.60
C1
C 2 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
17.61
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
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Benzene and Aromatic Compounds 17–19
O
O
O
O
b.
HO
OH
OCH3
HO
OCH3
OH
OCH3
The enol O–H proton is more acidic than an
alcohol O–H proton because the conjugate
base is resonance stabilized.
O
O
OCH3
HO
OH
OCH3
OCH3
c. Curcumin is colored because it has many conjugated 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 6 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
H
p
H
440
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Chapter 17–20
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
H
OH
OH
H
N
H
N
H
H
H
H
N
N
H
H
H
N
H
N
H
H
H
H
OH
H
H
N
N
H
d.
H
H
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 A second resonance structure for A shows that the ring is completely conjugated and has 6 electrons, making it aromatic and especially stable. A similar charge-separated resonance
structure for B makes the ring completely conjugated, but gives the ring 4 electrons, making it
antiaromatic and especially unstable.
O
O
O
+
A
6 electrons
aromatic
stable
O
+
B
4 electrons
antiaromatic
not stable
17.64 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.
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Compounds
441
Benzene and Aromatic Compounds 17–21
O
O
HSO4–
O
O
HSO4–
H
H OSO3H
O
H
H OSO3H
H OSO3H
(R)-carvone
OH
OH
H2SO4 +
H
HSO4–
carvacrol
17.65 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 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.
442
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Chapter 17–22
17.66
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.
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443
Electrophilic Aromatic Substitution 18–1
C
Chhaapptteerr 1188:: EElleeccttrroopphhiilliicc A
Arroom
maattiicc SSuubbssttiittuuttiioonn
M
Meecchhaanniissm
m ooff eelleeccttrroopphhiilliicc aarroom
maattiicc ssuubbssttiittuuttiioonn ((1188..22))
• 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
TThhrreeee rruulleess ddeessccrriibbiinngg tthhee rreeaaccttiivviittyy aanndd ddiirreeccttiinngg eeffffeeccttss ooff ccoom
mm
moonn ssuubbssttiittuueennttss ((1188..77––1188..99))
[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.
SSuum
mm
maarryy ooff ssuubbssttiittuueenntt eeffffeeccttss iinn eelleeccttrroopphhiilliicc aarroom
maattiicc ssuubbssttiittuuttiioonn ((1188..66––1188..99))
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 (+ or +)
444
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Chapter 18–2
FFiivvee eexxaam
mpplleess ooff eelleeccttrroopphhiilliicc aarroom
maattiicc ssuubbssttiittuuttiioonn
[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:
H
[1] with alcohols
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
ROH
H2SO4
R
H
[2] with alkenes
CH2 CHR
H2SO4
CH3
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445
Electrophilic Aromatic Substitution 18–3
[5] Friedel–Crafts acylation–Replacement of H by RCO (18.5)
O
H
RCOCl
AlCl3
C
•
R
ketone
The reaction does not occur on benzene rings
substituted by meta deactivating groups or
NH2 groups (18.10B).
O
Otthheerr rreeaaccttiioonnss ooff bbeennzzeennee ddeerriivvaattiivveess
[1] Benzylic halogenation (18.13)
Br
R
Br2
h or or
NBS
h or ROOR
R
benzylic bromide
[2] Oxidation of alkyl benzenes (18.14A)
R
COOH
KMnO4
•
A benzylic C–H bond is needed for reaction.
benzoic acid
[3] Reduction of ketones to alkyl benzenes (18.14B)
O
C
R
Zn(Hg), HCl
or
NH2NH2, –OH
R
alkyl benzene
[4] Reduction of nitro groups to amino groups (18.14C)
NO2
H2, Pd-C
or
Fe, HCl
or
Sn, HCl
NH2
aniline
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Chapter 18–4
C
Chhaapptteerr 1188:: A
Annssw
weerrss ttoo PPrroobblleem
mss
18.1
The 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 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
H
E
E
B
+
+
18.3
E
B
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
[1]
O
S
O
O
H OSO3H
+
S
O + O H
H
SO3H
H
+
[2]
=
SO3H
R
R
+
SO3H
+
electrophile
H
SO3H
H
SO3H
R
R
resonance-stabilized carbocation
–
HSO4
H
SO3H
[3]
SO3H
H2SO4
R
R
B
HSO4–
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Electrophilic Aromatic Substitution 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.
18.5
CH(CH3)2
(CH3)2CHCl
+
a.
O
O
AlCl3
c.
+ CH3CH2
C
C
Cl
Cl
b.
CH2CH3
AlCl3
+
AlCl3
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.
18.6
O
C
a.
O
CH2CH2CH(CH3)2
O
C
Cl
O
O
C
C
c.
CH2CH2CH(CH3)2
Cl
O
C
Cl
b.
C
18.7
+
[1] CH3CH2 Cl
+
_
CH3CH2 Cl AlCl3
AlCl3
electrophile
H
+
[2]
H
CH2CH3
_
H
CH2CH3
H
CH2CH3
CH3CH2 Cl AlCl3
+
resonance-stabilized carbocation
AlCl4–
_
[3]
H
CH2CH3
Cl AlCl3
CH2CH3
To be reactive in a Friedel–Crafts alkylation reaction, the X must be bonded to an sp3 hybridized
carbon atom.
18.8
Br
a.
sp2
Br
Br
b.
unreactive
18.9
AlCl3
HCl
c.
sp3
reactive
sp2
d.
unreactive
Br
sp3
reactive
The product has an “unexpected” carbon skeleton, so rearrangement must have occurred.
CH3
[1] CH3 C CH2 Cl
H
AlCl3
CH3
CH3 C
+
_
CH2 Cl AlCl3
1,2-H shift
H
Rearrangement
CH3
CH3 C
CH3
+ AlCl4–
448
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Chapter 18–6
H
H
H
C(CH3)3
+
C(CH3)3
[2]
H
C(CH3)3
C(CH3)3
_
Cl AlCl3
H
C(CH3)3
[3]
C(CH3)3
HCl
AlCl3
18.10 Rearrangements do not occur with acylium ions formed in a Friedel–Craft acylation because the
acylium ion is resonance stabilized.
R C O
R C O
18.11 Both alkenes and alcohols can form carbocations for Friedel–Crafts alkylation reactions.
OH
H2SO4
a.
+
b.
+ (CH3)2C
C(CH3)3
H2SO4
CH2
H2SO4
+
c.
OH
d.
H2SO4
+
18.12
[1] Cl
Cl
Cl
O
A
Cl
Cl
Cl
Cl
Cl
O
Cl
Cl3Al Cl
Al Cl
+
AlCl4–
O
Cl
O
Cl
H
[2] Cl
H
Cl
Cl
O
H
O
Cl
Cl
H
O
Cl
_
[3] Cl
O
Cl
Cl
O
Cl
B
O
Cl
Cl AlCl3
H
Cl
+
HCl
+
AlCl3
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Electrophilic Aromatic Substitution 18–7
18.13 In parts (b) and (c), a 1,2-shift occurs to afford a rearrangement product.
a.
Cl
c.
OH
b.
18.14
a.
b.
CH2CH2CH2CH3
alkyl group
electron donating
c. OCH2CH3
electronegative O
electron withdrawing
Br
halide
electron withdrawing
18.15 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.
b.
O
CH3
O
CH3
O
O
O
C
C
C
CH3
CH3
O
O
C
C
CH3
O
CH3
CH3
O
CH3
O
CH3
C
CH3
O
C
CH3
CH3
18.16 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, +, or
halogens are electron withdrawing.
I
OCH3
a.
b.
lone pair on O
electron donating
halogen
electron withdrawing
C(CH3)3
c.
R group
electron donating
450
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Chapter 18–8
18.17 First classify the substituents in the starting material as: ortho, para activating, ortho, para
deactivating, or meta deactivating. Then draw the products.
CH2CH3
CH3CH2Cl
OCH3
a.
OCH3
OCH3
CH3CH2
AlCl3
lone pair on O
o,p activating
ortho product
para product
NO2
b.
HNO3
Br
Br
H2SO4
halogen
o,p deactivating
O2N
ortho product
Br
para product
Cl
c.
NO2
Cl2
NO2
FeCl3
meta deactivating
meta product
18.18 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
OH
c.
+
NO2
lone pairs on O
electron donating
reacts faster
R group
electron donating
reacts faster
O2N
+
CH2CH3
O2N
18.19 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
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Electrophilic Aromatic Substitution 18–9
18.20
Cl
NO2
OCH3
CH3
b.
a.
halogen
electron withdrawing
least reactive
intermediate
reactivity
lone pairs on O
electron donating
most reactive
electron withdrawing
least reactive
R group
electron donating
most reactive
intermediate
reactivity
18.21 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
18.22
Cl
H
ortho
attack
Cl
Cl
E
H
E
+
+
+
Cl
Cl
H
E
+
+
H
E
Cl
H
E
E
preferred
especially good
All atoms have an octet. product
Cl
meta
attack
Cl
E+
H
Cl
Cl
+
H
E
+
H
E
+
Cl
H
E
E
452
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Chapter 18–10
+
Cl
Cl
Cl
Cl
Cl
Cl
+
para
attack
+
E+
H
H
+
H
E
E
H
E
H
E
E
preferred
product
especially good
All atoms have an octet.
18.23 Polyhalogenation occurs with highly activated benzene rings containing OH, NH2, and related
groups with a catalyst.
OH
OH
Cl2
a.
Cl
OH
Cl
OH
CH3
Cl
Cl2
b.
FeCl3
OH
c.
Cl
CH3
Cl2
CH3
Cl
FeCl3
Cl
Cl
18.24 Friedel–Crafts reactions do not occur with strongly deactivating substituents including NO2, or
with NH2, NR2, or NHR groups.
a.
SO3H
CH3Cl
no reaction
c.
AlCl3
N(CH3)2
CH3Cl
AlCl3
no Friedel–Crafts
reaction
strongly deactivating
CH3
CH3
b.
Cl
CH3Cl
AlCl3
Cl is an o,p
director.
Cl
d.
CH3
Cl
NHCOCH3
CH3Cl
NHCOCH3
AlCl3
+
CH3
NHCOCH3
18.25 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.
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Electrophilic Aromatic Substitution 18–11
o,p
o,p
OCH3
OCH3
a.
CH3
NO2
HNO3
c.
H2SO4
COOCH3
CH3
NO2 HNO3 O2N
CH3
NO2
H2SO4
meta
COOCH3
NO2
o,p
meta
o,p (strong)
OCH3
OCH3
Br
b.
HNO3
O2N
Cl
OCH3
Br
Cl
Br
H SO
Cl
O 2N
HNO3
d.
Br H2SO4
2
4
o,p
oppose
products due to
OCH3 directing effects
NO2
Br
Br
NO2
NO2
o,p
18.26
Cl
a.
OH
Cl2, FeCl3
SO3, H2SO4
c.
SO3H
OH
CH3Cl, AlCl3
CH3
(+ ortho isomer)
O
ClCOCH3
CH3
Br goes ortho to
the stronger activator.
O
HNO3, H2SO4
C
AlCl3
Br
Br2
SO3H
Put meta director on first.
b.
OH
C
CH3
CH3
O2N
18.27 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.
CH2CH3
CHCH3
CH2CH2
or
A
2o and benzylic
1o
B
Br
CHCH3
only product
CH2CH2Br
not formed
454
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Chapter 18–12
18.28 Radical substitution occurs at the carbon adjacent to the benzene ring (at the benzylic
position).
CH(CH3)2
a.
CH(CH3)2
Br2
CH(CH3)2
+
FeBr3
Br
Br
conditions for electrophilic
aromatic substitution
CH(CH3)2
c.
Br
CH(CH3)2
b.
+
FeCl3
C(CH3)2
Br2
h
CH(CH3)2
Cl2
Cl
CH(CH3)2
conditions for electrophilic
aromatic substitution
conditions for radical
substitution
Cl
18.29
Br
Br2
Br2
h
a.
Br
K+ OC(CH3)3
FeBr3
Br
Br
(+ para isomer)
Br
NaOH
Br2
h
b.
OH
Br
c.
O
mCPBA
K+ OC(CH3)3
Br2
h
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.
Cl
C
O
C
CH2CH2CH2CH3
CH2CH2CH2CH3
CH2CH2CH2CH2CH3
Zn(Hg) + HCl
AlCl3
O
b.
Cl
C
O
C
C(CH3)3
CH2C(CH3)3
Zn(Hg) + HCl
C(CH3)3
AlCl3
18.31
O
O
Cl
Zn(Hg)
O
AlCl3
HCl
O
Cl
AlCl3
p-isobutylacetophenone
(+ ortho isomer)
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Electrophilic Aromatic Substitution 18–13
18.32 When R = CH3 in C6H5CH2R, the product can be made by two different Friedel–Crafts reactions.
O
Cl
C
O
CH3
Zn(Hg)
AlCl3
CH3CH2Cl
CH3
AlCl3
HCl
18.33
CH3Cl
CH3
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)
18.34
CH3
O
ClCH2CH3
a.
C
Cl
O
C
CH3
CH2CH3
O
SO3
CH3
AlCl3
AlCl3
C
H2SO4
HO3S
CH2CH3
CH2CH3
(+ ortho isomer)
Br
o,p director
O
o,p director
C
b.
Cl
Br
CH3
AlCl3
CH2CH3
Both are o,p directors, but they are meta to each
other. The alkyl group must be obtained by
reduction of a carbonyl.
Br
Br2
C
CH3
Zn(Hg)
FeBr3
C
O
CH3
HCl
CH2CH3
O
Br
c.
Cl2
FeCl3
CH3CH2Cl
Cl
AlCl3
K+ –OC(CH3)3
Br2
Cl
h
Cl
Cl
(+ ortho isomer)
CHO
Cl
[1] BH3
[2] H2O2, –OH
OH
PCC
Cl
456
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Chapter 18–14
18.35 OH is an ortho, para director.
OH
OH
HNO3
a.
OH
OH
OH
NO2
Cl
Cl2
g.
H2SO4
FeCl3
NO2
OH
OH
Cl
OH
OH
OH
SO3H
SO3
b.
h.
(+ ortho
isomer)
H2SO4
OH
OH
NH2
OH
OH
OH
(+ ortho
isomer)
AlCl3
O
OH
C
(+ ortho
isomer)
HCl
CH2CH(CH2CH3)2
OH
OH
CH(CH2CH3)2
AlCl3
Zn(Hg)
CH(CH2CH3)2
O
C
(CH3CH2)2CHCOCl
d.
(+ ortho
isomer)
HCl
NO2
i.
OH
Sn
SO3H
CH3CH2Cl
c.
Cl
NH2NH2
(+ ortho
isomer)
j.
OH
O
C
(+ ortho
isomer)
OH
CH2CH(CH2CH3)2
CH(CH2CH3)2
OH
O
C
CH(CH2CH3)2
(+ ortho
isomer)
k.
OH
OH
Br
Br2
e.
(+ ortho
isomer)
h
Br
OH
OH
Br
OH
l.
Br
Br2
f.
Br2
Br
FeBr3
OH
OH
OH
(+ ortho
isomer)
KMnO4
(+ ortho
isomer)
O
C
OH
Br
18.36 CN is a meta director that deactivates the benzene ring.
CN
CN
CN
Br2
a.
c.
FeBr3
CN
CN
SO3H
CN
CN
H2SO4
H2SO4
Br
d.
NO2
CH3COCl
e.
CH3CH2CH2Cl
HNO3
b.
CN
SO3
No reaction
AlCl3
AlCl3
No
reaction
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457
Electrophilic Aromatic Substitution 18–15
18.37
CH(CH3)2 CH3CH2COCl
a.
CH(CH3)2
AlCl3
O
CH3CH2
O
C
O
CH3CH2COCl
C
b.
CH2CH3
C
CH(CH3)2
+
CH(CH3)2
N(CH3)2
No reaction
AlCl3
CH3CH2COCl
c.
No Friedel–Crafts reaction
AlCl3
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
NO2
HNO3
a.
NO2
HO
CH3
b.
O2N
NO2
H2SO4
SO3
NO2
CH3
OH
H2SO4
OH
HO
HO
CH3
HO3S
OH
SO3H
CH2CH3
c. Cl
OCOCH3
CHO
d.
CH3
CH3CH2Cl
Cl
AlCl3
OCOCH3
CHO
Br2
FeBr3
CHO
Br
CH3
CH3
Br
COCH3
e.
O
CH3COCl
NHCOCH3
NO2
NO2
NHCOCH3
AlCl3
CH3O
f.
O
O2N
HNO3
H2SO4
CH3O
NO2
NO2
NO2
NO2
NO2
Cl
g.
CH3O
COOCH3
Cl2
FeCl3
CH3O
COOCH3
C
O
CH3
458
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Chapter 18–16
OCH3
Br
Br
SO3
h.
OCH3
Br
SO3H
HO3S
H2SO4
OCH3
18.39 Watch out for rearrangements.
a.
Cl
b.
c.
Cl
d.
Cl
Cl
rearrangement
2° carbocation
rearrangement
3° carbocation
18.40
O
O
O
Cl
Cl
OCH3
AlCl3
OCH3
A
AlCl3
18.41
a.
CH3
KMnO4
C(CH3)3
HOOC
C(CH3)3
Br
Br2
b.
h
CH3
Zn(Hg), HCl
c.
C
CH3
CH2CH2CH3
C
H H
O
Br2
d.
FeBr3
Br
O
NH2NH2
e.
–
OCH2CH3
f.
CH2CH2CH3
OH
Br
Br2
OCH2CH3
FeBr3
Br
Br
Br
OCH3
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Electrophilic Aromatic Substitution 18–17
18.42
C bonded to 2 H's
must use acylation
followed by reduction.
O
O
Cl
Zn(Hg)
a.
AlCl3
HCl
C bonded to 1 H
can be added directly
by alkylation.
Cl
b.
AlCl3
O
O
CH2CH3
Cl
c.
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
18.43
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
460
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Chapter 18–18
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
NO2
Zn(Hg), HCl
OCH3
HNO3
H2SO4
CH3CH2CH2CH2
B
18.44 Use the directions from Answer 18.19 to rank the compounds.
OH
NO2
a.
CHCl2
CH2Cl
d.
least reactive
intermediate
reactivity
CHO
most reactive
least reactive
Cl
intermediate
reactivity
CH2NH2
b.
most reactive
CH3
NH2
e.
least reactive
intermediate
reactivity
most reactive
least reactive
intermediate
reactivity
most reactive
NH2
NO2
c.
least reactive
intermediate
reactivity
most reactive
18.45 Electron-withdrawing groups place a positive charge in the benzene ring. Draw the resonance
structures to show how NO2 puts a positive charge in the ring, giving it an electron-withdrawing
resonance effect. Electron-donating groups place a negative charge in the benzene ring. Draw
the resonance structures to show how F puts a negative charge in the ring, giving it an electrondonating resonance effect.
a.
O
O
O
O
N
N
N
N
O
O
O
O
O
O
N
N
O
O
O
N
O
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Electrophilic Aromatic Substitution 18–19
F
F
F
F
F
b.
18.46
Br
C N
[1]
a.
b.
c.
d.
O
[2]
withdraw
donate
less
deactivate
a.
b.
c.
d.
C
[3]
withdraw
withdraw
less
deactivate
a.
b.
c.
d.
CH3
O
withdraw
donate
more
activate
18.47
O
O
C
a.
b.
O
c.
CH3
more electron rich
due to CH3 group
more reactive
more electron rich
due to O atom
more reactive
more electron rich
due to C atom
more reactive
18.48
E
a.
N
more electron rich
due to N atom
faster
O2N
c.
N
E
electron
withdrawing
N(CH3)3
less electron rich
due to (+) charge on N and
electron-withdrawing NO2 group
slower
N
O2N
N(CH3)3
E
E
b.
NH(CH3)2
less electron rich
due to (+) charge on N
slower
E
NH(CH3)2
d. O2N
N
electron electron donating
withdrawing
Effects cancel out.
similar in reactivity to benzene
O2N
N
462
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Chapter 18–20
18.49
E
a.
E+
+
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.
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 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
H E
H E
E+
Para attack:
E+
H
E
H
H
E
H
E
H
E
H
E
H
E
E
b.
O N
ortho, para
director
E+
O N
+
O N
E
Ortho and para products are isolated.
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.
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Electrophilic Aromatic Substitution 18–21
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
H
E
O N
H
E
O N
H
E
O N
O N
especially stable
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
HNO3
N(CH3)2
+
H2SO4
acts as a base
N(CH3)2
HNO3
H2SO4
now a meta director
NO2
H
E
464
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Chapter 18–22
18.52
a.
H
AlCl3
Cl
Cl
1,2-H shift
+ _
AlCl3
3° carbocation
+ AlCl4–
H
H
H
H
_
Cl AlCl3
HCl + AlCl3
Cl
b.
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
18.53
OCH3
H OSO3H
H
OCH3
+
OCH3
OCH3
H
H
OCH3
HSO4–
HSO4–
OCH3
H2SO4
+
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Electrophilic Aromatic Substitution 18–23
18.54
OH
H OSO3H
CH3O
OH2
CH3O
CH3O
+
CH3O
CH3O
HSO4–
H
CH3O
1,2-CH3 shift
CH3O
H
+ H2O
CH3O
H
CH3O
H2SO4
+
H
HSO4–
18.55
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
466
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Chapter 18–24
18.56
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
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.
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.
18.57
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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Electrophilic Aromatic Substitution 18–25
18.58
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
H2O
+ AlCl3
18.59 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
+ CH3OH
O2N
CH2Br
electron-withdrawing group
destabilizing
slower reaction
+ Br
CH2Br
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.
468
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Chapter 18–26
18.60 Addition of HBr will afford only one alkyl bromide because the intermediate carbocation leading
to its formation is resonance stabilized.
H
H
CH3
CH3
H Br
H
H
CH3
H H
CH3
H H
CH3
H H
H H
resonance
stabilized
or
H
Br
H H
CH3
Br
CH3
H H
H
only
no resonance
stabilization
As a result, this carbocation
does not form.
18.61
O
H
O
+ H OSO3H
+
+
OH
H
HSO4
H
H
O
+
resonance-stabilized
cation
HSO4
OH
OH
(+ 3 additional
resonance structures)
H OSO3H
H
HO
H
OH
bisphenol A
H2SO4
HO
OH
+
HSO4
18.62
Cl
a.
AlCl3
O
O
b.
Zn(Hg), HCl
Cl
AlCl3
O
c.
Cl2
FeCl3
Cl
Cl
Zn(Hg), HCl
Cl
AlCl3
O
(+ para isomer)
OH
H2O
(+ 5 additional
resonance structures)
(+ 3 additional
resonance structures)
O
H
OH
HO
O
Cl
HSO4
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Electrophilic Aromatic Substitution 18–27
d.
NO2
NO2
HNO3
Br2
FeBr3
H2SO4
Br
f.
NO2
HNO3
Br2
e.
(+ para isomer)
H2SO4
FeBr3
Br
CH3Cl
CH3
AlCl3
Br
CH3
SO3
H2SO4
COOH
KMnO4
SO3H
SO3H
(+ para isomer)
CH3
CH3Cl
g.
COOH
KMnO4
AlCl3
COOH
HNO3
H2SO4
NO2
CH3Cl
h.
AlCl3
O
i.
Cl
C
Br2
CH3
KMnO4
Br
CH3
FeBr3
Br
COOH
(+ ortho isomer)
O
NO2
O
Br2
CH2CH3
CH2CH2CH3
Zn(Hg), HCl
FeBr3
AlCl3
CH2CH2CH3
HNO3
H2SO4
Br
Br
Br
(+ isomer)
CH3
j.
CH3
CH3
CH3Cl
SO3
Cl2 (excess)
AlCl3
H2SO4
FeCl3
Cl
COOH
Cl
SO3H
KMnO4
SO3H
Cl
Cl
SO3H
(+ ortho isomer)
Br
k.
(CH3)2CHCl
HNO3
AlCl3
H2SO4
Cl2
NO2
FeCl3
(+ para isomer)
Br2
Cl
NO2
(+ isomer)
h
K+ –OC(CH3)3
Cl
NO2
Cl
NO2
470
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Chapter 18–28
18.63
NO2
a.
NH2
Br2
HNO3
Sn
FeBr3
H2SO4
HCl
Br
Br
Br
(+ ortho isomer)
b.
CH3
CH3
CH3
Br
CH3Cl
HNO3
Br2
AlCl3
H2SO4
FeBr3
NO2
NO2
(+ ortho isomer)
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
Cl
e.
AlCl3
Zn(Hg)
SO3
NaOH
HCl
H2SO4
SO3– Na+
SO3H
(+ ortho isomer)
O
O
Cl
f.
AlCl3
O
Br
Br2
NH2NH2
FeBr3
Br
HNO3
Br
H2SO4
–OH
NO2
(+ isomer)
H2
Pd-C
Br
NH2
O
g.
CH3CH2Cl
AlCl3
AlCl3
Br2 Br
Cl2
Cl
O
FeCl3
O
h
O
Cl
(+ ortho isomer)
Cl
K+ –OC(CH3)3
NH2NH2
–
Cl
O
OH
Cl
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Electrophilic Aromatic Substitution 18–29
18.64
CH3
CH2Br
Br2
a.
h
CH2OH
b.
product in (a)
c.
product in (a)
d.
product in (b)
NaOH
CH2OC(CH3)3
OC(CH3)3
CHO
PCC
Cl2
CH3
e.
CH3
FeCl3
Cl
KMnO4
COOH
(+ para isomer)
Cl
O
f.
Cl
CH3
C
(CH2)4CH3
O
C
CH3
AlCl3
Zn(Hg), HCl
CH3
(CH2)5CH3
(CH2)4CH3
(+ ortho isomer)
O
g.
CH3
CH3
O
Cl
C
O
C
CH3
COOH
(+ ortho isomer)
CH3
HNO3
KMnO4
CH3
CH3
AlCl3
CH3
h.
C
CH3
COOH
HNO3
NO2
H2SO4
NO2
KMnO4
NO2
H2SO4
(+ para isomer)
NO2
NO2
O
CH3 Br
2
i.
FeBr3
CH3
AlCl3
Br
CH3
Cl
h
Br
(+ ortho isomer)
O
O
CH3
–OH
Br
OH
Br
Br
O
O
O
O
HNO3
j.
Br2
O2N
NH2NH2
O2N
Cl
AlCl3
H2SO4
–
OH
H2
Pd-C
(+ ortho isomer)
H2N
472
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Chapter 18–30
O
Cl
CH3
k.
O
AlCl3
Br
Br2
Cl2
CH3
CH3
FeCl3
h
O
O
Cl
(+ ortho isomer)
Cl
–OH
PCC
CHO
CH2OH
O
O
Cl
Cl
18.65
Cl
[1] NaH
a.
[2] CH3Br
HO
AlCl3
CH3O
CH3O
(+ ortho isomer)
b.
HO
CH3
[1] NaH
CH3Cl
[2] CH3CH2Br CH3CH2O
AlCl3 CH3CH2O
h
(+ ortho isomer)
OCH3
[1] NaH
[2] CH3I
HO
CH3
CH3Cl
c.
AlCl3
CH3O
CH3O
(+ ortho isomer)
HNO3
H2SO4
(excess)
CH3CH2O
–OH
CH2OH
[1] NaH
[2] CH3I
CH3CH2O
CH2Br
Br2
CH3CH2O
O2N
CH3
H2
H2N
CH3
Pd-C CH O
3
CH3O
NO2
NH2
18.66
HO
Br
SOCl2
a.
K+ OC(CH3)3
Br2
Cl
Br
Br
Br2
2 NH2
C CH
h
AlCl3
Br
HBr
b.
ROOR
(from a.)
C CH
c.
C C
NaH
O
C CCH2CH2O
C CCH2CH2OH
H2O
(from a.)
HNO3
d.
O2N
H2SO4
(from a.)
Br2
h
K+ –OC(CH3)3
O2N
O2N
Br
(+ ortho isomer)
HC CH
NaH
O2N
CH2CH2 C CH
HBr
ROOR
Br
C CH
O2N
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Electrophilic Aromatic Substitution 18–31
C CH
e.
C CCH2CH3
[1] NaH
[2] CH3CH2Cl
(from a.)
Na
NH3
(from a.)
Br
Br2
f.
FeBr3
(from a.)
K+ OC(CH3)3
Br2
h
Br
Br
OH
[1] OsO4
Br
[2] H2O, NaHSO3
Br
OH
(+ ortho isomer)
g.
Cl2
HNO3
FeCl3
H2SO4
(from a.)
Cl
Br2
Cl
K+ –OC(CH3)3
Cl
h
NO2
(+ ortho isomer)
Cl
Br
NO2
mCPBA
NO2
Cl
O
NO2
18.67
Cl
O
O
AlCl3
Br
NH2NH2
Br2
–OH
h
K+ –OC(CH3)3
(E + Z)
Br2
Br
2 NaNH2
Br
18.68 One possibility:
Cl
Br
Br2
O
Cl
AlCl3
h
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
474
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Chapter 18–32
18.69
HO
CH3
CH3Cl
[1] NaH
[2] CH3CH2CH2Br
O
AlCl3
COOH
KMnO4
O
O
(+ ortho isomer)
HNO3
H2SO4
COOH
O2N
X
O
18.70 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
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
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
3 H’s adjacent to 2 H’s
2 H’s adjacent to 3 H’s
para disubstituted benzene
Explanation
2 H’s adjacent to 2 H’s
2 H’s adjacent to 2 H’s
monosubstituted benzene
Structure:
Br
Structure:
Br
18.71 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
Structure:
O
monosubstituted benzene
18.72
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
Explanation
6 H’s adjacent to 1 H
1 H adjacent to 6 H’s
monosubstituted benzene
Structure:
O
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475
Electrophilic Aromatic Substitution 18–33
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:
18.73
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
HSO4–
CH3 C CH3
C CH2
CH3
CH3
OH
OH
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
18.74
Molecular formula (Z): C9H9ClO
IR absorption at 1683 cm–1: C=O
1
H NMR spectral data:
Absorption ppm
# of H’s
triplet
1.2
3
quartet
2.9
2
multiplet
7.2–8.0
4
Explanation
3 H’s adjacent to 2 H’s
2 H’s adjacent to 3 H’s
disubstituted benzene
Structure:
O
Cl
Z
476
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Chapter 18–34
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
OH
OH
(–) charges on the ortho and para positions
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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477
Electrophilic Aromatic Substitution 18–35
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:
Attack at C3:
+ E+
N
H
N
H
N
H
N
H
N
H
E
H
+ E+
H
E
N
H
N
H
more reactive than benzene
E
N
H
H
H
E
N
H
E
N
H
H
E
3-position
N
H
fewer resonance structures
Attack at C3 does not occur.
N
H
E
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
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Chapter 18–36
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
+ H2O
RO
+ A
RO
(+ 1 resonance structure)
A
H
+
H A +
RO
A
RO
(+ 3 resonance structures)
18.79
a. The reaction could follow a two-step mechanism: [1] addition of the nucleophile to form a carbanion,
followed by [2] elimination of the leaving group.
O2N
+
Cl
OCH3
[1]
Cl
O2N
[2]
OCH3
O2N
resonance-stabilized carbanion
OCH3
+
Cl–
b. The NO2 group stabilizes the negatively charged intermediate by an electron-withdrawing inductive
effect and by resonance.
Cl
O2N
OCH3
O
Cl
O2N
OCH3
O
Cl
N
N
OCH3
O
The negative charge can be delocalized
onto the electronegative O atom.
O
O
N
O
Cl
OCH3
Cl
OCH3
c. m-Chloronitrobenzene does not undergo this reaction because no resonance structure can be drawn
that delocalizes the negative charge of the reactive intermediate onto the O atom of the NO2 group.
Cl
OCH3
O2N
meta NO2 group
O2N
Cl
Cl
OCH3
OCH3
O2N
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the Acidity of the O−H
Bond
479
Carboxylic Acids and the Acidity of the O–H Bond 19–1
C
Acciiddiittyy ooff tthhee O
O––H
HB
Boonndd
Chhaapptteerr 1199:: C
Caarrbbooxxyylliicc A
Acciiddss aanndd tthhee A
G
Geenneerraall ffaaccttss
• Carboxylic acids contain a carboxy group (COOH). The central carbon is sp2 hybridized and
trigonal planar (19.1).
• Carboxylic acids are identified by the suffixes -oic acid, carboxylic acid, or -ic acid (19.2).
• Carboxylic acids are polar compounds that exhibit hydrogen bonding interactions (19.3).
SSuum
mm
maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ((1199..44))
IR absorptions
C=O
O–H
~1710 cm–1
3500–2500 cm–1 (very broad and strong)
1
O–H
C–H 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
G
Geenneerraall aacciidd––bbaassee rreeaaccttiioonn ooff ccaarrbbooxxyylliicc aacciiddss ((1199..99))
• Carboxylic acids are especially acidic
O
O
+
because carboxylate anions are
R C
B
R C
+
H B
O H
O
resonance stabilized.
• For equilibrium to favor the products,
carboxylate anion
pKa 5
the base must have a conjugate acid
with a pKa > 5. Common bases are
listed in Table 19.3.
FFaaccttoorrss tthhaatt aaffffeecctt aacciiddiittyy
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).
480
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Chapter 19–2
Substituted benzoic acids.
• Electron-donor groups (D) make a substituted benzoic acid less acidic than benzoic acid.
• Electron-withdrawing groups (W) make a substituted benzoic acid more acidic than benzoic
acid.
COOH
COOH
COOH
D
W
less acidic
higher pKa
pKa = 4.2
pKa > 4.2
more acidic
lower pKa
pKa < 4.2
Increasing acidity
O
Otthheerr ffaaccttss
• 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).
• A sulfonic acid (RSO3H) is a strong acid because it forms a weak, resonance-stabilized conjugate
base on deprotonation (19.13).
• Amino acids have an amino group on the 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).
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481
Carboxylic Acids and the Acidity of the O–H Bond 19–3
C
Chhaapptteerr 1199:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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
Number the chain to put COOH at C1.
5 carbon chain = pentanoic acid
4-chloropentanoic acid
4
1 OH
8
9
Number the chain to put COOH at C1.
9 carbon chain = nonanoic acid
4-isopropyl-6,8-dimethylnonanoic acid
19.2
a. 2-bromobutanoic acid
c. 3,3,4-trimethylheptanoic acid
O
O
e. 3,4-diethylcyclohexanecarboxylic
acid
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
a. -methoxyvaleric acid
OH
c. ,-dimethylcaproic acid
OCH3
O
O
b. -phenylpropionic acid
OH
d. -chloro--methylbutyric acid
OH
Cl
OH
O
482
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Chapter 19–4
19.4
C5 or carbon
OH
a.
OH
C2 or carbon
b.
CO2H
CO2H
HO
C3 or carbon
IUPAC: 2-hydroxypropanoic acid
common: -hydroxypropionic acid
IUPAC: 3,5-dihydroxy-3-methylpentanoic acid
common: ,-dihydroxy--methylvaleric acid
19.5
O
O
a.
O Li+
lithium benzoate
b. Na+ O
O
O
c.
H
sodium formate
or
sodium methanoate
O
d.
K+
O
Br
potassium 2-methylpropanoate
sodium 4-bromo-6-ethyloctanoate
19.6
COO–Na+
COOH
C2
sodium 2-propylpentanoate
2-propylpentanoic acid
19.7
19.8
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
Na+
CH3CH2CH2CH2
C
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
O
alcohol
1 strong absorption
~3600–3200 (OH) cm–1
19.9
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
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Carboxylic Acids and the Acidity of the O–H Bond 19–5
19.10
O
H
O
O
OH
HO
OH
H H
Ha
Hb
Hc
Hc
Hd Hd
2 singlets
2 singlets
1:1 ratio
1:1 ratio
Although both compounds have an absorption at 10–12 ppm in their 1H NMR
spectra (due to Hb and Hc), Ha, which is bonded directly to the carbonyl
carbon, is much farther downfield than Hd because it is more deshielded.
19.11
HO
HO
COOH
HO
There are five tetrahedral stereogenic
centers. Both double bonds can
exhibit cis–trans isomerism.
Therefore, there are
27 = 128 stereoisomers.
COOH
HO
OH
PGF2
a prostaglandin
OH
enantiomer
19.12 1° Alcohols are converted to carboxylic acids by oxidation reactions.
O
a.
OH
c.
OH
COOH
CH2OH
O
b.
(CH3)2CH
C
(CH3)2CH
OH
CH2 OH
19.13
a.
CH2OH
Na2Cr2O7
A
b.
CH3C CCH3
B
c. O2N
COOH
H2SO4, H2O
KMnO4
CH3
O2N
COOH
C
(Any R group with benzylic H's
can be present para to NO2.)
OH
[1] O3
[2] H2O
2° OH
d.
CH3COOH
(2 equiv)
D
O
CrO3
OH
CO2H
H2SO4, H2O
1° OH
19.14
a.
b.
COOH
CH3
CH3
NaOH
OH
NaOCH3
COO
CH3
Na+
+ H2 O
O
Na+
+ HOCH3
d.
CH3
NaH
c. CH3 C OH
CH3
COOH
CH3
C O Na+ + H2
CH3
NaHCO3
COO
Na+
+ H2CO3
484
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Chapter 19–6
19.15 CH3COOH has a pKa of 4.8. Any base with 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
19.16
Ha
H OHb
Increasing acidity: Ha < Hb < Hc
OHc
H OHb
– Ha
O
OHc
negative charge on C
unstable conjugate base
O
mandelic acid
Ha
H O
– Hb
OHc
O
Ha
Ha
H OHb
– Hc
negative charge on O
more stable conjugate base
H OHb
O
O
O
O
negative charge on O,
resonance stabilized
most stable conjugate base
19.17 Electron-withdrawing groups make an acid more acidic, lowering its pKa.
CH3CH2 COOH
least acidic
pKa = 4.9
ICH2
COOH
CF3 COOH
one electron-withdrawing group
intermediate acidity
pKa = 3.2
three electron-withdrawing F's
most acidic
pKa = 0.2
19.18 Acetic acid has an electron-donating methyl group bonded to the carboxy group. The CH3 group
both stabilizes the acid and destabilizes the nearby negative charge on the conjugate base,
making CH3COOH less acidic (with a higher pKa) than HCOOH.
O
electron-donating CH3 group
CH3
+
C
OH
acetic acid
CH3 stabilizes the
partial positive charge.
O
CH3
C
O–
conjugate base
CH3 destabilizes the
negative charge.
a less stable conjugate base
19.19
a. CH3COOH
least acidic
HSCH2COOH
HOCH2COOH
intermediate
acidity
most acidic
b. ICH2CH2COOH
least acidic
ICH2COOH
I2CHCOOH
intermediate
acidity
most acidic
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485
Carboxylic Acids and the Acidity of the O–H Bond 19–7
19.20
a.
CH3
COOH
least acidic
COOH
Cl
intermediate
acidity
COOH
most acidic
O
b. CH3O
COOH
CH3
COOH
C
COOH
CH3
least acidic
intermediate
acidity
most acidic
19.21
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
19.22 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.
19.23 To separate compounds by an aqueous extraction technique, compounds must have different
solubility properties. CH3CH2COOH and CH3CH2CH2OH are low molecular weight organic
compounds that can hydrogen bond to water, so they are water soluble. They also both dissolve
in organic solvents. As a result, they are inseparable because of their similar solubility
properties.
19.24
weaker conjugate base
better leaving group
CF3SO3H
CF3 is electron withdrawing.
stronger acid
lower pKa
CF3SO3–
stronger conjugate base
worse leaving group
CH3SO3H
CH3 is electron donating.
weaker acid
higher pKa
CH3SO3–
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Chapter 19–8
19.25
phenylalanine
COOH
C
H2N
methionine
COOH
H
H
COOH
C
C
NH2
H2N
COOH
H
CH2CH2SCH3
H
CH3SCH2CH2
C
R
R
NH2
S
S
19.26 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.
19.27
COOH
H3N C H
COO
COO
H3 N C H
H2 N C H
H
H
H
pH = 1
glycine
pH = 11
neutral form
19.28
COO
pI =
pKa(COOH) + pKa(NH3+)
(2.58) + (9.24)
=
2
2
= 5.91
H3N C H
CH2
19.29
+
+
H3N CH COOH
H
electron-withdrawing group
H3N CH COO–
H
The nearby (+) stabilizes the conjugate base by an
electron-withdrawing inductive effect, thus making
the starting acid more acidic.
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487
Carboxylic Acids and the Acidity of the O–H Bond 19–9
19.30 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.
–
+
O
lithium butanoate
CH3CH2CH2COO Li
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
19.31
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
O
K+
O
h. sodium -bromobutyrate
Cl
Cl
e. m-hydroxybenzoic acid
C
O
OH
Na+
Br
O
d. ,-dichloropropionic acid
Cl
g. potassium acetate O
i. 2,2-dichloropentanedioic acid
O
O
O
HO
OH
HOOC
OH
Cl Cl
j. 4-isopropyl-2-methyloctanedioic acid
OH
O
HO
OH
O
19.32
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
488
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Chapter 19–10
19.33
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
19.34
O
OH
a.
CrO3
OH
[1] O3
c.
KMnO4
CH3
+ CO2
[2] H2O
H2SO4, H2O
b. (CH3)2CH
COOH
C C H
HOOC
d. CH3(CH2)6CH2OH
COOH
Na2Cr2O7
CH3(CH2)6COOH
H2SO4, H2O
19.35
[1] BH3
a.
CH2
[2] H2O2, OH
B
[1] NaNH2
HC CH
HC CCH3
[2] CH3I
C
[2] CH3CH2I
CH(CH3)2
(CH3)2CHCl
c.
COOH
H2SO4, H2O
A
[1] NaNH2
b.
CrO3
CH2OH
–
CH3CH2C CCH3
[1] O3
CH3CH2COOH
[2] H2O
D
CH3COOH
E + F
COOH
KMnO4
AlCl3
G
H
19.36
Bases: [1] –OH pKa (H2O) = 15.7; [2] CH3CH2– pKa (CH3CH3) = 50; [3] –NH2 pKa (NH3) = 38;
[4] NH3 pKa (NH4+) = 9.4; [5] HCC– pKa (HCCH) = 25.
a.
CH3
COOH
pKa = 4.3
All of the bases
can deprotonate this.
b.
–OH,
Cl
c. (CH3)3COH
OH
pKa = 9.4
CH3CH2–, –NH2, and HCC–
can deprotonate this.
pKa = 18
CH3CH2–, –NH2, and HCC–
can deprotonate this.
19.37
a.
COOH
K+
OC(CH
COO K+
3)3
+ HOC(CH3)3
pKa = 18
Reaction favors products.
pKa = 4.2
OH
b.
pKa 16
NH3
O
NH4+
pKa = 9.4
Reaction favors reactants.
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489
Carboxylic Acids and the Acidity of the O–H Bond 19–11
Na+ NH2
+
OH
c.
+ NH3 + Na+
O
pKa = 10
d.
Reaction favors products.
pKa = 38
CH3–Li+
COOH
COO Li+
CH3
CH4
Reaction favors products.
pKa = 50
CH3
pKa 4
e.
Na+H
OH
Na+
O
Reaction favors products.
pKa = 35
pKa 16
CH3
f.
H2
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
19.38 The stronger acid has a lower pKa and a weaker conjugate base.
COOH
CH2OH
a.
or
c.
carboxylic acid
stronger acid
lower pKa
weaker conjugate base
CH3
or
COOH
Cl
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
or
ClCH2COOH
FCH2COOH
F is more electronegative.
weaker acid
stronger acid
higher pKa
lower pKa
stronger conjugate base
weaker conjugate base
b.
d.
or
NCCH2COOH
CN is electron withdrawing.
stronger acid
lower pKa
weaker conjugate base
CH3COOH
weaker acid
higher pKa
stronger conjugate base
19.39
Br
a.
COOH
least acidic
COOH
OH
b.
CH3
COOH
Cl more electronegative
most acidic
OH
Cl
least acidic
OH
O2N
most acidic
intermediate acidity
COOH
c.
Cl
Br is electronegative
intermediate acidity
COOH
CH3
COOH
CF3
least acidic
intermediate acidity
OH
most acidic
OH
OH
d.
Br
least acidic
O2N
intermediate acidity
O2N
NO2
most acidic
COOH
490
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Chapter 19–12
19.40
O
a.
BrCH2COO–
BrCH2CH2
weakest base
COO–
(CH3)3
intermediate basicity
CCOO–
strongest base
O2N
weakest base
b.
O
NH
weakest base
O
O
c.
intermediate basicity
strongest base
CH2
intermediate basicity
strongest base
19.41
Increasing acidity
ICH2COOH
pKa values
BrCH2COOH
least acidic
3.12
FCH2COOH
2.86
2.66
F2CHCOOH
F3CCOOH
most acidic
0.28
1.24
19.42 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
[1] KOH
H
an alcohol
H
HO
N
[2] CH3I
H
H
HO
codeine
N
CH3
HO
CH3
O
H
H
HO
N
CH3
O
O
O
H
H
N
Many resonance structures
stabilize the conjugate base.
O
O
O
HO
morphine
CH3O
O
H
CH3
O
H
N
CH3
H
HO
O
H
N
CH3
H
HO
H
N
CH3
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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 +
N
O
p-nitrophenol
pKa = 7.2
O
O
O +
N
O +
O
O
OH
O
N
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
O
O2N
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
492
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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
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
CO2H
O
A
pKa = 3.2
O
N
H
B
pKa = 3.9
C
CO2H
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.
19.46
CO2H
CO2H
O2N
O2N
D
C
E
–
H+
–
O
CO2H
H+
– H+
O
O
O
O
O2N
O2N
O
Since the benzene ring is bonded
to the carbon (not the carbonyl
carbon), this compound is not
much different than any alkylsubstituted carboxylic acid.
least acidic
Since the NO2 group is bonded to a
The electron withdrawinginductive 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
O
N
O
O
O
unlike charges nearby
stabilizing
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Carboxylic Acids and the Acidity of the O–H Bond 19–15
19.47
O
CH3
C *
OH
NaOH
CH3
O
O
C *
O
C *
O
CH3
+
H3O+
H3O
labeled O atom
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.
19.48
O
a.
Ha
O
Hc
Hb
loss of Hb:
Ha
Hc
The most acidic proton forms the
most stable conjugate base.
O
O
1,3-cyclohexanedione
increasing acidity: Hb < Ha < Hc
O
one Lewis structure
least stable conjugate base
O
loss of Ha:
Hc
Hb
O
Hc
O
Hb
O
loss of Hc: H
a
O
O
Ha
Hb
O
2 resonance structures
intermediate stability
Ha
Hb
O
Hb
O
3 resonance structures
most stable conjugate base
Hb
N
N
b.
Hc
O
Ha
N
Hc
loss of Hb:
O
Ha
O
Ha
acetanilide
N
N
Hc
O
Ha
Ha
Hb
Hb
N
N
N
O
one Lewis structure
least stable conjugate base
Ha
N
Hc
O
Hb
Hc
Hc
O
7 resonance structures
most stable conjugate base
increasing acidity: Ha < Hc < Hb
loss of Ha:
N
Hc
Ha
N
loss of Hc:
Ha
O
Ha
2 resonance structures that
delocalize the negative charge
intermediate stability
Hc
O
O
Ha
Hc
O
494
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Chapter 19–16
19.49
O
O
O
H
H
H
H H
O
O
H
H
H
HO
OH
O
O
H
H
H
H
HO
O
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
19.50 As usual, compare the stability of the conjugate bases. With RSO3H, loss of a proton forms a
conjugate base that has three resonance structures, all of which are equivalent and place a
negative charge on a more electronegative O atom. With the conjugate base of RCOOH, there
are only two of these resonance structures. Thus, the conjugate base RSO3– is more highly
resonance stabilized than RCOO–, so RSO3H is a stronger acid than RCOOH.
O
O
O
O
R S O
R S O
R S O
O
O
O
O
O
base
R S O H
O
O
R
C
base
O H
R
C
O
R
C
three resonance structures
for the conjugate base
two resonance structures
for the conjugate base
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
C
N
CH3
N
CH3
C
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.
O
OH
C
somewhat less stable
with the (–) charge on N
O
CH3
O
H
CH3
C
O
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495
Carboxylic Acids and the Acidity of the O–H Bond 19–17
19.53
COOH
A
•
•
B
•
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
•
•
•
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
496
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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)
C NMR data: 30, 36, 128, 130, 133, 139, 179 = 7 different types of C’s
1
H NMR data:
13
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
19.60
Molecular formula C6H12O2 (1 double bond due to COOH)
1H
NMR: 1.1 (singlet), 2.2 (singlet), and 11.9 (singlet) ppm
CH3
CH3 C CH2COOH
CH3
D
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Carboxylic Acids and the Acidity of the O–H Bond 19–19
19.61
Molecular formula: C8H6O4: 6 degrees of unsaturation
IR 1692 cm–1 (C=O)
1H NMR 8.2 and 10.0 ppm (singlets)
O
O
HO
OH
COOH
aromatic H
19.62
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
19.63
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
19.64
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
19.65
NH
NH
COOH
NH2
COOH
proline
enantiomer
O
O
COO
zwitterion
19.66
a. methionine
O
H3N CH C OH
b. serine
O
O
H3N CH C 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
H2N CH C O
pH = 6
form at isoelectric point
pH = 11
498
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Chapter 19–20
19.67
pKa(COOH) + pKa(NH3+)
a. cysteine pI =
= (2.05) + (10.25) / 2 = 6.15
2
b. methionine pI =
pKa(COOH) + pKa(NH3+)
= (2.28) + (9.21) / 2 = 5.75
2
19.68
O
O
H2N
C
C
OH
lysine
This lone pair is localized
on the N atom, making it a
base.
H2N H
N
H
H2N H
OH
tryptophan
This lone pair is delocalized in the system to give 10 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.
19.69 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
R
OH
X
O
O
NH3
acid–base reaction
NH3
O– NH4+
SN2 reaction
X
R
O– NH4+
NH3
19.70
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
COOH
H
base (1 equiv)
NH3
C
HOOCCH2CH2
COO–
H
base (2nd equiv)
NH3
C
OOCCH2CH2
COO–
H
base (3rd equiv)
NH2
C
OOCCH2CH2
H
NH3
C
OOCCH2CH2
H
C
HOOCCH2CH2
c. monosodium glutamate
COO–
COO– Na+
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499
Carboxylic Acids and the Acidity of the O–H Bond 19–21
19.71 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.
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
19.72
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
COOH
C
O
o-hydroxybenzoic acid
more acidic than benzoic acid
Intramolecular hydrogen bonding stabilizes the
conjugate base, making the acid more acidic than
benzoic acid.
19.73
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 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.
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Chemistry
501
Introduction to Carbonyl Chemistry 20–1
C
Chheem
miissttrryy
Chhaapptteerr 2200:: IInnttrroodduuccttiioonn ttoo C
Caarrbboonnyyll C
R
Reedduuccttiioonn rreeaaccttiioonnss
[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 ,-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 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
•
A single enantiomer is
formed.
(S) 2o alcohol
[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
502
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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
O
Oxxiiddaattiioonn rreeaaccttiioonnss
Oxidation of aldehydes to carboxylic acids (20.8)
O
R
C
O
CrO3, Na2Cr2O7, K2Cr2O7, KMnO4
H
or
Ag2O, NH4OH
C
R
OH
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.
•
carboxylic acid
PPrreeppaarraattiioonn ooff oorrggaannoom
meettaalllliicc rreeaaggeennttss ((2200..99))
[1] Organolithium reagents:
R X
[2] Grignard reagents:
•
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
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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
R
Reeaaccttiioonnss w
wiitthh oorrggaannoom
meettaalllliicc rreeaaggeennttss
[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'
OH
R C R"
[2] H2O
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
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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 ,-unsaturated aldehydes and ketones (20.15B)
[1] R'Li or R'MgX
OH
R
[2] H2O
C
C
C
C
•
More reactive organometallic reagents—
R'Li and R'MgX—react with ,unsaturated carbonyls by 1,2-addition.
•
Less reactive organometallic reagents—
R'2CuLi— react with ,-unsaturated
carbonyls by 1,4-addition.
R'
O
R
C
allylic alcohol
C
O H
C
C
R
[1] R'2CuLi
[2] H2O
C
R'
ketone
PPrrootteeccttiinngg ggrroouuppss ((2200..1122))
[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
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Chemistry
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Introduction to Carbonyl Chemistry 20–5
C
Chhaapptteerr 2200:: A
Annssw
weerrss ttoo PPrroobblleem
mss
20.1
[1] Csp2–Csp2
a.
[3]
Csp2–Csp2
20.2
[2] : C –O
: Cp–Op
sp2
O
-sinensal
sp2
A carbonyl compound with a reasonable leaving group undergoes substitution reactions. Those
without good leaving groups undergo addition.
O
a.
CH3
C
O
b.
CH3
CH3CH2CH2
no good leaving group
addition reactions
20.3
b. The O is sp2 hybridized.
Both lone pairs occupy sp2 hybrid orbitals.
H
C
O
O
c.
Cl
CH3
Cl–good leaving group
substitution reactions
C
C
d.
OCH3
OCH3–reasonable leaving group
substitution reactions
H
no good leaving group
addition reactions
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
20.4
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
C
O
H
or
CH3CH2CH2
C
O
CH3
c.
O
CH3CH2
C
O
or
CH3
less hindered carbonyl
more reactive
C
Cl
or
CH3
C
OCH3
better leaving group
more reactive
less hindered carbonyl
more reactive
b.
CH3CH2
O
CH3CH(CH3)
C
O
d.
CH2CH3
CH3
C
O
or
OCH3
better leaving group
more reactive
CH3
C
NHCH3
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Chapter 20–6
NaBH4 reduces aldehydes to 1° alcohols, and ketones to 2° alcohols.
20.5
O
a.
CH3CH2CH2
OH
NaBH4
C
H
CH3CH2CH2 C H
CH3OH
NaBH4
c.
H
CH3OH
O
OH
NaBH4
O
b.
OH
CH3OH
1° Alcohols are prepared from aldehydes and 2° alcohols are from ketones.
20.6
H
OH
O
a.
OH
b.
O
OH
O
c.
20.7
OH
3° Alcohols cannot be made by reduction of a carbonyl group,
because they do not contain a H on the C with the OH.
1-methylcyclohexanol
20.8
O
a.
O
OH
[1] LiAlH4
d.
[2] H2O
O
Pd-C
OH
NaBH4
O
e.
b.
O
O
H2 (1 equiv)
c.
NaBH4 (excess)
NaBD4
f.
CH3OH
Pd-C
20.9
O
OH
NaBH4
a.
OH
CH3OH
NaBH4
CHO
b.
CH3OH
c. (CH3)3C
O
OH
NaBH4
CH3OH
(CH3)3C
OH
(CH3)3C
20.10
O
HO H
Cl
A
Cl
[1] (S)-CBS reagent
[2] H2O
OH
CH3OH
CH3OH
O
OH
H2 (excess)
B
OH
D OH
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Chemistry
Introduction to Carbonyl Chemistry 20–7
20.11
Part [1]: Nucleophilic substitution of H for Cl
O
O
C
CH3
Cl
[1]
O
CH3 C Cl
[2]
H
H3Al H
C
CH3
H replaces Cl.
aldehyde
+ AlH3
Cl–
+
H
Part [2]: Nucleophilic addition of H– to form an alcohol
O
O
CH3
C
H
[3]
H3Al H
OH
H OH
CH3 C H
[4]
H
+ AlH3
+ –OH
CH3 C H
H
1o alcohol
20.12 Acid chlorides and esters can be reduced to 1° alcohols. Keep the carbon skeleton the same in
drawing an ester and acid chloride precursor.
O
O
CH2OH
a.
C
or
Cl
OH
CH2OH
OCH3
Cl
b.
C
O
C
O
CH3O
or O
c.
OCH3
CH3O
or
Cl
C
O
OCH3
CH3O
20.13
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
d.
NH
[1] LiAlH4
NH
[2] H2O
20.14
O
CH2NH2
a.
C
O
NH2
c.
O
N
b.
CH2CH3
CH2CH3
C
O
N
CH2CH3
CH2CH3
or
N
CH2CH3
N
H
N
H
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Chapter 20–8
20.15
O
COOCH3
a.
[1] LiAlH4
O
b. CH3O
[1] LiAlH4
CH3O
OH
CH3O
OH
[2] H2O
+ HOCH3
OH
NaBH4
COOCH3
CH3OH
O
O
c.
OH
[2] H2O
NaBH4
CH3O
OH
CH3OH
[1] LiAlH4
OH [2] H2O
NaBH4
CH3OH
HO
OH
+ HOCH3
Neither functional group reduced
20.16
CO2CH2CH3
a.
O
b.
CO2CH2CH3
O
CO2CH2CH3
c.
H2
CO2CH2CH3
(1 equiv)
Pd-C
O
H2
CO2CH2CH3
(2 equiv)
Pd-C
OH
[1] LiAlH4
OH
[2] H2O
O
d.
+ CH3CH2OH
OH
CO2CH2CH3
NaBH4
CO2CH2CH3
CH3OH
O
OH
20.17 Tollens reagent reacts only with aldehydes.
Ag2O, NH4OH
a.
b.
CH2OH
Na2Cr2O7
H2SO4, H2O
COOH
Ag2O, NH4OH
OH
No reaction
OH
O
C
CHO
O
Na2Cr2O7
H2SO4, H2O
OH
O
C
OH
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Introduction to Carbonyl Chemistry 20–9
20.18
O
OH
CHO
c. B
CHO
PCC
O
HO
B
H
OH
OH
CH2OH
NaBH4
a. B
d. B
O
Ag2O, NH4OH
C
HO
CH3OH HO
O
O
OH
[1] LiAlH4
b. B
[2] H2O
OH
CH2OH
e. B
C
CrO3
H2SO4, H2O
HO
OH
O
OH
20.19
a. CH3CH2Br + 2 Li
c. CH3CH2Br + 2 Li
CH3CH2Li + LiBr
b. CH3CH2Br + Mg
CH3CH2MgBr
CH3CH2Li
2 CH3CH2Li + CuI
+ LiBr
LiCu(CH2CH3)2 + LiI
20.20
HC CCH2CH2CH2CH2CH2CH3 + NaH
Na+ C CCH2CH2CH2CH2CH2CH3 + H2
hydrogen gas
HC CCH2CH2CH2CH2CH2CH3
BrMgC CCH2CH2CH2CH2CH2CH3 + CH4
methane gas
+ CH3MgBr
20.21
+ LiOH
Li + H2O
a.
CH3
H2O
CH
3 C MgBr +
b.
CH3
MgBr
c.
CH3
CH3 C H
+ HOMgBr
+ HOMgBr
+ H2O
d. CH3CH2C C Li
+ H2O
CH3CH2C CH + LiOH
CH3
20.22 To draw the product, add the benzene ring to the carbonyl carbon and protonate the oxygen.
OH
O
a.
H
C
OH
H C H
[1]
H
O
MgBr
c.
[2] H2O
CH3CH2
C
[1]
H
MgBr CH3CH2 C H
[2] H2O
OH
O
b. CH3CH2
C
[1]
CH2CH3
MgBr CH3CH2 C CH2CH3
d.
[2] H2O
O
[1]
[2] H2O
MgBr
OH
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Chapter 20–10
20.23 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.
C C
Na+
[1] CH2=O
C C CH2OH
[2] H2O
H C H
[2] H2O
20.24 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
20.25
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
20.26
OH
O
a.
+ CH3Li
linalool
(three methods)
OH
Li
O
Li
lavandulol
O
+ CH2=O
Li
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Introduction to Carbonyl Chemistry 20–11
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
20.27
N(CH3)2
N(CH3)2
OH
O
BrMg
OCH3
OCH3
venlafaxine
20.28
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
20.29
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
20.30
O
OH
a.
C
CH3
CH3O
C
+
MgBr
CH3
(2 equiv)
O
b. (CH3CH2CH2)3COH
CH3O
C
CH2CH2CH3 + CH3CH2CH2MgBr
(2 equiv)
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Chapter 20–12
O
OH
c. CH3 C CH2CH(CH3)2
C
CH3O
CH2CH(CH3)2
CH3
+ CH3MgBr
(2 equiv)
20.31 The R group of the organocuprate has replaced the Cl on the acid chloride.
O
a. CH3CH2
C
Cl
CH3CH2
[2] H2O
O
O
C
C
Cl
c. CH3CH2
CH3
O
[1] (Ph)2CuLi
Cl
[2] H2O
Ph
O
[1] [CH3CH2CH(CH3)]2CuLi
O
b.
[1] (CH3)2CuLi
[2] H2O
20.32
O
a. (CH3)2CHCH2
C
O
O
[1] LiAlH[OC(CH3)3]3
Cl [2] H O
2
C
(CH3)2CHCH2
H
c. (CH3)2CHCH2
O
b.
(CH3)2CHCH2
C
C
[2] H2O
O
Cl
HO
[1] CH3CH2Li (2 equiv)
d.
(CH3)2CHCH2
O
[1] (CH3CH2)2CuLi
Cl
C
[1] LiAlH4
Cl
[2] H2O
[2] H2O
(CH3)2CHCH2CH2OH
20.33
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
20.34
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
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Introduction to Carbonyl Chemistry 20–13
20.35
OH
a.
O
O
c.
b.
OH
OH
O
OH
O
d.
(+ enantiomer)
20.36 The characteristic reaction of ,-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
a.
CH3
[1] (CH3)2CuLi
O
[1] (CH3)2CuLi
c.
[2] H2O
[1] H C C Li
[2] H2O
O
CH3
[2] H2O
CH3
CH3
HO C CH
CH3
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
20.37
[1] (CH2=CH)2CuLi
a.
O
[2] H2O
O
O
(from a.)
O
b.
[1] CH2=CHLi
c.
[1] CH2=CHLi
[2] H2O
OH
[2] H2O
OH
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Chapter 20–14
20.38
a. CH3CH2OH
HBr or
PBr3
PCC
OH
Mg
CH3CH2Br
O
CH3CH2MgBr
[1] CH3CH2MgBr
OH
[2] H2O
HBr
OH
b.
Br
(from a.)
OH
c.
H2SO4
H2
+
Pd-C
(from a.)
OH
d.
HBr or
PBr3
Mg
Br
H2O
CH3CH2OH
O
e.
MgBr
PCC
O
OH
CH3CHO
OH
H2O
MgBr
PCC
(from d.)
H2SO4
CH3CH2OH
CH2=CH2
O
mCPBA
20.39
O
a.
O
NaBH4
H
CH3OH
OH
[1] LiAlH4
OH
h.
[1] C6H5Li
H
[2] H2O
O
H2
OH
[1] (CH3)2CuLi
i.
H
Pd-C
H
PCC
O
Na2Cr2O7
e.
H
H
O
No reaction
Ag2O
NH4OH
j.
[1] HCCNa
H
O
O
OH
H2SO4, H2O
O
f.
No reaction
[2] H2O
O
d.
OH
H
[2] H2O
O
c.
H
O
H
OH
[2] H2O
O
b.
[1] CH3MgBr
g.
k.
OH
l.
[2] H2O
[1] CH3CCLi
H
O
OH
OH
[2] H2O
TBDMSCl
OH
imidazole
O–TBDMS
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Introduction to Carbonyl Chemistry 20–15
20.40
O
a.
O
OH
NaBH4
CH3OH
[2] H2O
O
O
OH
[1] LiAlH4
b.
OH
[1] CH3MgBr
g.
[2] H2O
[2] H2O
O
O
OH
H2
c.
[1] (CH3)2CuLi
i.
Pd-C
O
PCC
O
No reaction
Na2Cr2O7
e.
No reaction
[2] H2O
O
d.
C6H5 OH
[1] C6H5Li
h.
OH
[1] HCCNa
j.
[2] H2O
O
No reaction
H2SO4, H2O
OH
[1] CH3CCLi
k.
[2] H2O
f.
O
OH
Ag2O
No reaction
NH4OH
O–TBDMS
TBDMSCl
l.
imidazole
20.41
Li (2 equiv)
a.
Br
b.
Br
c.
Br
d.
Li
MgBr
e.
MgBr
(CH3CH2CH2CH2)2CuLi
f.
Li
+
Li
Mg
LiBr
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
H2O
+ LiOH
D2O
D
CH3CCH
+
+ DOMgBr
LiCCCH3
20.42
a.
MgBr
b.
MgBr
CH2 O
H2O
OH
O
HO
H2O
OH
c.
MgBr
CH3CH2COCl
H2O
g.
MgBr CH3COOH
h.
MgBr
HC CH
i.
MgBr
CO2
+ CH3COO
HC C
O
H3O+
OH
O
d.
MgBr CH3CH2COOCH3
e.
MgBr
f.
MgBr
H2O
OH
H2O
+
j.
MgBr
k.
MgBr
l.
MgBr
OH
H2O
D2O
D +
OD
OH
O
CH3CH2OH
+ CH3CH2O
H2O
OH
516
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Chapter 20–16
20.43
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
20.44 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
R
H OH
c.
[1] (R)-CBS reagent; [2] H2O
S
20.45
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
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Chemistry
Introduction to Carbonyl Chemistry 20–17
20.46
O
C
a. (CH3)2CHCH2CH2
Cl
b. (CH3)2CHCH2CH2
[1] (CH2=CH)2CuLi
Cl
c. (CH3)2CHCH2CH2
[1] C6H5MgBr
(2 equiv)
Cl
O
C
d. (CH ) CHCH CH
3 2
2
2
(CH3)2CHCH2CH2
[2] H2O
O
C
(CH3)2CHCH2CH2
[2] H2O
C
H
O
O
C
O
[1] LiAlH[OC(CH3)3]3
CH=CH2
OH
(CH3)2CHCH2CH2
[2] H2O
C C6H5
C6H5
OH
[1] LiAlH4
Cl
C
(CH3)2CHCH2CH2
[2] H2O
C H
H
20.47
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
[2] H2O
CH3CH2
C
H
20.48
OH
a. HO
CHO
O
CrO3
CHO
PCC
O
CHO
COOH
HO
NH4OH
H
b. HO
Ag2O
CHO
OH c. HO
H2SO4, H2O O
Na2Cr2O7
CHO
d. HO
COOH
HOOC
H2SO4, H2O
20.49
a.
NaBH4
O
COOCH3
OH
O
[1] LiAlH4
c.
CH3OH
COOCH3
OH
(CH3)2N
[2] H2O
OH
(CH3)2N
O
b. O
[1] LiAlH4
COOCH3 [2] H2O
OH
d.
CH2OH
[1] LiAlH[OC(CH3)3]3
O
OH
Cl
O
[2] H2O
O
OH
H
O
518
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Chapter 20–18
20.50
CH3
a. CH3
CH3
[1] CO2
MgBr
CH3
COOH
[1] CH2=O
MgBr
f.
[2] H3O+
CH3
CH3
O
OH
O [1] CH3CH2MgBr
b.
CH2OH
[2] H2O
O
[1] (CH3)2CuLi
g.
[2] H2O
[2] H2O
OH
O
c.
OH
[1] C6H5Li
CHO
[1] CH3MgBr
h.
[2] H2O
COCl
d.
O
i.
[1] C6H5MgBr
(excess)
OH
[1] C6H5Li
[2] H2O
OH
OH
C
O
[2] H2O
[1] (CH3)2CuLi
j.
C6H5
e.
COOCH2CH3
[1] CH3MgCl
(excess)
CH3
[2] H2O
OH
[2] H2O
C CH3
20.51
O
[1] C6H5MgBr
a.
HO C6H5
+
HO C6H5
[2] H2O
b. (CH3)3C
O
[1] CH3Li
[2] H2O
c.
(CH3)3C
OH
[1] CH3CH2MgBr
O
OH
+
CH2CH3
[2] H2O
CH3
O
[1] (CH2=CH)2CuLi
OH
[1] Mg
COOH
[2] CO2
H
[3] H3O+
O
OH
CH=CH2
CH=CH2
Br
H
CH2CH3
+
[2] H2O
e.
+ (CH3)3C
CH3
OH
d.
[1] (S)-CBS reagent
H OH
f.
[2] H2O
O
H OH
[1] (R)-CBS reagent
g.
[2] H2O
CH3
[2] H2O
OH
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Chemistry
Introduction to Carbonyl Chemistry 20–19
O
h.
[1] LiAlH4
OCH2CH3
CH2OH
+ HOCH2CH3
[2] H2O
H
H
20.52 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
O
OH
[2] H2O
H
endo OH group
from
exo OH group below
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.
20.53 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
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
A
OH
OTBDMS
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Chapter 20–20
20.54 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.
[1] O3
[2] CH3SCH3
O
C
CHO
O
OH
[1] Mg
H2SO4
[1] C6H5MgBr
HCl
Cl
D
[2] H2O
[2] CH2=O
CH2OH F
[3] H2O
A
B
mCPBA
[1] LiAlH4
[1] (CH3)2CuLi
O
E
[2] H2O
OH
Br
OH
MgBr
PBr3
G
OH
CH3
[2] H2O
[1] C6H5CHO
Mg
[2] H2O
I
H
J
K
20.55
O
O
R
a.
Cl
Si
R
H OH
[1] (R)-CBS
reagent
R
Si
[2] H2O
R
O
O
H OH
Cl
O
NaI
I
O
R
Si
A
R
O
B
(CH3CH2)3SiCl
imidazole
H OH
HO
H
N
KF
O
R
H OSi(CH2CH3)3
NHCH(CH3)2
Si
HO
R
O
D
O
R
(CH3)2CHNH2
Si
R
O
H OSi(CH2CH3)3
I
C
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Chemistry
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O
b.
O
N(CH3)2
O
N(CH3)2
[1] C6H5MgBr
N(CH3)2
H2SO4
[2] H2O
E
HO
O
F
(Z and E isomers)
O
O
[1] LiCu(CH CH CHC5H11)2
COOCH3
c.
COOCH3
OR'
RO
[2] H2O
RO
G
OR'
O
several
steps
COOH
HO
OH
PGE1
20.56
H OH
O
O
O
a.
O
O
O
CH3 MgBr
O
+
+ MgBr
O
H OH
+
+ MgBr
CH3 MgBr
OH
HOMgBr +
OH
H OH
O
b.
O
O
O
O
CH2CH2CH2CH2
BrMg CH2CH2CH2CH2 MgBr
MgBr
BrMg
H OH
O
O
O
+
+ MgBr
+
+ MgBr
OH
HOMgBr +
OH
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Chapter 20–22
20.57
O
O
CH3 MgBr
O
OCH3
O
O
OCH3
O
OCH3
+
O
O
(CH3)3COH + OH
O
CH3 MgBr
O
O
OCH3
O
CH3 MgBr
H OH CH3 C CH3
CH3
O
H OH
OH
O
O
OH
CH3 MgBr
O
OH
O
+ OH
+ OH
+
OCH3
H OH
H OH
CH3OH + OH
20.58
MgBr
+ CH O
3
O
O
C
CH3O C OCH3
OCH3
O
CH3O
C
+ CH3O
MgBr
MgBr
HO H
O
O
O
C
C
CH3O C
MgBr
MgBr
+ CH3O
MgBr
H2O
OH
C
+ CH3OH + HOMgBr
20.59
CH2OH
a.
MgBr
O
H
C
b.
H
H
MgBr
OH
or
H
O
MgBr
O
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OH
c.
BrMg
(C6H5)2C=O
(C6H5)3COH
O
MgBr
e.
BrMg
d.
HO
O
or
O
MgBr
or
O
CH3MgBr
20.60
O
OH
a.
C
CH3O
O
C
MgBr
c. (CH3CH2CH2CH2)2C(OH)CH3
(2 equiv)
CH3O
b. CH3 C CH2CH2CH(CH3)2
CH3O
C
CH3
BrMg
(2 equiv)
O
OH
C
CH2CH2CH(CH3)2
+ CH3–MgBr
CH3
(2 equiv)
20.61
OH
Li
O
c.
H
(two ways)
C CH2CH3
O
or
(three ways)
H
+
Li
O
OH
b.
OH
+
a.
O
C
+
or
Li
+
O
or
(three ways)
+
or
Li CH2CH3
O
Li
CH2CH3 +
Li
O
or
+
Li
CH3CH2
O
C
Li
OCH3
+
(2 equiv)
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Chapter 20–24
20.62
OH
a.
O
+
MgBr
HO
H
H
c.
O
C6H5
b.
H
+ BrMg C6H5
O
H
MgBr
+
OH
20.63
C6H5
C6H5
Br
Mg
Br
KOC(CH3)3
C6H5
H2
C6H5
C6H5
Pd-C
LiAlH4
Br
C6H5
H2O
MgBr
C6H5
C6H5
20.64
PBr3
a.
OH
Mg
Br
O
PCC
H
[1] BrMg CH2CH3
[2] H2O
OH
CH3OH
PBr3
CH3Br
Mg
O
b.
[1] CH3MgBr
H
OH
O
PCC
[2] H2O
[1] CH3MgBr
OH
[2] H2O
(from a.)
OH
PBr3
Br
c.
CH3OH
OH
OH
[2] H2C=O
[3] H2O
(from b.)
d.
[1] Mg
H2SO4
PCC
CH2 CH2
mCPBA
O
[1] BrMg CH2CH3
OH
(from a.)
[2] H2O
20.65
Br
Mg
O
MgBr
OH
H2O
H
CHO
MgBr
Mg
Br
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Chemistry
525
Introduction to Carbonyl Chemistry 20–25
20.66
OH
Br
PBr3
a.
MgBr
Mg
CH2OH
[1] CH2=O
[2] H2O
MgBr
b.
COOH
[1] CO2
[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.
OH
Pd-C
+
OH
[1] CH3MgBr
H2SO4
[1] O3
CH3
[2] H2O
[2] (CH3)2S
(from c.)
O
CHO
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
(from c.)
C6H5
H2
(from c.)
f.
C6H5
H2SO4
[2] H2O
O
O
PCC
major product
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Chapter 20–26
20.67
OH
OH
f.
(from c.)
O
PCC
b.
MgBr
Cl
SOCl2
a.
c.
PBr3
[1] CH3CHO
(CH3)2CHD
[1]
O
MgBr
h.
MgBr
(from b.)
[2] H2O
MgBr
OH
[2] H2O
[1]
O
i.
[1] O
MgBr
CHO
(from c.)
Mg
OH
D2O
g.
Br
PCC
[2] H2O
MgBr
OH
OH
[1] CH2=O
(from c.)
OH
d.
[2] H2O
[2] H2O
(from c.)
[1] 2 Li
j.
OH
[(CH3)2CH]2CuLi
Br
[2] CuI (0.5 equiv)
MgBr
[1] CO2
e.
+
[2] H3O
(from c.)
(from c.)
COOH
[1] O
[2] H2O
O
20.68
OH
H
O
H
H
H
PCC
H
HO
O
[1] NaH
H
H
HO
H
H
[2] CH3I
H
CH3O
estradiol
[1] HC CLi
[2] H2O
OH
C CH
H
H
H
CH3O
mestranol
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Chemistry
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20.69
Br
a.
CH3CH2Cl
Br2
AlCl3
h
Br
Br2
K+ –OC(CH3)3
[2] H2O2, –OH
MgBr
Mg
O
OH
[1]
FeBr3
[2] H2O
CH3Cl
Br2
AlCl3
h
OH
MgBr [1] H2C=O
Br Mg
[2] H2O
O
OH
MgBr [1] CH3CH=O
b.
OH
[1] BH3
PCC
[2] H2O
(from a.)
O
CH3
O
Cl
CH3Br
AlCl3
O
Br NaOH
OH PCC
H
O
OH
Mg
[1] CH3MgBr
PCC
[2] H2O
(from a.)
20.70
HBr
a.
[1] Mg
ROOR
Br
[2] CO2
[3] H3O+
[1] Mg
b.
Br
(from a.)
OH
c.
Br2
Br
[2] H2C=O
[3] H2O
h
O
[1] Mg
OH
[2] O
[3] H2O
20.71
Br
Br2
a.
MgBr
Mg
[2] H3O+
FeBr3
MgBr
b.
[1] CH2=O
[2] H2O
(from a.)
COOH
[1] CO2
CH2OH
PCC
CHO
CH3OH
PCC
CH2=O
OH
528
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Chapter 20–28
OH
O
[1] PhMgBr
CHO
PCC
(from a.)
c.
[2] H2O
(from b.)
PBr3
d. CH3CH2CH2OH
Mg
CH3CH2CH2Br
PCC
CH3CH2CH2OH
CH3CH2CH2MgBr
[1] CH3CH2CH2MgBr
CH3CH2CHO
HO
[1] PhMgBr
PCC
[2] H2O
OH
O
(from a.)
[2] H2O
O
OH
e.
PCC
H
OH
O
O
H2O
O
Br
Br2
PCC
FeBr3
MgBr
(from a.)
20.72
PCC
a. HO
[1] CH3CH2MgBr
H
O
OH
O
CH3CH2MgBr
CH3CH2Br
O
PCC
b.
OH
Mg
PBr3
CH3CH2OH
PCC
[2] H2O
OH
[1] CH3CH2CH2MgBr
[2] H2O
OH
c.
Mg
PBr3
CH3CH2CH2OH
CH3CH2CH2MgBr
CH3CH2CH2Br
PCC
H
[1] CH3CH2CH2CH2MgBr
PCC
[2] H2O
O
CH3CH2CH2CH2OH
H
d.
OH
O
Mg
PBr3
CH3CH2CH2CH2MgBr
CH3CH2CH2CH2Br
[1] (CH3)2CHCH2MgBr
PBr3
Mg
[2] H2O
O
OH
Br
(from c.)
(CH3)2CHCH2OH
PBr3
(CH3)2CHCH2Br
MgBr
[1] CO2
Mg
(CH3)2CHCH2MgBr
[2] H3O+
CO2H
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Chemistry
Introduction to Carbonyl Chemistry 20–29
e.
HBr
OH
Br
Mg
MgBr
H2O
K+ –OC(CH3)3
PBr3
+
OH
H
(from c.)
Br
O
20.73
a.
[1] (CH3)2CuLi
O
[1] CH3MgBr
[2] H2O
O
[2] H2O
H2SO4
OH
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
O CH3CH2I
OH NaH
Br
Br
O
Br
(+ ortho isomer)
MgBr
[1] Br2, FeBr3
d.
H2O
[2] Mg
OH
mCPBA
[1] NaH
e. HC CH
O
HO
CH3CHC CH
[2] CH3CH=O
[3] H2O
TBDMS–Cl
imidazole
OTBDMS
CH3CHC CH
NaH
OTBDMS
CH3CHC C
[1] O
[2] H2O
OTBDMS
OH
CH3CHC C
OH
(CH3CH2CH2CH2)4NF
OH
20.74
Hb
Hb
O
CH3
Ha
H
CH3
Hb
IR peak: 1716 cm–1 (C=O)
1H NMR: 2 signals (ppm)
CH3
doublet 1.2 (Hb)
H
CH3
Ha
septet 2.7 (Ha)
Hb
C7H14O
A
Hd
Hd
Hb
NaBH4
H
CH3
CH3OH
H
Hc
OHa
CH3
H
CH3 CH3
Hd
Hd
C7H16O
B
Hc
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)
530
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Chapter 20–30
20.75
Hb
H
Hb
H
O
CH3
CH3
C4H8O
Ha
Ha
Hc
Hc
H
H
IR peak 3600–3200 cm–1 (OH)
1H 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
CH3 CH3
[2] H2O
Ha Ha
C10H14O
C
NMR: 2 signals (ppm)
singlet (6 H) 1.3 (Ha)
singlet (2 H) 2.4 (Hb)
D
1H
20.76
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)
20.77 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
CH3CH2CH2C N
O
[1] CH3MgBr
H H
CH3
[2] H3O+
Hc
b
CH3
H H
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)
20.78 Molecular ion at m/z = 86: C5H10O (possible molecular formula).
Hb Hd Hd
[1] (CH3)3CLi
[2] CH2=O
[3] H2O
CH3 H H
H
OH
He H
Hf
H H
Hc Hc
H
Ha
IR peaks: 3600–3200 cm–1 (OH)
1651cm–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
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531
Introduction to Carbonyl Chemistry 20–31
20.79
Hb
He
[1] CH3MgBr
O
[2] H2O
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
HOMgBr
OH
O
H OH
20.80
O
OH
Br
[1] (R)-CBS
reagent
[2] H2O
HO
OTBDMS
Br
imidazole
(2 equiv)
HO
COOCH3
COOCH3
[1] NaH
HO
C6H5
Br
TBDMSCl
[2] Br
A
TBDMS O
COOCH3
Br
O
C6H5
Br
NH3
H2N
O
B
A
+
OTBDMS
H
N
B
TBDMS O
COOCH3
OTBDMS
H
N
TBDMS O
O
C6H5
[1] LiAlH4
[2] H2O
O
C6H5
(CH3CH2CH2CH2)4NF
OH
(R)-salmeterol
C6H5
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Chapter 20–32
20.81
larger group
equatorial
axial
OH
[1] L-selectride
(CH3)3C
O
(CH3)3C
[2] H2O
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
OH axial
H2O
H
(CH3)3C
R3B H
H equatorial
(CH3)3C
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.
R3B H
Axial H's hinder
H
axial attack.
H
(CH3)3C
H
O
(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.
20.82 The carbon of an ,-unsaturated carbonyl compound absorbs farther downfield in the 13C
NMR spectrum than the carbon, because the carbon is deshielded and bears a partial positive
charge as a result of resonance. Since three resonance structures can be drawn for an ,unsaturated carbonyl compound, one of which places a positive charge on the carbon, the
decrease of electron density at this carbon deshields it, shifting the 13C absorption downfield.
This is not the case for the carbon.
122.5 ppm
O
150.5 ppm
mesityl oxide
O
O
O hybrid:
+
+
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Introduction to Carbonyl Chemistry 20–33
20.83
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
[3] H2O
W
H
H C
CN
–
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
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535
Aldehydes and Ketones 21–1
C
—N
Nuucclleeoopphhiilliicc A
Addddiittiioonn
Chhaapptteerr 2211:: A
Allddeehhyyddeess aanndd K
Keettoonneess—
G
Geenneerraall ffaaccttss
• 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).
RC
CH
HO
O aanndd R
R222C
CO
O ((2211..44))
SSuum
mm
maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ooff R
–1
C=O
~1715 cm 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.
Csp –H of CHO
~2700–2830 cm–1 (one or two peaks)
CHO
C–H to C=O
9–10 ppm (highly deshielded proton)
2–2.5 ppm (somewhat deshielded Csp –H)
C=O
190–215 ppm
2
1
H NMR absorptions
13
C NMR absorption
3
N
Nuucclleeoopphhiilliicc aaddddiittiioonn rreeaaccttiioonnss
[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
1o
•
•
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.
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.
–
536
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Chapter 21–2
[4] Wittig reaction (21.10)
R
R
C O
+ Ph3P
C C
•
alkene
•
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.
C
(R')H
(R')H
Wittig reagent
The reaction forms a new C–C bond and a
new C–C 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
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.
O
Otthheerr rreeaaccttiioonnss
[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].
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Aldehydes and Ketones 21–3
[2] Conversion of cyanohydrins to aldehydes and ketones (21.9)
OH
O
–OH
R C H(R')
CN
C
H(R')
+ H2O
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
H2O
H+ or –OH
R C H(R')
COOH
-hydroxy
carboxylic acid
[4] Hydrolysis of imines and enamines (21.12)
NR
O
NR2
H
(R')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')
OR"
•
O
H+
+ H2O
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.
538
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Chapter 21–4
C
Chhaapptteerr 2211:: A
Annssw
weerrss ttoo PPrroobblleem
mss
As the number of R groups bonded to the carbonyl C increases, reactivity towards nucleophilic
attack decreases.
21.1
a.
(CH3)2C
O
CH3CH O
2 R groups
CH2 O
b.
O
O
O
1 R group 0 R groups
Increasing reactivity
decreasing alkyl substitution
Increasing reactivity
decreasing steric hindrance
More stable aldehydes are less reactive towards nucleophilic attack.
21.2
CHO
CHO
cyclohexanecarbaldehyde
This aldehyde has no added
resonance stabilization.
benzaldehyde
Several resonance structures delocalize the partial
positive charge on the carbonyl carbon, making
it more stable and less reactive towards
nucleophilic attack.
O
O
O
C
C
C
H
H
O
O
C
C
H
O
H
C
H
O
C
H
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.
21.3
CHO
a.
(CH3)3CC(CH3)2CH2CHO
c.
4
3 2 1 H
5
H
O
5 C chain = pentanal
O
3,3,4,4-tetramethylpentanal
8
CHO
b.
6
8 C chain = octanal
1
7
CHO
5
4
3
2
2,5,6-trimethyloctanal
Cl
Cl
2 1
CHO
Cl
3 4
Cl
3,3-dichlorocyclobutane4 C ring =
carbaldehyde
cyclobutanecarbaldehyde
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Aldehydes and Ketones 21–5
Work backwards from the name to the structure, referring to the nomenclature rules in Answer
21.3.
21.4
a. 2-isobutyl-3-isopropylhexanal
c. 1-methylcyclopropanecarbaldehyde
CHO
3 carbon ring
O
6 C chain
CH3
H
b. trans-3-methylcyclopentanecarbaldehyde
5 carbon ring
CHO
d. 3,6-diethylnonanal
CHO
9 C chain
H
or
O
CH3
CH3
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
O
4
c.
5
O
8 C chain =
octanone
(CH3)3C
3
b.
1
8
O
5-ethyl-4-methyl-3-octanone
CH3
(CH3)3C
O
5 C ring =
cyclopentanone
(CH3)3CCOC(CH3)3
5 C chain =
pentanone
CH3
3
2
3
4
5
O
2,2,4,4-tetramethyl-3-pentanone
2
1
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
e. 6,6-dimethyl-2-cyclohexenone
6 C ketone
benzyl group: benzoyl group: 5 C ketone
O
b. methyl vinyl ketone
O
O
CH2
C
f. 3-ethyl-5-hexenal
O
O
c. p-ethylacetophenone
O
CHO
O
CH2
540
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Chapter 21–6
Compounds with both a C–C double bond and an aldehyde are named as enals.
21.7
a. (2Z)-3,7-dimethyl-2,6-octadienal
3
7
6
b. (2E,6Z)-2,6-nonadienal
2
1
3
7
CHO
6
2
cucumber aldehyde
1 CHO
neral
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 electrondeficient 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.
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
a.
CHO
or
conjugated C=O
lower wavenumber
CHO
b.
or
O
higher wavenumber
O
smaller ring
higher wavenumber
21.10 Since a charge-separated resonance structure can be drawn for a carbonyl group, more electron
donor R groups stabilize the (+) charge on this resonance form. The two R groups on the ketone
C=O thus help to stabilize it.
O
R
C
O
R
With an aldehyde, only one electron-donor R:
O
CH3CH2CH2CH2CH2
C
R
CH3CH2CH2CH2CH2
C
R
With a ketone, two electron-donor R groups:
O
H
C
O
H
The H of RCHO does not stabilize the charge-separated
resonance structure, so it contributes less to the hybrid.
The C=O has more double bond character.
higher wavenumber
CH3CH2
C
O
CH2CH2CH3
CH3CH2
C
CH2CH2CH3
The 2 R groups stabilize the charge-separated resonance
structure, so it contributes more to the hybrid.
The C=O has more single bond character.
weaker bond
lower wavenumber
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21.11 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
21.12
[1] DIBAL-H
a. CH3CH2CH2COOCH3
CH3CH2CH2CHO
[2] H2O
PCC
b. CH3CH2CH2CH2OH
CH3CH2CH2CHO
[1] BH3
c.
HC CCH2CH3
CH3CH2CH2CHO
[2] H2O2, HO–
d. CH3CH2CH2CH CHCH2CH2CH3
[1] O3
[2] Zn, H2O
CH3CH2CH2CHO
21.13
O
Cl
a.
O
O
CH3
c.
AlCl3
H2O
C CH
H2SO4
HgSO4
O
O
[1] (CH3)2CuLi
Cl
b.
[2] H2O
21.14
OCH3
CH3O
CH3O OCH3
[1] O3
OHC
CHO
[2] (CH3)2S
Cleave this C=C with O3.
21.15
LiAlH4
or
NaBH4
O
O
C
C
H
H
weaker base
Equilibrium favors the weaker base.
The H– nucleophile is a much stronger base than
the alkoxide product.
stronger base
21.16 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.
H OH
NaBH4
CH3OH
new stereogenic
center
O
b.
(CH3)3C
H OH
H OH
Add
OH
CH=CH2 stereochemistry:
[1] CH2 CHMgBr
[2] H2O
Add
stereochemistry:
CH=CH2
OH
(CH3)3C
(CH3)3C
(CH3)3C
CH=CH2
OH
542
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Chapter 21–8
21.17 Treatment of an aldehyde or ketone with NaCN, HCl adds HCN across the double bond. Cyano
groups are hydrolyzed by H3O+ to replace the 3 C–N bonds with 3 C–O bonds.
OH
CHO
a.
CHCN
NaCN
OH
b.
HCl
OH
H3O+, CN
COOH
21.18
HO
HO
HO
O
O
OH
HO
HO
O
O
CN
HO
enzyme
OH
C
O
CN
enzyme
H
C
H
+
HCN
toxic
by-product
amygdalin
21.19
CH3
a.
CH3
+
C O
Ph3P=CH2
C CH2
CH3
b.
CH3
O +
Ph3P=CHCH2CH2CH2CH3
CHCH2CH2CH2CH3
21.20
BuLi
a. Ph3P + BrCH2CH3
Ph3PCH2CH3
Br–
b. Ph3P + BrCH(CH3)2
Ph3PCH(CH3)2
Ph3P=CHCH3
BuLi
Ph3P=C(CH3)2
Br–
c. Ph3P + BrCH2C6H5
BuLi
Ph3PCH2C6H5
Ph3P=CHC6H5
Br–
21.21
CHO
a.
CH2CH3
+
Ph3P CHCH2CH3
CH2CH3
CHO
+
b.
c.
CHO
+
Ph3P CHC6H5
COOCH3
Ph3P CHCOOCH3
COOCH3
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Aldehydes and Ketones 21–9
21.22 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.
CH3
CH2CH3
C C
CH3
H
a.
CH3
C
CH3
CH2CH3
+
PPh3
O C
or
H
CH3
C O + Ph3P
CH3
2° halide precursor
(CH3)2CHX
b.
C C
H
CH2CH3
+
C PPh3
H
H
1° halide precursor
XCH2CH2CH3
preferred pathway
CH3CH2
CH2CH3
CH3CH2
CH2CH3
C
(only one route possible)
O C
H
H
(cis or trans)
C6H5
c.
CH3
C6H5
C C
H
C PPh3
H
C6H5
CH3
+
H
or
O C
H
(cis or trans)
1° halide precursor
C6H5CH2X
CH3
C O
+
Ph3P
C
H
H
(both routes possible)
1° halide precursor
XCH2CH3
21.23
a.
Two-step sequence:
O
HO CH3
[1] CH3MgBr
H2SO4
[2] H2O
One-step sequence:
O
b.
minor product
Ph3P=CH2
tetrasubstituted
major product
(E and Z
isomers)
only product
Two-step sequence:
O
[1] C6H5CH2MgBr
OH
[2] H2O
CH2C6H5
H2SO4
trisubstituted
conjugated C=C
One-step sequence:
O
CHC6H5
Ph3P=CHC6H5
CH2C6H5
trisubstituted
CHC6H5 only product
21.24 When a 1o amine reacts with an aldehyde or ketone, the C=O is replaced by C=NR.
a.
CHO
O
b.
CH3CH2CH2CH2NH2
CH3CH2CH2CH2NH2
CH
NCH2CH2CH2CH3
NCH2CH2CH2CH3
c.
O
CH3CH2CH2CH2NH2
NCH2CH2CH2CH3
544
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Chapter 21–10
21.25 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
b.
O + NH2CH2CH2CH3
H
H
CH3
N
CH3
O
NH2
21.26
overall
reaction
H2O
H
H
O
H
rhodopsin
N
H
OPSIN
N
OPSIN
11-cis-retinal
NH2 OPSIN
H3O
nucleophilic
attack
proton
transfer
OPSIN
OPSIN
NH2 C O
NH
H
OPSIN
C OH
H
21.27
N(CH3)2
N(CH3)2
CH3
N
H
+
CH3
21.28
O
H
This carbon has four bonds to C's. To make an enamine, it needs
a H atom, which is lost as H2O when the enamine is formed.
21.29 • Imines are hydrolyzed to 1° amines and a carbonyl compound.
• Enamines are hydrolyzed to 2° amines and a carbonyl compound.
O
H2O
a.
CH N
+
C
H+
H2N
H
imine
1° amine
H2O
b.
CH2 N
CH3
enamine
H+
CH2 N H
CH3
2° amine
C OH2
H
H OH2
O
NH
+
O
H2O
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Aldehydes and Ketones 21–11
c.
(CH3)2NCH
C(CH3)2
O
H2O
H+
enamine
+
(CH3)2NH
C CH(CH3)2
H
2° amine
21.30 • 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
21.31
H
O H
H OH2
O H
O H
O H
O H
+
O
+ H2O
+ H3O
(+ 1 resonance
structure)
H2O
21.32 Treatment of an aldehyde or ketone with two equivalents of alcohol results in the formation of an
acetal (a C bonded to 2 OR groups).
O
a.
O
+
2 CH3OH
OCH3
TsOH
+ HO
b.
OH
TsOH
OCH3
21.33
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
O
O
546
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Chapter 21–12
21.34 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
TsOH
+ HOCH CH OH
2
2
+ H2O
overall reaction
O H
TsO
H
HO O CH2CH2OH
O H
HO O CH2CH2OH
+ TsOH
+ TsO
hemiacetal
HOCH2CH2OH
H
HO O CH2CH2OH
H O
O CH2CH2OH
O CH2CH2OH
O CH2CH2OH
TsOH
+ H2O
+ TsO–
O CH2CH2OH
O
TsO
O H
O
O
+ TsOH
carbocation
re-drawn
acetal
21.35
CH3O OCH3
a.
b.
+
H2O
O
H2SO4
CH3
O
O
O
CH3
+ 2 CH3OH
+
c.
H2O
H2SO4
CH3
O
O
OH
+
C
CH3
OH
O
H2SO4
+ H2O
C
CH3
C
CH3
+ HOCH2CH2OH
21.36
HO
O
O
H2O
O
H2SO4
safrole
+
H
H
HO
formaldehyde
21.37 Use an acetal protecting group to carry out the reaction.
O
O
O
O
H3
O+
[2] H2O
TsOH
COOCH2CH3
O
O
[1] CH3Li (2 equiv)
HOCH2CH2OH
COOCH2CH3
OH
OH
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Aldehydes and Ketones 21–13
21.38
OH
O
C4
C1
OH
O
a. HO
O
H
b.
C1
O
H
C4
OH
C5
C1
C5
C1
21.39 The hemiacetal OH is replaced by an OR group to form an acetal.
OH
a.
OCH2CH3
O
H+
+ CH3CH2OH
O
OH
HO
OCH2CH3
O
b.
H+
+ CH3CH2OH
HO
HO
O
HO
21.40
acetal
O
HO
H
O
O
H
OH
O
O
H
OCH3
H
acetal
HO
O
H
O
HO
COOH
HO
monensin
O
H
O
HO
O
H
O
OH
acetal
OH
acetal
O
OH
H
digoxin
hemiacetal
Ether O atoms are indicated in bold.
21.41
HO OH
a.
HO OH
O
*
*
*
HO
*
OH *
OH
5 stereogenic centers
(labeled with *)
OH
d.
HO
CHO
OH
HO OH
b.
HO OH
O
HO
hemiacetal C
OH
OH
HO OH
O
OH
HO
OH
-D-galactose
O
OCH3
HO
OH
-D-galactose
c.
e.
HO OH
+
O
HO
OH
OCH3
O
548
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Chapter 21–14
21.42 Use the rules from Answers 21.3 and 21.5 to name the aldehydes and ketones.
O
a. (CH3)3CCH2CHO
O
O
re-draw
H
4 C = butanal
3,3-dimethylbutanal
h. (CH ) C
3 3
re-draw
C
CH(CH3)2
5 C = pentanone
2,2,4-trimethyl3-pentanone
(common name:
tert-butyl isopropyl ketone)
O
5 C = pentanone
2-chloro-3-pentanone
b.
Cl
O
8 C = octanone
8-phenyl-3-octanone
c. Ph
d.
CHO
5 C ring
2-methylcyclopentanecarbaldehyde
O
O
j.
f. (CH3)2CH
CH3
O
CHO
O
6 C ring = cyclohexanone
5-isopropyl-2-methylcyclohexanone
k.
CHO
6 C = hexanal
3,4-diethylhexanal
E
8 C = octenone
(5E)-2,5-dimethyl-5-octen-4-one
CHO
trans-2-benzylcyclohexanecarbaldehyde
l.
CH2Ph
o-nitroacetophenone
NO2
6 C ring = cyclohexanone
5-ethyl-2-methylcyclohexanone
e.
g.
CH3
i.
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. -methoxypropionaldehyde
O
H
i. 2-sec-butyl-3-cyclopentenone
O
OCH3
O
j. 5,6-dimethyl-1-cyclohexenecarbaldehyde
e. 3-benzoylcyclopentanone
CHO
O
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Aldehydes and Ketones 21–15
21.44
O
O
H
O
H
H
2-ethylbutanal
hexanal
O
O
H
2,2-dimethylbutanal
O
H
H
(2R)-2-methylpentanal
(2S)-2,3-dimethylbutanal
O
3,3-dimethylbutanal
O
H
(2R)-2,3-dimethylbutanal
H
4-methylpentanal
O
H
O
(2S)-2-methylpentanal
O
H
H
(3S)-3-methylpentanal
(3R)-3-methylpentanal
21.45
H
= C6H5CH2CHO
phenylacetaldehyde
O
a. NaBH4, CH3OH
b. [1] LiAlH4; [2] H2O
g.
C6H5CH2CH2OH
C6H5CH2CH2OH
h.
C6H5CH2CH(OH)CH3
NaCN, HCl
e. Ph P CHCH
3
3
OCH2CH3
CH3CH2OH (excess), H+
OCH2CH3
C6H5CH2CH(OH)CN
H
NH
C6H5CH2CH=CHCH3
mild acid
H
(CH3)2CHNH2
j.
O
OH
HO
NCH(CH3)2
mild acid
(E and Z isomers)
N
i.
(E and Z isomers)
f.
(E and Z isomers)
N(CH2CH3)2
c. [1] CH3MgBr; [2] H2O
d.
H
(CH3CH2)2NH, mild acid
O
H+
21.46
O
2-heptanone
a. NaBH4, CH3OH
OH
d.
[1] LiAlH4; [2] H2O
e.
OH
c. [1] CH3MgBr; [2] H2O
NC
OH
CHCH3
OH
b.
NaCN, HCl
Ph3P CHCH3
+
f. (CH3)2CHNH2, mild H
(E and Z isomers)
NCH(CH3)2
550
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Chapter 21–16
N(CH2CH3)2
(CH3CH2)2NH, mild H+
g.
N(CH2CH3)2
(E and Z isomers)
CH3CH2O OCH2CH3
CH3CH2OH (excess), H+
h.
NH
N
i.
N
mild H+
(E and Z isomers)
OH
HO
j.
O
O
H+
21.47
H
Ph3P CHCH2CH3
O
a.
CHO Ph3P CHCOOCH3
c.
CHCH2CH3
COOCH3
H
+
COOCH3
H
CHO
b.
Ph3P
O
d.
Ph3P CH(CH2)5COOCH3
H
CH(CH2)5COOCH3
H
21.48
a. CH3CH2Cl
[1] Ph3P
[1] Ph3P
CH3CH=C(CH3)2
c.
CH2Cl
[2] BuLi
[3] (CH3)2C=O
CH=CHCH2CH2CH3
[2] BuLi
[3] CH3CH2CH2CHO
(E and Z isomers)
c. Ph3P CHCH=CH2
BrCH2CH=CH2
[1] Ph3P
b.
CH2Br
[2] BuLi
[3] C6H5CH2CH2CHO
CH=CHCH2CH2C6H5
(E and Z isomers)
21.49
a.
Ph3P CHCH2CH2CH3
b.
Ph3P C(CH2CH2CH3)2
BrCH2CH2CH2CH3
BrCH(CH2CH2CH3)2
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Aldehydes and Ketones 21–17
21.50
HO
CHO
O
[2] H2O2,
–OH
CH HO C CH
[1] LiC
A
HO
COOH
C Ag2O
B [1] BH3
NH4OH
O
O
HO
[2] H2O
TBDMSO
E TBDMS–Cl
D H2O
TBDMSO
OH
F [1] CH3Li
[2] H2O
H2SO4
HgSO4
N
NH
(CH3CH2CH2CH2)4N+F
HO
G
OH
21.51
a. CH3CH2CHO +
b.
mild
H2N
O
CH3CH2CH N
acid
HOCH2CH2OH
O
H+
O
H3O+
c.
H2N
N
O
O
d.
mild
+
C6H5
N
H
N
acid
(E and Z isomers)
C6H5
HO
CN
C
e.
C6H5
H3O+, HO
COOH
C6H5
C6H5
C6H5
CH3CH2OH
OH
f.
O
H3O+
N
g.
OCH2CH3
H+
O
OCH3
h. CH3O
O
H3O+
OCH3
+
HN
O + HOCH3
CH3O
21.52
CH3CH2O
OCH2CH3
O
+ HOCH2CH3
a.
CH3O
O
OCH3
OCH3
OCH3
b.
HO
O
+ HOCH3
OCH2CH3
OCH3
OCH3
H
c.
O
+ HOCH2CH3
552
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Chapter 21–18
21.53 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
H2O
O
PCC
O
H+
G
F
O
OH
O
E
21.54
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.
OH
CH3OH
CH3CH2
d. HO
+
OH
CH3CH2
O
CH3OH
HO
CH3CH2
+
O
O
HO
HCl
OCH3
OH
OCH3
21.55
new stereogenic center
O
OH
O
OH
O
OH
CHO
HO
A
achiral
S
new stereogenic center
H
CHO
chiral
H
O
S
HO H B
An equal mixture of enantiomers
results, so the product is optically
inactive.
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.
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Aldehydes and Ketones 21–19
21.56
OH
H
HO
a.
acetal O
O
O
H
H
acetal
acetal
O
O
etoposide
O
O
H
CH3O
O
OCH3
OH
b. Lines of cleavage are drawn in.
OH
H
HO
O
H
O
O
O
H
O
H+
O
O
H
CH3O
OH
H
HO
H2O
O
OCH3
O
+
HO
H
CH3O
OH
HO
OH
H
O CH
HO H
O
OH
OH
+ CH3CHO
+ CH2=O
OCH3
OH
21.57 Use the rule from Answer 21.1.
O
O
H
a.
O
O
O
b.
O
Increasing reactivity
decreasing steric hindrance
Increasing reactivity
decreasing steric hindrance
21.58
O
H2O
HO OH
Less stable carbonyl compounds give a higher percentage of hydrate.
Cyclopropanone is an unstable carbonyl compound because the bond angles
around the carbonyl carbon deviate considerably from the desired angle. Since
the carbonyl carbon is sp2 hybridized, the optimum bond angle is 120°, but the
three-membered ring makes the C–C–C bond angles only 60°. This destabilizes
the ketone, giving a high concentration of hydrate when dissolved in H2O.
21.59 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 charge-separated
resonance form, giving the C=O more single bond character.
554
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Chapter 21–20
O 2N
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
21.60 Use the principles from Answer 21.22.
CH3CH2
a.
CH2CH2CH3
CH3CH2
H
CH3CH2
C C
CH3CH2
CH3CH2
CH2CH2CH3
C O
Ph3P C
C PPh3
or
H
CH2CH3
CH3
C C
b.
H
CH3
or
Ph3P C
H
CH2CH3
C PPh3
H
O C
H
1° alkyl halide precursor
(XCH2CH2CH3)
H
(Both routes possible.)
1° alkyl halide precursor
(XCH2CH3)
O
PPh3
c.
Ph3P CH2
or
methyl halide precursor
(CH3X)
preferred pathway
C6H5
d.
H
2° alkyl halide precursor
[(CH3CH2)2CHX]
CH2CH3
C O
H
O C
CH3CH2
1° alkyl halide precursor
(XCH2CH2CH2CH3)
preferred pathway
CH3
CH2CH2CH3
C6H5
O
Ph3P
CH2 O
2° alkyl halide precursor
(CH3CH2CH2CHXCH3)
or
C6H5
PPh3
H
1° alkyl halide precursor
(C6H5CH2X)
preferred pathway
2° alkyl halide precursor
X
21.61
a.
N
O
O
O
H
+ HOCH2CH2OH
c.
O
+
NH2
b.
N
O
+
HN
d. CH3O
OCH3
OCH3
CH3O
+ HOCH3
H
O
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Aldehydes and Ketones 21–21
21.62
PCC
a. C6H5–CH2OH
b. C6H5–COCl
C6H5–CHO
[1] LiAlH[OC(CH3)3]3
f. C6H5–CH=CH2
[2] Zn, H2O
C6H5–CHO
g. C6H5–CH=NCH2CH2CH3
[2] H2O
c. C6H5–COOCH3
[1] DIBAL-H
[1] O3
h. C6H5–CH(OCH2CH3)2
C6H5–CHO
[2] H2O
[1] LiAlH4
d. C6H5–COOH
e. C6H5–CH3
[2] H2O
KMnO4
PCC
C6H5–CH2OH
C6H5–COOH
H2O
C6H5–CHO
H+
H2O
C6H5–CHO
H+
C6H5–CHO
[1] LiAlH4
[2] H2O
C6H5–CHO
PCC
C6H5–CH2OH
C6H5–CHO
21.63
PCC
a.
OH
Cl
b.
c. CH3COCl (CH3CH2)2CuLi
e. CH3C CCH3
O
O
(CH3)2CuLi
d. CH3CH2C CH
O
O
H2O
H2SO4
HgSO4
O
H2O
H2SO4
HgSO4
O
21.64
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:
CHO
Ph3P
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
21.65
a.
CH3CH2CH2CH CHCH3
PCC
One possibility:
CHO
OH
PBr3
OH
Br
[1] Ph3P
[2] BuLi
CHO
Ph3P CHCH3
CH3CH2CH2CH CHCH3
(E and Z)
556
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Chapter 21–22
b.
C6H5CH CHCH2CH2CH3
CH3OH
One possibility:
SOCl2
CH3Cl
Br2
AlCl3
h
CHO
Br [1] Ph3P
[2] BuLi
C6H5CH PPh3
C6H5CH CHCH2CH2CH3
(from a.)
(E and Z)
c.
PCC
OH
OH
O
Br
PBr3
O
[1] Ph3P
Ph3P CHCH(CH3)2
[2] BuLi
21.66
H2O
a.
PCC
OH
H2SO4
Ph3P CHCH2CH3
H+
O
CHCH2CH3
O
b.
O
HOCH2CH2OH
O
(from a.)
PBr3
CH3CH2CH2OH
O
c.
CH3CH2CH2Br
H2NCH2CH2CH3
[1] Ph3P
[2] BuLi
Ph3P CHCH2CH3
NCH2CH2CH3
mild acid
NH3 (excess)
(from a.)
BrCH2CH2CH3
PBr3
HOCH2CH2CH3
OH
d.
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
e.
[1] OsO4
[2] NaHSO3, H2O
OH
OH
O
TsOH
PCC
OH
OH
O
O
[2] H2O
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Aldehydes and Ketones 21–23
21.67
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
h
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)
21.68
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
558
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Chapter 21–24
21.69
O
a. CH3CH2OH
PCC
CH3
C
H
PBr3
Mg
CH3CH2Br
CH3CH2MgBr
H2SO4
b. CH3CH2OH
O
[1]
C
CH3
H
[2] H2O
Br2
CH2=CH2
H OH
PCC
C
CH3
CH2CH3
Br
2 NaNH2
Br
O
CH3
CH2CH3
[1]
CH3
O
O
OH
O
C
[2] H2O
H
(from a.)
OH
PCC
CH2CH3
HC C Na+
[2] NaHSO3, H2O
HO
CH3 C OCH2CH3
TsOH
NaH
HC CH
[1] OsO4
O
OCH2CH3
CH3CH2OH
C
TsOH
21.70
O
O
O
mCPBA
O
O
OH
NHC(CH3)3
O
(CH3)3CNH2
NHC(CH3)3
O
H2O
O
O
O
H3O+
X
OH
NHC(CH3)3
HO
+ (CH3)2C=O
HO
albuterol
21.71
Br
Mg
O
Br
CHO
O
[1]
HOCH2CH2OH
TsOH
BrMg
O
O
O
OH
O
O
[2] H2O
H2O
H+
OH
A
21.72
a.
O H
Na+H–
O
Cl
+ H2 + Na+
b.
O
O
acetal
O
CH3
O
+ Cl–
O
methoxy methyl ether
CHO
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Aldehydes and Ketones 21–25
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
21.73
H
H
N
N
H2O
H2O
H
OH
H
HO
N
H
N
HO
H OH2
H H
N
a.
H
H OH2
H2O
(+ 1 resonance
structure)
O
O
H2O
NH3
O
NH2
H OH2
H
H2O
NH2
(+ 1 resonance
structure)
560
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Chapter 21–26
NH2
H
O
NH2
H
O
H
H
H H
N
O
b.
NH2 O
N
proton
transfer
N
H OH2
OH
NH2 O
NH2 O
H2O
N
N
N
N
OH2
N
H
OH2
H2O
H
N
N
NH2 O
NH2 O
proton
OH
transfer
H OH2
H
N
H H
O
H3O+
H2O
N
N
+ H3O+
N
H2O
N
H
21.74 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
21.75
O
H218O
18O
H+
overall
O
OH
H+
OH
18
OH
+
H218O
H
H218O
+
OH H
18
OH
+ H318O
OH2
18
OH
18O
+ H2O
H
H218O
+ H318O
18
O
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Aldehydes and Ketones 21–27
21.76
H H OH
a.
O
OCH2CH3
O
OCH2CH3
O
H
H OH2
+
OH
O
H
H2O
H OH
H OH2
+ H OCH2CH3
H2O
O
H3O+
O
O
HO
H O
HO
H
OH + H2O
O
H
H
O
b.
H OH2
H
OH
OH
+
H
H
O
O
OH
H
H OH2
O
OH
H OH
OH2
H
OH
O
O
O
H
H OH
H OH
H
O
H2O
H
O
O
O
O
H OH
+ H3O+
O
21.77
OCH3
OCH3
+
CH3O O
HO
H OTs
enol ether
CH3O
OH
H
O
O
O
acetal
O
OH
OH
+ OTs
H OTs +
H OTs
OTs
H
O
O
H
CH3O
O
OH
OTs
+ CH3OH
OH
+ OTs
562
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Chapter 21–28
21.78
H O
HO
O
NH2
+ CH3
HO
C
N C H
H CH3
H
HO
H OH2
H OH
proton
HO
CH3
transfer
HO
N C H
HO
dopamine
H OH2
HO
HO
HO
NH
HO
NH
HO
H
CH3
H3O
HO
CH3
N H
HO
CH3
CH3
H2O
salsolinol
N C H
(+ 3 more resonance
structures)
C
HO
H2O
H
H2O
21.79
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
21.80 Hemiacetal A is in equilibrium with its acyclic hydroxy aldehyde. The aldehyde can undergo
hydride reduction to form 1,4-butanediol and Wittig reaction to form an alkene.
OH
O
a.
OH
O
OH
C
A
OH
H
This can now be reduced with NaBH4.
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)
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21.81
O
H OH2
CH3–MgI
O
HO
OCH3
HO
OCH3
OCH3
OCH3
OCH3
OCH3
H2O
5,5-dimethoxy-2-pentanone
OCH3
OCH3
H2O
H
H OH2
H
HO
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
OH
O
H3O
OH
H
(+ 1 resonance
structure)
Y
H2O
H O CH3
21.82
cyclopropenone
(1640 cm–1)
O
O
2-cyclohexenone
(1685 cm–1)
O
O
These three resonance structures include an
aromatic ring; 4n + 2 = 2 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
O
There are three resonance structures for 2cyclohexenone, 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.
21.83
CHO
a.
and
c.
O
aldehyde
ketone
The sp2 hybridized C–H bond of the aldehyde
absorbs at 2700–2830 cm–1.
b.
and
O
higher wavenumber
for C=O
O
conjugated with a
benzene ring
lower wavenumber
O
and
O
smaller ring
higher wavenumber
for C=O
564
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Chapter 21–30
21.84
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
2 H's adjacent to 3 H's
CH2
a monosubstituted benzene ring
O
21.85
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
Ha Hc
Hb
21.86
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
–1
IR absorption at ~2700 cm–1 CH of RCHO
IR absorption at ~2700 cm 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
H
CH2 CH3
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Aldehydes and Ketones 21–31
21.87
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)
21.88
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)
CHO
21.89
OH
OH
OH
a. HO
b.
CHO
HO
HO
HO
OH
OH
CHO
OH
21.90
OH
OH
O
HO
HO
OH
H Cl
OH
O
HO
HO
OH
OH
-D-glucose
CH3OH
HO
HO
above
OH
O
OH
OH
H
O
below
OH
Cl
+
HCl
acetal
OH
O
HO
HO
OCH3
CH3
OH
CH3OH
O
HO
HO
OH
O
CH3
H
The carbocation is trigonal planar, so CH3OH attacks from two
different directions, and two different acetals are formed.
HO
HO
O
OH
+
acetal
OCH3
O
OH
H2O
Cl
O
HO
HO
HO
HO
OH
Cl
OH
OH
O
HO
HO
OH2
OH
HCl
566
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Chapter 21–32
21.91
H+
H
O
H
O
OH
O
OH
O
OH
O
O
OH
O
OH
O H
O
O
H2O
H3O
21.92
O
a.
O
brevicomin
O
HO
b.
O
OH
Br
O
[1] Ph3P
O
O
Br [2] BuLi
H+
PPh3
CH3CH2CHO
brevicomin
O
H3O+
OH
H3O+ O
OH
O
[1] OsO4
O
PCC
O
CH=CHCH2CH3
[2] NaHSO3, H2O
OH
CH3CH2CH2OH
(E and Z isomers)
OH
21.93
OH
a.
HO
HO
acetal carbon
OCH3
O
c.
OH
HO
O
HO
O
CH3O
CH3O
CH3O
OH
O
HO
(OH can be up or down.)
OH
OH
[2] CH3OH, HCl
HO
HO
O
OH
HO
O
HO
O
HO
OCH3
(OCH3 can be up or down.)
OCH3
[3] NaH (excess)
CH3I (excess)
O
CH3O
OH
(OH can be up or down in both
products.)
hemiacetal carbon
HO
OH
HO
CH3O
OH
HO
+
b. [1] H3O HO
OCH3
O
CH3O
CH3O
O
CH3O
O
CH3O
OCH3
O
OCH3
OCH3
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Their Derivatives —
Nucleophilic Acyl
Substitution
567
Carboxylic Acids and Their Derivatives 22–1
C
Deerriivvaattiivveess—
—N
Nuucclleeoopphhiilliicc A
Accyyll SSuubbssttiittuuttiioonn
Chhaapptteerr 2222:: C
Caarrbbooxxyylliicc A
Acciiddss aanndd TThheeiirr D
SSuum
mm
maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ooff R
RC
CO
OZZ ((2222..55))
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
13
C NMR absorption
•
•
C–H to the C=O absorbs at 2–2.5 ppm.
N–H of an amide absorbs at 7.5–8.5 ppm.
•
C=O absorbs at 160–180 ppm.
SSuum
mm
maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ooff R
RC
CN
N ((2222..55))
•
•
IR absorption
13
C NMR absorption
CN absorption at 2250 cm–1
CN absorbs at 115–120 ppm.
––
weeeenn tthhee bbaassiicciittyy ooff ZZ– aanndd tthhee pprrooppeerrttiieess ooff R
SSuum
mm
maarryy:: TThhee rreellaattiioonnsshhiipp bbeettw
RC
CO
OZZ
• 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)
G
Geenneerraall ffeeaattuurreess ooff nnuucclleeoopphhiilliicc aaccyyll ssuubbssttiittuuttiioonn
• The characteristic reaction of compounds having the general structure RCOZ is nucleophilic acyl
substitution (22.1).
• The mechanism consists of two basic 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).
568
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Substitution
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Chapter 22–2
N
Nuucclleeoopphhiilliicc aaccyyll ssuubbssttiittuuttiioonn rreeaaccttiioonnss
[1] Reactions that synthesize acid chlorides (RCOCl)
[a] From RCOOH
(22.10A):
O
R
O
+
C
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
O
C
Cl–
+
R'
O
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
H2O
2
R
C
Cl–
OH
O
+
OR'
H2O
(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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Their Derivatives —
Nucleophilic Acyl
Substitution
569
Carboxylic Acids and Their Derivatives 22–3
[b] From (RCO)2O
(22.9):
O
R
C
O
O
O
C
+
R'OH
R
O
[c] From RCOOH
(22.10C):
R
C
OH
+
+
RCOOH
OR'
O
H2SO4
R'OH
C
R
R
C
+ H2O
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):
O
R
C
O
O
C
[c] From RCOOH
(22.10D):
R
R
NH3
R
(2 equiv)
R
OH
R
[2] 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
C
+
NH2
O
+
O
C
C
+ H2O
NHR'
O
+
NH3
OR'
R
C
NH2
+
R'OH
N
Niittrriillee ssyynntthheessiiss ((2222..1188))
Nitriles are prepared by SN2 substitution using unhindered alkyl halides as starting materials.
R X
+
–CN
R C N
X–
+
SN2
R = CH3, 1o
R
Reeaaccttiioonnss ooff nniittrriilleess
[1] Hydrolysis (22.18A)
R C N
O
H2O
(H+ or –OH)
R
C
O
OH
(with acid)
or
R
C
O–
(with base)
570
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Chemistry, Third Edition
Substitution
Chapter 22–4
[2] Reduction (22.18B)
[1] LiAlH4
[2] H2O
R CH2NH2
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
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Text
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Their Derivatives —
Nucleophilic Acyl
Substitution
571
Carboxylic Acids and Their Derivatives 22–5
C
Chhaapptteerr 2222:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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
CH3 NH2
H
C
O
Cl
CH3
O
O
C
C
NH2
H
C
Cl
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
572
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Chapter 22–6
O
c. CH3CH2CON(CH3)CH2CH3
e.
re-draw
CH3CH2
O
C
O
C
benzoic propanoic anhydride
O
N
acyl group =
propanoic
N-ethyl-N-methyl
N-ethyl-N-methylpropanamide
acyl group =
propanamide
1
O
d.
C
H
acyl group =
benzoic
OCH2CH3
CN
f.
3
alkyl group = ethyl
acyl group =
formate
3-ethylhexanenitrile
6 carbon chain =
hexanenitrile
ethyl formate
22.5
a. 5-methylheptanoyl chloride
d. N-isobutyl-N-methylbutanamide
g. sec-butyl 2-methylhexanoate
O
O
O
N
Cl
b. isopropyl propanoate
O
e. 3-methylpentanenitrile
h. N-ethylhexanamide
O
O
CN
O
O
H
O
O
O
N
H
f. o-cyanobenzoic acid
c. acetic formic anhydride
OH
CH3
CN
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.
22.6
22.7
a. CH3
O
O
O
C
C
C
OCH2CH3
and CH3
N(CH2CH3)2
c. CH3CH2CH2
amide: C=O at
lower wavenumber
O
b.
and
O
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
O
O
O
anhydride:
2 C=O peaks
NH2
1° amide: 2
N–H absorptions
O
and
C
Cl
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Their Derivatives —
Nucleophilic Acyl
Substitution
573
Carboxylic Acids and Their Derivatives 22–7
22.8
Hb
O
Hc signal from the 5 H's
on the aromatic ring
Ha
O
H H
CH3
A
Ha
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)
O
H H
O
Hb
Hb signal from the
10 H's on the two
aromatic rings
B
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)
22.9
NH2
H2N
H H H
N
S
O
HO
CH3
N
H H H
S
N
O
CH3
N
O
O
COOH
H COOH
cephalexin
(Trade name: Keflex)
amoxicillin
a. 4 stereogenic centers
b. 24 = 16 possible stereoisomers
c. enantiomer
a. 3 stereogenic centers
b. 23 = 8 possible stereoisomers
c. enantiomer
H2N
NH2
H H H
N
S
O
HO
H H H
S
N
CH3
N
CH3
O
N
O
O
H COOH
COOH
22.10 To draw the products of these nucleophilic acyl substitution reactions, find the nucleophile and
the leaving group. Then replace the leaving group with the nucleophile and draw a neutral
product.
nucleophile
O
a.
CH3
C
nucleophile
CH3OH
Cl
CH3
O
O
C
C
OCH3
+ HCl
b. CH3
O
NH3
OCH2CH3
CH3
C
NH2 + HOCH2CH3
leaving group
leaving group
22.11 More reactive acyl compounds can be converted to less reactive acyl compounds.
a.
CH3COCl
CH3COOH
more reactive
YES
b. CH3CONHCH3
less reactive
NO
c.
less reactive
CH3COOCH3
more reactive
CH3COOCH3
less reactive
d.
(CH3CO)2O
more reactive
CH3COCl
NO
more reactive
YES
less reactive
CH3CONH2
574
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Substitution
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Chapter 22–8
22.12 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
OCH
strongest base
least reactive
b.
2
CH3CH2
C
OCH3
NH2
a.
NH
O
Cl
weakest base
most reactive
3
intermediate
O
O
O
C
C
C
CH3CH2
NHCH3
NHCH
OH
CH3CH2
OH
intermediate
3 strongest base
least reactive
Cl
O
O
C
CH2CH3
OOCCH
2CH3 weakest base
most reactive
22.13
CH3
O
O
O
O
C
C
C
C
O
CH3
Cl3C
O
The Cl atoms are electron withdrawing,
which makes the conjugate base (the
leaving group, CCl3COO–) weaker
and more stable.
CCl3
acetic anhydride
trichloroacetic anhydride
22.14
O
H2O
a.
O
O
OH
+
pyridine
Cl
O
b.
CH3COO
N
–
H Cl
c.
NH2 + NH4+Cl–
NH3
excess
O
O
O
CH3 + Cl–
(CH3)2NH
N(CH3)2
d.
+ (CH3)2NH2 Cl–
excess
22.15 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
O
OH
+
OCH3
O Cl
OCH3
O
Cl
O Cl
O
H
N
H
OCH3
OCH3
N
OCH3
OCH3
O
O
OCH3
A
Cl
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Their Derivatives —
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Substitution
575
Carboxylic Acids and Their Derivatives 22–9
22.16
O
O
O
O
O
OH
H2O
O
NH3
HO
a.
c.
O
b.
CH3OH
NH2
O
O
NH4
excess
O
CH3
O
O
(CH3)2NH
HO
d.
O
N(CH3)2
excess
O
(CH3)2NH2
22.17 Nucleophilicity decreases across a row of the periodic table so an NH2 group is more
nucleophilic than an OH group.
more nucleophilic
Acetylation occurs here.
H
N
NH2
HO
C
CH3
O
HO
acetaminophen
(active ingredient in Tylenol)
less nucleophilic
22.18 Reaction of a carboxylic acid with thionyl chloride converts it to an acid chloride.
O
a. CH3CH2
C
O
SOCl2
OH
CH3CH2
C
O
C
b.
Cl
O
[1] SOCl2
OH
O
[2] (CH3CH2)2NH (excess)
C
Cl
C
N(CH2CH3)2
(CH3CH2)2NH2 Cl–
22.19
COOH + CH3CH2OH
a.
H2SO4
C
OCH2CH3 + H2O
O
O
COOH
b.
+
OH
C
H2SO4
+ H2 O
O
O
COOH
c.
+
C
NaOCH3
O
Na+
+ CH3OH
576
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Chapter 22–10
O
O
H2SO4
d. HO
+ H2O
OH
O
22.20
O
C
O
CH318OH
OH
C
18
OCH3 + H2O
22.21
O
C
HO
OH
O
HOSO3H
H
C
HO
HO OH
O
H
O
HO OH
H
HSO4–
+
O
–OSO
O
+ H2SO4
3H
HSO4–
O H
O
HO OH2
O
O
H2O +
HOSO3H
+ HSO4–
22.22
O
a. O
CH3NH2
O H3NCH3
OH
b.
c.
+
O
O
CH3NH2
DCC
O
O
[1] CH3NH2
OH
NHCH3
OH
NHCH3 + H2O
[2] 22.23
R
OH
OH
OH
O H
O H
C
C
C
C
C
OR'
R
OR'
R
OR'
R
product of step [1]
OH
R
O H
OH
R
product of step [5]
22.24
O
[1] H3O+
OH
+
HO
O
+
HO
O
a.
O
CH3O
CH3O
[2] H2O, –OH
octinoxate
CH3O
O
C
OH
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–11
O
[1] H3O+
O
b.
OH
O
+
HO
+
HO
OH
O
OH
[2] H2
octyl salicylate
O, –OH
O
OH
22.25
O
a.
CH3CH2
–OH,
H2O
C 18
OCH3
18
O
CH3CH2
C
O
+ H18OCH3
b.
O–
CH3CH2
–OH,
C
18
O
H2O
CH3CH2
OCH3
C
O
O
CH3CH2
C 18
O
+ HOCH3
22.26
HOCH2
O
OH
HOCH2
HO
RCOOH, H2SO4
O
O
HO
RCOOCH2
OH
OH
RCOO
CH2OH
a long chain fatty acid
RCOO OOCR
OOCR
CH2
O
O
CH2OOCR
OOCR
O
RCOO
sucrose
olestra
22.27
CH2OCO(CH2)15CH3
CH2 OH
Na+ OOC(CH2)15CH3
CH OH
Na+ OOC(CH2)15CH3
CH2 OH
Na+ OOC(CH2)7CH=CH(CH2)7CH3
hydrolysis
CHOCO(CH2)15CH3
CH2OCO(CH2)7CH=CH(CH2)7CH3
cis
glycerol
soap
cis
22.28
O
O
NH
H2O, OH
OH
proton
transfer
O
N H
H
O
NH
O
O
NH2
22.29 Aspirin has an ester, a more reactive acyl group, but 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.
578
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Chapter 22–12
ester
O
C
amide
H
N
CH3
O
COOH
CH3
O
HO
acetylsalicylic acid
C
acetaminophen
22.30
"Regular" amide is not hydrolyzed.
H H H
N
S
R
O
H3
H H
N
R
O+
O
N
O
H
S
HN
OH
O
COOH
COOH
22.31
O
H
N
O
H
N
N
H
N
H
O
O
nylon 6,10
O
Cl
Cl
H2N
+
NH2
O
22.32
OH
+
HOOC
COOH
HO
1,4-dihydroxymethylcyclohexane
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.
terephthalic acid
O
O
O
O
O
O
O
O
Kodel
22.33
O
O
OH
OH
+
OH
OH
Reaction occurs here (–H2O).
O
O
O
O
O
O
O
O
PLA
polyl(lactic acid)
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–13
22.34 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
HN
HO
glucosamine
O
NAG
22.35
a. CH3CH2CH2 Br
NaCN
CH3CH2CH2 CN
H2O, –OH
c.
CN
CN
H2O, H+
b.
COO
COOH
CN
COOH
22.36
a.
CH3
O
OH
NH
C
C
C
NH2
CH3
O
b.
C
c.
NH
CH3CH2
NH2
OH
CH3CH2
C
O
OH
C
NHCH3
NCH3
22.37
a. CH3CH2 Br
[1] NaCN
[2] LiAlH4
[3] H2O
[1] DIBAL-H
b. CH3CH2CH2 CN
CH3CH2 CH2NH2
O
CH3CH2CH2 C
[2] H2O
H
22.38
OCH3
a.
[1] CH3CH2MgCl
CN
[2] H2O
OCH3
O
C
CH2CH3
O
CN
C
[1] C6H5Li
b.
[2] H2O
22.39
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
[2] H2O
O
CH2 C
CH2 CN
C(CH3)3
d.
CH2 C
H3O+
OH
580
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Chapter 22–14
22.40
CH3 C N
O
O
[1] CH3CH2MgBr
[1] CH3MgBr
CH3CH2 C N
[2] H2O
[2] H2O
22.41
O
Cl
a.
g.
2,2-dimethylpropanoyl chloride
N
O
N-cyclohexylbenzamide
H
O
b.
h.
cyclohexyl pentanoate
m-chlorobenzonitrile
O
Cl
CN
c.
O
O
O
i.
Cl 3-phenylpropanoyl chloride
isobutyl 2,2-dimethylpropanoate
d.
2-ethylhexanenitrile
O
j.
C
CN
O
e.
Cl
O
cis-2-bromocyclohexanecarbonyl chloride
Br
O
O
k.
N
cyclohexanecarboxylic anhydride
N,N-diethylcyclohexanecarboxamide
O
O
f.
cyclopentyl
cyclohexanecarboxylate
l.
phenyl phenylacetate
O
O
22.42
a. propanoic anhydride
O
e. isopropyl formate
H
b. -chlorobutyryl chloride
O
O
O
O
i. benzoic propanoic anhydride
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
C
l. N,N-dibenzylformamide
h. vinyl acetate
O
O
O
O
NH2
O
CH3
H
N
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–15
22.43 Rank the compounds using the rules from Answer 22.12.
O
a.
O
O
NH2
OCH2CH2CH3
NH
OCH
2 strongest base
least reactive
Cl
2CH2CH3
intermediate
O
b.
Cl
O
O
O
O
ester
least reactive
anhydride
intermediate
O
weakest base
most reactive
O
F3C
O
O
CF3
anhydride with electronwithdrawing F's
most reactive
O
O
c.
OH
SH
OH
Cl
SH
intermediate
strongest base
least reactive
Cl
weakest base
most reactive
22.44
a. Better leaving groups make acyl compounds more reactive. A has an electron-withdrawing NO2
group, which stabilizes the negative charge of the leaving group, whereas B has an electron-donating
OCH3 group, which destabilizes the leaving group.
O
CH3
C
O
O
NO2
CH3
A
C
OCH3
O
B
an electron-withdrawing substituent an electron-donating substituent
O
O
O
N
O
N
O
O
OCH3
O
OCH3
O
one possible
resonance structure
leaving group from A
one possible
resonance structure
leaving group from B
Adjacent negative charges destabilize
the leaving group.
Delocalizing the negative charge on the
NO2 stabilizes the leaving group
making A more reactive than B.
resonance structures for the leaving group
b.
O
O
C N
N
R
imidazolide
Nu
R C N
Nu
O
N
R
C
Nu
N
N
N
N
N
N
N
N
N
The leaving group is both resonance stabilized and aromatic
(6 electrons), making it a much better leaving group than
exists in a regular amide.
N
582
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Chapter 22–16
22.45
Reaction as an acid:
O
CH3
C
O
B
NH2
CH3
O
C
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.
22.46
CH3CH2CH2CH2COCl
NH3
H2O
a.
CH3CH2CH2CH2COOH
pyridine
b.
d.
N Cl–
H
CH3CH2OH
c.
O
CH3COO
CH3CH2CH2CH2
O
CH3
CH3CH2CH2CH2CON(CH2CH3)2
excess
N Cl–
H
O
(CH3CH2)2NH2 Cl–
C6H5NH2
f.
Cl–
excess
CH3CH2CH2CH2CONHC6H5
C6H5NH3 Cl–
22.47
O
O
O
a.
SOCl2
d.
no reaction
H2O
2
NaCl
no reaction
O
O
b.
(CH3CH2)2NH
OH
e.
excess
N(CH2CH3)2
O
O
CH3OH
c.
O H2N(CH2CH3)2
OCH3
O
O
f.
OH
NH4 Cl–
(CH3CH2)2NH
e.
CH3CH2CH2CH2COOCH2CH3
pyridine
CH3CH2CH2CH2CONH2
excess
CH3CH2NH2
excess
NHCH2CH3
O
O
H3NCH2CH3
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Nucleophilic Acyl
Substitution
583
Carboxylic Acids and Their Derivatives 22–17
22.48
C6H5CH2COOH
a.
O
NaHCO3
O Na
b.
NaOH
O
c.
SOCl2
O
O
CH3OH
H2SO4
+ H2CO3 g.
OCH3
O
CH3OH
h.
+ H2 O
–OH
O Na
i.
O
[1] NaOH
[2] CH3COCl
O
Cl
d.
NaCl
no reaction
j.
O
CH3NH2
DCC
e.
NH3
O
k.
(1 equiv)
O NH4
f.
O
[1] NH3
[2] l.
NH2
O
O
NHCH3
[1] SOCl2
O
[2] CH3CH2CH2NH2
NHCH2CH2CH3
O
[1] SOCl2
[2] (CH3)2CHOH
OCH(CH3)2
22.49
c.
CH3CH2CH2CO2CH2CH3
O
H2O, –OH
HO
O
O
a.
SOCl2
no reaction
d.
NH3
NH2
O
b.
HO
O
H3O+
HO
OH
e. CH3CH2NH2
NHCH2CH3
HO
22.50
NH2
a.
H3O+
CH2COOH
b.
CH2COO
H2O, –OH
O
22.51
a.
H3O+
COOH
d.
[1] CH3CH2Li
[2] H2O
C6H5CH2CN
b.
H2O, –OH
COO
[2] H2O
C
[2] H2O
O
[1] LiAlH4
CH2NH2
f.
O
H
e. [1] DIBAL-H
CH3
c. [1] CH3MgBr
O
[2] H2O
584
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Chapter 22–18
22.52
O
COOH
C
SOCl2
a.
OH
g.
O
h. C6H5CH2COOH
+
+
C6H5
N
H
(excess)
N
C6H5
[2] H2O
d. (CH3)2CHCOOH +
H3O+
C6H5CH2CH2CH2COOH
HOOC
O
COOH
(CH3CO)2O +
NH2
O
H2O
–OH
NHCOCH3
(excess)
+
NH2
O
O
+ CH3COO– H3N
O
O
f.
C6H5CH2CH2CH2CN
O
k.
NHCOCH3
O
C6H5(CH2)2NH(CH2)3CH3
OCH(CH3)CH2CH3
CH3
e.
CH3CH2CH2CH2
O
j.
C
(CH3)2CH
[1] SOCl2
CH2CH2CH3
O
H2SO4
CH3CH2CHOH
i.
C
[2] H2O, –OH
[2] CH3CH2CH2CH2NH2
[3] LiAlH4
[4] H2O
N Cl–
H H
O
[1] CH3CH2CH2MgBr
c. C6H5CN
CH3CH2CH2CH2Br
OH
b. C6H5COCl
O
[1] NaCN
Cl
H3O+
OH
l.
C6H5CH2CH2COOCH2CH3
OH
O
H2O
–OH
C6H5CH2CH2COO–
+
O
HOCH2CH3
22.53 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
NaOH
b.
O
O
O
O
X
H2O
O
CO2 Na+
OH
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–19
22.54
Br NaCN
CN
H3O+
COOH SOCl2
A
COCl [1] (CH3)2CuLi
C
[2] H2O
O
D
B
[1] LiAlH4
CH3OH,
[2] H2O
CH3
E
H+
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
J
K
L
[1] CH3MgBr
[2] H2O
O
M
22.55
O
H D
C CH3
a.
CH3COCl
OH
O
c.
pyridine
CH3
CH3
Br
b.
CH3
NaCN
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)
C6H5
CH3
H
NH
C
O
COOCH2CH3
+
CH3
22.56 Hydrolyze the amide and ester bonds in both starting materials to draw the products.
a.
O
O
CO2CH2CH3
H2O
O
CO2H
O
HO
N
H
OH
H2N
NH2
oseltamivir
NH2
C6H5
C
H
NH3
+
Cl–
586
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Chapter 22–20
NH2
H
N
HOOC
b.
NH2
H2O
O
CH3O
HOOC
H2N
+ CH3OH
OH
COOH
O
O
aspartame
phenylalanine
22.57
O
C
Cl
a.
O
C Cl
C Cl
H
O
HO
O
b.
NHNH2 + Cl
+ HN
N
O
C
NH2NH
NH2NH
NH2NH2
O
O
O
O
O
proton
O
OH
transfer HO
OH
H+
O
OH
22.58
H+
O
CH3
C
OH
OH
CH3
C
OH
OH
18
O H
H
18
CH3 C OH2
18
18
O H
+ H2 O
OH
CH3 C OH
CH3 C OH
O H
OH
CH3 C
18
+ H2O
O H
O H
CH3
C
O
H2O
18
CH3
OH
C
18
+ H3O+
OH
A
OH
CH3 C
18
O H
A
OH
CH3
18
O H
A
+ H3O+
+ H2 O
Two possibilities for A:
OH
CH3 C
C 18
O H
OH
H2O
CH3
C
18
O
+ H3O+
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–21
22.59
H2O
H
H2O
O H
O
O
OH
OH
OH
HOH2
O
O
OH
O
OH
O
H2O
OH
H
-butyrolactone
HOH2
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
CH2O C
CH3
inactive enzyme
A
O H
OH
O
CO2H
OH
CO2H
+
CH3
salicylic acid
A
588
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Substitution
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Chapter 22–22
22.61
O
R
O
C
O R' +
OH
This bond is not broken.
H CH3
O
CH3
C
O R CH2CH3
X
C
mechanism
CH3
C
+ R'OH
O
H CH3
O
accepted
C
R
H2O
SN2
O
H CH3
O
HO
C
CH2CH3
(2R)-2-butanol
CH3
C
O
X
C
CH2CH3
According to the accepted
mechanism, the stereochemistry
around the stereogenic center is
retained in the product.
This bond is cleaved.
H CH3
O
CH3
C
C
O
CH2CH3
CH3
C
H2 O
SN2 alternative
X
CH3 H
O
OH
O
CH3CH2
C
OH
(2S)-2-butanol
This SN2 mechanism would form the
product of inversion leading to (2S)-2butanol. Since (2R)-2-butanol is the
only product formed, the SN2
mechanism does not occur during
ester hydrolysis.
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
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–23
22.63
H OCH2CH3
O
OH
O
O
H–Cl
A
Cl
H
O
O CH2CH3
OCH2CH3
OH
OH
O
Cl
H–Cl
OCH2CH3
OH2
OCH2CH3
O
OH
OCH2CH3
O
O
OH
OCH2CH3
OH
O
H
Cl
H
H–Cl
Cl
Cl
CO2CH2CH3
OCH2CH3
OCH2CH3
O
Cl
+
Cl
Cl
=
O
H2 O
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
+ H2O
O
O
O
O
HO
O H
HO
HO
+ H3O+
H2O
OH2
OH
+ H2O
590
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Chapter 22–24
22.65
(any base) B
O
NH2
OH
+
CH3CH2O
X
C
H N
OCH2CH3
O
C
OCH2CH3
O
OCH2CH3
C
H N
OH
OCH2CH3
OCH2CH3
OH
+ HB+
diethyl carbonate
O
O
HN
H
HN
+ CH3CH2O–
O
Y
O
OCH2CH3
H N
O
O
H N
H
C
OCH2CH3
OH
B
+ HB+
HB+
O
OCH2CH3
+ CH3CH2O–
CH3CH2OH + B
22.66
sp3 C
sp2 C
O
RCH2 Cl
R
less electrophilic C
more crowded C since it is
surrounded by four atoms
C
Cl
O
R
C
Cl
This resonance structure illustrates
how the electronegative O atom
withdraws more electron density
from C.
The sp2 hybridized C of RCOCl is much less crowded,
and this makes nucleophilic attack easier as well.
more electrophilic C
due to electron-withdrawing O
more reactive
22.67
O
H3
HO
CN
O+
Ha
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 H c
A
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591
Carboxylic Acids and Their Derivatives 22–25
proton
HO
C N
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
NH2
O H
H O
H OH2
O
H O
NH3
O
O
NH3
O
H3O
H2O
H2O
O
22.68 Fischer esterification is 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
O
b.
OH
O
HO
O
OH
O
O
d.
HO
OH
HO
O
O
O
22.69
NH3
Br excess
a.
NH2
H3O+
NaCN
Br
H
N
OH
NH2
CN
O
DCC
O
NaOH
b.
Br
OH
H2SO4
OH
OH
(from a.)
(from a.)
O
O
[1]
c.
O
CN
[2] H2O
MgBr
Mg
Br
O
MgBr
592
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Chapter 22–26
O
d.
OH
(from b.)
CrO3
OH
H2SO4, H2O
H2SO4
OH
O
O
(from c.)
[1]
MgBr
e.
OH
(2 equiv)
O
[2] H2O
O
(from d.)
(from c.)
[1]
f.
MgBr
OH
(2 equiv)
O
[2] H2O
O
(from b.)
22.70
a.
Br
–CN
H2O,
CN
b.
CN
H+
COOH
(from a.)
c.
COOH
SOCl2
COCl
(from b.)
d.
COOH
(from b.)
e.
HOCH2CH3
CO2CH2CH3
H2SO4
[1] CH3MgBr
CN
(from a.)
C
[2] H2O
CH3
O
[1] DIBAL-H
f.
CN
(from a.)
g.
CHO
[2] H2O
[1] LiAlH4
CN
(from a.)
h.
CH2NH2
(from g.)
CH2NH2
[2] H2O
CH3COCl
CH2NHCOCH3
22.71
a. CH3Cl
+ NaCN
CH3 CN
H3O+
CH3 COOH
[1] CO2
CH3 Cl + Mg
Br
+ NaCN
b.
sp2
Br
+ Mg
CH3 MgCl
[2] H3O+
CH3 COOH
This method can't be used because an SN2 reaction
can't be done on an sp2 hybridized C.
MgBr
[1] CO2
[2] H3O+
COOH
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Substitution
Carboxylic Acids and Their Derivatives 22–27
This method can't be used because an SN2 reaction
can't be done on a 3° C.
c. (CH3)3CCl + NaCN
(CH3)3C
(CH3)3C
Cl + Mg
d. HOCH2CH2CH2CH2Br
[1] CO2
MgCl
(CH3)3C
[2] H3O+
H3O+
HOCH2CH2CH2CH2 CN
+ NaCN
COOH
HOCH2CH2CH2CH2 COOH
This method can't be used because you can't make
a Grignard reagent with an acidic OH group.
HOCH2CH2CH2CH2 Br + Mg
22.72
CH3OH
O
SOCl2
CH3
CH3Cl
HNO3
H2SO4
AlCl3
CH3
COOH
KMnO4
O2N
H2SO4
O 2N
C
CH3CH2OH
OCH2CH3
O2N
H2
Pd-C
(+ ortho isomer)
O
C
OCH2CH3
H2N
22.73
O
NH2
HO
N
H
N
H
HO
CH3COCl
NaH
N
H
O
N
H
SOCl2
serotonin
O
N
H
CH3Cl
CH3COOH
SOCl2
CH3OH
O
CrO3
H2SO4, H2O
N
H
CH3O
CH3CH2OH
N
H
melatonin
22.74
O
O
CH3CH2Cl
a.
OH
COH
KMnO4
AlCl3
OH
CNH2
NH3
OH
OH
OH
salicylamide
NO2
HNO3
Cl
(2 equiv)
OH
(+ para isomer)
b.
SOCl2
O
H2, Pd-C
H2SO4
HO
(+ ortho isomer)
HO
NH2
H
N
CH3COCl
HO
(More nucleophilic
NH2 reacts first.)
C
CH3
O
acetaminophen
594
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Chapter 22–28
H
N
c.
C
CH3
NaH
O
HO
H
N
C
CH3
H
N
CH3CH2Br
O
O
CH3
O
CH3CH2O
acetaminophen
(from b.)
C
p-acetophenetidin
22.75
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
C
CH3OH
O2N
H
O2N
O2N
(+ ortho isomer)
H2, Pd-C
O
C
O
c.
FeBr3
Br
HNO3
H2SO4
O
O
OCH3
N
H
Br2
OCH3
+
C
Cl
[1] CrO3, H2SO4,
H2O
[2] SOCl2
Br
O2N
(+ ortho isomer)
H2N
CH3CH2CH2OH
O
Br
H2, Pd-C
Br
Cl
H2N
OCH3
[1] CrO3, H2SO4,
H2O
[2] SOCl2
CH3CH2OH
H
N
O
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Their Derivatives —
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Substitution
Carboxylic Acids and Their Derivatives 22–29
(CH3)3COH
CH3OH
HCl
SOCl2
(CH3)3CCl
d.
CH3Cl
AlCl3
KMnO4
SOCl2
HO
AlCl3
Cl
(+ ortho isomer)
Br
Br2
FeBr3
Br
CH3Cl
O
O
Br
KMnO4
Br
NaOH
HO
AlCl3
(+ ortho isomer)
O
O
O
O
O
O
(CH3)3C
Br
22.76
CH3Cl
a.
CH3
AlCl3
CH3OH
SOCl2
Br2
CH2Br
h
NaCN
CH2CN
H2O
H+
CH2COOH
HOCH2CH3
H2SO4
CH3Cl
CH2COOCH2CH3
ethyl phenylacetate
NO2
(from a.)
b.
CH3
CH3Cl
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
H2SO4, H2O
O
benzyl acetate
CH3COOH
22.77
[1] LiAlH4
[2] H2O
Cl
a. Cl
–CN
NC
excess
CN
O
H2O
H2SO4
NH2
H2N
OH
HO
O
b.
Br2
h
Br
KOC(CH3)3
[1] O3
CHO
CrO3
COOH
[2] Zn, H2O
CHO
H2SO4, H2O
COOH
596
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Chapter 22–30
22.78
O
a. CH3
C
O
CH313CH2OH
OH
d. CH3 CH2OH
–
13
CH3 CH2Br
13C
CH3
H2SO4
OH
O
CH3
18
OH
OCH2CH3
CH3COCl
CH3CH218OH
+ base (H18O–)
PBr3
C
CH3
H218O
13
O
CH3CH2OH
13
H2SO4, H2O
c. CH3CH2Br
13
OCH2CH3
O
CrO3
b. CH313CH2OH
C
CH3
H2SO4
CH3
CH218OH
C
18
OCH2CH3
18O
CrO3
13
H2SO4, H218O
18O
13 C 18
CH3
OH
CH3CH2OH
H2SO4
13 C
CH3
+
OCH2CH3
H218O
22.79
a. HO
OH
HO
and
C
C
O
O
O
O
O
OH
O
O
O
O
O
O
NH
b. ClOC
COCl
and
NH2
H2N
O
HN
NH
O
HN
22.80
O
a.
O
O
O
O
O
O
O
OH
HO
O
O O
O O
b.
O
O
HO
OH
O
O O
O
O
HO
OH
HO
OH
O
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Their Derivatives —
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Substitution
597
Carboxylic Acids and Their Derivatives 22–31
22.81
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
=
RO
OH
HO
O
docetaxel
carbamate
H
O
NHR
O
O
O
O
RO
O
NHR
RO
1
most stable
O
NHR
O
RO
4
least stable
NHR
RO
3
2
More basic N atom allows N to
donate electron density more
than O, so this structure
contributes more than 3 to the
hybrid.
Increasing stability: 4 < 3 < 2 < 1
H OH2
O
c.
OH
O
O
NHR
O
NHR
H
O
C
H
H2O
NHR
H2O
H2O
NHR
H OH2
O
C
+ NH2R
H OH2
O
H3O
H2O
H3NR
OH
O
d.
O
O
OH
O
N
H
O
H3O+
OH
docetaxel
HO
O
O
H
O
O
O
CO2
COOH
OH
H3N
OH
OH
O
O
OH
CH3CO2H
HO
HO
H
HO
HO
O
598
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Chapter 22–32
22.82
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
C=O at < 1700 cm–1
due to the stabilized amide
NH2
2 NH absorptions at 3200–3400 cm–1
C=O absorption higher wavenumber
O
O
and
b.
O
O
d. O
Cl
OH
O
and
CH3
Cl
Acid chloride CO absorbs at
much higher wavenumber.
OH absorption at
3500–3200 cm–1 + C=O
ketone
only C=O
22.83
O
a. C6H5COOCH2CH3
CH3CH2COOCH2CH3
O
b.
most resonance
stabilized
least resonance
stabilized
CH3COOCH3
CH3CONH2
Increasing wavenumber
CH3COCl
Increasing wavenumber
22.84
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
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
O
Cl
O
b.
C4 H7 N
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
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
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
O
O
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
N
H
CH3
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
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Their Derivatives —
Nucleophilic Acyl
Substitution
599
Carboxylic Acids and Their Derivatives 22–33
22.85
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
22.86
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
22.87
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
22.88
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) – CH3 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) – CH2 adjacent to the electronegative O atom and another CH2
600
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Chapter 22–34
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
22.89
O
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.
O
ClCH2 C
ClCH2 C
NH2
N H
4.02 ppm
H
different
environments
7.35 and 7.60 ppm
This resonance structure gives a significant
contribution to the resonance hybrid.
22.90
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.
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Their Derivatives —
Nucleophilic Acyl
Substitution
601
Carboxylic Acids and Their Derivatives 22–35
22.91
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
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of Carbonyl Compounds at
the a Carbon
603
Substitution Reactions of Carbonyl Compounds 23–1
C
Caarrbboonnyyll C
Coom
mppoouunnddss aatt tthhee C
Chhaapptteerr 2233:: SSuubbssttiittuuttiioonn R
Reeaaccttiioonnss ooff C
Caarrbboonn
K
Kiinneettiicc vveerrssuuss tthheerrm
mooddyynnaam
miicc eennoollaatteess ((2233..44))
O
Kinetic enolate
• The less substituted enolate
• Favored by strong base, polar aprotic solvent, low temperature:
LDA, THF, –78 oC
R
kinetic enolate
O
Thermodynamic enolate
• The more substituted enolate
• Favored by strong base, protic solvent, higher temperature:
NaOCH2CH3, CH3CH2OH, room temperature
R
thermodynamic
enolate
H
Haallooggeennaattiioonn aatt tthhee ccaarrbboonn
[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 carbon.
•
The reaction occurs via enolate
intermediates.
Polysubstitution of X for H occurs on the carbon.
-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 bond.
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Chapter 23–2
R
Reeaaccttiioonnss ooff --hhaalloo ccaarrbboonnyyll ccoom
mppoouunnddss ((2233..77C
C))
[1] Elimination to form ,-unsaturated carbonyl compounds
O
R
Li2CO3
Br
LiBr
DMF
O
R
•
Elimination of the elements of Br and H
forms a new bond, giving an ,unsaturated carbonyl compound.
•
The reaction follows an SN2 mechanism,
generating an -substituted carbonyl
compound.
[2] Nucleophilic substitution
O
O
Nu
Br
R
R
Nu
A
Allkkyyllaattiioonn rreeaaccttiioonnss aatt tthhee ccaarrbboonn
[1] Direct alkylation at the carbon (23.8)
O
•
O
C H
C
[1] Base
C R
C
[2] RX
+
X–
The reaction forms a new C–C bond to the
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
[2] RX
[3] H3O+, H
•
R CH2COOH
•
H C COOEt
COOEt
diethyl malonate
[1] NaOEt
[1] NaOEt
[2] RX
[2] R'X
[3] H3O+, The reaction is used to prepare carboxylic
acids with one or two alkyl groups on the carbon.
The alkylation in Step [2] follows an SN2
mechanism.
R CHCOOH
R'
[3] Acetoacetic ester synthesis (23.10)
O
CH3
C H
C H
[1] NaOEt
[2] RX
[3] H3O+, •
O
CH3
C
CH2 R
•
COOEt
ethyl acetoacetate
O
[1] NaOEt
[2] RX
[1] NaOEt
[2] R'X
[3] H3O+, CH3
C
CH R
R'
The reaction is used to prepare
ketones with one or two alkyl
groups on the carbon.
The alkylation in Step [2] follows
an SN2 mechanism.
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Substitution Reactions of Carbonyl Compounds 23–3
C
Chhaapptteerr 2233:: A
Annssw
weerrss ttoo PPrroobblleem
mss
23.1 • To convert a ketone to its enol tautomer, change the C=O to C–OH, make a new double bond to
an 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
H
C6H5
C6H5
H
c.
C6H5
C6H5
O
O
O
OH
O
e.
O
O
O
OH
OH
f.
C6H5
Draw mono enol
tautomers only.
OH
C6H5
O
OH
OH
(Conjugated enols are preferred.)
23.2
O
OH
OH
(E and Z)
2-butanone
23.3
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.
The mechanism has two steps: protonation followed by deprotonation.
OH
OH
H H OH2
O H
H
H
H2O
H
H
O
H
H
H3O+
O
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Chapter 23–4
23.4
Cl
H
H
H
O
O D
D
D Cl
H
H
O D
D Cl
Cl
D H
D Cl
H
O D
O D
H Cl
H Cl
D D
D D
H
H
D Cl
O D
O
Cl
23.5
a.
CH3CH2O
O
O
O
O
O
C
C
C
C
C
C
OCH2CH3
CH3CH2O
H
b.
CH3
C
OCH2CH3
CH3CH2O
H
O
O
O
O
C
C
C
C
C
OCH2CH3
CH3
C
c.
CH3
O
C
C
CHC N
CH3
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
23.6
The indicated H’s are to a C=O or CN 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
OCH2CH3
d. O
O
CH3
23.7
O
O
O
–H+
O
O
–H+
O
O
–H+
O
O
O
no resonance stabilization
least acidic
O
O
O
O
O
O
Two resonance structures
stabilize the conjugate base.
intermediate acidity
Three resonance structures
stabilize the conjugate base.
most acidic
O
O
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Substitution Reactions of Carbonyl Compounds 23–5
23.8
In each of the reactions, the LDA pulls off the most acidic proton.
O
O
O
LDA
a.
c.
THF
CHO
b.
O
LDA
CH3 C
OCH2CH3
CHO
LDA
CN
d.
THF
CH2 C
THF
OCH2CH3
CN
LDA
THF
23.9
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.
23.10 • 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
O
H2O
O
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.
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Chapter 23–6
b.
O
O
O
NaOH
or
H
H
CH3
O
CH3
H2O
H
CH3
H
CH3
(3R)-3-methylcyclohexanone
This stereogenic center is not located at the carbon,
so it is not deprotonated with base. Its configuration is retained in
the product, and the product remains optically active.
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
I
I
O
O
I2, –OH
c.
O
+ HCI3
23.14
O
O
O
a.
Br
Li2CO3
LiBr
DMF
O
CH3SH
c.
Br
O
SCH3
O
CH3CH2NH2
b.
Br
NHCH2CH3
23.15 Bromination takes place on the carbon to the carbonyl, followed by SN2 reaction with the
nitrogen nucleophile.
CH3
Br
O
O
O
N
O
O
O
O
NHCH3
Br2
LSD
CH3CO2H
N
COC6H5
N
COC6H5
N
M
COC6H5
(Section 18.5)
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Substitution Reactions of Carbonyl Compounds 23–7
23.16
O
a. CH3CH2
C
CH2CH3
O
O
[1] LDA, THF
CH3CH2
[2] CH3CH2I
C
O
[1] LDA, THF
CHCH3
O
c.
O
[2] CH3CH2I
CH2CH3
O
O
[1] LDA, THF
[1] LDA, THF
b.
CH2CH2CN
d.
CH2CHCN
[2] CH3CH2I
[2] CH3CH2I
CH2CH3
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
b.
NaOCH2CH3
O
CH3Br
O
NaOCH2CH3
CH3CH2OH
O
O
LDA
c.
(from a.)
CH3CH2OH
THF
O
CH3I
O
CH3Br
O
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Chapter 23–8
O
d.
O
O
LDA
CH3Br
THF
(from b.)
23.20
LDA
THF
O
O
Br
CH3I
O
O
O
O
A
Br2
O
CH3CO2H
O
C
B
CH2
Li2CO3
LiBr
DMF
O
O
-methylene-butyrolactone
23.21 Decarboxylation occurs only when a carboxy group is bonded to the C of another carbonyl
group.
O
COOH
a.
COOH
b.
COOH
O
c.
COOH
d.
COOH
COOH
YES
NO
YES
NO
23.22
H3O+
[1] NaOEt
a. CH2(CO2Et)2
[2]
Br
CH2 CH2COOH
b. CH2(CO2Et)2 [1] NaOEt
[1] NaOEt
H3O+
[2] CH3Br
[2] CH3Br
CH3
CH3 C COOH
H
23.23
COOH
a.
Cl
b.
Cl
Br
O
Br
O
COOH
23.24 Locate the 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
H3O+
[1] NaOEt
CH2(CO2Et)2
[2] (CH3)2CHCH2CH2CH2CH2Br
b.
(CH3)2CHCH2CH2CH2CH2CH2COOH
COOH
[1] NaOEt
[1] NaOEt
[2] CH3CH2CH2Br
[2] CH3CH2CH(CH3)CH2CH2Br
CH2(CO2Et)2
H3O+
COOH
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611
Substitution Reactions of Carbonyl Compounds 23–9
c.
(CH3CH2CH2CH2)2 CHCOOH
H3O+
[1] NaOEt
[1] NaOEt
[2] CH3CH2CH2CH2Br
[2] CH3CH2CH2CH2Br
(CH3CH2CH2CH2)2CHCOOH
CH2(CO2Et)2
23.25 The reaction works best when the alkyl halide is 1° or CH 3X, since this is an SN2 reaction.
a.
(CH3)3C–CH2COOH
COOH
b.
(CH3)3CX
3° alkyl halide
(too crowded)
c.
(CH3)3C–COOH
This compound has 3 CH3 groups on the 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.
X
aryl halide
(leaving group on an
sp2 hybridized C)
Aryl halides are unreactive
in SN2 reactions.
23.26
O
O
a.
CH3
C
[1] NaOEt
CH2CO2Et
[2] CH3I
[3] H3O+, C
CH3
O
O
CH2CH3
b.
C
CH3
[1] NaOEt
CH2CO2Et
CH3
[2] CH3CH2CH2Br
[3] NaOEt
[4] C6H5CH2I
[5] H3O+, C
CH CH2
CH2CH2CH3
23.27 Locate the C. All alkyl groups on the C come from alkyl halides, and the remainder of the
molecule comes from ethyl acetoacetate.
O
a.
CH3
C
O
CH2 CH2CH3
CH3
C
CH2
COOEt
O
H3O+
[1] NaOEt
[2] CH3CH2Br
CH3
C
CH2CH2CH3
O
O
b. CH3
C
C
CH3
CH(CH2CH3)2
CH2
COOEt
[1] NaOEt
[1] NaOEt
[2] CH3CH2Br
[2] CH3CH2Br
O
H3O+
CH3
C
CH(CH2CH3)2
O
O
O
c.
CH3
C
CH2
[1] NaOEt
[1] NaOEt
[2] CH3CH2Br
[2] CH3(CH2)3Br
COOEt
23.28
O
CH3
C
CH2CO2Et
+
Br
H3O+
Br
O
NaOEt
(2 equiv)
CH3
X
CO2Et
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Chapter 23–10
23.29
O
O
O
a.
CH3
C
H3
[1] NaOEt
CH2CO2Et
[2]
CH3O
Br
CO2Et
b. CH3
C
O
[1] LDA, THF
CH3
[2]
CH3O
CH3O
nabumetone
CH3O
O
O+
Br
CH3O
nabumetone
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Substitution Reactions of Carbonyl Compounds 23–11
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
HO
O
HO
O
d.
H
(mono enol form)
O
OH
conjugated enol
(more stable)
e.
b.
O
conjugated enol
(more stable)
O
OH
O
c.
OH
OH
O
OH
O
OH
O
f.
OEt
OH
O
OH
OEt
OEt
unconjugated enol
(less stable)
O
O
OH
OH
OH
OEt
OEt
23.31
O
O
O
OCH2CH3
O
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.
OCH2CH3
ethyl acetoacetate
23.32
a. CH3CH2CH2CO2CH(CH3)2
c.
e. NC
O
OH
O
C
CH2CH3
O
b.
O
d. CH3O
CH2CN
f.
HOOC
23.33
O
CH2
a. CH3CH2 C
OH
Ha Hb
Hc
Ha is part of a CH3 group =
least acidic.
Hb is bonded to an C =
intermediate acidity.
Hc is bonded to O = most acidic.
b.
C
H
Hb O
CH3
Ha
Hc
Hc is bonded to an sp2 hybridized C = least acidic.
Ha is bonded to an C = intermediate acidity.
Hb is bonded to an C, and is adjacent
to a benzene ring = most acidic.
O
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Chapter 23–12
Hc
O
O
O
c.
d.
H
CH3
H
H
Hc
e. HO
Hb
Ha
Ha
H
Ha Hb
Hc is bonded to an C =
least acidic.
Ha is bonded to an C, and is
adjacent to a benzene ring =
intermediate acidity.
Hb is bonded to an C between two
C=O groups = most acidic.
O
COOH
sp3
H
H
Hb
Hc
sp3
Hb is bonded to an
hybridized
C = least acidic.
Hc is bonded to an C =
intermediate acidity.
Ha is bonded to O = most acidic.
hybridized
Ha is bonded to an
C = least acidic.
Hb is bonded to an C =
intermediate acidity.
Hc is bonded to O = most acidic.
23.34
CN
LDA
a.
d.
THF
O
CN
LDA
THF
O
O
OCH3
b.
LDA
THF
O
O
O
LDA
e.
THF
O
f.
LDA
c.
O
OCH3
LDA
O
O
THF
O
O
THF
23.35 Enol tautomers have OH groups that give a broad OH absorption at 3600–3200 cm–1, which
could be detected readily in the IR.
23.36
O
O
Hb
Ha
Ha
remove Ha
Hb
Ha
Ha
Hb
Ha
O
O
Hb
Ha
O
Hb
Hb
remove Hb
Ha
Ha
Hb
Hb
Removal of Ha gives two
resonance structures. The
negative charge is never on O.
O
Hb
Ha
Ha
O
Hb
Ha
Ha
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).
Hb
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Substitution Reactions of Carbonyl Compounds 23–13
23.37
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
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.
O
2,2-dimethyl-1,3cyclohexanedione
23.38 In the presence of acid, (R)--methylbutyrophenone enolizes to form an achiral enol.
Protonation of the enol from either face forms an equal mixture of two enantiomers, making the
solution optically inactive.
O
OH
O
H3O+
O
H2O
H
H
H
achiral
(E and Z isomers)
(R)--methylbutyrophenone
In the presence of base, (R)--methylbutyrophenone is deprotonated to form an achiral enolate, which
can then be protonated from either face to form an optically inactive mixture of two enantiomers.
O
O
O
O
H2O
–OH
H
H
H
achiral
(R)--methylbutyrophenone
23.39 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
+
O
S isomer
active enantiomer
H
O
H
[4]
O
H
H
A
(+ one more resonance structure)
O
OH
H
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Chapter 23–14
23.40
O
CH3
C
O
O
CH3
CH2
O
C
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 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 H's of the ketone more acidic.
23.41 LDA reacts with the most acidic proton. If there is any H2O present, the water would
immediately react with the base:
[(CH3)2CH]2N
Li
[(CH3)2CH]2NH
H OH
Li
OH
LDA
23.42
O
O
2,4-pentanedione
base (1 equiv)
O
O
[1] CH3I
O
O
[2] H2O
A
One equivalent of base removes
the most acidic proton between
the two C=O's, to form A on
alkylation with CH3I.
base
(2nd equiv)
O
O
[1] CH3I
O
O
[2] H2O
more nucleophilic site
B
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.
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Substitution Reactions of Carbonyl Compounds 23–15
23.43
O
NaOH
(CH3)3C
H2O
O
(CH3)3C
A
(CH3)3C
B
COCH3
O
H
H
(CH3)3C
C
COCH3
(CH3)3C
one axial and one equatorial
group, less stable
Both groups are equatorial.
more stable
This isomerization will occur since it makes
a more stable compound.
(CH3)3C
COCH3
Both groups are equatorial.
Compound C will not isomerize since it
already has the more stable arrangement
of substituents.
Isomerization occurs by way of an intermediate enolate, which can be protonated to either reform A, or give B. Since B has two large groups equatorial, it is favored at equilibrium.
COCH3
COCH3
H
(CH3)3C
OH
(CH3)3C
planar enolate
23.44 Protons on the carbon of an ,-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.
O
X
Remove the H on this 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), or from
above to form Y.
H
Y
618
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Chapter 23–16
23.45 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
OH
CH3COOH
OH
Br2
O
O
Br
2-pentanone
A
C=C has
2 bonds to C
more stable
(E and Z isomers)
C=C has
1 bond to C
Br
B
major product formed
from the more stable enol
23.46 • 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
O H
O
H
H
a.
O
H
H
O H
O
O
Br
O
H Br
O
C
b.
C
H H
H
–
OH
+
Br
H
H
I
I
H
I
+ I–
H
Repeat
[1] and [2]
two times.
C
–CI
+
H
C
O
O
O
Br
[2]
H
O
O
O H
H
+ H2O
O
CI3
CI3
OH
3
CHI3
–OH
23.47 Use the directions from Answer 23.24.
a.
CH3OCH2 CH2COOH
CH3OCH2Br
b.
C6H5
c.
COOH
C6H5
H
O
C
H
Br Br
Br
O
[1]
H
Br
O H
H
O
H
Br
Br
COOH
and
Br
O
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Substitution Reactions of Carbonyl Compounds 23–17
23.48
a. CH3CH2CH2CH2CH2 CH2COOH
H3O+
[1] NaOEt
CH2(CO2Et)2
[2] CH3CH2CH2CH2CH2Br
CH3CH2CH2CH2CH2CH2COOH
b.
[1] NaOEt
[1] NaOEt
[2] CH3CH2CH2CH2CH2Br
[2] CH3Br
CH2(CO2Et)2
COOH
H3O+
COOH
c.
COOH
[1] NaOEt
[1] NaOEt
[2] (CH3)2CHCH2CH2CH2CH2Br
[2] CH3Br
CH2(CO2Et)2
H3O+
COOH
23.49
O
O
O
O
[1] NaOEt
O
O
O
[2] CH3CH2CH2Br
O
[1] NaOEt
O
[2] CH3CH2CH2Br
O
O
O
H3O+
O
O
valproic acid
23.50
H3O+
[1] NaOEt
a.
CH2(CO2Et)2
[2] BrCH2CH2CH2CH2CH2Br
[3] NaOEt
b.
COOH
(from a.)
c.
COOH
[1] LiAlH4
COOH
CH2OH
[2] H2O
CH3OH
H2SO4
CO2CH3
CH3
[1] CH3MgBr (2 equiv)
C OH
[2] H2O
CH3
(from a.)
d.
COOH
CH3CH2OH
H2SO4
(from a.)
COOCH2CH3
[1] LDA
[2] CH3I
CH3
COOCH2CH3
H
620
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Chapter 23–18
23.51
a.
O
[1] Na+ –CH(COOEt)2 HO
[2] H2O
CH3CHCH2 CH(COOEt)2
O
CH3
c.
C
CH3
O
[1] Na+ –CH(COOEt)2
C
[2] H2O
CH3
CH(COOEt)2
Cl
nucleophilic attack here
b.
[1] Na+ –CH(COOEt)2
[2] H2O
CH2 O
d.
HOCH2CH(COOEt)2
CH3
O
O
C
C
O
[1] Na+ –CH(COOEt)2
[2] H2O
CH
CH3
3
O
C
CH(COOEt)2
+ CH3COOH
23.52 Use the directions from Answer 23.27.
O
O
a.
CH3
O
CH3
COOEt
C
CH2
CH3
C
[2] Br
[1] NaOEt
[1] NaOEt
[2] CH3CH2Br
[2] Br
COOEt
O
c.
O
CH2
O
b.
O
C
CH2
H3O+
H3O+
[1] NaOEt
COOEt
O
H3O+
[1] NaOEt
O
[2] Br
O
d.
CH3
C
[1] NaOEt
CH2
COOEt
O
H3O+
[2]
Br
Br
[3] NaOEt
23.53
O
a.
CH3
C
[1] NaOEt
CH2
COOEt
b.
CH3
C
CH2
c.
CH3
CH3
[1] NaOEt
[1] NaOEt
H3O+
[2] CH3Br
[2] CH3Br
COOEt
O
C
[2] CH3CH2Br
O
O
H3O+
O
LDA
CH(CH3)2
THF
CH2
C
C
CH2CH2CH3
O
CH3
C
CH(CH3)2
O
CH3I
CH3CH2
CH(CH3)2
C
CH(CH3)2
(from b.)
O
d.
CH3
C
CH(CH3)2
(from b.)
NaOCH3
CH3OH
O
CH3
C
O
CH3I
C(CH3)2
CH3
C
C(CH3)3
O
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Substitution Reactions of Carbonyl Compounds 23–19
23.54
O
O
O
Li2CO3
a.
LiBr
DMF
Br
O
f.
(E and Z)
[2] Li2CO3, LiBr, DMF
O
O
COOH
b.
O
[1] Br2, CH3CO2H
O
I2 (excess)
g.
O
+ CHI3
–OH
COOH
[1] LDA
c. CH3CH2CH2CO2Et
COOH
CH3CH2
h.
CH3CH2CHCO2Et
Cl
NaH
CN
C N
[2] CH3CH2I
O
O
Br
d.
O
NHCH(CH3)2
(CH3)2CHNH2
O
Br2 (excess)
i.
–OH
O
O
O
Br Br
O
[1] LDA
e.
NaI
j.
[2] CH3CH2I
I
Cl
23.55
O
O
O
O
[1] LDA
O
[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
23.56
CHO
a.
OH
NaBH4
CH3OH
p-isobutylbenzaldehyde
[1] PBr3
CN
[2] CH3I
[2] NaCN
A
CN
[1] LDA
B
C
H3O+
COOH
ibuprofen
622
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Chapter 23–20
COOH
b.
removal of the
most acidic H
D
[2] CH3I
COO
[1] LDA
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.
23.57
CO2CH3
a.
CO2CH3
Cl
NH
+
N
K2CO3
S
Cl
S
Cl
sp2
Cl on
hybridized
C does not react.
Cl on sp3
hybridized C reacts.
H COOH
The N atom acts as a
nucleophile to displace Cl–.
H COOCH3
CH3OH
b. HO
H COOCH3
TsCl
HO
H2SO4
Cl
NH
B
N
inversion of
configuration
SN2
Cl
C
CO2CH3
S isomer
S
TsO
pyridine
Cl
A
clopidogrel
(racemic)
S
Cl
clopidogrel
(single enantiomer)
23.58
O
O
O
O
LDA
Br
NaOCH2CH3
THF
Br
–78 oC
A
B
CH3CH2OH
room temperature
A
C
23.59
O
O
a.
O
+
CO2
COOH
In order for decarboxylation to occur readily, the COOH
group must be bonded to the C of another carbonyl
group. In this case, it is bonded to the carbon.
[1] NaOEt
(CH3CH2)3CCH(CO2Et)2
b. CH2(CO2Et)2
[2] (CH3CH2)3CBr
The 3° alkyl halide is too crowded to react with the
strong nucleophile by an SN2 mechanism.
c.
H
O
[1] LDA
[2] CH3CH2I
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 C.
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623
Substitution Reactions of Carbonyl Compounds 23–21
23.60
O
O
O
CH3
LDA
O
I
+
O
I–
O
CH3
I
I–
+
23.61
O
O
O
O
O
O
O
EtO O
O
OEt
EtO
OEt
EtO
OEt
H H
H
EtO
O
OEt
O
+ HOEt
–
OEt
–
OEt
O
CO2Et H3O+
O
O
O
H
OEt
O
O
OEt
O
O
+ HOEt
23.62
LDA = B
O
C6H5
H
Br
B
Br
C6H5
C6H5
O
+
Br
+ HB+
O
Br
+ HB+
C6H5 H
B
O
H
Br
C6H5
Br
–OC(CH
Br
C6H5
3)3
C6H5
+ HB+
C6H5
O
H C6H5
O
Br
C6H5
O
Br
O
Br
C6H5
+
C6H5
B
O
O
O
+ HOC(CH3)3
O
+ Br
This reaction occurs
with both bases [LDA
and KOC(CH3)3].
+ Br
624
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Chapter 23–22
23.63
1st new C–C bond
O
H
H
OCH2CH3
O
LDA
THF
–78 oC
Remove
proton here.
H
OCH2CH3
[3]
Cl
Cl
B
O
OCH2CH3
LDA
THF
–78 oC
[2]
H
[1]
A
O
OCH2CH3
Cl
Cl
[4]
O
Cl
OCH2CH3
C
2nd new C–C bond
Cl
23.64
O
COCH3
a.
O
Cl2
Cl
NHC(CH3)3
H2NC(CH3)3
H2O, HCl
COO–
COCH3 –OH, I
2
excess
b.
H3O+
COOH
+ CHI3
O
COCH3
[1] LiAlH4
[2] H2O
[1] LDA, THF
c.
OH
[2] CH3CH2CH2Br
O
COCH3
Br2
[1] LDA, THF
d.
Li2CO3
CH3COOH
[2] CH3Br
Br
23.65
O
O
Br
Br2
a.
CH3COOH
O
O
Br
b.
OCH3
NaOCH3
(from a.)
O
OH
O
[1] LDA, THF
c.
d.
[1] LiAlH4
[2] CH2=CHCH2Br
[2] H2O
O
O
O
O
O
[1] LDA, THF
[1] LDA, THF
[2] CH3Br
[2] CH3CH2Br
LiBr
DMF
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Substitution Reactions of Carbonyl Compounds 23–23
O
O
Br
e.
(from a.)
CH3CH2Br
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
Li2CO3
LiBr
DMF
[2] H2O
O
O
O
Br
[1] LDA, THF
f.
O
[1] (CH3CH2)2CuLi
NaOCH2CH3
[2] BrCH2CH2CH2CH2Br
O
O
O
[1] LDA, THF
g.
[1] LDA, THF
[2] CH3CH2Br
O
Li2CO3
CH3COOH
[2] CH3CH2Br
O
Br
O
[1] NaOCH2CH3
h.
O
O
Br2
O
Br
Br2
Li2CO3
LiBr
DMF
CH3COOH
[2] CH3CH2Br
LiBr
DMF
(from g.)
23.66
O
O
O
Cl
AlCl3
O
Cl2
Br2
FeCl3
CH3CO2H
Br
Cl
Cl
H2NC(CH3)3
O
NHC(CH3)3
Cl
bupropion
23.67
O
a.
OH
PBr3
Br
O
O
[1] NaOEt
COOEt
COOEt
H3O+
[2]
mCPBA
Br
O
O
b.
O
O
O
COOEt
[1] NaOEt
[2] HC CCH2Br
H3O+
HOCH2CH2OH
TsOH
O
COOEt
O
O
[1] NaH
[2] CH3I
H3O+
O
O
H2
Lindlar catalyst
O
O
CH3
626
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Chapter 23–24
O
O
COOEt
c.
[1] NaOEt
OH
O
COOEt
[2]
H3O+
NaBH4
CH3OH
CH2Br
CH3
Br2
h
23.68
O
O
Br2
a.
O
CH3CH2NH2
CH3COOH
NH3 (excess)
Br
CH3CH2Br
NHCH2CH3
PBr3
CH3CH2OH
O
b.
O
Li2CO3
LiBr
DMF
Br
OH
NaBH4
CH3OH
(from a.)
Br2
c.
Br
FeBr3
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
O
(from b.)
O
[1] (C6H5)2CuLi
[2] H2O
C6H5
SOCl2
d.
CH2Br
Br2
h
CH3
CH3OH
CH3Cl
AlCl3
[1] Li (2 equiv)
[2] CuI (0.5 equiv)
O
(from b.)
[1] (C6H5CH2)2CuLi
[2] H2O
O
C6H5
NaBH4
CH3OH
OH
C6H5
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Substitution Reactions of Carbonyl Compounds 23–25
23.69
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
23.70
CH2(COOEt)2
NaOEt
OH
PBr3
Br
–
CH(COOEt)2
CH(COOEt)2
H3O+, Cl
Y
SOCl2
COOH
O
23.71
O
a.
O
O
Y =
[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)
628
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Chapter 23–26
O
O
b.
CH3 Li
O
O
O
O
O
O
+ I–
Y
CH3
I
23.72 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+
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Text
Substitution Reactions of Carbonyl Compounds 23–27
23.73
CH3 Li
O
OCH2CH3
H OH2
O
CH3
OCH2CH3
OH
H3O+
OH2
OCH2CH3
CH3
+ Li+
OCH2CH3
CH3
+ H2O
CH3
OCH2CH3
+ H2O
+ H2O
X
H
CH3
OH
OCH2CH3
H OH2
OH
OCH2CH3
CH3
CH3
+ H2O
OH
OCH2CH3
CH3
OCH2CH3
H
+ H2O
O H
+ H2O
O
+ H2O
H3O+
HOCH2CH3
23.74
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
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Reactions
631
Carbonyl Condensation Reactions 24–1
C
Reeaaccttiioonnss
Chhaapptteerr 2244:: C
Caarrbboonnyyll C
Coonnddeennssaattiioonn R
TThhee ffoouurr m
maajjoorr ccaarrbboonnyyll ccoonnddeennssaattiioonn rreeaaccttiioonnss
Reaction type
Reaction
[1] Aldol reaction
(24.1)
O
2
RCH2
OH
–OH
C
H
H2O
[2] Claisen reaction
(24.5)
RCH2
C
C
or
R
OR'
[1] NaOR'
O
[2] H3O+
C
RCH2
H
H3O+
CHO
C
R
(E and Z)
,-unsaturated
carbonyl compound
-hydroxy carbonyl
compound
O
2
CHCHO
H
aldehyde
(or ketone)
RCH2
–OH
RCH2 C
O
CH
ester
C
OR'
R
-keto ester
O
O
[3] Michael reaction
(24.8)
–OR'
+
R
–
,-unsaturated
carbonyl compound
[4] Robinson
annulation (24.9)
R
1,5-dicarbonyl compound
carbonyl
compound
–OH
H2O
O
O
,-unsaturated
carbonyl
carbonyl compound compound
2-cyclohexenone
U
Usseeffuull vvaarriiaattiioonnss
[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"
-hydroxy carbonyl
compound
–
OH
or
H3O+
C
R
C
H
R"
C
R'
(E and Z)
,-unsaturated
carbonyl compound
632
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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)
[a] With 1,6-diesters:
OEt
O
[1] NaOEt
O
O
O
C
[2] H3O+
OEt
OEt
[b] With 1,7-diesters:
OEt
O
O
O
[1] NaOEt
[2] H3
OEt
O+
O
C
OEt
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Reactions
Carbonyl Condensation Reactions 24–3
C
Chhaapptteerr 2244:: A
Annssw
weerrss ttoo PPrroobblleem
mss
24.1
O
CH2CHO
H
a.
OH
c. CH3
C
OH
O
CH3 C CH2 C CH3
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
H
no H
no aldol reaction
C
d.
(CH3)3C
H
no H
no aldol reaction
yes
C
CHO
e.
H
CH3
H
yes
yes
24.3
base
O
a.
O
O
OH
base
O
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
C C
H
+
+ HSO4
O
H
O
HSO4
CH3CH CH C
H
H
+
H2O
H2SO4
Locate the and C’s to the carbonyl group, and break the molecule into two halves at this
bond. The C and all of the atoms bonded to it belong to one carbonyl component. The C
and all of the atoms bonded to it belong to the other carbonyl component.
24.5
O
OH
CHO
a.
b. C6H5
c.
C6H5
OH
H
C6H5
O
O
CHO
H
CHO
CHO
O
C6H5
O
634
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Chapter 24–4
24.6
–OH
O
H
H H
H
C
H OH
O
C
O
H C
C
H
H
C H
O
O
+ H2O
OH
OH
H C
+
H C
–OH
C H
CHO
+ H2O
C H
H
O
O
–OH
24.7
CHO
a. CH3CH2CH2CHO and CH2=O
CHO
or
O
c. C6H5CHO
OH
and
O
O
b. C6H5COCH3 and CH2=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
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24.10 Find the and 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
N
CH3O
Pd-C
X
CH3O
donepezil
CH3O
24.12 All enolates have a second resonance structure with a negative charge on O.
O
O
H
O
EtO
O
O
O
H
H
H
EtO
O
OH
O
+ OH
H
H
OH
O
O
O
EtO H
EtOH
24.13
O
O
–
a.
CHO
O
O
OH
O
b.
H2O
[1] O3
[2] (CH3)2S
CHO
O
–OH
H2O
–
OH
H2O
24.14
1-methylcyclopentene
EtOH
EtOH
H H
O
2-cyclohexenone
636
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Chapter 24–6
24.15 Join the C of one ester to the carbonyl C of the other ester to form the -keto ester.
O
a.
OCH3
O
O
O
b.
O
OCH2CH3
OCH3
O
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 -dicarbonyl compound.
O
a. CH3CH2CO2Et and HCO2Et
H
O
OEt
Only this compound
can form an enolate.
b.
O
and
CO2Et
HCO2Et
H
C
OEt
O
Only this compound
can form an enolate.
O
CH3
c.
C
O
and
CH3
CH3
C
O
O
OEt
The ketone
forms the enolate.
d.
O
and
O
O
C
C
OEt
O
The ketone
forms the enolate.
24.17 A -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
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Carbonyl Condensation Reactions 24–7
24.18
O
O
O
[1] NaOEt
a.
OEt
[2] (EtO)2C=O
O
b.
C
C6H5CH2
O
[1] NaOEt
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
EtO
B
H3O+
CO2H
ibuprofen
24.20
O
CH3O2C
CO2CH3
base
CH3O
CH3O2C
CO2CH3
O
OR
OR
X
1,6-Diester forms a five-membered ring.
24.21
O
O
O
O
O
OEt
EtO
EtO
OEt
O
24.22 A Michael acceptor is an ,-unsaturated carbonyl compound.
O
a.
,-unsaturated
yes - Michael acceptor
O
O
O
d.
c.
b.
O
not ,-unsaturated
not ,-unsaturated
CH3O
,-unsaturated
yes - Michael acceptor
638
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Chapter 24–8
24.23
O
a.
+
CH2 CHCO2Et
CH3
C
[1] NaOEt
[2] H2O
CH2CO2Et
CH3
O
O
C
C
OEt
COOEt
[1] NaOEt
+ CH2(CO2Et)2
b.
[2] H2O
O
O
EtO2C
O
O
c.
+
CH2
CH3
C
CH2CN
O
[1] NaOEt
[2] H2O
CO2Et
CN
O
24.24
O
O
O
O
EtO2C
a.
CO2Et
b.
O
O
O
O
24.25 The Robinson annulation forms a six-membered ring and three new carbon–carbon bonds: two bonds and one bond.
new C–C bond
O
O
O
OH
re-draw
+
a.
O
O
O
H2O
O
O
new and bonds
new C–C bond
O
O
COOEt +
b.
COOEt
re-draw
O
COOEt
OH
H2O
O
O
new and bonds
new C–C bond
O
c.
O
+
re-draw
OH
O
O
H2O
O
new and bonds
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Reactions
Carbonyl Condensation Reactions 24–9
new C–C bond
EtO2C
O
+
d.
O
COOEt
EtOOC
re-draw
O
OH
H2O
O
O
new and bonds
24.26
a.
O
O
O
b.
c.
O
O
O
O
O
O
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Chapter 24–10
24.27 The product of an aldol reaction is a -hydroxy carbonyl compound or an ,-unsaturated
carbonyl compound. The latter type of compound is drawn as product unless elimination of H2O
cannot form a conjugated system.
a. (CH3)2CHCHO only
O
OH
–
OH
H2O
–
(CH3)2CHCHC(CH3)2
d.
(CH3CH2)2C=O only
CHO
OH
H2O
O
CH3
–OH
b. (CH3)2CHCHO + CH2=O
e. (CH3CH2)2C=O + CH2=O
CH3 C CHO
H2O
H2O
CH2OH
O
CHO
–OH
c. C6H5CHO + CH3CH2CH2CHO
–OH
O
–
H2O
+ C6H5CHO
f.
OH
H2O
(E and Z isomers)
(E and Z isomers)
24.28
HO
CHO
O
H
OH
CHO
H
O
OH
OH
CHO
CHO
24.29
O
O
a.
CH3
C
OH
[1] LDA
O
CH3CH2CH2 C CH2 C
CH3 [2] CH3CH2CH2CHO
H
[3] H2O
b. CH CH
3
2
CH3
C
OH
[1] LDA
OEt
[2]
O
O
[3] H2O
CHO
O
C
CH C
H
CH3
O
O
24.30
O
O
CHO
a.
O
c.
O
b. OHC
CHO
CHO
O
OEt
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24.31 Locate the and C’s to the carbonyl group, and break the molecule into two halves at this
bond. The C and all of the atoms bonded to it belong to one carbonyl component. The C
and all the atoms bonded to it belong to the other carbonyl component.
O
O
a.
OH
b.
O
d.
c.
C6H5
O
O
CH CHCN
C6H5
O
H
e.
C6H5
O
CH3
O
O
O
CH3
C6H5
CH2=O
O
C6H5
CH3CN
CHO
C6H5
24.32
O
CHO
(CH3)2C=O
NaOEt
EtOH
CH3O
CH3O
X
24.33
O
O
O
O
c.
b.
a.
d.
HO
O
O
O
H
H
O
O
O
CH3
CH3
H
O
O
24.34 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
24.35
O
a. C6H5CH2CH2CH2CO2Et
O
C6H5CH2CH2CH2
OEt
CH2CH2C6H5
O
b. (CH3)2CHCH2CH2CH2CO2Et
(CH3)2CHCH2CH2CH2
O
OEt
CH2CH2CH(CH3)2
642
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Chapter 24–12
CH3O
c. CH3O
O
O
CH2COOEt
OEt
OCH3
24.36
O
O
CH3CH2CH2CH2CO2Et + CH3CH2CO2Et
O
O
OEt
O
OEt
O
O
O
OEt
OEt
24.37
O
O
a. CH3CH2CH2CO2Et only
O
OEt
f. CH3CH2CO2Et + (EtO)2C=O
CH2CH3
O
O
b. CH3CH2CH2CO2Et + C6H5CO2Et
g.
O
c. CH3CH2CH2CO2Et + (CH3)2C=O
O
h.
O
O
H
O
O
+
Cl
C
OEt
OEt
O
O
O + HCO2Et
CH2CH3
O
EtO
O
OEt
O
O
EtO
O
OEt
d. EtO2CC(CH3)2CH2CH2CH2CO2Et
O
O
e. C6H5COCH2CH3 + C6H5CO2Et
24.38
O
O
CH3O
CH3O
a.
O
OEt
(EtO)2C=O
or
O
CH3O
O
OEt
EtO
OEt
C6H5
b.
O
O
O
C6H5
O
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Reactions
Carbonyl Condensation Reactions 24–13
O
CHO
c.
d. C6H5CH(COOEt)2
H
CH3
O
O
O
O
C6H5
OEt
OEt
EtO
OEt
or
O
CH3CHO
OEt
24.39
O
To form bond (b):
To form bond (a):
O
O
O
O
EtO
bond (a)
EtO
bond (b)
CH3
O
24.40 Only esters with 2 H’s or 3 H’s on the carbon form enolates that undergo Claisen reaction to
form resonance-stabilized enolates of the product -keto ester. Thus, the enolate forms on the
CH2 to one ester carbonyl, and cyclization yields a five-membered ring.
This is the only carbon with 2 H's.
O
O
NaOCH3
OCH3
CH3O
CH3O
O
OCH3 [1] nucleophilic attack
CH3O
CH3OH
O
CH3O
O
O
CO2CH3
CH3O2C
[2] loss of CH3O–
[3] deprotonation
O
B
O
H3
CO2CH3
CH3O2C
O+
CH3O2C
OCH3
CO2CH3
CH3O2C
H
O
O
O
highly resonance-stabilized enolate
Formation of this enolate drives
the reaction to completion.
acidic H between 2 C=O's
24.41
O
a.
O
+
C6H5
C6H5
–OEt,
O
O
EtOH
C6H5
C6H5
O
O
CO2Et
b.
–OEt,
+
O EtOOC
EtOH
O
O
O
c.
+ CH2(CN)2
–OEt,
EtOH
CN
CN
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Chapter 24–14
O
O
O
d.
+
–OEt,
CO2Et
EtOH
O
EtO2C
24.42
O
O
O
a.
c.
b.
O
O
CO2Et
O
EtO2C
O
d.
CN
CO2Et
O
O
O
O
O
O
O
C6H5
CO2Et
EtO2C
CN
CO2Et
C6H5
24.43
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
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Carbonyl Condensation Reactions 24–15
24.44
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
24.45
O
O
c.
a.
O
O
O
O
O
O
O
O
d.
O
b.
O
O
O
24.46
CH3CH2CH2CHO
a.
–OH
CH3CH2CH2
H2O
b.
–OH
CHO
(E and Z)
H
CH2
CH2=O, H2O
CH2CH3
CHO
C
CH2CH3
g.
h.
CHO
c.
i.
[1] LDA
[2] CH3CHO; [3] H2O
d.
e.
H
NaOEt, EtOH
CO2Et
j.
[1] CH3Li
k.
NaBH4
CH3OH
HOCH2CH2OH
TsOH
CH3NH2
mild acid
O
(CH3)2NH
mild acid
CrO3
CH3CH2CH2CH2OH
l.
O
H
CH3CH2CH2
C
CH3
N
H
CH3CH2CH=CHN(CH3)2
(E and Z)
CH3CH2CH2COOH
H2SO4
[2] H2O
f.
CH3CH2CH2CH2OH
OH
CH3CH2CH2
CH2(CO2Et)2
EtO2C
OH
H2
Pd-C
O
Br2
H
CH3COOH
Br
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Chapter 24–16
m.
Ph3P=CH2
CH3CH2CH2
O
OH
n.
C CH2
NaCN, HCl
o.
CH3CH2CH2 C H
H
[1] LDA; [2] CH3I
H
CN
24.47
O
O
O
–OH
a.
e.
H2O
C
O
O
NaOEt, EtOH
f.
b.
O
(CH3)2C=O
O
+
CHO
CH3
C
C6H5
NaOCH3
C6H5
+
CH3OH
O
O
O
CO2Et
OH
H2O
O
C6H5
+
–OH
C6H5
H2O
g.
–
CH3
CHO
CN
O
d.
[2] CH3CH2CHO
O
NaOEt, EtOH
c. NCCH2CO2Et
CH3
O
[3] H2O
O
O
CH3
OH
[1] LDA
CH2CO2Et
[1] NaOEt, EtOH
CH2CO2Et
[2] H3O+
(E and Z)
h.
O
(E and Z) O
O
O
OEt
24.48
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+
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
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Carbonyl Condensation Reactions 24–17
24.49
H
H
B
B
CH CHCl
C CHCl
H
CH2 CH2Cl
CH2 CH2Cl
sp2
sp3
Vinyl halides undergo elimination by an E1cB mechanism more readily than alkyl halides because
the carbanion intermediate formed from a vinyl halide has an sp2 hybridized C, while the carbanion
derived from CH3CH2Cl is sp3 hybridized. The higher percent s-character of the sp2 hybridized
anion makes it more stable, and therefore, it is formed more readily.
24.50 The final step in the reaction sequence involves an intramolecular crossed Claisen reaction
between a ketone and an ester to form a -dicarbonyl compound.
[1] O3
CrO3
O
[2] (CH3)2S
CHO
EtOH
O
H2SO4
H2O
O
H2SO4
COOH
A
COOCH2CH3
B
C
[1] NaOEt, EtOH
[2] H3O+
O
O
D
C13H20O2
24.51
O
[1] O3
O
O
[2] (CH3)2S
O
O
O
NaOH
EtOH
H
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
H H
CO32–
[1]
H
H
C
O
O
C
C
H
H
+ HCO3–
+ H
O
HH
H
[2]
O
C
C
C
H H
H OCO2
H
HO
C
O
O
HH
[3]
C
C
H H
HOCH2
H
C
C
H
HOCH2 CH2OH
+ CO32–
Repeat steps [1]–[3] with these
two H's and CO32–.
648
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Chapter 24–18
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
OH
H O
OH
O
A
O
most stable enolate
less hindered carbonyl
O
H2O
H
OH
O
H2O
–
OH
1-acetylcyclopentene
major product
O
H H
–OH
O
–
H
O
H
O
H
O
O
OH
H O
H OH
H
B
The more hindered ketone
carbonyl makes nucleophilic
attack more difficult.
OH
OH
O
H
H
H2O
CHO
–OH
H2O
O
H
H
H H
–
O
H OH
O
O
H
OH
H2O
H
H
H
–OH
OH
O
O
O
O
OH
C
less stable enolate
H2O
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.
O
–OH
24.54 All enolates have a second resonance structure with a negative charge on O.
O
O O
O
O
a.
HH
Na+ –OEt
O
O
O
O
O
H
+
O
H
+ EtOH
O
H2O
+
O
O
H3
O+
EtO
O
+ EtOH
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Na+ –OCH3
H H
H
CH3O
b.
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
O
CH2 N
O
CH2 N
O
O
–OH
O
C6H5
C
O
O
CH2 N
H
H OH
C6H5 C CH2
O
OH
NO2
OH
C6H5 C CH NO2
H
C6H5CH CHNO2
C6H5 C CH NO2
H H
H
+
–OH
+ H2O
–OH
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
OH
O
HO
O
H
O
H OH
O
650
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Chapter 24–20
24.57 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
24.58
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
H
O
CH3COO
CH2
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
24.59
C6
CO2Et
a.
O
+ EtOH
H
ethyl 2,4-hexadienoate
CO2Et
OEt
EtO O
OEt
CO2Et
CO2Et
O
EtO
O
OEt
diethyl oxalate
O
CO2Et
EtO2C
O
OEt
CO2Et
+
OEt
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Carbonyl Condensation Reactions 24–21
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 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.
24.60
O
O
–OH,
a.
H2O
C6H5
C6H5CHO
O
b.
O
–
OH, H2O
O
CH2OH
O
PCC
CHO
H2C=O
O
O
–OH,
c.
HCO2Et
–OEt,
CH3CH2OH
O
H2O
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)
H2O
Pd-C
24.61
CHO
[1] O3
CHO
[2] (CH3)2S
CHO
NaOEt
EtOH
A
B
24.62
O
O
[1] LDA, THF
a.
[2] O
H
[3] H2O, –OH
O
O
–OH
b.
O
H2O
O
652
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Chapter 24–22
O
O
O
[1] NaOEt
[1] Br2, CH3COOH
c.
O
[2] Li2CO3, LiBr, DMF
O
H3O+
O
O
O
OEt
COOEt
[2] H2O
O
CN
NaOEt, EtOH
d.
CH2(CN)2
CN
O
–
e.
[1] (CH3)2CuLi
OH, H2O
O
O
O
[2] H2O
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
O
OH
[1] LDA
c.
CH3OH
O
C6H5
[2] C6H5CHO
SOCl2
– H2O
unstable
PCC
[1] Br2, h
CH3Cl
C6H5
OH
[2] –OH
AlCl3
HO
O
C6H5
d.
H2 (excess)
C6H5
Pd-C
(from c.)
O
[1] CH3MgBr
e.
O
OH
H2SO4
[1] O3
[2] H2O
[2] (CH3)2S
Mg
CH3OH
PBr3
CH3Br
major
product
O
CHO
–OH,
H2O
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Carbonyl Condensation Reactions 24–23
24.64
[1] NaOEt
OEt
a.
CO2Et
[2] H3O+
O
b.
O
[1] NaOEt
CO2Et
[2]
O
CO2Et
Br
(from a.)
O
PBr3
OH
C6H5
[1] NaOEt
c.
CO2Et
[2]
O
CO2Et
Br
C6H5
C6H5
[1] LiAlH4
OH
[2] H2O
O
OH
(from a.)
CH3OH + SOCl2
[1] CH3Cl, AlCl3
Br
[2] Br2, h
C6H5
OEt
d.
[1] NaOEt
O
O
C6H5
[2] H3O+
O
O
AlCl3
O
Cl
SOCl2
O
CrO3
OH
OH H2SO4
H2O
24.65
a.
[1] O3
CHO
–
[2] (CH3)2S
CHO
H2O
CHO
OH
NaBH4
CHO
b.
OH
CH3OH
(from a.)
O
c.
CHO
CrO3
COOH
CHO
H2SO4
COOH
(from a.)
H2SO4
O
CO2Et [1] NaOEt
[2] CH3I
(from c.)
CO2Et
[1] NaOEt
CO2Et
[2] H3O+
H2O
O
d.
OH
O
CO2Et
H3O+
CO2Et
654
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Chapter 24–24
24.66
O
O
CHO
O
a.
CH3O
+
O
CH3O
octinoxate
[1] NaH
b.
[2] CH3Cl
HO
Br
Br2
CH3O
[2] H2C=O
CH3O
SOCl2
CH2OH
[1] Mg
(+ ortho isomer)
CH3O
PCC
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
BrMg
Mg
Br
PCC
PBr3
HO
HO
24.67
O
O
[1] HCO2Et
a.
[2]
CHO
H3O+
NaOEt, EtOH
O
O
CHO
b.
O
NaOEt, EtOH
CHO
O
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Carbonyl Condensation Reactions 24–25
O
O
O
O
O
H
OEt
H
OEt
c.
O
H OEt O
O
+
O
O
+
O
HCO2Et
OEt
+
O
H OH2
d.
=
O
O
O
O
+OH
+ H
OH
H2O
H2O
+
OH
H OH2
OH
no H
OH
O
OH
OH
+
H OH2
H2O
+O H
O
-hydroxy ketone
+
OH2
H
+
H2O
H2O
H2O
H3O+
O
O
O
H2O
This reaction is an acid-catalyzed aldol that proceeds by way of enols not enolates. The -hydroxy
ketone initially formed cannot dehydrate to form an ,-unsaturated carbonyl because there is no H on
the 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.
656
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Chapter 24–26
24.68 Rearrangement generates a highly resonance-stabilized enolate between two carbonyl groups.
OEt
O
O
O OEt O
O
OEt
O
OEt
O
OEt
O
H
OCH2CH3
OCH2CH3
OCH2CH3
OCH2CH3
CH3CH2O
H OCH2CH3
OCH2CH3
O
O
H
CO2CH2CH3
O
OEt
O OEt
CO2CH2CH3
CO2CH2CH3
O
OCH2CH3
(+ 2 resonance structures)
H OCH2CH3
This product is a highly
resonance stabilized enolate.
This drives the reaction.
H3O+
O
H
CO2CH2CH3
24.69 All enolates have a second resonance structure with a negative charge on O.
O
O
O
O
B
[2]
[1]
+
H OH
O
O
H
OH
[3]
B
O
[4]
OH
[5]
+ OH
HB+
+ HB+
+ –OH
O
OH
O
O
HO
+
O
Repeat steps
[1]–[5] by
deprotonating the
indicated CH3.
O
H
isophorone
+ HB+
(+ 2 resonance
structures)
H
B
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Carbonyl Condensation Reactions 24–27
24.70 All enolates have a second resonance structure with a negative charge on O.
B
O
O
COOEt
O
H
H
O
CO2CH3
CO2CH3
CO2CH3
new bond
new bond
H2O
+ –OH
O
CO2CH3
24.71
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
OH
C6H5 C CH
H
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
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659
Amines 25–1
C
Chhaapptteerr 2255:: A
Am
miinneess
G
Geenneerraall ffaaccttss
• 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).
SSuum
mm
maarryy ooff ssppeeccttrroossccooppiicc aabbssoorrppttiioonnss ((2255..55))
Mass spectra
Molecular ion Amines with an odd number of N atoms give an odd
molecular ion.
IR absorptions
N–H
3300–3500 cm–1 (two peaks for RNH2, one peak for R2NH)
1
NH
CH–N
0.5–5 ppm (no splitting with adjacent protons)
2.3–3.0 ppm (deshielded Csp3–H)
C–N
30–50 ppm
H NMR absorptions
13
C NMR absorption
mppoouunnddss ((2255..1100))
C
Coom
mppaarriinngg tthhee bbaassiicciittyy ooff aam
miinneess aanndd ootthheerr ccoom
• 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 electron-withdrawing
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).
PPrreeppaarraattiioonn ooff aam
miinneess ((2255..77))
[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.
660
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Chapter 25–2
[2] Gabriel synthesis (25.7A)
•
•
O
R X
+
CO2–
–OH
N
R NH2
H2O
+
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
[1] LiAlH4
[2] H2O
NR'2
RCH2 N R'
R'
R' = H or alkyl
1o, 2o, and 3o amines
[4] Reductive amination (25.7C)
R
C O
R2"NH
+
•
R
NaBH3CN
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.
R
Reeaaccttiioonnss ooff aam
miinneess
[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
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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] 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
+
R N N
Cl–
NaNO2
R N H
HCl
alkyl diazonium salt
R N N O
HCl
R
R
N-nitrosamine
R
Reeaaccttiioonnss ooff ddiiaazzoonniiuum
m ssaallttss
[1] Substitution reactions (25.14)
With CuX:
With H2O:
N2
+
Cl–
F
X
OH
phenol
With HBF4:
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
662
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Chapter 25–4
C
Chhaapptteerr 2255:: A
Annssw
weerrss ttoo PPrroobblleem
mss
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
C6H5
H
N
N
H
NH2
b. CH3CH2O
2o amine
N CH3
O
1o amine
3o amine
25.2
3o amine
CH3
a.
b.
HO C CH2NH2
CH3 N CH2CH2OH
CH3
3o alcohol
25.3
CH3
1o amine
1o alcohol
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.
OH
+
CH3
HO
CH3
+
a. CH3 N CH2CH2 N CH2CH3
CH3
H
N
b.
HO
H
N has 3 similar groups.
25.4
CH3 NH2
NH2
NH2
+ 3 more
resonance
structures
140 pm
partial double
Because the lone pair on N can be delocalized
bond character
on the benzene ring, the C–N bond has partial double
bond character, making it shorter. Both the C and N atoms
must be sp2 hybridized (+ have a p orbital) for
delocalization to occur. The higher percent s-character
of the orbitals of both C and N shortens the bond as well.
147 pm
The C–N bond is formed from two
sp3 hybridized atoms and the lone
pair is localized on N.
25.5
NHCH2CH3
a.
CH3CH2CH(NH2)CH3
2-butanamine
or
sec-butylamine
c.
N(CH3)2
N,N-dimethylcyclohexanamine
e.
N-ethyl-3-hexanamine
CH3
b.
(CH3CH2CH2CH2)2NH
dibutylamine
d.
f.
NHCH2CH2CH3
NH2
2-methyl-5-nonanamine
2-methyl-N-propylcyclopentanamine
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Amines 25–5
25.6
An NH2 group named as a substituent is called an amino group.
a. 2,4-dimethyl-3-hexanamine
c. N-isopropyl-p-nitroaniline
g. N-methylaniline
e. N,N-dimethylethylamine
NHCH(CH3)2
NHCH3
N
O2N
NH2
b. N-methylpentylamine
d. N-methylpiperidine
f. 2-aminocyclohexanone
h. m-ethylaniline
O
NH2
NH2
NHCH3
N
CH2CH3
25.7
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.
O
a. (CH ) CHCH CH CH C CH CH
3 2
2
3
3
2
3
alkane
lowest
boiling point
25.8
ketone
intermediate
boiling point
(CH3)2CHCH2NH2
amine
N–H can hydrogen
bond.
highest boiling point
CH3
alkane
lowest
boiling point
O
CH3
ether
intermediate
boiling point
NH2
amine
N–H can hydrogen
bond.
highest boiling point
1o Amines show two N–H absorptions at 3300–3500 cm–1. 2o Amines show one N–H absorption
at 3300–3500 cm–1.
molecular weight = 59
one IR peak = 2° amine
25.9
b.
CH3 N CH2CH3
H
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)
NH
CH3 adjacent to CH2
(CH3)3C
CH2 adjacent to CH3
N
H
664
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Chapter 25–6
25.10 The atoms of 2-phenylethylamine are in bold.
O
N
(CH3CH2)2N
a.
CH3
b.
H
CH3O
O
H
N
H
LSD
lysergic acid diethyl amide
H
HO
N
CH3
codeine
25.11 SN2 reaction of an alkyl halide with NH3 or an amine forms an amine or an ammonium salt.
NH3
Cl
a.
NH2
excess
NH2
b.
CH3CH2Br
N(CH2CH3)3
Br–
excess
25.12 The Gabriel synthesis converts an alkyl halide into a 1o amine by a two-step process:
nucleophilic substitution followed by hydrolysis.
CH3O
NH2
a.
b.
Br
(CH3)2CHCH2CH2NH2
CH3O
NH2
Br
c.
(CH3)2CHCH2CH2Br
25.13 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.
NH2
b.
aromatic
An SN2 does not occur
on an aryl halide.
cannot be made by
Gabriel synthesis
can be made by Gabriel
synthesis
NH2
N
H
c.
d.
2° amine
cannot be made by
Gabriel synthesis
N on 3° C
An SN2 does not
occur on a 3° RX.
cannot be made by
Gabriel synthesis
25.14 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
CH3
CH3CHC N
CH3CHCNH2
CH3
CH3
O
b.
CH2NH2
CH2NO2
C N
C
NH2
c.
NH2
NO2
C
N
O
NH2
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Amines 25–7
25.15 Primary (1°), 2o, and 3o amides are reduced to 1o, 2o, and 3o amines respectively, using
LiAlH4.
CONH2
O
CH2NH2
a.
NHCH3
c.
NHCH3
O
b.
N
N
25.16
General reaction:
(CH3)2CHNH2
[H]
R C N
isopropylamine
RCH2NH2
The amine needs 2 H's here.
The C bonded to the N must have 2 H's for the amine to be formed by reduction of a nitrile.
25.17 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 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
N bonded to a 3° C
cannot be made by
reduction of an amide
25.18 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
CH3NH2
NHCH3
c.
NaBH3CN
O
b.
NH3
NaBH3CN
O
(CH3CH2)2NH
NaBH3CN
NH2
d.
O
+
25.19
NH2
a.
O
NH3
OH
b.
OH
NHCH3
O
NH2CH3
or
OH
NH2
CH2=O
NH2
NaBH3CN
N(CH2CH3)2
NHCH(CH3)2
666
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Chapter 25–8
25.20
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
a.
NH2
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
25.21 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 like HCl and H2SO4, and by carboxylic acids as well.
a. CH3CH2CH2CH2 NH2
CH3CH2CH2CH2 NH3 + Cl–
+ HCl
b. C6H5COOH
+
c.
pKa 10
weaker acid
products favored
pKa = –7
N
H
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
25.22 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
X
b.
(CH3CH2CH2CH2)3N
and
+
H Cl
CH3
NH3 Cl–
CH3
X
• insoluble in H2O
• soluble in CH2Cl2
• soluble in H2O
• insoluble in CH2Cl2
(CH3CH2CH2CH2)2O
+
H Cl
Y
(CH3CH2CH2CH2)3NH Cl–
• soluble in H2O
• insoluble in CH2Cl2
(CH3CH2CH2CH2)2O
Y
• insoluble in H2O
• soluble in CH2Cl2
25.23 Primary (1°), 2o, and 3o alkylamines are more basic than NH3 because of the electron-donating
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
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Amines 25–9
25.24 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
25.25 Amides are much less basic than amines because the electron pair on N is highly delocalized.
CONH2
NH2
NH2
amide
least basic
arylamine
intermediate basicity
alkylamine
most basic
25.26
sp2 hybridized
more basic
This N is also sp2 hybridized
but the electron pair
CH3
occupies a p orpital so it can
N
N
delocalize onto the aromatic
a. Electron pair on N
CH3
occupies an sp2
ring. Delocalization makes
hybrid orbital.
this N less basic.
DMAP
4-(N,N-dimethylamino)pyridine
b.
sp3 hybridized N
stronger base
N
H
CH3
N
nicotine
sp2 hybridized N
higher percent s-character
weaker base
25.27
This electron pair is delocalized,
making it a weaker base.
Br
2
H
N
N
stronger base
sp3 hybridized N
25% s-character
HO
H
NH2
CH3O
a.
N
stronger base
This compound is similar to
DMAP in Problem 25.26a.
b.
sp2 hybridized N
33% s-character
sp hybridized N
33% s-character
N
c.
N(CH3)2
stronger base
sp3 hybridized N
25% s-character
668
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Chapter 25–10
25.28 Amines attack carbonyl groups to form products of nucleophilic addition or substitution.
CH3CH2CH2NH2
O
a.
b. CH3
O
O
C
C
O
NCH2CH2CH3
CH3CH2CH2NH2
CH3
COCl
c.
CH3
O
O
O
C
C
C
NHCH2CH2CH3
CH3
O
CONHCH2CH2CH3
CH3CH2CH2NH2
(CH3CH2)2NH
O
N(CH2CH3)2
O
(CH3CH2)2NH
CH3
COCl
CH3
C
N(CH2CH3)2
CON(CH2CH3)2
(CH3CH2)2NH
25.29 [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.
C
CH3
NH2
O
C CH3
(CH3)3CCl (CH ) C
NH
3 3
AlCl3
Cl
(+ ortho isomer)
O
NH2
b.
C
CH3
O
C CH3
H3O+
NH
(CH3)3C
H
N
Cl
Cl
C
CH3
H
N
O
AlCl3
O
C
CH3
NH2
H3O+
O
CCH2CH3
O
O
(+ para isomer)
25.30
transition
state
+
Energy
transition state:
N(CH3)3
H
Ea
OH
starting
materials
(no 3-D geometry shown here)
H°
products
Reaction coordinate
25.31
a. CH3CH2CH2CH2 NH2
b.
(CH3)2CHNH2
[1] CH3I (excess)
[2] Ag2O
[3] [1] CH3I (excess)
[2] Ag2O
[3] CH3CH2CH=CH2
CH3CH=CH2
c.
NH2
[1] CH3I (excess)
[2] Ag2O
[3] NH2
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Amines 25–11
25.32 In a Hofmann elimination, the base removes a proton from the less substituted, more accessible carbon atom, because of the bulky leaving group on the nearby carbon.
[1] CH3I (excess)
a.
CH2CHCH3
CH=CHCH3
[2] Ag2O
[3] NH2
major product
[1] CH3I (excess)
b.
H2N
major product
+
[1]
CH3I (excess)
c.
[2] Ag2O
[3] CH2CH=CH2
N
H
[2] Ag2O
[3] (3 C's)
least substituted 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 C
25.33
K+ –OC(CH3)3
a.
K+ –OC(CH3)3
c.
Cl
Br
[1] CH3I (excess)
b.
NH2
[1] CH3I (excess)
d.
[2] Ag2O
[3] (E and Z)
[2] Ag2O
[3] NH2
25.34
N2+ Cl–
NH2
NaNO2
a.
HCl
CH3
b.
H
CH3CH2 N CH3
HCl
HCl
N
H
CH3
NaNO2
NaNO2
c.
N
NO
NaNO2
CH3CH2 N CH3
d.
HCl
NO
N2 Cl–
NH2
25.35
a.
NH2 [1] NaNO2, HCl
[2] CuBr
b.
NH2
Br
[1] NaNO2, HCl
OH
c. CH3O
NH2
d.
N2+ Cl–
[2] H2O
O2N
O2N
Cl
[1] NaNO2, HCl
[1] CuCN
CH2NH2
[2] LiAlH4
[3] H2O
Cl
25.36
NH2
NO2
a.
HNO3
H2
H2SO4
Pd-C
F
[1] NaNO2, HCl
[2] HBF4
CH3O
[2] HBF4
F
670
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Chapter 25–12
NO2
b.
NO2
HNO3
H2
H2SO4
Pd-C
(from a.)
OH
NH2
[1] NaNO2, HCl
[2] H2O
NO2
OH
NH2
CH3
[1] NaNO2, HCl
c.
CH3Cl
[2] NaI
NH2
I
(+ para isomer)
AlCl3
I
(from a.)
NH2
NH2
Cl
Cl
Cl2 (excess)
d.
Cl
Cl
[1] NaNO2, HCl
[2] H3PO2
FeCl3
(from a.)
Cl
Cl
25.37
OH
NH2
N N
a.
NH2
c.
HO
OH
N N
HO
b.
HO
N N
OH
25.38 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
CH3
O2N
H2N
Cl
HO
Cl– N2
Cl– N2
CH3
25.39
a.
N N
OH
b. O2N
NO2
alizarine yellow R
para red
+
N N
Cl–
N N
OH
COOH
+
NO2
O2N
N N
Cl–
OH
COOH
OH
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Amines 25–13
25.40
O
CH3
O
O
O
O
O
–
O3S
N N
O
Dacron
O
methyl orange
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.
N
CH3
672
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Chapter 25–14
25.41
a. CH3NHCH2CH2CH2CH3
N-methyl-1-butanamine
(N-methylbutylamine)
b.
i.
N
H
tripropylamine
CH2CH3
2-ethylpyrrolidine
f. (C6H5)2NH
diphenylamine
NH2
1-octanamine
(octylamine)
c.
(CH3CH2CH2)3N
e.
CH3CH2CH2CH(NH2)CH(CH3)2
2-methyl-3-hexanamine
N C(CH3)3
g.
NH2
j.
NH2
k.
CH2CH3
4,6-dimethyl-1-heptanamine
3-ethyl-2-methylcyclohexanamine
N-tert-butyl-N-ethylaniline
CH3
N
d.
h. O
l.
NH2
N(CH2CH3)2
CH2CH2CH3
4-aminocyclohexanone
N-methyl-N-propylcyclohexanamine
N,N-diethylcycloheptanamine
25.42
h. 3-methyl-2-hexanamine
e. N-methylpyrrole
a. cyclobutylamine
NH2
NH2
N
b. N-isobutylcyclopentylamine
CH3
i. 2-sec-butylpiperidine
f. N-methylcyclopentylamine
N
NHCH3
H
c. tri-tert-butylamine
N
H
g. cis-2-aminocyclohexanol
N[C(CH3)3]3
NH2
NH2
or
d. N,N-diisopropylaniline
OH
N[CH(CH3)2]2
j. (2S)-2-heptanamine
H NH2
OH
25.43
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
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Amines 25–15
25.44 [* denotes a stereogenic center.]
a.
*
N
CH3
b.
CH2CH3
*
*
CH3CH2CHCH
CH
CH
2
2
2 N CH2CH2CH2CH3
CH2CH3
CH3
1 stereogenic center
2 stereoisomers
N
H
H CH3
CH3
CH3CH2
CH2CH2CH2
H CH3
N
Cl
CH3
N
Cl
CH2CH2CH2
N
Cl
CH3 CH2CH3
C
CH3CH2
CH2CH2CH2CH3
CH2CH2CH2
H CH3
CH3 CH2CH3
C
CH3CH2
CH2CH3
H CH3
CH3 CH2CH3
C
CH2CH3
H
CH3
2 stereogenic centers
4 stereoisomers
CH2CH2CH2CH3
CH3 CH2CH3
C
CH3CH2
CH2CH2CH2CH3
N
Cl
CH2CH2CH2
N
Cl
CH2CH2CH2CH3
25.45
a.
c. (CH3CH2)2NH
(CH3CH2)2NH or
N
sp3 hybridized N
stronger base
b.
2° alkylamine
stronger base
sp2 hybridized N
weaker base
HCON(CH3)2 or (CH3)3N
or
(ClCH2CH2)2NH
2° alkylamine
Cl is electron withdrawing.
weaker base
or
d.
NH
N
H
amide
alkylamine
weaker base stronger base
weaker base
(delocalized electron
pair on N)
stronger base
25.46
a.
NH2
NH2
NH3
arylamine intermediate
least basic
basicity
alkylamine
most basic
O2N
b.
N
H
delocalized
electron pair on N
least basic
d.
N
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
674
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Chapter 25–16
25.47 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
CH2NH3
pKa = 10.7
NH3
pKa intermediate
NH2
alkylamine
cyclohexanamine
pKa = 4.6
CH2NH2
electron-withdrawing inductive
effect of the sp2 hybridized C's
benzylamine
NH2
resonance-stabilized
aromatic amine
aniline
25.48 The most basic N atom is protonated on treatment with acid.
O
O
O
O
NH
a.
NH2
CH3CO2H
N
N
more
basic
O
CH2COOH
+ CH3CO2–
O
CH2COOH
benazepril
N
N
b.
NH
CH3CO2H
N
NH2 + CH3CO2–
N
varenicline
most
basic
25.49
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.
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Amines 25–17
25.50 The most basic N atom is protonated on treatment with acid.
a 3° alkylamine with an
sp3 hybridized N
most basic
O
O
Cl
a.
N
Cl
Cl
b.
HCl
N
NH
N
H
Cl–
NH
aripiprazole
a delocalized electron
part of an amide
pair on the benzene ring
least basic
intermediate
O
Cl
N
O
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:
O2N
NH2
O2N
NH2
O2N
NH2
O2N
NH2
O2N
NH2
meta
NH2
NH2
O2N
O2N
NH2
O
para
N
NH2
O2N
NH2
O2N
O
NH2
O
N
NH2
O
O
N
O
25.52
N
A
N
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
B
N
Geometry makes it difficult
to have a double bond here.
676
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25.53
B
NH
N
N
pyrrole
pKa = 23
stronger acid
N
N
N
B
NH
stronger conjugate base
The electron pair is not
delocalized on the ring.
The N atom is sp3
hybridized.
pyrrolidine
pKa = 44
weaker acid
weaker conjugate base
The electron pair is delocalized, decreasing the
basicity. The N atom is sp2 hybridized.
N
25.54
a. C6H5CH2CH2CH2Br
b. C6H5CH2CH2Br
NH3
excess
NaCN
c. C6H5CH2CH2CH2NO2
[1] LiAlH4
C6H5CH2CH2CN
H2
Pd-C
d. C6H5CH2CH2CONH2
C6H5CH2CH2CH2NH2
[2] H2O
[1] LiAlH4
[2] H2O
C6H5CH2CH2CH2NH2 e. C6H5CH2CH2CHO
NH3
NaBH3CN
C6H5CH2CH2CH2NH2
C6H5CH2CH2CH2NH2
C6H5CH2CH2CH2NH2
25.55
a. (CH3CH2)2NH
NH2
b.
N(CH3)2
c.
H
N
d.
O
CH3
C
NHCH2CH3
H
N
N(CH3)2
NH2
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
b.
H
N
C6H5
H
C6H5
or
NH2
O
(CH3CH2CH2)2NH
H
N
H
H
or
H
C
O
CH2CH3
O
d.
NH2
C6H5
O
O
c. (CH3CH2CH2)2N(CH2)2CH(CH3)2
H2N
H
CH3CH2CH2
N
H
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Amines 25–19
25.57
a.
NH2
C6H5
O
b.
O
C6H5
c.
C6H5
NH3
CHO
NH
NaBH3CN
C6H5
CH2NH2
NaBH3CN
NH2
(CH3)2NH
d.
N(CH3)2
NaBH3CN
NaBH3CN
O
NH
25.58
a.
CH2NH2
Br
NH3
excess
CN
CH2NH2
[1] LiAlH4
b.
[2] H2O
CONH2
CH2NH2
[1] LiAlH4
c.
[2] H2O
CHO
CH2NH2
NH3
d.
NaBH3CN
CH3
CH2Br
e.
NH3
h
excess
COOH
CONH2
[1] SOCl2
f.
[2] NH3
NH2
g.
CH2NH2
[1] LiAlH4
[2] H2O
CN
[1] NaNO2, HCl
CH2NH2
[1] LiAlH4
[2] H2O
[2] CuCN
h.
CH2NH2
Br2
NO2
HNO3
H2
Pd-C
H2SO4
NH2
Then as in (g).
25.59 Use the directions from Answer 25.22. Separation can be achieved because benzoic acid reacts
with aqueous base and aniline reacts with aqueous acid according to the following equations:
COOH
NaOH
benzoic acid
• soluble in CH2Cl2
• insoluble in H2O
COO–Na+
+ 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
678
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Chapter 25–20
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
+
Cl–
NH3
CH3
25.60
H
N
N-ethylaniline
a.
b.
HCl
H H
N
f.
CH3I (excess)
Cl–
+ CH2=CH2
O
CH3CH2COCl
CH3COOH
N
g.
H H
N
CH3COO–
O
h. The product in (g)
c.
N(CH3)2
Ag2O
N
(CH3)2C=O
HNO3
O
N
N
H2SO4
O2N
NO2
CH3
N
d.
e.
CH2O
i. The product in (g)
[1] LiAlH4
NaBH3CN
CH3I
CH3 CH3
N
(excess)
I–
N
[2] H2O
O
j. The product in (h)
H2
O
N
Pd-C
NH2 H2N
N
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Amines 25–21
25.61
NH2 p-methylaniline
CH3
CH3COCl
f.
HCl
a.
CH3
NH3
CH3COCl
CH3COOH
g.
h.
NaNO2
i. Step (b), then
excess
e.
N2+Cl–
CH3
HCl
C CH3
NH
c. (CH3CO)2O CH
3
CH3I
NH3 CH3COO–
CH3
C CH3
NH
CH3
O
d.
NH2
Cl–
O
b.
AlCl3
AlCl3
CH3
O
CH3COCl
C CH3
NH
CH3
AlCl3
N(CH3)3 I–
CH3
(CH3)2C=O
CH3
C O
CH3
N C(CH3)2
j.
CH3CHO
NaBH3CN
CH3
25.62
a. CH3CH2CH2CH2NH2
b. CH3CH2CH2CH2NH2
c. CH3CH2CH2CH2NH2
d. CH3CH2CH2CH2NH2
e. CH3CH2CH2CH2NH2
f. CH3CH2CH2CH2NH2
ClCOC6H5
CH3CH2CH2CH2NHCOC6H5
O=C(CH2CH3)2
[1] CH3I (excess)
[2] Ag2O
[3] C6H5CHO
NaBH3CN
NaBH3CN
CH3CHO
CH3I
excess
CH3CH2CH2CH2N=C(CH2CH3)2
CH3CH2CH=CH2
CH3CH2CH2CH2NHCH2C6H5
CH3CH2CH2CH2NHCH2CH3
[CH3CH2CH2CH2N(CH3)3]+I–
25.63
a. CH3(CH2)6NH2
[1] CH3I (excess)
[2] Ag2O
[3] CH3(CH2)4CH=CH2
[1] CH3I (excess)
b.
NH2
[2] Ag2O
[3] (E + Z)
major product
NHCH2CH3
680
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Chapter 25–22
c.
[1] CH3I (excess)
N
H
CH2=CH2
e.
major product
+ (CH3)2CHN(CH3)2
[2] Ag2O
[3] 1
2
N
H
[2] Ag2O
[3] 3
CH2=CHCH3 + CH3CH2N(CH3)2
CH3
[2] Ag2O
[3] NH2
CH3
N(CH3)2
+
major product
2
(CH3)2N
[1] CH3I (excess)
d.
1
[1] CH3I (excess)
CH3
CH3
+
(E + Z)
(CH3)2N
CH3
CH3
(E + Z)
CH2
major product
CH3
25.64
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] O
NH3
NaBH3CN
OH
O
CH3NH2
mCPBA
NHCH3
e.
(from d.)
O
O
O
Br2
f.
CH3COOH
25.65
CH3
N
a.
one stereogenic center
benzphetamine
Br
NH2CH2CH2CH2CH3
NHCH2CH2CH2CH3
3
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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
1
H
N
NH
CH2=O
CH3
N
[1] CH3I
2
[2] Ag2O
[3] d.
(CH3)2NCH2
major product
elimination across , 1
(elimination across , 2)
25.66
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] CH3CH=CH2
(CH3)2NCH(CH3)2
CH3CH2CH2N(CH3)2
682
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Chapter 25–24
25.67 NH2 and H must be anti for the Hofmann elimination. Rotate around the C–C bond so the NH2
and H are anti.
C6H5
a.
NH2
H
C6H5
NH2
CH3
b.
H
CH3
C6H5
CH2CH3
CH3
c.
H
(CH3)3C
C(CH3)3
NH2
CH2CH3
CH2CH3
C(CH3)3
rotate 120°
counterclockwise
three steps
(CH3)3C
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
25.68
N2+ Cl–
Cl
HO
Cl
Br
a. H2O
b. H3PO2
A
Cl
h. C6H5NH2
d. CuBr
H
Cl
Cl
NH2
N N
OH
Cl
NC
Cl
F
Cl
e. CuCN
i. C6H5OH
c. CuCl
N N
Cl
Cl
f. HBF4
I
I
Cl
Cl
j. KI
g. NaI
25.69 Under the acidic conditions of the reaction, aniline is first protonated to form an ammonium salt
that has a positive charge on the atom bonded to the benzene ring. The –NH3+ is now an
electron-withdrawing meta director, so a significant amount of meta substitution occurs.
NH2
H OSO3H
NH3
+ HSO4–
This group is now a meta director.
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Amines 25–25
25.70
CH3
R
C
CH3CH2CH2
[1] CH3I (excess)
H
NH2
H
C C
[2] Ag2O
[3] A
H
B
O
[1] O3
H
[2] CH3SCH3
CH2CH2CH3
H
C
O
H
H
C
CH2CH2CH3
25.71
a.
Br
Br
H H Na+
N
Br
OH
H
N
Br
Br
Na+
+ CH3CH2NH2
Br
N
H CH2CH3
+ H2O
Na+
OH
+ H2O + Na+
N
CH2CH3
H A
H
O
O
H
N
O
H
b.
N
proton
transfer
NH2
OH2
H
N
H3B–H
H
H
N
+ H2O
proton
source
+ A
H
N
+ BH3
25.72
OH
O
OH
HN
H N
H2C=O
H A
OH
OH
N
proton
transfer
A
OH
OH
OH
H
N
OH2
OH
H2C
N
N
N
A
H2O
H A
OH
H
OH
H
N
OH
H
N
N
684
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Chapter 25–26
25.73
N N
N N
N N
N N
N N
X
R N N
X
alkyl diazonium salt
The N2+ group on an aromatic ring is stabilized by
resonance, whereas the alkyl diazonium salt is not.
aryl diazonium salt
25.74
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
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Amines 25–27
25.75 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.
H Cl
O N O
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
25.76
O
a.
NH2
NH3
[1] CH3I
NaBH3CN
[2] Ag2O
[3] O
OH
NaBH4
b.
(Hofmann elimination, less
substituted C=C favored)
H2SO4
(more substituted C=C favored)
CH3OH
(E + Z isomers formed)
25.77
NO2
HNO3
H2SO4
a.
b.
NO2
Br2
FeBr3
NH2
H2
Pd-C
ClCOCH3
H2
Br
NO2
Pd-C
Br
NH2
Br
NHCOCH3
(from a.)
CH3Cl
AlCl3
c.
CH3
HNO3
O2N
H2SO4
CH3
H2
Pd-C
H2N
CH3
[1] NaNO2, HCl
[2] NaI
I
(+ ortho isomer)
H2N
[1] NaNO2, HCl
d.
NC
Br2
FeBr3
[2] CuCN
NC
Br
(from a.)
e.
Br2
FeBr3
HNO3
Br
H2SO4
NO2
Br
(+ para isomer)
H2
Pd-C
NH2
Br
[1] NaNO2, HCl
[2] H2O
OH
Br
CH3
686
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Chapter 25–28
f.
O2N
CH3
KMnO4
O2N
COOH
H2
H2N
Pd-C
COOH
[1] NaNO2, HCl
HO
[2] H2O
COOH
(from c.)
O
O
ClCOCH3
g.
HNO3
H2SO4
AlCl3
h.
NH2
[1] NaNO2, HCl
O
O2N
O
H2N
H2
[1] NaNO2, HCl
[2] NaI
Pd-C
C6H5N2+Cl–
OH
I
N N
OH
[2] H2O
(from a.)
(from d., Step [1])
25.78
NH2
a.
CN
[1] NaNO2, HCl
COOH
H3O+
[2] NaCN
[2] NH2CH3
Br
Br
Br2
CH3COCl
NH2
b.
CONHCH3
[1] SOCl2
Br
–
CH3Cl
NHCOCH3 FeBr3
OH
NHCOCH3 H2O
NHCOCH3 AlCl3
(+ ortho isomer)
NH2
[1] NaNO2, HCl
[2] H3PO2
CH3
Br
COOH
COOCH2CH3
HOCH2CH3
c.
H2SO4
Br
(from a.)
d.
COOH
[1] LiAlH4
[2] H2O
CH2OH
Br2
Br
FeBr3
CH2OH
(3x)
(from a.)
Br
HO
H2N
e.
HO
HO
[1] NaNO2, HCl
[2] H2O
CH3Cl
AlCl3
C6H5N2
CH3
(+ ortho isomer)
+Cl–
(from a., Step [1])
N N
CH3
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Amines 25–29
25.79
CH3Cl
[1]
Br2
AlCl3
h
H2C=O
NH2
OH
N
H
NaBH3CN
(excess)
–OH
Br
[2]
NH3
Br
CHO CH NH
3
2
PCC
N
H
NaBH3CN
(from [1])
[1] LiAlH4
[2] H2O
O
COOH [1] SOCl
2
KMnO4
[2] CH3NH2
[3]
NCH3
[1] LiAlH4
H
[2] H2O
N
H
(from [1])
[4]
CHO
CN [1] DIBAL-H
[1] HNO3, H2SO4
Then route [2].
[2] H2O
[2] H2, Pd-C
[3] NaNO2, HCl
[1] LiAlH4
[4] CuCN
[2] H2O
NH2
Br
Br2
[5]
Then route [1].
COOH
MgBr [1] CO
2
Mg
Then route [3].
[2] H3O+
FeBr3
[1] H2C=O
[2] H2O
OH
Then route [2].
25.80
O
O
mCPBA
O
O
O
[1] LiAlH4
O
[2] H2O
O
PBr3
OH
O
PCC
O
NHCH3
O
CH3NH2
O
NaBH3CN
O
CH3NH2
O
O
Br
a.
CH3O
MDMA
MDMA
[part (b)]
[part (a)]
CH3O
CN [1] LiAlH
4
NaCN
[2] H2O
CH3O
CH3O
NHCH3
O
25.81
CH3O
Br
O
CH3O
CH3O
NH2
CH3O
CH3O
mescaline
688
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Chapter 25–30
CH3O
Br
CH3O
CH3O
HBr
b.
CH3O
CH3O
ROOR
CH3O
excess
CH3O
CH3O
CH3O
Br
CH3O
Mg
MgBr
c.
CH3O
NH2
NH3
O
[1]
CH3O
[2] H2O
CH3O
CH3O
CH3O
OH
PBr3
CH3O
CH3O
CH3O
Br
CH3O
CH3O
CH3O
NH3
excess
CH3O
NH2
CH3O
CH3O
25.82
NO2
HNO3
a.
H2SO4
NH2
H2
CN
[1] NaNO2, HCl
COOH
H3O+
[2] CuCN
Pd-C
COOH
Cl2
FeCl3
Cl
Cl2
HNO3
b.
NO2
CH3Cl
NH2
CH3
OH
H2
NO2
CH3
Pd-C
CH3CH2Cl
d.
AlCl3
HNO3
CH3CH2
H2SO4
CH2NH2
[1] LiAlH4
[2] H2O
H2
NO2
Pd-C
CH3CH2
NH2
[1] NaNO2, HCl
[2] NaCN
(+ ortho isomer)
CH3
CH3
CH3CH2
CH3
Br2
O2N
CH3
(from c.)
Br
Cl
[1] NaNO2, HCl
N N
[2]
NH2
(from b.)
(from a.)
Cl
NH2
H3O+
CH3
Br
Cl
NH2
Cl
H2N
NH2
[1] NaNO2, HCl
[2] CuCN
COOH
Pd-C
FeBr3
f.
H2
CN
CH3CH2
(+ ortho isomer)
e. O2N
Cl
[2] H2O
Pd-C
H2SO4
Cl
[1] NaNO2, HCl
NO2
HNO3
CH3
AlCl3
Cl
H2
FeCl3
(2X)
H2SO4
c.
Cl
Cl
Cl
CH3CH2
[1] NaNO2, HCl
[2] H2O
CN
HO
CH3
Br
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Amines 25–31
25.83
O
NO2
HNO3
a.
NO2 H
2
Cl2
FeCl3
H2SO4
NH2
O
H
N
O
O
Pd-C
Cl
Cl2
FeCl3
H
N
Cl
Cl
O
Cl
(+ isomer)
Cl
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
ClCOCH2CH3
c.
O
O
Cl2
AlCl3
OH
NHCH3 NaBH4
Cl NH2CH3
NHCH3
CH3OH
H2O, HCl
25.84
CH3OH
OH
NH2
a.
[1] HNO3, H2SO4
[1] NaNO2, HCl
[2] H2, Pd-C
[2] H2O
SOCl2
OH
OH
CH3Cl
Br2
AlCl3
FeBr3
(2X)
Br
OCH3
Br
[1] NaH
CH3
CH3
CH3
[1] Br2, h
Br
NH2
CH3O
Br
CH3CH2OH
CrO3, H2SO4, H2O
CH3COOH
SOCl2
CH3COCl
b.
O
H
N
O
(from a.)
H
N
Cl
–OH,
O
AlCl3
O
(+ ortho isomer)
H2O
Br
[2] CH3Cl
(+ ortho
isomer)
NH2
Br
O
NH2
[2] NH3
(excess)
690
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Chapter 25–32
CH3OH
SOCl2
[1] CH3Cl, AlCl3
Cl2
c.
Cl
FeCl3
HOOC
[2] KMnO4
[1] LiAlH4
[2] H2O
Cl
HO
Cl
PCC
O
Cl
H
(+ ortho isomer)
NH2
(from a.)
mild acid
N CH
Cl
CH3OH
SOCl2
CH3
CH3Cl
d.
AlCl3
CH3
HNO3
H2SO4
COOCH3
[1] KMnO4
[2] CH3OH, H+
O2N
COOCH3
H2
Pd-C
O2N
H2N
PCC
(+ ortho isomer)
O
(CH3)2CHNH
CO2CH3
OH
NaBH3CN
NH2
e. O2N
CH3
(from d.)
[1] H2, Pd-C
F
CH3
[2] NaNO2, HCl
[3] HBF4
O
O
[1] KMnO4
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
[2] NaNO2, HCl
(from d.)
CH3
[1] H2, Pd-C
[3] NaI
I
O
[1] KMnO4
Cl
[2] SOCl2
I
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691
Amines 25–33
25.85
molecular weight = 87
C5H13N
two IR peaks = 1° amine
H
H
H
H
H
N
N
N
N
H
H
H
H
H
N
H
H
H
N
H
H
N
H
N
25.86
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
7.2 (multiplet, 5 H) benzene ring
6.8–7.2 (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
25.87
HO
C N
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.
1H 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
25.88 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
H2N
C
HA
NH2
guanidine
H2N
NH2
NH2
NH2
C
C
C
NH2
pKa = 13.6
H2N
NH2
H2N
NH2
NH2
H2N
C
NH2
692
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Chapter 25–34
25.89
CH3
N
[1] CH3I (excess)
CH3
N
CH3
I
[2] Ag2O
[2] Ag2O
[1] CH3I (excess)
[3] [3] N(CH3)2
C8H10
N(CH3)3
I
Y
25.90
One possibility:
O
O
O
CH3COCl
a.
O
O
AlCl3
O
Br2
H2NC(CH3)3
CH3CO2H
O
O
Br
O
O
O
(+ isomer)
NaBH4
CH3OH
HO
O
H3O+
HO
OH
albuterol
b.
O
+
HN(CH2CH3)2
H2SO4 O N
2
OH
N
H
A
[1] NaNO2, HCl
H2
Pd-C
O
N
H
N(CH2CH3)2
HO
HNO3
N
H
[2] H2O
H2N
[1] NaH
HO
O
Cl
[2]
Br2
O2N
COOH
HNO3
O
SOCl2
O2N
O
Br
COOH
H2SO4
H3O+ CO2
O
O
O
COCl
A
O2N
O
O
N(CH2CH3)2
H2
H2N
Pd-C
O
Mg
O
O
N(CH2CH3)2
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C–C Bond-Forming Reactions 26–1
C
miinngg R
Reeaaccttiioonnss iinn O
Orrggaanniicc SSyynntthheessiiss
Chhaapptteerr 2266:: C
Caarrbboonn––C
Caarrbboonn B
Boonndd--FFoorrm
C
Coouupplliinngg rreeaaccttiioonnss
[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
•
[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–
•
C
Cyycclloopprrooppaannee ssyynntthheessiiss
[1] Addition of dihalocarbenes to alkenes (26.4)
C
X X
C
CHX3
C
KOC(CH3)3
C
CH2I2
C C
Zn(Cu)
H H
C
C C
+ ZnI2
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.
•
•
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)
R'X is most often a vinyl halide
or aryl halide.
With vinyl halides, coupling is
stereospecific.
M
Meettaatthheessiiss ((2266..66))
2 RCH CH2
Grubbs
RCH CHR
catalyst
Grubbs
catalyst
diene
+ CH2 CH2
+ CH2 CH2
693
694
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Chapter 26–2
C
Chhaapptteerr 2266:: A
Annssw
weerrss ttoo PPrroobblleem
mss
26.1 A new C–C bond is formed in each coupling reaction.
a.
(CH2=CH)2CuLi
Cl
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
OH
A=
B=
(CH3CH2)2CuLi
several
steps
I
(CH3)2CuLi
OH
several
steps
OH
O
CO2CH3
C18 juvenile
hormone
26.3
Br
a. (CH3)2CHCH2CH2I
[1] Li
[(CH3)2CHCH2CH2]2CuLi
CH2CH2CH(CH3)2
[2] CuI
or
Br
b.
[1] Li
[2] CuI
[1] Li
Cl
(CH3)2CHCH2CH2I
CuLi
CH2CH2CH(CH3)2
2
CuLi
Br
[2] CuI
2
Cl
c.
Cl
[1] Li
[2] CuI
CuLi
2
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C–C Bond-Forming Reactions 26–3
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
O
O
H B
O
a. CH3CH2CH2 C C H
b.
I
Li
Li
CH3O
O
B(OCH3)3
Br
Br2
c.
I
B
H
CH3O
Pd(PPh3)4, NaOH
B(OCH3)2
Li
Li
I
NaOH
B(OCH3)2
B(OCH3)3
Br
Pd(PPh3)4
CH3O
CH3O
Pd(PPh3)4
NaOH
(+ ortho isomer)
CH3O
26.6
Br
Heck
a.
+
b.
I
reaction
OCH3
+
OCH3
Heck
reaction
Br
c.
CN
+
OCH3
d.
Heck
CN
reaction
Br
+
O
Heck
OCH3
reaction
O
26.7 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
a.
Br
OCH3
O
OCH3
695
696
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Chapter 26–4
Br
b.
O
c.
O
CO2CH3
CO2CH3
Br
26.8 Add the carbene carbon from either side of the alkene.
H
CH3
a.
CHCl3
C C
CH3 C
H
KOC(CH3)3
H
H
Cl Cl
C
C
H
+
CH3
H
C
H
enantiomers
H
H
C
C
Cl Cl
CH3CH2
b.
CH2CH3
CHCl3
C
KOC(CH3)3 CH3CH2
H
C C
H
H
Cl Cl
C
C
+
CH2CH3
H
CH3CH2
H
c.
CH3
CHCl3
CCl2
KOC(CH3)3
C
CH2CH3
H
C
identical
CH3
C
Cl Cl
CH3
+
CCl2
H
H
enantiomers
26.9
CHCl3
a.
Cl
Cl
KOC(CH3)3
Br
Br
(from b.)
CHBr3
b.
LiCu(CH3)2
c.
Br
Br
KOC(CH3)3
26.10
a.
CH2I2
+
ZnI2
Zn(Cu)
H
b.
CH2I2
+
Zn(Cu)
H
ZnI2
c.
CH2I2
Zn(Cu)
+
ZnI2
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C–C Bond-Forming Reactions 26–5
26.11 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
26.12
Grubbs
catalyst
a.
c.
(E and Z)
Grubbs
catalyst
OCH3
Grubbs
catalyst
b.
(CH2=CH2 is also formed in each reaction.)
(E and Z)
OCH3
OCH3
26.13
+
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.
26.14
O
a.
Ph
+
O
O
Ph
CO2CH3
O
CO2CH3
CO2CH3
b.
CO2CH3
26.15 High dilution conditions favor intramolecular metathesis.
O
Y =
O
O
Under high dilution conditions, an intramolecular reaction is favored, resulting in Y.
Two different molecules do not often come into contact, so two double bonds in the
same molecule react.
697
698
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Chapter 26–6
long chain
1
O
Alkenes 1 and 2 differ in proximity to the ether side chain. Three products
are possible because alkene 1 can react with alkene 1 in another molecule
(forming Z); alkene 1 can react with alkene 2 in another molecule (forming
Z'); and alkene 2 can react with alkene 2 in another molecule (forming Z").
At usual reaction concentrations, the probability of two molecules
approaching each other is greater than the probability of two sites
(connected by a long chain) in the same molecule reacting together, so the
intermolecular reaction is favored.
O
O
2
long chain
O
Z
O
=
O
O
O
O
O
Z' =
O
O
O
O
O
O
Z" =
O
O
O
O
O
26.16 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.
O
b.
CH3O
CH3O
OH
OH
CO2CH3
CO2CH3
CHO
CHO
O
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C–C Bond-Forming Reactions 26–7
26.17
Cl
CH2CH2CH3
+
a.
(CH3CH2CH2)2CuLi
N
B(OCH3)2
b.
Pd(PPh3)4
Br
+
N
NaOH
O
O
Pd(OAc)2
Br
+
c.
P(o-tolyl)3
(CH3CH2)3N
N
d.
Cl
[1] Li
[2] CuI
[3]
Br
O
e.
Br
Pd(PPh3)4
B
+
O
f.
NaOH
Br +
CH3O
N
CO2CH3
Pd(OAc)2
CH3O
P(o-tolyl)3
(CH3CH2)3N
g.
Br
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
26.18
Br
a.
c.
CO2CH3
Br
CO2CH3
Br
d.
b.
Br
CH3O
CH3O
699
700
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Chapter 26–8
26.19
Each coupling reaction uses Pd(PPh3)4
and NaOH to form the conjugated diene.
I
A
I
O
H
H B
O
O
C C H
B
B
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 therefore be trans.
D
26.20 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
26.21 Inversion of configuration occurs with substitution of the methyl group for the tosylate.
a.
H
(CH3)2CuLi
OTs
H
(CH3)2CuLi
CH3
d.
CH3
OTs (CH3)2CuLi
26.22
HBr
1-butene
ROOR
Br
+
CuLi
2
[1] Li
[2] CuI
CuLi
2
H
CH3
H
H
(CH3)2CuLi
OTs
H
H
OTs
b.
c.
CH3
octane
H
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C–C Bond-Forming Reactions 26–9
26.23 Add the carbene carbon from either side of the alkene.
H
Cl
CHCl3
a.
CHCl3
e.
KOC(CH3)3
Cl
KOC(CH3)3
Cl Cl
C
H
H
H
CH2I2
b.
H
C
H
CH3
Br
CHBr3
Br
KOC(CH3)3
H
Cl Cl
CH3
Br
+
Br
Cl Cl
C
H
H
H
CH2I2
C
C
H
H
d.
C
H
Zn(Cu)
c.
H
+
C
CHCl3
f.
KOC(CH3)3
+
H
C
C
H
Zn(Cu)
H
H
26.24 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
26.25 High dilution conditions favor intramolecular metathesis.
H
OH
H
Grubbs
catalyst
N
a.
OH
N
OH
c.
Grubbs
catalyst
CO2CH3
OH
O
O
O
b.
CO2CH3
O
Grubbs
catalyst
26.26 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
c.
O
CO2CH3
702
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Chapter 26–10
26.27 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.
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)
+
26.28
O
O
O
M
O
O
26.29 All double bonds can have either the E or Z configuration.
a.
c.
b.
26.30
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
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C–C Bond-Forming Reactions 26–11
O
O
Grubbs
f.
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
26.31
O
Cl
Cl3C C
Cl
O Na+
C Na+ +
O C O
Cl
Cl
C
+
CO2
+
NaCl
Cl
26.32 This reaction follows the Simmons–Smith reaction mechanism illustrated in Mechanism 26.5.
Br
CHBr2
Zn(Cu)
[1]
CH
Zn
[2]
Br
+
ZnBr2
26.33
Ph2S
O
OCH3
Ph2S
a sulfur ylide
1,4-addition of the
sulfur ylide to the
carbon
O
OCH3
OCH3
a resonance-stabilized
enolate
O
methyl trans-chrysanthemate
Ph2S
704
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Chapter 26–12
26.34
+ N2
A
N
N
CO2CH3
N
N
CO2CH3
CO2CH3
diazo compound
CO2CH3
B
26.35
I
PAr3
[1]
a.
Ar3P Pd
Pd(PAr3)2
oxidative addition
I
PAr3
[2]
+
I
Ar3P Pd
syn addition
PAr3
Ar3P Pd
I
PAr3
[3]
H
H
H
The H cis to Pd
is eliminated.
+
syn elimination
H
E
This H
is trans
to Pd.
Ar3P Pd
[4]
I
reductive
elimination
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, and therefore it cannot be removed if elimination occurs in a syn fashion.
26.36
O
H B
Br NaNH2
C CH
O
Br
O
(– HBr)
B
O
(Z)-2-bromostyrene
Pd(PPh3)4
NaOH
A
26.37
Br2
FeBr3
O
O
H
H B
O
B
O
Br
Pd(PPh3)4, NaOH
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C–C Bond-Forming Reactions 26–13
26.38
HO
SOCl2
OH
HO
Cl
TBDMS–Cl
imidazole
HC CH
NaH
HC C
TBDMSO
Cl
O
B
OH
O
O
Bu4NF
BH
O
O
HC C
B
OTBDMS
OTBDMS
O
26.39
a. CH3O
CH3O
HNO3
NO2
H2SO4
H2
NH2
Pd-C
Br2
[1] NaH
OH
[2] CH3Br
PBr3
Synthesize these two components, and
then use a Heck reaction to synthesize the product.
Br
CH3O
NaNO2
N2+Cl–
HCl
CH3O
H2O
OH
Br
CH3OH
CH3CH2OH
SOCl2
CH3CH2Cl
CH2CH3
AlCl3
h
Pd(OAc)2
Br
CH3O
Br2
CHCH3
KOC(CH3)3
Br
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
706
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Chapter 26–14
26.40
CH2I2
a.
Zn(Cu)
Br Br
b.
OH
OH
Br Br
NBS
Br
NaOH
CHBr3
OH
h
OH
KOC(CH3)3
Br
c.
Br
CH2I2
(CH3CH2CH2CH2)2CuLi
Zn(Cu)
(from b.)
Br
d.
CuLi
CH2I2
2
Br
Zn(Cu)
(from b.)
26.41
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 26.39.
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C–C Bond-Forming Reactions 26–15
26.42
[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
707
708
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Chapter 26–16
26.43
Either compound can be used to
a. CH3
OCH3
CH3
Br
Br
OCH3 synthesize the organoborane, so
two routes are possible.
Possibility [1]:
CH3Cl
Br2
CH3
AlCl3
Br2
[1] Li
CH3
FeBr3
HNO3
Br
FeBr3
Br
[2] B(OCH3)3
Br
NO2
H2
CH3
Br
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
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C–C Bond-Forming Reactions 26–17
Br
O
c.
Br
O
O
O
This can't be converted to an organoborane
reagent via an organolithium reagent.
three
B(OCH3)2
(as in 26.43b)
steps
Br2
FeBr3
HNO3
Br
H2SO4
[1] NaNO2, HCl
H2
Br
NO2
Pd-C
Br
NH2
[2] H2O
Br
OH
CH3COCl
pyridine
B(OCH3)2
Br
O
Br
Pd(PPh3)4
NaOH
O
O
O
26.44
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
O
O
MgBr
TBDMS–Cl
imidazole
OTBDMS
710
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Chapter 26–18
26.45
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
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C–C Bond-Forming Reactions 26–19
26.46
Cl
CH3
CH3O
N
Cl
CH3
CH3O
CO2CH3
LiCu
N
[1] (CH3CH2CH2CH2)4N+ F—
CO2CH3 [2] PCC
Cl
CH3
CH3O
N
CO2CH3
OTBDMS
OTBDMS
I
CHO
2
directed aldol
[1] CH3CHO + LDA CH2CHO
[2] H3O+
O
N
O
O
Cl
CH3O
O
Cl
O
N
CH3
CH3O
N
several steps
CO2CH3
O
CH3O
OH
N
H
O
CHO
A
maytansine
26.47
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)
OH
b.
[1] OsO4
OTBDMS
TBDMS–Cl
HO
Br
H2SO4
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
712
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Chapter 26–20
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)
26.48
O
a.
O
O
O
b.
O
O
O
O
[1] LiAlH4
2
CO2Et
EtO2C
CO2Et
CO2Et
OH
[2] H2O
OH
[1] NaH (2 equiv)
Cl (2 equiv)
[2]
O
O
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C–C Bond-Forming Reactions 26–21
26.49
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
N
O
O
OCH2CH3
O
+ H2O
resonance-stabilized
H3O+
H OCH2CH3
26.50
O
CH3O
CH3O
H
O
X
O PPh3
PPh3
CH3O
H
O
Y
O
Ph3P
CH3O
CH3O
O
+ PPh3
O
Ph3P
+ Ph3P=O
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715
Carbohydrates 27–1
C
Chhaapptteerr 2277:: C
Caarrbboohhyyddrraatteess
IIm
mppoorrttaanntt tteerrm
mss
• Aldose
A monosaccharide containing an aldehyde (27.2)
• Ketose
A monosaccharide containing a ketone (27.2)
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 (27.2C)
• Epimers
Two diastereomers that differ in configuration around one stereogenic center only
(27.3)
• Anomers Monosaccharides that differ in configuration at only the hemiacetal OH group (27.6)
• Glycoside An acetal derived from a monosaccharide hemiacetal (27.7)
A
Accyycclliicc,, H
Haaw
woorrtthh,, aanndd 33--D
D rreepprreesseennttaattiioonnss ffoorr DDD--gglluuccoossee ((2277..66))
CHO
Haworth
projection
CH2OH
O
H
H
OH
H
H
H
HO
OH
HO
H
OH
CH2OH
O
H
H
OH
OH
H
H
H
OH
H
OH
H
HO
H
CH2OH
anomer
CHO
H
3-D representation
O
HO
HO
H
HO
HO
H C OH
OH
OH
OH
H
HO C H
H
H
H
H C OH
H
OH
anomer
acyclic form
OH
OH
OH
H
H C OH
O
H
OH
H
CH2OH
R
Reeaaccttiioonnss ooff m
moonnoossaacccchhaarriiddeess iinnvvoollvviinngg tthhee hheem
miiaacceettaall
[1] Glycoside formation (27.7A)
OH
HO
HO
OH
O
OH
OH
ROH
HCl
OH
O
HO
HO
+ HO
OR
HO
OH
OR
-D-glucose
O
•
•
Only the hemiacetal OH reacts.
A mixture of and glycosides
forms.
•
A mixture of and anomers
forms.
OH
glycoside
glycoside
[2] Glycoside hydrolysis (27.7B)
OH
OH
HO
HO
O
OH
OR
H3O+
HO
HO
OH
O
HO
+
HO
OH
OH
OH
OH
anomer
anomer
+
O
ROH
716
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Chapter 27–2
R
Reeaaccttiioonnss ooff m
moonnoossaacccchhaarriiddeess aatt tthhee O
OH
H ggrroouuppss
[1] Ether formation (27.8)
OR
OH
O
HO
HO
OH
OH
O
RO
RO
Ag2O
RX
•
•
All OH groups react.
The stereochemistry at all stereogenic centers
is retained.
•
•
All OH groups react.
The stereochemistry at all stereogenic centers
is retained.
OR
OR
[2] Ester formation (27.8)
OH
HO
HO
O
Ac2O or
OH
AcCl
OH
OAc
O
AcO
AcO
OAc
AcO
pyridine
R
Reeaaccttiioonnss ooff m
moonnoossaacccchhaarriiddeess aatt tthhee ccaarrbboonnyyll ggrroouupp
[1] Oxidation of aldoses (27.9B)
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 (27.9A)
CHO
CH2OH
NaBH4
CH3OH
CH2OH
CH2OH
aldose
alditol
[3] Wohl degradation (27.10A)
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.
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717
Carbohydrates 27–3
[4] Kiliani–Fischer synthesis (27.10B)
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
O
Otthheerr rreeaaccttiioonnss
[1] Hydrolysis of disaccharides (27.12)
O
H3O+
This bond is cleaved.
O
+
O
O
O
OH
OH
OH
A mixture of anomers is formed.
[2] Formation of N-glycosides (27.14B)
CH2OH
O
H
H
OH
H
H
OH
OH
RNH2
mild H+
CH2OH
O
H
H
OH
H
H
+
NHR
OH
CH2OH
O
H
H
OH
NHR
H
H
OH
•
Two anomers are formed.
718
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Chapter 27–4
C
Chhaapptteerr 2277:: A
Annssw
weerrss ttoo PPrroobblleem
mss
27.1
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.
CHO
CH2OH
a. a ketotetrose
C O
b. an aldopentose
H C OH
COOH
=
CH3
CHO
HO
CH3
OHC
OH
H C OH
CH2OH
HO
H
C
CH3
CHO
CH2CH3
re-draw
H
HOCH2 H
CHO
re-draw
C
C
c.
CH2CH2OH
CH2CH2OH
CHO
= HO
CH3
H
H
CHO
CH2OH = H
CHO
re-draw
HO
CH3
C
CH3
CH2OH
CH2CH3
CH2CH3
OH
d. H C CHO
C
CHO
=
HO
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).
CH2NH2
a. Cl
[1]
H
[1]
Cl C H
CH2NH2
[2]
H
[1]
CHO
CH2NH2
[3]
1
H
H
CHO
CHO
[3]
1 Cl C H 4
1
3
2
2
CHO
CHO
[3]
1 Cl C H 4
1 Cl C H
CH2OH
CH2OH
COOH
CH2Br
H
[1]
Cl C CH2Br
H
[2]
1 Cl C CH2Br 2
H
4
S configuration
H forward
S configuration
3
3
3
COOH
H forward
3
3
d. Cl
Cl C H
CH2NH2
CH2NH2
[2]
S configuration
2
2
[2]
CH2OH
COOH
Cl C CH2Br 2
4
Cl C H
CH2OH
CH2NH2
1 Cl C CH2Br 2
CH2NH2
CHO
Cl
CHO
3
3
H
CHO
b. Cl
CH2NH2
Cl C CH2Br
CH2Br
H
c.
H C OH
CH2OH
COOH
27.3
c. an aldotetrose
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.
a. CH3 C OH
b.
H C OH
H C OH
CH2OH
27.2
CHO
H C OH
COOH
[3]
1
Cl C CH2Br 2
H
S configuration
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719
Carbohydrates 27–5
27.4
R
CHO
S
H C OH
R
HO C H
H C OH
R
H C OH
CH2OH
D-glucose
27.5
a. aldotetrose: 2 stereogenic centers
b. a ketohexose: 3 stereogenic centers
CH2OH
CHO
C O
H C * OH
H C* OH
H C * OH
[• = stereogenic center]
H C* OH
H C* OH
CH2OH
CH2OH
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.
27.6
CHO
a.
H
OH
H
OH
HO
CHO
HO
H
H
HO
CH2OH
CHO
H
HO
OH
HO
H
OH
CH2OH
B
A
H
H
CH2OH
OH group on the left: L sugar
H
C
OH group on the left: L sugar
OH group on the right: D sugar
b. A and B are diastereomers.
A and C are enantiomers.
B and C are diastereomers.
27.7
The D- notation signifies the position of the OH group on the stereogenic carbon farthest from
the carbonyl group, and does not correlate with dextrorotatory or levorotatory. The latter terms
describe a physical phenomenon, the direction of rotation of plane-polarized light.
27.8
There are 32 aldoheptoses; 16 are D sugars.
C2 R
CHO
CHO
CHO
CHO
C3 R H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
HO
H
H
OH
H
OH
HO
H
OH
H
CH2OH
H
HO
H
H
OH
HO
OH
H
OH
H
CH2OH
CH2OH
H
OH
CH2OH
720
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Chapter 27–6
27.9
Epimers are two diastereomers that differ in the configuration around only one stereogenic
center.
epimers
CHO
H
H
CHO
OH
HO
OH
H
H
OH
CH2OH
CH2OH
D-erythrose
D-threose
CHO
and
H
HO
OH
H
CH2OH
L-threose
27.10 a. D-allose and L-allose: enantiomers
b. D-altrose and D-gulose: diastereomers but not epimers
c. D-galactose and D-talose: epimers
d. D-mannose and D-fructose: constitutional isomers
e. D-fructose and D-sorbose: diastereomers but not epimers
f. L-sorbose and L-tagatose: epimers
27.11
a.
HO
CH2OH
CH2OH
CH2OH
C O
C O
C O
H
H
H
H
OH
HO
OH
HO
OH
CH2OH
D-fructose
HO
H
H
HO
H
H
H
CH2OH
L-fructose
enantiomers
27.12
CH2OH
S
C O
CH2OH
S
C O
H
HO
H
OH
HO
H
OH
H
HO
CH2OH
D-fructose
b.
H
H
OH
CH2OH
D-tagatose
OH
CH2OH
D-tagatose
CH2OH
C O
c.
HO
H
HO
H
OH
H
CH2OH
L-sorbose
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721
Carbohydrates 27–7
27.13
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 anomer of:
CH2OH
O
H
[1]
CHO
[2]
OH
H
OH
D sugar, CH2OH
H
OH
is drawn up.
H
OH
H
H
H
HO
[1]
O
CH2OH
[2]
is drawn down.
H
farthest away C,
OH on left
= L sugar
c. Draw the anomer of:
OH
H
OH
OH
CH2OH
O
CH2OH
OH
[2]
D sugar, CH2OH
is drawn up.
farthest away C,
OH on right = D sugar
H
[3]
HO
O OH
CH2OH
OH OH
H
H
H
The 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.
CH2OH
O
H
OH
H
H
H
H
[1]
CH2OH
O
H
OH
OH
H
H
L sugar, CH2OH
H
First three substituents are
on the right so they are
drawn down.
H
CH2OH
HO
anomer
OH is down
for a D sugar.
H
OH
CHO
HO
farthest away C,
OH on right = D sugar
CHO
HO
H
[3]
CH2OH
O
H
H
H
OH
b. Draw the anomer of:
HO
H
OH
H
CH2OH
CH2OH
O
H
anomer
OH is up for
a D sugar.
[3]
H
HO
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.
722
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Chapter 27–8
27.14 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
CHO
[1]
CHO
[2]
H
H
H
OH
CHO
[3]
H
OH
H
OH
HO
OH
H
CH2OH
"down" groups
on right
OH
H
H
CH2OH
OH
CH2OH
OH on right =
D sugar
CH2OH is down =
L sugar
H
O
H
CH2OH
OH
H
OH
H
b.
HO
OH
"down" groups
on right
CHO
[1]
CHO
[2]
CHO
[3]
HO
H
HO
"up" group
on left
H
CH2OH
CH2OH
H
H
OH
H
OH
HO
H
CH2OH
OH on left =
L sugar
27.15 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
[1]
O
H
[2]
OH
H
H
OH
H
H
OH
HO
OH
H
H
b.
O is an "up" atom.
O
H
H
CH2OH
OH
H
H
[1]
O
[2] HO
OH
H
H OH
H
OH
O
H
H
OH
HO
O
H
H
HO
OH
OH
With so many axial groups,
this is not the more stable
conformation of this sugar.
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723
Carbohydrates 27–9
27.16 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
O
H
H
OH
OH
OH
H
CH2OH
H
OH
O
H
H
OH
OH
H
H
D-erythrose
27.17
HO
HO
CH3CH2OH HO
OH
HO
HCl
HO O
a. HO
HO
HO
HO O
H
H
HO O
+ HO
OCH2CH3
HO
OCH2CH3
H
-D-mannose
OH
OH
OH
CH3CH2OH
O
H
b.
OH
OH
OH
OH
O
OH
O
+
H
HCl
OH
OH
HO
OCH2CH3
OCH2CH3
OH HO H
-D-gulose
c.
CH2OH O
OH
CH3CH2OH
HO
H
HCl
H
CH2OH
H
OH
OCH2CH3
CH2OH O
+
HO
H
H
CH2OH
H
OH
CH2OH O
CH2OH
HO
H
H
OCH2CH3
H
OH
-D-fructose
27.18
H OH2
H
+
HOCH2
H
H
OH
OCH2CH3
O
HOCH2
H
H
OH
+
OCH2CH3
O
H
H
OH
H
OH
H
+ H2O
HOCH2
O
H
H
OH
H
+
H
+ CH3CH2OH
OH
resonance-stabilized
carbocation
HOCH2
+
O
H
H
H
OH
OH
H
724
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Chapter 27–10
H
HOCH2
H2O
above
HOCH2
O
+
H
H
OH
OH
O
H
H
OH
H
OH
H
H
H2O
+
O H
HOCH2
OH
O
+ H3O+
H
H
OH
H
OH
H
H
planar carbocation
HOCH2
H2O
H
below
H
H
OH
HOCH2
H
O
HO
H
O
H
+
H
O H
HO
OH
H
+ H3O+
H
OH
H2O
27.19
a. All circled O atoms are part of a glycoside.
b. Hydrolysis of rebaudioside A breaks each bond
indicated with a dashed line and forms four
molecules of glucose and the aglycon drawn.
OH
HO
OH
OH
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
rebaudioside A
Trade name: Truvia
H
O
O
HO
O
OH
HO
+
OH
O
H
+
H
O
OH
HO
OH
HO
OH
O
HO
OH
aglycon
HO
Both anomers of glucose
are formed, but only the anomer is drawn.
27.20
OH
CH3O
OH
Ag2O
O
a.
OH
HO
OH
H
OH
c.
H
OCH2C6H5
NaH
OH
C6H5CH2O
OCH3
CH3O
C6H5CH2O
OH
HO
C6H5CH2O
CH3I
O
CH3O
O
b.
OCH3
C6H5CH2Cl
OH
H
OCH2C6H5
O
OCH2C6H5
C6H5CH2O
H
O
C6H5CH2O
OCH2C6H5
C6H5CH2O
C6H5CH2O
H
OCH2C6H5
O
H3O+
OH
C6H5CH2O
C6H5CH2O
H
+ anomer + HOCH2C6H5
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Text
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725
Carbohydrates 27–11
OH
OH
d.
Ac2O
OH
HO
OH
H
C6H5COO
O
+
C6H5
OH
HO
C
Cl
pyridine
OOCC6H5
C6H5COO
O
product in (c)
OOCC6H5
O
C6H5COO
OH
H
f.
OAc
AcO
AcO
O
e.
O
pyridine
OH
H
OH
OAc
OAc
O
+
C6H5
C
C6H5CH2O
H
OCH2C6H5
O
Cl
pyridine
C6H5CH2O
+ anomer
OCOC6H5
C6H5CH2O
H
27.21
CH2OH
HO
HO
H
CH2OH
CH2OH
O
H
H
NaBH4
HO
H
CH3OH
HO
OH
H
CH2OH
OH
HO
H
H
HO
H
H
HO
OH
CH2OH
CH2OH
D-tagatose
H
H
OH
D-galactitol
D-talitol
27.22 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.
hemiacetal
hemiacetal
HO
a.
CH2OH
O
H
H
OH
OH
b.
H
H
H
OH
reducing sugar
CH2OH O
H
H
OH
H
H
OH
OH
reducing sugar
acetal
CH2OH O
H
c.
H
OH
hemiacetal
HO
HO
H
H
O
O
d. HO
OCH2CH3
OH
O
HO
nonreducing sugar
OH
OH
HO
lactose
reducing sugar
27.23
CHO
a.
HO
H
H
OH
H
OH
COOH
Ag2O
NH4OH
HO
b.
H
OH
H
OH
H
OH
CH2OH
H
OH
H
OH
CH2OH
COOH
Br2
H2O
H
OH
CHO
H
CHO
c. HO
H
CH2OH
HO
H
HO
H
H
OH
H
OH
CH2OH
CH2OH
COOH
HNO3
H2O
HO
H
H
H
OH
OH
COOH
OH
726
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Chapter 27–12
27.24 Molecules with a plane of symmetry are optically inactive.
CHO
CHO
a.
COOH
H
OH
H
OH
H
OH
H
CH2OH
D-erythrose
H
HO
H
HO
OH
OH
HO
H
OH
HO
H
HO
H
COOH
H
OH
H
CH2OH
OH
COOH
optically inactive
D-galactose
COOH
HO
H
H
optically inactive
HO
H
c.
COOH
OH
HO
CHO
b.
H
H
H
H
OH
CH2OH
COOH
D-lyxose
optically active
27.25
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
27.26
CHO
HO
a.
H
H
OH
CHO
HO
H
H
HO
H
HO
H
CH2OH
OH
H
CH2OH
b.
H
OH
H
OH
H
OH
CH2OH
D-ribose
H
OH
H
OH
CH2OH
D-threose
CHO
CHO
CHO
c.
OH
HO
H
HO
H
H
OH
CH2OH
D-galactose
CHO
HO
CHO
H
H
OH
H
OH
H
OH
H
OH
H
OH
OH
H
H
CH2OH
OH
CH2OH
CHO
HO
H
HO
HO
H
CHO
H
H
OH
OH
H
OH
H
HO
H
HO
OH
CH2OH
H
H
H
OH
CH2OH
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727
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Carbohydrates 27–13
27.27
Possible optically inactive D-aldaric acids:
CHO
COOH
H
OH
H
H
H
OH
OH
H
OH
OH
H
OH
CH2OH
COOH
H
plane of
symmetry
HO
H
H
H
COOH
A'
CHO
OH
OH
HO
OH
H
H
OH
COOH
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.
This is A.
Product of Wohl degradation:
CHO
CHO
H
OH
H
OH
H
OH
CHO
Wohl
H
OH
H
OH
CH2OH
COOH
[O]
COOH
H
OH
HO
H
OH
H
CH2OH
COOH
A'
optically inactive
CHO
[O]
H
HO
OH
H
COOH
Wohl
H
OH
H
H
H
CH2OH
optically active
OH
HO
OH
CH2OH
A"
B
Since this compound has no plane of symmetry, its
precursor is B, and thus A'' = A.
27.28
CHO
HO
CH2OH
H
H
HO
OH
HO
H
H
H
H
HO
OH
CHO
H
HO
OH
rotate
H
H
180°
HO
OH
CH2OH
CHO
H
OH
D-idose
H
H
OH
CH2OH
identical
27.29
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
optically inactive
A'
H
HO
H
H
CH2OH
OH
CH2OH
optically inactive
CHO
K–F
COOH
A''
COOH
H
OH
H
OH
H
OH
HO
H
H
OH
H
OH
H
OH
HO
H
H
OH
H
OH
H
OH
H
A'
NaBH4
OH
HO
This is A.
This is B.
OH
CH2OH
H
OH
CHO
H
CHO
OH
CH2OH
B'
[O]
COOH
optically inactive
CHO
CHO
[O]
OH
COOH
optically active
HO
H
HO
H
H
OH
H
K–F
OH
HO
H
HO
H
H
OH
CH2OH
CH2OH
B''
A''
728
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Chapter 27–14
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.
27.30
OH
OH
O
HO
HO
HO
HO
OH
OH
O
HO
OH
HO
H OH2
O+
OH
+
OH
OH
HO
+ H2O
HO
HO
OH +
O
H O
HO
OH
O
HO
HO
OH
OH
O
+
OH
O
HO
O
OH
HO
HO
-D-glucose
OH
H2O
above
OH
H
O
HO
HO
H2O
+
OH
HO
OH
O
HO
HO
OH
HO
O
HO
HO
-D-glucose
+
OH
+ H3O+
OH
planar carbocation
below
OH
O
HO
HO
H2O
HO
O
HO
HO
+
OH
HO
H
OH
-D-glucose
H2O
27.31
OH
O
HO
HO
OH
O
HO
HO
HO
O
O
OH
OH
HO OH
OH
OH
H3O+
OH
-D-glucose
+
HO
HO
O
OH
HO
-D-glucose
anomer
27.32
glycoside bond
OH
O
HO
HO
4
1 O
OH
O
HO
OH
cellobiose
OH
OH
Two possible anomers here. OH is drawn.
The same products are formed
on hydrolysis of the and anomers of maltose.
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Text
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729
Carbohydrates 27–15
27.33
a.
4
OH
HO
4
O
O HO
OH
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
27.34
OH
O
OH
O
O
HO
O
HO
NHAc
HO
NHAc
OH
O
OH
O
OHO
NHAc
O
NHAc
chitin—a polysaccharide composed of NAG units
OH
O
OH
O
HO
O
NH2
HO
NH2
O
HO
OH
O
OH
O
chitosan
OHO
O
NH2
NH2
27.35
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
H
OH
OH
NHC6H5 +
H+
OH
+ H2O
H
O
HO
NHCH3
H
OH
O
HO
OH
NHCH3
H
OH
H
OH
H
O
HO
H
OH
OH
+ H2O
NHC6H5
730
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Chapter 27–16
27.36
OH
OH
OH
O
HO
HO
O
HO
HO
OH
OH
+
OH2
OH
OH
H OH2
OH
O
HO
HO
HO
HO
+
O+
+
H2O
OH
+ H2O
OH
OH
CH3CH2NH2
HO
HO
above
H
H2O
+
NHCH2CH3
O
O
HO
HO
NHCH2CH3 +
OH
HO
OH
H3O+
O
HO
HO
+
OH
OH
OH
HO
HO
below
CH3CH2NH2
O
HO
O
HO
HO
+
OH
+
NHCH2CH3
H
H3O+
NHCH2CH3
H2O
27.37
O
O
NH
a.
N
HO CH2
O
N
HO CH2
O
N
N
NH2
O
H
H
OH
OH
NH
b.
OH
27.38
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
O
N
N
H
H
O
N
H
guanine
G
H
cytosine
C
H
O
NH
N
N
N
O
N
N
N
H
N
N
H
N
H
guanine
G
O
H
CH3
N
N
H
thymine
T
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Carbohydrates 27–17
27.39 Label the compounds with R or S and then classify.
CHO
H
H
OH
b.
a. CH3CH2 C OH
CH2CH3
c.
A
R
R
identical
d.
C
CH3CH2
H OH
CHO
CHO
OH
H
CHO
R
identical
C
CH2CH3
H
HO
CHO
S
enantiomer
S
enantiomer
27.40 Use the directions from Answer 27.2 to draw each Fischer projection.
COOH
=
a. CH3 C Br
CH3 S
Br
Br
b. CH3 C Cl
H
C Br = H S
re-draw CH3
CH3
CH2CH3
C
C
CH3CH2
f.
S
CH2CH3
OCH2CH3
OCH2CH3
Cl S
Br
H
Cl
S
R
C
g. H
S
CH2CH3
S
Cl
Cl C
H
Br
H
C
R
Br
=
Br
Cl
S
R
re-draw
CH3
H
re-draw
CH3CH2
H
Br
C
H
Cl = CH3CH2 S
CH3
h. HO
Br
Br
CHO
CHO
H
27.41 Epimers are two diastereomers that differ in the configuration around only one stereogenic
center.
CHO
CHO
OH
HO
H
H
H
HO
OH
HO
CH2OH
C4
OH
H
H
CH2OH
D-xylose
L-arabinose
27.42
CHO
HO
a.
H
H
H
CHO
H
OH
HO
OH
HO
CH2OH
D-arabinose
CHO
OH
H
H
CH2OH
enantiomer
HO
b. HO
C3
H
CHO
H
H
H
c. HO
OH
CH2OH
epimer
H
OH
H
OH
OH
O
d. HO
OH
OH
CH2OH
diastereomer
(but not epimer)
H
H
OH
CH2OH
CH2OH
H
H
S
HO
=
S
HO C H
HO
R
H C OH
HO H
Br
CH3
CH3
HO H H OH re-draw HO C H
Cl
Br
H C Br = H S
Br C H
Br S
CHO
H
H
H
CH3
Br
C
Br
Cl
re-draw Br C H
C
CH3
OCH3
H
CH2CH3
Br
Cl
CH2CH3 = CH3
Cl
d.
Br
OCH3
CH3O OCH2CH3
C
H
CH3
Cl
Cl
H
=
Cl C H
CH3
re-draw
CH3
H C Br
e.
H
H
c.
CH3
COOH
constitutional
isomer
732
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Solutions Manual to
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Chapter 27–18
27.43
CHO
HO
CHO
H
HO
CHO
H
H
CH2OH
OH
C O
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
H
OH
CH2OH
OH
D
HO
OH
OH
HO
O
OH
HO
F
H
E
c. B and C enantiomers
d. A and D constitutional isomers
e. E and F diastereomers
27.44
OH
H
H
OH
OH
H
a. anomers, epimers, diastereomers, reducing sugars
b.
CHO
O
H
H
H
A
H
HO
OH
OH
OH
O
H
H
OH
H
OH
B
OH
H
This is the acyclic form of
both A and B.
OH
CH2OH
D-xylose
27.45 Use the directions from Answer 27.13.
a. -D-talopyranose
[1]
CHO
HO
H
HO
H
HO
H
H
OH
CH2OH
CH2OH
O
H
[2]
CH2OH
O
H
OH
[3]
OH
H
H
CH2OH
O OH
H
OH
OH
H
H
H
anomer
OH is up for
a D sugar.
D sugar, CH2OH
is drawn up.
farthest away C,
OH on right = D sugar
D-talose
b. -D-mannopyranose
CHO
HO
HO
H
H
[1]
CH2OH
O
H
[2]
CH2OH
O
H
H
H
H
OH
OH
CH2OH
D-mannose
OH
D sugar, CH2OH
is drawn up.
farthest away C,
OH on right = D sugar
anomer
OH is up for
a D sugar.
[3]
CH2OH
O OH
H
OH OH
H
OH
H
H
H
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733
Carbohydrates 27–19
c. -D-galactopyranose
CH2OH
O
H
[1]
CHO
H
CH2OH
O
H
[2]
H
OH
OH
HO
H
D sugar, CH2OH
HO
H
is drawn up.
H
OH
CH2OH
anomer
OH is down.
CH2OH
O H
[3] OH H
OH H
OH
H
H
OH
farthest away C,
OH on right = D sugar
D-galactose
d. -D-ribofuranose
CH2OH
O
[1]
[2]
H
CHO
H
OH
H
OH
H
OH
CH2OH
CH2OH
O
H
H
OH
[3]
CH2OH
H O
H
OH
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. -D-tagatofuranose
CH2OH
O
[1]
CH2OH
H
C O
HO
H
HO
H
H
OH
CH2OH
[2]
CH2OH
O
CH2OH
H
OH
[3]
D sugar, CH2OH
is drawn up.
anomer
OH is down.
farthest away C,
OH on right = D sugar
27.46
CHO
a. HO
H
HO
H
HO
H
HO
OH
anomer OH
D sugar
HOOH OH
D sugar
HOOH
HO
O
O
HO
OH
anomer
CH2OH
farthest away C,
OH on right = D sugar
CHO
b.
H
H
HO
H
OH
OH
H
OH
CH2OH
farthest away C,
OH on right = D sugar
D sugar
OH
HO
O
OH OHOH
anomer
D sugar
OH
HO
O
OH
OH OH
anomer
CH2OH
OH
H
OH
H
D-tagatose
H
CH2OH
OH O
H
734
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Chapter 27–20
c.
CHO
HO
D sugar
HOOH OH
H
H
O
OH
HO
H
H
OH
OH
D sugar
HOOH
HO
O
OH
OH
OH
anomer
anomer
CH2OH
farthest away C,
OH on right = D sugar
27.47 Use the directions from Answer 27.14.
"up" group
on left
CH2OH
OH
a.
CH2OH is up =
D sugar
O
H
OH
CHO
[2]
"up" group
on left
H
OH
b.
H
H
HO
OH
HO
"down" group
on right
[1]
CHO
H
H
H
OH =
H
CH2OH
"up" group
on left
H
H
OH
HO
H
HO
H
CH2OH
CH2OH
OH on left =
L sugar
CH2OH is up =
D sugar
O
[1]
OH
CHO
CHO
[2]
CHO
[3]
HO
OH
H
OH
OH
H
OH
CHO
[3]
HO
CH2OH
H
CH2OH
CHO
[2]
HO
OH
H
CH2OH
H
"down" group
on right
OH
CH2OH
OH
OH
OH on right =
D sugar
O
H
CH2OH
OH
H
OH
HO O
H
"up" group
on left
CH2OH is down =
L sugar
OH
CHO
[3]
H
H
HO
CHO
H
H
c.
[1]
OH
H
H
CH2OH
OH
CH2OH
OH on right =
D sugar
H
H
OH
H
OH
H
OH
CH2OH
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Text
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition
735
Carbohydrates 27–21
CHO
d.
OH
H
OH
O
H
OH
HO
OH
OH
H
OH
OH
H
OH
CHO
=
HO
HO
CH2OH
OH
O
H
H
OH
= HO
H
H
OH
CH2OH
OH
C O
CH2OH
OH
OH
H
OH
OH
OH
H
H
OH
CH2OH
D sugar
e. HOCH2
H
O
f. HO
O
H
OH
H
OH
H
OH
CH2OH
27.48
CHO
HO
H
H
H
a.
OH
OH
H
CH2OH
b.
CH2OH
OH CH2OH
H
H O HO
H O HO
H
OH
H
OH
OH
anomer
D-arabinose
H
H
H
O OH
OH
OH
H
OH
H
anomer
H
H
H
O H
OH
OH
H
OH
OH
two anomers in the pyranose form
27.49
Two anomers of D-idose, as well as two conformations of each anomer:
anomer
axial
O
HO
OH
OH OH
OH
O
OH
OH
HO
OH
4 axial substituents
anomer
OH
equatorial CH2OH group
OH OH
OH
O
4 equatorial OH groups
More stable conformation for the anomer—the
CH2OH is axial, but all other groups are equatorial.
equatorial CH2OH group
OH
OH
3 axial substituents
OH
O
HO
axial
OH
axial
OH
HO
3 equatorial OH groups
The more stable conformation for the anomer—the
CH2OH is axial, as is the anomeric OH, but three
other OH groups are equatorial.
736
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Chapter 27–22
27.50
CH3O
HO
O
D-gulose
g. The product in (a), then H3O+
O
HO
C6H5CH2O
OCH3
OCH2C6H5
O
c. C6H5CH2Cl, Ag2O
OOCC6H5
O
O OOCC H
6 5
COC6H5
OCH3 OCH
3
O
+ anomer
+ anomer
HO
OH
C6H5COO
OCH3
OH
b. CH3OH, HCl
HO
OH
f. C6H5COCl, pyridine
CH3O
CH3O
HO
OH
C6H5COO
OCH3
O
a. CH3I, Ag2O
CH3O
CH3O
OH
OAc
h. The product in (b), then Ac2O, pyridine
OAc
O
+ anomer
O
OCH2C6H5
CH2C6H5
C6H5CH2O
HO
OAc
i. The product in (g), then C6H5CH2Cl, Ag2O
OH
O
CH3O
d. C6H5CH2OH, HCl
OCH3
O
+ anomer
HO
OH
OAc
OCH2C6H5
CH3O
OAc
CH3O
CH3O
OCH3
O
OAc
OAc
OAc
CH3O
CH3O
27.51
OH
OH
O
a. CH3OH, HCl
HO
CHO
HO
OH
COOH
d. Br2, H2O
+ anomer
OCH3
H
H
OH
H
OH
H
OH
b. (CH3)2CHOH, HCl
HO
CH2OH
OH
OH
O
OH
D-altrose
c. NaBH4, CH3OH
HO
HO
H
H
OH
H
OH
H
OH
CH2OH
+ anomer
OCH(CH3)2
COOH
e. HNO3, H2O
HO
H
H
OH
CH2OH
H
OH
H
H
H
OH
H
OH
H
+ anomer
OCH2C6H5
j. The product in (d), then CH3I, Ag2O
O
e. Ac2O, pyridine
OAc
OCH3
OH
CH2OH
OH
COOH
+ anomer
OCH2C6H5
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Text
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Carbohydrates 27–23
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
+
AcO
OH
HO
OAc
OAc
O
i. Ac2O, pyridine
CHO
H
g. [1] NaCN, HCl
[2] H2, Pd-BaSO4 HO
H
[3] H3O+
H
OH
H
OH
H
OH
H
OH
H
OH
+ anomer
OAc
OAc
H
CH2OH
+ anomer
j. C6H5CH2NH2, mild H+ OH
OH
O
HO
+ anomer
CH2OH
NHCH2C6H5
OH
27.52
a. CH3OH, HCl
CH2OH
OCH3
OH
H
CHO
H
HO
H
OH
H
O
OH
H
H
OCH(CH3)2
OH
OH
CH2OH
g. [1] NaCN, HCl
[2] H2, Pd-BaSO4
[3] H3O+
HO
H
OH
HO
H
+
H
H
H
OH
H
OH
HO
OH
H
H
CH2OH
CH2OH
H
CHO
H
HO
+ anomer
OH
c. NaBH4, CH3OH
H
H
CHO
CH2OH
H
D-xylose
HO
+ anomer
H
CH2OH
CHO
f. [1] NH2OH
[2] (CH3CO)2O, NaOCOCH3
[3] NaOCH3
H
H
b. (CH3)2CHOH, HCl
H
O
OH
OH
CH2OH
OH
H
h. CH3I, Ag2O CH2OCH3
H
O
OH
OCH3
H
H
OCH3
OCH3
H
CH2OH
+ anomer
COOH
d. Br2, H2O
H
HO
H
OH
H
OH
i. Ac2O, pyridine
CH2OAc
O
OAc
H
H
H
OAc
OAc
H
+ anomer
CH2OH
COOH
e. HNO3, H2O
H
HO
H
OH
H
OH
COOH
j. C6H5CH2NH2, mild H+
CH2OH
H
O
OH
H
H
NHCH2C6H5
OH
H
+ anomer
738
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27. Carbohydrates
Text
Study Guide/Solutions Manual to accompany: Organic Chemistry, 3rd Edition
Solutions Manual to
accompany Organic
Chemistry, Third Edition
© The McGraw−Hill
Companies, 2011
Chapter 27–24
27.53
OH
H
H
OH
H
O
H
OH
H
H
H
H
O
OH
OH
H
N
H
O
O
H
H
HO
H3O+
OH
OH
O
solanine
HO
H
H
HO
HO
OH
OH
O
HO
+
HO
H
aglycon
monosaccharide
(both anomers)
OH
N
H
+
O
HO
O
O
H
OH
H
OH
aglycon
H
OH
HO
HO
HO
+
monosaccharide
(both anomers)
H
HO
O
HO
HO
H3O+
salicin
OH
OH
H
O
HO
HO
OH
OH
O
HO
OH
OH
OH
monosaccharide
(both anomers)
monosaccharide
(both anomers)
27.54
CHO
H
CHO
OH
HO
H
H
H
CHO
HO
H
HO
HO
OH
H
OH
H
CH2OH
OH
H
OH
H
CH2OH
D-glucose
H
H
OH
OH
CH2OH
D-arabinose
D-mannose
27.55
CHO
CHO
HO
a. H
OH
H
OH
CH2OH
H
H
H
OH
H
OH
CH2OH
H
OH
H
OH
CH2OH
HO
HO
H
CHO
CHO
HO
H
H
H
HO
H
HO
H
HO
H
HO
OH
CH2OH
H
OH
CH2OH
H
OH
H
H
OH
CH2OH
c.
CHO
CHO
HO
H
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
H
HO
H
HO
OH
CHO
b.
CHO
CHO
HO
H
OH
CH2OH
H
OH
CH2OH
H
OH
H
OH
CH2OH
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Text
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739
Carbohydrates 27–25
27.56
OH
OH
CH3OH
O
a. HO
OH
HO
HO
HO
HCl
OH
+ anomer
H
H
NaBH4
HO
H
OH CH3OH
H
OH
H
CH3I
H
OH
Ag2O
H
OH
c.
H
H
OCH3
H
OCH3
CH2OCH3
COOH
H
COOH
OH
H
Br2
HO
H
OH
H2O
H
OH
H
OH
H
OH
HO
OCH3
CH3O
CH2OH
OH
H
Ac2O
H
CH2OH
OCH3
OCH2CH3
+ anomer
H
H
CHO
O
CH3CH2O
CH3CH2O
CH2OCH3
OH
CH2OH
H
Ag2O
OH
CH2OH
OH
HO
OCH3
OCH2CH3
+ anomer
CHO
b.
CH3CH2I
O
OAc
AcO
pyridine
H
H
OAc
H
OAc
CH2OH
CH2OAc
27.57 Molecules with a plane of symmetry are optically inactive.
CHO
H
H
NaBH4
OH
H
CHO
CH2OH
OH
CH3OH
OH
H
H
H
CH2OH
OH
H
OH
HO
NaBH4
H
H
OH
CH2OH
OH
CH3OH
OH
H
H
H
CH2OH
CH2OH
OH
HO
OH
CH2OH
D-xylose
D-ribose
27.58
OH
OH
O
a. HO
+
OCH3
H3O
HO
OH
OH
OH
HOCH2
c.
OH
O
OH
OH
OCH3
HO
O
HO
O
HO
OH
OH
b.
OH
O
H3O+
OCH3
HO
O
HO
OCH2CH3
NHCH2CH3
OH
OH
H3O+
HOCH2
OH
O
HO
OH
OH
OH
OCH3
HO
O
OH
OH
+ CH3OH
OH
+ CH3CH2OH
OH
HOCH2
OH
O
+ NH2CH2CH3
OH
OH
740
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© The McGraw−Hill
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Chapter 27–26
27.59
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
OH
OH
OH +
O H
H OH2
H
H OH
H OH
27.60
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
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Text
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Carbohydrates 27–27
27.61
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
OCH3
OH
O
HO
HO
HO
HO
B
HO
HO
OCH3
H
OH
H OH2
+
O+
HO
HO
OH
A
+ H2 O
+
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
27.62
COOH
H
OH
H
OH
H
OH
H
OH
H
=
CH2OH
OH O
H
H
H
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
O
H
O
H
H
H
OH
OH
O
OH
OH
+
H A
OH
OH
OH
OH
A
H
H
H A
OH
OH
OH
OH
OH
H
H
OH
OH
O
A
O+
OH
OH
+
CH2OH
H
OH H
OH
O
H
H
H
H
O H
A
O
H
H
H
OH
+ H2O
OH
OH
+
OH2
OH
OH
OH
+
A
742
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© The McGraw−Hill
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Chapter 27–28
27.63
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 OH
C O
C O
C O H
HO
OH
H
H
H
HO
OH
H
OH
H
H
CH2OH
H
OH
H
CH2OH
enediol
A
Deprotonation of the OH at C2
of the enediol forms a new
enolate that goes on to
form the ketohexose.
HO
OH
H
H
H C OH
C O
H
HO
OH
H
OH
H
CH2OH
H
OH
+
OH
CH2OH
+ H2O
27.64
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]
HO
H
H
H
HO
H
OH
HO
H
OH
H
COOH
optically active
OH
COOH
optically active
CHO
[O]
HO
H
HO
H
H
OH
CH2OH
A"
D-lyxose
OH
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Text
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Carbohydrates 27–29
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
Wohl
H
H
COOH
COOH
optically
inactive
CHO
CHO
OH
HO
H
HO
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.
27.65
CHO
HO
CH2OH
H
HO
CH2OH
H
HO
H
OH
H
OH
HO
H
OH
H
OH
H
CH2OH
CH2OH
CHO
H
HO
H
HO
OH
H
H
CH2OH
OH
CH2OH
identical
(by rotating 180°)
D-arabinose
H
D-lyxose
27.66
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"
744
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© The McGraw−Hill
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Chapter 27–30
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'.
27.67
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
OH
HO
H
OH
H
CH2OH
HO
[H]
H
H
H
CH2OH
optically
active
H
HO
OH
CH2OH
A'
CHO
H
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''.
27.68
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
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Text
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745
Carbohydrates 27–31
27.69 A disaccharide formed from two galactose units in a 14--glycosidic linkage:
glycoside bond
OH
O
HO
HO
OH
O
1 O
OH 4
OH
HO
OH
27.70 A disaccharide formed from two mannose units in a 14--glycosidic linkage:
glycoside bond
HO
HO
O
HO
HO
1
O
HO
4
OH
O
HO
OH
27.71
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
Ag2O
H3O+
OCH3
O
OCH3
O
OH
+
CH3O
OCH3
OH
E
OCH3
O
HO
CH3O
CH3O
+
OCH3
O
OH
F + CH3OH
(Both anomers of E and F are
formed, but only one is drawn.)
OCH3
D
OH
OCH3
+ CH3OH
C
OCH3 OCH
3
O
H3O+
OCH3
O
HO
CH3O
OH
CH3O
OCH3
O
CH3O
OCH3
O
CH3O
CH3O
OCH3
OCH3
A
B
OH
OCH3
O
27.72
OH
a.
glycoside bond
OH
O
1
HO
OH
b.
c.
4
O
OH
OH
OH
O
O
OH
HO
OH
1 4--glycoside bond
reducing sugar
(hemiacetal)
HO
HO
1
O
HO
HO
reducing sugar
(hemiacetal)
glycoside bond
6
OH
O
1 6-glycoside bond
OH
746
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Chapter 27–32
27.73
a and b.
OH
OH
CH2OH
O
HO
O CH2
O
OH OH
OH
1 6--glycoside bond
HO
OH
stachyose
1 2--glycoside bond
OH
HOCH2 O
OH O
c.
CH2OH
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
OH
CH2OH
OH
O
O
=
O
OH
OH
O
HO
O CH2
H3O+
OH
1 6--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
OCH3
O
O
OCH3
CH3O
CH3O
f.
product
in (e)
H3O+
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
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747
Carbohydrates 27–33
27.74
OH
O
HO
HO
isomaltose + anomer
OH
O
Isomaltose must be composed of two glucose units
in an -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.
O
HO
HO
OH
the hemiacetal
OH
[1] CH3I, Ag2O
[2] H3O+
OCH3
O
CH3O
CH3O
1
CH3O
OH
6
O
CH3O
CH3O
OH
CH3O
(Both anomers are present.)
OH
27.75
OH
O
HO
HO
CH3O
CH3O
CH3I
trehalose
OH
O
Ag2O
OH
OCH3
O
CH3O
H3O+
O
OH
OH
O
HO
OCH3
CH3O
CH3O
OCH3
OCH3
O
CH3O
OCH3
O
CH3O
OH
(both anomers)
Trehalose must be composed of D-glucose units only, joined in an -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.
27.76
HO
a. HO
O
H2N
CH2OH
1
O
HO
HO
c.
6
O
H
O
H
HO
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
748
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Chapter 27–34
27.77
Ignoring stereochemistry along the way:
A
HO
OH
OH H
H A
O
O
[1]
OH
OH
OH
A
OH
A
O
O
OH
HO
OH
[4]
OH
OH
HO
H A
HO
OH
HO
C
[5]
O
CH3
OH
O
CH3
O
OH
C
CH3
[8]
HO
OH
H
[7]
HO
HO
O
[6]
OH2
HO
OH
HO
[3]
OH
OH
OH
O
OH
[2]
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
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Carbohydrates 27–35
27.78 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 -glycosidic linkage. D-Fructose
must be joined to its adjacent sugar by an -glycosidic linkage.
HO
CH3O
OH
O
HO
OH
galactose
O
HO
HO
HO
O
HO
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.)
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751
Amino Acids and Proteins 28–1
C
Chhaapptteerr 2288:: A
Am
miinnoo A
Acciiddss aanndd PPrrootteeiinnss
SSyynntthheessiiss ooff aam
miinnoo aacciiddss ((2288..22))
[1] From -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, 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+, [3] Strecker synthesis
O
R
C
NH4Cl
H
NaCN
NH2
NH2
H3O+
R C CN
R C COOH
H
H
-amino nitrile
PPrreeppaarraattiioonn ooff ooppttiiccaallllyy aaccttiivvee aam
miinnoo aacciiddss
[1] Resolution of enantiomers by forming diastereomers (28.3A)
• 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 (28.3B)
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
752
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Chapter 28–2
[3] By enantioselective hydrogenation (28.4)
R
NHAc
H
AcNH
H2
C C
Rh*
COOH
C
H2O, –OH
COOH
H CH2R
C
COOH
H CH2R
S enantiomer
Rh*
H2N
S amino acid
= chiral Rh hydrogenation catalyst
SSuum
mm
maarryy ooff m
meetthhooddss uusseedd ffoorr ppeeppttiiddee sseeqquueenncciinngg ((2288..66))
• 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 28.2).
A
Addddiinngg aanndd rreem
moovviinngg pprrootteeccttiinngg ggrroouuppss ffoorr aam
miinnoo aacciiddss ((2288..77))
[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
CO2H
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
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753
Amino Acids and Proteins 28–3
[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,
H2N
H+
C
C
H2N
OCH3
H R
H R
O
O
C
C
C6H5CH2OH, H+
OH
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
SSyynntthheessiiss ooff ddiippeeppttiiddeess ((2288..77))
[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.
SSuum
mm
maarryy ooff tthhee M
Meerrrriiffiieelldd m
meetthhoodd ooff ppeeppttiiddee ssyynntthheessiiss ((2288..88))
[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
754
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Chapter 28–4
C
Chhaapptteerr 2288:: A
Annssw
weerrss ttoo PPrroobblleem
mss
28.1
H CH3 O
S C
S
OH
R
H2N H
O
H CH3 O
S C
R C
H
CH3
OH
S
H2N H
CH3
OH
H O
R
H NH2
R C
OH
H NH2
L-isoleucine
28.2
a.
b.
NH3+
C COO–
(CH3)2CH
c.
NH3+
(CH3)2CHCH2
C
COO–
+
HOOCCH2CH2 C COO–
N
H
H
NH3+
d.
COO–
H
H H
28.3
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.
28.4
The most direct way to synthesize an -amino acid is by SN2 reaction of an -halo carboxylic
acid with a large excess of NH3.
a.
NH3
Br CH COOH
large excess
H
b. Br CH COOH
CH CH3
NH3
large excess
H2N CH COO– NH4+
H
c. Br CH COOH
CH2
glycine
NH3
large excess
H2N CH COO– NH4+
CH2
H2N CH COO– NH4+
CH CH3
CH2
phenylalanine
CH2
CH3
CH3
isoleucine
28.5
CH3I
a.
CH3
H2N C COOH
alanine
c.
H
(CH3)2CHCH2Cl
b.
CH2CH(CH3)2
H2N C COOH
leucine
28.6
H
H
C N C COOEt
CH3
COOEt
CH(CH3)CH2CH3
H2N C COOH
H
H
O
CH3CH2CH(CH3)Br
[1] NaOEt
[2] CH2=O
[3] H3O+, CH2OH
H2N C COOH
H
serine
isoleucine
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Amino Acids and Proteins 28–5
28.7
O
a. H2N CHCOOH
(CH3)2CH
CH CH3
O
b. H2N CHCOOH
C
H
C
(CH3)2CHCH2
CH2
O
c. H2N CHCOOH
H
C6H5CH2
CH2
C
H
CH CH3
CH3
valine
CH3
leucine
phenylalanine
28.8
a. BrCH2COOH
NH3
large excess
H
b. CH3CONH C COOEt
COOEt
NH2CH2COO– NH4+
H
[1] NaOEt
[2] (CH3)2CHCl
CH(CH3)CH2CH3
H
d. CH3CONH C COOEt
CH(CH3)2
H2N CHCOOH
[2] H3O+
H
H2N C COOH
[3] H3O+, [1] NH4Cl, NaCN
c. CH3CH2CH(CH3)CHO
COOEt
[1] NaOEt
[2] BrCH2CO2Et
H2N C COOH
CH2CO2H
[3] H3O+, A chiral amine must be used to resolve a racemic mixture of amino acids.
28.9
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)
28.10
NH2
C
NH2
COOH
+
R
S
Ac2O
AcNH
C
AcNH
COOH
+
H CH2CH(CH3)2
C
proton transfer
H CH2CH(CH3)2
S
C
C6H5
(R isomer only)
CH3 H
COO–
enantiomers
R
H2N
Step [1]:
React both enantiomers with the
R isomer of the chiral amine.
COOH
C
(CH3)2CHCH2 H
S
AcNH
enantiomers
(CH3)2CHCH2 H
H CH2CH(CH3)2
To begin:
Convert the amino acids into N-acetyl
amino acids (two enantiomers).
COOH
C
+
H3N
C
CH3 H
R
C6H5
AcNH
C
(CH3)2CHCH2 H
R
COO–
+
H3N
C
CH3 H
C6H5
diastereomers
R
These salts have the same configuration around one stereogenic center,
but the opposite configuration about the other stereogenic center.
756
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Chapter 28–6
separate
Step [2]:
Separate the diastereomers.
AcNH
+
COO–
C
H3N
H CH2CH(CH3)2
C
C6H5
AcNH
CH3 H
S
C
C
C6H5
C
CH3 H
R
R
H2O, –OH
NH2
H3N
(CH3)2CHCH2 H
R
Step [3]:
Regenerate the amino acid
by hydrolysis of the amide.
+
COO–
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.
28.11
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
28.12
H
a. H2N CHCOOH
H
CH3
(CH3)2CH
NHAc
b. H2N CHCOOH
C C
COOH
H2NCOCH2
NHAc
C C
H
CH2
c. H2N CHCOOH
COOH
CH2
CH CH3
CH2
CH3
CONH2
NHAc
C C
H
COOH
28.13 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
H2N
CH CH3
CH3
Leu
His
H H
C
C
O
N-terminal
(CH3)2CHCH2
H
O
H
N
C
C
OH
C
N
C
H
O
H CH2
C-terminal
amide
NH
Gly–His–Leu
N
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Amino Acids and Proteins 28–7
O
O
O
O
c. H2N CH C OH H2N CH C OH H2N CH C OH H2N CH C OH
CH2
CH3
CH2SCH3
M
A
CH OH
CH OH
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
28.14
N
HN
CONH2
a.
H2N
C
O
CH2 H
C
C
N
H
C
H
N
O
H CH2
O
C
C
H2N
b.
OH
H CH(CH3)2
C
Arg–Asn–Val
R–N–V
NH
C
N
H
C
O
O
H
N
C
C
OH
H CH2
CH2
CH2
CH2
CONH2
CH2
C
NH2
NH2
HN
CH2 H
C
H CH2
CH2
CH2
O
Lys–His–Gln
K–H–Q
28.15 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.
28.16 The s-trans conformation has the two C’s oriented on opposite sides of the C–N bond.
The s-cis conformation has the two C’s oriented on the same side of the C–N bond.
O
H2N
C
C
H H
O
H H
C
OH
N
C
H
H2N
C
C
HH
O
H
s-trans
N
C
H
H
C
OH
O
s-cis
28.17
O
H2N
HO
N
H
H
N
O
O
N
H
leu-enkephalin
H
N
O
O
OH
758
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Chapter 28–8
28.18
O
HO
HS
O
a. H2N
H
H
N
O
O
H2N
H2N
COOH
H
O
N
N
COOH
H2N
H
S
H
N
N
COOH
HS
O
NH2
O
O
glutathione
b.
H
OH
COOH
N
S
N
N
COOH
O
H
H
O
OH
O
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 COOH.
O
This comes from the amino acid glutamic acid.
O
OH
This carboxy group is used to form the amide bond in the peptide, not the COOH, as is usual. That's what makes glutathione's structure unusual.
COOH
glutamic acid
28.19
O
C6H5
a.
S
O
C6H5
N
CH3
N
H
b.
S
from Ala
N
N
H
from Val
28.20 Determine the sequence of the octapeptide as in Sample Problem 28.2. 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
28.21 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
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759
Amino Acids and Proteins 28–9
28.22
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
28.23
a.
O
(CH3)2CHCH2 H
H2N
C
Leu
OH
C
O
(CH3)2CHCH2 H
[(CH3)3COCO]2O
C
OH
Boc N
C
(CH3CH2)3N
H
O
(CH3)2CHCH2 H
Boc N
H
C
OH +
C
O
O
H2N
C
C
H2N
C
O
C
C6H5CH2OH, H+
OH
H CH(CH3)2
H CH(CH3)2
C
Val
new amide bond
DCC
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
C
OCH2C6H5
H CH(CH3)2
(CH3)2CHCH2 H
OCH2C6H5
H2N
O
C
C
OH
H CH(CH3)2
Leu–Val
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Chapter 28–10
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
O
C
C
CH3 H
H2
Pd-C
OCH2C6H5
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
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Amino Acids and Proteins 28–11
O
CH3 H
c.
H2N
C
CH3 H
C
OH [(CH3)3COCO]2O
(CH3CH2)3N
Ala O
C
Boc N
H
H2N
OH
C
C
C
H H
O
O
C6H5CH2OH, H+
OH
H H
Gly
new amide bond
+
DCC
B
Boc N
H
CH3 H
H2N
C
C
C
CH3 H
O
H
N
O
C
C6H5CH2OH,
OH
H2N
+
H2N
C
C
H2
OCH2C6H5
Boc N
H
Pd-C
C
C
O
H H
O
H
N
C
C
OH
H H
C
C
C
OCH2C6H5
new amide bond
C
C
CH3 H
DCC
OCH2C6H5
BOC N
H
O
C
C
H
N
O
O
C
C
CH3 H
N
H
H H
C
C
H2
OCH2C6H5
Pd-C
O
CH3 H
Boc N
H
C
CH3 H
B
DCC
Boc N
H
C
C
O
H
N
O
C
C
H H
CH3 H
N
H
C
C
O
H
N
C
O
new amide bond
+
B
O
CH3 H
D
OCH2C6H5
CH3 H
H+
Ala O
C
C
C
A
CH3 H
A
H2N
O
H
N
C
C
C
C
N
H
C
OH
O
H H
D
O
C
CH3 H
HBr
OCH2C6H5
H H
CH3COOH
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
762
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Chapter 28–12
28.24
All Fmoc-protected amino acids are made by the following general reaction:
R H
C
H2N
C
R H
Fmoc–Cl
OH
Na2CO3, H2O
O
H H
The steps:
Fmoc N
H
C
OH
C
Fmoc N
H
C
C
OH
O
[1] base
[2] Cl CH2 POLYMER
O
H H
Fmoc N
H
C
C
O CH2 POLYMER
O
[1]
[2] DCC
N
H
H
Fmoc N
O
C
C
OH
H CH(CH3)CH2CH3
(CH3)2CHCH2 H
Fmoc N
H
C
C
H
N
O
H
C
O
H H
C
C
N
H
C
[1]
O CH2
O
CH(CH3)CH2CH3
[1]
N
H
CH3 H
[2] DCC
C
OH
Fmoc N
C
H
O
N
H
POLYMER
[2] DCC
H
Fmoc N
H
(CH3)2CHCH2 H
Fmoc N
H
C
C
H H
O
C
C
C
N
H
C
O CH2 POLYMER
O
CH(CH3)CH2CH3
OH
O
[1]
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
(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
3
2
3
Ala–Leu–Ile–Gly
+ F CH2 POLYMER
28.25 Antiparallel -pleated sheets are more stable then parallel -pleated sheets because of geometry.
The N–H and C=O of one chain are directly aligned with the N–H and C=O of an adjacent chain
in the antiparallel -pleated sheet, whereas they are not in the parallel -pleated sheet. This
makes the latter set of hydrogen bonds weaker.
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Amino Acids and Proteins 28–13
28.26 In a parallel -pleated sheet, the strands run in the same direction from the N- to C-terminal
amino acid. In an antiparallel -pleated sheet, the strands run in the opposite direction.
H
H
O CH3 H
N
C
C
C
N
H CH3 H
H
C
H
N
O
O CH3 H
C
C
N
H CH3 H
C
C
HO
OH
O
H CH3 O
H H CH3 O
H
C
C
N
C
C
N
C
N
C
C
N
C
H
H
O
H CH H O
H CH
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
CH
O H CH3
H
3
O
H
C
O
C
H CH3
3
CH3 H
H
C
N
N
H
C
O
O CH3 H
C
C
H CH3
N
H
C
C
OH
O
antiparallel
parallel
28.27
a. Ser and Tyr
b. Val and Leu
H2N CH COOH
H2N CH COOH
CH2
CH2
OH
side chains with
OH groups
H2N CH COOH
c. 2 Phe residues
H2N CH COOH
CH CH3
CH2
CH3
CH CH3
H2N CH COOH
H2N CH COOH
CH2
CH2
CH3
OH
hydrogen bonding
side chains with only
C–C and C–H bonds
van der Waals forces
van der Waals forces
28.28 a. The R group for glycine is a hydrogen. The R groups must be small to allow the -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 -pleated sheets are stacked one on top of another so few polar
functional groups are available for hydrogen bonding to water.
764
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Chapter 28–14
28.29 All L-amino acids except cysteine have the S configuration. L-Cysteine has the R configuration
because the R group contains a sulfur atom, which has higher priority.
1
2
The S atom gives the R group
a higher priority than COOH,
resulting in the R configuration.
1
NH2
NH2
C
HSCH2
3
COOH
H
2
C
R
COOH
H
3
With all other R groups, the COOH has a higher
priority than R, giving it the S configuration.
28.30
a.
R COOH
S
H2N C H
COOH
CH3 C SH
COOH
b.
H C NH2
H2N C H
CH3 C SH
CH3
H2N C H
CH3 C S
CH3
(R)-penicillamine
COOH
CH3
(S)-penicillamine
S C CH3
CH3
28.31 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–
amino acid, a salt
H2O soluble and ether insoluble
C
COOH
N-acetyl amino acid
ether soluble
28.32 The electron pair on the N atom not part of a double bond is delocalized on the five-membered
ring, making it less basic.
H2N CH COOH
CH2
HA
H2N CH COOH
CH2
+
NH
NH2
N
N
When this N is protonated...
H2N CH COOH
CH2
NH
N
sp3 hybridized N atom
...the ring is no longer aromatic.
H2N CH COOH
CH2
HA
preferred
path
When this N is protonated...
H2N CH COOH
CH2
+
NH
NH
+N
H
N
H
...the ring is still aromatic.
6 electrons
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765
Amino Acids and Proteins 28–15
28.33
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 electrons), making it less
available for donation, and thus less basic.
28.34 At its isoelectric point, each amino acid is neutral.
a.
+
COO–
H3N C H
b.
+
COO–
H3N C H
CH3
CH2CH2SCH3
alanine
methionine
c.
+
COO–
H3N C H
CH2COOH
aspartic acid
d.
COO–
H2N C H
+
CH2CH2CH2CH2NH3
lysine
28.35
[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.
a.
28.36
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
766
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Chapter 28–16
28.37
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
28.38 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.
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Amino Acids and Proteins 28–17
28.39
H CH2CH(CH3)2
H2N
C
COOH
leucine
C
H2N
g. C6H5COCl, pyridine
C6H5
COOCH3
CH3
C
N
H
C
C
OH
C
H2N
N
H
C
C
O
H CH2CH(CH3)2
h. [(CH3)3COCO]2O
Boc N
H
(CH3CH2)3N
C
COOH
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
AcNH
e.
C
H3N
C
H2N
C
H CH2CH(CH3)2
H
C
N
COOCH
Boc N
C
C 3
H
O H H
COO
H CH2CH(CH3)2
k. Fmoc–Cl, Na2CO3, H2O
Fmoc N
H
O
COOH
H CH2CH(CH3)2
f. NaOH (1 equiv)
H H
j. The product in (h), then NH2CH2COOCH3 + DCC
COOH
H CH2CH(CH3)2
HCl (1 equiv)
OH
O
H CH2CH(CH3)2
c. C6H5CH2OH, H+
C
H CH2CH(CH3)2
O
b. CH3COCl, pyridine
H CH2CH(CH3)2
O
H CH2CH(CH3)2
+
a. CH3OH, H
l. C6H5N=C=S
C6H5
S
N
C
COOH
CH2CH(CH3)2
N
H
768
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Chapter 28–18
28.40
H CH2C6H5
H2N
C
COOH
phenylalanine
C
H2N
g. C6H5COCl, pyridine
COOCH3
CH3
C
N
H
C
C
OH
C
H2N
h.
N
H
[(CH3)3COCO]2O
C
O
(CH3CH2)3N
Boc N
H
COOCH2C6H5
C
COOH
i. The product in (d), then
NH2CH2COOCH3 + DCC
H CH2C6H5
H
C
N
COOCH3
AcNH
C
C
O
H CH2C6H5
d. Ac2O, pyridine
C
AcNH
C
H3N
f.
H CH2C6H5
H
C
N
COOCH3
Boc N
C
C
H
O
H H
COOH
H CH2C6H5
k. Fmoc–Cl, Na2CO3, H2O
H2N
C
Fmoc N
H
O
COOH
l. C6H5N=C=S
H CH2C6H5
NaOH (1 equiv)
C6H5
COO
S
28.41
a. (CH3)2CHCH2CHCOOH
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+, 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+
H H
j. The product in (h), then NH2CH2COOCH3 + DCC
H CH2C6H5
e. HCl (1 equiv)
OH
C
H CH2C6H5
O
H CH2C6H5
c. C6H5CH2OH, H+
C6H5
C
H CH2C6H5
O
b. CH3COCl, pyridine
H CH2C6H5
O
H CH2C6H5
a. CH3OH, H+
COOH
HO
[1] NaOEt
[2] ClCH2CH2CH2CH2NHAc
[3] H3O+, NH2
CH2CH2CH2CH2NH2
H2N C COOH
H
N
C
COOH
CH2C6H5
N
H
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769
Amino Acids and Proteins 28–19
28.42
a. Asn
b. His
O
H2N CH C OH
c. Trp
O
H2N CH C OH
CH2
CH2Br
C O
C O
NH2
NH2
CH2Br
CH2
O
CH2Br
H2N CH C OH
CH2
NH
NH
HN
N
N
HN
28.43
OH
O
H
H
C N C COOEt
CH3
CHCH3
[1] NaOEt
H2N C COOH
[2] CH3CHO
[3] H3O+, COOEt
H
threonine
28.44
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
28.45
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+, C COOEt
CH2CH2SCH3
O
C
770
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Chapter 28–20
28.46
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
O
COOEt
H
C N C COOEt
CH3
CH2COOH
CH2CH2COOEt
glutamic acid
+ H O
2
28.47
CH3
CH3
COOH
C
R
CH3
H2N
C6H5
C
(R isomer only)
CH3 H
C
+
H3N
C
C6H5
CH3 H
H OH
R
enantiomers
S
proton transfer
COO–
COOH
HO H
H OH
Step [1]:
React both enantiomers with the
R isomer of the chiral amine.
C
+
CH3
COO– H3N
C
C6H5
diastereomers
CH3 H
HO H
R
C
R
S
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.
CH3
Step [3]:
Regenerate lactic acid
by protonation.
+
COO– H3N