9-2
n H
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----­
NOS \f3d
Lh -E -O N8SI
LUT ONS MANUAL
Jan William Simek
California Polytechnic State University
OR
C CHEMISTRY
SIXTH EDITION
L. G. Wade, Jr.
".�JL.-':'
Prentice
Hall
Upper Saddle River, NJ
07458
Assistant Editor: Carole Snyder
Project Manager: Kristen Kaiser
Executive Editor: Nicole Folchetti
Executive Managing Editor: Kathleen Schiaparelli
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Production Editor: Kathryn O'Neill
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Manufacturing Buyer: Ilene Kahn
Cover Image Credit: Joseph Galluccio (2004)
PEARSON
© 2006 Pearson Education, Inc.
Prentice
Hall
Pearson Prentice Hall
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Printed in the United States of America
10
9
ISBN
8
7
6
5
0-13-147882-6
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TABLE OF CONTENTS
Preface
........................................................................................................................................
v
Symbols and Abbreviations ...................................................................................................... vii
Chapter 1
Introduction and Review
Chapter 2
Structure and Properties of Organic Molecules
Chapter 3
Structure and Stereochemistry of Alkanes
Chapter 4
The Study of Chemical Reactions
Chapter 5
Stereochemistry
Chapter 6
Alkyl Halides: Nucleophilic Substitution and Elimination
Chapter 7
Structure and Synthesis of Alkenes
Chapter 8
Reactions of Alkenes
Chapter 9
Alkynes
Chapter 10
Structure and Synthesis of Alcohols
Chapter 11
Reactions of Alcohols
Chapter 12
Infrared Spectroscopy and Mass Spectrometry
Chapter 13
Nuclear Magnetic Resonance Spectroscopy
Chapter 14
Ethers, Epoxides, and Sulfides
Chapter 15
Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy
Chapter 16
Aromatic Compounds
Chapter 17
Reactions of Aromatic Compounds
Chapter 18
Ketones and Aldehydes
Chapter 19
Amines
Chapter 20
Carboxylic Acids
Chapter 21
Carboxylic Acid Derivatives
Chapter 22
Condensations and Alpha Substitutions of Carbonyl Compounds
Chapter 23
Carbohydrates and Nucleic Acids
Chapter 24
Amino Acids, Peptides, and Proteins
Chapter 25
Lipids
Chapter 26
Synthetic Polymers
Appendix I:
Summary of IUPAC Nomenclature
Appendix 2:
..... . ...
.
.
... . ... . ................ . ....... ................. . ...
.......... ............
.
......................... ........................
....................................
.
.....................
.
. . . .................... . . ............................... .............
.
.... . ... .......... . ... . . . ........................
.
....... . ...
.
...
.
. .. .
. ... ..
..
.....
.
................. . ....... .......
.
.................................. ...........
.
...... . . ........................ . ......................
.................
.............. .................
.
.
..... . .......
.
.
..........
......... ..........
..... . .... . ......................
.................. . ................................ ............ ............
........ . ............ ...........
....... . . .............
..
.
....
. ..
.. .
.
..
.........
.... . ...... .....
.. . ...
..
.
.
. . . . ....
.
................................. . .......... .............. ..............
...... ........
.
. ..........
... . ..... . ........ . . ..............................................
. . . . ......... . . . ....... . . .. . .
.................. . ...
..
.
........ . ...........
.
............
............... . . . ................................
............................................ ......... . ............
.............. .... . ................. ............................. ..................
. . . ........... .......................................................................
.
....... .....................................................................................
... . .................
. .
..
...................................................... ......................
... . ... . . ...........
................................ . ........ ........ ...................
.
.......
.
...........
............................... ............
.
.
.............................. . .......... ...................................... ....
. ... . ....
.
. . ............... .
.
.......... .......
.
.. . ....... .................................
. . . . ........ .......... . ................. .......... ................
Summary of Acidity and Basicity
... . . . .......
iii
.
......
.
.
.
... ........ .................. . . . . ............
1
25
45
59
79
99
135
159
191
211
229
259
271
299
317
341
365
401
439
469
499
537
585
617
645
659
675
689
PREFACE
Hints for Passing Organic Chemistry
Do you want to pass your course in organic chemistry? Here is my best advice, based on over
thirty years of observing students learning organic chemistry:
Hint #1: Do the problems. It seems straightforward, but humans, including students, try to take
the easy way out until they discover there is no short-cut. Unless you have a measured IQ above 200
and comfortably cruise in the top 1 % of your class, do the problems. Usually your teacher (professor or
teaching assistant) will recommend certain ones; try to do all those recommended. If you do half of
them, you will be half-prepared at test time. (Do you w ant your surgeon coming to your appendectomy
having practiced only ha l/the procedure?) And w hen you do the problems, keep this Solutions Manual
CLOSED. A void looking at my answer before you write y our answer-your trying and struggling with
the problem is the most valuable part of the problem. Discovery is a major part of learning. Remember
that the primary goal of doing these problems is not just getting the right answer, but understanding the
material well enough to get right answers to the questions you haven't seen yet.
Hint # 2 : Keep up . Getting behind in your work in a course that moves as quickly as this one is
the Kiss of Death. For most students, organic chemistry is the most rigorous intellectual challenge they
have faced so far in their studies. Some are taken by surprise at the diligence it requires. Don't think
that you can study all of the material the couple of days before the exam-well, you can, but you won't
pass. Study organic chemistry like a foreign language: try to do some every day so that the freshly­
trained neurons stay sharp.
Hint #3: Get h elp when y ou need it. Use your teacher's office hours w hen you have difficulty.
Many schools have tutoring centers (in which organic chemistry is a popular offering). Here's a secret:
absolutely the best way to cement this material in your brain is to get together with a few of your fellow
students and make up problems for each other, then correct and discuss them. When you write the
problems, you will gain great insight into what this is all about.
Purpose of this Solutions Manual
So w hat is the point of this Solutions Manual? First, I can't do your studying for you. Second,
since I am not leaning over your shoulder as you write your answers, I can't give you direct feedback on
what you write and think-the print medium is limited in its usefulness. What I can do for you is:
I) provide correct answers; the publishers, Professor Wade, Professor Kantorowski (my reviewer), and
I have gone to great lengths to assure that what I have written is correct, for we all understand how it can
shake a student's confidence to discover that the answer book flubbed up; 2) provide a considerable
degree of rigor; beyond the fundamental requirement of correctness, I have tried to flesh out these
answers, being complete but succinct; 3) provide insight into how to solve a problem and into where the
sticky intellectual points are. Insight is the toughest to accomplish, but over the years, I have come to
understand where students have trouble, so I have tried to anticipate your questions and to add enough
detail so that the concept, as well as the answer, is clear.
It is difficult for students to understand or acknowledge that their teachers are human (some are
more human than others). Since I am human (despite what my students might report), I can and do
make mistakes. If there are mistakes in this book, they are my sole responsibility, and I am sorry. If
you find one, PLEASE let me know so that it can be corrected in future printings. Nip it in the bud.
What's New in this edition?
Better answers! Part of my goal in this edition has been to add more explanatory material to
clarify how to arrive at the answer. The possibility of more than one answer to a problem has been
The IUPAC Nomenclature appendix has been expanded to include bicyclics, heteroatom
noted.
replacements, and the Cahn-Ingold-Prelog system of stereochemical designation.
Better graphics! The print medium is very limited in its ability to convey three-dimensional
structural i�formation, a problem that has plagued organic chemists for over a century. I have added
some graphiCS created .1I1 the software, Chem3D®, to try to show atoms in space where that information
is a key part of the solution.
In drawing NMR spectra, representational line drawings have replaced
rudimentary attempts at drawing peaks from previous editions.
Better jokes? Too much to hope for.
v
Some Web Stuff
Prentice-Hall maintains a w eb site dedicated to the Wade text: try www.prenhall.com/wade.
Two essential web sites providing spectra are listed on the bottom of p. 270.
Acknowledgments
No project of this scope is ever done alone. These are team efforts, and there are several people
w ho have assisted and facilitated in one fashion or another w ho deserve my thanks.
Professor L. G. Wade, Jr., your textbook author, is a remarkable person. He has gone to
extraordinary lengths to make the textbook as clear, organized, informative and insightful as possible.
He has solicited and followed my suggestions on his text, and his comments on my solutions have been
perceptive and valuable. We agreed early on that our primary goal is to help the students learn a
fascinating and challenging subject, and all of our efforts have been directed toward that goal. I have
appreciated our collaboration.
My new colleague, Dr. Eric Kantorowski, has reviewed the entire manuscript for accuracy and
style. His diligence, attention to detail and chemical w isdom have made this a better manual. Eric
stands on the shoulders of previous reviewers who scoured earlier editions for errors: Jessica Gilman,
Dr. Kristen Meisenheimer, and Dr. Dan Mattern. Mr. Richard King has offered numerous suggestions
on how to clarify murky explanations. I am grateful to them all.
The people at Prentice-Hall have made this project possible. Good books would not exist
without their dedication, professionalism, and experience. Among the many people who contributed
are: Lee Englander, w ho connected me with this project; Nicole Fo1chetti, Advanced Chemistry Editor;
and Kristen Kaiser and Carole Snyder, Project Managers.
The entire manuscript was produced using ChemDraw®, the remarkable software for drawing
chemical structures developed by CambridgeSoft Corp., Cambridge, MA . We, the users of sophisticated
software like ChemDraw, are the beneficiaries of the intelligence and creativity of the people in the
computer industry. We are fortunate that they are so smart.
Finally, I appreciate my friends who supported me throughout this project, most notably my
good friend of almost forty years, Judy Lang. The students are too numerous to list, but it is for them
that all this happens.
Jan William Simek
Department of Chemistry and Biochemistry
Cal Poly State University
San Luis Obispo, CA 93407
Email: jsimek@calpoly.edu
DEDICATION
To my inspirational chemistry teachers:
Joe Plaskas, w ho made the batter;
Kurt Kaufman, w ho baked the cake;
Carl Djerassi, w ho put on the icing;
and to my parents:
Ervin J. and Imilda
B.
Simek,
who had the original concept.
vi
SYMBOLS AND ABBREVIA TIONS
Below is a list of symbols and abbreviations used in this Solutions Manual, consistent with those used in
the textbook by Wade. (Do not expect all of these to make sense to you now. You will learn them
throughout your study of organic chemistry.)
BONDS
a single bond
a double bond
a triple bond
a bond in three dimensions, coming out of the paper toward the reader
a bond in three dimensions, going behind the paper away from the reader
a stretched bond, in the process of forming or breaking
-
111111111111
ARROWS
in a reaction, shows direction from reactants to products
signifies equilibrium (not to be confused with resonance)
signifies resonance (not to be confused with equilibrium)
shows direction of electron movement:
the arrowhead with one barb shows movement of one electron;
the arrowhead with two barbs shows movement of a pair of electrons
shows polarity of a bond or molecule, the arrowhead signifying the more negative end of the
dipole
SUBSTITUENT GROUPS
Me
a methyl group, CH3
Et
an ethyl group, CH2 CH3
Pr
a propyl group, a three carbon group (two possible arrangements)
Bu
a butyl group, a four carbon group (four possible arrangements)
the general abbreviation for an alkyl group (or any substituent group not under scrutiny)
R
Ph
a phenyl group, the name of a benzene ring as a substituent, represented:
-
..
Ar
..
<»
or
the general name for an aromatic group
0-
continued on next page
vii
Symbols and Abbreviations, continued
SUBSTITUENT GROUPS, continued
Ac
an acetyl group:
Cy
a cyclohexyl group:
Ts
o
II
CH3 - C -
0-
o
tosyl, or p-toluenesulfonyl group: CH3 �
"==1
- So
o
II
Boc a t-butoxycarbonyl group (amino acid and peptide chemistry): (CH3)3C -0 - C II
Z,
or a carbobenzoxy (benzyloxycarbonyl) group (amino acid and peptide chemistry):
Cbz
REAGENTS AND SOLVENTS
DCC
dicyclohexylcarbodiimide
DMSO dimethylsulfoxide
ether
o
o- N= C= N -Q
II
S
H3C ....... 'CH3
diethyl ether, CH3CH20CH 2 CH3
Cl
<>
MCPBA meta-chloroperoxyhenzo;c ac;d
MVK
NBS
o
methyl vinyl ketone
H3C
N
-bromosuccinimide O
continued on next page
.......
g�
O
�
N
I
Br
viii
0
g - 0 - OH
0
< )- CH2-0 - g -
Symbols and Abbreviations, continued
REAGENTS AND SOLVENTS, continued
pyridinium chlorochromate, Cr03· HCI N
PCC
•
Sia2BH
disiamylborane
THF
tetrahydrofuran
� ')
CH3 H H H CH 3
I
I
I
I
I
H-C-C-B-C-C-H
I
I
I
I
CH3 CH 3
CH3 CH3
o
SPECTROSCOPY
infrared spectroscopy
IR
nuclear magnetic resonance spectroscopy
NMR
mass spectrometry
MS
ultraviolet spectroscopy
UV
parts per million, a unit used in NMR
ppm
hertz, cycles per second, a unit of frequency
Hz
megahertz, millions of cycles per second
MHz
tetramethylsilane, (CH3) 4Si, the reference compound in NMR
TMS
s, d, t, q singlet, doublet, triplet, quartet, referring to the number of peaks an NMR absorption gives
nm
nanometers, 10-9 meters (usually used as a unit of wavelength)
mJz
mass-to-charge ratio, in mass spectrometry
8
in NMR, chemical shift value, measured in ppm
A
wavelength
frequency
v
OTHER
a,
ax
e,eq
HOMO
LUMO
NR
0,
m,p
axial (in chair forms of cyclohexane)
equatorial (in chair forms of cyclohexane)
highest occupied molecular orbital
lowest unoccupied molecular orbital
no reaction
artha, meta, para (positions on an aromatic ring)
when written over an arrow: "heat"; when written before a letter: "change in"
partial positive charge, partial negative charge
energy from electromagnetic radiation (light)
specific rotation at the D line of sodium (589 nm)
ix
CHAPTER I-INTRODUCTION A ND REVIEW
1-1
1-2
(a)
(e)
P 1 s 2 2s 2 2p6 3s 2 3px 1 3py 1 3pz 1
1 s 2 2s2 2p6 3s 1
S 1 s 2 2s 2 2p6 3s 2 3px 2 3py 13pz 1
Mg 1 s 2 2s2 2p6 3s 2
Is 2 2s2 2p6 3s 2 3px 1
CI
1 s 2 2s2 2p6 3s 2 3px2 3py2 3pz 1
AI
Ar
1 s 2 2s 2 2p6 3s 2 3px 2 3 py 2 3 pz 2
Si 1 s 2 2s 2 2p6 3s 2 3px 1 3py I
In this book, lines between atom symbols represent covalent bonds between those atoms. Nonbonding
electrons are indicated with dots.
H H H
Na
H - N- H
I
H
(b)
H H
I
I
H - C - C-N - H
I
I
I
H H H
(c)
H-O-H
H
H
I
I
(f) H -C - O - C - H
I
•• I
H
H
1-3
(a)
(i)
H :0: H
I
I
I
H - C - C - C-H
I
I
I
H H H
:N
N:
(g)
• •
H
I
(h)
+
H - ••
O-H
I
H
(d)
I
I
I
H-C-C-C-H
I
I
I
H H H
H
H
I
I
H- C - C - F:
I
I
H
H
(j ) : F- B - F:
I
-:F:
H -B- H
I
H
.
.
The compounds in (i) and (j ) are unusual in that boron
does not have an octet of electrons-normal for boron
because it has only three valence electrons.
(b)
H-C
N:
(c)
H - O - N= O
(d)
O == C = O
:0 :
(e)
(i)
1-4
II
H - C =N - H
(h) H - N=N- H
(f) H - C - O - H
(g) H - C == C - C I :
I
I
I
••
H
H H
H
H
I
I
H - C == C - C - H
(j )
H - C == C == C - H
(k) H - C C - C - H
I
I
I
I
I
I
H
H H
H
H
H
(a) G}J -NQ)
(e)
8
H - C =N - H
I
H
(b)
(f)
H-C
NeD
C)J(D
•
·8
8
II
H - C-O - H
88(.:)
(c)
H - O - N= O
(g)
H - C == C - C leD
I
I
8
H H
0)
There are no unshared electron pairs in parts (i), (j ) , and (k).
1
0)
8
(d)
0)
0)
o == c=o
(h)
H - N=N- H
0)
0)
88
1-5
The symbols "8+" and "8-" indicate bond polarity by showing partial charge. (In the arrow symbolism,
the arrow should point to the partial negative charge.)
�
�
(a) C -C 1
�
�
8-
8+
(b)
C- O
�
�
(c)
C-N
(g)
N-O
8+
8-
(h)
N-S
1-6
8-
8+
(i) N - B
Non-zero formal charges are shown beside the atoms.
H
H
H H H
1
1+ 1
1
C1:
H
N
H
H
C
N
- C - H :Cl:
(b)
(a) H - C - O-H
:
(c)
1
1 1+
1 1
1
H H
H
H H H
In (b) and (c), the chlorine is present as chloride ion. There is no covalent
bond between chlorine and other atoms in the formula.
H
H
I
1(g) Na+ H - B - H
(e) H - C - H
(d) Na+ : O - C - H
(f) H - C-H
I
1
1
I
H
H
H
H
1+
-
•
. .
. .
+
•
1+
-
H
H
H,
H
1- /
(h) Na+ H - B - C N:
(i) H -/C - O - C - H
U) H - O-N-H
I
I
H
H
H
H
:F-B-F:
H ,H1/ H
:F:
As shown in (d), (g), (h), and (k),
alkali
metals like sodium and
C
potassium form only ionic bonds,
1 /H
never covalent bonds.
· O - C - C '- H
H - C = O+- H
(I)
1
I
H
H
/C1'
H H H
1-7 Resonance forms in which all atoms have full octets are the most significant contributors. In
resonance forms, ALL ATOMS KEEP THEIR POSITIONS-ONLY ELECTRONS ARE SHOWN
IN DIFFERENT POSITIONS.
:0:
:0:
: 0:
1
"
1
:O - C=O
O = C - O:
(a) : O - C - O:
•
•
1 _+ '
1
•
•
. .
-
•
•
•
• •
. . -
-
..
:0 :
"
(b) : O - N - O:
+
. . -
..
. .
. .
•
. . -
..
..
. .
•
-
..
:0 :
1
: O - N=O
-
-
+
: 0:
I
O=N - O:
...
2
+
•
•
-
1-7
continued
(c) :O-N=O
(d)
..
H-C=C-C-H
I
I
. .
O=N-O:
..
I
+
H H H
..
-
.. H-C-C=C-H
I
I
+
I
H H H
H - C = C - C - H .. .. H - C - C=C - H
I
I
I
I
I
I
H H H
H H H
(f) Sulfur can have up to 12 electrons around it because it has d orbitals accessible .
:0 :
: 0:
: 0:
I
I
II
O==S-O: ..
:O - S - O: .. .. :O - S=O
(e)
. .
. . -
-
II
II
:0 :
: 0:
: 0:
:0:
II
:O - S=O
I
:0:
-
(g)
H
I
H-C-H
H :0I :
I
I
H - C - C+
\
I
:0 :
H
I
H - C-H
I
H
II
..
..
..
..
•
•
H
I
H-C-H
I
H + 0:
1/
I
..
H- C- C
\
I
H :0 :
I
H-C-H
I
H
:0:
I
O==S==O
I
:0 :
.
..
..
..
.
-
..
/
:0 :
II
O=S- O:
I
:0 :
• •
.
-
.
H
I
H - C -H
H :0I :
I
I
H-C-C
\\
I
H + 0:
I
H- C- H
I
H
1-8 Major resonance contributors would have the lowest energy. The most important factors are:
maximize full octets; maximize bonds; put negative charge on electronegative atoms; minimize charge
separation.
+
(a) H - C - N=O .. .. H - C - N-O: .. .. H - C=N - O: (negative charge on
electronegative
I
I
I
II
I
I
H :0 :
H :0 :
H :0 :
atoms)
minor
mInor
major
•
•
-
. .-
+
+
3
1-8 continued
H
H
H
+
+
+
\ +
\
\
C - C = N - O:
C = C - N - O:
C=C - N=O
(b)
/
/
/
1
1
1
11
1
1
H
H
H
H :0 :
H :0 :
H :0 :
minor
major
major
These two forms have equivalent energy and are major because they have full octets, more
bonds, and less charge separation than the minor contributor.
+
(c)
H - C- O-H
H - C = O - H ..
1
1
H
H
mmor
major
(octets, more bonds)
•
(e)
(f)
+
+
H - C=N=N:
1
H
major (negative charge
on electronegative atom)
..
H - C - C N:
1
H
minor
..
.-
•
H - C - N N:
1
H
minor
H - C=C=N:
1
H
major (negative charge
on electronegative atom)
+
+
H - N - C - C=C - N - H "
1
1
1
1
1
H H H H H
mmor
t
+
•
H-N=C - C=C - N - H
1
1
1
1
1
H H H H H
major
these two forms are major contributors
because all atoms have full octets
+
H - N - C = C - C - N - H ..
1
1
1
1
1
H H H H H
mmor
• •
•
H- N - C= C - C = N - H
1
1
1
1
1
H H H H H
major
:0: :0 :
:0 : :0 :
:0 : :0 :
II
II
1
II
II
1
H - C - C-- C - H----- H - C = C - C - H ----- H - C - C = C - H
1
1
1
H
H
H
major
mITIor
major
these two have equivalent energy and are major because the
negative charge is on the more electronegative oxygen atom
. . -
(g)
.­
•
+
(d)
..
..
• •
4
1 -8 continued
:0 :
II
(h) H - C - N - H
I
major
..
:0 :
+
I
H - C =N H
-
....
..
II----I
..
�
H
(no charge separation)
-
I
minor
H
1-9 Your Lewis structures may appear different from these. As long as the atoms are connected in the
same order and by the same type of bond, they are equivalent structures. For now, the exact placement of the
atoms on the page is not significant. A Lewis structure is "complete" with unshared electron pairs shown.
H H H
H H H H H H
I
I
I
I
I
I
I
I
I
(b) H - C - C - C - Cl:
(a) H - C - C - C - C - C - C - H
I
I
I
I
I
I
I
I
I
H /C H
H H H H /C H
H H1 ' H
H H1 ' H
H H :0:
H H :0 :
H
I
II
I
I
I
II
/
(d) H - C - C - C-H
(c) H-C - C - C - C=C
\
I
I
I
I
I
H
H H
H
H H
Always be alert for the implied double or triple bond. Remember that C has to have four bonds,
nitrogen has three bonds, oxygen has two bonds, and hydrogen has one bond. The only
exceptions to these valence rules are structures with formal charges.
H :0:
II
I
(e) H - C - C - C N:
I
H
(f)
H ,H1 / H
H , CI :0:
II
H -/ C - C - C - O - H
I
H
C
/
H H1 ' H
H H :0: H H
I
I
II
I
I
(g) H-C - C - C - C - C - H
I
I
I
I
H H
H H
1 - 10 Complete Lewis structures show all atoms, bonds, and unshared electron pairs.
(a)
H
H H
(b)
(c) H H
H ,HI
H /H I/H
1/
I
I
H , /, C
C-H
H
C
C
H
C
C
,C /
II \\
H - C , C /- H
C
H
H
,
I
I
/ , N. . /C '
C-H
I
1/
H -/ C�' N '�C ,- H
H
H
H -C - C
I'H
I
H
H
H
H
H I
\
I
..../.. C , ,..... C ,...... H
H
0
H
H
• •
_
• •
5
(e )
1-10 continued
· ·
H ,, / H · 0I I·
H " /C /C,
H
H - C ,C
II
I
H-C
, ...... C,
H / / C"- H
H H
(f)
H H
H " /,, C /.... ,...-0
H - C ' C/··
I
I
H -/ C, /'
/.C,
C
H
H
I
H
•
•
H
H :0 : H H H
(h)
:0 : H
I
II
I
I
II
I
I
I
H ' /C:--.., /,C-C - H
H-C - C-C-C-C-H
c
"c
I
I
I
I
I
I
H
II
H
H H H
C
C,
H/ , C 'i H
I
H
1-1 1 There is often more than one correct way to write condensed structural formulas. You must often make
inferences about what a condensed formula means according to valence rules, especially in structures with
C=O as shown in parts (a) and (d).
(b) (CH3hCHCH2 CH20 H
(c) (CH3hCHCH2 CH(OH)CH3
(a) CH3COCH2 CH2 CH3
0
or CH3CH(CH3)CH2 CH(OH)CH3
or CH3CH(CH3)CH2 CH20H
(g)
(the
has a double bond to the
(d) CH3CH2CH(CH3)CH2 CHO
c arbon preceding it)
(the 0 has a doubl e bond to the
c arbon preceding it)
1-1 2 If the percent values do not sum to 100%, the remainder must be oxygen. Assume 100 g of sample;
percents then translate directly to grams of each element.
There are usually many possible structures for a molecular formula. Yours may be different from the
examples shown here.
some possible structures:
(a) 40.0 gC = 3.33 moles C 3.33 moles = 1 C
1 2.0 g/mole
H ° H
I
I
II
6.67 g H
HO
C-C-C
- OH
3.33
moles
=
1.98
2
H
=
6.60
moles
H
1.0 1 g/mole
I
I
H
H
53.33 gO = 3.33 moles ° 3.33 moles = 1 °
16.0 g/mole
OH
empirical formula
c::::::> empirical weight
30
==
�
molecular weight = 90, three times the empirical weight
=
three times the empirical formula = molecular formula
6
=
=
c:::::>
I C3H603 I
A
Ho
OH
MANY other
structures possible.
1 - 1 2 continued
(b) 32.0 g C
1 2.0 g/mole 2.67 moles C 1.34 moles 1.99 2 C
6.67 g H
5H
4.93
6.60 moles H -:- 1.34 moles
l.01 g/mole
IS.7 g N
14.0 g/mole 1.34 moles N -:- 1.34 moles 1 N
42.6 g a
.
16.0 g/mole - 2.66 moles a -;- 1.34 moles 1 .99 2 a
empirical formula I C2H sN02 I c:::::::> empirical weight 75
=
=
some possible structures:
::=
H
=
molecular weight
empirical formula
(c)
molecular formula
=
C2HsN02
=
37.2 g C
1 2.0 g/mole 3. 1 0 moles C -:- 1 .55 moles 2 C
7.7 5 g H
1.0 1 g/mole 7.67 moles H -:- 1 .55 moles 4.95
55.0 g CI
.
35 .45 g/mole - 1 .55 moles Cl -;- 1.55 moles - 1 CI
=
=
=
=
empirical formula
molecular weight
=
·
1 .� ��/�:l e
3��4� ��ole
=
=
=
I C2H sCl I
=
molecular formula
3.20 moles C
4.75 moles H
1 .60 moles CI
empirical formula
=
=
-:-
I C2H 3Cl I
1 .60 moles
1 .60 moles
c:::::::>
=
2.97
I
H
I
5H
There is only one structure
possible with this molecular
formula:
c:::::::>
=
64.46
H
I
I
H - C- C==C- C- H
7
I
I
I
H C l Cl H
Cl
C]
1 Cl
=
I
H
some possible structures:
H
H
I
=
I
H - C - C - Cl
I
3H
empirical weight
H
I
H
=
62.45
molecular weight 1 25, twice the empirical weight c:::::::>
twice the empirical formula molecular formula ""-C4
-H
- -6- C-12--,-o.
=
I
MANY other
structures possible.
2C
=
H
/
I C2HsCI I
1.60 moles
-:-
a
II
H
=
c:::::::> empirical weight
64, same as the empirical weight
=
empirical formula
3S.4 g C
12.0 g/mole
::=
\
N- C- O - C- H
_
_
(d)
c:::::::>
75, same as the empirical weight
=
H
::=
=
I
I
H H
=
_
I
H- C- C-N02
=
=
H
I
Cl
U
�
MANY other
Cl
structures possi ble.
1-13
1 mole HEr
(a) 5 .00 g HBr x 80.9
g HBr
=
0.06 1 8 moles HEr
0.0618 moles
1 00 mL
pH
(b)
=
-
H30 +
0.06 1 8 moles HBr
0.06 1 8 moles H 30 + (100% dissociated)
x
10gIO [H30+]
1000
mL
1L
=
-
0.6 1 8 moles H30 +
1 L solution
=
logl o (0.6 1 8)
=
�
0.0375 moles NaOH
1.50 g NaOH x 1 mole NaOH
40.0 g NaOH
0.0375 moles -OH (100% dissociated)
0.0375 moles NaOH
=
1 000 mL
0.75 moles -OH
1 L solution
lL
14 1 x 1 0-14
1.33 x 10-14
[H 3 0+] 1 x 1 0- =
0.75
[-OH]
pH -loglO [H30+] -logl o (1.33 x 10- 14 )
0.0375 moles -OH
50. mL
x
=
=
=
=
=
0.75 M
(the number of decimal places
in a pH value is the number of
significant figures)
=
1-14
(a) By definition, an acid is any species that can donate a proton. Ammonia has a proton bonded to
nitrogen, so ammonia can be an acid (although a very weak one). A base is a proton acceptor, that is, it
must have a pair of electrons to share with a proton; in theory, any atom with an unshared electron pair can
be a base. The nitrogen in ammonia has an unshared electron pair so ammonia is basic. In water, ammonia
is too weak an acid to give up its proton; instead, it acts as a base and pulls a proton from water to a small
extent.
(b) water as an acid: H 2 O + NH3
-OH + NH4+
--
H 2 O + HCl
water as a base:
(c) methanol as an acid: CH30H + NH3
methanol as a base: CH30H
+
H 3 O+
ClCH3O+
NH
4+
CH3OH2 + + HS04-
H2 SO4
8
+
1-15
(a) HCOOH
stronger
acid
pKa 3.76
(b) CH3COOweaker
base
(c) CH30H
stronger
acid
pKa 1 5.5
(f)
1-16
NaNH2
stronger
base
+
..
--
--
HCN
FAVORS
weaker PRODUCTS
acid
pK a 9.22
+
CH 3 COOH
stronger
acid
pKa 4 . 74
+
CH30- Na+
weaker
base
.-
--
HCN
stronger
acid
pK a 9.22
+
HCOOweaker
base
.-
--
CH30H
weaker
acid
pK a 1 5 .5
+
(d) Na+ -OCH3
stronger
base
(e)
-CN
stronger
base
+
.-
HOCH 3
weaker
acid
pK a 15.5
+
+
CH3O- FAVORS
stronger REA CTANTS
base
FAVORS
NH3
weaker PRODUCTS
acid
pK a 33
NaCN
weaker
base
FAVORS
PRODUCTS
FAVORS
HCl + H2O
H3O+ + CIPRODUCTS
stronger
stronger
weaker
weaker
acid
base
base
acid
The first reaction in Table 1-5 shows the Keq for this reaction is 160, favoring products.
--
.-
H30+ + CH3OH2O + CH30H
FAVORS
PRODUCTS
stronger
weaker
stronger
weaker
acid
base
base
acid
pK a -1 .7
pK a 1 5.5
The seventh reaction in Table 1-5 shows the Keq for the reverse of this reaction is 3 .2 x 1 0-16.
Therefore, Keq for this reaction as written must be the inverse, or 3. 1 x 1 0 15, strongly favoring
products.
--
.-
:0:
II
CH 3 - C - O - H
:0:
II
CH 3 - C - O-H
+\
H
Protonation of the double-bonded oxygen gives three resonance forms (as shown in Solved Problem
1-5(c)); protonation of the single-bonded oxygen gives only one. In general, the more resonance
forms a species has, the more stable it is, so the proton would bond to the oxygen that gives a more
stable species, that is, the double-bonded oxygen.
• •
9
1-l7 In Solved Problem 1-4, the structures of ethanol and methylamine are shown to be similar to
methanol and ammonia, respectively. We must infer that their acid-base properties are also similar.
(a) This problem can be viewed in two ways. 1 ) Quantitatively, the pKa values determine the order of
acidity. 2) Qualitatively, the stabilities of the conjugate bases determine the order of acidity (see Solved
Problem 1-4 for structures): the conjugate base of acetic acid, acetate ion, is resonance-stabilized, so
acetic acid is the most acidic; the conjugate base of ethanol has a negative charge on a very
electronegative oxygen atom; the conjugate base of methylamine has a negative charge on a mildly
electronegative nitrogen atom and is therefore the least stabilized, so methylamine is the least acidic.
acetic acid > ethanol > methylamine
pK a 4.74
pK a "" 1 5.5 pK a 33
weakest acid
strongest acid
(b) Ethoxide ion is the conjugate base of ethanol, so it must be a stronger base than ethanol; Solved
Problem 1-4 indicates ethoxide is analogous to hydroxide in base strength. Methylamine has pKb 3.36.
The basicity of methylamine is between the basicity of ethoxide ion and ethanol.
ethoxide ion > methylamine > ethanol
weakest base
strongest base
""
1-1 8 Curved arrows show electron movement, as described in text section 1-14.
stronger acid
__
stronger base
.. CH3CH2-�:
conjugate base
weaker base
equilibrium favors PRODUCTS
'0'
'11' � �
(b) CH3CH2 - C - 0 - H + CH3 - N - CH3
I
stronger acid
H
stronger base
.
•
•
•
(c)
• •
•
•
equilibrium favors
PRODUCTS
i-·· :�: ..
:O - S - O - H
II
:0 :
• •
•
•
. .
CH3 - �+- CH3
H
conjugate acid
weaker acid
1
--
equilibrium favors PRODUCTS
O'
.. � 0. '11'
CH3 - O - H + H - O - S - O - H
II
stronger base
:0·:
stronger acid
CH 3-� -H
H
conjugate acid
weaker acid
H
•
•
+
--
H
1+
.. CH 3 - O-H
conjugate acid
weaker acid
•
conjugate base
weaker base
+
•
:0 :
II
O==S - O-H
I
: 0:
conjugate base, weaker base
..
�
--I..
l-.
...
....
. .
10
.
.
•
•
:0:- r
,
..
O==S - O - H
II
:0:
• •
1-1 8 continued
(d)
- .. � 0·
Na+ :O - H + H-S-H
..
stronger acid
stronger base
(e)
---
_
equilibrium favors PRODUCTS
__
H
�
/'
I
• •CH3 - �� H + CH3-O:
stronger base
H
stronger acid
"
+
..
.. H - O - H + Na+ :S-H
conjugate acid conjugate base
weaker base
weaker acid
CH3 - � - H
H
conjugate base
weaker base
---
equilibrium favors PRODUCTS
i
+
stronger base
stronger acid
conjugate acid
weaker acid
equilibrium favors PRODUCTS
(g)
:0:
II . �
CH 3 - C - O - H
..
weaker acid
:0:
- . . II
:O-S-CH 3
• • II
:0 :
~
+
�
...
.
......f---l
equilibrium favors REACTANTS
+
:0:
II • CH 3 -C - •.0:
.
:0 :
II
0==S-CH3
••
I
:0 :
weaker base
:0:
: 0:
II
I
CH 3 -C -�: ...... ... CH 3-C = �
conjugate base
stronger base
CH 3 - O-H
••
conjugate acid
weaker acid
•
:•0:
�
I ••
CH
-C
=O
....
.
..
.
3
conjugate base
••
weaker base
_
-...
.
......f--
-
:0:-
•• I
0==S-CH3
• • II
:0 :
}
:0 :
II
H - O-S-CH3
• • II
:0 :
conjugate acid
stronger acid
1-1 9 Solutions for (a) and (b) are presented in the Solved Problem in the text. Here, the newly formed
bonds are shown in bold.
H
I
••
/" - CH 3
H-B-H
(c)
H - B - H + CH 3-O
"
1'1+
CH 3-O - CH 3
acid H �base
f---l�
_
(d)
oI I :
c
CH 3-C�
-H
:O - H
+
acid
••
••
base
••
..
:0:
I
CH 3-r - H
:O - H
11
t
1 - 19 continued
(e) Bronsted-Lowry--c-proton transfer
� :�:
:�:
i
_}
H :0:
H - C - C - H +/..
: O - H �==� H --..�C - C - H ....- H - b == b - H
)
k � base
acid
.
(f)
..�
CH3 - � - H
H
base
+
()
..
.
CH3 - S=!:
acid
+
•
.
•
.-
+
..
H - O-H
:Cl :
1-20
Learning organic chemistry is similar to learning a foreign language: new vocabulary, new grammar (the
reactions), some new concepts, and even a new alphabet (the symbolism of chemistry). This type of
definition question is intended to help you review the vocabulary and concepts in each chapter. All of the
definitions and examples are presented in the Glossary and in the chapter, so this Solutions Manual will
not repeat them. Use these questions to evaluate your comprehension and to guide your review of the
important concepts in the chapter.
1-21 (a) CARBON!
(b) oxygen
1-22
valence e- --- 1
H
Li
2
Be
3
B
(c) phosphorus
(d) chlorine
4
5
6
7
C
N
0
S
F
Cl
Br
I
p
8
He (2e-)
Ne
1-23
(a) ionic only
(b) covalent (H-O-) and ionic (Na+ -OH)
(c) covalent (H-C and C-Li), but the C-Li bond is strongly polarized
(d) covalent only
(e) covalent (H-C and C-O-) and ionic (Na+ -OCH3)
(f)
covalent (H-C and C=O and C-O- ) and ionic (HC02 - Na+)
12
(g) covalent only
1-24
: Cl
Cl:
:..
Cl "
:.Cl.
Cl:
:.Cl.
Cl:
N
(b)
P
(a)
N ...... . .
p ........
: Cl I Cl:
I
I
: .c!'"
I Cl:
. :CI:
:Cl:
:Cl:
:Cl:
CANNOT EXIST
NCls violates the octet rule; nitrogen can have no more than eight electrons (or four atoms) around it.
Phosphorus, a third-row element, can have more than eight electrons because phosphorus can used
orbitals in bonding, so PCls is a stable, isolable compound.
1-25 Your Lewis structures may look different from these. As long as the atoms are connected in the
same order and by the same type of bond, they are equivalent structures. For now, the exact placement
of the atoms on the page is not significant.
H ,H1/ H
H , C1 / H
(a) H - N - N-H
(c) H - C - N+- C "- H :CI :
(b) H - N=N - H
I
I
H
H / CI
H H
/
H H1" H
.. ,../ ..
•
/ ..
Cl:
,../ ..
•
•
,
/ ..
•
/'
. •
. . -
(d)
H
I
H - C - C N:
I
H
(g)
:0:
II
H - O-S-O-H
II
:0 :
(e)
H :0:
II
I
H-C - C - H
I
H
(h)
H
I
H - C - N=C = O
I
H
H
H :N H
I
II
I
H-C-C-C-H
I
I
H
H
/
1-26
(a)
(k)
H
H
:0:
I
II
I
(i) H - C-O-S-O-C - H
I
II
I
H
H
:0 :
H H
H ,,1/
H " CI
H -/ C-C - N=O
I
H
C
H /H1" H
H
H
:0:
I
I
II
H - C-C = C-C-C = C-C - O-H
I
I
I
I
I
I
H H H H H H
.
H :0: H
I
I
II
(f) H - C-S-C - H
I
I
H
H
H :0: H :0:
I
II
I
II
(b) :N C-C - C - C - C-H
I
I
H
H
•
13
1-26 continued
H - O: H :0:
.
(c)
.
I
I
I
I
II
H-C = C-C-C-C-O-H
I
I
H
H
••
H H
1-27 In each set below, the second structure is a more correct line formula. Since chemists are
human (surprise!), they will take shortcuts where possible; the first structure in each pair uses a
common abbreviation, either COOH or CHO. Make sure you understand that COOH does not
stand for C-O-O-H. Likewise for CHO.
(a)
� /'... �
/' � �
.....
OR
� COOH
OH
� OH
OH
1 -28
(a)
OR
o
I
0
o
CHO
OR
H
0
H H H
H H H H
1
1
1
1
1
I
I
H - C - C - C - C - H and H-C - C - C-H
1
1
I
I
1
I
I
H C H
H H H H
H /H1 ' H
H
(b)
� CHO
H
N - C TIY
N C
o
OH
OR
(c)
(b)
COOH
H
I
H
1
H - C - C - N:
I
I
I
H H H
and
H
H
I
H-C-N-C-H
I
I
H
H
H
I
I
14
these are the only two possibilities,
but your structures may appear
different-making models will
help you visualize these structures
these are the only two possibilities,
but your structures may appear
different-making models will
help you visualize these structures
1 -28 continued
(c) There are several other possibilities as well. Your answer may be correct even if it does not appear
here. Check with others in your study group.
H H
H
H H H
H H H
I
I
I
I
I
I
I
I
I
:O - C-C - O - C - H
:O - C - C-C - O:
:O-C-C - C - H
I
I
I
I
I
I
I
I
I
I
I
I
I
H H H
H
H H H H H
H H :0: H
I
H
H :0:
:0 :
These are the only three
I
II
/ \
(d) H - C-C - H H-C = C-O - H
H - C - C - H structures with this molecular
formula.
I
I
I
I
I
H
H H
H H
H H
H
H H H
H H H
1 -29
I
I
I
I
I
I
I
I
I
(a) only three O - C-C-C - H
H - C - C - O - C - H H - C-C - C - H
possible
I
I
I
I
I
I
I
I
I
I
structures
H
H H H H
H H
H 0 H
I
HOCH2 CH2 CH 3
CH3CH2 0CH 3
H
CH3 CH(OH)CH3
(b) There are several other possibilities as well.
H
H H 0
H 0 H
H
I
I
I
I
II
II
I
I
H - C= C - C - H
H -C-C-C- H H - C - C - C-H
H - C=C-C - O - H
I
I
I
I
I
I
I
I
I
I
H H
H 0 H
H
H
H H H
I
CH3 CH2 CHO
CH 3 COCH 3
H2 C = CHCH 2 0H
H
H
H 2 C =C(OH)CH3
I
H - C = C - O-C - H
I
I
I
H H
H
H2 C = CHOCH 3
1-30 General rule: molecular formulas of stable hydrocarbons must have an even number of hydrogens.
The formula CH2 does not have enough atoms to bond with the four orbitals of carbon.
H H
one carbon:
I
I
H
C
-H
H
C
H
C
=
C
H
H
C-C
-H
two
carbons:
I
I
I
I
I
H-C-H
C2 H 4 H H
C2 H6 H H
I
H CH4
H
H
H H H
I
I
I
I
I
H - C = C-C - H
H-C - C - C - H
three carbons: H -C C - C - H
I
I
I
I
I
I
I
C3H6 H H H
C 3 HS H H H
C3 H 4 H
15
H H
(b) H ..... � _ � _H
\
H I
H
..../. C ,0 0/ C "....
N
H
H
I
H
�
H H H :0:
I I I "
H - N - C-C-C-C-O - H
I I I I
H H H H
(e)
H
H
(g)
I
\
C-C \ :0:
H,
"
II
\
H -/C-C
C-S==O
\
/
I
H
C = C :0:
I
\
H HI
H
H H
\ I
/C, ,....H
C,
:N
II
H
H\ C I HI H
/
H - C , C - H H HH
I I,....
I
\
C
H - C-C
/H
I
1 " 0O0- C/- C-H
H H
\
"
C_HH
I \
H H
:0: ·0 0, , ·0
(h)
II H
H , /C, I /C,oo
C
C
O-H
I
\\
/C-C-H
I
H
H
(c) C4H9NO
(h) C6H60 3
1 -32 (a) CsHsN
(b) C4H9N
(g) C7Hs0 3 S
(f) C9H 1SO
1 -33
(a) 1 00% - 62.0% C - 1 0.4% H 27.6% oxygen
62.0 g C
.
1 2.0 g/mole - 5. l7 moles C -;- l.73 moles - 2.99 3 C
1 0.4 g H
l.01 g/mole 1 0.3 moles H -;- l.73 moles 5.95 6 H
(c) some possible structures-MANY
other structures possible:
H H
H , "" C, ' H
C-O-H
H-CI
I
H-C
" 'c" C-O-H
,
H '"
H
H H
H H H H H °
1��06g7�ole l.73 moles ° -;- l.73 moles 1 °
H-C-C-C-C-C-C-O-H
(b) empirical formula C 3H60 c:::::> empirical weight 58
H H H H H
H H H H H 0
molecular weight = 1 17, about double the empirical weight
H-O-C-C-C-C-C - C-H
q double the empirical formula
molecular formula
H H H H H
C 6H120 2
H H H H 0
H
H-C-C-C-C-C-O-C-H
H H H H
H
=
_
_
=
=
=
=
=
I
I
_
==
=
=
I
16
I
I
I
I
I
I
I
I
I
I
I
1
I
I
I
I
I
I
I
1
I
I
I
I
II
I
I
I
I
=
II
II
I
I
1-34 Non-zero formal charges are shown by the atoms.
+
+
(a) H - C = N = N :
H - C - N - N:
H ,H1 / H
(b) H
C
1
1
,
+1
H
H
H -/ C - N - O:
H
H
I
H
H
C
I
1
+
+
1
(c) H - C = C-C - H
(d) H - C-N=O
(e) H - C-O
-C-H
H /1'
H H
1
I +1
1 1 1
1 1
H C H
H H H
H :0:
H /H1 ' H
1-35 The symbols "8+" and "8-" indicate bond polarity by showing partial charge. Electronegativity
differences greater than or equal to 0.5 are considered large.
• •
....
..I----I..
�
-
• •
. . -
8+
8-
8-
8+
8+
8-
8+
8+
(c)
C-Li
large
8-
8+
(h)
N-H
large
(g) C - Mg
large
(f) C-B
large
8-
8-
(b) C - H
small
(a) C - CI
large
8+
8-
8-
8+
(d) C - N
small
(i) O - H
large
8+
8-
8+
8-
(e) C - O
large
(j )
C - Br
small
1-36 Resonance forms must have atoms in identical positions. If any atom moves position, it is a
different structure.
(a) different compounds-a hydrogen atom has changed position
(b) resonance forms-only the position of electrons is different
(c) resonance forms-only the position of electrons is different
(d) resonance forms-only the position of electrons is different
(e) different compounds-a hydrogen atom has changed position
(f) resonance forms-only the position of electrons is different
(g) resonance forms-only the position of electrons is different
(h) different compounds-a hydrogen atom has changed position
(i) resonance forms-only the position of electrons is different
(j ) resonance forms-only the position of electrons is different
1-37
(a)
H :0:
H :0:
I
II
H - C-C - C - H
1
I
H
H
(b) :0:
II
H - C - C = C-C - H
I
I
I
H H H
..
..
i
I
. .
H - C - C=C - H
1
I
H
H
:0:
:0:
II
�
....
..
I-.
--J..
H - C-C - C = C - H
1
I
I
H H H
17
I
....
..
f-.
--J
�
..
H - C = C-C = C - H
1 1 I
H H H
1-37 continued
(c)
<>
C H2
0� /;
(d)
4
•
C H2
(f)
(g)
-
+
( )
�:
4
CH,
0/
+
�
CH2
..
.
0---0
0+
<>
•
�
�
.
0
..
0.
----
-:
0=
/-
.0
.
< > �: -o�
+
( >
CN- - CN+-H
+
0 .. 0
0
-H
+
4
Q
:0 :
..
o
..
+
..
_
H
•
•
0
(h)
.
..
CH,
+
+
(e)
+
O
.�_
�
0+
----
Q Q
+
:0 :
:0
:
..
-
(i) CH 3 - C = C - C = C - C - CH3
I
I
I
I
I
H H H H H
+
+
CH3 - C = C - C - C = C-CH 3
/ HI HI HI HI HI
-----
CH3 - C C = C - C = C -CH3
I
I
I
I
I
H H H
H H
(j) no resonance fonns-the charge must be on an atom next to a double or triple bond, or next to a non­
bonded pair of electrons, in order for resonance to delocalize the charge
-
18
1-38
..
(a) O==S-O:
+
(b) 0=0-0:
+
..
•
..
•
.
.
:O-S==O
+
..
O==S==O
•
..
:0-0=0
+
(c) The last resonance form of S0 2 has no equivalent form in 03. Sulfur, a third row element, can
have more than eight electrons around it because of d orbitals, whereas oxygen, a second row
element, must adhere strictly to the octet rule.
1-39
(a)
#3
� ..
#2
NH
�
II
/
CH 3 - N - C - NH 2
I
H I
, H+to#3 �H+to#2
�
#1
•
H NH
I II
CH3 - N - C - NH 2
I
H
no other resonance forms
•
• •
NH
II
CH3-N - C - NH3
I
H
no other resonance forms
+
NH2
II
CH 3 - � - C - NH2
H
+
+
t
NH2
NH2
NH2
I
I +
I
CH3 - N = C - NH2
CH 3 - N - C - NH 2
CH3 - N - C = NH2
I
I
I
H
H
H
(b) Protonation at nitrogen#3 gives four resonance forms that delocalize the positive charge over all three
nitrogens and a carbon-a very stable condition. Nitrogen#3 will be protonated preferentially, which we
interpret as being more basic.
+
.......
t---'l...
1 -40
(a) CH 3 - C - C N:
I
H
minor
:0:
......
t----I
....
I
(b) CH3 - C==C - C - CH3
I I
H H
minor
+
.......
t---'l....
+
CH3 -C == C = N:
I
H
major (negative charge
on electronegative atom)
. :0 :
I
CH 3 -C - C ==C - CH 3
I I
H H
mmor
• •
.
.... ..
t----I...
+
19
......
t----I...
:0 :
II
CH3 - C - C ==C - CH 3
I I
H H
major-full octets,
no charge separation
1 -40 continued
(c)
:0 :
:0:
II -. . II
CH3 - C - ? - C - CH3
.. : 0:
...
....
.
1-----1
..
.
I
II
I
..
....1-----1
.
..
.
H
major
II
I
I
CH3 - C - C == C - CH3
H
�-----------
minor
.. :0:
:0:
CH3 - C == C - C - CH3
\
H
:0:
}
------�--_/
major
Y
negative charge on electronegative atoms-equal energy
+
+
• - +
CH3 - C - C == C - N == 0 - CH3 - C == C - C• - N = 0 - CH3 - C == C - C == N - 0 :
(d)
I
I
H
I
H
I
••
I
H :0:
I
H
I
H
I
H :0 :
••
I
H
I
H
I
••
:0 :
major-negative charge on
electronegative atoms
minor
minor
I
H
NOTE: The two structures below are resonance forms, varying from the first two structures in part (d) by
the different positions of the double bonds in the N02. Usually, chemists omit drawing the second form of
the N02 group although we all understand that its presence is implied. It is good idea to draw all the
resonance forms until they become second nature. The importance o/ understanding resonance forms
cannot be overemphasized.
+
••
CH3 - C - C == C - N - 0:
I
H
I
H
I
II
H :0:
II
••
NH2
II
CH3CH2 - C - NH2 .......I---J.�
+
\..major-full octets
y
I +
CH3CH2 - C == NH2
major-full octet:;
equal energy
+
CH3 - C - CH3
I
+
I
H
no resonance stabilization
+
CH2 == C - C - CH3 ......f---l
.
.,
�
I
H
I
H
+
••
I ••
H
more stable-resonance stabilized
CH3 - C - 0 - CH3 ......I---l.,� CH3 - C == 0 - CH3
H
�)
I
H :0:
+
CH3CH2 - C - NH2 ......I---J.,..
1 -4 1
(a)
I
H
NH2
I
mmor
I
H
NH2
(e)
- +
CH3 - C == C - •C• - N - 0:
..
....f-.
-.
�
+
H
CH2 - C == C - CH3
I
H
more stable-resonance stabilized
I
I
H
H
I +
I
CH2 == C - C - CH2
H
no resonance stabilization
20
1-4 1 continued
(c) H - C - CH3
\
H
H - C - C - N:
\
H
...
H - C = C = N:
\
H
..
more stable-reson ance stabi l i zed
no resonance stabi li zati on
+
CH
c './ 2
\ H
C
...:/
C
C 'H
\
H
no resonance stabilization
H
\
CH3 - C - CH3
I
CH3 - C - CH 3
more stable-resonance stabilized
(e)
CH3 - N - CH 3
1
CH3 - C - CH 3
CH3 - � - CH 3
CH3 - C - CH3
+
....
...I-.
--I..
�
+
+
no resonance stabilization
more stable-resonance stabilized
1-42 These pK a values from the text, Table 1-5, and Appendix 5 provide the answers. The lower the
pK a ' the stronger the acid.
most acidic
least acidic
<
<
NH3 <
33
1-43 Conjugate bases of the weakest acids will be the strongest bases. The pKa values of the conjugate
acids are listed here. (The relative order of the first two was determined from the pK a values of sulfuric acid
and protonated acedic acid in Appendix 5 of the textbook.)
least basic
most basic
from �6. 1
from �5
from 1 5.5
from 4.74
from 1 5.7
from 33
1 -44
(a) pK a
loglO K a = log lO (5.2 x 10 5 ) 4.3 for phenylacetic acid
for propionic acid, pKa 4.87: K a 10 4 .87 1 .35 x 1 0 5
(b) phenylacetic acid is 3.9 times stronger than propionic acid
5 .2 x 1 0 5
3.9
1.35 x l0 - S
(c)
CH2 COO- + CH3CH 2 COOH ...
CH 2 COOH + CH3CH2COO­
weaker acid
stronger acid
Equ i l i bri u m favors the weaker acid and base. In this reaction, reactants are favored.
=
�
�
=
-
=
-
=
-
-
=
<>
-- < >
21
1-45 The newly formed bond is shown in bold.
.. ('. .
.. CH 3 - � - CH3
+
-CI
CH
(a)
CH 3 - .O:
3
. .:
. �
electrophile
nucleophile
Lewis acid
Lewis base
(b) CH3 - O - CH 3 + H - O - H
+1 )
••
H 3 c �ucleoPhile
Lewis base
e I ectrophOI Ie
Lewis acid
(c)
c
�:
H - C-H
nucleophile
Lewis base
1
H-C-H
1
H - N+- H
1
H
H
1
CH 3 +-N - CH2CH3
1
H
electrophile
Lewis acid
�
' ,, '
�
"
+
..:CI :
+
:0:
(e) CH 3-C - CH 3 + H-O-S-OH
�. " . .
nucleophile
:0:
Lewis base
electrophile
Lewis acid
(g )
H +� - H
CH3
.. -
�
·0·
+
:0:
:N - H
'----/ 1
H
electrophile
nucleophile
Lewis acid
Lewis base
+
CH 3 - ? - CH3
..:CI :
+
--t..�
-
:O - H
:0:
"
"
CH3 -C-CH 3 + :O-S-OH
.. " ..
:0:
CI
1+
:Cl - Al - Cl
1
nucleophile
electrophile
Cl
Lewis base
Lewis acid
This may also be written in two steps: association of the CI with AI, and a second step where
the C-Cl bond breaks.
.. : 0:
I
..
CH 3 - C - CH2 + : O - H
.. CH 3 - C == CH 2 + H - O - H
••
� J nucleophile
'-- I �
H
Lewis base
electrophile
Lewis acid
:R� -
---t�
22
1 -45 continu�
(h)
F-B - F
I
F
electrophile
Lewis acid
CH2 = CH2
nucleophile
Lewis base
+
�
(i) BF3 - CH2 - CH2
electrophile
Lewis acid
1 -46
(a) H2S04
+
+
<)
o
HS0 4-
g- O- H
(d) (CH3hN - H
+
-OH
(e) HO - C - OH
+
2 -0H
oII
(f)
H20
+
(g) HCOOH
NH3
+
CH 30-
CH3COOH
+
CH 3 COO-
(CH3hN:
+
+
.. BF3 - CH2 - CH2 - CH2 - CH2
CH2 = CH2
nucleophile
Lewis base
CH3COO-
+
(b) CH3 COOH
(e)
•
F
+
-I
F - f - CH2 - CH2
F
+
+
(CH 3 hN
+
< )
�OH
(CH3hN:
oII
+
+
H
o
g- o-
+
H20
H20
-O-C - OHO-
-
+
2 H20
+NH4
HCOO-
CH30H
+
1 -47
CH3CH2 - 0- Li+ + CH4
(a) CH3 CH2 - O - H + CH3-Li
(b) The conjugate acid of CH3Li is CH4 Table 1 -5 gives the pKa of CH4 as > 40, one of the weakest acids
known. The conjugate base of one of the weakest acids known must be one of the strongest bases known.
-----l.�
23
1 -48
From the amounts of CO2 and H20 generated, the milligrams of C and H in the original sample can be
determined, thus giving by difference the amount of oxygen in the 5.00 mg sample. From these values, the
empirical formula and empirical weight can be calculated.
(a) how much carbon in 1 4.54 mg CO2
14.54 mg CO2 x
1 mmole C
1 mmole CO2
x
1 mmole CO2
44.0 1 mg CO2
X
12.0 1 mg C
1 mmole C
how much h)::drogen in 3.97 mg H2Q
1 .008 mg H
2 mmoles H
1 mmole H2O
X
x
3.97 mg H20 x
1 mmole H
1 8.016 mg H2O 1 mmole H2O
=
3.968 mg C
=
0.444 mg H
how much ox):: gen in 5.00 mg estradiol
5.00 mg estradiol 3 .968 mg C 0.444 mg H = 0.59 mg °
calculate empirical formula
3.968 mg C
= 0.3304 mmoles C
0.037 mmoles = 8.93 9 C
1 2.01 mg/mole
-
-
==
0.444 mg H
1.008 mg/mole
=
0.440 mmoles H
0.037 mmoles
0.59 mg °
16.00 mg/mole
=
0.037 mmoles °
0.037 mmoles
empirical formula
=
�
=
=
1 1 .9 "" 1 2 H
1°
empirical weight
(b) molecular weight = 272, exactly twice the empirical weight
twice the empirical formula = molecular formula =
24
=
1 36
CHAPTER 2-STRUCTURE AND PROPERTIES OF ORGANIC MOLECULES
2- 1 The fundamental principle of organic chemistry is that a molecule's chemical and physical properties
depend on the molecule's structure: the structure-function or structure-reactivity correlation. It is essential
that you understand the three-dimensional nature of organic molecules, and there is no better device to
assist you than a molecular model set. You are strongly encouraged to use models regularly when reading
the text and working the problems.
(a) requires use of models
H H
(b)
The wedge bonds represent bonds coming out of the plane of the paper
,f
H"
C """
H toward you.
C /"
C /" ./ The dashed bonds represent bonds going behind the plane of the paper.
J\
J\./
H H
H H
(b) The electrostatic potential map for
2-2 (a) The hybridization of oxygen is sp3 since it has
water shows that the hydrogens have
two sigma bonds and two pairs of nonbonding electrons.
low electron potential (blue), and the 3
The reason that the bond angle of 1 04.5° is less than the
area of the unshared electron pairs in sp
perfect tetrahedral
angle
of
1
09.5°
is
that
the
lone
pairs
in
orbitals has high electron potential (red).
the two sp3 orbitals are repelling each other more
strongly than the electron pairs in the sigma bonds,
high electron
thereby compressing the bond angle.
potential (red)
O
repulsion
low electron
O
,�) potential
O
(blue)
O
"" H
.
H ) compression
2-3 Each double-bonded atom is sp2 hybridized with bond angles about 1220°; geometry around sp2 atoms
is trigonal planar. In (a), all four
carbons and the two hydrogens on the sp carbons are all in one plane.
3
Each carbon on the end is sp hybridized with tetrahedral geometry and bond angles about 109°. In (b), the
two3 carbons, the nitrogen, and the two hydrogens on the sp2 carbon are all in one plane. The CH3 carbon is
sp hybridized with tetrahedral geometry and bond angles about 109°.
(0'
(0
(b)
2-4 The hybridization of the nitrogen and the triple-bonded carbon are sp, giving linear geometry
(C-C-N are linear) and a bond angle around the triple-bonded carbon of 1 80°. The CH3 carbon is
sp 3 hybridized, tetrahedral, with bond angles about 109°.
25
2-5
(a) linear, bond angle 1 80°
• •
• •
o == C == o
+2 +
Sp
Sp Sp+ 2
• •
• •
(b) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09° around each atom
not a bqnd-;-shows
not a bond-s how s
� � lon� pair gomg
'.
H
H
lone Rair coming \ • ,
behmd paper
"'--- /
out 01 paper
I
I
H
H
0
,
H-C-O-C-H
....... , .....
C
C"
I
I
A '"
l ,
H
H
H H H H
(c) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09° around each atom
H
H
H
.
.
H
�' ..... H
/
C
H -/" C - N - C "- H
not a bond-shows
I
I ........-- lone
H
H
pair going behind paper
C
H
'/
N
..
/ 1'
'-...
H H H
C /' , " "
J '=:.,"
C-H
H H J \H
H
(d) trigonal planar around the carbon, bond angles 1 20°; tetrahedral around the single-bonded oxygen,
bond angle 109°
:0:
3
II
,.....-- ·. 0 ·.
sp
all atoms in
C
r
I
H
II
one plane
/
"°/
H
sp2 H - C - O - H
• •
• •
�
• •
• •
..
• •
(e) carbon and nitrogen both sp, linear, bond angle 1 80°
H - C N:
all three atoms in a line
(0
trigonal planar around the sp2 carbons, bond angles 1 20°; around the sp3 carbon, tetrahedral geometry
and 1 09° angles
sp2
H 1\ H
H
\C == C/
I
H - C - C == C - H
I
H
I
H
I
H
/
H/C "
\,' H
sp3
H
\
H
(g) trigonal planar, bond angle about 1 20°
(the other resonance form
.0. == .0.- .0.-·
• •
+
• •
-
of ozone shows that BOTH
end oxygens must be sp 2_
see Solved Problem 2-8)
26
2-6 Carbon-2 is sp hybridized. If the p orbitals making the pi bond between C-l and C-2 are in the plane of
the paper (putting the hydrogens in front of and behind the paper), then the other p orbital on C-2 must be
perpendicular to the plane of the paper, making the pi bond between C-2 and C-3 perpendicular to the paper.
This necessarily places the hydrogens on C-3 in the plane of the paper. (Models will surely help.)
model of perpendicular n bonds
H
H " '" 1
2
3/
'C == C == C
'
"'
H
t
H
sp
2-7 For clarity, electrons in sigma bonds are not shown.
(a) carbon and oxygen are both sp2 hybridized
One pair of electrons on oxygen is
always in an sp2 orbital. The other
pair of electrons is shown in a p
orbital in the first resonance form, and
in a pi bond in the second resonance
form.
empty
orbital
(b) oxygen and both carbons are sp2 hybridized
H 1 200
H
\\
C == .O.
. 1..
H -· C
1 2�\H
....
..I---l..
�
\C - O:
.. ..
II
H-C
\H
27
2-7 continued
(c) the nitrogen and the carbon bonded to it are sp hybridized; the left carbon is sp2
1 800
H
\
1 200
c
/
H
� .N:.
(
H
II
�/
�
H
c-c
N:
oil
(d) the boron and the oxygens bonded to it are sp2 hybridized
H - O:
H
\
I
B-O:
I
H-O
. .:
• •
P
..
II
H
H - O:
\
I
B=O:
I
H - O. :
+
Oe Q(
(j"" " " '" 0
O�o 06
�
O
SP'
H-O
'' B
P
H-O
sp2
P
-
P
0
II
�
.
empty p
P
H
sp'
-
H
B - O· : ..
I
H-O
. .:
oG�
" " ",,:
:n
(.:) p
VH
H_
+
_
sp2
�
•
sP2
{J'Q"" '" " "
,\JD
P
..
',
_
�QD
O
!
Vt
-
• •
28
H - O:
H
\
I
B-O:
II
H - .O.
+
�'
B - O ""'"
sp'
' ' �O
"
\\
H-O
" �
�
�/ O
"
+
Oe
/.:\ Sp 2
H-o
I
H-O
H
sp
• •
2-8 Very commonly in organic chemistry, we have to determine whether two structures are the same or
different, and if they are different, what structural features are different. In order for two structures to be
the same, all bonding connections have to be identical, and in the case of double bonds, the groups must
be on the same side of the double bond in both structures. (A good exercise to do with your study group is
to draw two structures and ask if they are the same; or draw one structure and ask how to draw a different
compound.)
(a) different compounds; H and CH3 on one carbon of the double bond, and CH3 and CH2 CH3 on the
other carbon-same in both structures; drawing a plane through the p orbitals shows the H and CH3 are on
the same side of the double bond in the first structure, and the H and the CH2CH3 are on the same side in
the second structure, so they are DIFFERENT compounds
H 3 e'
\
" CH3
/
- - - - - - - e ::E-C- - - - - - - - - -
I
H"
.. ... .. ... ..
\
" CH2CH3
these are DIFFERENT
(b) same compound; in the structure on the right, the right carbon has been rotated, but the bonding is
identical between the two structures
(c) different compounds; H and B r on one carbon, F and Ci on the other carbon in both structures; H and
CI on the same side of the plane through the C=C in the first structure, and H and F on the same side of
the plane through the C=C in the second structure, so they are DIFFERENT compounds
(d) same compound: in the structure on the right, the right carbon has been rotated 1 200
2-9
(b)
CH3 - N :
".. CH3
:N
H
H H
"
(a)
NOT INTER­
"
"
,
C
CONVERTI
LE
C
B
H-C-C=N-C-H
CH3 ",.. ....... H
CH3 ",.. ....... H
/ , � I
, '-. 3
sp
two CH3's on opposite
two CH3's on the same
H \ 2 H
sides of the C=N
side of the C=N
sp
compare
compare
(c) the CH3 on the N is on the same side
as another CH3 no matter how it is
drawn-only one possible structure
and
2- 10
(a)
".. CH3
:N
"
C
",.. " CH3
CH3
H"
F
/
C = C "/
H
F
. . groups on one
(b) no cis-trans iSOmeriSm
.
.
two IdentIcal
(c) no cis.. -tram, Isomensm
.
. } carb on 0 f th e d ou bl e b on d
(d) no cis-trans Isomensm
trans
(e)
QC = C CH3
H
...
,
...
CIS
H
and
H
QC
�
c,
...
CH3
H
"-
trans
29
"cis " and "trans " not
defined for this example
2-1 1 Models will be helpful here.
(a) cis-trans isomers-the first is trans, the second is cis
(b) constitutional isomers-the carbon skeleton is different
(c) constitutional isomers-the bromines are on different carbons in the first structure, on the
same carbon in the second structure
(d) same compound-just flipped over
(e) same compound-just rotated
(f) same compound-just rotated
(g) not isomers-different molecular formulas
(h) constitutional isomers-the double bond has changed position
(i) same compound-just reversed
(j) constitutional isomers-the CH3 groups are in different relative positions
(k) constitutional isomers-the double bond is in a different position relative to the CH3
2- 1 2
(a) 2.4 D
4.8 x 8 x 1 .2 1 A
0.4 1 , or 4 1 % of a positive charge on carbon and 4 1 % of a negative charge on oxygen
:0:
(b)
I
..
..... C .......
R + R
B
Resonance form A must be the major contributor. If B were the major contributor, the value of the charge
separation would be between 0.5 and 1.0. Even though B is "minor", it is quite significant, explaining in
part the high polarity of the C=O.
8
=
=
..
.....
2- 1 3
B oth NH3 and NF3 have a pair of nonbonding electrons on the nitrogen. In NH3, the direction of
polarization of the N-H bonds is toward the nitrogen; thus, all three bond polarities and the lone pair
polarity reinforce each other. In NF3 , on the other hand, the direction of polarization of the N-F bonds is
away from the nitrogen; the three bond polarities cancel the lone pair polarity, so the net result is a very
small molecular dipole moment.
polarities reinforce;
large dipole moment
polarities oppose;
small dipole moment
1 O·
� N�
\\ ' H
H
2- 1 4 Some magnitudes of dipole moments are difficult to predict; however, the direction of the dipole
should be straightforward, in most cases. Actual values of molecular dipole moments are given in
parentheses. (Each halogen atom has three non bonded electron pairs, not shown below.) The C-H is
usually considered non-polar.
(a)
LCi
H
(b)
H �, +-­
H it"
/
large dipole ( 1 . 8 1 )
C-F
H � C � large dipole ( 1 .54)
/
XX: CI
H
..
..
I
I
net
net
H
....
30
2-14 continued
(c)
F
(d)
�" cL
/ net dipole 0
��
�F
=
each end oxygen has
one-half negative
charge as it is the
composite of two
resonance forms; see
net solution to 1 -38(b)
small dipole (0.52)
Qo 1
(e)
T
0�,
.
�
0
'x�o " " o�
(g)
(i)
ro or
0; 13
1
�
C1
/
' CH3 net
H
1 \)
1
\," ' CH 3
/ N ,�
H3C
(k)
,/
CH3
net
1
net
(f)
(I)
\\ �
Il-F
F
-
..
net
large dipole (2.95)
large dipole
net
/
net
large dipole (l .45)
----+
CI
N CD
I
\
small dipole (0.67)
F
� �
H-C
(h)
large dipole (2.72)
large dipole ( 1 .70)
e C
�
B
I
-
net dipole
=
0
(m)
tt�
� N" "
H "'" \\ " H
H
net dipole = 0
net dipole 0
In (k) through (m), the symmetry of the molecule allows the individual bond dipoles to cancel.
=
2- 1 5 With chlorines on the same side of the double bond, the bond dipole moments reinforce each other,
resulting in a large net dipole. With chlorines on opposite sides of the double bond, the bond dipole
moments exactly cancel each other, resulting in a zero net dipole.
large net dipole
C� �CI
H
/
C == C
'"
H
net dipole
1
31
=
0
(hydrogen bonds shown as wavy bond)
2- 1 7
(a) (CH3hCHCH2CH2CH(CH3h has less branching and boils at a higher temperature than
(CH3hCC(CH3h .
(b) CH3(CH2)sCH2 0 H can form hydrogen bonds and will boil at a much higher temperature than
CH3(CH2) 6CH3 which cannot form hydrogen bonds.
(c) HOCH2(CH2)4CH2 0 H can form hydrogen bonds at both ends and has no branching; it will boil at a
much higher temperature than (CH3hCCH(OH)CH3 .
Cd) (CH3CH2CH2hNH has an N-H bond and can form hydrogen bonds; it will boil at a higher
temperature than (CH3CH2hN which cannot form hydrogen bonds.
(e) The second compound shown (B) has the higher boiling point for two reasons: B has a higher molecular
weight than A ; and B , a primary amine with two N-H bonds, has more opportunity for forming hydrogen
bonds than A, a secondary amine with only one N-H bond.
2- 1 8
(a) CH3CH2 0 CH2CH3 can form hydrogen bonds with water and i s more soluble than
CH3CH2CH2CH2CH3 which cannot form hydrogen bonds with water.
(b) CH3CH2NHCH3 is more water soluble because it can form hydrogen bonds; CH3CH2CH2CH3 cannot
form hydrogen bonds.
(c) CH3CH20 H is more soluble in water. The polar O-H group forms hydrogen bonds with water,
overcoming the resistance of the non-polar CH3CH2 group toward entering the water. In
CH3CH2CH2CH2 0 H, however, the hydrogen bonding from only one OH group cannot carry a four-carbon
chain into the water; this substance is only slightly soluble in water.
(d) B oth compounds form hydrogen bonds with water at the double-bonded oxygen, but only the smaller
molecule (CH3COCH3) dissolves. The cyclic compound has too many non-polar CH2 groups to dissolve.
32
2-19
(a)
H H H H H
I
I
I
I
I
H-C-C-C-C-C-H
I
I
I
I
I
H H H H H
(b)
alkane
(Usually, we use the tenn "alkane"
only when no other groups are present.)
(d)
H
I
H-C-C
I
H-C
H
(g)
'
"
H
=
I
C-C-H
/
I
(e)
C-H
H
C-C
I
I
'H
H
H
H
H H H
I
I
I
I
H,
C
C-C-C-H
H - C � "C � I
I
I
I
I H
C H
H - C .. . C
I
H'
H
C� I
H
H
I H
H
(h)
•
(a)
0
H
H
I
I
"
H-C-C-C-H
I
(d)
I
aldehyde
H
H H
H H
I
I
I
I
H-C-C-O-C-C-H
I
I
I
I
H H
H H
(b)
(e)
ketone
I
I
H
H
H
" C ..
H
I
,
C - C :: C - C - H
H-C�
I
I
I
H-C
C .. H
H
" C � ... H
H
' "
H
C=C-H
I
I
H H
H
,
0
H
H H
H
I
I
I
I
H -C-C-C-C-H
I
I
I
I
H H H H
alcohol
0
H"
H
H
C ..
,
" -O-H
C-C
H-C�
I
H -, C ..
C .- H
�
H
,c.
H
H
H
H
,
I
carboxylic acid
(h)
(g)
H H H
I
I
I
H-C-C-C-C-H
I
I
H H
H-C-C'
I
I
H
H
H
I
(f)
H
H
"
H'
c;
H
"C ,
/
H
H H H
I
I
I
H - C - C :: C - C - C - C - H
I
I
I
I
H H H
H
I
alkyne
H
H H
I
I
I
C ..
H,
C=C-H
C� " C�
I
"
C
C
�
H , "C � ' H
I
H
aromatic hydrocarbon
and alkene
(i)
alkyne, alkene, cycloalkane
H
ether
H
•
cycloalkene
2-20
alkene
(c)
cyc10alkane
cycloalkyne
H'
H
H H
I
I
I
H-C-C=C-C-C-H
I
I
I
I
I
H H H H H
I;I
H
aldehyde
33
H
aromatic hydrocarbon
and cycloalkene
(c)
(f)
0
H
H H
I
I
I
"
H-C-C-C-C-H
I
I
I
H
H H
ketone
H
H -' C �
I
0
H
" C '- H
I
H-C
C-H
H , ..., c .� . H
H
H
ether
o
" -H
C
�.
H
.C - C .
I H
H I
H H
'c '
,H
' C o:. " C � " C - H
I
I
II
C
C
C
H ' " c � " C �_ , H
I
I
H
H
C
H
alcohol
2-2 1
(a)
H
H
0
1
H-C-C-C-N-C-H
1
1
1
1
H
H
H
11
1
1
H
H
(b)
H
H
1
1
1
H
1
H-C-C-N-C-C-H
1
1
H
H
1
H
1
H
H
1
�)
H
H
(e)
H
I
H
I
I
H-C-C-O-C-C-H
H'
I
I
1
I
H
H
H
(f)
H
1
11
1
1
H
H
H
I
0
1
H-C-C-C-O-C-H
amine
amide
H
1
H
C
H
'
1
H
H
ester
H
1
1
1
H
H - C - C - C - C :: N
1
1
1
H
H
H
H
nitri Ie
ether
I;I H
H
(g)
H 'C ' H
I
I
H
I
0
I
U)
(i)
II
,H H
C.
1
N-C- H
H-C
'
I
1
H
H -, C .... ... C .- H
C
H
H-C-C-C-C-C-O-H
I
H
H
'
I
C
H
'
I
I
H
carbox ylic acid
(k)
o
H
11
H
" ... C ..
N - C '- H
H-C
I
I
"
H
H
H
H,
H
1
H -, C .... ... C - H
H
"C
H
H
H
cyclic ester
(I)
H
H
H,
H -' c '
C
ketone and ether
,H
11
1
I
H
0
H
1
.N-C-C-H
H -, C ....
,
H
... C .- H
C
H
H
H
1
H
"
cyclic amide
(m)
"
H
amine
a
H
H -'C '
C
I
H-C
'
H
H
' C '- H
.�'
1
C-H
'
H
H-C-H
1
amide
H
ketone and amine
(0 )
(n)
I
H"
H
H-C
I
1
H -, C .... ... C - H
H
"C
H
H
H
,
C
.H
H
" C - C :: N
1
H -, C ' ' C - H
"C
H
H
H
H
,
cyclic ester
,
nitrile
ketone
2-22 When the identity of a func ti onal group depends on several atoms, all of those atoms should be
circled. For example, an ether is an oxygen between two c arbons, so the oxygen and both c arbons should
be circled. A ketone is a carbony l group between two other carbons , so all those atoms should be circled.
(a)
€2 §9
alkene
H CH3
2
(b)
€ Y
0
ether
34
3
(c)
CH3
(g
Il
C-H
aldehyde
2-22
continued
g
(e)
6 c:3)
(f)
H
II
H C - NH
(d)
CH
�
amine
q u i te a few ways that carbon and hydrogen ato m s c an go together to form alkyl and
a ry l groups . So when you see this s y m bo l , you s h ould know that it represents ONl.. Y
some combination of carbon and h ydrogen atoms-except w h en it includes other
ato m s .
(g)
(i)
(h)
alkene
CH3
Please refer to solution
(
8
ketone
aromatic
(k)
2-23
2-24
carboxylic acid
represent alkyl and ary l groups . As you w i l l s ee in t h e c o urse of your s tudy , there are
(this also looks like an aldeh yde, b u t
an amide h as higher "priority " as y o u
wi l l s e e l ater)
(j)
~
S uggested by student R i c h ard King: R i s the symbol that organic chemists use to
amide
aromatic
II
R C - O- H
CH3
�
R
1-20, page 12 of this Solutions Manual .
The examples here are representative. Your examples may be di fferent and stil l c orrect. What is
important in thi s problem i s to h ave the same functional group.
(a) alkane: hydrocarbon
wi th all single bonds ; can be
acyclic (no ring) or cyclic
(b) alkene : contains a
c arbon-carbon double bond
(d) alcohol : contai ns an
(e) ether: contains an
ox ygen between two carbons
H H H
I
I
I
H - C-C - C - H
I I I
H H H
OH group on a carbon
H H
I
I
H - C - C - O- H
I
I
H H
H
I
H - C - C == C - H
I
I
I
H H H
H
H
I
I
H - C - O-C - H
I
I
H
H
35
(c) alkyne: contains a
c arbon-carbon triple bond
H
I
H - C-C -C - H
I
H
(f) ketone : conatins a carbonyl
group between two carbons
H 0 H
I II I
H-C-C-C-H
I
I
H
H
2-24 continued
II
I
I
H
H
I
I
(j ) ester: contains a carbonyl
group with an O-C on one side
a
I
1/
I
I
H
I
I
I
I
H
(I ) amide : contains a carbonyl
group with a nitrogen on one side
I
a
I
II
H
I
I
H
H-C-C-N-H
H
H
(m) nitrile: contains the carbon-ni trogen triple bond:
2-25
2-26
0
H-C-C-O-H
H - C - N - H or R group
H
II
H
(k) ami n e : contains a nitrogen
bonded to one , two, or three
carbons
H H or R group
H-C-C-O-C-H
H
I
o
H-C-C-C-H
H
(i ) carboxylic acid: contains a
c arbonyl group with an OH
group on one side
(h) aromatic hydrocarbon :
a cyclic hydrocarbon with
alternating double and single
bonds
(g) aldehyde: contains
a carbonyl group with a
hydrogen on one side
H H a
H3 C - C
N
Models show that the tetrahedral geometry of CH2 CI 2 precl udes stereoisomers .
H
(a)
/
"
H
C
I
H - _\ - H
C
C
/
\
(b) Cyclopropane must have 60° bond angles compared with the usual sp 3
bond angle of 109 . 5° in an acyclic molecule.
H
H
(c) Li ke a bent spring, bonds that deviate from their normal angles or positions are highly strained.
Cycl opropane i s reactive because breaking the ring relieves the strain .
2-27
f)
(a)
(b)
H
0_
(c)
?:}'0
+
./ 0 " " "
.....
\ H
H
H
H
/' H
H '::,
C
H
""
C
/
J '\.'
H
""
", N
I
--
�/
H
N
H
Sp 3 , no bond angle because
oxygen is bonded to only one atom
(d)
(J
both sp 3 , all
�
�
==
1 09°
beh i nd the p l ane
of the paper
��
)
\�
J '\.H
C, - H
H
H
3
a l l sp , al l
==
1 09°
angles around sp 3 atoms
angles around sp2 carbon
==
36
==
109°
120°
2-27 continued
(h)
beh i nd the p l ane
plane of the
of the pape
p aper
� (j {)
H
angles around sp 3 atom
angles around sp 2 atoms
�
�
;
i n front o f the
/
a
"'"
I
(i)
...-/
�H
H
,
,
C
H
1 09°
1 20°
2-28
For cl ari ty in these picture s , b o n d s between hydrogen and a n sp 3 atom are n o t l abeled; these bonds are
s-sp 3 overlap.
(b)
I ;;\'..
A\(A/· ·
H-C-C-O-H
1 09°
(e)
(f)
2
( )
g
H
109°
37
7/I-(�Yfp:
�
C- 1'H
t t
sp 3 _ sp 2
H
sp 2 _ sp 3
2
H
1 09°
sp 3 _sp 2
sp 2 -s
� OJ �
�<
f"' ; o,, � �
(i)
I
sp 3 _sp 3
H
___
C
1 09°
I
II
120°
2 2
--- sp -sp
1 20 °
\ H / C � C. / C �
H
/ ,,\ � 2
�
' H\ Sp -S
H
+
p-p
sp 2 _ sp 3
2-29 The second resonance form of formamide is a minor but significant resonance contri butor. It shows
that the nitrogen-carbon bond h as some double bond character, requiring that the nitrogen be sp 2
h ybridized with bond angles approaching 1 20°.
:0:
II
°
°
°
+
/
sp 2
H-C=N-H
I
H
H-C-N-H
I
H
2-30
01
° °
°
(a) The major resonance contributor shows a carbon-carbon double bond, suggesting that both carbons are
sp 2 hybridized with trigonal p l an ar geometry . The CH3 carbon i s sp 3 h ybridized with tetrahedral geometry .
� :R:
H-C-C-C-H
I
I
H
H
:
�
sp 3 / 1
...
..I-----;.
�
?�
Sp 2
H-C-C=C-H
I
H
H
m�or
�nor
(b) The major resonance contri butor shows a carbon-nitrogen double bond, suggesting that all three
carbons and the nitrogen are sp 2 h ybridized wi th tri gonal pl anar geometry .
+
H-N-C=C-C-H
I
I
I
I
H H H H
..
• •
+
.. H - N - C - C = C - H ..
I
I
I
I
H H H H
minor
minor
+
•
H-N=C-C=C-H
I
I
I
I
H H H H
major
2-3 1 In (c) and (d), the un shadowed p orbitals are vertical and parallel. The s hadowed p orbitals are
perpendicula; and hori zontal .
(aJ
Q\2
H3C ,,,
H3C
M
"-
(bJ
. KN V
n
h�
�
(d)
_QOYoQ?o
O[Q[J riJJ
C H3
2-32
(b) The cop l an ar atoms i n the structures to the left and
below are marked w i th asterisks.
(a)
cis
(c)
H
*
HJC
"
QD
M
(d)
,, ,
�C - C ,
"
,,
eH,CH]
*
trans
H
There are sti l l six
copl anar atoms.
2-33
Collinear atoms are marked with asteri sks.
2-34
(a) no cis-trans i someri sm
(b)
(c) no cis-trans i someri sm
0
If/
(d) Theoretical l y , cyclopentene could show c i s-trans i somerism. In reality, the trans form is too unstable
to exist because of the necessity of stretched bonds and deformed bond angles. trans-CycJopentene has
never been detected ·
.
c ls
"trans "--not possible bec ause of ring strain
/J
V
these are cis- trans isomers , but
the designation of cis and trans
to spec i fic structures is not
defined because of four
different groups on the double
bond
(e)
39
2-35
(a) consti tutional i somers-the c arbon skeleton s are different
(b) constitutional i somers-the position of the c h l orine atom has changed
(c) c i s-trans i somers-the first is cis, the second i s trans
(d) constitutional i somers-the c arbon skeleton s are different
(e) c i s-trans i somers-the first is trans, the second is cis
(f) same compound-rotation of the first structure gi ves the second
(g) c i s-trans isomers-the fi rst is cis, the second is trans
(h) consti tutional isomers-the position of the double bond relative to the ketone h as changed (while it
i s true that the first double bond i s cis and the second is trans, in order to have cis-trans i somers, the rest
of the structure must be i dentical)
CO2 i s linear; its bond dipoles c ancel, so i t has no net dipole . S02 i s bent, s o i ts bond dipoles do not
2-36
c ance l .
t: o��"o:
..
O == C == O
net di pole moment
=
..
..
0
..
net dipole moment
2-37 Some magni tudes of dipole moments are di fficult to predict; however, the direction of the dipole
should be straightforw ard in most cases. Actual values of molecular dipole moments are given in
parentheses. (The
bond is usually consi dered non-pol ar. )
C-H
(a)
�
'x/
...... CH3
'--J
N
U1
H / 'CH3 \
/
or
net
net
l arge dipolemoment
l arge dipole moment
(b)
� �
CH3 - C
I
N CJ
�
net
l arge dipole moment ( 3 . 96)
/Br
��
B r IllIII /
C
B
�
(c )
(d)
Br
net dipole moment
=
I
I
net
CH2 CH2
" /
2
CH
moderate dipole moment
net dipole moment
=
1
net
l arge dipole moment (2 .89)
(0 4N"
�b
", � /
2
R1
C
CH3/ 'CH3
0
e lectron pairs on
bromines are not shown
2
0; /0
(g)
CH2
/CI
CH; 'C /
I 2 /C1 1 /
net
CH
""
'CH2 H
moderate dipole moment
electron pairs on
c h l or i n e are not shown
0
40
2-38 Diethyl ether and I -butanol each have one oxygen , so each c an form hydrogen bonds with w ater
(water supplies the H for hydrogen bonding with dieth yl ether) ; their water solubi lities should be similar.
The boiling point of I -butanol is much higher because these molecules can hydrogen bond with each
other, thus requiring more energy to separate one molecule from another. Diethy l ether molecules cannot
hydrogen bond with each other, so it is rel ati vely easy to separate them .
CH3CH 2
-
0
-
CH 2CH3
CH3CH2CH2CH2 - O H
I -butanol
can hydrogen bond with water
can hydrogen bond with i tself
diethy l ether
can hydrogen bond with water
cannot h ydrogen bond with i tself
2-39
C
N - CH]
N-meth y I pyrro Ii di ne
b.p. 8 1 °C
tetrahydropyran
b.p. 8 8 DC
piperi di ne
b . p . 1 06DC
o-
OH
cyc!opentanol
b.p. 14 1 °C
(a) Piperi dine has an N-H bond, so it can hydrogen bond with other molecules of i tself.
N-Methylpyrrolidine h as no N-H, so it cannot hydrogen bond and will require less energy (lower boi ling
point) to separate one molecule from another.
(b) Two effects need to be explained: 1 ) Why does cyclopen tanol h ave a higher boi ling point than
tetrahydropyran ? and 2) Why do the oxygen compounds have a greater difference i n boi ling points than
the analogous nitrogen compounds ?
The answer to the first question i s the same as in (a) : cyclopentanol c an h y drogen bond with its
neighbors while tetrahydropyran c annot .
The answer to the second questi on l i es in the text, Table 2-1, that shows the bond dipole moments for
C-O and H-O are much greater than C-N and H-N; bonds to oxygen are more pol arized, with
greater charge separati on than bonds to ni trogen .
How i s thi s reflected in the data? The boi ling points of tetrahydropyran ( 8 8 DC ) and
N-methy l pyrrolidine (8 1 °C) are close ; tetrahydropyran molecules would h ave a slightly stronger dipole­
dipole attraction , and tetrahydropyran is a little less " branched" than N-methy l pyrrolidine, so it is
reasonable that tetrahydropyran boi l s at a s l i ghtly higher temperature. The l arge difference comes when
comparing the boi ling points of cyclopentanol (141 DC) and piperidine ( 1 06°C). The greater polari ty of
O-H versus N-H is refl ected in a more negati ve oxygen (more electronegati ve than ni trogen) and a
more posi ti ve hydrogen , resulti n g i n a much stronger intermolecular attraction . The conclusion i s that
hydrogen bonding due to O-H i s much stronger than that due to N-H.
2 40
(a) can hydrogen bond with i tself and with water
(b) can hydrogen bond onl y with w ater
(c) can hydrogen bond wi th i tself and with water
(d) can hydrogen bond only with w ater
(e) cannot hydrogen bond
(f) cannot hydrogen bond
-
2-4 1
(g)
(h)
(i)
(j)
(k)
(I)
can
can
can
can
can
can
hydrogen bond only with water
hydrogen bond w i th i tself and wi th water
hydrogen bond only with water
hydrogen bond onl y with water
hydrogen bond only with water
hydrogen bond with itself and with water
Higher-boi ling compounds are li sted.
( a) CH3CH(OH)CH3 can form h ydrogen bonds with other i dentical molecules
(b) CH3CH 2 CH2 C H2 CH3 has a h i gher molecular weight than CH3CH2C H 2CH3
(c) CH3CH2CH 2 CH2 CH3 has less branching than (CH3hCHCH 2 CH3
(d) CH3CH 2 CH 2 CH 2 CH 2 Cl h as a hi gher molecular weight AND dipole-dipole interaction compared with
CH3CH 2 CH 2 C H2CH3
41
2-42
(a)
(b)
ether
ether
alkene
aldehyde
(c)
(d)
ketone
(e)
(f)
ester (cyclic)
aromatic
amide (cyclic)
alkene
(h)
(g)
amme
2-43
.
:0:
:0:
II
CH3
/C'
t
CH3
CH3
"
S .......
/ ••
/
ester
.
:0 :
CH3
2
sp -planar
f----J
..
..
....
..
CH 3
I
S .......
/ ••
/
+
CH 3
tsp3 -tetrahedral
The key to thi s problem is understanding that sulfur has a lone pair of electrons . The second resonance
form shows four pairs of electron s around the sulfur atom, an electroni c configuration requiring sp 3
2
hybri dizati on . S u l fur in DMSO c annot be sp l i ke carbon in acetone, s o we would expect sulfur's
geometry to be pyramidal (the four electron pairs around sulfur require tetrahedral geometry , but the
three atoms around sulfur define i ts shape as pyrami dal).
42
2-44
(a) penicillin G
CQ)
([)
_
II
'1
thioether ("thio" means
sulfur repl aces oxygen)
amide
\\
\
II
C -N
CH 2
__
aromatic
(b) dopamine
aromatic
�
)
©
CH 2
I
H2
(In l ater c h apters , you wi l l learn that the
OH group on a benzene ring is a speci al
functional group called a "phenol ". For
now, it fi ts the broad definition of an
alcohol . )
amIne
CH 2
)
(c) thyroxine
aromatic
�
CH 2
@e
.
amme
c arbox ylic acid
aromatic
(d) testosterone
43
CHAPTER 3-STRUCTURE AND STEREOCHEMISTRY OF ALKANES
3- 1
(a) C nH2n+2 where n
=
25 gives C 2sHs2
Note to the student: The IUP AC system of nomencl ature has a wel l defined set of rules determining how
structures are named. You w i l l fi nd a summary of these rules as Appendix 1 i n thi s Solutions Manual .
3-2 Use hyphens to separate letters from numbers. Longest chains may not a l w ays be written left to right.
(a) 3 -methylpentane (always fi nd the longest chai n ; it may not be written in a straight line)
(b) 2-bromo-3-methylpentane (al w ays find the longest chai n)
(c) 5-ethyl-2-methyl-4-propylheptane ("When there are two longest chains of equal length, use the chain
w i th the greater number of substituents . " )
(d) 4-isopropyl-2-methyldecane
3-3 Thi s Solutions Manual w i l l present l ine formulas where a question asks for an answer including a
structure. If you use condensed structural formulas instead, be sure that you are able to " translate" one
structure type into the other.
(a)
(b)
(c)
(d)
3 -4 Separate numbers from numbers with commas .
(a) 2-methylbutane
(b) 2,2-di methylpropane
(c) 3 -ethyl-2-methylhexane
(d) 2,4-di methylhexane
(e) 3 -ethyl-2,2,4,5-tetramethylhexane
(f) 4-t-butyl-3-methylheptane
3-5 (Hints: systematize your approach to these problems. For the i somers of a six carbon formula, for
example, start with the i somer containing all six carbons in a strai ght chain , then the i somers containing a
fi ve-carbon chain, then a four-carbon chain, etc. Carefully check your answers to AVOID DUPLICATE
STRUCTURES . )
(a)
n-hexane
2,2-di methylbutane
2-methylpentane
3-methylpentane
2,3 -di methylbutane
45
3-5 continued
(b)
�
n-heptane
2-methylhexane
T�
3 , 3 -di methylpentane
2,3-di methylpentane
2 ,2-dimethylpentane
3-methylhexane
��
2,4-di methylpentane
3 -ethylpentane
2,2,3 -trimethylbutane
3-6 For thi s problem, the carbon numbers in the substituents are indicated in italics.
(a)
CH3
(b)
CH3
1
-CH2CHCH3
-CHCH3
1
(d)
1
2
I-methylethyl
common name
(c)
I
=
2
1
-CHCH2CH3
3
2-methylpropyl
common name isobutyl
isopropyl
CH3
2 0/
=
/1C
CH3
30
l,l- di methylethyl
common name = r-butyl or rerr-butyl
3-7
(b)
3 -8 Once the number of c arbons is determined, C nH2n+2 gi ves the formula.
(a) C IOHn
(b) C 1sH32
3-9
(a) (lowest b.p.) hexane
<
(b) (CH3hC-C(CH3h
<
octane
<
decane (highest b.p.) -molecular weight
CH3CH2C(CH3hCH2CH2CH3
(b) octane
<
(lowest m . p . )
octane -branching
(highest b.p.)
(lowest b.p.)
3-10
(a) (lowest m.p . ) hexane
<
<
octane
<
decane (highest m.p.) -molecu l ar weight
CH3CH2C(CH3hC H2CH2CH3
<
(CH3hC-C(CH3h -branching
(hi ghest m . p . )
46
2
3
I-methylpropyl
common name = sec-butyl
CH3
.-! � CH3
(a)
1
3- 1 1
------------------------., --------------------------1 5 .0 kJ/mole
: (3.6 kcal/mole)
%��l
H
H
4.2 kJ/mole + 2
H-H
x
x
%��l
H
5 .4 kJ/mole
H-CH3
dihedral
angle 8
1 20 °
CH3
H
staggered
ec lipsed
1
�
60°
0°
eclipsed
1 5 .0 kJ/mole ( 3 . 6 kcaI/mole)
=
3-1 2
A l l energy values
are per mole .
,,
1 2.
0°
42H
H
e
&:
H
5
54 H
H3 C
ec lipsed 5
H
�2
H
'fJ H
CH3
staggered
H3
H3C
�
H
e
300°
dihedral
360° angle 8
�
,
H3 C 1 2.5
CH3
1 2.5
1
&
H
HC .H4 3
,,
240°
1 80°
J!I4. 2
4. 2
ifH.4 3
ecli psed 5
ecli psed
� CH3
H 'fJ H
3.8
CH3
3.8
H3C
H
'fJ H
H
staggered
CH3
staggered
(Note that the lowest
energy conformers at
60° and 1 80° have at
l east one CH3-CH3
interaction = 3.8 kJ
(0.9 kcal) higher than
ethane.)
Relative energies on the graph above were calculated using these values from the text:
3 .8 kJ/mole (0.9 kcaI/mole) for a CHrCH3 gauche (staggered) interaction ; 4 . 2 kJ/mole (l.0 kcaI/mole)
for a H-H ec l ipsed i nteraction ; 5 .4 kJ/mole ( 1 . 3 kcaI/mole) for a H -CH} eclipsed interaction; 1 2.5
kJ/mole (3 . 0 kcaI/mole) for a CHrCH} ecli psed interaction . These values in kJ/mole are noted on
each structure and are summed to give the energy value on the graph . S l i gh t variation between thi s
graph and t h e one i n the text are due t o rounding.
47
all bonds are staggered
- bold bonds are coming toward the
reader from the plane of the paper
H
dashed bonds are coming toward the
reader from the plane
3- 1 4
(a) 3-sec-butyl-I,I-dimethylcyc l opentane (the 5-membered ring gi ves the base name)
(b) 3-cyclopropyl-I,I-dimethylcyclohexane (the 6-membered ring gi ves the base name)
(c) 4-cyc l obutylnonane (the chain is l onger than the ring)
3- 1 6
o
(a) no cis-trans isomeri sm possible
(b)
and
�H 3
�H
CH3
trans
cis
----.Y
�
(d)
and
CHJ
2
of
the paper
----.Y
�
H
and
2
H
CH 3
cis
trans
3- 1 7 In (a) and (b), numbering of the ring is determined by the first group alphabetically being assigned to
ring c arbon 1 .
CIS
trans
(a) cis-I-methyl-3-propylcyclobutane ("m" comes before "p"-practice that alph abet!)
(b) tral1s-I-t-butyl-3-ethylcyclohexane (the prefi xes t, s, and 11 are ignored in assigning alphabetic al priority)
(c) trans- l ,2-dimethylcyclopropane (either carbon with a CH3 could be c arbon-I; the same name results)
3- 1 8 Combustion of the cis i somer gives off more energy, so cis- 1 ,2-di methylcyclopropane must start at a
higher energy than the trans i somer. The Newman projecti on of the c i s i somer shows the two methyls are
eclipsed with each other; in the trans i somer, the methyls are sti ll eclipsed, but with hydrogens, not each
other-a lower energy.
"'\ more strain ==
CHj ) hi gher energy
CH3
�
cis
HH
48
�
3-19 trans-1 ,2-Dimethylcyc l obutane i s more stable than cis because the two methyls can be farther apart
when trans , as shown in the New m an projections .
""\ more strain =
CH ; ) hi gher energy
CH3
cis
HH
In the 1 ,3 -dimethylcyc l obutanes, however, the cis allows the methyls to be farther from other atoms and
therefore more stable than the trans.
H
A
CHl
CH3 � H
trans
3-20
axi al only
equatori al only
showing both axial
and equatori a l
3 - 2 1 The abbrevi ation for a methyl group, CH3 , is "Me ". Ethyl i s "Et", propyl is "Pr", and butyl is "B u".
�
(a)
Me
�
(b)
H
H
H
Me
H
Me
Me
al l methyls axial
(all H's equatori al )
H
H
all methyls equatorial
(al l H's axial)
Note that axial groups alternate
up and down around the rin g .
3 -22 Carbons 4 a n d 6 are b e h i n d the circ les.
49
3-23 The isopropyl group can rotate so that i ts hydrogen i s near the axial hydrogen s on c arbons 3 and 5,
s i milar to a methyl group's hydrogen, and therefore simi l ar to a methyl group i n energy. The t-butyl group,
however, must point a methyl group toward the hydrogens on carbon s 3 and 5 , giving severe diaxial
interactions, causing the energy of thi s conformer to j ump dramatic al l y .
}JJ
5=H 3
C Hl
i sopropylcyclohexane
t-buty Icyclohexane
3-24 The most stable conformers have substituents equatori al.
�
CH3
H
(b)
CH]
H
3-25
(a) cis
-
equatori al , axial
(b) trans
H
H
EQUAL ENERGY
p:J
H
axi al , equatori al
H
�
�
H3C
�
CH 3
H
equatorial , equatori al
lower energy
axi al , axial
h igher energy
(c) The trans i somer is more stable because BOTH substituents can be in the preferred equatori al positions.
3-26
Positions
cis
trans
1,2
(e,a) or (a,e)
(e,e) or (a,a)
1,3
(e,e) or (a,a)
(e,a) or (a,e)
1 ,4
(e,a) or (a,e)
(e,e) or (a,a)
50
3-27 The more stable confonner p l aces the l arger group equatori al .
(a)
CH2CH3
c;tH7
4�H]
CH3
�CH2CH]
-----
H
(b)
CH3
CH2CH3
H
(c)
(CH 3hCH
H
more stable
H
�
CH3 CH2CH'
-----
more stable H
H]
H 3C ,
C -CH
...
--
H
H
� CH2CH]
H
more stable
H
(d)
more stable H
3-28 The key to detennining c i s and trans around a cyclohexane ring i s to see w hether a substituent group
is "up " or "down" relative to the H at the same carbon. Two "up" groups or two "down" groups will be c i s ;
one "up " and one "down" w i l l be trans. Thi s works independent o f t h e confonnati on t h e molecule i s in !
(a) cis- l ,3-dimethylcyclohexane
(d) cis- 1 ,3-di methylcyc l ohexane
(b) cis- l ,4-dimethylcyclohexane
(e) cis- l ,3-dimethylcyclohexane
(c ) trans-I,2-dimethylcyclohexane
(f) trans-I,4-dimethy l c yclohexane
H
(b)
;------f- CCCH3h
� CH3
H
( c ) B ulky substi tuents like t-butyl adopt equatorial rather than axial posi tions, even i f that means altering the
c onformation of the ring. The tw i st boat conformati on allows both bulky substituents to be "equatorial".
H
H
CH 3
I
C -CH 2 CH 3
I
CH 3
51
3-30 The nomenclatur e of bicyclic alkanes is summarized in Appendix l .
(a) bicycl o[3 . 1 .0]hexane (b) bicyc l o[3 . 3 . 1 ]nonane (c) bicycl o[2 . 2 . 2]octane
3-3 1 Using models is essential for thi s problem.
H
H
rotate
picture
----
(d) bicyclo[3 . 1 . 1 ]heptane
•
H
from text Fi gure 3-29
3-32 Please refer to solution 1 -20, page 1 2 of this Solutions Manual .
3-33
(a) The third structure i s 2-methylpropane (i sobutane). The other four structures are all n-butane.
Remember that a compound's i dentity is determined by how the atoms are connected, not by the position
of the atoms when a structure is drawn on a page.
(b) The two structures at the top-left and bottom-left are both cis-2-butene. The two structures at the top­
center and bottom-center are both I-butene. The unique structure at the upper right i s trans-2-butene .
The unique structure at the lower right i s 2-methylpropene.
(c) The first two structures are both cis- l ,2-dimethylcyclopentane. The next two structures are both
trans-l,2-di methylcyc lopentane. The l ast structure i s different from all the others, cis- l ,3dimethylcyc lopentane.
(d)
Anal ysi s of the structures shows that
some double bonds begin at c arbon-2 and
some at c arbon-3 of the longest chain .
The three structures l abeled A are the
same, with the double bond trans; B is a
geometric i somer (cis) of A. C and Dare
constituti onal i somers of the others.
B
A
D
(e) Naming the structures
shows that three of the
structures are trans- l ,4dimethylcyclohexane, two are
the cis i somer, and one is cis1,3-dimethylcyclohexane.
Al though a structure may be
shown in two different
conformations , it sti l l
represents only one compound.
H3C
('(.�
�
H
:
��h
3
�CH3H
cis- l A-dimethyl
�
H
H3C
trans- l ,4-di methy I
H
M
CH3
cis- I,3-dimethyl
H3C H
trans-I,4-di methyl
52
-PH
H
CH3
CH3
cis- l ,4-dimethy I
®
CH3 H
trans- l ,4-di methyl
3-34 Line formul as are shown.
( a)
�
(b)
(d)
(g)
(Cl
;or:
H"
(j)
"
H
�
(h)
(k)
8
(Cl
I)
(
�
01
0-0
"
(i )
d
6
CH3
or
,
""
"CH2CH3
otxj
-
CI
(I)
B
'H
"
H'
3 - 3 5 There are many possible answers to each of these problems. The ones shown here are examples of
c orrect answers. Your answers may be different AND correct . Check your answers i n your study group.
(a)
or
3-methyl heptane
2-methyl heptane
4-methyl heptane
(b)
4 , 5 -diethyldecane
3 , 5-diethyldecane
(Any combination is
correct except using
posi tion numbers 1 or 2
or 9 or 1 0 . Why won't
these work?)
6
CH3
(c)
Q
CH2 CH3
CH2 CH3
cis- 1 ,2-diethylcycloheptane
cis- 1 ,3 -diethylcycloheptane
(d) only two possible an swers
CH3
O
,,,,CH3
trans-l,2-dimethy l c y c l opentane
[)
CH2CH3
cis- 1 ,4-di ethylcycloheptane
""
CH3
trans- 1 ,3 -dimethylcyclopentane
53
NOT 1 ,5
NOT 1 ,6
NOT 1 ,7
3 - 3 5 continued
Other ri n g sizes are possible,
although they must have 6
or more c arbons to be longer
than the 5 c arbons of the
substituent chain .
(e)
(2,3 -dimethylpentyl )cycloheptane
(2,3-di methylpenty I )cyc looctane
(f)
Any combination where the number of c arbons
in the bridges sums to 8 will work. (Two
carbons are the bridgehead carbons . )
bicyclo[4.4.0]decane
bicyclo[3 . 3 .2] decane
3-36
HO - CH 2CH 2CH 3
HO - CH 2CH 2CH 2CH 2CH 3
HO - C H2CH 2CH 2CH 3
HO - CH 2CH 2CH 2CH 2C H 2CH 3
3-37
(a)
(b)
(c)
(d)
3 -ethyl-2,2,6-trimethylheptane
3-ethyl-2,6,7-trimeth y loctane
3 ,7 -diethyl-2,2,8-trimethyldec ane
2-ethyl-l, l -di methylcyclobutane
(e)
(f)
(g)
(h)
bicyclo[4 . 1 .0] h eptane
cis- l -ethyl-3-propylcyc lopentane
( 1 , I-di ethylpropyl)cyclohexane
cis-l-ethyl-4-isopropylcyclodecane
3 - 3 8 There are ei ghteen i somers of C sH IS. Here are ei ght of them. Yours may be different from the ones
shown . An easy w ay to compare is to n ame yours and see if the n ames match.
�
2-meth y Iheptane
Il-octane
�
2 , 2,4-tri methylpentane
3-39
(a)
�
�
~
2 , 3 ,4-tri methy Ipentane
2 , 3 -dimethylhexane
T
T
*
3 -ethylhexane
3-ethyl-3 -methy l pentane 2,2,3,3-tetrameth y I butane
(b)
4
4
6
correct name : 3 -methylhexane
(longer chain)
(c)
correct n ame : 3 -ethyl-2-methy lhexane
(more branching with this numbering)
�I
(d)
CI
correct name: 2 ,2-dimethylbutane (include
a position number for each substituent,
regardless of redundancies)
correct name: 2-ch l oro-3 -methy l hexane
(begin numbering at end c l osest to substi tuent)
54
3-39 continued
(f)
(e)
c orrect name: sec-butylcyclohexane or
( I-methylpropyl)cyclohexane
(the l onger chain or rin g is the base name)
c orrect n ame: 1 ,2-diethylcyclopentane
(position numbers are the l owest possible)
3 -40
(a) n-Octane has a h i gher boi l i n g point than 2,2,3 -trimethylpentane because linear molecules boil h igher
than branched molecules of the s ame molecular weight (increased van der Waals interaction).
(b) 2-Methylnonane has a h igher boi ling point than n-heptane because it has a signifi c antly higher
molecular wei ght than n-heptane.
(c) n-Nonane boi l s hi gher than 2,2,5 -tri methylhexane for the s ame reason as in (a).
3-4 1 The point of attach ment i s shown by the bold bond at the left of each structure .
-CH 2CHCH 2CH3
I
1°
C H 3 2-methylbutyl
-CH2CH2CH 2CH 2CH 3
1°
ll-pentyl
CH 3
-CH2CH2CHCH3
I
1°
CH3 3-methy lbutyl
(i sopenty l )
-CHCH2CH 3
2° I
CH 2CH 3 l -ethylpropyl
I
CH3
-CHCHCH 3
I
2° CH3
1 ,2-di meth y l propyl
I
-C -CH 2CH3
3° I
CH 3 l , l -dimethylpropyl
(t-pentyl )
CH3
I
-CH 2 - C -CH 3
I
1°
CH3
2 , 2-dimethylpropyJ
(neo-pentyl)
3 -42 In each case, put the l argest groups on adjacent c arbons in anti positions to make the most stable
conformations.
(a) 3 -methylpentane
2
carbon-3 cannot b e seen;
it is behind carbon-2
�
H
C-2 i s the front c arbon with H, H , and CH3
C-3 i s the back c arbon with H , CH 3, and CH2CH 3
55
3-42 continued
(b) 3,3-dimethylhexane
carbon-4 cannot be seen;
it is behind carbon-3
C-3 is the front carbon w i th CH3, CH3, and CH2CH3
C-4 i s the back c arbon w i th H, H, and CH2CH3
3-43
(a)
M:
�
axial
H
ax al
equatorial
CH3
equatori al
CH3
axi al
CH3
equatori al
(b)
I�
more stable
(lower energy)
H equatorial
CH3
axial
less stable
(higher energy)
(c) From Section 3- 1 4 of the text, each gauche interaction raises the energy 3.8 kJ/mole (0.9 kcallmole),
and each axial methyl has two gauche i nteractions, so the energy is:
2 methyls x 2 interactions per methyl x 3.8 kJ/mole per interaction = 1 5 .2 kJ/mole (3.6 kcallmole)
(d) The steric strain from the 1 ,3-di axi al interaction of the methyls must be the difference between the
total energy and the energy due to gauche i nteractions:
23 kJ/mole - 1 5 .2 kJ/mole = 7.8 kJ/mole
(5.4 kcallmole - 3.6 kcal/mole = 1 .8 kcallmole)
3-44 The more stable conformer p l aces the l arger group equatori a l .
(a)
�7C
a
CH(CH3 )2
H'
��
a CH2CH3
-�
H
H
(b)
H3 )2
fE:f
a CH CH3
2
H
H
H
CH(CH3h
e
more stable
�
H
-�
CH
e I li
CH3
more stable
a CH2CH3
(c)
�
CH(CHJh
e
� �
--
CH
e
H
H
more stable
56
CH2CHJ
e
3-44 continued
(d)
H-- �
H2 CH ]
H
a
H
CH 3
more stable
�
a CH3
(e)
H
more stable H
H
�H 2C H 3
3-45 (Usi ng models is essential to this problem.)
In both cis- and trans-decalin, the cyclohexane rings can be i n chair conformati ons. The relati ve energies
w i l l depend on the number of axial substi tuents.
trans
no axial substituents
MORE STABLE
cis
one axial substituent
3-46 chair form of glucose-al l substituents equatori al
I
OH
CH2
HO
HO
OH
OH
H
H
(without ri ng H's shown)
57
CHAPTER 4-THE STUDY OF CHEMIC AL REACTIONS
4- 1
(a)
H H H
I
I
I
I
I
I
H-C -C-C-H
I
C I'
CI - C Q Q
+
IV
---
:�
CI-C·
I
..
:I .
H
H
I
I
(d)
H H H
H H H
(a)
I
I
H-C-C-C·
4-2
H
H
(c)
(b)
+
C I -C I
HU U u
---
C
CI- ·
I
+
H-CI
H
H
I
CI-C-C1
I
+ CI'
H
(b) Free-radical halogenati on substitutes a halogen atom for a hydrogen . Even i f a molecule has only one
type of hydrogen, substituti on of the first of these hydrogens forms a new compound. Any remaining
hydrogens in this product c an compete with the initi al reactant for the avai l able h a logen . Thus, chlorinati on
of methane, CH4, produces a l l possible substitution products: CH3C I , C H2 C I 2 , CHCI3 , and CCI4.
If a molecule has different types of hydrogens, the reaction can generate a mixture of the possible
substitution products.
(c) Production of CCI4 or C H3CI can be controlled by alteri ng the rati o of C H4 to C12. To produce CCI4,
use an excess of CI 2 and let the reaction proceed unti l all C-H bonds h ave been repl aced with C-C1
bonds . Producing CH3CI is more chal lenging because the reaction tends to proceed past the first
substi tution. By using a very l arge excess of CH4 to C1 2 , perhaps 1 00 to 1 or even more, a chlorine atom i s
more likel y t o fi nd a CH4 molecule than it i s t o find a CH3C I , s o o n l y a s m a l l amount of CH4 i s transformed
to CH3C1 by the time the C I 2 runs out, with almost no CH 2 Cl 2 bei ng produced.
4-3
(a) Thi s mechani sm requires that one photon of l i ght be added for each CH3 CI generated, a quantum yield
of 1. The actual quantum yield is several hundred or thousand. The h i g h q uantum yield suggests a chain
reaction, but thi s mechani sm is not a chain; it has no propagation step s .
( b ) Thi s mechan i sm conflicts w i th a t least t w o experimental observations. First, t h e energy o f l i ght
requi red to break a H-CH3 bond i s 4 3 5 kJ/mole ( 1 04 kcallmole, from Table 4-2) ; the energy of l i ght
determined by experi ment to initiate the reaction i s onl y 251 kJ/mole of photons (60 kcallmole of photons),
much less than the energy needed to break thi s H-C bond. Second, as in (a), each CH3Cl produced would
require one photon of l i ght, a quantum yield of 1 , i nstead of the actual number of several hundred or
thousand. As in (a), there i s no provi sion for a chain process, since al l the radicals generated are also
consumed in the mechan i s m .
4-4
(a) The twelve hydrogens of cyclohexane are all on equivalent 2° c arbons. Replacement of any one of the
twelve will lead to the same product, c h lorocyclohexane. n-Hexane, h owever, h as h ydrogens in three
different posi tions: on c arbon- l (equi valent to carbon-6) , carbon-2 (equi valent to c arbon-5), and carbon-3
(equi valent to carbon-4). Monoch lorination of n-hexane w i l l produce a mi xture of a l l three possible
isomers : 1 - , 2-, and 3-chlorohexane.
59
4-4 continued
(b) The best conversion of cyclohexane to chlorocyclohexane would require the ratio of cyclohexanel
chlorine to be a l arge number. If the ratio were small , as the concentration of c h lorocyclohexane increased
during the reaction, chlorine would begin to substitute for a second hydrogen of chlorocyclohexane,
generating unwanted products. The goal is to have chlorine attack a molecule of cyclohex ane before it ever
encounters a molecule of c h lorocyclohexane, so the concentration of cyclohexane should be kept hi gh.
4-5
(a)
- 2.1 kJ/mole - 2100 Ilmole
='
e -IlGOIRT
=
e
�
-(-2100 J/mole)/«8.314 J/K-mole)· (298 K»
e
2100 I 2478
=
e
0.847
Keq
=
2.3
=
degrees kelvin
[ill
=
[CH3S H] [HBr]
(b)
K
=
[CH3B r] [H 2 S]
[CH3B r]
initi al concentrati ons:
final concentrations
Keq
o
=
=
2.3
1 -x
x 0 x
=
1
x
=
(l- x) (l- x )
1 .3 x
2
-
-
4.6 x + 2.3
2
1 - 2x + x 2
o
o
x
x
x
c::::> x 2
> x
(using quadratic equation)
=
2.3 x
2
0.60 ,
=
-
1
4.6 x + 2.3
-
x
=
0 .40
4-6 2 acetone � di acetone
Assume that the initial concentration of acetone is 1 molar, and 5% of the acetone is converted to diacetone.
NOTE THE MOLE RATIO.
[acetone]
[diacetone]
1 M
0 . 95 M
initial concentrations:
final concentrations:
6.Go
[di acetone]
2
[acetone]
=
=
I
-
=
2 .303 RT 10glO Keq
+ 8.9 kJ/mole
o
0.025 M
0.025
(0.95 )
=
-
2
2.303 «8 . 3 1 4 J/K-mole) 0 (298K)) o l og(0 .028)
(+ 2. 1 kc al/mole)
I
4-7 6.5° w i l l be negative since two molecules are combined into one, a loss of freedom of motion .
Since 6.so is negati ve, - T6.So is positi ve; but 6.Go is a l arge negati ve n umber since the reaction goes to
completion . Therefore, /'li{0 must also be a large negative number, necessari ly l arger in absol ute value
than 6.Go. We can explain this by formation of two strong C-H bonds (4 1 0 kJ/mole each ) after breaking
a strong H-H bond (435 kJI mole) and a WEAKER C=C pi bond.
60
4-8
(a) 6.5° i s positive--one molecule became two smaller molecules w i th greater freedom of motion
(b) 6.5° is negative-two smaller molecules combined into one l arger molecule w i th less freedom of
motion
(c) 6.5° cannot be predic ted since the number of molec ules i n reactants and products i s the same
4-9
(a)
h�
Cl - C 1
2
hv
(2) +
ii (3(4)) Cl+ +
(5) + +
(1)
initi ation
-
"�,,
Cl·
propagati on
h
H-CH 2 CH3
� h• CH
Cl -
C l·
-
2 CH3
H-Cl +
-
+
C H2 CH3
•
Cl - CH2 C H3
Cl·
Cl - CI
Cl
Cl·
Cl- C H 2 CH3
• CH2CH3 Cl·
• CH2 CH 3 - CH 3CH 2 CH 2 CH3
CH3CH 2•
•
termination
(b)
(2)
(3)
(6)
step ( 1 )
break Cl-Cl
step
break H-CH 2 CH3
make H-Cl
step
!ili0
!ili0
!ili0
break Cl-Cl
make Cl-CH2CH3
step (3 )
!ili0
!ili0
/�lr
step
!ili0
(2)
+ 242
+ 431
+242339
=
=
-
=
kJ/mole
(+ 58
(+ 98
(+ 5881
kcal/mole)
4 1 0 kJ/mole
kcal/mole)
kJ/mol e (- 1 03 kcal/mole)
-2 1 kJ/mole (-5 kcaUmole)
=
=
kJ/mole
kJ/mole (-
-
=
kcallmole)
kcal/mol e)
-97 kJ/mole (-2 3 kcaUmole)
(c) 6.HO for the reaction is the sum of the !ili0 val ues of the individual propagation steps:
(
- 2 1 kJ/mole
- 5 kcal/mole
4- 1 0 (a)
(2)
i
(1)
initiation
propagati on
(3)
++ -23
- 97 kJ/mole
=
kc allmole
hv
Br - B r
,,��
H-CH 3
B r·
h�
h
+
+
�h
B r- Br
•
=
CH3
break B r-Br
!ili0
=
step (2)
break H-CHl
make H-B r step (2)
!ili0
!ili0
M-I0
=
(3)
(3)
break B r-Br
make B r-CH3
step
(++
!ili0
!ili0
M-I0
+ + -24
2
H-Br +
Br-CH3 +
+ 192 (+ 46
+ 435368 (+ 88
+ 192293 (+
B r·
• CH3
-
step ( 1 )
step
1 1 8 kJ/mole
- 28 kcaUmole)
=
B r·
-
kJ/mole
=
kcallmole)
-
1 04 kcal/mole)
kJ/mole
kcallmole)
kJ/mole ( -
-
46 kcal/mole)
kJ/mole
kJ/mole ( - 70 kcal/mole)
+67 kJ/mole (+16 kcaUmole)
=
=
=
=
-1 0 1 kJ/mole (-24 kcaUmole)
(b) 6.HO for the reaction is the sum of the I'IJJO val ues of the individual propagation steps:
67 kJ/mole
1 6 kcallmole
-
101
- 34 kJ/mole
kJ/mole
kcallmole = - 8 kcaUmole)
=
61
4- 1 1
(a) first order: the exponent of [ (CH3hCCI ] in the rate l aw = 1
(b) zeroth order: [CH30H ] does not appear in the rate law ( i ts exponent is zero)
(c) first order: the sum of the exponents in the rate law = 1 + 0 = 1
4- 1 2
(a) first order: the exponent of [cyclohexene] i n the rate l aw = 1
(b) second order: the exponent of [ B r2 ] in the rate law = 2
(c) third order: the sum of the exponents in the rate law = 1 + 2 = 3
4- 1 3
(a) the reaction rate depends o n neither [ethylene] nor [hydrogen] , s o i t i s zeroth order in both species. The
overall reaction must be zeroth order.
(b) rate = kr
(c) The rate law does not depend on the concentration of the reactants. It must depend, therefore, on the
only other chemical present, the catal yst. Apparently, whatever is happening on the surface of the catalyst
determines the rate, regardless of the concentrations of the two gases. Increasing the surface area of the
catalyst, or simply adding more c atalyst, would accelerate the reaction .
4- 1 4
(a)
t
+ 1 3 kJ/mole
HCI
•
+
CH 3
CH 4
+
CI ·
....................... �-----
reaction coordinate
(b) Ea = + 1 3 kJ/mole ( + 3 kcal/mole)
(c) !ili0 = - 4 kJ/mole ( - 1 kcal/mole)
4- 1 5
Cl 2 +
•
CH 3
_____
(a)
t
./......
..........
.
i
....... .,. +
k ca IImole
- 1 09 kJ/mole
reaction coordinate
(b) reverse: CH 3C I + CI ·
(c) reverse: Ea
4
=
-----
C l2
+
----•
CH 3
+ 1 09 kJ/mole + + 4 kJ/mole = + 1 1 3 kJ/mole
( + 26 kcallmole + + 1 kcallmole = + 27 kcaVmole)
62
4- 1 6
numbers are kJ/mol e
(a)
t
+ 1 92
reaction coordinate -(b) The step leading to the hi ghest energy transition state i s rate-l i mi ti n g . In this mechanism, the first
propagation step is rate- l i mi ting:
CH4
[t-------�; 1
(2) [H-�tt'- -H---- -�; r
.
H
8
+
.
8
/
[H-�-------- ]
B r· +
(c)
HBr +
--
1
( )
+
(3)
Br - - - -- - B r
•
CH3
"8· " mean s parti al radical character on the atom )
( d ) W O for the reaction i s the sum of the WO values o f the individual propagation steps (refer t o the
solution to 4- 1 0 (a) and (b»:
+ 67 kJ/mole + - 1 0 1 kJ/mole = - 34 kJ/mole
( + 1 6 kcallmole + kcallmole = - 8 kcaUmole)
24
4- 1 7
(a)
initiation
2
(2) I H-CH3 - H CH3
- I-CH3
CH3
1
IH-CH3
-I
H-I
II--CIH3
(1)
hv
hn
I- I
r\�n
•
propagation
(3)
(2)
step ( 1 )
break
step
break
make
step (3 )
break
make
I·
-I
+
h� h
+'
I�I
+ I·
•
Mfo
step (2)
step (3 )
WO
WO
MIo
WO
WO
MIo
+ .
=
=
=
=
=
=
=
+ 1 5 1 kJ/mole ( + 36 kcal/mole)
+ 4 35 kJ/mole (+ 1 04 kcal/mole)
- 297 kJ/mole ( - 7 kcallmole)
1
+ 138 kJ/mole ( + 3 3 kcaUmole)
+ 1 5 1 kJ/mole ( + 36 kcallmole)
- 234 kJ/mole (- 56 kc al/mole)
-8 3 kJ/mole ( -20 kcaUmole)
63
4- 1 7 continued
(b) Mia for the reaction is the sum of the Mia values of the i ndividual propagation steps :
+ 1 3 8 kllmole + - 8 3 kllmole =
(+ 33 kcallmole + - 20 kcallmole
55 kJ/rnole
+ 1 3 kcaVrnole)
+
=
(c) Iodination of methane is unfavorable for both kinetic and thermodynamic reasons . Kinetical l y , the rate
of the first propagation step must be very slow because it is very endothermi c ; the acti vation energy must be
at least + 1 3 8 kllmole. Thermodynamically, the overall reaction is endothermic, so an equil ibri um would
favor reactants, not products ; there i s no energy decrease to dri ve the reaction to products .
4-18 Propane has six primary hydrogens and two secondary hydrogen s , a rati o of 3 : 1 . I f primary and
secondary hydrogens were repl aced by chlorine at equal rates , the chloropropane i somers would reflect the
same 3: 1 ratio, that is, 7 5 % l -c h loropropane and 25% 2-chloropropane.
H -- 30
(b)
H
(d)
CH3
t
lo
H
H
H
H
H
CH3- �-CH3
t
H
(c)
I
H
H
--
I
30
CH3- C-CH 2CH3
t CH3 t t
lo
t 20 1 0
t
10
10
10
H
H
H
OR H
(e)
H
H
H
all are 20 H
all are 20 H
30
al l are 20 H
except bri dgehead
H's (labeled 30)
4-20
3 a H abstraction
Cl- +
H
I
H -Cl
CH3-C-CH 3
I
+
CH3-C-CH3
I
CH3
CH3
break 30 H -C(CH3h
make H-Cl
overall 3 a H abstraction
10 H abstraction
Mia
Mia
Mr
=
=
=
+ 3 81 kllmole ( + 9 1 kcallmole)
- 43 1 kllmole ( - 1 03 kcal/mol e)
-50 kJ/rnole ( -1 2 kcaVrnole)
H
H
I
I
CI- + H -CH -C-CH 3
2 I
CH3
break 10 H-CH2CH(CH3h
make H-CI
overall 10 H abstraction
H -CI
+ -CH -C-CH3
2
I
CH3
Mia
!J..Ho
/).H0
=
=
=
+ 4 1 0 kllmole (+ 98 kcallmole)
- 43 1 kllmole ( - 1 03 kcallmole)
-2 1 kJ/mole ( -5 kcaVrnole)
(Note: Mia for abstraction of a 1 0 H from both ethane and propane are + 4 1 0 kllmole (+ 98 kcallmole).
It i s reasonable to use this s ame value for abstracti on of the 10 H in i sobutan e . )
64
4-20 continued
t l
H
C o +
H-C H2-
1° radi= c-21al; kJ/mole (- 5 kcal/mole)
I
� -CH3
CH3
DJ{0
3° radi= c-50al; kJ/mole (- 12 kcaUmole)
reaction coordinate ..
3° radiclealadiisnmore
cal, it ilseadireasonabl
totSihenince1f°erradithatcfalothr.efoactrmiivnatgitohneenergy
g to thnegat
e 3° iradive tchalanis lowerforthfoanrmtheingactthiev1at°ioradin energy
ng to e
4-21Methylbutane can produce four mono-chloro isomers. To calculate the relative amount of each in the
2-product
xtuofre,hydrogen.
multiply tEach
he numbers
ofamount
hydrogensdividwhied cbyh coulthe sumd leadoftalolththate product
tiwimesl provi
the reactde tihveity
forpercentthat tofmiypeeach
rel
a
t
i
v
e
amount
s
in the product mixture.
I'1Ho
Mlo
DJ{0
(6 6.1°H0 rel) xat(irveacte amount
ivity 1.0)
(3 3.0 relxat(irveeactamount
ivity 1.0)
23.6.50 x 100= 3° ) xl (reactivity 5.5) (2 2°H) x (reactl ivity 4.5) 23.3.05 x 100
= :5.5 relative amount = 9.0 relative amount
� � x 100
� :� x 100
2
2
4-(a)22When n-heptane is burned, only 1° and 2° radicals can be formed (from either or bond
clisooct
eavageane).(2These
areimethihgylhpenergy,
unstoawbl)eisradiburned,
cals whi3° radich rapicalsdlcany forbemfotormed
her product
st.herWhen or
,
2
,
4
t
r
ent
a
ne,
bel
f
r
om
ei
The 3°ionraditranslcalastaresetololweress "knocki
in energyng.th"an 1° or 2°, relatively stable, with lowered
reactivbond
ity. Slcloeweravage.combust
isooctane (2,2,4-tnmethylpentane)
Any indicated bond cleavage produce a 3° radical.
2
CH3
CH3
I
I
CICH2 - CHCH2CH3
CH3
CH3
I
I
=
CH3-C-CH2CH3
CH3· CH· CHCH3
I
(l
26%
CH3 - CHCH2CH2Cl
lOH)
=
I
C
C
=
H
1 3%
=
23%
=
38%
C-H
C-C
C-H
C-C
CH3
I
JV\
H
I
JV'
C - � -CH - C -CH 3
C H 3 - �? -+
CH 3
will
I
CH3
65
4-22 continued
CH3
CH3
I
(b)
H 3 C - C- O - H
I
I
H 3 C- C- O-
o R -an alkyl
radi cal
+
CH3
I
R-H
+
CH3
When the alcohol hydrogen i s abstracted from t-butyl alcohol , a rel ati vely stable t-butoxy radical
( (CH3)3C- O 0 ) is produced. Thi s low energy radical i s slower to react than alkyl radicals , moderating
the reaction and producing less " knocking."
4-23
(a) 1 ° H abstraction
F0
CH3CH 2 CH3
+
break 1 ° H - CH2 CH 2 CH3
make H-F
overall 1 ° H abstraction
2° H abstraction
F0
M-[0
M-[0
Mlo
=
=
=
+ 4 1 0 kJ/mole ( + 98 kcaJ/mole)
- 569 kJ/moJe (- 1 36 kcal/mole)
-159 kJ/mole ( -38 kcallmole)
H-F
CH3CH2 CH3
+
+
H-F
break 2° H - CH(CH3)2
make H-F
overal l 2° H abstraction
MiO
Mio
Mlo
=
=
=
+ 397 kJ/mole ( + 95 kcaJ/mole)
569 kJ/mole (- 1 36 kcal/mole)
-
- 172 kJ/mole ( -41 kcallmole)
(b) Fluorination is extremely exothermic and is likely to be i ndi scrimi nate in which hydrogens are
abstracted. (In fact, C- C bonds are also broken during fluorination.)
(c) Free radical fl uorination is extremely exothermi c . In exothermi c reacti ons, the transition states
resemble the starting material s m ore than the products, so while the 1 ° and 2 ° radicals differ by about 1 3
kJ/mole (3 kcal/mole) , the trans i ti on states w i l l differ by only a tiny amount. For fluorination, then , the rate
of abstraction for 1 ° and 2° h ydrogen s wil l be virtuall y identical. Produc t rati os will depend on statistical
factors onl y .
Fl uorination is difficult to contro l , but if propane were monofluorinated, t h e product mixture would
reflect the rati o of the types of h y drogens : six 1 ° H to two 2° H , or 3 : 1 ratio, giving 75% 1 -fluoropropane
and 25% 2-fluoropropane .
H
4-24
I
H
H
I
I
CH 3- C- C- CH 3
I
in i ti ation
propagation
B r Br
I
CH 3- C-C-CH 3
I
I
CH3CH 3
Mechan i sm
Br
I
CH3CH3
nn
Br-Br
hv
--
I
I
2 Br o
���-->�
H
I
CH3CH3
H
CH3- C- C- CH 3 + B r o � HB r +
I
I
CH 3- C- C-CH 3
I
CH 3 C H 3
I
.� nn
I
I
CH 3- C- C- CH 3 + Br-Br
H
�
I
•
I
I
CH 3- C- C- CH 3
I
CH3CH 3
Br
I
CH 3 - C-C- CH 3
I
I
C�C�
C� C�
66
+
Br 0
4-25 (a) Mechanism
initiation
nn
H
Br 0
�
IU
H
2
---
a
propagation
i
i
hv
B r- B r
..
-
r-'�
�H
UH
6=: C(
B,·
+
..
•
\n
B,-B,
+
H
H
A
U
H
-
+ HBr
H
Br
B,·
+
(b) Energy calcul ation uses the value for the al lylic C-H bond from Table 4-2.
.
First
propagatIOn
step
Second
propagation
step
break allylic H-CH[ri ng]
make H-B r
overall allylic H abstraction
WO
WO
M-/0
bre ak B r-Br
make 2° C-B r
overall C-B r formati on
MfD
WO
M-/0
[
=
=
=
=
=
+ 364 kJ/mole (+ 8 7 kcallmole)
- 368 kJ/mole ( - 88 kc allmole)
-4 kJ/mole ( -1 kcaUmole)
+ 1 92 kJ/mole (+ 46 kcallmole)
28 5 kJ/mole ( - 68 kcallmole)
- 93 kJ/mole ( -22 kcaUmole)
-
The first propagation step is rate-limiting.
slmclu;c of Ihc transHi on slateo
H
0-
80
8
OJ
- - - H - - - - - - - - - B,
t
(c) The Hammond postul ate tel ls us that, in an exothermic reaction, the transition state is closer to the
reactants in energy and in structure . Since the first propagation step is exothermic (although not by
much), the transition state is c loser to cyclohexene + bromine radic al . This is indicated in the transi tion
state structure by showing the H closer to the C th an to the Br.
(d) A bromine radical wi l l abstract the hydrogen with the lowest bond dissoc i ation energy at the fastest
rate . The al lylic hydrogen of cyc lohexene is more easi ly abstracted than a hydrogen of cyclohexane
because the radical produced is stabilized by resonance. (Energy values below are per mole . )
6 6
aliphatic: 397 kJ (95 kcal ) - H H
SLOWER
4-26
. .
H H
. .
-
FASTER
3 64 kJ (87 kcal) : allylic
. .
. .
¢x ¢x ~ if
-----
OCH3
B HA radical
I
•
-----
-----
h-
•
:::--..
OCH3
OCH3
67
I
OCH3
4-27
R- O· J
J¢Ct
()
�-0
H3C
I
h
0
R
..
R- O
,H
·0
+
II
H3C
"TT
�
I
l-R
CH3
stabilized
by resonance­
Vitamin E
l ess reactive than R O·
4-28 The triphenylmethyl c ati on is so stable because of the delocalization of the charge. The more
resonance forms a species h as, the more stable it will be.
CH3
�
o
O
+0
------c
c
+O O
�
b Q= b
b
/'
< }-co ---- < }-c0 ---- o-c0 ---- < }-c0
+
+U
b
b
b+
+
�
< }-cQ+ ---- < }-cP ---- < }-+0
c
b
b
b
C
II
(Note : these resonance forms do not include the simple benzene resonance forms as shown below; they
are signifi cant, but repetiti ve, so for simplicity, they are not drawn here . )
'I �
0-+/
C
_
4-29
most stabl e (c)
>
,
Ph
----
Ph
(b)
+
II
C
,
----
(a) least stable
+
(b) CH3-CHCHCH3
I
many more
Ph
+
(c ) CH3 -CCH2CH3
4-30
>
�
0-+/
Ph
(a) CH3 -CHCH2CH2
I
I
CH3
CH3
CH3
3°
2°
1°
most stable (c)
.
(c) CH3 -CCH2CH3
I
>
(b)
>
(a) least stable
.
.
(b) CH3-CHCHCH3
(a) CH3 -CHCH2CH2
I
CH3
CH3
3°
2°
I
CH3
68
1°
4-3 1
O: H
:
II ( I
�
II
-
H 3 C- C- C- C- C H3
I
..
:
O-H
+
•
-- H20 +
•
base
H
II
..
:0:
:0 :
-
: 0:
:0 :
I
II
II
..
:0:
-
:0:
II
I
C /C
C ....
____
/ C ....
", C , ", C
___
H 3 C/ ' C / ' CH 3
H 3 C ..... ' C ..... ' CH 3
H 3 C ..... ' c/ ' CH 3
-
•
•
I
I
H
H
--
I
H
H-B ase +
4-34 Please refer to sol ution 1 -20, page 1 2 of this Soluti ons Manual .
4-35
transiti on states
/�
(a) and (c)
t
t
-
---------
-------- -
l
r
(b) Mio is negati ve (decreases), so the reaction is
exothermic .
(d) The fi rst transi tion state detennines the rate
since it is the hi ghest energy point. The structure
of the fi rst transi ti on state resembles the structure
of the intennedi ate since the energy of the
transition state is c losest to the energy of the
intennediate .
E
Q
products
reacti on c oordi nate
4-36
t
acti vation
energy
--
� tranSitIOn
.. state
reactants
products
reaction coordinate -69
4-37
t
�
energy of h i ghest transition state
detennines rate
t
--- --------------------
reaction coordinate
t-Jl0 is posi tive
-
4-38
The rate law is first order with respect to the c oncentrations of hydrogen ion and of t-butyl alcohol , zeroth
order with respect to the concentration of c h lori de ion, second order overall .
rate = kr [(CH3)3COH 1 [ H + 1
1°
(c)
H
H
H
1°
3°
�
(d)
H
�
CH3
H
1°
�
�--+-- H
......t--'r--
H
(e)
1°
all are 2° H except for the
two types l abeled
�
H
H
all are 2° H except as l abeled
4-40
(a) break H-CH2CH3 and I-I, make I-CH2CH3 and H-I
kllmole: ( + 4 1 0 + + 1 5 1 ) + ( - 222 + - 297) = + 42 kJ/mole
kcallmole: ( + 98 + + 36) + ( - 53 + - 7 1 ) = + 1 0 kcallmole
(b) break CH3CH2-CI and H-I, make CH3CHrI and H-CI
kllmole: ( + 3 3 9 + + 297) + ( - 222 + - 43 1 ) = -1 7 kJ/mole
kcal/mole: ( + 8 1 + + 7 1 ) + ( - 53 + - 1 03) = - 4 kcallmole
70
3°
H
all are 2 ° H except for the
two types l abeled
H
3°
---
4-40 continued
(c) break (CH3hC-OH and H-C l , m ake (CH3hC-Cl and H-OH
kJ/mol e : ( + 381 + + 431) + ( - 331 +
498) = -1 7 kJ/mole
kc al/mole: ( + 9 1 + + 103) + (-79 + -119)= -4kcallmole
-
(d) break CH3CH2-CH3 and H-H, make CH3CHrH and H-CH3
kJ/mole: ( + 356 + + 435) + ( - 410 + -435)= -S4kJ/mole
kcallmole: ( + 85 + + 104) + ( - 9 8 + -104) = - 1 3 kcallmole
(e) break CH3CHrOH and H-B r, make CH3CHrB r and H-OH
kllmole: ( + 381 + + 368) + ( -285 +
498) =
34kJ/mole
kcal/mole: ( + 91 + + 88) + ( - 68 + - 119)= 8 kcallmole
-
-
-
4-41 Numbers are bond dissoci ation energies in kcallmole in the top line and kl/mole in the bottom line.
< >- C
H2
CH2 = CH C H2
>
85
356
(a)
(b)
(CH3hC
87
364
most stable
4-42
>
o
ey
_
ey
ey
_
>
CH3CH2
91
95
98
381
397
410
>
CH3
104
435
least stable
Only o n e product; chlorination would work. B romination o n a 20
carbon would not be predicted to be a h i gh-yielding process .
Cl
C HP
CH3
(CH3)zCH
>
Ch�
+
Hl
Q
+
CH1
Cl
Chlorination would produce four constituti onal isomers and would not be a good method to make only one
of these. Monobromination at the 30 carbon would gi ve a reasonab l e yield.
H
H
H
CI
H
H
(c)
�
CH3 - -� -CH3
I
I
I
I
I
I
CH 3 -C-C-CH 3
-
+
CH3 CH3
CH3 CH3
I
I
I
I
CH 3 -C-C-CH 2 Cl
CH3 CH3
Chlori nation would produce two consti tuti onal i somers and would not be a good method to make only
one of these . Monobromination would be selective for the 30 c arbon and would give an excellent yield.
CH3 CH3
(d)
I
CH 3 -C-C-CH 3
I
CH3 CH3
I
-
I
CH3 C H3
4-43
initiati on
(1)
(2)
propagation
I
I
I
I
Only one product; chlorination would gi ve
a high yield. Mon obromination would be
very difficult si nce all hydrogens are on 1°
c arbons .
CH 3-C-C-CH 2 CI
CH3 CH3
�l
CI
�
� 6)
�
:
H
(3)
Cl
71
O
;C
+ C l·
Termination steps are any
two radicals combining.
4-44
(a)
CH2=CH-CH2
.
. f----l.�
......
.
:0:
II
(c)
CH3 -C-O·
(d)
H
O
I
• •
CH3-C=O
.�
......f----l
.
• •
H
H
H
:0:
I
....
..f----l
.
.�
4-45
(a) Mechanism
nn
initiation
Br- B r
propagation
(e)
�H
UH
hv
H
H
H
�H
U
f----l.�
.......
H
cY
H
2 B r·
---
JH3C yy' CH3
� -1
WId·)
H3C � CH3
�
H � '=Jj.J
_
The 3 ° allylic H is abstracted selecti vely (faster
than any other type In the molecule) , fonrung an
intermedi ate represented by two non-equivalent
resonance forms. Parti al radic al c h aracter on
two different carbons of the allylic radical leads
to two different product s .
Br
_
+ HBr
H3C� CH3
�
j
�
·
W
W
l
f
second propagation step
+
Br·
(b) There are two reasons why the H shown i s the one that is abstracted by bromine radical: the H is 3° and
it is allylic, that i s , neighbori n g a double bond. Both of these factors stabi l i ze the radical that is created by
removing the H atom.
4-46 Where mixtures are possible, only the major product is shown .
(a)
(b )
(c)
0
�
�
rY B r
V
� CH�
U
CH3H
C H3 -C-C-CH3
CH3 CH3
3 ° hydrogen abstracted selecti vely, faster than 2° or 1 °
�
I
I
I
I
only one product possible
-
CH3 Br
CH3-C-C-CH3
CH3 CH3
I
I
I
I
72
3 ° hydrogen abstracted selectively,
faster than 1 °
4-46 continued
(d)
Br
00
-
decalin
(e)
�
(f)�
-
CO
�
CHCH3
I
I
U
3 ° hydrogen abstracted selectively ,
faster than 2° or 1 °
+
Br
�
Br
Q¢-OQ+ cb
Br
CH3C\
4-47
(a) As
is produced, it c an c ompete with
generate CHCI3, etc.
+
+
CH4CH3 CCl2·
CCH2C\
C 3 CCl2
CH2Cl2 Cl·
Cl2
CHCl3
Cl
C 3 Cl2
C 3C
RroRagation steRs
+
•
l H
•
l
l
+
•
+
+
•
•
CHCl2 +
C l
+
•
.....
..
..
...
both 2°-formed
in equal amounts
from resonance-stabi l i zed benzylic radical
Br
h
(h)
3' hydrogen abstracted selectively. faster than 2'
CH4
l +
key to answering this question correctly.
for avai l able Cl ., generatin g
+++ CH3CC ·2C
+ +CHCl2CCC··
+·
HCl
ClCH3
HCl
CH2C\2
HCl
CHCl3
HC
CCl4
H
A l l hydrogens at the starred positions are
equi valent and allylic. The from the lower
ri ght carbon has been removed to make an
intermedi ate w i th two non-equivalent
resonance forms, giving the two products
shown . Drawing the resonance fonns is the
CH2Cl2.
This can
•
•
l
H
l
l
•
+
•
l
C l3
Cl
(b) To maxi mize H l and minimize formation of polychloromethanes, the ratio of methane to chlorine
must be kept high (see problem 4-2).
To guarantee that all hydrogens are replaced with chlorine to produce CCI4, the ratio of chlorine to
methane must be kept high.
73
4-48
(a) n-Pentane c an produce three monochloro isomers. To c alculate the relative amount of each in the
product mixture, multiply the numbers of hydrogens whic h could lead to that product times the reactivity for
that type of hydrogen . Each relative amount divided by the sum of all the amounts will provide the percent
of each in the product mixture.
CI
CI
CI- CH2CH2CH2CH2CH3
( 6 10 H) x (reactivity 1 .0)
6.0 relative amount
=
I
CH3-CHCH2CH2CH3
( 4 20 H) x (reactivity 4 . 5 )
1 8 .0 relative amount
=
total amount
(b)
6.0 x 1 00
3 3 .0
=
1 8%
i�:�
x 1 00
=
=
I
CH3CH2 - C HCH2CH3
( 2 20 H) x (reactivity 4 . 5 )
9.0 relative amount
=
3 3 .0
55%
9 .0
3 3 .0
x 1 00
=
27%
4-49
(a) The second propagation step in the chlorination of methane is highly exothermic (i1Ho = 1 09 kJ/mole
( - 26 kc al/mole». The transition state resembles the reactants, that is, the CI-CI bond will be slightly
stretched and the CI-CH3 bond will j ust be starting to form.
8·
8·
CI - - - - - - C I - - - - - - - - - - - - - - - CH 3
weaker
stronger
-
(b) The second propagation step in the bromination of methane is highly exothermic (i1Ho = - 1 0 1
kJ/mo\e ( - 24 kcal/mole». The transition state resembles the reactants, that is, the B r-Br bond will be
slightl y stretched and the B r-CH3 bond will j ust be starting to form.
Br - - - - - - Br stronger
- - - - - - - - - •..
weaker
- CH3
74
4-50
1 1)
1<
Two mech anisms are possib l e depending on whether HO' reacts with c h l orine or with cyc!opentane.
Mech an ism
initi ation
(2)
HO-OH
HO' + Cl -Cl
f)
Cl· +
(4)
propagation
f(2))
Mechanism 2
initi ation
propagation
C l -Cl
:>0
'0
H
C l -Cl +
210
30
(24
Cl
2
'0
+
>O
H
----
H-OH
H
---
CI
+
>O
50
2
+
'O
+ C l·
H
H-Cl
----
Cl'
+
HO'
H
:>0
Cl' +
----
---
:>0
'0
Cl'
+
H
H-Cl
----
H
+
HO' +
f)
HO'
HO-Cl
----
HO-OH
(4)
2
---
D
2,
(2()2)
1
The energies of the propagation steps determine which mech anism is fol l owed . The bond dissociation
energy of HO-Cl is about
kJ/mole (about
kcallmole), making initiation step
in mechanism
is exothennic by
endothennic by about
kJ/mo l e (about 8 kcallmole) . In mechanism i n i ti ation step
about
kJ/mole
kcallmol e ) ; mechanism is preferred.
101
4-51
��
Cl -C-H
I
4-52
H
1400
+
:O-H
U·
---
H2O
+
�l_
C
Cl -C:
I
c=�>
1;
0
1400
fj, G
Assuming fj, H is about the same at
=
I
. .
+ :Cl :
H
a c arbene
K, the equi l i brium constant is
=
'c:
H
This critical equation is the key to this problem: 11 G
At
Cl
---
1400
-137 kJ/mole
K
=
therefore:
=
-137,1400000
11 H - T 11 S
>
K as i t is at calorimeter temperature:
=
JIK-mole
-98 J/K-mole (-23 callK-mole)
This is a l arge decrease in entropy, consistent with two molec ules combi ning into one.
75
4-53
Assume that chlorine atoms (radic als) are sti l l generated in the initi ati on reaction . Focus on the propagation
steps. B ond dissoci ation energies are gi ven below the bonds, in kJ/mol e (kcal/mole).
Cl o
+ H-CH3
435 ( 1 04)
CI - C1
242 ( 5 8 )
+
-
o CH3
H-CI + 0 CH3
43 1 ( 1 03)
C I - CH3 + Cl o
3 5 1 (84)
-
+ 4 kJ/mole ( + 1 kcal/mole)
!1H
=
!1H
= -
1 09 kJ/mole ( - 26 kcal/mole)
What happens when the different radical species react with iodi ne?
Cl o
+ I-I
1 5 1 (36)
+
I-I
1 5 1 (36)
o CH3
-
I - CI
2 1 1 (50)
01
+
1 - CH3 + I 0
234 (56)
-
!1H
=
- 60 kJ/mole ( - 14 kcal/mole)
!1H
=
- 8 3 kJ/mole ( - 20 kcal/mole)
Compare the second reaction in each pair: methyl radical reacting with c h lorine is more exothermic
than methyl radical reacting with i odine ; this does not explain how iodine prevents the chlorination
reaction . Compare the first reaction in eac h pair: chlorine atom reacti n g with i odine is very exothermic
whereas c h l orine atom reacting with methane is slightly endothermic . Here is the key: chlorine atoms
w i l l be scavenged by i odine before they h ave a chance to react with methane. Without chlorine atoms,
the reaction comes to a dead stop.
4-54 (a)
initiation
fl
nn
hv
B r- B r
f' � "
(2)
Br 0 +
(3)
�
propagation
H
2 Br 0
-
H-SnB u3
�
B'
H-Br
-----
O'
o SnBu3
H
+ 1 . snBU j
O·
(j-
-
H
(4)
+
+
B r - SnBu3
H
--.. ":":\
+
H-SnB u 3
-
n
H
+
o
SnBu3
(b) A l l energies are in kJ/mole. The abbreviation "cy" stands for the cyclohexane ring .
Step 2 : break H-S n , make H-Br: +3 1 0 + - 368
=
- 5 8 kJ/mol e
Step 3 : break cy-Br, make B r-Sn: +28 5 + - 5 5 2
=
- 2 6 7 kJ/mol e WOW !
S tep 4 : break H-S n , make c y-H: + 3 1 0 + - 397
=
- 87 kJ/mole
The sum of the two propagation steps are : - 267 + - 87
76
=
- 354 kJ/mole -a h ugely exothermic reaction .
4-55
Mechanism 1 :
(a) CI
°
03 ---
+
(b) 2 CIO
°
---
CIO
+
C I - O - O-Cl
hv
--- 02
(c) CI - O - O - Cl
°
2 CI -
+
The biggest problem i n Mechanism 1 lies in step (b).
The concentration of CI atoms is very smal l, so at any
given time, the c oncentrati on of C I O wi l l be very smal l .
The probabi lity o f two C I O radicals finding each other to
form CIOOCI i s virtual l y zero. Even though thi s
mechani sm shows a catalytic cycle with Clo (starting the
mechanism and being regenerated at the end), the middle
step makes it highly unl i kely.
Mechanism 2 :
�
(d) 03
(e) CI
03
+
°
(f) CIO
02
°
°
+
°
+
CIO
---
---
02
°
+
+
02
Cl o
S tep (d) i s the " light" reaction that occurs naturally in
dayl i ght. At night, the reacti on reverses and regenerates
ozone.
Step (f) i s the crucial step . A l ow concentration of CIO
will find a rel ati vely high concentration of ° atoms
because the " l i ght" reaction i s producing ° atoms in
relative abundance. Clo i s regenerated and begins
propagation step (e), continuing the c atalytic cycle.
Mechanism 2 i s beli eved to be the dominant mechani sm in ozone depletion . Mechanism 1 can be
discounted because of the l ow probability of step (b) occurring, because two species i n very low, catalytic
concentrati on are required to fin d each other in order for the step to occ ur.
4-56
(a)
[
H_
�
S.
I_
_ __ - H -
_ _ _ _ _
�:
---
r
In each c ase, the bond
from carbon to H (D) i s
breaking and the bond
from H (D) to CI is
forming.
\ C2Hs-C I + DCI ) + \C2H4DCI + H C I ;
�
7 o
1
9 %
D repl acement: 7 % 7 1 D = 7 (reacti vity factor)
H repl acement : 9 3 % 7 5 H = 1 8 .6 (re activity factor)
relative reacti vity of H : D abstraction = 1 8 .6 7 7 = 2 . 7
Each hydrogen i s abstracted 2 . 7 times faster than deuterium.
(c) In both reactions of chlorine wi th ei ther meth ane or ethane, the first propagation step i s rate-limiting.
The reaction of chlorine atom with methane is endothennic by 4 kJ/mole ( 1 kcal/mole), while for ethane
thi s step is exothennic by 2 1 kJ/mole (5 kcallmole). By the Hammond Postulate, differences in activation
energy are most pronounced in endothennic reactions where the transition states most resemble the
products. Therefore, a c hange in the methane molecule causes a greater c h ange in its transition state energy
than the same change in the ethane molecule causes in its transition state energy . Deuteri um wi l l be
abstracted more slowly in both meth ane and ethane, but the rate effect wi l l be more pronounced in methane
than in ethane .
77
C H A PTER 5-STEREOCHEMISTRY
Note to the student: S tereochemi stry is the study of molecular structure and reactions i n three dimension s .
Mol ecu l ar models wi l l b e especially hel pful in t h i s chapter.
The best test of whether a h o u se h o l d object is chiral is whether it would be used equally wel l by a left­
or ri ght-handed person . The c h i ral obj ects are the corkscrew, the writing des k , the screw-c ap bottle (only
for refi l ling, however; i n use, it would not be chiral ) , the rifle and the knotted rope; the corkscrew, the
bottle top, and the rope each have a twi st in one direction, and the ri fle and desk are c learly made for ri ght­
handed users. A l l the other objects are achiral and would feel equi valent to ri ght- or left-handed users .
5-1
5-2
(a) cis
achiralidentical mirror i mages
(b) trans
H3
�
H
mirror
chiral­
enantiomers
CH3
mirror
(c) cis first, then trans
achiralidentical mirror i mages
mirror
achiralidentical mirror i mages
mIrror
CH3
chiralenan ti omers
I
(d)
" ,C
H " � '-... CH2CH3
Br
mIrror
79
5-2 continued
(e)
chiral­
enantiomers
a
a
mirror
(f)
\J(j
0"
'"
°
'"
chiral­
enantiomers
mirror
5 - 3 Asymmetric c arbon atoms are starred.
Br
I
(a)
(b)
OH
I
no asymmetric carbons­
s ame structure
OH
I
no asymmetric c arbons­
same structure
(c)
enanti omers
(d)
(e)
OH
I
CI
OH
I
H
6
no asymmetri c c arbons­
same structure
no asymmetri c c arbon­
same structure
80
5-3 continued
COOH
(f)
*
/
*
"' H
~
*
CI
~
*
H
CI
CI
(i )
~
*
same structure
*
CI
CI
�
*
l
*
Q
" '"
*
H
H
"
/
C == C
"
H
CH3
enantiomers
H
CI
CI
jL7i �
CI
U)
enantiomers
*
" ,C
H " � " CH3
H2 N
NH2
CI
(h)
I
JC ,"
H3C
(g)
COOH
Q
H
'
,..
H3C
/
C == C
no asymmetric c arbonssame structure
CI
/
"
H
enantiomers
H
(k)
CH3
enantiomers
H3C
5-4 You may have chosen to in erchange two groups different from the ones shown here . The type of
isomer produced w i l l sti l l be the same as listed here .
Interchanging an y t w o groups around a c hirality center wi l l create a n enantiomer of t h e first structure.
Interc h anging the B r and the H
creates an enantiomer of the
structure in Fi gure 5 - 5 .
Interchanging the ethyl and
the isopropyl creates an
en anti orner of the structure
i n Figure 5 - 5 .
O n a double bond, interc hanging the two groups o n O NE o f the stereocenters w i l l create the other
geometric (cis-trans) i somer. However, interc h anging the two groups on B OTH of the stereocenters w i l l
gi v e the origi nal structure.
H " ", CH3
C
original
interchange H and
structure
II
CH3 on top stereocenter
is cis
to produce trans
'" C .......
H
CH 3
81
5-5
(a)
(b)
H
""
" ,
H
(f)
CHO
1
CH2 Cl
1
(d)
H'
H" " C,
J
CH3
Cl
H
Br
Br
chiral-no
plane of
symmetry
chiral-no
plane of
symmetry
plane o f
symmetry
plane of
symmetry
(e)
m
Br
(c)
H " "JC ' CH OH
2
HO
COOH
1
(h)
p l ane of
symmetry
I
H" I C,
J
H 2N
chiral-no
pl ane of
symmetry
CH3
chiral-no
pl ane of
s ymmetry
thi s view is from
the right side of the
structure as drawn
in the text
plane of
symmetry
5-6 AL WAYS place the 4th priority group away from you. Then determine if the sequence 1 �2�3 i s
c l ockwise (R) o r counter-clockwi se (S). (There i s a Problem-S olving Hint near the e n d of section 5-3 in the
text that descri bes what to do when the 4th priority group is c l osest to you . )
rotate CH3 u p
(a)
(b)
R
only one asymmetric carbon
3
CH 3
1
*1
C
S
2
(c)
Br � :: � CH 2 C
H
4
(d)
S
(g)
SU2
- *:
H
: *
Cl
....-
�
1
3h
1
*C 3
C == C ..... -: � CH 3
/
I
H
H H
H
"
R
4
(f)
S
-
-
Cl
Cl
(i) S
(h)
2
R
3
H
82
4
O�
H
C"
�
��
I
( H3COhHC
3 /- CH2
'/
C -. H
CH(CH3h
" '"
4
I,
see next page for an
explanation of part (i)
5 -6 continued
� - CH3
Part (i) deserves some explan ation . The di fference between groups 1 and 2 h i n ge on what is on the "extra"
oxygen .
CH(OCH3h
I
�
O�
H - C - O - CH3
I
t
C
H
,/
�
higher priority
*
s
OH
OH
1
./ C " ,
CH3CH 2 CH 2..... " H
CH3
(f)
./JC " ,
CH 3 ./ �" H
NH 2
(g)
(h)
1
R
COOH
1*
'Y C
H"
NH2'"
R
" CH3
I;I�
�
S
�
I;I
R *� * S
C;: l �
* � S
� C;: l
R �*
R *
S
*
" ,C "
H "�
CH2 CH 2 CH3
H3C
COOH
S
Cl
Cl
H
Cl
(j )
II
H
,
/
R
0
/(
c == c
"-
H-C
I
_
lower priority
5 -7 There are no asymmetric c arbons in 5-3 (a), (b) , (d), (e) , or (i).
(c)
o-c
�
I
0/
Cl
Cl
Cl
Q�
'" "
H
CH3
H
CH3
CH3
*
H
S
/
H
C == C "
C 3
H
(k)
R
83
R
I m ag in a ry
.
5-8
2.0 g 1 1 0 m L
=
0 . 20 g/mL ; 1 00 mm
+ 1 .74°
(0.20) ( 1 )
5-9
0 . 5 g I 10 mL
=
=
=
1 dm
+ 8 . 7 ° for (+ )-gl yceraldehyde
0 .05 g/mL ; 20 cm
- 5 .0°
(0.05) (2)
=
-
=
2 dm
50° for (-)-epinephrine
5- 1 0
Measure using a solution of about one-fourth the concentration of the first. The value w i l l be either + 45° or
45°, which gives the s i gn of the rotation.
-
5-1 1
Whether a sample i s dextrorotatory (abbrevi ated " ( +)" ) or levorotatory (abbrevi ated " ( -)") i s determined
experi mental ly by a pol arimeter. Except for the molecule glyceraldehyde , there is no direct, universal
correlati on between direction of optical rotation (( +) and (-)) and designation of confi guration (R and S). In
other words, one dextrorotatory compound might have R configuration while a differen t dextrorotatory
compound might h ave S configuration.
(a) Yes , both of these are determined experimental l y : the (+) or (-) by the polari meter and the smell by the
nose.
(b) No, R or S cannot be determined by either the pol arimeter or the nose.
(c) The drawings show that (+)-carvone from caraway has the S confi g uration and (-)-carvone from
spearmint has the R confi guration.
o
o
(For fun , ask your instructor if you can
smell the two enantiomers of carvone.
S ome people are unable, presumably
for genetic reasons , to di stinguish the
fragrance of the two enantiomers . )
3
(+)-carvone (caraway seed)
(-)-carvone (spearmint)
CH3
CH3
I
C"
/ "
"" H
CH3CH2
OH
+
I
./ C " ,
� " OH
CH3CH2 ./
H
(R)-2-butanol
(S)- 2-butanol
one-third of mi xture
two-thi rds of mi xture
Chapter 6 will explain how these mixtures come about. For thi s problem, the S enantiomer accounts for
66 . 7 % of the 2-butanol i n the mi xture and the rest, 3 3 . 3 % , is the R en anti orner. Therefore, the excess of one
enantiomer over the racemic mixture must be 3 3 . 3 % of the S, the enantiomeric exces s . (All of the R i s
"canceled" b y an equal amount of the S, algebraic ally a s wel l a s in optical rotation . )
(R)-2-bromobutane
The optical rotation o f pure (S)-2-butanol i s + 1 3 . 5 ° . The optical rotation o f thi s mi xture i s :
3 3 . 3 % x ( + l 3 . 5 °) = + 4 . 5 °
84
5- 1 3 The rotation of pure (+)-2-butanol i s + 1 3 . 5 °
observed rotation
-----rotation of pure enantiomer
+ 0.45 °
x 1 00%
+ 1 3 .5 °
=
To calculate percentages of (+) and (-) :
-
(+) + ( )
(+)
(-)
=
=
( -)
3 .3 %
(+ )
(+)
(+)
(-)
5-14
(a)
=
=
=
=
=
3 . 3 % optical purity
3 . 3 % e.e. = excess o f (+) over (-)
(two equations in two unknowns)
1 00%
2
.
=
1 03 . 3 %
- (+)
( 1 00% - (+))
1 00 %
=
3 .3%
5 1 .6% (rounded)
48 .4%
Drawing Newman projections is the clearest way to determine symmetry of conformations.
,,-h H
Br � CI
H
H
chiral­
optically acti ve
}
''
Br
',C I
,
(b)
H
Plane includes
Br and CI
(c)
I
chiral­
optically acti ve
I
H
I
I
I
H
I
B r - C - C - Cl
(d)
H
� k�
C 2
H
B,
l(:y
�: 20
H
H
Br
plane of symmetry-not optically
acti ve despite the presence of two
asymmetric c �rbons
,
,
Br
*
C I CH2 - C - CI
H H
no asymmetric carbons
(e)
,,-h H
B r � CI
CI
H
plane of symmetry containing
B r-C-C-CI ; not optically active
H
H
H
�CH2 ,,-h H
H �CH20 B r
H
H
H
no plane of symmetry­
optically active
(other chair form is
equi valent-no pl ane of
symmetry)
Br
(f)
�
Br
1
H
�
- - "'
H
4
Br
plane of symmetry through
C - l and C-4-not
optically acti ve
no asymmetric c arbons
Part (2) Predictions of optic al acti v i ty based on asymmetri c centers give the same answers as predictions
based on the most symmetric conformation .
85
5- 1 5
(a)
H
'-
"
CI
/
" H
(b)
\,\
C == C == C '
'CI
no asymmetric carbons , but the
molecule i s chiral (an allene ) ; the
drawing below i s a three­
dimensional picture of the allene
in (a) showing there is no plane
of symmetry because the
substituents of an al lene are in
different planes
no asymmetric carbons , but
the molecule i s chiral (an
allene)
(d)
H
"-
CI
(f)
(e)
/
C == C
n o asymmetric carbons ; thi s
allene has a plane of symmetry
between the two methyls (the
plane of the paper), including all
the other atoms because the two
pi bonds of an allene are
perpendicular, the Cl i s i n the
plane of the paper and the pl ane
of symmetry goes through it; not
a chiral molecule
/
H
/
C == C
"
/
H
"-
H
H
planar molecule-no
asymmetric c arbons; not a
chiral molecule
(g)
o
0
H
" '-j
�JB r
�/
p l ane o f symmetry bi sectin g
t h e molecule; no asymmetric
c arbons ; not a chiral molecule
Br
no asymmetric carbons , but
the molecule is chiral due to
restricted rotation; the
drawing below is a three­
dimensional picture showing
that the rings are
perpendicular (hydrogens are
not shown)
(h)
two asymmetric c arbon s ;
a chiral compound
86
H H
no asymmetric carbons , and
the groups are not l arge
enough to restrict rotation ; not
a chiral compound
5- 1 6
(a)
H
+
OH
HO
+
H
CH3
+
H
Br
CH2CH3
s ame
(c)
+
HO
+
CH3
CH3
H
CH2CH3
(R)-2-butanol
H
CH3
+
enantiomer
enantiomer
+
+
+
CH3
1I
H
HO
CH2CH3
H
CH2CH3
CH2CH3
same
e n anti omer
Br
CH2CH3
CH3
OH
H
COOH
CH2CH3
CH3
+
same
enantiomer
enantiomer
Br
COOH HO
OH
CH3
CH3
+
Rules for Fischer
projections:
CH3
H
COOH
COOH
H
OH
1 . Interchanging any
two groups an odd
number of times (once,
three times, etc . )
makes a n enantiomer.
Interchanging any two
groups an even
number of times (e.g.
twice) returns to the
original stereoi somer.
2. Rotating the
structure by 90 0 makes
the enantiomer.
Rotating by 1 800
returns to the original
stereoisomer.
(The second rule is an
application of the first.
Prove this to yourself.)
CH3
same
5- 1 7
(a)
HO
+
CH20H
H
CH3
5- 1 8
1 80 0 rotation of the right structure does not give left
structure ; n o plane of symmetry: chiral�nantiomers
(a)
CH20H
- - - - -Br
+
H f' on the left ; also has plane of symmetry: same structure
1 80 0 rotation of the ri ght structure gi ves same structure as
CH20H
(c )
Br +��
,
Br
plane of symmetcy same structure
(:H 3
,
87
mirror
5- 1 8 continued
CHO
CHO
H
(d)
H
+
+
OH
HO
OH
HO
H
H
Ho
H
H
$
CH20H
i:
i�
H n 1 80 0 rotation of the right structure gi ves same structure as
HO
- - - - - - - - - - - - - - - - - - - on the left; also has p l ane of s y mmetry : same structure
H
HO
OH
CH20H
CH20H
(f)
1 80 0 rotation of the right structure does not give left
structure; no plane of symmetry : chiral--enantiomers
H
CH20H
CH20H
(e)
+
+
H
HO
OH
+�:
+
1 80 0 rotation of the right structure does not gi ve left
structure; no plane of symmetry : chiral--enantiomers
H
CH20H
CH20H
5 - 1 9 If the Fischer projection is drawn correctly, the most oxidi zed c arbon w i l l be at the top; thi s is the carbon
with the greatest number of bonds to oxygen . Then the numbering goes from the top down.
(d) 2R, 3R
(e) 2 S, 3R (numbering down)
(a) R
(b) no chiral center
(c) no chiral center
(f) 2 R, 3R
(g) R
(h) S
(i) S
5 -20
(a) enantiomers--confi gurations at both asymmetri c carbons inverted
(b) diastereomers--confi guration at only one asymmetric c arbon inverted
(c) diastereomers--con fi guration at only one asymmetric c arbon inverted (the left c arbon)
(d) constitutional i somers-C=C shifted position
(e) enantiomers--c h iral , mirror i mages
(f) diastereomers--configuration at only one asymmetri c carbon inverted (the top one)
(g) enanti omers--confi guration at all asymmetric carbons inverted
(h) same compound-superimposable mirror images (hard question ! use a mode l )
(i) diastereomers--c onfiguration a t o n l y one chirali ty center (the nitrogen) inverted
5 -2 1
(a)
' --
'
:;
u
l �:
CH 3
"
CI
'
U
plane of
s y mmetry
CH3
meso structure
not optical l y active
, l -f
,
H
:
B
*
�
r
,
'
,
Cl
*
H3C
CH3
plane of
symmetry
88
5-2 1 continued
CH3
Br
(b)
Cl
+
+
Cl
Cl
Br
B,
CH3
+
Br
Cl B
lrCl
C�
-
+:
*
C
�
H3C
CH3
-
f I Br
H
H
B
�
*
*
H3C
CH3
=*
CH3
50: 5 0
racemic mixture-not opti c al l y acti ve, although each enanti omer by itself would be optically acti ve
:
H
(c)
H
(
� planes of
=t=
tH3
,
H
i HH
1 i f
'---LJ/
/ : \
H3C
symmetry �
H
CC H3
:
not chiral
not opti c al l y acti ve
(d)
OH
(a)
H
+
H
CH3
A
(f)
�/
H
HOH2 C
*
(g)
+
Cl
Cl
COOH
CH3
+
+
B
C
H
H
H
Cl
CH3
enantiomers : A and B ; C and D
diastereomers : A and C ; A and D ; B and C; B a nd D
89
OH
H
*
CH3 1 CH3
not optically active­
superimposable on its
mirror image­
technical l y , thi s i s
a meso structure
(may require models ! )
COOH
HO
CH 2 0H
*
CH3
OH
'\
plane of
s ymmetry
*
HO
COOH
H?
H
�
H
CH3
optical l y acti ve
5-22
*
B
H �
optically active
H0 -T- H
CH 2 0H
Br
*
CH 2 0H
H � OH
H � OH HO
H � OH
HO
CH3
CHO
(e)
t
CH 2 0 H
H
mesonot optically active
COOH
HO
H
+
+
CH3
D
H
Cl
5-22 continued
(b)
-�
-$�:-- =;��rY
H
HO
+
+
HO
OH
H
H
enantiomers: F and G
diastereomers : E and F; E and G
+
+
H
OH
COOH
COOH
COOH
E
G
F
COOH
COOH
COOH
H ---If-- OH
HO -I-- H
HO --ll-- H
OH
HO -I-- H
H ---If-- OH
HO ---lf-- H
H --I- Br
B r --lf-- H
H --II-- Br
B r --ll-- H
COOH
H
COOH
COOH
COOH
I
J
K
COOH
COOH
COOH
COOH
COOH
(c)
COOH
COOH
COOH
H
HO -f-- H
H --If-- OH
HO
H
H
H --I- Br
OH
H ---lf-- OH
H --I- OH
HO --ll-- H
Br --lf-- H
HO r--I-- H
H --II-- OH
H --I- Br
Br --lf-- H
COOH
COOH
COOH
M
N
L
enantiomers: H and I ; J and K ; L and M; N and 0
di astereomers : any pair which is not enantiomeric
COOH
o
cm
(d)
p
Q
enantiomers : P and Q
diastereomers : P and T; P and U;
Q and T; Q and U; T and U
rti
H
'
H3
H
plane of
symmetry
MESO
plane of
symmetry
MESO
T
U
90
5-23 Any diastereomeric pair could be separated by a physical process l i ke disti l l ation or crystal lization .
Di astereomers are found in p arts (a) , (b) , and (d) . The structures in (c) are enantiomers ; they could not be
separated by normal physical means .
5 -24
0
H
HO
(
o
+CH3��
r enantiomers
'
�
H
H �' 0
CH3
HO H
H OH
COOH
+
OH
H
COOH
(S)-2-butyl (R,R)-tartrate
mirror
(R)-2-butyl (S,S)-tartrate
CH2CH3
H+O
CH3
H--If--- O H
HO--l�H
COOH
di astereomers
\...
�
di astereomers
o
o
HO--lf--- H
H 0H
COOH
----i...--
(R)-2-butyl (R,R)-tartrate
mirror
(S)-2-butyl (S,S)-tartrate
5 -2 5 Please refer t o solution 1 -20, page 1 2 of t h i s Solutions Manual .
5-26
H
CH20H
I*
H+OH
,C",-"
"
C / CH3
CH3
HO
(a)
(b)
l
''
......
B
j
H :,
R
chiral
(e)
S
H
- - - ... - - -
R
- - - - - _ .. .
* Br
/C H 2 B r
meso ; achiral
R
c h i ral
(f)
plane of
symmetry
iB:,H H/i OH
sH
(g)
s
/*
B
S
CH2
Br
R
chiral
91
B'
*
CH3
chiral
(h) S
CH3
H'" * OH
s�
H
R
*
OH
OH
/CH2CH3
*
chiral
5 -26 continued
(i) Br
"
(
"
C=C=C'
�
(j )
Br
CI
C
chiral molecule, but
no chiral centers
(Y
*
I
(k)
Br
"- s
o-
Bf
achiral
chiral
plane of symmetry
(m)
(I)
pl ane of symmetry
meso; achiral
C n)
chiral
(0)
R
s
R
(NH2 i s group 1 ,
CH3 i s group 4)
chiral
chiral
c hiral
5 -27
(a)
<;:H3
en anti orner
* =
H - C - CI
c::====�>
CH2CH3
no plane of symmetry
no diastereomer
chiral structure
<;:H3
en anti orner
* =
CI - C - H
>
CH2CH3
chiral structure
no plane of symmetry
no diastereomer
chiral structure
chiral structure
(c)
* -
Br - C - H
*I
Br - C - H
CH2CH2CH3
en anti orner
>
<;:H3
=
H - C - Br
*
*
I
H - C - Br
CH2CHzCH3
chiral structure
*
-
H - C - Br
*
I
Br - C - H
CH2CH2CH3
no plane of symmetry
chiral structure
chiral structure
c::====�> (inverting two groups on the
diastereomer
bottom asymmetric c arbon
i nstead of the top one would
also gi ve a diastereomer)
92
continued
(d) Br
Br
enanti omer H
chistructralure
H not chistrrauctl ; ure
H
di
a
st
e
reomer
(i
n
ver
t
i
n
g
t
w
o
gr
o
ups
on
t
h
e
nochiplralanestructof symmet
r
y
r
i
g
ht
asymmet
r
i
c
car
b
on
i
n
st
ure
ofdiatstheereomer
left one) would also giveeada
(invertitom nasymmet
g two groups
onbonthe
HO-C-H
di
a
st
e
reomer
a
e
pl
of
n
bot
r
i
c
car
�
one woul) d
H-C-OH ialnsstoegiadvofe athdieatstopereomer
H .: C OH symmetry
CH1CH3
CH1CH3
chiral structure
noa mesoenantstioructmerure, not chiml
racemic mixture of enantiomers; each is chiml with no plane of symmetry
(f) ( H2CH3
�
H2CH�
HO-C-H
H-C-OH
H-C-OH
HO-C-H �stereomer
H2CH3 diastereCH2CH3
omer
H-C-OH symmet
pane
H .:. C OH this mesorystructure is a diastereomer of each of the
CH2CH enantiomers; it is not chiml
(d� 1I
(a) CH70H (b) CHO
O +�{
11 + 011 H + Br
H + OH
CH3
CH3
CH3
5-27
c::=====�>
>
* -
� ':': � ��
_
>
_ _ _ _
_
_
-------------
*
*
1
------------
�
�
*
* -
*
1
*
+
�
� :
�
�
- - - - -
* ::
�
t
1
- - - - - - - - -
I
0
_
3
5 -2 8
93
f
m eso
5-29 Your drawings may look different from these and sti l l be correct. Check configuration by assigning
R and S to be sure .
(a)
CH3
COOH
�
CHO
(b)
"H
NH2
HOCH 2
�
(d)
" OH
H
5-30
(a) s ame (meso)-pl ane of symmetry , superimposable
(b) enantiomers--configuration inverted at both asymmetric carbon atoms
(c) enanti omers--configuration i nverted at both asymmetric carbon atoms
(d) en anti omers-solve this problem by switching two groups at a time to put the groups in the same
positions as in the first structure; it takes three switches to make the i dentical compound, so they are
enantiomers; an even n umber of switches would prove they are the same structure
(e) enanti omers--confi guration inverted at both asymmetric carbon atoms
(f) diastereomers--configuration i n verted at onl y one asymmetric carbon
(g) enantiomers--configuration inverted at both asymmetric carbon atoms
(h) same compound-rotate the right structure 1 80° around a hori zontal axi s and it becomes the left
structure
5-3 1 Drawing the enantiomer of a chiral structure i s as easy as drawing i ts mirror i mage.
(b)
Br
+
CHO
(c)
CHO
H
CH2 0 H
(e)
HO
H
HO
H
HO
H
(d)
CH3
,H
""
C == C == C '
'
/
Br
"
H
o
(f) plane of symmetryno en anti orner
5-32
(a) 1 .00 g / 20.0 mL
=
0.050 g/mL ;
(0.0500) ( 2.00)
(b)
0.050 g / 2.0 mL
=
=
20.0 c m
2 .00 dm
=
- 1 2.5°
0 . 0 2 5 g/mL ;
2.0 cm
=
0 . 20 dm
+ 0.043°
(0.025) (0.20)
5 - 3 3 The 32% of the mixture that is (-)-tartaric aci d wi ll cancel the optical rotation of the 32% of the
m i x ture that i s (+)-tartaric acid, leavi ng only (68 - 32) = 36% of the mi xture as excess (+)-tartari c aci d to
gi ve meas urabl e optical rotation . The specific rotation w i l l therefore be only 36% of the rotation of pure
(+)-tartaric aci d : (+ 1 2 . 0°) x 3 6 % = + 4 . 3 °
94
5-34
(b) Rotation of the enantiomer will be equal in magnitude, opposite in sign:
(c) The rotation
- 7 95°
--' - 15.90°
x
-7.95°
100%
is what percent of
=
50% e.e.
-
- 15.90°.
15.900?
There is 50% excess of (R)-2-iodobutane over the racemic mixture; that is, another 25% must be R and
25% must be S. The total composition is 75% (R)-(-)-2-iodobutane and 25% (S)-(+)-2-iodobutane.
5-35 All structures in this problem are chiral.
C al
H
H
+
OH
OH
CH20H
+
A
CHO
HO
H
H
HO
CH20H
+
+
B
CHO
H
OH
HO
H
CH20H
+
+
C
CHO
H
HO
OH
H
CH20H
+
+
D
enantiomers: A and B; C and D
diastereomers: A and C; A and D; Band C; Band
D
(b)
E
F
G
enantiomers: E and F; G and H
diastereomers: E and G; E and H; F and G; F and H
95
H
5-35 continued
(c) This structure is a challenge to visualize. A model helps. One way to approach this problem is to
assign R and 5 configurations. Each arrow shows a change at one asymmetric carbon.
CH3
H3C
CH3
H
CH3
H
H \ S
H
H
� L
H
H
�
H3C
H
.0.
H
H
�
.0.
CH3
H
H
H
H
CH3
H
H
CH3
H3C
H
H3C
H3C
H
Summary
R5S5 (K) is the enantiomer of RRR5 (M)
5555 (L) is the enantiomer of RRRR (P)
The structures with two R carbons and two 5 carbons (J, N, and 0) have
special symmetry. J is a meso structure; it has chirality centers and is
superimposable on its mirror image-see Problem 5-20(h). Nand 0 are
enantiomers, and are diastereomers o f all of the other structures. Give
yourself a gold star if you got this correct!
5-36
s� .Ix. ��
S*
A
meso
diastereomers
15,3R
equivalent
to 1R,35
B
meso
C
R*
chiral
1R,3R
15,3R
equivalent
to 1R,35
enantiomers
D
chiral
15,35
C and Dare enantiomers. All other pairs are diastereomers.
*Structures A and B are both meso structures, but they are clearly different from each other. How can
they be distinguished? One of the advanced rules of the Cahn-Ingold-Prelog system says: When two
groups attached to an asymmetric carbon differ only in their absolute configuration, then the neighboring
(R) stereocenter takes priority. Now the configuration of the central asymmetric carbon can be assigned:
5 for structure A, and R for structure B. This rule also applies to problem 5-37. (Thanks to Dr.
Kantorowski for this explanation.)
96
5-37 This problem is similar to 5-36.
(a)
*
H
2
H
3
5
*
H
4
R
s CH3
H
Br
Br
Br
H
*
5
*
R
Br
Br
H
H
Br
H
CH3
�
'-diastereomers
*
H
H
Br
Br
Br
Br
*
CH3
B
C
meso
meso
chiral
25,4R
equivalent
to 2R,45
2 5,4R
equivalent
to 2R,45
2R,4R
A
(b)
Br
CH3
CH3
CH3
[CH3
'--enantiomers�
*
Br
H
*
H
CH3
D
chiral
2 5,45
The configuration of carbon-3 in A and B can be assigned according the rule described above in the
solution to problem 5-36: C-3 in A is 5, and C-3 in B is R.
(c) According to the IUPAC designation described in text Section 5-2B, a chirality center is "any atom
holding a set of ligands in a spatial arrangement which is not superimposable on its mirror image." An
asymmetric carbon must have four different groups on it, but in A and B, C-3 has two groups that are
identical (except for their stereochemistry). C-3 holds its groups in a spatial arrangement that is
superimposable on its mirror image, so it is not a chirality center. But it is stereogenic: in structure A,
interchanging the Hand Br at C-3 gives structure B, a diastereomer of A; therefore, C-3 is stereogenic.
(d) In structure C or D, C-3 is not stereogenic. Inverting the H and the Br, then rotating the structure 180°,
shows that the same structure is formed. Therefore, interchanging two atoms at C-3 does not give a
stereoisomer, so C-3 does not fit the definition of a stereogenic center.
5-38 The Cahn-Ingold-Prelog priorities o f the groups are the circled numbers in (a).
(a)
H
CH3
H
CH3
H
"
H2
..
H
C
CH
CH
H
C
" .......::
'
'
Pt
CH 3
C '/" "c/
CH3
"c/ "c/
/"
'
H
/
''
H' CH3
H CH3
H
H
10 (2)1
I
'
'
lev 0.1
\
o 0
(R)-3,4-dimethyl-l-pentene
I
\
0 0
(5)-2,3-dimethylpentane
(b) The reaction did not occur at the asymmetric carbon atom, so the configuration has not changed­
the reaction went with retention o f configuration at the asymmetric carbon.
(c) The name changed because the priority o f groups in the Cahn-Ingold-Prelog system of nomenclature
changed. When the alkene became an ethyl group, its priority changed from the highest priority group
to priority 2. (We will revisit this anomaly in problem 6-21(c).)
(d) There is no general correlation between R and 5 designation and the physical property of optical
rotation. Professor Wade's poetic couplet makes an important point: do not confuse an object and its
properties with the name for that object. (Scholars of Shakespeare have come to believe that this quote
from Juliet is a veiled reference to designation of R,5 configuration versus optical rotation of a chiral
molecule. Shakespeare was way ahead of his time.)
97
5-39
(a) The product has no asymmetric carbon atoms but it has three stereocenters: the carbon with the OH,
plus both carbons o f the double bond. Interchange of two bonds on any o f these makes the enantiomer.
(b) The product is an example of a chiral compound with no asymmetric carbons. Like the allenes, it is
classified as an "extended tetrahedron"; that is, it has four groups that extend from the rigid molecule in
four different directions. ( A model will help.) In this structure, the plane containing the COOH and
carbons of the double bond is perpendicular to the plane bisecting the OH and H and carbon that they are
on. Since the compound is chiral, it is capable of being optically active.
COOH
H
HO
H
(c) As shown in text Figure 5-16, Section 5-6, catalytic hydrogenation that creates a new chirality center
creates a racemic mixture (both enantiomers in a 1: 1 ratio). A racemic mixture is not optically active. In
contrast, by using a chiral enzyme to reduce the ketone to the alcohol (as in part (b)), an excess of one
enantiomer was produced, so the product was optically active.
98
In
CHAPTER 6-ALKYL HALIDES: NUCLEOPHILIC SUBSTITUTION AND ELIMINATION
6-1
problems like part (a), draw out the whole structure to detect double bonds.
(c) aryl halide
(d) alkyl halide
(a) vinyl halide
(b) alkyl halide
6-2 (a)
,,
H
H
(b)
,,
Br
J-C-I
Br -C- Br
�
Br
(f)
(e) vinyl halide
(f) aryl halide
� I-C-H
,,
~
FX '-rH2F +
(d)
(Cl
(e)
I
Br
Br
(g)
H
(hl
Cl
Br
or (CH3hCCI
6-3 IUPAC name; common name; degree of halogen-bearing carbon
l-chloro-2-methylpropane; isobutyl chloride; 1°
diiodomethane; methylene iodide; methyl
1,I-dichloroethane; no common name; 1°
2-bromo-l,1,l-trichloroethane; no common name; all 1°
trichloromethane; chloroform; methyl
(f) 2-bromo-2-methylpropane; t-butyl bromide; 3°
(g) 2-bromobutane; sec-butyl bromide; 2°
(h) l-chloro-2-methylbutane; no common name; 1°
(i) cis-l-bromo-2-chlorocyclobutane; no common name; both 2°
U) 3-bromo-4-methylhexane; no common name; 2°
(k) 4-fluoro-l,1-dimethylcyclohexane; no common name; 2°
(I) trans-l,3-dichlorocyclopentane; no common name; all 2°
(a)
(b)
(c)
(d)
(e)
6-4
Cl
CI
Cl
CI
CI
Cl
Cl
>
Cl
0
Kepone®
6-5 From the text, Section 2-9B, the bond dipole moment depends not only on bond length but also on
char ge separation, which in tum depends on the difference in electronegativities of the two atoms
connected by the bond. Because chlorine's electronegativity (3.2) is si gnificantly higher than iodine's (2.7),
the C-Cl bond dipole is greater than that of C-I, despite C-I being a longer bond.
6-6
(a) n-Butyl bromide has a hi gher molecular weight and less branching, and boils at a higher temperature
than isopropyl bromide.
(b) t-Butyl bromide has a higher molecular weight and a larger halogen, and despite its greater branching,
boils at a higher temperature than isopropyl chloride.
(c) n-Butyl bromide has a hi gher molecular weight and a larger halogen, and boils at a higher temperature
than n-butyl chloride.
6-7 From Table 3-2, the density of hexane is 0.66; it will float on the water layer (d 1.00). From Table 6-2,
the density of chloroform is 1.50; water will float on the chloroform. Water is i mmiscible with many
organic compounds; whether water is the top layer or bottom layer depends on whether the other material is
more dense or less dense than water. (This is an i mportant consideration to remember in lab procedures.)
6-8
(a) Step (1) is initiation; steps (2) and (3) are propagation.
(1)
-
nn
Br
(2) H2C
(3) H2C
Br
--
2 Br·
hv
= � Cr\
- + Br � HBr + { = � - C
n
Br - Br �
= HC - . � \
= HC - I +
CH2
CH2
H2C
•
H2C
+
..
H2 "
CH2
Hi - �
=
CH2
}
Br·
Br
- (+ 368) - 4 kJ/mole
kcallmole: + 87 - (+ 88) - 1 kcallmole
Step (3): break Br-Br, make allylic C-Br: kJ/mole: + 192 - (+ 280) - 88 kJ/mole
kcallmole: + 46 - (+ 67) - 2 1 kcal/mole
overall
- 4 + - 88 - 92 kJ/mole (- 1 + - 2 1 - 22 kcal/mole )
This is a very exothermic reaction; it is reasonable to expect a s mall activation energy in step ( 1),
(b) Step (2): break allyJic C-H, make H-Br: kJ/mole: + 364
=
=
=
=
M!
=
=
=
so this reaction should be very rapid.
6-9
(a) propagation steps
H3C
,
H3C
/
C��)
C=C
/CH2
,
+ Bre
CH3
repeats chain
mechanism
100
Br· +
+
6-9
continued
The resonance-stabilized allylic radical intermediate has radical character on both the 1 ° and 3° carbons, so
bromine can bond to either o f these carbons producing two isomeric products.
(b) Allylic bromination of cyclohexene gives 3-bromocyclohex-l-ene regardless of whether there is an
allylic shift. Either pathway leads to the same product. If one of the ring carbons were somehow
marked or labeled, then the two products can be distinguished. (We will see in following chapters how
labeling is done experimentally.)
o
6- 10
(a)
o-
N BS
Br
+
y
the second structure from an
allylic shift is identical to the
first structure-only one
compound is produced
Br
CH3
CH3
I
I
CH3-C-CH3
I
CH3
CH3-C-CH2CI
I
hv
CH3
This compound has only one type of hydrogen-only one monochlorine isomer can be produced.
CH3
CH3
CH3- � -H
CH3- �- Br
I
I
(b)
CH2CH3
hv
CH2CH3
Bromination has a strong preference for abstracting hydrogens (like 3°) that give stable radical
intermediates.
hv
(or NBS)
Bromine atom will abstract the hydrogen giving the most stable radical; in this case, the radical
intermediate will be stabilized by resonance with the benzene ring.
(d)
CO
Br
hv
(or N BS)
OJ
the second structure from an
allylic shift is identical to the
first structure-only one
compound is produced
+
Br
Bromine atom will abstract the hydrogen giving the most stable radical; in this case, the radical
intermediate will be stabilized by resonance with the benzene ring.
6- 1 1
(a) substitution-Br is replaced
(b) elimination-H and OH are lost
(c) elimination-both Br atoms are lost
6- 12
(a) CH3 (CH2)4CH2 - OCH2CH3
(b)
CH3 (CH2)4CH2 - eN
101
6- 13 The rate law is first order in both 1-bromobutane, C4H9Br, and methoxide ion. If the concentration of
C4H9Br is lowered to one-fifth the original value, the rate must decrease to one-fifth; if the concentration of
methoxide is doubled, the rate must also double. Thus, the rate must decrease to 0.02 molelL per second:
rate
=
( 0.05 molelL per second ) x
original rate
( 0. 1 M )
( 0.5 M )
x
( 2.0 M )
( l.0 M )
=
0.02 molelL per second
new rate
change in
NaOCH3
change in
C4H9Br
A completely different way to answer this problem is to solve for the rate constant k, then put in new values
for the concentrations.
rate = k [C4H9Br] [NaOCH3] � 0.05 mole L-I sec -I = k (0.5 mol L-I) (l.0 mol L-I) �
rate constant k = 0.1 L mol-I sec-I
rate = k [CH3IJ [NaOH] = (0. 1 L mol-I sec-I) (0.1 mol L-I) (2.0 mol L-I)
=
0.02 mole L-I sec -I
6- 14 Organic and inorganic products are shown here for completeness.
+
(a) (CH3)3C - 0 - CH2CH3
KEr
+
+
Br
+
NaI
(f)
+
-
(c) (CH3hCHCH2 - NH3
(e)
NaCI
�I
�F
+
-
NH4 Br
+ NaCI
+
K CI
(lS-crown-6 is the catalyst and does not change; CH 3CN is the
solvent)
6- 15 All reactions in this problem follow the same pattern; the only difference is the nucleophile (-:Nuc).
Only the nucleophile is listed below. (Cations like Na+ or K+ accompany the nucleophile but are simply
spectator ions and do not take part in the reaction; they are not shown here.)
� CI
l-chlorobutane
(a) HO-
(f) -O CH2CH3
(b)
F-
+
-:Nuc
----
� Nuc
from K F/lS-crown-6
(d)
+
CI-
-CN
(e) H C_C-
(g) excess NH3 (or - NH2)
6- 1 6
(a) ( CH3 CH2)2NH is a better nucleophile-Iess hindered
(b) (CH3hS is a better nucleophile- S is larger, more polarizable than 0
(c) PH:, is a better nucleophile-P is larger, more polarizable than N
(d) CH3 S- is a better nucleophile-anions are better than neutral atoms of the sa me element
(e) ( CH3)3N is a better nucleophile-Iess electronegative than oxygen, better able to donate an electron pair
(f) CH3S- is a better nucleophile-anions are generally better than neutral atoms, and S is larger and more
polarizable than 0
(g) CH3CH2CH20- is a better nucleophile-Iess branching, less steric hindrance
(h) 1- is a better nucleophile-Iarger, more polarizable
102
..
6- 17 A mechanism must show
electron movement.
H
H
3 +
CH3-O-CH
• •
H+
�
1",
+..
CH3-0-CH3
+ :Br:
• •
1
-
CH3-�: + CH3-!3[:
---
�
Protonation converts OCH3 to a good leaving group.
6- 18 The type of carbon with the halide, and relative leaving group ability o f the halide, determine the
reactivity.
methyl iodide> methyl chloride> ethyl chloride> isopropyl bromide»
.
neopentyl bromide,
t-buty I'IOd'd
1 e
most reactive
}
least.
reaCllve
Predicting the relative order of neopentyl bro mide and t-butyl iodide would be difficult because both would
be extremely slow.
6- 19 In all cases, the less hindered structure is the better SN2 substrate.
(a) 2-methyl- 1-iodopropane ( 1° versus 3°)
(b) cyclohexyl bromide (2° versus 3°)
(c) isopropyl bromide (no substituent on neighboring carbon)
(d) 2-chlorobutane (even though this is a 2° halide, it is easier to attack than the 1° neopentyl type in 2,2dimethyl-1-chlorobutane-see below)
CH3
a neopentyl halide1
(e) isopropyl iodide (same reason as in (d))
hindered to backside attack
I
/c""
- C, � CH 3
by neighboring methyl groups
H
, "\. C
Lt13
H
H
"'--- -:Nuc
6-20 All SN2 reactions occur with inversion of configuration at carbon.
(a)
8-
Br
--P<
:
H
H
H08-
trans
H
CH,
transition state
HO:
(b)
:
Br:
inversion
cis
-:CN
�
+
R
103
..
-
:Br:
6-20 continued
H
H
.-
1
H
;···
H3C
_
(d)
CH3CH2
c� I �
,
H"
�
CH3 F
:SH
Br
CH3 � :
(f)
S
H- �:
I
H
�/ �
H3CO
.
�-
-
H
,Br
..
-
:
Br:
c::::::>
H
H
CH3
+
+
I
CH}
CH2CH}
.
,
..
H
+
:
Br:
• •
(e)
H
�I
� ,�
+
r.
.B
• •
,
�
R
c::::::>
H
��
� ID
C;CI
C
-
+
:CJ :
---II"�
6-2 1
(a) The best leaving groups are the weakest bases. Bromide ion is so weak it is not considered at all basic;
it is an excellent leaving group. Fluoride is moderately basic, by far the most basic of the halides. It is a
terrible leaving group. Bromide is many orders of magnitude better than fluoride in leaving group ability.
(b)
inverted, but
still named S
H"�
"'" : CH}
.
F
H
:
, I CH}
��
I
OCH3
I
+
..
:
Br:
transition state
(c) As noted on the structure above, the configuration is inverted even though the designations o f the
configuration for both the starting material and the product are S; the oxygen of the product has a lower
priority than the bromine it replaces. Refer to the solution to problem 5-38, p. 97, for the caution about
con fusing absolute configuration with the designation of configuration.
Cd) The result is perfectly consistent with the SN2 mechanis m. Even though both the reactant and the
product have the S designation, the con figuration has been inverted: the nomenclature priority of fluorine
changes from second (after bromine) in the reactant to first (before oxygen) in the product. While the
designation may be misleading, the structure shows with certainty that an inversion has occurred.
104
6-22
.. -
�
:Br:
slow
planar
carbocation
+
+
CH3CH20H2 Be
(or si mply HBr)
6-23 The structure that can form the more stable carbocation will under go SN1 faster.
(a) 2-bromopropane: will form a 2° carbocation
(b) 2-bromo-2-methylbutane: will form a 3° carbocation
(c) allyl bromide is faster than n-propyl bromide: allyl bromide can for m a resonance-stabilized
intermediate.
(d) 2-bromopropane: will form a 2° carbocation
(e) 2-iodo-2-methylbutane is faster than t-butyl chloride (iodide is a better leaving group than chloride)
(f) 2-bromo-2-methylbutane (3°) is faster than ethyl iodide (1°); although iodide is a somewhat better
leaving group, the difference between 3° and 1° carbocation stability dominates
6-24 Ionization is the rate-determining step in SN l. Anything that stabilizes the intermediate will speed
the reaction. Both of these compounds form resonance-stabilized intermediates.
(x
H
I
benzylic
+
6-25
CH3
-T
H
I
T
H
�H
VH
---
( )=
CH2 ---
allylic
O
CHz
+
H
t:..
- HCH3
H
H
•
Br-
+
+ I
CH3 - C - CHCH3
+
C
I
I
CH2CH3
---
H
- R-
�
H:1
H
CH3 - � - CHCH3 ......f'CH3 - � - CHCH3 /
.. ..---... ('I.
:?: CH3 CH3CH2- 9: H-+? CH3
( Br
H
� < }- �
I
CH2CH3
105
CH2CH3
H2
6-26 I t is i mportant to analyze the structure of carbocations to consider if migration of any groups from
adjacent carbons w i l l lead to a more stable carbocation . As a general rule, if rearrangement would lead to a
more stable carbocation , a carbocation w i l l rearrange. (Beginning with this problem, only those unshared
electons pairs involved in a partic ular step w i l l be shown. )
-
CH3
CH3
I
(a) CH3-C
I
1\
CH3
y y
I
1- +
CH-CH3
+
CH3- - -CH3
CH3 H
I�
nucleophilic attack on unrearranged carbocation
c��
CH3 :O-CH3
? ?
!
I
CH3- - -CH3
. .
CH3 :O-CH3
HRCH3
�
-----l..
I
I
I
I
CH3- C - C-CH3
CH3 H
CH3 H
unrearranged product
nucleophilic attack after carbocation rearrangement
CH3
I
+
CH3 -C-C-CH 3
�
I�I
CH3 H
Methyl shift to the
2° carbocation forms a
more stable 3° carbocation.
cj:
nucleophilic attack on unrearranged carbocation
rl<�",--;:
H
H
H
:R - CH,CH
CH2CH3
d=�:
I
;
C H3
CH 3
unrearranged
product
106
6-26(b) continued
nuc leophilic attack after carbocation rearrangement
H
C)) H
��
H :..b�CH2CH]
rf-
hydride
shift
�
___
30
�
CH3
Hydride shift to the
2° carbocation forms a
more stable 3° carbocation .
rearranged
product
H
(c )
H
H
H
:(j
H H H
H
OH
�O·=C-CH3
I
H
... .
3
o-H
nucleophi lic attack on unrearranged carbocation
H
0I
unrearranged
product
I
• •
:0 :
II
O-C-CH
III
• •
?-H
O=C-CH3
+
Note: braces are used to indicate
the ONE chemical species represented
by multiple resonance forms.
o-.
H
....
..I----J
..
�
I
•
•
?
: -H
O-C-CH3
.
+
...
:0:
II
3
HO-C-CH
•
•
The most basic species in a mixture i s the most likely to remove a proton .
I n this reaction , acetic acid is more basic than iodide ion.
107
H
H H
H
H
Cj H
�-C-CH3H C
D
D
H
y)
6-2 6(c) continued
nucleophilic attack after carbocation rearrangement
H
H
H
I
H
+
H
H
hydride
shift
..�
---....
H
HO : Cc;?: H
-
H
H
H
-
II
:0 :
O..
II
-
H+
CH 3
.....
f--------
rearranged product
3
..
:I
O-
: 0:
I
allylic-resonance-stabilized
(removes
as on p. 107)
H
H
H �==C-CH3
I
�
plus two other
resonance forms
as shown on p. 107
Comments on 6-26(c)
(1) The hydride shift to a 20 carbocation generates an allylic, resonance-stabilized 20 carbocation.
(2) The double-bonded oxygen o f acetic acid is more nucleophilic because of the resonance forms it can
have after attack. ( See Solved Problem 1-5 and Problem 1- 16 in the text.)
(3) Attack on only one carbon of the allylic carbocation is shown. In reality, both positive carbons would
be attacked in equal amounts, but they would give the identical product in this case. In other compounds,
however, attack on the different carbons might give different products. ALWAYS CONSIDER ALL
POS SIBILITIES.
(d)
ct
H
0
CH2-I
The 1 0 carbocation initially formed is very
unstable; some chemists believe that
rearrangement occurs at the same time as
the leaving group leaves. At most, the 10
carbocation has a very short lifetime.
---l"�
r+CH2 N CH :�� CH2CH,;
hydride shift followed by nucleophilic attack
'------'
3
V
---l"�
V
-
108
6-26 (d) continued
alkyl migration (ring expansion) followed by nucleophilic attack
H
2°
+ �I
H
CH 2
\
: 0 -CH2CH3
-.-.------�.�
CH 2
/ :..0-CH2CH3
I
H
6-27
CH3
0
(a) CH3 - �-0-C-CH3
I
"
(e) CH
(b) CH3-CHCH20CH3
I
CH 2CH3
CH3
SN2 , 1 °
S N 1 , 3°
o
CH2CHl
SN 1 , weak
nucleophile
SN 1 , 3°
6-28
CH 3 CH2
CH3
/
I
C
CH3
"'H
Br
(R)-2-bromobutane
I
H 2 0,�
�
SN2-
CH 3 CH 2
inversion
C""
/ ,'OH
H
(S)-2-butanol
H2 0,�
S N l­
CH3
I
racemization
+
\..
S N2, strong nucleophile
CH 3 CH2
(R)-2-butanol
, OH
/C"",
H
�
(S)-2-butano
y
50 : 50 mixture-racemic
If SN 1 , which gives racemization, occurs exactly twice as fast as SN2, which gives inversion, then the
racemic mixture (50 : 50 R + S) is 66.7% of the mixture and the rest, 3 3 . 3%, is the S enantiomer from SN2.
Therefore, the excess o f one enantiomer over the racemic mixture must be 3 3 . 3%, the enantiomeric excess.
(In the racemic mixture, the Rand S "cancel " each other algebraically as well as in optical rotation.)
The optical rotation o f pure (S)-2-butanol is + l 3.5° . The optical rotation o f this mixture is:
3 3 . 3% x + l 3.5° = + 4.5°
109
6-29
(a) Methyl shift may occur simultaneously with ionization. The lifetime of 1 ° carbocations is exceedingly
short; some chemists believe that they are only a transition state to the rearranged product.
CH3 1 °
methyl
CH3
30
+
I
I
shift
CH3- -CHFHl
•
CH3-C
H2
Ag I +
�
�J
H3
I�
H 2 0:
CH3
I
CH3 - C-CH 2CH3
I
...
:O-H
(b) Alkyl shift may occur simultaneously with ionization.
CH 2
\I
� Ag+
---I.�
Ag I
CH2
+
alkyl shift­
ring expansion
..
..
6-30
UBf
(a)
H
:Br:
+ CH30H2
+
The SN 1 mechanism begins with ionization to form a carbocation, attack of a nucleophile, and in
the case of ROH nucleophiles, removal of a proton by a base to form a neutral product.
(b) In the E l reaction, the solvent (methanol, in this case) serves two functions: it aids the ionization
process by solvating both the leaving group (bromide) and the carbocation; and second, it serves as a base
to remove the proton from a carbon adjacent to the carbocation in order to form the carbon-carbon double
bond. The SN1 mechanism adds a third function to the solvent: the first step is the same as in E l ,
ionization to form the carbocation; the second step has the solvent acting as a nucleophile-this step is
different from E l ; third, the solvent acts like a base and removes a proton, although from an oxygen (SN1)
and not a carbon (El). Solvents are versatile!
110
6-3 1 Thi s solution does not show
complete mechanisms. Rather, it
shows the general sequence of a l l
four pathw ays leading t o fi ve
different products.
;on;z.
2-bromo-3-methylbutane
;/
_ H
i\
________
- - - - - - - - -...- .
- - - - - - - - - �.
--
-
-
-
-
-
-
---
----
-
-
-
-
-
Five unique products
shown in the boxes. One
alkene can be produced
from either the r or the
30 carbocation.
-
.......... -_..............-_..............
¥
CH3
hydride shift
...
.
.. .. ___ .............................
.
..... __ I
�
H3C
+
rearranged
30 carboc ation
CH3
..
"
CH3
H3C
� �"'C H2
T
H
H1C
_________
two possible E 1 products
6 32
111
:
----------------______ 1
H
CHl
____________
two possible
E1 products
6-33 Substi tution products are shown first, then elimination products .
(a)
(c )
� )
+
+
with rearrangement
without rearrangement
o
(d)
6-34 Et is the abbreviation for ethy l , so EtOH is ethanol .
0\3 Cf � r �Et
: Br :
··
EtOH
-----l....
�
•
• •
H
0 ..,J
+
•
�? :
V
Et
Et
+ EtOH2
+
E l -MINOR product­
more substituted alkene
H
H
" /
�
C -H • •
J
HOEt
cr
+ EtOH2
+
without rearrangement
··
I
HOEt
E I -MAJOR product­
more substituted alkene
6-35
Cll3
CH
CH 3
• •
S N 1 product:
with rearrangement
112
. .
..
+ EtOH2
+
6-36
yl
NaOCH2CH3
..
+
E2-minor
product-less
substituted alkene
(monosubstituted)
Br
6-37
NaO C H 3
Br
(a)
�
�
..
�
+
NaOMe
Br
�
NaO H
Br
(c )
(d)
~
H"'
0
Br
'
"
...
�
+
�
+
minor al kenemonosubsti tuted
CH]
NaOEt
H
...
H atoms that are
removed are shown
a
E2-major
product-morc
substituted alkene
(trisubstituted)
�
+
�
~
O'
mi nor alkenedisubstituted
CH]
major alkenetrisubstituted
Br
E2-minor
113
OCH2CH3
SN2-probably
very small amount
because of steric
hindrance
OCR,
�
No SN2 product wi l l be
formed; S N2 c annot
occ ur at a 3° carbon.
+
OH
~
S N2 product-very
small amount because
2° carbon is hindered
major al kenetrisubstituted
CH]
yl
S N2 product-some
w i l l be formed on a
2° c arbon
mi nor alkenedisubstituted
major alkenetrisubsti tuted
(cis + trans)
...
+
major alkenedisubstituted
(cis + trans)
mi nor alkenemonosubsti tuted
(b)
�
(rCH]
OEt
-
+
S N2 product showing
inversion at the
c arbon-some of this
product wil l be formed
on a 2° carbon
6-38 In systems where free rotation is possible, the
to be abstracted by the base and the leaving
group (Br here) must be anti-coplanar. The E2 mechanism is a concerted, one-step mechanism, so the
arrangement of the other groups around the carbons in the starting material is retained i n the product;
there is no intermediate to allow t i me for rotation of groups . (Models w i l l help.)
H
8-
r:RCH3
Ph
H
H'
� r/
C-C,
/
3
""CH
�
Ph
: 9CH3
• •
:t:
, CH3.
- Ph "" "'C- C"'
H�
-
Br
transition state
H and Br anti -coplanar
�
�----------
(
- - - - - - -
=
:0
trans
Ph
only geometric isomer
possible from a
concerted mechanism
�----------�
stretched bond being broken or forme
orbital picture of this reaction
H
•••
H� : OCH3
,,
H
I
• •
...--
--..�
H
P h --+---=--+--
Ph
trans
Br
Br
transition state
The other diastereomer has two groups interchanged on the back c arbon of the New man projection ,
where it must give the cis-alkene.
H
Ph
H
H
Ph
•
Ph
CH3
Ph
Br
cis
114
CH3
pi bond
6-39 lin this problem, all new internal alkenes fonn cis i somers as wel l as the trans i somers shown.
Section 1
Br
�
Br
�
C
� �
OH
H3
.
NaOCH3
·
CH30H
OCH3
�
S N2
El
=<
+�
E2
+�
E2
CH3
NaOCH3
----j..
�
+� +�
El
SN1
H2C
CH30H
E2
CH3
Section 2
+�
�
Br
E2
A gN03
�
Br
+�
CH30H
..
�
El
+�
El
Section 3
+ Br
CH3
H3C
CH3
+ Br
CH3
H3C
C H3
•
=<
CH3
NaOCH3
H2C
CH30H
E2
CH3
+ OCH3
CH3
CH30H
•
�
H3C
CH3
+
SN 1
=<
CH3
H2C
CH3
El
Section 4
Br
�
Br
�
NaOCH3
"
CH30H
�
CH30H
OCH3
�
SN2
+�
.� +
E2
+�
E2
OCH3
SN1
115
El
+�
E1
6-39 continued
Section 5
Br
�
I
�aI
-.
�
Br
�
�
+
E2
�
E2
6-40
(a) Ethoxide is a strong base/nucleophi le-second-order conditions. The I ° bromide favors substitution
over e l i mination, so SN2 will predominate over E2.
�
substitution-major
elimination-minor
(b) Methoxide i s a strong base/nucleophi l e-second-order conditions. The 2° chloride will undergo SN2 by
backside attack as well as E2 to make a mixture of alkenes.
OCH3
�
e l i mination­
minor alkene
substi tution
elimination­
major alkene
(cis + trans)
(c) Ethoxide is a strong base/nucleophile-second-order conditions. The 3° bromide is h indered and
c annot undergo SN2 by backside attack. E2 is the only route possible.
CH3 - C=CH2
I
CH3
(d) Heating in ethanol i s conditions for solvolysis, an SN I reaction.
OCH2 CH3
I
CH 3 -C-CH3
I
CH 3
(e) Hydroxide is a strong base/nucieophile-second-order condi tions. The 1° iodide i s more l i kely to
undergo SN2 than E2, but both products wi ll be observed.
CH 3 - CHCH20H
I
CH3 -C = CH2
I
C�
C�
substitution-major
elimination-minor
(f) S i l ver ni trate i n ethanol/water is ionizing conditions for I ° alkyl halides that w i ll lead to rearrangement
followed by substitution on the 3° carbocation.
OCH2CH3
C H3 -
?
I
-
CH3
CH3
from ethanol as
n ucieophile
OH
?
I
CH3- -CH3
116
CH3
from w ater as
n ucieophile
6-40 conti nued
(g) S i l ver nitrate in ethanoVwater is ionizing conditions for 1 ° alkyl halides that w i l l lead to rearrangement
fol lowed by substi tution on the 30 c arbocation.
OH
from ethanol as
OCH2CH 3
nucleophile
I
CH3-C -CH2CH3
I
from w ater as
nucleophile
I
CH3-C-CH2CH3
I
CH3
CH3
CH6OCH3
( h ) Heating a 30 halide in methanol is quintessential first-order conditions, ei ther E l or S N I (sol volysis).
\.
substitution
(S N l )
C 6
V
(!,"cc)
/
elimination (E l )
(i) Ethoxide in ethanol on a 30 halide will lead to E2 elimination; there w i l l be no s ubstituti on.
CH3
A
U
major alkene
tri substituted
minor alkene
disubstituted
6-41 Please refer to solution 1 -20, page 1 2 , of this Sol utions Manual .
(a)
6-42
(e)
I""
�
CI
0-
CI
(d)
CI-C - CH20H
I
CI
I
I
(f)
CI
H-C-H
I
CI
I
6-43
(a) 2-bromo-2-methylpentane
(b) l -chloro- l -methylcyc lohexane
(c) 1 , I -dichloro-3-f1uorocyc loheptane
(g)
CI
(h)
C1 -C H
I
-
CI
I
CI
�
o
'
(i)
I-
CH2CH3
I
� -CH3
CH3
(d) 4-(2-bromoethy 1)- 3-(fI uoromethy 1)-2-methy I heptane
(e) 4,4-dichloro-5-cyc lopropyl-l- iodoheptane
(f) cis- I ,2-dich loro- l -methylcyc lohexane
6-44 Ease of backside attack (less steric hindrance) decides which undergoes S N2 faster i n all these
examples except (b).
(a)
�CI
faster than
(b)
�I
faster than
�
CI
Primary R-X reacts faster than 2° R-X.
�CI
117
Iodide a better leaving group than chloride.
6-44
(c)
continued
Cl
faster than
�
Br faster than
(d)
�
(e)
OCH2Ci
faster than
(f)
�Br
faster than
Less branching on a neighboring carbon.
�
Cl
Br
00
Same neighboring branching, so 1° faster than 2°.
Primary R-X reacts faster than 2° R-X.
OCI
XBr
Less branching on a neighboring carbon.
Formation of the more stable carbocation decides which undergoes SN 1 faster in all these examples
except (d).
Cl
Cl
(a)
faster than
6-45
(b)
(c)
30+
�
�
2° Cl
OBr
faster than
2°
� Cl
1°
faster than
OCH2Br
2°
(d)
(e)
OJ
X
r
1°
faster than
faster than
3°
(t)
Q
Br
2° allylic
faster than
OCI
113
~
Br
Q
2°
Br
1 18
2°
(leaving group ability)
6-46 For SN2, reactions should be designed such that the nucleophile attacks the least highly substituted
alkyl halide. ("X" stands for a halide: Cl, Br, or I.)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Q-
+
HO-
-
-
CL
OH
Q-
X
HO-
--�
..
-
CH2X
+
0
O:.A
X
�
NH3
excess
-
-
-
Q-
0
CH20H
SCH2CHl
some
0
0
Q-
U
OH
also produced
OH
CH2NH2
H2C = CHCH2CN
-
HC:::C
6-47
(a)
CH2X
(1)
CH3
CH - 0
/
CH3
\
(2)
CH3
�
-
;."
/ U
CH3 2°
+
CH-O-CH2CH3
\
�
__
CH3
t
/
1°
�-
CH-X
\
CH3
cfCH2CH3
this bond
formed
CH3
I
CH2CH3
CH-O-CH2CH3
\
..
CH3
/
t
this bond
formed
Synthesis ( 1 ) would give a better yield of the desired ether product. ( 1) uses SN2 attack of a nuc\eophilc on
a 1° carbon, while (2) requires attack on a more hindered 2° carbon. Reaction (2) would give a lower yield
of substitution, with more elimination.
(b) CANNOT DO S N2 ON A 3° CARBON!
CH3 � umps into H
'\
I
CH3-C-Cl
+ -OCH3
I
C�
6-47 (b) continued on next page
b
CH2
II
CH3-C
I
C�
before it can find C
..
119
elimination
(E2) competes
6-47 (b) continued
Better to do SN2 on a methyl carbon:
CH 3
CH3 -
?
I
-O
�
+
CH 3
I
CH3 - C-O - CH3
C H3 - X
U
I
CH3
CH3
6-48
(a) SN2-second order: reaction rate doubles
(b) SN2-second order: reaction rate increases six times
(c) Virtually all reaction rates, including this one, increase with a temperature increase.
6-49 This is an SN I reaction; the rate law depends only on the substrate concentration, not on the
nucleophile concentration.
(a) no change in rate
(b) the rate triples, dependent only on [ t-butyl bromide ]
(c) Virtually all reaction rates, including this one, increase with a temperature increase.
6-50 The key to this problem is that iodide ion is both an excellent nucleophile AND leaving group.
Substitution on chlorocyclohexane is faster with iodide than with cyanide (see Table 6-3 for relative
nucleophilicities). Once iodocyclohexane is formed, substitution by cyanide is much faster on
iodocyclohexane than on chlorocyclohexane because iodide is a better leaving group than chloride. So
fast reactions involving iodide replace a slower single reaction, resulting in an overall rate increase.
0-
o-
+
Cl
Cl
+
-CN
-I
moderate rate
fast
----
�..
�
0-
..
0-
1
+
-CN
CN
fast
�
�
..
----
0-
CN +
-
1
recycles-only a small amount needed (catalyst)
6-51
(a)
(c)
(d)
rearrangement
6-52
(a)
120
rearrangement
two
(b)
6-52
continued
(i)
·
0:
H�(
H
equivalent resonance fonns
H
H
H
.��"
equivalent resonance fonns
}
(iv)
H
H
�
H
i
loss of bromide
gives unstahle 10
H 2 carhocation Ihat
quickly rearranges
to a 2° allylic
carbocation
1/��· .Jy�,,}
equivalent resonance fonns-same as from part (iii)
(c) Only one substitution product arises from equivalent resonance fonns.
(ii)
(i)
(iii)
(iv)
OCH2 CH3
�
cis + trans
OCH 2 CH3
�
cis + trans
same as part (iii)
+
6-53
most
stable
al\ylic
(+ on 3°
and 2° C)
>0>0>
al\ylic
(+ on 1 °
and 2° C)
12 1
least
stable
6-54
CHX H H
-Z 2,0
X
H
0
I
t
U
I
t
CH3
CH3
+
I"
hydride
shift
hydride
shift
H
(
C
+ H2
C(
OR
I alkyl shiftt nng expansIOn
H
6 (j
6-55 Reactions would also give some elimination products; only the substitution products are shown here.
(a)
<;:H3
N::C-C-H
CH2CH3
SN2 gives inversion:
only product:
6-56
(a)
(d)
(g)
CH3CH20CH2CH3
C:::CH
CH3(CH2)8CH2
I
Q
(b)
<;:H3
H-C-OH
H-<;:-CH3
CH2CH3
S2
I
(c)
<;:H2CH3
CH3CH20 <;:-CH3
CH(CH3h
-
-
solvolysis ,
N only
with inversion
+
SNl,
<;:H2CH3
CH3-<;:-OCH2CH3
CH(CH3h
-
racerruzation
(b)
< }-CH2CH2CN
(c)
< }-SCH2CH3
(e)
< �N+- CH3
(f)
(CH3hC -CH2CH2NH2
(h)
HO "'0"" CH3
122
1-
substitution
6-57
H
Br
+
+
CH 3
B,
H
B'
H
+
+
H
+
racemic mixture
Br
CH3
Regardless of which bromine is substituted
on each molecule, the same mixture of
products results.
-----y---�
25,35
"----
HO
Br
+
+
CH 3
2R,35
elimination
H
Br -:.
Br
::: H
'-Y
/ '"
KOH
H
H
+
H
H
2R,3R
+
Each of the substitution products has one
chiral center inverted from the starting
material. The mechanism that accounts for
inversion is SN2 . If an SN 1 process were
occurring, the product mixture would also
contain 2R,3R and 25,35 diastereomers.
Their absence argues agai nst an SN 1
process occurring here.
OH
+
racemic mixture
rotate..
-)
Br
CH 3
25,3R
B
S=HB,3
�
:M
H
H
CH 3
H and Br
anti -coplanar
HO
'
H'-')
�
trans
The other enantiomer gives the same product (you should prove this to yourself).
(a)
The absence of cis product is evidence that only the E2 elimination is occurring in one step through an
anti-coplanar transition state with no chance of rotation. If El had been occuring, rotation around the
carbocation intermediate would have been possible, leading to both the cis and trans products.
6-58
+ 15.58°
x 100%
=
. .
.
98% of ongmal optIcal activity
=
98% e.e.
Thus, 98% of the 5 enantiomer and 2% racemic mixture gives an overall composition of 99% 5 and 1% R.
+15.90°
(b) The 1% of radioactive iodide has produced exactly 1 % of the R enantiomer. Each substitution must
occur with inversion, a classic S N2 mechanism.
(a) An SN 2 mechanism with inversion will convert R to its enantiomer, 5. An accumulation of excess 5
does not occur because it can also react with bromide, regenerating R. The system approaches a racemic
mixture at equilibrium.
6-59
��Be
f\ ::
Br -C-H
CH2CH3
R
+
S N2
inversion
�H3
..
123
H-C-Br
CH2CH3
5
+ Br-
6-59 continued
(b) In order to undergo substitution and therefore inversion, HO- wouid have to be the leaving group, but
H O- is never a leaving group in SN2. No reaction can occur.
(c) Once the OH is protonated, it can leave as H20. Racemization occurs in the SN1 mechanism because
of the planar, achiral carbocation intermediate which "erases" all stereochemistry of the starting material.
Racemization occurs in the SN2 mechanism by establishing an equilibrium of Rand S enantiomers, as
explained in 6-59(a).
HO-C-H
.. ::
H+
--
\:::)
HO-C-H
+
H-C-OH
racemic mixture
The carbocation produced in this E l elimination is
and will not rearrange. The product ratio follows
the Zaitsev rule.
30
minor
(b)
(c)
trace-from
unrearranged
20 carbocation
trace amount
from either
unrearranged
20 carbocation
or rearranged
carbocation
30
The carbocation produced in this E l
elimination is 20 and can either eliminate to
give the first two alkenes , or can rearrange by
from rearranged a hydride shift to a 30 carbocation which
30 carbocation
would produce the last two products. The
amounts of the last two products are not
predictable as they are both trisubstituted, but
the first product will certainly be the least.
¥+�+Jy
trace-from
unrearranged
20 carbocation
major
from rearranged
30 carbocation
minor
from rearranged
30 carbocation
124
The carbocation produced in this E l
elimination is 2 0 and can either eliminate to
give the first alkene, or can rearrange by a
methyl shift to a 30 carbocation which would
produce the last two products. The middle
product is major as it is tetrasubstituted
versus disubstituted for the last structure and
monosubstituted for the first structure.
6-61 The allylic carbocation has two resonance forms showing that two carbons share the positive
charge. The ethanol nucleophile can attack either of these carbons, giving the S N 1 products; or loss
of an adjacent H will give the El product.
O'
n
CH2-B,
H
O' +
SNI
..
t
a
CH2
H
1
:�-CH2CH3
•
+H
H
---
H
1
:O-CH2CH3
..
�H2
H
El
- Br-
['t CH2
�H
6-62
H
•
H
"--./
H
U
(a)
SN I
aI
�
H
..
.OH
.
...
H
...
+H
H
(1+
H
�
'---- : 0
-CH2CH3
..
...
_
CH2
'
OcH 2CH3
('y
+
�
. Z-CH2CH1
H I
H
•
/ ' ..
C;(
}
O'
CH2OCH2CHl
H
('yCH2
� OCH 2CHl
H
�
H2
H
H
H
H
-
CH2
CH2 - OCH2CH3
-H2O:
...
OH
H
-H2O:
...
OH
H
··�H+ OSh
U
OH
..
a
..
··/'"W C);,.+1
H
S N2
I
w
/'"
r:!
C);
9:
+1
0'
H
(b)
:O-CH2CH3
• •
1 O'
H2
+OH
1
H
..
'�
."!3r.
�
125
H
Q �: 0':
(J
U
H
O
�
H
H
...
• •
H
:ii,:
"--./ ..
B'
+
H
...
H20:
B'
6-63 NBS generates bromine which produces bromine radical. Bromine radical abstracts an allylic
hydrogen, resulting in a resonance-stabilized a\lylic radical. The a\lylic radical can bond to bromine at
either of the two carbons with radical character.
H
H2C=C-C-CH3
H CH3
I
I
I
+ Br-
--
HEr +
continues
propagation
•
1H2C=C-C-CH3
H CH3
I
I
Br
I
H2C=C-C-CH3
H CH3
Br +
I
I
a\lylic
f----I�....
..
� Br2
+
H2C-T ��H CH3
=
CH,
Br
I
H2C-C=C-CH3
H CH3
I
I
6-64 The bromine radical from NBS will abstract whichever hydrogen produces the most stable
intermediate; in this structure, that is a benzylic hydrogen, giving the resonance-stabilized benzylic radical.
< }�HCH)
HBr +
�
1< }CHCH).
�
•
....
..f----I�_
Br
< }CHCH3
�
Br 2
+
Br·
<
OCHCH)
1>= CHCH3
(Even though three carbons of the ring have
some radical character, these are minm
resonance contributors. The product is most
stable when the ring has all three double
bonds intact, necessitating that the bromine
bond to the benzylic carbon.)
6-65 Two related factors could explain this observation. First, as carbocation stability increases, the
leaving group will be less tightly held by the carbocation for stabilization; the more stable carbocations
are more "free" in solution, meaning more exposed. Second, more stable carbocations will have longer
lifetimes, allowing the leaving group to drift off in the solvent, leading to more possibility for the
incoming nucleophile to attack from the side that the leaving group just left.
The less stable carbocations hold tightly to their leaving groups, preventing nucleophiles from
attacking this side. Backside attack with inversion is the preferred stereochemical route in this case.
6-66
a CH3 ----..
ionization
�
Br
(t� CH3 _--=---H
Br
+
rearrangement
2° carbocation
126
�
3° carbocation
mechanisms continued
on next page
}
6-66 continued
substitution on 20 carbocation
6
H
�H
CH3
:6-CH
• •
3
CH3
..
:O-CH
3
• •
-I
...
�
----
substitution on rearranged 30 carbocation
a
_
CH3
H
I
/
�-
:O-CH
3
• •
....
elimination from rearranged 30 carbocation
H
6-67
(a)
one step
E2-
(� r
CH3-CH-CH3
'-= :OH
--
OH
I
CH3-CH-CH3 + Br-
(b) In the E2 reaction, a C-H bond is broken. When D is substituted for H, a C-D bond is broken,
slowing the reaction. In the SN2 reaction, no C-H (C-D) bond is broken, so the rate is unchanged.
(c) These are first-order reactions. The slow, rate-determining step is the first step in each mechanism.
Br
Explanation on next page
H
l
+
slow... .. I,
fast
:Br:
H2C-CH-CH3 -- H2C=CH-CH3
E l H2C-CH-CH3
(
H
�
I
SN I
(BI r
H3C-CH-CH3
�
o:
+
slow
... H3 C-CH-CH3 -fast
+ Br127
6-67 (c) continued
The only mechanism of these two involving C-H bond c leavage is the E l , but the C-H c leavage does
NOT occur in the slow, rate-determining step. Kinetic isotope effects are observed only when C-H (C--­
D) bond cleavage occurs in the rate-determining step. Thus, we would expect to observe no change in rate
for the deuterium-substituted molecules in the E l or S N I mechanisms. (In fact, this technique of measuring
isotope effects is one of the most useful tools chemists have for determining what mechanism a reaction
fol lows . )
6-68 Both products are formed through E 2 reactions. The difference is whether a D o r a n H is removed
by the base. As explained in Problem 6-67 , C-D cleavage can be up to 7 times slower than C-H
cleavage, so the product from C-H c leavage should be formed about 7 times as fast. Thi s rate preference
is reflected in the 7 : 1 product mixture . ("Ph" is the abbreviation for a benzene ring.)
D
H"
"
.:-!f
�H
" ,C - C
- D, Br
H
..
"
"
/
/
C=C
Br
Ph
Ph
requires C-D bond cleavage; slow; minor product
H
"
,H
.:-� H
" C-C
Ph " 'I
- H, Br
'
'
H
H
...
Br
D
requires C-H bond c leavage; 7 times faster; major product
6-69 The energy, and therefore the structure, of the transition state determines the rate of a reaction. Any
---I"'� [
l
factor which lowers the energy of the transition state will speed the reaction.
R
N
R" "� :
R
\
�
n
R' - X
R
N��---- R , - -R" ��
R
__ _
8x
transition statedeveloping charge
R
N+_ '
R" �� R
R
x
-
This example of S N2 is unusual in that the nucleophi le is a neutral molecule-it is not negatively charged.
The transition state is beginning to show the positive and negative charges of the products (ions), so the
transition state is more charged than the reactants. The polar transition state will be stabilized in a more
polar solvent through dipole-dipole interactions, so the rate of reaction will be enhanced in a polar solvent.
t
,
reaction coordinate
\,
---
polar sol vent­
lower energy,
faster reactIon
___
128
The problem is how to explain this reaction:
: NEt2
Et2 N :
I
I
HO- �
H2C - CH - CH2CH3
H2C - CH - CH2CH3
6-70
Solution
I
Cl
I
OH
1
+
Cl-
2
facts
1) second order, but several
thousand times faster than similar
second order reactions without the
NEt2 group
2) NEt 2 group migrates
Clearly, the NEt2 group is i nvolved. The nitrogen is a nucleophile and can do an internal nucleophi lic
substitution (S N i), a very fast reaction for entropy reasons because two different molecules do not have to
come together.
very fast
3
The slower step is attack of HO- on intermediate 3; the N is a good leaving group because it has a
positive charge. Where will HO- attack 3? On the less substituted carbon, in typical SN2 fashion.
:NEt2
I
H 2C - CH - CH2CH3
OH
I
2
This overall reaction is fast because of the neighboring group assistance in fonning 3. It is second order
because the HO- group and 3 collide in the slow step (not the only step, however). And the NEt2 group
"migrates", although in two steps.
6-71 The symmetry of this molecule is crucial.
(a)
Ph H
Ph
I I I
CH3 - C - C - C - CH3
I I I
H
Br H
Regardless of which adjacent H is removed by t-butoxide, the
product will be 2,4-diphenyl-2-pentene.
(b) Here are a Newman projection, a three-dimensional representation and a Fischer projection of the
required diastereomer. On both carbons 2 and 4, the H has to be anti-coplanar with the bromine while
leaving the other groups to give the same product. Not coincidentally, the correct diastereomer is a meso
structure.
H
Ph
Br H
CH3
H
2
H3C
H
Ph
Ph
Br
H and Br are anti-coplanar
Ph
�
129
H3C
CH3
Ph
Br
Ph
3
CH3
H
H
6-72
"
(a)
Ph "
phenyl
Ph
I
CH3CHCHCH3
I
=
NaOCH3
...
E2
Ph
I
CH3CHCH = CHz
Ph
CH3-C = CHCH3 +
I
major
Br
__
(b) H and Br must be anti -coplanar in the transition state
,H
H
2R,3R
f
K
Ph
CH3
\
\\
Ph
Ph
H
Br
Br
H
�
�
H
CH3
CH3
------
Ph I
II C"
C
CH3""- - . 'CH3
\\
\\
" 1
methyls
CH3
Br
th
�
CH3
Na OCH3
\\
\\
H
(c)
CH3
�
CH3 ---{
Ph
V
-+
r..
rru nor
Ph
'r
H
H
';(
H
CIS
CH3
CH3
Ph "
\\ CH3
II :=
C
C"
CH3
CH3.......
'H
'1
,\
methyls trans
2S,3 R
(d) The 2S, 3S is the mirror i mage of 2R, 3R; it would give the mirror image of the alkene that 2R, 3R
produced (with two methyl groups cis). The alkene product i s planar, not chiral , so i ts mirror image is the
same : the 2S, 3S and the 2R, 3 R g i ve the same alkene.
130
6-73 All five products (boxed) come from rearranged carbocations. Rearrangement, which may occur
simultaneously with i on ization, can occur by hydride shift to the 3° meth y lcyclopentyl cation, or by ri ng
6
expansion to the cyclohexy l cation.
H
n
C H 2 - Br
-
Br
..
6
+
�H� �
6 (y
�
hydride shift
H
�� H 2
H2
�··
H
H�CH3
..
H
___
&H
\
6.
b:
:\
X' �
H
CH3
C
CH3
A
��CH3..
U
U
+
C H2
-
H CH3
H6
6
CH3
alkyl shift (ri ng expansion)
)q �
--
(J'-- U
H
H
H
�
O
o
H
CH3
H
�
O
+
H
�
H C H3
..
..
II
(t.�
H3 �
qC
HOCH 3
•
• •
-
O
CH3
6-74 Begin with a structure of (S)-2-bromo-2-tluorobutane . Since there i s no H on C-2, the lowest
priority group must be the CH3 . The Br has highest priority, then F, then C H2 CH3, and CH3 i s fOUlth .
Sodium methoxide i s a strong base and nucleophi le, so the reaction must be second order, E2 or SN2 .
(a)
H
H3C
�
Br
CH3 NaOCH3
----l..
�
F
H
H3C
Br
ffi
H3C '<;Y- F
H3C
­
( \/ ,
H
H
13 1
H
NaOCH 3
-----
H 3C
H3C
( '> / )
-
6-74 continued
In regular structural formulas, the reaction would give three products i nc l uding the stereoisomers shown
above .
�
Br
+
F
(5)
(b)
3
4
l�
Br
3
4
l�
NaOCH3
F
F
OCH3
In these structures, the numbers 1 to 4 indicate the
group's priori ty in the Cahn-Ingold-Prelog system.
(5)
(5)
A cursory analysis of the designation of configuration would suggest to the uncritical mind that this
reaction proceeded w ith retention of configurati on-but that would be w rong ! You know by now
that a careful analysis i s required. In the Cahn-Ingol d-Prelog system, the F i n the starting material
was priority group 2, but i n the product, because Br has left, F is now the fi rst priority group. So
even though the designation of configuration suggests retention of configuation, the molecule has
actual ly undergone invers i on as would be expected with an S N 2 reaction . (See the solution to
problem 6-2 1 for a s i mi l ar example . )
6-75
(a) Only the propagation steps are shown . NB S provides a low concentratin o f Br2 which generates
bromine radical in ultraviolet l i ght. In the starting material, all 8 allyJic hydrogens are equ ivalent.
H
H
I
Br ·
recycles
I
Br - Br
'---I
Br
+
Br
+
H
Br ·
H
+ Br ·
(b) The first step in thi s first-order solvolysis is ionizati on , followed by rearra ngement.
H
H
hydride
C 30H
H
shift
+
---....... B r
+
H
II
0?Cr CH2 H
_
mechanism continued on next page
-!CrCH2
132
allylic !
6-75 (b) continued
El
--:
H
H3 H +
H
�
�
SN I
H
�
�
,
- CH2
t
&
H
CH2
CH2
+
(j-
CH30H
• •
�H2
{y
H3C
� �
H .O - H �
CH3� H
CH3� H
CH 2
.
.
··
�
1 I
Cf
H
..
H
•
(:Y
H
H3C
}
�
•
CH2
'3
_ CH 2
U+
CH \�H2
OCH3
\
CH3•OH
•
+
X
� H�
1+
CH30H2
•
� C1 H2
U
+
CH1 0H2
. +
(c) This first-order solvolysis generates an allylic carbocation in the ionization step.
EtOH
..
Br
•
•
�
H
! Et�H
Et ,• • / H
H
+
�
��
Et H
The El is shown on the next page.
H
• •
?H)
,..,
CD
! Et�H
Et
OEt
OEt
l 33
+
EtOH2
+
co
These are the SN 1 products.
+
Et OH2
+
6-75 (c ) continued
El
H
H
+
EtO H
EtOH 2
+
..
Q<> :_ (0)
(d) The first step in this first-order solvolysis is ionization, fol lowed by rearrangement.
C 30H
_
__
�
H
Br
Br
.
rearrangement by alkyl shift
-
Q(>�� Q<>
CH3�:
�
� � Qj
'
H
CH3
H
cp
�
H3C
:O
•
•
�
)
CP
OCH3
E l on rearranged carbocation
G;1
H
�
�
CH3 H
H�
((\
+
�
H
134
H
CH30H2
+
OCH 1
SN 1 product from
unrearranged
carbocation
H3C
CH3� H
H�
O- H
j 'Y
SN 1 on rearranged carbocation
�
�
H
�
+ �.�.
:
+
CH30H 2
+
CHAPTER 7-STRUCTURE AND SYNTHESIS OF ALKENES
7- 1 The number of elements of unsaturation in a hydrocarbon formula is given by:
2(#C) + 2 - (#H)
2
2(6) + 2 - (1 2)
2
=
1 element of un saturation
(b) Many examples are possible. Yours may not match these, but all must have either a double bond or a
ring, that is, one element of unsaturation .
o
2(4) + 2 - (6)
2
7-2
=
2 elements of un saturation
H
H3 C
/ C = C = CH2
1:
7 -3 Hundreds of examples of C4�NOCI are possible. Yours may not match these, but all must contain
two elements of unsaturation .
�
CI � O y NH2
\d
CI
N - C = C - CI
�
OH
CI
�
C
U
N =O
N
OH
7-4 Many examples of these formulas are possible. Yours may not match these, but correct answers must
have the same number of elements of unsaturation.
�
(a) C3H4CI2
=>
C3�
=
C4HgO
=>
CI � CI
Cl
Cl
(b)
1
C4H g
=
�
o
1
o
OH
�
13S
7 -4 continued
(c) C4H402 => C4H4
HC
(d) C s H s N0 2
=>
3
=
II
C - C - OCH3
C s sH s
[r
°
4
=
0 , + ,.... 0
'
°
(e) C6H3NClBr
HC
=>
C
C - N H,
C6 s H S
=
N
N
°
5
o
Br
C - C == C - C
I
I
B r NH
I
C
CH
Cl
)U
U
N == C == CH2
Br
Cl
O
C-N
A
Cl
Note to the student: The IUP AC system of nomenclature is undergoing many changes, most notably
i n the placement of position numbers. The new system places the position number close to the
functional group designation, which is what this Solutions Manual will attempt to follow ; however,
you should be able to use and recognize names in either the old or the new style. Ask your instructor
which system to use.
7-5
(a)
(b)
(c)
(d)
(e)
4-methylpent- l -ene
2-ethylhex- l -ene
penta- l ,4-diene
penta- l ,2,4-triene
2,5-di methy\cyclopenta- l ,3 -diene
(f) 4-vinylcyclohex- l -ene (" 1" is optional)
(g) 3-phenylprop- l -ene (" 1 " i s optional)
(h) trans-3 ,4-dimethy\cyclopent- l -ene (" 1 " is optional)
(i) 7-methy\enecyclohepta- l ,3 , 5-triene
7-6 (b), (e), and (f) do not show cis, trans isomerism
(a)
�
(Z)-hex-3-ene
(cis-hex -3 -ene)
( c)
\d\
(2Z,4Z)-hexa-2,4-diene
cis, cis-hexa-2,4-diene
(d)
;=C
(Z)-3-methylpent-2-ene
�
\
(E)-hCX -enc
(trans-hex-3-ene)
(2Z,4E)-hexa-2,4-diene
cis, trans-hexa-2,4-diene
(2E,4E)-hexa-2,4-diene
trans, trans-hexa-2,4-diene
" Cis" and "trans" are not clear for this example;
"E' and Z are unambiguous.
"
(E)-3-methylpent-2-ene
136
'
7-7
(a)
�
IL
-
2,3 -dimethylpent-2-ene
(neither cis nor trans)
CI
(d)
'0
5 -chlorocyclohexa- l ,3 -diene
(positions of double bonds need
to be specified)
(c)
(b)
3 -ethylhexa- l ,4-diene
(cis or trans not specified; the vinyl
group is part of the main chain)
I -methylcyclopentenc
(f)
(e )
cis-3 ,4-dimethylcyclohexene
(could also have drawn trans)
(E)-2,5-dibromo-3 -ethylpent-2 -ene
(cis does not apply)
7-8
(a)
'x
CI
Br
(E)-3-bromo-2-chloropent-2-ene
(b)
(2E,4E)-3-ethylhexa-2,4-diene
Br
/=C
(2)-3 -bromo-2-chloropen t-2-ene
(2Z,4E)-
(2E,42)-
(c ) no geometric i somers
(d)
(2)-penta- l ,3 -diene
(E)-penta- l ,3-diene
(e )
(E)-4-t-butyl-5-methyloct-4-ene
(2)-4-t-butyl-5 -meth yloct -4-ene
1 37
(2Z,42)-
7 -8 continued
(f)
CI
CI
CI
CI
(2E,5£)-3 ,7-dichloroocta- 2 , 5 -diene
(2Z, 5 £)-3, 7 -dich loroocta-2 , 5 -diene
CI
CI
(2E,5Z)-3, 7 -dich loroocta-2 , 5 -diene
(2Z,5Z)-3,7 -dichloroocta-2 ,S -diene
(g) no geometric i somers (an E double bond would be too highly strained)
(h)
W
W
(£)-cyc lodecene
(Z)-cyc Iodecene
(i)
W
( I E,S£)-cyc Iodeca- l , S-diene
C)
W
( 1 Z,S£)-cycI odeca- l ,5-diene ( I Z,5Z)-cyclodeca- l ,5 -diene
7-9 From Table 7- 1 , approximate heats of h ydrogenation can be determined for similarly substituted
alkenes. The energy difference is approxi mately 6 kJ/mole ( 1 .4 kcal/mole), the more highly substituted
alkene being more stable.
gem-disubstituted
1 1 7 kJ/mole
(28.0 kcal/mole)
tetrasubstituted
1 1 1 kJ/mole
(26.6 kcal/mole)
7- 1 0 Use the relati ve values i n Figure 7-7 .
(a)
2 x (trans-disubstituted - cis-di substituted)
=
2 x (22 - 1 8) = 8 kJ/mole more stable for trans, trans
(2 x (S . 2 - 4.2) = 2 kcal/mole)
(b)
gem-di substituted - monosubstituted
2-methylbut- l -ene i s more stable
=
20
1 1 = 9 kJ/mole
(4. 8 - 2.7 = 2. 1 kcallmole)
-
138
7- lO continued
(c)
trisubstituted - gem-disubstituted
=
25 - 20 = 5 kJ/mole
(5 .9 - 4.8 = 1 . 1 kcallmole)
2-methylbut-2-ene i s more stable
(d)
tetrasubstituted - gem-disubstituted
o(CH3
CH3
=
26 - 20 = 6 kJ/mole
(6.2 - 4 . 8 = 1 .4 kcallmole)
2,3-di methylbut-2-ene i s more stable
7- 1 1
(a)
strained but stable
(b) could not exi st-ring s i ze must be 8 atoms or greater to include trans double bond
(c )
Despite the ambiguity of this name, we must assume that the trans refers to the
two methyls since thi s compound can exist, rather than the trans refeni ng to the
alkene, a molecule which could not exist.
(d)
stable-trans in l O-membered ring
(e) unstable at room temperature�annot have trans alkene i n 7-membered ring (possibly i solable at very
low temperature-thi s type of experiment is one of the challenges chemi sts attack with gusto)
(f) stable-alkene not at bridgehead
(g) unstable-violation of B redt's Rule (alkene at bridgehead in 6-membered ring)
(h) stable-alkene at bridgehead i n 8-membered ring
(i) unstable-violation of B redt's Rule (alkene at bridgehead i n 7-membered ring)
7- 1 2
(a) The di bromo compound should boi l at a higher temperature because of its much l arger molecular weight.
(b) The cis should boil at a h i gher temperature than the trans as the trans has a zero dipole moment and
therefore no dipole-dipole interactions .
(c) 1 ,2-Dichlorocyclohexene should boi l a t a hi gher temperature because o f i ts much l arger molecular
weight and larger dipole moment than cyclohexene.
7- 1 3
(a)
0
O�3
NaOH
•
acetone
mmor
E2
major
E2
139
7 - 1 3 continued
(b)
Cl
(c)
( d)
0
E2
E2
major
minor
20
NaOCH3
..
CH30H
hindered base gives Hofmann product as
major isomer
6] oE2
+
major product i s difficult to predict
for 2° halides
S N2
NaOC(CH3h
..
(CH3hCOH
0
NaOC(CH3h i s a bulky base and a poor
nucleophi le, minimizing SN2
E2
-
7-14
B ase:
=
-,
H
� Ph
H
"'�
" C-C
/
H3C "
Ph
�Br
-
H3C " " "
"" H
C === C "
"'"
'
Ph
Ph
cis
7- 1 5 (a )
substitution product-since the substrate is a
neopentyl halide and highly hindered,
the SN2 substitution is slow, and elimination is favored
1 40
E2
elimination
':r--f(
7 - 1 5 continued
(b)
H
jPhH
""
Ph "
Br
(CH3 CH2) 3 N :
Br
meso
���
(c)
Ph
Ph
C=C
"
/
H
Br
only alkene isomer
E2
Br
/
"
Cb
HO
(d)
NaOH
..
acetone
+ minor substi tution products
Ph
C=C
"
/
H
Ph
only alkene i somer
E2
B
Br
Ph
one enantiomer
of the d, l pair
(c)
/
"
+
minor s ubstitution products
H
SN2 i s the only mechanism possi ble because
no H can get coplan ar w i th C l ; an eli mination
product would violate B redfs Rule
See Appendix 1 i n thi s manual for
numbering and naming bicyc l ic systems.
Cl
H
H
Models show that the H on C-3 cannot be anti-coplanar with the Ci on C-2. Thus, this E2 elimi nation must
occ ur with a syn-coplanar orientation: the D must be removed as the Cl leave s .
7- 1 6 As shown i n Solved Problem 7 - 3 , the H and the Br must have trans-diaxial orientation for the E 2
reaction t o occur. I n part (a), t h e c i s i somer h a s the methyl i n equatorial position, t h e NaOCH3 can
remove a hydrogen from either C - 2 or C-6, giving a mixture of alkenes where the most highly substituted
i somer is the maj or product (Zaitsev). In part (b), the trans i somer has the methyl in the axial posi tion at
C-2 , so no elimination can occur to C-2. The only possible elimination orientation i s toward C-6.
�l
Br
(a)
I
� H
..
eIther H
can be removed
( b)
6
NaOCH3
..
C H3 0H
cj(.
Br
.
H3C
only this H can be removed
6
H
+
major
H
1
c5 c5
NaOCH3
..
CH3 0H
c5
Zaitsev orientation
minor
only alkene formed; stereochemistry of E2 precludes
other i somer from forming
14 1
7- 1 7 E2 elimination requires that the H and the leaving group be anti-coplanar; in a chair cyclohexane,
this requires that the two groups be trans diaxial. However, when the bromine atom is in an axial
position, there are no hydrogens in axial positions on adj acent c arbons, so no elimination c an occur.
c�
hydrogens lrans diaxial lo lhe Br
CH3
H
7- 1 8
B r -- axial
(X)'"
�
H
'
(D
H
Br
(a)
=> �
� T -1
H
Br
C:::::>
H
KOH
Br axial
+
�
= Br
� T � atOrial
H
H
(b) S howing the chair form of the decalins makes the answer clear. The top i somer locks the H and the O r
i nto a trans-diaxial conformation--optimum for E 2 elimination. The bottom i somer h a s B r equatori al where
it is exceedingly slow to eli minate .
7-19
c{(;
Mode ls are a b i g help for this problem.
Br
(a)
•
H
NaOCH3
..
CH3
the only H
trans diaxial
(b)
the only elimination
product
H
H
D
(from elimination of H and B r)
H
+
both trans diax ial­
gives two products
H
H
(from elimination of D and B r)
142
=
h
W
:
H
H
H
7-20
(
-
Ph
H
�-C �/
C,
B
0
�
o
" " " Ph
--- Ph " ," " CC
H """ - ' H
---
H
Ph
cis
transition state
both Br anti-coplanar
�
(-
- - - - - -
=
:0
only geometric i somer
possible from a
concerted mechanism
�----------�
�----------
stretched bond being broken or forme
Br - - - - - - - r
Br
H
0
:I:
Br
"':'" "
/
/,
0
:.10:
o
.....-
Ph
H
Ph
H -+----;;---1-- Ph
�
--
pi bond
..
�
Ph
.
Br
cis
Br
transition state
7-2 1
(5, 5)
H
( :.10:
o
o
r
o o
0
•
Br
H
H3C �
"
, C ::..::..: C , "
,H
8- , "
CH3
Br
H
- �/ B r
3C �
C-C
/
X"�" H
/,
B
CH3
0
, II , C-C '\\" , H
___ H """"
- 'CH3
H3C
cis
transition state
both Br anti -coplanar
------------��---------r.
stretched bond being broken or forme:0
- - - - - -
=
H
H3C
Br
:Qx
Br - - - - - - - I
Br
H3C
..
only geometric isomer
possible from a
concerted mechanism
H3C
H3C
,
,
H
:j:
.......
H3C
H3C
H
cis
Br
transition state
143
pi bond
7-22
(a)
�H
Br
NaI
acetone
Br
o
cyclohcxene
(b )
cis-hept-3 -ene
CH3
r"fyH
CH2CH3
Br
(d)
NaI
acetone
C-l
The two bromi ne atoms must be anti­
coplanar in order to e l i minate . Onl y the
bromines at
and C-2 fit that
requirement; the bromines at
and C-3
cannot eliminate .
C-2
(5)- 3-bromo- l -ethyl-3-meth y Icyc lohexene
<D
�· � H
(e)
rotation
Br
Br
NaI,
acetone
.
9
H
�
H
b
In the first conformation shown , the bromine atoms are not coplanar and c annot
eli minate. Rotati on in this l arge ring can place the two bromines anti-coplanar,
generating a trans al kene in the ring. (Use models ! )
trans-cyclodecene
7 -23 The stereochemical requirement of E2 elimination is anti-coplanar; in cyc lohexanes , thi s translates
to trans-diaxial . B oth dibromides are trans, but because the t-butyl group must be in an equatorial
position , only the left molecule c an h ave the bromines diaxial. The one on the right h as both bromines
loc ked into equatori al positions, from which they cannot undergo E2 elimination.
(CH3hC
#
�
H
Br
---
trans-diaxialcan do E2
(CH3hC
1 44
H
�
Br
r HI
trans-diequatorial--cannot do E2
CH
CH
3
��42U CH_3_ .. �
3
�
:
o
3
CH
2
CH
�
:
-CH
+
2
CH
?
.
?
0
cj..
_
H
CH:2CH3
(Br
II
O-CH
2
CH
3
H
H
H
2
CH
2
C
"
�
�
r'
CH
2
CH
:
R
·
H
�
(j
H�:�H -CH2CH3
]
CH
.. ()
C
CH
3
H
7
�:�-CH
2
CH
3
+
..
rY
3
+
3
.
.
:
Gf
C
O-CH
2
CH
3
:
(t,
�
�
d
H
H �- 2
H
H :O-CH2CH3 CH2CH}
°
H
CH3 � .
]
C
()
H
H
7
�-CH
2
CH
3
�
+ H
H 2CH3 CH]
H H �:�-CH
a
CH3H
H
°
H
3
]
C :O-CH2CH..3
r:f;
/
V h<
O
�
\.
•
I
�. I
(b)
...
Br
�
,
(I
�
J
�
I
..
c,
�
I
I
..
carbocation rearrangement
°2 _
H
I
same product mi xture as in PaIt (a)
• •
145
�
0H
• • Hf -OS03H
(J __
(a)
�
__
__
•
(f
H
..0 -H
1+
___
/).
+
G
/).
___
cis + trans
major isomer
(X •oH• •• _H-OS03H CfR-H
(c)
H
1+
�
t_�__
___
A
Ll
__
CH3
•
from unrearranged 2° carbocation
-.
HS04
H20
H/-OS03H
...�
H
H
H
1
C
� I)
H ,\
�H
-OS03o/
}
OS03H
/
7 -25 In these mechanisms, the base removing the final proton is shown as
that water removes this proton.
f
It i s equally possible
•
+
!
�
mi nor isomer
rearrangeh ydride shift
•
CH3
H
H
ct H
from rearranged 3° carbocation
greatest amount
146
least amount
7-26
(a) f..Go
tJ0J{
=
=
=
-
Tf..So
+
116,000 Ilmol - 298 K ( Il7 1IK·mol)
+
81 ,100 Ilmol
=
+
8l. l kl/mol (+19.3 kcaVmole)
f..Go is positive, the reaction is disfavored at 25°C
(b)
f..GIOOO
=
=
+ 116,000 Ilmol - 1273 K (117 lIK·mol )
- 3 2,900 Ilmol
=
- 3 2 .9 klima I (- 8.0 kcal/mole)
f..G is negati ve , the reaction is favored at 1000°C
7-27
(a) basic and nucleophilic mechanism: B a(OH}z is a strong base
(b) ac idic and electrophilic mechanism: the catalyst is H+
(c) free radical chain reacti on: the catalyst is a peroxide that ini tiates free radical reactions
(d) acidic and electrophi l i c mechanism: the catalyst B F3 i s a strong Lewis aci d
7-28
(a)
H
�
� H - OS03H
I
..
CH3 - C - C - C-OH
I
I
I
H
H
H
I
H
I
I
I
H
•
•
•
•
I
I
I
I
H
I
H
H
H
H
H
I
H
I
CH3-C-C - C-OH
I
....
I
I
+
\.::;
I
+
H
H
I
1+
CH3-C==C - CH3 ..
CH 3-C - C-C- H ..I--CH3-C-C-C
I...) +
I
H ) :\.
I
I-.-A I
E+Z
'----..- H
H
H
H
H
This 1 ° c arbocation may or may
not exist. It i s shown for clarity.
H
(b)
H
H
i
H
H
H
hydride
shift -
H
(j-
""
147
OCH3
+
NaBr
7-29
(a)
( b)
()
H
9.H H - OP03H2
�
------;..
6 \..
1H
rl!
vo
H H
.. �"
(J
�
+
c5-�� c5
H00S03H
.
Two possible rearrange ments
H
1
( CH2
H-C-H
I) H
C f7 �
H
+
H H,o:.
0 CJ
+
• •
H
H20
H
..
H H20:
•
•
(X
I
o+ c5
2. Alkyl shift-ring expansion
+
CH2
H- �J
H
1
H�"\
C rl
+ C
'H H,o�
O O
......
-
(c ) carbocation formation
�
HO: �
H 0oS03�
H
H
• •
H
�
,/'-..
�I�'H
T
H
'
contin ued on next page
148
H
H
Thi s 10 carbocation mayor may not
exist. It is shown for clarity
�20·
�
1. Hydride shift
......
"
+
H
("'1+
CH2-�H
G
c
+
H2 0
7-29 (c) continued
without rearrangement
H HI
r:-.
H2O:
��?H :
�H�
C H H2O:
+
H
1<H
H
H�
H
~ 1<H
H
+
-
H
H
..
H
w ith hydride shift
�
..
E+Z
H
H
H
H�
+(�
..
H20:
H HH -- .-.------- H
H �:
./"'--r�
H
H
El product from
unrearranged
carbocation
rearrangement by alkyl shift
El on rearranged c arboc ation
H
149
7-29 (d) continued
E1 on rearranged carbocation
�
�H20 :.
<:;LJ
00
+
H,o'
+
�
7-30 Please refer to solution 1 -20, page 1 2 of this Solutions Manual.
�
7 -3 1
(a)
(c)
(d)
Br
(h)
(g)
(f)
Br
�
r;
Br
0
1
(")
(e)
o
Z
E
�
7 -32
(a) 2-ethylpent-l-ene (number the longest chain containing the double bond)
(b) 3-ethylpent-2-en e
(c ) (3E,6E)-octa-l,3,6-triene
(d) (E)-4-ethylhept-3 -ene
(e) l-cyc lohexy lcyclohexa-l,3-diene
(f) (3Z, 5E)-6-ch loro-3-(chloromethyl)octa-l ,3,5 -triene
7 -33
(a) E
(b) nei ther-two methyl groups on one carbon
�
7-34
(a)
F
(E)-l-fluoroprop-l-ene
(Z)-l-fluoro­
prop-l-ene
0
F
A
�F
(c) Z
F
3-fluoroprop-l-ene
2-fluoroprop- l -ene
(d) Z
[>-F
fluorocyclopropane
(b) C4H 7 B r has one element of unsaturation, but no rings are permitted in the problem, so all the isomers
must have one double bond. Only four i somers are possible with four c arbons and one double bond, so
Y
y
Br
H
��
H �
Br
Br
.
B r - �/'o...'�
Br
�"
�
�
�Br
Br
(b) Cholesterol,
C27H460,
four must be rings"
Br
� Br
Br
�
�
Br
has fi ve elements of unsaturation . If only one of those is a pi bond, the other
150
7-3 5
(a)
(b)
~ ~
cis,trans
trans, trans
2£,4£
;=\'
trans,cis
�
2£,4Z
2Z,4Z
2Z,4£
cis,cis
This problem is i ntended to show the difficulty of using cis-trans nomenclature with any but the
simplest alkenes. Cis and trans are ambi guous: the first alkene in part ( a) is cis if the two similar
substituents are considered, but trans if the chain is considered. The £-Z nomenc l ature i s
unambiguous and i s preferred for a l l four o f these i somers.
7-36 (a) and (d) have no geometric i somers
(b)
(c)
trans-pent-2-ene
(E)-pent-2-ene
cis-pent-2-ene
(Z)-pent-2-ene
y
B r cis- 1 ,2-dibromopropene
�
1(Z)- 1 ,2-dibromopropene
trans- 1 ,2-dibromopropene
(E)- 1 ,2-di bromopropene
Br
Br
(f)
trans,trans-hexa-2 ,4-diene
(2£,4E)-hexa-2,4-diene
7-37
F
(a)
1
>===<
H
(c)
Cl 1 Cl
A
Br
7-3 8
( a)
H
o
Br
cis,trans-hexa-2,4-diene
(2Z,4E)-hexa-2,4-diene
H
F
>===<
cis,cis-hexa-2,4-diene
(2Z,4Z)-hexa-2 ,4-diene
F
F
H
di pole moment
Br
1
Br
Br
H
CH3
>===<
(b)
=
cis-hcx-3-ene
(Z)-hex-3-ene
trans-hex -3-ene
(E)-hex -3-ene
Br
(e)
�
H
0
CH3
>===<
Br
dipole moment
=
0
Cl 1 Cl
>===<
H
H
larger dipole moment
(no bromines opposing
the dipole of the chlorines)
( b)
maj or
+�
minor
15 1
(c)
major
mi nor
7 - 3 8 continued
(d)
mmor
major
major
H H
CH3-C-C-CH3
H OH
7-39 Only
(a), (b) and (d).
(a)
alkene i somers are shown. Minor alkene i somers would also be produced in parts
I
I
I
I
E+Z
II II
CH3 CH3
II II 3
CH3-C-C-CH
H H
CH3-C-C-CH3
(c)
Br
(d)
H
7 -40
(a)
�
-
hindered base
Br
Br
+ Zn
�
---....,..
E+Z
NaOH,
N aI, acetone
Zn,
tl
(d)
7-41
0
�
only product
HOC(CH3h
+ ZnBr2
tl
..
Br
+ NaBr +
Br
Br
Q'
H20
H H
CH3COOH CH3-C=C-CH3
I
I
OR CH3COOH
Q'0H H2S04,
(c)
en
Br
'"
(b)
+
H
NaOC(CH3h
(b)
(a)
H H
I
I
CH3-C=C-CH3
..
KOC(CH3h
..
2
Br ' hv
..
Q'
(b)
Br
0
0
0
KOC(CH3h
..
0
+
� +
J
\
�---�,,�---�
major
1\
152
minor
7-4 1 continued
(c)
�
+
�
(d)
�
)l
U + U
major
(e)
� �
U +U
maj or
rrunor
rrunor
The E2 mechanis m requires anti-coplanar orientation of R and B r.
7-42 The bromides are shown here . Chlorides or iodides would also work.
�
(a)
Br
7-43
(b)
Br
Y
(c)
�
(d)
Br
cSf
Br
( a) There are two reasons w h y alcohols do not dehydrate with strong base. The potential leaving
group, hydroxide, i s i tself a strong base and therefore a terrible leaving group . Second, the strong base
deprotonates the -OR faster than any other reaction can occur, consuming the base and making the
leaving group anionic and therefore even worse.
(b) A halide i s already a decent leaving group. S i nce halides are extremely weak bases , the halogen
atom is not easily protonated, and even if it were , the leaving group abi l i ty is not significantly
enhanced. The hard step is to remove the adj acent R, something only a strong base can do-and
strong bases will not be present under strong acid condi tions.
7-44
(a)
�
(b)
(c)
(d)
without rearrangement
)==I
�H
rearrangement
maJor­
Zaitsev
153
5 continued
7-4with
rearrangement
hydride
shift
7- 4
(a)
major­
Zaitsev
61,
/' ......
/'
major
(c )
+
"y
� +�
(d)
mInor
maj or
minor
maj or
minor
only product from E2anti -coplanar is
possi ble only from
carbon wi thout CH3 by
rem
+
(b)
(e)
C
3
)
D
minor
major
N
y
��
ax
l
(CH3hC -.£:::::i CI
eq
The conformation must be considered because the leaving
group must be in an axi al position. The t-butyl group is so large
that it must be equatori al , locking the conformation into the
chair shown. As a result, only the Cl at position is axial, so
that is the one that must leave, not the one at position 3. Two
isomers are possible but it is difficult to say which would be
formed in greater amount.
7- 4 7
H
� .�H�
.
V
etcH 0I
4
H
H'
,
'-
H
�
- H20
HSO4- ..
-
..
--�
_
OSO
ct
3H
H
154
()+C
H
/
�
_ _
+
'\ H
H
...
'
•
H 20 .
"
�
El works well because only one carbocation and
only one alkene are possible. Substitution i s not a
problem here. The only nuc leophiles are water,
which would s i mply form starting material by a
reverse of the dehydration , and bisulfate anion.
Bisulfate anion is an extremely weak base and poor
nucleophile; if it did attack the carbocation , the
unstable product would quickly re-ionize, with no
net change, bac k to the carbocation.
7-48
I
I
I
I
I
�
CH 3-C-C-CH3
HO
:9
o
I
CH3 CH3
CH3 CH3
I
1+
CH 3-C-C-CH 3
----
H+
HO
--I"�
"'9oH2
I
l
I
+
CH3 -C-C-CH3
HO
C H3
II
I�I
CH3 CH3
-H20
methyl shift
CH3
+
I
CH 3 -C-C-CH 3
I
CH 3-C-C-CH 3
I
: 0 : CH3
I
: OH CH3
The dri ving force for thi s rearrangement i s the great stabil ity of the resonance-stabi lized, protonated
c arbonyl group.
7 -49 NBS generates bromine w hich produces bromine radical. B romine radical abstracts an allylic
hydrogen, resulting i n a resonance-stabi l ized aUylic radical. The allylic radical can bond to bromine at either
of the two carbons with radical character. See the soluti on to problem 6-63.
NBS c::=>
CC
\.-I
H
hv
B r2
H2
Br
��
+
0
1a
2 Br
____
0
CH 2
HBr
<;:, H
� a
reCYCI
B ro
Br
7-50
Ph
� CHCH3
H
X
,
CH {" "
Ph
H
�
25,3R
\
H
Br
CH 2
+
NaOEt
-----l.�
CH3
+
H
Ph
+
'-r----/
�
CH2 CH3
Ph
B
H -CHCH 3
Ph
!
�
JYJ
A = E,Z mixture
�
Br
+
Ph
Ph
X
,
CH {" "
Ph 25 5
,3
Ph
Ph
A
=
Ph
+
E,Z mi xture
Ph
E2 dehydrohalogenation requires anti-coplanar arrangement of H and Br, so specific ci s-trans I s ome rs (8 or
C) are generated depending on the stereochemistry of the starting material . Removing a h ydrogen from C-4
(achiral) will give about the s ame mixture of cis and trans (A) from either diastereomer.
155
�
7 - 51
by 4 kJ/mote
more stable than
~
more stable than
by 16 kJ/mole
Steric crowdi ng by the (-butyl group is responsible for the energy difference. In cis-but-2-ene, the two
methyl groups h ave only s l i ght i n teraction. However, in the 4,4-dimethylpent-2-enes , the l arger size of
the t-butyl group crowds the methyl group in the cis isomer, increasing its strai n and therefore i ts energy.
7-52
() C1
endocyclic
\. tri substituted
V
V ()'
endocyclic
\tri substituted
exocyclic
disubstituted )
9 kJ/mole
V
exocyclic
tri substituted)
5 kJ/mole
A standard pri nciple of science is to compare experiments which differ by only one vari able. Changing
more than one vari able c louds the interpretation, possibly to the poi nt of invalidating the experi ment.
The first set of structures compares endo and exocyclic double bonds, but the degree of substitution on the
alkene i s also di fferent, so this comparison is not valid-we are not isolating si mply the exo or endocyclic
effect.
The second pai r is a much better measure of endo versus exocyclic stability because both alkenes are
tri substituted, so the degree of substitution plays no part in the energy values. Thus , 5 kJ/mole is a better
value.
7-53
(a) CH3CH 2CH =CH 2
(b) No reaction-the two bromines are not trans and therefore cannot be trans-diaxial .
(e)
(d)
0
(X)
bromines are diequatorial­
cannot undergo E2
H
H
No reaction-the bromines are trans , but they are diequatorial because of the
locked conformation of the trans-decalin s ystem. E2 can occur only when
the bromines are trans diaxial.
156
E2, the two groups to be eliminated must be coplanar. In confonnationally mobile systems like
7-54 Inmolecules,
acyclic
or in cyclohexanes, anti coplanar is the preferred orientation where the H and leaving
group are 1 80°apart. In rigid systems like-norbornanes, however, SYN-coplanar (angle 0°) is the only
possible orientation and E2 will occur, although at a slower rate than anti-coplanar.
The structure having the H and the Cl syn-coplanar is the trans, which undergoes the E2 elimination.
(It is possible that the other H and Cl eliminate from the trans isomer; the results from this reaction cannot
distinguish between these two possibilities.)
H
Cl
Cl
H
Cly'/syn­
�oplanar
H
Cl
extremely slow to eliminate­
H and C\ not coplanar
It is interesting to note that even though three-membered rings are more strained than four membered
7-55 three
rings,
-membered rings are far more common in nature than four-membered rings. Rearrangement
from a four-membered ring to something else, especially a larger ring, will happen quickly.
.
trans
without rearrangement
H�
+ 2 H20:..
� CH
y--
unstable 1 ° carbocation-­
short lifetime if it exists
at all
mmor
with rearrangement-hydride shift
H
H20:
(P�H2
3° carbocation, but
still in a strained
4-membered ring
•
157
minor
minor
7-55 continued
with rearrangement-alkyl shift-ring expansion
��H2
H
-�.
C "......-A
H2
("�-H�. O[ H
H2C-C-H
H20:
I
H
H
2° carbocation on
5-membered ring­
HOORAY!
H
('xCH/'RiI
. .
.�
�
t
-H 0
2
MAJOR
carbocation-terrible!-will rearrange; can't do
hydride shift, must do alkyl shift ring expansion
1°
=
of adjacent
abstraction
gives-bridgehead
Halkene
violates
Bredfs Rule
though it is 3°;
this is an unstable carbocation even
bridgehead carbons cannot be sp 2 (planar-try to make
a model), so this carbocation does a hydride shift to
2°, more stable carbocation
a
t hydride shift
this 2° carbocation loses an adjacent H to form an
alkene; can't form at bridgehead (Bredt's Rule)­
only one other choice
158
8-1 Major products are produced in greatest amount; they are not necessarily the only products produced.
Br
(c)
(b) Cl
(d)
(a) Br
H
B'
+H
�H
<r ¢Til
CHAPTER 8-REACTIONS OF ALKENES
8- 2 H
H
H
H
H H
H+
---
H
H
O
H
+
allylic
produced in about equal amounts
(mixture of cis and trans)
H
H
-t�
H
H-C+ H
H
.Q
H H
H H
:Br:
1 .. H H
H -O- H
Br H H H
·
\
+
1-bromobut-2-ene
3-bromobut-l-ene
Because the allylic carbocation has partial positive charge at two carbons, the bromide nuc\eophile can
bond at either electrophilic carbon, giving two products.
8-3
�
II
II
II
L1
2 H3C-C-O·
(a) initiation steps H3C-C-0-v' O-C-CH3 ••
°
o
+
H3C-C-O·
H-Br
"�JG
°
II
propagation steps
�
~
C�
_)
+
• Br
(this radical has
another resonance
fonn but it is not
gennane to this
mechanism)
°
-
+
• Br
the 3° radical is more stable
than the 1° radical
-
���
II
H3C-C-OH
••
°
-
159
B'
• Br
k
1-bromo-2-methylpropane
+
8-3 continued
(b) initiation steps
H2
H 3C-C -0·
• •
• •
+
+ • Br
H-Br
,,--JG
propagation steps
the 3° radical is more stable
than the 2° radical
+
•
Br
I-bromo-2-methylcyclopentane (cis + trans)
2
�
(c) initiation steps
-
+
H
propagation steps
I
-
/C
Ph' ·
y Br
I
�
I
H
Ph
H
•
Br
the benzylic radical is
stabilized by resonance, and
more stable than the aliphatic
radical
Br
+
• Br
2-bromo-l-phenylpropane
(recall that "Ph" is the abbreviation for "phenyl")
8-4
(a)
(c)
�
(y
OH
(d)
OH
�
HBr
ROOR
H2SO4
�
H2SO4
�
•
•
•
(b)
Br�
0
�
HBr
HBr
ROOR
•
160
..
�
Br
�
C)'
~
Br
HBr
..
Note: A good synthesis uses
major products as intermediates,
not minor products. Knowing
orientation of addition and
elimination is critical to using
reactions correctly.
CH3 H
�
CH3 H H
*
�
H ----"" HH
CH3 CH3 �'--O-H CH3 CH3 +
"'" +
1
H 2°
H
HI
8-5
CH3
+ I CH3
C CH3
CH3/
H
CH3
tf
CH3
CH3
't -H.. CH3
\[)oo
CH3
8-6
(a)
:O
I
H
\f
'-..
:O- H
H� I
H
>=<.H
CH
(a)
CH3
(b)
CH3
CH3
CH3tr
H
CH2CH3
3°
CH3
jf
CH3
CH3
U
0
1
3
,CH3
CH3/ "1'
CH3
O
H30+
H30+
+
0d:J
from 3° benzylic carbocation
CH3
(
CH3
Af
CH3
gOAC
CH2CH3
:
/ I
CH3
1f
0 0
CH3
i
�
H\
ACO :
NaBH4
+
H
(c)
..
..
HI
2, 3-dimethyl-2-butene
+
Hg(OAc)
CH3
I
0
�H
H'
( b)
from 3° carbocation
8-7
+
..
4 ....
'6H
CH3
+ I CH3
/ C � CH3
.. CHJ
°
O
-H
O
HO
/ ')H
I
/
H
H
k� H 2,3-dimethy-1 2-butanol
CH3
I CH3
;C
CH3
proton removal
..
•
nucleophilic attack by water
CH3
methyl
shift
16 1
CH3
gOAC
H
CH2CH3
o
I
H
H3C-CII
°
"Ac" = acetyl
II
H3C-C'-O
o
"OAc" or "AcO" = acetate
8-8
CH 3
(f'::
HO
(a)
(d)
"
(OAC)
(b)
�
&,
(b)
(c)
Hg(OAc}z
CH30H
KOH,�
.-
.-
Hg(OAc}z
8-10
�
(a) , (b)
6
(e) , (f)
8-11
(a)
(b)
(c)
..
�
�
�
Br
..
..
B H3
°
THF
-----;.-�
H2O
N aB H 4
..
OH
C
B H2
H202
.-
Hg(OAc}z
NaBH 4
..
KOH,�
..
.-
Hg(OAc)
o.
H2 02
HO-
,
H
O
� OH
.-
HO-
HaB:
H202
I
l
Hg(OAC)
"
OCH3
Using an acid-catalyzed hydration In part
(c) would initially form a 2° carbocation
that would quickly rearrange to 3" on
carbon-3. The desired product would not
be synthesized.
�
�
HO-
H2 O
�
.-
B H2
BH3 0TH F
O
:
OCH 3
Hg(OAc}z
�
B H3 0T HF
OCH3
II
OCH J
NaBH4
6
CI
(c)
H
B H3 0THF
�
(c), (d )
U
O-
CI
8-9
(a)
X
H
H20 2
.-
�
HaO:
OH
C
.-
HO-
HO�
�
OH
..
�
B H3
°
THF
.-
H2 0 2
H O-
162
.-
~
OH
8-12 Instead of borane attacking the bottom face of I-methylcycJopentene, it i s equally likely to attack the
top face, leading to the enantiomer.
s=<:.
6
BH3 · T HF
..
H
8-13
(a)
CH3
(b)
,'QH
H
V �
+
OH
OH
mi nor-steric
hindrance to
attack of B H 3
major
-
CH3
S
8-14
(a)
Et
Et
>=< H
Et
B H3 · THF
Me
H
Z
S
� �
K
''
Me
Et
"
OH
H
)
enantiomers
R
( b)
Et
H
>=< Et
Me
S
� �
�K�
Et
B H3 • THF
H
Me'
Et
�---------..
E
H
V
enantiomers
The enantiomeric pair produced from the Z-a l kene i s diastereomeric with the other enantiomeric pair
produced from the E-al kene. Hydroboration-oxidation is stereospeci fic , that i s, eac h a l kene gives J specific
set of stereoi somers, not a random mi xture.
8- 15
(a)
(b)
CO
CO
Hg (O Ach
NaBH 4
...
...
H2 O
CO
OH
B H3 · THF
..
H 202
HO -
..
cp
OH
163
continued
C
(c) o
8 -15
8-16
c5
CH3 CH3
��
empty p orbital in \bottom
"'-- :Br:
planar carbocation
�
CH3 Br
The planar carbocation is responsible for non-stereoselectivity. The bromide nuc\eophile can attack from
the top or bottom, leading to a mixture of stereoisomers. The addition is therefore a mixture of syn and anti
addition.
8-17 During bromine addition to either the cis- or trans-alkene, two new chiral centers are being formed.
Neither alkene (nor bromine) is optically active, so the product cannot be optically active.
The cis-but-2-ene gives two chiral products, a racemic mixture. However, trans-but-2-ene, because of
its symmetry, gives only one meso product which can never be chiral. The "optical inactivity" is built into
this symmetric molecule.
This can be seen by following what happens to the configuration of the chiral centers from the
intermediates to products, below. (The key lies in the symmetry of the intermediate and inversion of
configuration when bromide attacks.)
IDENTICAL-SYMMETRIC
.
�Br
" H
H"'�M
H3 GH3 �'
a Be nuc\eophile could : r OR :Br
:
, �:
(
��--------�
"\
�------�
attack at either carbon
CH3 H-
.
+
l
J
+
H",:\
Br+
H CH3
Br
Br
Br
H
f
+
'---<'
Br + H
'" ",
H
'H
�
/
Br
Br
CH3
CH3
CH3
2S,3S
ENANTIOMERS 2R,3R
continued on next page
-+
':r-(
H
�·
·
I
-+
H
Br
H + Br
CH3
--------------------------------------------------______________________________________ u __________ u __ •
164
8-17
continued
ENANTIOMERIC
�
TRANS
\
CH � CH3
( B!
H
U
"" ,U "
\B/ bromide attack on this
1V\""11"'CH+
H'/CH3
3
r bsame
romonium
ion gives the
+
results
a Be nucleophile could :Br: OR :Br:
attack at either carbon
�------�
,
�(
t
�---------
'
••
J
H
H CH H
Br
Br )1-J,
Br
Br
H CH3 �
+
H + Br
Br + H
\'\�
/
�"
'
H
CH
3
Br
Br
H
CH3
CH3
CH3
2R,3S
ID ENTICAL-MESO 2R,3S
CONCLUSION: anti addition of a symmetric reagent to a symmetric cis-alkene gives racemic product,
while anti addition to a trans-alkene gives meso product. (We will see shortly that syn addition to a cis­
alkene gives meso product, and syn addition to a trans-alkene gives racemic product. Stay tuned.)
+
+
Enantiomers of chiral products are also produced but not shown.
Br
..
� + :Br:
(a)
Br-Br
8-18
(b)
(c)
(Hex)
+
+
0�.
Br\
+
Br-Br --- , U )
-
___
Br
0 Br
"'Br
�e1-I
'>---f/
:
:Br
'" Br
H"'/
.
�
"
It
"1
H" Et
�' Hex . Et
Bromide will attack
the other carbon of the
bromonium ion as well.
165
�
continued
(d) Three new asymmetric carbons are produced in this reaction. All stereoisomers will be produced with
the restriction that the two adjacent chlorines on the ring must be trans.
Cl
Cl
8-18
-
+
--G--I<
+
Cl-Cl
..
l
:Cl:
j
:Cl:
h r�Cl I
Cl�
CH3
C
CH �
8-19 The trans product results from water attacking the bromonium ion from the face opposite the
bromine. Equal amounts of the two enantiomers result from the equal probability that water will attack
either C-l or C-2.
B, - �r <:v
+
Q
H
H
Q�
water will do nucleophilic
attack at either carbon
..
H20:
a
+
equal amounts of enantiomers
enantiomer of product
H
8-5
°\O+_ H �\ H in Solved Problem
:OH
O
from Solved Problem 8-5
H
�O
O-H
the bromonium ion
: -H �
��
"",CH3
CH
...
shown here is the
CH3
..
3
�+
U
enantiomer of the one
Br
\
\
Br
>
Br>
shown in Solved
\
Problem 8-5
H
H
from Solved Problem 8-6: the bromonium ion shown is meso asHit has a plane of symmetry; attack
by the
nucleophile at the other carbon from what is shown in the text will create the enantiomer
H
H
:Cl:
"'Br
+
..
,Br
8-20
0
a',
-
//� �
show products from
HI;I �
both nucleophiles
attacking at this carbon
oR
0
I
-
H
U
0 0
a
�;
Br
'
"
a
H
OH
'"' H
H
these are the enantiomers of the
structures shown in the text
166
8-21 The chiral products shown here will be racemic mixtures.
H
(b) 1
(C)H3C
-;;61
plus enantiomer
(e) CH3 ,OH
Rr +
8-22
(a)
(y
�
O
H
HO
CH3 ,Cl
plus enantiomers
CH
(c)
Br
(y Rr
CI2
H2O
H2SO4
l
(b)
C
..
~
..
6
C1
H
�
plus enantiomer
_
C
CI
6 0
..
..
KOH
(y
(c) C):)
CI2
(d)
CH3 ,OH
!t,.
,----< '
H3C l-!
Cl
\"H"",'CH-�
HO/
plus enantiomer
.
6
.
..
CI
CI2
H2O
0H
Cl
8-23
(a) �
(d) -0-<
(b) �
8-24 Limonene, CIOH16, has three elements of unsaturation. Upon catalytic hydrogenation, the product,
CIOH20, has one element of unsaturation. Two elements of unsaturation have been removed by
hydrogenation-these must have been pi bonds, either two double bonds or one triple bond. The one
remaining unsaturation must be a ring. Thus, limonene must have one ring and either two double bonds
or one triple bond. (The structure of limonene is shown in the text in Problem 8-23(d), and the
hydrogenation product is shown above in the solution to 8-23(d).)
8-25 The BINAP ligand is an example of a conformationally hindered biphenyl as described in text
section
5 -9A and Figure 5 -1 7 . The groups are too large to permit rotation around the single bond
connecting the rings, so the molecules are locked into one chiral twist or its mirror image.
"rear"
!t,.
H2O
"front"
"front"
naphthalene
naphthalene
ring
nng
These simplified three-dimensional drawings of the enantiomers show that the two naphthalene rings arc
twisted almost perpendicular to each other, and the large -P(Phh substituents prevent interconversion of
these mirror images.
167
8- 26
(a)
(b)
(c)
8- 27
Methylene inserts into the circled bonds.
S
c5
+
� + :CH2
�H
~
d
+
( b)
(a)
BOTTOM VIEW
site of origi nal
double bond
(c)
__
the original 6-membered ring
is shown in bold bonds
SIDE VIEW
8- 28
(a)
( b)
(c)
(a)
8- 29
F
0
OH
6
0
CH21 2
Zn(Cu)
CH2Br2
50% NaOH (aq)
•
R
•
H2SO4
L!
•
(b)
Br
�H
0
0
Br
CHCl3
50% NaOH (aq)
•
(c)
168
cjCl Cl
CHo
(d)
CQ
H
"
,
o
H
(a)
Rc
' =o
8-30
fa . H
V ' 0l...''\
H "C ( Ct" H
Et" cis 'Et
"
'1
•
,
\"
ENANTIOMERS
A"Et" H
H""Et,
r--------A.......
'
J-<Ofl fl0j---P -(Et
Et �"
(b)
:O: �
HO
'
Rc
, =o
EtH
I,
II/
+
HEt
\"
'o'
H ,v
('\t "Et
c" "
"c
Et" trans
- 'H
,
OH
,\\
fa l.... H
'I
\
•
:O:�
,
"
H' A�" t
Et
IDENTICAL-MESO
"
Et
�Ofl floyfl
HO�H" Et HEt"'/ �OH
stereochemistry shown in Newman projections:
+
H'�f\]\
E;H
t
H-O: OR :O-H
H tI H
"
H+
Etf!
"
I
Ofl
,---<" H
H O�Et
+
2;.
fl0
+
Et
•
• •
:O-H
)(
'>-,!-I ---Y
HEt"7 "OH2
+
---
E
r�--------A------�\
�
(
)
"'
I
-H +
"
"
-
•
H+
+
---
,
----�
:O
u-
U
(\�" Et
H'�1\
t
H-O: OR :O-H
H + H
"
"
.
,
.
I
,
"
..
I
El
f!
H
Et
flO
Ofl
�
- H + ,--<
+
H 0. / "HEt HEt"'/ "O H 2
+
•
2
-
• •
"
Q)H
• •
OH
OH
Et
H rotate
Et
H
..
H
Et
M
H
�
:
£
Et
H
Et
H � Et
OH
trans
MESO
Remember the lesson from Problem 8-17: anti addition of a symmetric reagent to a symmetric cis-alkene
gives racemic product, while anti addition to a trans-alkene gives meso product. This fits the definition of a
stereospecific reaction, where different stereoisomers of the starting material (cis and trans) are converted
into different stereoisomers of product (a dl-pair and meso form).
v
x:
---
---
H�
H-
169
Et
8-31
R
G0,
c t )�
----/'
H
"
"
"
8-32
(a)
'
C=O
t...H
0'
trans
==
I
C
•
''''CH1
�
8-33
�
H
/
"" ,
� \-
o
H
CH 3 CH2 �
OH
Ho
�
'-/
/
��CH2CH3
HO
�
O
HO",
H " 'O H
H+
t
CH2CH 3
H
�
�
\. or )
H
�
H
OH
- H+
HO
M
C-O
II
�
(d) � rotate.
cj
\
o
<
R=
C-O
o
�
o
,
II
All chiral products are racemic mixtures.
(b)
o
(c)
R
g
2
+
anti addition
trans-alkene gives
meso product
to
H
H
CH2CH 3
t�is is the (e) � trans
product � OH
OH
H
H
CIS
�
+�
--- 1-Y
enantiomers
8-34 All these reactions begin with achiral reagents; therefore, all the chiral products are racemic.
(c) H
(b)
(a)
H
H
---
H
• •
H
--
CH 3 -O:
CH3 0
H
H
I
H
(d)
():
CH 3
HO
OH
H
OH
H
'"
o:�:
H
CH2CH3
=f��
HO
H
OH
H
(e)
CH 3
�
OH
OH
H
same as (d)!
HO
H
HO
(f)
'"
H
CH2CH 3
=f��
H
Refer to the observation in the solution to Problem 8-35 on the next page.
CH3
HO
H
H
CH3
170
H
CH3
H0
Hl
+
OH
same as (c)!
HO
+�
CH3
X
�
OCH3
HO
/
. � "'CH2CH 3 ==:>
HO
CH1
a
=f�H
� CH2CH l
HO
H
CH3
l
H
8-35
�
(a)
HO
HO
(c)
(d)
�
�
OH
�
(b)�
CH3C03H
..
H2 O
rotate
Os04
rotate
H20 2
..
..
..
meso
OH
�
HO
racemic (d,l)
OH
�
HO
meso
racemic (d,l)
OH
Have you noticed yet? For symmetric alkenes and symmetric reagents (addition of two identical X groups):
cis-alkene + syn addition
cis-alkene + anti addition
meso
----7
----7
trans-alkene + syn addition
trans-alkene + anti addition
racemic
----7
racemic
----7
meso
Assume that cis/synlmeso
are "same", and trans/anti!
racemic are "opposite".
Then any combination can
be predicted, just like math!
+1
+1
-1
-1
+1
x
+1
=
x
-
1
=
-1
x
+1
=
-1
=
+1
x
-1
8-36 Solve these ozonolysis problems by working backwards, that is, by "reattaching" the two carbons of
the new carbonyl groups into alkenes. Here's a hint. When you cut a circular piece of string, you still have
only one piece. When you cut a linear piece of string, you have two pieces. Same with molecules. If
ozonolysis forms only one product with two carbonyls, the alkene had to have been in a ring. If ozonolysis
gives two molecules, the alkene had to have been in a chain.
(a) two carbonyls from
ozonolysis are in a chain, so
alkene had to have been in a
nng
(b) two carbonyls from
ozonolysis are in two different
products, so alkene had to have
been in a chain, not a ring
(c) two carbonyls from ozonolysis
are in two different products, so
alkene had to have been in a chain,
not a ring
E or Z of alkene cannot be
determined from products
8 - 37
(a)
�
03
�
Mc,�
�o
(b)
o�
H
+
�
OH
O
17 1
8-37 continued
0=0
(c )
U
(d)
(e)
(f)
03
Me2S
Me2S
03
•
•
=0
0
�H
0
•
•
U
U
KMn04
t!.
cold
�OH
•
0
0
•
0
0
0
KM n04
+
d
·aH
1I110H
8-38 The representation for a generic acid w i l l be H-B, where B is the conj ugate base.
:x:r+}
+/
J. C
/ B:-
H
P Te\ � /, -
8-39 Catalytic B F3 reacts with trace amounts of water to form the probable catalyst:
-
F
I
+1
dimer
_
_
catalyst
etc.
...
tetramer
trimer
172
H
I
8-40 H-B is used here to symbolize a generic acid.
-
:B
�
alkenes
polymers
�
(colored)
Note : the dashed bond symbol is used here to i ndicate the continuation of a polymer chai n .
8-41
H
I
------
H
I
+
C-C ·
I
H
\
H
/
n/)\...( \H
�
H
I
H
I
H
H
Ph
Ph
H
I
------ C-C-C-C ·
C=C
I �
Ph
I
H
H
I
I
I
I
10 radi cal , and not resonance-stabilized­
this orientation is not observed
Orientation of addition always generates the more stable intermediate; the energy difference between a 10
radical (shown above) and a benzylic radical is huge. The phenyl substituents must necessarily be on
alternating carbons because the orientation of attack is always the same-not a random process.
8-42 For clarity, the new bonds formed in this mechanism are shown in bold .
\
H
RO ·
+
H
2 RO·
RO - OR
/
C=C
.!)\...( \H
��
---t.�
H
C
C
�'"o"+-",,
, /V \
I (
H
H
I
H
H
H
H
H
/
\
..
RO-C-C.
I
I
H
I
H
H
H
H
H
I
I
I
I
I
c
/
\
l
/
I
H
I
H
I
H
I
H
I
H
H
H
H
H
H
H
I
I
I
I
I
I
I
etc. ---- RO-C-C-C-C-C-C·
173
)k)
RO-C-C-C-C ·
H
H
I
H
\H
8-43
II
o
\ n/
H
COCH3
C=C
HO :
.. �/
H
---
\
Me
. .-
:0 :
II
H
C-OCH3
I _I
11:
HO-C-C :
I
I
H Me
\
:0 :
I
H
C-OCH3
I
II
HO-C-C
I
I
H Me
.
o
II
COCH3
\ n/
C=C
\
/
H
Me
. .
· 0·
· ·
COOMe 11
H
H
C-OCH3
I
I _I
HO - C-C-C-C :
I
I
I
I
H Me H Me
I
•
•
!
· ·
· 0·
COOMe I
H
H
C-OCH3
I
I II
OO - C - C-C - C
I
I I I
H Me H Me
I
ele.
COOMe COOMe COOMe COOMe COOMe
Plexiglas
"
Me
Me
Me
Me
Me
D ashed bonds mean that the chain continues.
8 -44
0
0
II
H3C-C-00H
..
H30+
OH
a
"
<X ;Y
'
Na
-
OH
O
B r�
�
o
<X F
o
<X �
"'o
O
4
MCPBA
OH
174
OH
B r2
-
hv
�
8-45 In the spirit of this problem, all starting materials will have six carbons or fewer and only one carboncarbon double bond. Reagents may have other atoms.
(a)
Na .
::
O:::J O::�-:y
H BC�
major product from resonance
stabilized radical on 10 and 30 C
�
anti-Markovnikov
syn stereochemistry
�
0: �
CH,I, 0:
H
o
(b)
0
(c)
0
Bhvr ey Br
rB BC-p KOH
Br
2
(or
•
NBS)
2
C
�
3
Zn,
CuCI
Bhv·r 0- Br
( a)
(d)
KOH
2
•
-
NBS)
MCPS:
(or
only substit
is possible
,A/
° �
T
H
Ao 1
+
0
175
+ Na
-
0
-
H2 0
�:3 _aCH3
" "
,
BH
OH
2
H
intermediate
Cl
L5F
(g) r--peroxides do not
affect HCl addition
(c)
(f)
on
0H
./
�
Please refer to solution 1-20, page 12 of this Solutions Manual.
Br
(el e/'
:
o-
� \{ 0OH
hi
8-47
CH3
eyC::N
�
8-46
o
q
II
,
V
continued
CH3
'"
O'OH
""OH
8-47
(h )
CH3
(i) ('f. IIOH
V ""OH
QH
(I)
(0)
q;
HgOAc
intennediate
(a)
initiation
RO-OR
2 RO·
:JnBr
.. ROH
+ H
RO.0
propagation
8-48
r-,ri.
----
-
Br·
H
+�
r�
H
+
(n)
CO
OH
(p)
q;Cl
Br·
Br
HH
L"
+�
C
�H--C:S03; Q '--- :9-H
( b)
H
176
-..�
•
+
•
Br
8-48 continued
H
H-nBr
�
�
n
(c)
•
I
.I
(d)
H
1(..1- �
� :Br:
..
fH
'Y'
Br
----... Br
HO- H-C-Br --- :C-Br --- Br,
+
......
,.--.10..1
I
Br
I
f"
Br
I
�
+
(fit::
I
:B�:
.
.
:CBrz
B'
+
�
H
CI
H��CH3
X+C
�
CH3..O-H HOC;3 �
OCH3
H
�
•
I
•
_
B_r_-B" ; r:J
�
'--- .Br.
•
•
•
:
& .
�� :Ci:
+
F�'
•
Br
�
Br
Br
�
CI
.
�
CH30 -H H�CH3 Jy
OCH
• •
177
�
.
t
Br
'J'
I
OR
(f) �
�
•
aA
..
H
JY
:C1:
H--Q� HXH�
f/
C
H
H
H
hydride
shift
Br
�
00
� Brz
V
(e)
OR
'fH....,C... +
.
3
continued
(g) H � 0
. _H H-B
H J...
- <
T
8-48
H 3C
•
Ph
Recall that "Ph" is the abbreviation for phenyl.
(a)
8-49
(b)
(0)
(d)
(c)
(f)
(g)
(h)
(i)
(f
(f
(f
(f
(f
(f
(f
Hg(OAc)z
..
HEr
ROOR
Hg(OAc)z
..
NaBH4
NaBH4
..
.
OS04
H 20 2
or cold, dilute KMn04
..
Zn(Cu)
..
50%
NaOH (aq)
� OH
OH
V
� CI
OH
V
� Br
V 'Br
178
�H
H
Ph
Ph
HQ
"
8-50
(a)
CH3
-Q--("
�
OH
CH3
(e)
~
HO
H
(d)
(g)
CH3
,
OH
OH
B
-0-<
(h)
8
Br
OH
(j)
'
CH3
(m)
D--f
(c)
~
H
(f)
+ CO2
0H
OH
",
CH3
-b-\
(i)
Br
-b-f
Ci
Cl
"
�
CH,o
(I)
CH3
�
H
D-f
Br
C I�
(k)
+ CH2=O
HO
H
Cl
Br
HO "
D--f0
(b)
0
Br
OCH3
C 3
8-51 Each monomer h as two c arbons in the backbone, so the substi tuents on the monomer w i l l repeat
every two carbons in the polymer. Dashed bonds indicate continuation of the polymer chai n .
line formula representation
Cl
�
F
F
Cl
I
N
C
�
I
Cl
Cl
Cl
Cl
Cl
Cl
I
I
--- --- -CH2CHCH2 CHCH2CHCH2CH -- --
pol yvinyl chloride, PVC
0
F -;-III
Cl
F F
F
F
F
F
I
I
I
I
I
I
---- C - C - C - C - C -C---I
I
I
I
I
I
F
F
F
F
polytetrafluoroethylene, PTFE, Teflon
F
F
N
III
N
III
N
III
N
III
C
C
C
C
I
I
I
I
--- ----CH2CHCH2CHCH2CHCH2CH - - - -polyacry lonitrile , Orion
179
-
-
F
F
F
F
N
F
F
F
F
N
F
F
F
F
N
III
III
III
C
C
C
F
F
F
F
N
III
C
F
F
..
..
8-52 Without divinylbenzene, individual chains of polystyrene are able to sl ide past one another.
Addition of di vinylbenzene during polymerization forms bridges, or "crossl i nks", between chains, adding
strength and rigidity to the polymer. Divinylbenzene and similar molecu les are called crosslinkjng agents.
two polystyrene chains crossli nked
by a divinylbenzene monomer shown
in the dashed oval
H
CH
/ 3
\
H CH �\.. ,C/ vC\
I r
H
CH�
RO-CI -C.
I
HI CH
3
8-5 3 A peroxy radical i s shown as the initi ator. Newly formed bonds are shown in bold.
"-f--'
_-I.�
etc. ---
8-54
'C-OCH2CH3
I
H2C=CH
0,
ethyl acrylate
8 - 55 In each case, the compound (boxed) that produces the more stable c arbocation is more reactive.
(a)
(b)
0
3"
6
180
(c)
�
allylic
�
8-56 Once the bromonium ion i s formed, i t can be attacked by either nuc\eophile, bromide or chloride,
leadi ng to the mi xture of products.
Q
Br
--
~
Br
�
H
H
H
Br
��-�
�
'i---Y
�..
:Cl :
H
H
Cl
8-57 Two possible orientati ons of attack of bromi ne radical are possible:
(A) anti -Markovnikov
>=
+
•
Br
--
(B) Markovnikov
>=
+
•
Br
--
\ .
C,
/
Br
3 ° radical
'l-
r
� H2
HBr
.-
h
+
•
Br
Br
HBr
�
y;
+
•
Br
H
1° radical
The first step in the mechani s m i s endothermic and rate determi ning. The 3° radical produced in anti­
Markovnikov attac k (A) of bromine radical is several kl/mole more stable than the 1° radical generated
by Markovnikov attac k ( B ) . The Hammond Postulate tells us that it is reasonable to assume that the
activation energy for anti -Markovnikov addition is lower than for Markovnikov addition. This defines
the first half of the energy di agram.
The relative stabi l i ties of the final products are somewhat difficult to predict. (Remember that
stabi lity of final products does not necessari l y reflect rel ative stabi lities of intermedi ates; this is why a
thermodynamic product can be different from a ki netic product.) From bond dissoc i ation energies
(kJ/mole) in Table 4-2 :
anti -Markovnikov
Markovnikov
H t0 3°C
3 81
H to 1 ° C
410
Br to 1° C
285
666 kJ/mo\e
Br t0 3°C
272
682 kl/mole
If it takes more energy to break bonds in the Markovnikov product, it must be lower in energy, therefore,
more stable-OPPOSITE OF STABILITY OF THE INTERMEDIATES!
Now we are ready to construct the energy di agram; see the next page .
18 1
8-57 continued
t
Markovnikov
, ,'/
,
,
,
- ..
,
,
anti-Markovnikov
reaction
--
It is the anti-Markovnikov product that is the kinetic product, not the thermodynamic product; the anti­
Markovnikov product is obtai ned since its rate-determining step has the lower activation energy.
GS
8 - 58 Recall these facts about ozonolysis: each alkene c leaved by ozone produces two carbony l
groups; a n alkene i n a chain produces t w o separate products; a n alkene i n a ring produces one product
in which the two c arbonyls are connected.
(a)
O
H
+
(b)
CH2=O
O
+
0
�
H
o
CHO
0
H
0
8-59
(a)
(b)
CY
0
¢
(d)
H
0
CHO
CH3C0 3H
..
H2 O
CH3
U
'OH
OH
O
(or cold, dilute KMn04)
cis + syn
(cis double bond plus
syn stereochemistry of addition)
OH
OH
trans + anti
(trans double bond plus
anti stereochemistry of addition)
182
8-59 continued
B r2
(c )
� Br
� Br
..
�
1
This structure shows
a trans alkene in a
to-membered ring,
just the rotated view
of the structure to
the right.
Br2
(X
..
trans-cyclodecene
o
(d)
C l2
C!
tate around C-2
�'" :
H
� 2
Br
""
OH
CO
(e)
anti addition of B r 2 requires t ra ns alkene to
gi ve meso product
~
B H3 - THF
..
�
OH
(f)
�
V-/
CO
or
8 -60
or
�
�
Hg (OA c h� Na BH
_ _ _ _ _ -l..
t,
CH30H
A ) Unknown X, CSH9Br, h as one element of unsaturation. X reacts with neither bromi ne nor KMn04,
so the unsaturation in X cannot be an alkene; it must be a ri ng.
B ) Upon treatment with strong base (t-butoxide), X loses H and Br to give Y, CSHg, which does react
with bromine and KMn0 4 ; it must have an alkene and a ring. Only one isomer is formed.
C) Catalytic hydrogenation of Y gives methylcyc lobutane. This is a B IG c l ue because it gi ves the
carbon skeleton of the unknown. Y must have a double bond in the methylcyclobutane skeleton, and X
must have a B r on the methylc yc lobutane skeleton .
D ) Ozonolysis of Y gi ves a dialdehyde Z, CSHg0 2' which contains all the original carbons, so the
alkene c leaved in the ozonolysis had to be in the ring.
Let's consider the possible answers for X and see if each fits the information.
.---:f'
> U
KO-t-B u
1)
I
if this is X
ozonolysis
>
d'
°
+
O=CH2
DOES NOT FIT OZONOLYSIS RESULTS
so this c annot be X and Y
this must be Y;
only one product
more possi bilities on the next page
183
d KO-t-Bu
8-60 continued
2)
Br
>
1
if thi s i s X
c(
3)
i f this
Br
is X
4) P
Br
>
O t
-
-
�
Q
+
major
d
u
0(
Y would be a mi xture of alkenes , but the e l i mi nation
gives only one product. We already saw in ex ample 1
that the exocyclic double bond does not fi t the
ozonolysis resul ts so this structure cannot be X.
0(
Y would be a mixture of alkenes, but the e l i mination
gi ves only one product, so this structure of X i s not
consistent with the i n formation provided.
minor
major
KO-t- B u
K
d
+
minor
+Q
l
ozOnOIYSiS
SAME COMPOUND ;
this must be Y ;
on ly one product
if thi s i s X
>
r<
o
Z, a dialdehyde
The correct structures for X, Y, and Z are gi ven in
the fourth possibi l i ty. The on ly structural feature of
X that remains undetermined is whether it is the cis
or trans i somer.
trans
'(S)(
8-6 1 The clue to the structure of a-pinene is the ozonolysis. Worki ng backwards shows the alkene position .
o
bac kwards
o
>
� became carbony l carbons
a-pinene
'fit
�� T
{iX
�
�o {L:(
After ozonolysis, the two carbonyls are sti l l connected; the al kene must have been in a ring,
reconnect the two carbony l c arbons with a double bond.
~
~
�
BT
'
B r2
..
or
, I I CH3
Br
CH3
A
B r2
..
H2O
A
Br
BT
H2SO4
'OH
CH3
�
Zaitsev product
..
B
PhC03H
C
H 30+
..
"
OH
..
"
D
E
184
OH
so
8-62 The two products from pennanganate oxidation must have been connected by a double bond at the
carbonyl carbons. Whether the alkene was E or Z cannot be determi ned by this experiment.
CH3(CH2) 1 2 CH = CH(CH2hCH3
shown here as Z which is the naturally-occurri ng isomer
8-63
Unknown X
must have three alkenes
in this skeleton
Pt
CH2 = O +
Unknown X
�
o
H
0
o
+
A/ H
o
0
There are several ways to attac k a problem like this. One is the trial-and-error method, that is, put double
bonds in all possible positions unti l the ozonolysis products match. There are times when the trial-and-error
method is useful (as in s i mple problems where the number of possibilities is few), but thi s is not one of them.
Let's try logic. Analyze the ozonolysis products carefully-what do you see ? There are only two meth yl
groups , so one of the three termi nal carbons i n the skeleton (C-8, C-9, or C- l O) has to be a =CH2 . Do w e
know which terminal c arbon has the double bond? Yes, we can deduce that. If C- l O were double-bonded to
C-4, then after ozonolysis, C - 8 and C-9 must stil l be attached to C-7 . However, in the ozonolysis products ,
there is no branched chain, that i s , no combination of C-8 C-9 C-7 + C- l . What if C-7 had a double
bond to C- l ? Then we would have acetone, C 3COC 3, as an ozonolysis product-we don 't. Thus, we
can't have a double bond from C-4 to C- l O. One of the other terminal carbons (C- 8 ) must have a double
bond to C-7 .
H
8�
9
�
6
H
+
+
10
5
The other two double bonds have to be in the ring, but where? The products do not have branched chains,
so double bonds must appear at both C- l and C-4. There are only two possibi l i ties for this requirement.
I
H)---
or
}{}-
II
Ozonolysis of I would g i ve fragments contai ning one carbon, two carbons, and seven carbons. Ozonolysis
of II would gi ve fragments containing one carbon , four carbons, and five carbons. Aha! Our mystery
structure must be II.
(Editorial comment: Sc ience i s more than a collection of facts . The application of observation and logic to
solve problems by deduction and inference are critical scientific skills, ones that distinguish humans from
algae . )
185
8 -64 In this type of problem, begin by detennining which bonds are broken and which are fonned. These
w i l l always give c l ues as to w hat is happening.
H+
formed
1
goes to most
e lectronegati ve
atom
+
HOq� ..
H
protonated epoxide opens to gIve
the most stable carbocation (3()
\ 3 0 carbocation looks
for electrons, fi nds
them at nearby alkene,
formi ng a 6-membered
ring (yes ! )-leaves a
3 0 carbocation
+
8-65 See the solution to Problem 8 - 3 5 for simpl ified examples of these reactions .
(a)
(b)
(c)
(d)
HOOC J
HOOe
OH
HO
�
H
HooeH1 ""eOOH
OH
Hooe �
" '" IleOOH
HO
H
eOOH
trans
,,
fCOOH
7--<
�
OH
'---<
IH
HO/ �/eOOH
eOOH
eOOH
trans
Hooe�
(COOH
(COOH
syn � racemic
OS04
186
anti
cis + anti
OH
HO
�
"1
�/HlleOOH
Hooe
H
+
cis
+
�
syn
�
meso
racemic
�
meso
8-66
B03 - THF
u
..
HO
� H �nB
-
H 20 2
CH3
a
111 0
..
I I I OH
H
Cf.
8 -67 By now , these rearrangements should not be so " unexpected" .
U"
"":
H 20
I
�C
H
+
H
CH3
H
�
(Y
�
�
alkyl mi gration with ring expansion
gives 3° carbocation in 6-membered
ring--carbocation nirvan a !
H
CH3
•
Br
: !lr:
C H3
0
+
H
H
' CH]
You must be asking yourself, "Why didn't the methyl group migrate?" To which you answered by drawmg
the carbocation that would have been formed:
I;I
O\}
C+
2°
�
H
CH3 I "
QC<1''3°
H
--
H
C H3
CH3
The new carbocation is i ndeed 3 ° , but it is only in a 5-membered ri ng, not quite as stable as in a 6membered ring. In all probability, some of the product from methyl migration would be formed, but the 6membered ring would be the major product.
8-68 Each alkene w i l l produce two carbonyls upon ozonolysis or permanganate oxidation. Oxidation of
the unknown generated four c arbonyls, so the unknown must have had two alkenes . There is only one
possibi l i ty for their position s .
CO
H
the unknown
a�COOH
H
H
KMn04
..
/j.
:: COOH
H
187
+
C OOH
I
COOH
8-69
(a) Fumarase catalyzes the addition of H and OH, a hydration reaction .
(b) Fumaric aci d i s planar and cannot be chiral. Malic acid does have a chiral center and is chiral. The
enzyme-catalyzed reaction produces only the S enantiomer, so the product must be optically acti ve.
(c) One of the fundamental rules of stereochemistry i s that optical! y inacti ve starting materials produce
optically inactive products . Sulfuric-acid-catalyzed hydration would produce a racemic mixture of malic
acid, that is, equal amounts of R and S .
(d) If the product is opticall y active , then either the starting materi als or the catalyst were chiral . We
know that water and fumaric acid are not chiral, so we must infer that fumarase is chiral.
(e) The D and the OD are on the " s ame side" of the Fischer proj ection ( someti mes cal led the "erythro "
stereoi somer) . These are produced from either: (1) s y n addition t o cis alkenes, o r (2) anti addition to
tralls alkenes. We know that fumaric ac id is tralls , so the addition of D and OD must necessari l y be anti.
(f) Hydroboration i s a syn addition .
H ooe " "
H
1 . B D3 • THF
,H
' == e '"
....... e
'
eOOH
2. D 2 0 2 , DO-
D
..
H
,
DOOe
"
H
�
(note that OH exchanges
with D in DO- )
OD
""
H
eOOD
+
DOOe
+�
D
eOOD
eO D
"
D
eOOD
D
,>-1(, R
H
�
OD
eOOD
+�
eOOD
As expected, tralls alkene plus syn addition puts the two groups on the "opposite" side of the Fi scher
proj ecti on (sometimes cal led " threo" ) .
�.
0
8-70
(a)
Hg(OAc)
H
Hg(OAc )
+
Hg(OAc)
-. .
AcO :
mercunntum ion
Br
Br - Br
..
bromonium ion
188
Br
8-7 1 The addition of BH3 to an alkene is reversible. Given heat and time, the borane will eventually
" walk" its way to the end of the chain through a series of addition-elimination cycles. The most stable
alkylborane has the boron on the end c arbon; eventually, the series of equilibria lead to that product
which is oxidized to the primary alcohol.
..
..
...
�
----
most stable
189
H
H2
8-72 First, we explain how the mixture of stereoisomers results, then why.
We have seen many times that the bridged halonium ion permits attack of the nucIeophile only from the
opposi te side.
Q
expected:
H
Ci-CI
---
Ph
CI�
W
H
�:S:1:
�
~
H
Cl
trans only
Ph
A mixture of cis and trans could result only if attack of chloride were possible from both top and hottom,
somethi ng possible only if a carbocation existed at this carbon.
actual:
'"
0� H
H
--�
+�
CI
H
Ph
H
trans
Ph
cis
This picture of the p orbitals of benzene show
resonance overlap with the p orbital of the
carbocation. The chloride nucIeophile can form
a bond to the positive carbon from either the top
or the bottom.
Why does a carbocation exist here? Not only is it 3°, it is also next to a benzene ring (benzylic) and
therefore resonance-stabilized. This resonance stabilization would be forfeited in a halonium ion
intermediate.
0"::::
r'�/
�CI
CI
aCI
' Ph
IIi
H
//
:Ci :
'Ci
CI
CC"'H
Ph
+
. .
-
+
�
C
0:CI
190
�
�
CHAPTER 9-ALKYNES
9-1 Other structures are possible i n each c ase.
CH3CH2CH2CH2C:::CH
CH3CH2C:::CCH2CH3
0-
HC:::C - CH2CH = CHCH2CH2CH3
C:::CH
9-2 New IUPAC names are gi ven. The asterisk ( * ) denotes acetylenic hydrogens of terminal alkynes.
(a)
(b)
*
CH3 - C:::C-CH3
but-2-yne
CH3CH2 -C:::C - H
but- l -yne
*
CH3CH2 - C::: C - CH3
pent-2-yne
CH3CH2CH2 - C:::C - H
pent- l -yne
CH3
I
*
CH3CH - C:::C - H
3-methylbut-l-yne
9-3 The decomposition reaction below is exothermic (/)J{0 - 234 kJ/mole) as well as having an increase
in entropy. Thermodynamical l y, at 1 500°C, an increase in entropy will have a large effect on AG
(remember AG
/)J{ - TAS). Kinetically, almost any activation energy barrier will be overcome at
1 500°C. Acetylene would likely decompose i nto its elements:
=
=
HC::CH
1 500°
�
2 C
+ H2
9-4 Adding sodi um amide to the mixture will produce the sodium salt of hex- l -yne, leaving hex- l -ene
untouched. Distillation wi ll remove the hex- l -ene, leaving the non-volatile salt behind.
}
� CHCH2CH2CH2CH3 NaNH2.. J �H� CHCH2CH2CH2CH3
1 Na+ _ C C - CH2CH2CH2CH3
H -C C-CH2CH2CH2CH3
CH
=
•
=
non-volatile salt
9-5 The key to this problem is to understand that a proton donor will react only with the conjugate base of
a weaker acid. See Appendix 2 at the end of this Solutions Manual for an in-depth discussion of acidity.
(a) H -C:::C-H
+ NaNH2
--
(b) H-C:::C -H
+ CH3Li
--
H -C:::C : Na+ + NH3
H-C:::C : Li + + CH4
(c) no reaction: NaOCH3 i s not a strong enough base
(d) no reaction: NaOH is not a strong enough base
(e) H-C:::C : Na+ + CH30H -- H-C:::C - H + NaOCH3 (opposite of (c))
(0 H -C:::C : Na+ + H20
-- H-C:::C-H + NaOH (opposite of (d))
(g) no reaction: H - C :::C : Na+ is not a strong enough base
(h) no reaction: NaNH2 is not a strong enough base
191
H-C:::C-H NaNH2 H-C:::C : Na+ CH3CH2Br H-C:::C-CH2CH3
� NaNH2
CH3(CH2)S-C:::C-CH2CH3 CH3(CH2)SBr Na+ :C:::C-CH2CH3
(a) H-C:::C-H NaNH2
CH3CH2CH2CH2Br H-C:::C-CH2CH2CH2CH3
NaNH2.. CH3-C:::C-CH2 2CH3
(b) H-C:::C-H NaNH2
H-C:::
C
-CH
CH
CH
2
2
3
CH3I
CH3CH2CH2Br
NaNH2 CH3CH2-C:::C-CH2CH3
(c) H-C:::C-H 2) NaNH2
..
H-C:::
C
-CH2CH3
CH3CH2Br
CH3CH2Br
(d) cannot be synthesized by an reacti
o
n-woul
d
requi
r
e
at
t
a
ck
on
a
al
k
yl
hali
d
e
r/ 2°
��
_
CH3-C:::
C : Nale;+ CHCH2CH3 fr CH3-C:::C-CHCH2CH3
stmustrongbenucleophi
CH3
second order CH3
low yields; not practical
NaNH2
CH3-C:::
C
-H
(e) H-C:::C-H NaNH2
CH3I
BrCH2CHCH3 CH3-C::: C-CH2CHCH3
CH3
CH3
H-C:::C-H NaNH2
H-C:::: C-?H2
Br(CH2)8Br
Br�
� NaNH2
Na+ -,C-=C-CH2
H2C-C:::C-CH2
�
Br�
Intwilramol
e
cul
a
r
cycl
i
z
ati
o
n
of
large
ri
n
gs
must
be intramolecular and not intennolecular.be carried out in dilute solution so the last displacement
9-6
-
..
•
..
9-7
1)
..
2)
1)
1)
•
2)
CH
2)
1)
1)
•
2)
S N2
20
+
I
1)
2)
(f)
1)
..
1)
2)
..
I
I
..
2)
..
....!---.
r'.
�.
S N2
9-8
H20 H-C:::C-CH20H
(a) H-C:::C-H NaNH2
H2C=O
1)
2)
..
----l.�
192
9-8 contin ued
9H
H
0
NaNH
2
2
H3C-C-C::C-H
H-C:::C-H Ph yCH3
Ph
OH
NaNH
NaNH
2
2
H-C:::C-H CH3I CH3-C:::C-H 2) HCCH2CH2CH3.- CH3-C:::C-CHCH2CH2CH3
H20
OH
NaNH2
NaNH2
H-C:::C-H 2) CH3I CH3-C:::C-H 2) CH3CCH2CH3 CH3-C:::C -CCH2CH3
CH3
H20
H
(Br H
H H � -oo:�H
H-C=C-C-C�C�
H -C �C-CH2CH2CH3 :OH C=C
I)
Br H�oo Br{) \CH2CH2CH3
1
t: H
HI )
H
HI
H-C C-CH2CH3 HO-H H-C=C=C-CH2CH3 H-C=C-C-CH2CH3
U�
H 1t :OH
H
�
H-OH
H-C=C=C-CH2CH3 H-C-C=- C-CH2CH3
H-C-C=C-CH2CH3
H
H
H
1)
(b)
..
2)
----l.-�
I
o
1)
(c)
2)
1)
�
I
II
o
3)
1)
(d)
1)
�
I
�
II
o
3)
.
I
9-9
1
1
1
1
1
..
=(=
---
�
-
0 0
1
_
_
}
-0 0
�
1
-
=
R
o
Ii
8.3 1 4 JIK-mole
( !1G)
[internal]
[terminal]
193
Keg
=
75
e
U
�
(
}
/
_
�
0 0
9-1 0 To determine the equilibrium constant in the reaction:
!J.G =
RT In
tenninal alkyne � internal alkyne
!J.G
- 1 7.0 kJ/mole
Keg =
(- 4.0 kcallmole)
RT
e
=
1
0 0
0 0
1
_
�
-
0 0
i
,1
\
0 0
I
-
I
-(-17,000)
(8.314)(473)
)
98.7% i nternal
1 .3 % tenninal
=
e4.32
=
75
i
9-11 (a) This isomerization is the reverse of the mechanism in the solution to 9-9.
H �- H2 - .
H-C-C C-CH2CH3 .. H-C-C C-CH2CH3
H
H
H
H
H-C C-C-CH2CH3 H-C==C==C-CH2CH3
H -U2 �
H
H-C C-C-CH2CH3
H
KOH,
• •
I)
•
.
N
1
i
.
1
1-
�
• •
l
\""'
}
1
_
• •
NH
----
-
• •
I( H
�
.
HJH
H-( C==C==C-CH2CH3
:�
�H
• •
}
2
H-C==C==C-CH2CHJ
1
H
I
1
�
1
I
(b) All steps in part (a) are reversible. Wi th a weaker base like
an equilibrium mixture of pent-I­
yne and pent-2-yne would result. With the strong base NaNH2, however, the final tenninal alkyne is
deprotonated to give the acetylide ion:
H-C C -CH2CH2CH3
+
NaNH2
-
Na+
:C C-CH2CH2CH3 NH3
+
==
Because pent-l -yne is about l O pK units more acidic than ammonia, this deprotonation is not reversible.
The acetylide ion is produced and can't go back. Le Chatelier's Principle tells us that the reaction will try to
replace the pent- l -yne that is being removed from the reaction mixture, so eventually all of the pent-2-yne
will be drawn i nto the pent- l -yne anion "sink".
(c) Using the weaker base
pent-2-yne predominating.
KOH 200°C
at
will restore the equili brium between the two alkyne isomers wi th
9- 12
(a)
( 1)
Br
1
Br
1
H3C-T-T -CH3 KOH H3C-C C-CH3
H H
CH3CH==CHCH3 CCl4.. CH3CH-CHCH3
---
L1
Br
I
Br2'
(2)
(b)
(1)
(2)
Br
I
Br
I
1
1
Br
1
NaNH2
H-C-C-(CH2hCH3 1500 H-C C-(CH2hCH3
H H
CH2 CH(CH2)sCH3 Br2' CCl4 H2C-CH(CH2hCH3
==
•
Br
1
•
194
Br
1
9- 12
(c)
continued
KOH
(1)
Br Br
Br2' CCl4 CH3CH
-CH(CH2)4CH3
(2)
---
"
�
-I..
Br
(d)
at insteadcoulofdNaNH
have 2
been used
(1)
KOH
H
Br
(2)
H
9- 1 3
Lindlar catalyst
Na
(c)
Wcis
Br
NaNH2
H
Oneo convert
of the oneusefulstereoi
"tricsks"omerof iorgani
c chemiHavi
stry nisg tahepaiabilr ofity
treacti
n
t
o
another.
ons opposi
like thetetwsteoreochemist
reductionsryshown
inuseful
parts (a), because
and (b)a
that
give
i
s
very
intermediateThi(thes princi
alkyne)ple canis usedbe transformed
eicommon
ther stereoisomer.
again in partin(d).to
195
200°C
continued
9- 13
(d)
Br
~
Br
�
trans
KOH
-C
H2
I
,
C�
Licatalndlaryst
H
H
I
\
C == C
cis
/ L
to add onlcouly oned addequitovthealenttriofplebromine,
always avoiwasdiinngexcess.
an excessIf theof bromi
ne,is because
twothe bromi
molTheeculne,goaletsheofisfibromine
bond
i
f
bromine
alkyne
added toto
I
r
st
drops
of
al
k
yne
wi
l
encount
e
r
a
l
a
rge
excess
of
bromine.
nst
e
ad,
addi
n
g
bromine
the alkyne will always ensure an excess of alkyne and should give a good yield of dibromo product.
9- 14
H
-H
� :B�:
I
+
CH3CH2CH2 - C == C
2°
1°
betcarbocat
ter thanion
2°
carbocation and
�
0 0
resonance-stabilized
CI
9- 16
I
H
I
H3C - C - C - (CH2)4CH3
i
I
Cl
I
H
+
H
CI
H
}
I
I
H3 C - C - C- (CH2)4CH3
I
I
Cl
(b)of The second addition occurs to make the carbocation intermediate at the carbon with the halogen because
:ci:
�
-- k
y+ - - ---- -�-k· l. On
no stab·lIlzat
resonance
0
stabilization.
:C 1: H
I
I
C==C
0
:C1: H
I
I
I
+
-C-C H
H+
I
H
I
....
..
t-----��
-t
196
: I: H
II
I
C- -
resonance
stabilization
9-17
initi ation:
propagation:
Br.�
hv 2
H-Br
j -CH2CH2CH3
RO·
RO - OR
RO •
J
+
---
RO-
H Br
H-C C -CH2CH2CH3
Br
+
•
==
I
•
The
radi
c
al
i
s
more
stabl
e
than
The
anti
Markovni
k
ov
orientati
o
n
occurs
because
t
h
e
bromi
n
e
radi
c
al
fiatrtstacks(seefirtsthetosolmake
utiontheto most stable radical, which is contrary to electrophilic addition where the attacks
2°
1°.
H+
9-15).
9- 18
(b) C C-(CH2hCH3 HBr
HBr
H-
ROOR
H -C==C-(CH2hCH3
CI Cl
-C==C-(CH2hCH3
Br H
H C C -(CH2hCH3
Br
Br Br
H-C-C-(CH2)3CH3
Br Br
H
-
I
I
I
I
==
I
H
I
I
I
I
I
Licatnadlarlyst
(or Na, NH3) H Br
2HBr H-C-C-(CH2)3CH3
H Br
I
I
I
I
197
E
+ Z
E
+ Z
9- 19
r
+
CH3-
=c
)
!
H2 :
1
1+
;)(�(
CH2CII3
O
both 2° vinyl
!
0
H
!
)
!
H
0
O
H2 :
H2 :
)
'f 'f
CH3-
-CH2CH3
�
R
: H
H :
-CH2CH3
H
t
H+
H+
H
A
I)
I)
t
-CH2CH3
CH3-C == C-CH2CH3
1
+1
HO: H
�H
(
=
H2 :
:OH
'f 'f
CH3-
+
carbocations
CH3-C ==C - CH2CH3
H
CH3-
H
+
+
resonance1
CH3 - C - C-CH2CH3
CH3-C-C-CH2CH3 stabilized
carbocations
,
'",\
r' 1
1
H
t
:O-H
H-O : H
..; �
Hi
(resonance forms
not shown)
�. 0 t
H2
:
:
H
H
I
I
CH3-C-C-CH2 CH 3
CH3-C-C-CH2CH3
I
II
0
3-pentanone
II
1
0 H
2-pentanone
H
The role of the mercury catalyst is not shown in thi s mechanism. As a Lewis acid, it may act like the
proton in the first step, helping to form vinyl cations; the mercury is replaced when acid is added.
-H +
---I....
CH3-C ==C - CH2CH3
I
+Hg
198
I
OH
9-20
(a) But-2-yne is symmetric. Either orientation produces the same product.
CH3-C
Sia2BH
..
C-CH3
CH3
CH3
/
\
C=C
/
\
H
BSia2
H202
..
HO-
CH3
\
CH3
/
C=C
/
\
H
OH
HO-
---I"�
°
CH3CHz
II
-
C
-
(b) Pent-2-yne is not symmetric. Different orientations of attack will lead to different products on any
unsymmetrical internal alkyne.
CH3-C C-CH2CH3
�H �
S
CH3
CH2CH3
\
/
/
\
C=C
H
CH3
�
H202 , HOCH2CH3
CH3
�
CH2CH3
/
H
°
II
CH3 -C -CH2CH2CH3
II
°
CH3CCH2CH2CH2CH3
(2)
°
(b)
H202 , HO-
HO-
°
(1)
H
\
HO
II
CH3CH2-C -CH2CH3
(a)
\
C=C
/
°
9-2 1
�
\
C=C
\
H
OH
/
C=C
Sia2B
B Siaz
CH2CH3
/
\
/
/
\
CH3
II
( 1 ) CH3CCH2CH2CH2CH3
II
HCCH2CH2CH2CH2CH3
°
+
II
CH3CH2CCH2CH2CH3
(2) same mixture as in (b) ( 1)
°
(c)
II
( 1) CH3CH2CCH2CH2CH3
(d)
( 1)
�O
VV
°
(2)
II
CH3CH2CCH2CH2CH3
(2)
�O
VV
199
CH3
9-22
CH3 H
I
I
----1
..
1
H -C-C-BH)I
I
CH3 CH3
(a)
\
CH3
H
/
\
+
C==C
/
H3C
CH3
CH3 H H H CH3
I
I
I
I
I
H-C-C-B-C-C- H
I
I
I
I
CH3 CH3
CH3 CH3
disiamylborane, Sia2BH
(b) There is too much steric hindrance in Sia2BH for the third B-H to add across another alkene. The
reagent can add to alkynes because alkynes are linear and attack is not hindered by bulky substituents.
9-23
o
II
o
0
II
(a) (1) HO-C-C-(CH2hCH3
oxidation of a tenninal alkyne with
neutral KMn04 produces the ketone
and carboxylic acid without
cleaving the carbon-carbon bond
o 0
II
II
(b) (1) CH3 C-C - CH2CH2CH3
o
II
11
(2)
II
CO 2 + HO-C -(CH2hCH3
oxidation of a tenninal alkyne with
wann, basic KMn04 cleaves the
carbon-carbon bond, producing the
carboxylic acid and carbon dioxide
o
II
(d) (1) CH3CH-C-C-CH2CH3
1
CH3
(e) ( 1)
II
CH3 - C-O H + HO -C-CH2CH2CH3
o
(2)
0
II
0
II
0
(c) (1) CH3CH2-C-C-CH2CH3
o
(2)
(2 )
(2)
II
CH3CH2-C-O H
o
II
0
II
CH3CH-C-O H + HO -C -CH2CH3
1
CH3
o
OH
HO
o
9-24
(a) CH3 -C:::C- (CH2)4 -C:::C-CH3
200
9-25 When proposing syntheses, begin by analyzing the target molecule, looking for smaller pieces that
can be combined to make the desired compound. This i s especially true for targets that have more carbons
than the starting materials ; immediately, you will know that a c arbon-carbon bond forming reaction will be
necessary.
People who succeed at synthesis know the re a ctions-there i s no shortcut. Practice the reactions for
each functional group until they become automatic.
(a) analysis of target
from
acetylene
,,- -, -.
... I\:'��\�i;��>,:':
( ./'""'-.r.f
.
3° acetyl enic alcoho ls made
from acetylide plus ketones \/
"
forward direction:
H-C=:C-H
+ NaNH2
----
�'--'
-
,
•
'
OH
-
H-C=:C:
,
:
:
...
from alkylation
of acetyhde
' •• _---'
8r
put on less reactive
group first
�
Na+
......f------
�
H-C=C�
t
..
NaNH2
Na+ -:C=:C�
o
anal ysis of target: cyclopropanes are made by carbene insertion into alkencs;
to get cis substitution around cyclopropane, stereochemistry of alkene must be
cis; cis alkene comes from catalytic hydrogenation of an alkyne
NaNH2
H-C=:C-H
..
------I�
CH31
----
H3C-C =: C-H
NaNH2
..
CH3CH2Br
.. H3C-C=:C-CH2CH}
H2
CH212
..
(c)
)" 0 /(
H
CH3CH2
CH2CH2CH3
H
Z n(C u)
H
I
,
20 1
H
>=<
H3C
analysis of target: epoxides are made by direct epoxidation of
alkenes; to get trans substitution around epoxide,
stereochemistry of alkene must be trans; trans alkene comes
from sodium/ammonia reduction of an alkyne
9-26 Please refer to solution 1-20, page 12 of this Solutions Manual.
Lindlar
catalyst
CH2CH3
9-27
(a)
CH3CH2-C::C -(CH2)4CH3
o-
(d)
C::C-H
�'
(f)
(b)
H3C-C::C-(CH2)4CH3
(e)
CH3CH2-C:: C - CHCH2CH2CH3
( )-
C::C-H
I
CH3
I
CH3CH-C::C - (CH2hCH3
(g)
Br
(h)
H3C
C::C-CHCH2CH3
\
I
I
\
I
C=C
OH
H-C::C-CH=CH2
(k)
CH2CH3
H
H
(j)
H - C:: C - CH2 - C::C - CH2CH3
(i)
(c)
H-C::C
CH3
H
"
-(
C=CH2
,
H
9-28
(a) ethylmethylacetylene
(b) phenyl acetylene
(c) sec-butyl-n-propylacetylene
(d) sec-butyl-t-butylacetylene
9-29
(a) 4-phenylpent-2-yne
(b) 4,4-dibromopent-2-yne
(c) 2,6,6-trimethylhept-3-yne
(d) (E)-3-methylhept-2-en-4-yne
(e) 3-methylhex-4-yn-3-ol
(f) cycloheptylprop-I-yne
9-30
terminal alkynes
internal alkynes
(a)
CH3 - C::C-CH2CH2CH3
acetylide ions
H - C::C - (CH2hCH3
C:: C - (CH2hCH3
hex- l -yne
hex-2-yne
CH3CH2 - C:: C - CH2CH3
H - C::C - CHCH2CH3
- C:: C-CHCH2CH3.
I
I
CH3
hex-3-yne
CH3
3-methylpent- l -yne
CH3-C::C-CHCH3
NaNH2
C::C-CH2CHCH3
H - C :: C-CH2CHCH3
I
I
CH3
I
CH3
CH3
4-methylpent- l -yne
4-methylpent-2-yne
CH3
CH3
I
I
H -C::C - C - CH3
C::C-C-CH3.
I
I
CH3
CH3
3,3-dimethylbut- l -yn
�
(b) All four terminal alkynes will be deprotonated with sodium amide.
9-3 1
(R�-: �a
b
.C==CH
CH3CH2-C
H
"
CHCl
HC
OH
H20
---
•
I
NH2
CH
I
CH3CH2 - C-C-CH
202
HC
�
"
CHCI
ethchlorvynol
9-32
H-C:::C-H
NaNH2
CH3(CH2hBr
..
CH3(CH2h
(CH2)12CH3
I
\
C=C
/
\
H2
(a)
"1
Lindlar
catalyst
Cl
I
(b) H3C-C-CH2CH2CH3
Cl
I
CH2 = CCH2CH2CH3
CH3(CH2)12Br
CH3(CH2h -C::: C - (CH2) 12CH3
....
..
1-------
H
H
muscalure
9-33
NaNH2
CHiCH2h-C::: C -H
..
I
(c)
CH3CH2CH2CH2CH3
(f)
H - <? - <? - CH2CH2CH3
B r Br
Cl
H
(d)
(g)
G)
o 0
II
II
Na+
I
NaNH2
I
1500
Br
Br
(b)
I
KOH
I
2000
H3C - C -CH2CH3
Br
(c)
0
(k) H3C - C - CH2CH2CH3
: C ::: C-CH2CH2CH3
H3C - C -CH2CH3
_
CH3CH2-C=C-H
0
CO2 + HO - C - CH2CH2CH3
II
H2O
I
(i) H2C = CHCH2CH2CH1
H-C:::C -CH2CH3
H3C-C::: C-CH3
..
NaNH2
..
..
I
0
II
(I) H - C - CH2CH2CH2CH3
II
-
Br
B r Br
I
C=C
\
I
Br
CH2CH2CH3
(h)
HO - C-C-CH2CH2CH3
9-34
(a)
(e)
CH2 = CHCH2CH2CH3
Br
\
_
..
CH3CH2-C=C:
�
Na+
CH3CH2CH2CH2Br
CH3CH2 -C:: C -CH2CH2CH2CH3
H3C
(d)
\
B r2
C=C
I
\
\
I
H
H
(e)
H
I
CH2CH2CH3
H
C=C
I
H3C
---
\
CH2CH2CH3
CCl4
B r2
---
CCl4
Br B r
I
I
KOH
•
I
2000
CH3C -CCH2CH2CH3
H H
Br Br
I
I
•
I
CH3C -CCH2CH2CH3
H H
1 ) NaNH2
1 500 ..
2) H2O
203
..
H3C -C::: C -CH2CH2CH3
this product could contain minor
amounts of 3-hexyne from
rearrangement
H - C == C - CH2CH2CH2CH3
9-34 continued
Hz
(f)
..
Lindlar
catalyst
Na
(g)
NH3
..
H z0
(h) HC:: C - CHzCHzCHzCH3 ___---i"�
HZ S04
HgS04
[
W
cis
W
trans
��
CH2 =
H2CH2cH2cH3
unstable enol .
KOH
I
CH3C-CCH2CH2CH3
•
-
�
H3C - CH2CH2CH2C H]
this product could contain minor amounts of
3-hexyne from rearrangement
Br B r
I
1
I
---J"�
200°
H H
H3C - C:: C -CHzCH2CH3
major
Hz
H3C
Lindlar
•I catalyst
CHzCH2CH3
\
I
I
\
C=C
9-35
Br
�
KOH
..
200°
HC
�
\
Br
H
+
j
�
Y
'--------..
only the terminal alkyne reacts
with NaNH2 and acetone
OH
+
OH
CompoundD
b.p. 1 40- 1 50°C
under vacuum
204
H
from rearrangementcould contain 3-hexyne
J
/
�
o
t
MixtureA
2
�
t
Mhture B
J� �
CompoundC
b.p. 80-84°C
9-36
elimination
on 3° halide
(c)
CH3CH2-C:::C - CH20H
(after H20 workup)
OH
(e)
I
Na+
CH3CH2 -C::: C - CHCH2CH2CH3
(after H20 workup)
OH
(g)
-0-0
I
CH3CH2 - C::: C - C -CH2CH3
I
(after H20 workup)
CH3
9-37
NaNH2
(a)
HC::C-H
(b)
HC::C-H
•
HC::C:
NaNH2
•
CH3CH2CH2B r
HC::C:
--=----"'--..::...
..
.
�
HC::C -CH2CH2CH3
NaNH2
...
..f-----.
CH3I
(c)
H3C-C:: C - CH2CH2CH3
synthesized in part (b)
(d)
H3C -C:: C - CH2CH2CH3
H3 C
H2
\
C=C
•
Lindlar
catalys t
Na
H3C-C:: C - CH2CH2CH3
synthesized in part (b)
(f)
HC = C- CH2CH2CH2CH3
H
H
H3C
H
\
I
I
\
C=C
H
synthesized in part (b)
(e)
CH2CH2CH3
I
CH2CH2CH3
2 equiv.
H2
Pt
Br
I
2HBr
H3C -C -CH2CH2CH2CH3
I
synthesized in part (a)
Bf
205
continued
Sia2BH H-CCH2CH2CH2CH3
(g) H-C::fromC-CH2CH2CH3
(b)
H 0 H3C-CCH2CH2CH3
(h) from (b)
HgS04
(i) HC::C-H NaNH2 HC:: C:
9-37
o
1)
H-C::C-CH2CH2CH3
I I
2
°
•
II
H2S04
..
�
---t
catLindlal yastr
alproduce
kene mustthe be ciproduct
s to
from anti addition
NaNH2
HC::C: Na+
HC::C-H
Review the stereochemistry in the
solution to Problem 8-35, p. 171 of
this Solutions Manual.
(±)
U)
1)
�
..
-t
2)
Alternatively,attedrans-but-2-ene
d be c acid.
anti-hydroxyl
with aqueouscoulperaceti
Review the stereochemistry in the
solution to Problem 8-35, p. 171 of
this Solutions Manual.
9-38
NaNH2 H3C-C::C-CH3
CH31
H2 catLindlalyastr
1
H3C
OS04 C=C CH3
H H
almesokeneproduct
must befromcis tosynproduce
additiothen
---l"�
--
\
I
I
\
meso
dX PtH2 o-CH2CH2CH2CH3
� the fact that five equi v alents of hydrogen are consumed
says thatskelemust
2
carbon
ton have fi ve pi bonds in the above
O O I'�
0
\11O O
'I? 'I?
H-C-CH2CH2 -C-C-H H-C-C-H H-C-C-OH H-C-OH
from C::C
carboxylic acids
alkyne
carbonyls
alkenes
ene ithese
s cis orresultransts. cannot
o-CH=CH-C::CH beWhetdetheerrmithneedalkfrom
Compound
T�:�
o
II
6
5
X
II
I I
+
'-.
+
�
�
� 3
X
2
206
II
I I
+
'"'-__----.
)
y�----� 1
9-39 Compound
ozonolysis CH3(CH2)4-C-H CH3-C-CH2-C-OH HO-C-H
from alkene
from alkyne
Compound CH3(CH2)4CH C -CH2 -C C-H Whether
thedetalkeenerminised
cannot
be
CH3
from this information.
9-40 All four syntheses in this problem begin with the same reaction of benzyl bromide with acetylide ion:
HC::CH .. HC::C: � CH2Br .. PhCH2-C::CH .. PhCH2-C::C:
use this
benzyl
I bromide
Br �
PhCH2-C::C � 6-phenylhex-l-en-4-yne
allyl bromide
(b) PhCH2-C::C: Br....../ .. PhCH2-C::C� P aS04.. Ph '>={
ethyl bromide
H pentH - 2-ene
quinoline cis-l-phenyl
Lindlar catalyst
Na
Ph
(c) PhCH2-C::C: Br....../
ethyl bromide
X
phenylpent-2-ene
(d) Thestructure,
diol witalh tthough
he twothisOHonegroupsis notonmeso
the same
sidethine tthoep Fischer
projom egroup
ction isarethdifferent
e equival. eStntilof, iat
meso
because
and
bot
t
gives aorclueby anas tanto itsi addit
syntihonesitso. aThetrans"meso"
bond,
doublediolbond.can Webe formed
saw thebysameeitherthinga synin taddit
he solution itono atocis double
Ph '>={ OS04
Ph
H30+
Hcis H syn addition
anti addition product
product
CH2Ph
from part (b)
from part (c)
H + OH
H + OH
CH2CH3
racemic
Z
�
\
II
I I
y
Z:
N��
=
o
o
o
o
II
II
J
\.'-__......
,-__ -.;)
y
==
I
E or
Z
N��
_
6
helow
?'
�
+
+
\
dlB
+
trans-l-
9-37 (j).
..
trans
207
9-4 1
a2BH
(a) CH��-C�C-H SiH202,
HO(b) �.. reaction �H-O-Et
'"
R�C= \{ -==:-l
R-C C-H :O-Et
H :O Et � O�Et
H +O�E'
R-�-t react
R �-�
� 'c=/
ion2 R/
H H
H H
! H20 :
H
H : O�Et
H O-Et
H :o�H+
R-C-C-H
R-C-C-H .. R-C-C-H
H :O�H H20: H OH
H OH
ilJ
! - HOEt
H
H
R -C-C"+-H
R-C-C-H
R-?-? -H
II
H :OVH
H : O-H
H
(c) alkyne ROR-C=CH The
R-C CH
clorbiosertalthiats cleloeserctronsto aarenucltoeusthethannuclaeus,p orbithetalmoreis, asstapble.
An
s
2
orbitalsoaren iselmore
ongatstedablaway
from
the
nucl
e
us.
An
sp
2sp carbaniORon carbani
3
an
sp
carbani
o
n
because
t
h
e
2sp carbanion has es than
electron pair
alkene ROiwhis clcohseris tonlo thye positsivcharact
e nuclcharact
eeusr. The
tehranandspin2andancarbani
spthe3 carbanion
R-C-CH2 to form because of its relative stabil i ty. on easier
R-C=CH2
H
3sp HcarbaniORon
1)
•
2)
f\
{
+
-
•
1
-..
\ y\
/
1
...
/
\H
�
•
1
1
1
1
1
• •
1
1
1
1
1
•
H
1
-----<
....
..f----;.�
1
°
•
1
1
I)
1
1
1
+
..
}
1
33%
1
•
25%
1
I
208
is
Diols orareantmadei-dihydroxyl
by two react
ionsviafrom
Chapteusing
r anda peroxyacid
revisited in and water.
(d): either
syn-dihydroxyl
attioon
witusehinorgani
Os04'
a
t
i
on
an
epoxide
As
t
h
is
probl
e
m
says
c reagents, the solution shown here wil l use OS04.
Recall the stereochemical requirements of syn addition as outlined in this Solutions Manual, p. Problem
cis-alkene addition meso
citrsans--alkaenelkene additadditionion racemic
racemic
trans-alkene addition meso
Part
synthesis
on will .have to occur on the cis-alkene.
(b) wil(a)l asks
requiforre synthe addit
ion toofthethetrameso
ns-alisomer,
kene to sogivsyne thaddit
e iproduct
(a)
NaNH2
HC::C : Na+
HC::C-H
"!
i
OS04
Lindlalyastr -Zc::cJcat
H H
alproduce
kene mustthe meso
be cisproduct
to
from
synion addit
ion,withso
reduct
is
done
theLindlciasralcatkenealyst to produce
(b)
NaNH2
HC::C-H NaNH2 HC::C : Na+
"!
i
OS04
Na
NH3 -Zc::cJalkene mustthe be tproduct
rans to from
produce
syn addit
so reduction is
done
withion,Na!NH3
9-42
8-35:
8
9-4 1
17 1,
+ syn
+ anti
+ syn
+ anti
�
�
�
�
(±)
(±)
Part
(±)
..
..
f
\
..
..
..
(±)
209
Bf
Bf
This synthesis begins the same as the solution to problem
HC::CH NaNH2.. HC::C:
CH2Br .. PhCH2-C:::CH .. PhCH2-C:::C :
benzylde
bromi
The anion will add across the carbonyl group of the aldehyde:
H30+
H
--:
C
PhCH2-C:::
lf
acid-catalyzed dehydration
H2S04,
9-40:
9-43
01
NaNH2
_�
+
�
°
+
•
MCPBA
210
1
Ll
CHAPTER 10-STRUCTURE AND SYNTHESIS OF A LCOHOLS
10-1
Please seepropan-2-ol
the note on p. 136 of this Solutions Manual
regardi
nghplylcycl
acementohexan-l-ol
of positio("n 1"numbers.
(d)
t
ra
ns-2-met
is optional)
(a)(b) 2-phenyl
(e)
(E)
2-chl
o
ro-3-met
h
yl
p
ent-2-en-l-01
5-bromohept
a
n-2-ol
(c) 4-methylcyclohex-3-en-l-ol (" 1" is optional) (2R,3S)-2-bromohexan-3-ol
IUPAC name first, then common name.
10-2
(c)(d) l-cycl
oylbutbutylapn-l-01;
ropan-2-01;
noycommon
name
(a)(b) cyclopropanol
; cyclopropyl
3-met
h
i
s
opent
l
al
c
ohol
2-methylpropan-2-01;
t-butylalalcohol
cohol
(also isoamyl alcohol)
10-3 Only constitutional isomers are requested, not stereoiOHsomers, and only structures with an alcohol group.
(a) C3H O �OH propan-l-ol
A propan-2-01
(b) C4HlOO � �
�OH
�
OH
OH
OH
butan-I-ol
butan-2-01 2-methylpropan-I-ol 2-methylpropan-2-01
(c) C4H O has one element of unsaturati on, either a double bond or ring.
HO i>OH
OH
d
[>-Jmethanol l-methylcycl
�opropanol 2-methylcyclopropanol
cyclobutanol cyclopropyl
cis
or
t
ra
ns
OH * HO�
� OH
�
�
OH
but-3-en-I-ol but-3-en-2-01 but-I-en-2-01 but-I-en-I-ol (E or Z)
J:H
*
�OH � �OH
OH
but(E-2-en-l-ol
but-2-en-2-01
2-methyl2-methylprop-2-en-l-01
or Z)
(E or Z)
prop-l-en-I-ol
(d) C3H40
hasortwaothree-membered
elements of unsatrinugratandi on,asodoubleache bond.
structureAllmuststructures
have eimust
ther contai
a triplen bond,
or (Intwothe
doubl
e
bonds,
an
OH.
name, the "e" is dropped from "yne" because it follow by a vowel in "01".)
HO-C::C-CH3
HC::C-CHOH2 H2C=C=C!I
y. 0H
*
prop-l-yn-l-ol prop-2-yn-I-ol
OH
propa-l,2-dien-I-ol cycloprop-l-en-I-ol cycl'\Zoprop-2-en-l-ol
*Thesealcompounds
wistthructure
the OHwibonded
direct
ly to the carbon-carbon
doublis cale lbond
areynolcal. lThese
ed "enols"
or
"vinyl
c
ohol
s
.
"
The
t
h
OH
on
a
carbon-carbon
t
ri
pl
e
bond
e
d
an
are
unstable, although the structures are legitimate.
(f)
g
g
a
*
OH
211
1 0-4 (a) 8, 8-dimethy lnonane-2,7 -diol
(b) octane-l,8-diol
(c) cis-cyclohex-2-ene-l,4-diol
(d) 3-cyclopentylheptane-2,4-diol
(e) trans-cyclobutane-l,3 -diol
1 0-5 There are four structural features to consider when determining solubility in water: 1) molecules with
fewer carbons will be more soluble in water (assuming other things being equal); 2) branched or otherwise
compact structures are more soluble than linear structures; 3 ) more hydrogen-bonding groups will increase
solubility; 4) an ionic form of a compound will be more soluble in water than the nonionic form.
(a) Cyclohexanol is more soluble because its alkyl group is more compact than in I-hexanol.
(b) 4-Methylphenol is more soluble because its hydrocarbon portion is more compact than in I-heptanol ,
and phenols form particul arly strong hydrogen bonds with water.
(c) 3-Ethylhexan-3-ol is more soluble because its alkyl portion is more spherical than in octan-2-01.
(d) Cyclooctane-l,4-diol is more soluble because it has two OH groups which can h ydrogen bond with
water, whereas hexan-2-ol has only one OH group. (The ratio of carbons to OH is 4 to 1 in the former
compound and 6 to 1 in the latter; the smaller thi s ratio, the more soluble. )
(e) These are enantiomers and wil l have identical solubility.
10-6 Dimethylamine molecules can hydrogen bond among themselves so it takes more energy (higher
temperature) to separate them from each other. Tri methylamine has no N-H and cannot hydrogen bond,
so it takes less energy to separate these molecules from each other, despite its higher molecular weight.
1 0-7 See Appendix 2 at the back of thi s Solutions Manual for a review of acidity and basicity.
(a) Methanol is more acidic than t-butyl alcohol. The greater the substituti on, the lower the acidity.
(b) 2-Chloropropan-l-ol is more aci dic because the electron-withdrawing chlorine atom is closer to the
OH group than in 3-chloropropan-l-ol.
(c) 2,2-Dichloroethanol i s more acidic because two electron-withdrawing chlorine atoms increase acidity
more than just the one chlorine in 2-chloroethanol.
(d) 2,2-Difluoropropan-l-ol is more acidic because fluorine is more electronegati ve than chlorine; the
stronger the electron-withdrawing group, the more acidic the alcohol.
10-8
most
acidic
sulfuric acid
»
2-chloroethanol
>
>
water
ethanol
>
>
t-butyl alcohol
ammonia
>
least
acidic
hexane
(CH3hCOH
Sulfuric acid is one of the strongest acids known. On the other extreme, alkanes like hexane are the least
acidic compounds. The N-H bond in ammonia is less acidic than any O-H bond. Among the four
compounds with O-H bonds, the tertiary alcohols are the least acidic. Water i s more acidic than most
alcohols inc luding ethanol. However, if a strong electron-withdrawing substituent l i ke chlorine is near
the alcohol group, the acidity increases enough so that it is more acidic than water. (Determining exactly
where water appears in this list i s the most difficult part.)
10-9 Resonance forms of phenoxide anion show the negati ve charge delocalized onto the ring only at
carbons 2, 4, and 6:
HQ
66'
: 0:
5::::-'"
4
: 0:
: 0:
3
..
..
.
::::-...
..
..
6
CH
2 12
O
..
..
.
h-
6
: 0:
: 0:
H
...
..
10-9 continued
Nitro group at position 2
. .
:0: :0:
:0: :0:
Nitro at position 2
delocalizes
negative charge.
Nitro group at position 3
:0 :
Nitro at position 3 cannot delocalize negati ve charge at position 2 or 4no resonance stabilization.
:0 :
Nitro group at position 4
:0:
:0 :
..
..
Nitro at position 4 delocalizes negative charge.
-
• •
.."N ,
+
. •
-
:0'" 0:
Only when the nitro group i s at one of the negative carbons will the nitro have a stabilizing effect (via
resonance). Thus, 2-nitrophenol and 4-nitrophenol are substantially more acidic than phenol itself, but 3nitrophenol is only slightl y more acidic than phenol (due to the inducti ve effect).
OR
10- 10
A
�
�
(a) Structure A is a phenol because the OH i s bonded to a benzene ring. As a phenol, it will be acidic
enough to react with sodium hydroxide to generate a phenoxide ion that w ill be fairly soluble in water.
Structure B is a 2° benzylic alcohol, not a phenol, not acidic enough to react w ith NaOH.
(b) Both of these organic compounds will be soluble in an organic solvent like dichloromethane.
Shaking this organic solution with aqueous sodium hydroxide will ionize the phenol A, making it more
polar and water soluble; i t w i l l be extracted from the organic layer i nto the water l ayer, while the alcohol
will remain in the organic solvent. Separating these i mmiscible solvents w i l l separate the original
compounds. The alcohol c an be retrieved by evaporating the organic solvent. The phenol can be
i solated by acidifying the basic aqueous solution and filtering if the phenol i s a solid, or separating the
layers if the phenol is a liquid.
2 13
1 0- 1 1 The Grignard reaction needs a solvent containing an ether functional group: (b), (t), (g), and (h) are
possible solvents. Dimethyl ether, (b), i s a gas at room temperature, however, so it would have to be
liquefied at low temperature for it to be a useful solvent.
�Li
10- 1 2
(b)
(a) CH3CH2MgBr
(c) F
+ LiI
-D-
�+
Li
MgBr
(d)
LiCI
1 0- 1 3 Any of three halides-chloride, bromide, iodide, but not fluoride-can be used. Ether is the typical
solvent for Grignard reactions.
(a)
(b)
(c)
o
�
o-
+
MgCl
H
\
C=O
I
H
ether H30+
H
ether H30+
+
MgBr
\
C=O
I
..
--
..
--
H
H
MgI
\
+
C=O
I
ether H30+
...
--
H
Note: the alternative arrow symbolism
could also be used, where the two steps
are numbered around one arrow:
O
�
o-
CH2OH
,
'ether "
,
2) H30+
OH
NO! Me BAD!
This means that water is present with
ether during the Grignard reaction.
1 ) ether
...
CH2OH
OK
,
'ce
'
'
•
3 +'
,}'fo
"
1 0- 1 4 Any of three halides-chloride, bromide, iodide, but not fluoride-can be used. Grignard reactions
are always performed in ether solvent; ether is not shown here.
(a) two methods
I
H
In
MgCl
0
H
+
y
0
+ CH3MgI
H30+
...
--
H30+
--
...
�
(b) two methods
o
1.&
MgB,
�
0
+
Ib:
Ib:
OH
0
H
�
H30+
--
..
OH
H
+ IMg -......./
H30+
--
..
214
~
Where two methods can be
used to form the target
compound, the newly
formed bond is shown in
bold.
1 0- 1 4 continued
(c)
o
�
OH
o
MgCI
+
H
dD
10- 1 5 Grignard reactions are always performed in ether. Here, the ether i s not shown.
(a) Any of the three bonds shown in bold can be formed by adding a Grignard reagent across a ketone.
(ii)
(i)
(iii)
0
(i )
CH3C 2MgBr
CH3C 2CH2MgBr
:
O�
�
+ PhMgBr
Ph
� Ph
:Ph MgBr + "-- 'c = 0
I
Ph
(b)
:
�O
,\ ?:;01
'-("')
( Ph
111
Ph
Ph
H30+
--
(ii )
I
Ph -C-OH
�
I
Ph
(d) two methods
�MgCI
\ Cy
�
+
Cy
>= O
Cy
=
H30+
--
�
+ Cy-MgCl
cyclohexyl
:C!0
II
CH3 -C� MgBr
10- 16
---
?)
CH3-C-CI
:
I
(This is just a nucleophilic substitution where
CI is the leaving group. The unusual feature
is that it occurs at a carbonyl carbon.)
�
- CI-
�
Ph
:0
II
�
l
-Ph
Ph-MgBr
OH
I
CH3- -Ph
TPh
215
�:o:I
work-up
step
CH -C-Ph
3
I
Ph
ketone
intermediate
1 0- 17 Acid chlorides or esters will work as starting materials in these reactions. The typical solvent for
Grignard reactions i s ether; i t i s not shown here.
0
II
+ 2 PhMgBr
Ph-C-Cl
(a)
Jy
(b)
Ph
H30+
I
Ph-C-OH
•
�
I
Ph
H30+
+ 2 CH3CH2MgI
OCH]
�
•
0
OH
o
II
+ 2
Ph-C-Cl
(c)
10- 18
:qD
II
(a) H-C
o
�
o
./'0...
.
MgCI
. .
:?j
MgB r
---
+ 2 CH3CH2MgBr
0
II
(ii) HC-OEt
+ 2
0
II
(iii) HC-OEt
+ 2
��
<}
MgBr +
�
OH
H30+
•
�
MgB r
� MgBr
H30+
�
•
H30+
�
10- 19 Ether i s the typical solvent in Grignard reactions.
<}
aldehyde
intermediate
H-C-OEt
OH
I
�C�
I
H
II
(b) (i) HC-OEt
(a)
~
o
U
•
�
OR
� OR
V
216
10- 19 continued
(b)
(c)
�
MgCl
�
gl
10-20
(a)
HC
_
C:
0
U
+
0
U
+
�OH
H30+ � OH
H30+
�
----
----
�
�D
?'"O
(b)
10-2 1 There are more than one synthetic route to each structure; the ones shown here are representative.
The new bonds formed are shown here in bold. Your answers may be different and still be correct.
(a) � Br
Li
CuI (�C
- ----
2
(b)
� Br
Li
( �C
CuI
-----
2
(c)
�
Br
Li
�
UL i
�
I (�C
ULi
(�C
ULi
Cu
2
�
(
()l U
U Li
-----
Br
�
(
�
CJ"
Alternatively, coupli ng lithium dicyclohexy!cuprate with I-bromobutane would also work. As
this mechanism i s not a typical S N 2, it i s not as susceptible to steric hindrance like acetylide ion
substitution or a similar S N 2 reaction.
(d)
�
Br
Li
CuI
-----
2
10-22 These reactions are acid-base reactions in which an acidic proton (or deuteron) is transferred to a
basic c arbon in either a Grignard reagent or an alkyllithium.
(a)
+
CH3D Mg(OD)I
(b)
+
CH3CH2CH2CH3 LiOCH2CH3
0 CH3C-OLi
(d)
+
217
o
II
10-23 Grignard reagents are incompatible with acidic hydrogens and with electrophi lic, polarized multiple
bonds like C=O, N02, etc.
(a) As Grignard reagent is formed, it would instantaneously be protonated by the N-H present in other
molecules of the same substance.
(b) As Grignard reagent is formed, it would i mmediately attack the ester functional group present in other
molecules of the same substance.
(c) Care must be taken in how reagents are written above and below arrows. If reagents are numbered
" 1. . .. 2. ... etc.", it means they are added in separate steps, the same as writing reagents over separate
arrows. If reagents written around an arrow are not numbered, it means they are added all at once in the
same mixture. In thi s problem, the ketone is added in the presence of aqueous acid. The acid will
immediately proton ate and destroy the Grignard reagent before reaction with the ketone can occur.
(d) The ethy l Grignard reagent will be immediately protonated and consumed by the OH. Thi s reaction
could be made to work, however, by adding two equivalents of ethyl Grignard reagent-the first to consume
the OH proton, the second to add across the ketone. Aqueous acid will then protonate both oxygens.
6
OH
10-24 Sodium borohydride does not reduce esters.
(b) no reaction
(a) CH3(CH2) gCH20H
(c)
o
(e) HO·
Y
�OCH3
Y
(I) c,tcc
OH
no reaction (PhCOO- before
acid work-up)
1 lf
°
(d)
OH
6
OH
10-25 Lithium aluminum hydride reduces esters as well as other carbonyl groups.
(a) CH3(CH2) 8CH20H
(e) HO
+ HOCH3
(b) CH3CH2CH20H
� OH
y + HOCH3
(c)
PhCH20H
(d)
� OH
HO �
HO
(f)
OH
10-26
(a)
�H
o
OR
� OH
o
OR
(b)
�
o
�
OR
NaBH4
CH30H
�OH
•
OR
1) LiAlH4
2) H30+
1) LiAIH4
2) H30+
•
1) LiAIH4
2) H30+
NaBH4
CH30H
•
OH
•
�
218
OR
1) LiAIH4
2) H30+
•
•
1 0-26 continued
(c)
~
NaBH4
CH30H
..
OH
0
(d)
W
~
OR
O
NaBH4
CH30H
..
C�X:f
1) LiAIH4
2) H30+
..
LiAlH4 will NOT give the desired
product. LiAIH4 will also reduce the
ester in addition to the ketone.
OH
0
1 0-27 Approximate pKa values are shown below each compound. Refer to text Tables 1-5, 9-2, and
1 0-3, and Appendix 5 at the back of the text.
CH3S03H > CH3COOH > CH3SH > CH30H > CH3C:: CH > CH3NH2 > C H3CH3
< 0
4.74
"" 1 0.5
15.5
25
"" 35
50
least acidic
most acidic
1 0-28
(a) 4-methylpentane-2-thiol
(b) (Z)-2,3-dimethylpent-2-ene- l -thiol (" 1" is optional)
(c) cyclohex-2-ene- l -thiol (" 1" is optional)
10-29
�
HBr
ROOR
�
Br
� Br
�o
NBS
�
----l..
OR
�
NaSH
..
NaSH
..
c
�
� SH
(see Problem 8-2)
1 0-30 Please refer to solution 1-20, page 12 of this Sol utions Manual.
1 0-3 1
(a) 5-methyl-4-n-propylheptan-2-01; 2°
(b) 4-(I-bromoethyl)heptan-3-01; 2°
(c) (E)-4,5-dimethylhex-3 -en-l-01; 1°
(d) 3-bromocyclohex-3-en- l -01; 2° (" 1" is optional)
(e) cis-4-chlorocyclohex-2-en-l-ol; 2° (" 1" is optional)
(f) 6-chloro-3-phenyloctan-3-01; 3°
(g) (l-cyclopentenyl)methanol; 1°
1 0-32
(a) 4-chloro- l -phenylhexane- l ,5-diol
(b) trans-cyclohexane-l,2-diol
(c) 3-nitrophenol
(d) 4-bromo-2-chlorophenol
219
SH
3-methylbutane-I-thiol
2-butene-l-thiol
(but-2-ene-l-thiol)
OQ
-6
&
1 0-33
(a)
I
0
OH
OH
6
(c)
(b)
C-OH
(f)
OH
(g)
¢
�
� 6
OH
OH
(d)
HO
(h) H
I
SH
H
U)
(i )
�H
OH
(e)
OH
CH3S-SCH3
OH
(k)
~
10-34
(a) Hexan- l -01 will boil at a higher temperature as it is less branched than 3,3-dimethylbutan - l -01.
(b) Hexan-2-01 will boil at a higher temperature because its molecules hydrogen bond with each other,
whereas molecules of hexan-2-one have no intermolecular hydrogen bonding.
(c) Hexane- l ,5-diol will boil at a higher temperature as it has two OH groups for hydrogen bonding.
Hexan-2-ol has only one group for hydrogen bonding.
(d) Hexan-2-ol will boil at a higher temperature because it has a higher molecular weight than pentan-2-01.
All other structural features of the two molecules are the same, so they should have the same
intermolecular forces.
1 0-35
(a) 3-Chlorophenol is more acidic than cyclopentanol. In general, phenols are many orders of magnitude
more acidic than alcohols.
(b) 2-Chlorocyclohexanol is slightly more acidic than cyclohexanol; the proximity of the electronegati ve
chlorine to the OH increases its acidity.
(c) CyciohexanecarboxyJic acid is more acidic than cyclohexanol. In general , carboxylic acids are many
orders of magnitude more acidic than alcohols.
(d) 2,2-Dichlorobutan- l -ol is more acidic than butan-I-ol because of the two electron-withdrawing
substituents near the acidic functional group.
1 0-36
(a) Propan-2-ol is the most soluble in water as it has the fewest carbons and the most branching.
(b) Cyclohexane- l ,2-diol is the most soluble as it has two OH groups for hydrogen bonding. Cyclohexanol
has only one OH group; chlorocyc lohexane cannot hydrogen bond and is the least soluble.
(c) Cyclohexanol is the most soluble as it can hydrogen bond. Chlorocyclohexane cannot hydrogen bond,
and 4-methylcyclohexanol has the added hydrophobic methyl group, decreasing its water solubil ity.
1 0-37
(a)
(b)
�
U
OH
Hg(OAch
H2O
BH3' THF
•
•
NaBH4
H202
HO-
•
�
l
N
({
.-
"
"
OH
220
continued BH3 H202 OH
(c)�
HO- ~
OH
Hg(
O
Ach
NaBH4
(d)� H2O
�
OH
(a) OH (b)� JyPh
OH
V
(OHd) Thigroup,s prothbleefmirsconft thinugsetshata lhappens
ot of peoplis eth. atWhenthe GraiGrgnarigdnarredactresagentby reimovis addedng tthoeaH+compound
from thet0-.hat has an
et
h
er
0
CH3
M
g1
CH4
+
o�ly oned requieagentvaleaddednt of! �0- +MgI If a second equivalent (or excess)
00H (Gngnar
sAciis, dthhydriens addeditolysisadd
Iisf added
no morbefe Groriegnaracidd reagent H30+ CH3etherMg1 atbefofwitlGrhoergieigketvhydrnare othne.deorlediyagent
o
l
.
mathydreorilaylsiiss,rtehcoveren theed.starting
Mg 3 H H H
00H ( ' 0 p'. X 3
0+MgI
UOH
Ph
Ph
Ph
(e) Q--'OH Ph+OH
(
g
6
0H
)Ph
Ph
crtu
OH
�
HO�OCH3
HO�OH
(i) V
(k)�
V
OH
d:) HO OH (n) HO
1 0-37
- THF
�
�
�
..
10-38
(e)
o
o
1
+
�
will
a
(f)
0
(l)
(h)
U)
�
(m)
(0)
221
ndicating that the Grignard
ws are sacihownd isiadded.
reactionAlislalGrloiwedgnardtorepractoceed,ions arande runtheninietn haersescondolventst.ep,Twodiluarrte oaqueous
H
0
H3
+
?
(a) � H BrMg -..../
Mgether CH20..
�Br -OH
OH
0
Hp+
(c) )lH BrMg-Q
0
~
'
B
+
0
H3
Mgether U (YbH
(d) O -'
R
(e) O etMgher CH20 H30+ VOH
i l CH3
COCH2
O
OH
(g ) � H
~
0
(h ) �
XMgC C-CH3
agent
e
r
d
nar
g
i
Gr
a
s
i
XMgC=CCH3
,
y
l
al
c
Techni
of
compound
c
i
l
a
anomet
g
or
an
s
i
t
i
e
becaus
H-C=C-CH3
RMgX
usual
e
h
t
n
i
made
not
s
i
t
i
,
However
m.
u
i
magnes
d
nar
g
i
Gr
al
n
i
m
r
e
t
e
h
t
g
n
i
nat
o
t
o
depr
by
made
s
i
t
i
n;
o
i
h
s
a
f
alrekagentyl or wheralkenyle R is
alkyne as shown.
BH3 -THF
(a)
H
CH2
+
0
H3
0
Mg
Ph
(b )
Ph �OH
'CI ether
10-39
o
+
---
(b)
.. �
�
+
..
..
' h
..
---
..
..
--
o
(f)
o
+
��------�
'\
-
�------�
(
any
+
1 0-40
..
_
/'....
........",
--
..
..
222
1 0-40
continued
OH
o
eq.
H2
d
(c) 6
)
( 6 Pt .. 6
6
OH
o
�OEt
(e) �OEt
o
o
OH
o
H30+.. � OH
� OEt
o
OH
(a) V MgBr CH,O _e_th_e--;r.. �
V
OH
(b) V
HO� �
V
NaOH..
o
� OH
�
Mgether
( d)
(e )
'-- Br NaSH
�SH
� Br� ( 1 \CULi � Br
.�
j
:"�
I
the streacingtdhandof tbashe acie, sods torhethsiedebasofetsh. eThe
sequattrongerionTheaciwidtposihandthetiosweaker
tnroofngertheacibasequideandlwiiblrbasiualmwe canayswil berbeeactdetfatvoreormined
gievdeatthequibye weaker
librium. See Appendix in this Solutions
Manual for a review of acidity.
CH20H <weaker'}-o(a) CH3stroCngerH20� <stronger'}- OH CH3weaker
base acid
acid base
°
°
1
( f)
10-4 1
+
�
�
r-==\.
---f
+
�
(f)
'
+
1 0-42
2
+
+
223
products favored
KOH Cl-Q- OH
Cl
OH
CH3
O
c6
H20 Cl-Q- 0-K'
Cl
1 0-42 continued
(b)
+
h
a OH
stronger
base
KOH
+
p roducts favored
products favored
+
stronger
base
(CH3hCOH CH3CH20(CH3hCOH HO-
(CH3hCO- CH3CH2OH
reactants favored
+
+
stronger
acid
stronger
base
H20
weaker
acid
weaker
base
stronger
acid
weaker
base
weaker
acid
(e)
weaker
base
+
h
stronger
acid
(d)
weaker
acid
stronger
acid
stronger
base
(c)
+
weaker
acid
weaker
base
products favored
+
stronger
base
(g)
stronger
acid
KOH CH3CH20H
+
weaker
base
weaker
base
weaker
acid
products favored
stronger
acid
weaker
acid
stronger
base
�OH OR LiAIH4
(a) � H
OR
� OH LiAJH4
OR
� OR 1 ) LiAJH4
H30+
reactants favored
1 0-43
I)
°
1)
•
°
o
2)
•
224
•
continued0
OH
LiH3A0lH+ 4...
(b) � CH3NaBH40H
OR
�
0 NaBH, . OH
� CH30H ~ OR LiH3A0lH+ 4
OH
0
NaBH4
Li
A
l
H
4
(d)
...
OR
CH3
0
H
H3
0
+
OO
00
OH
NaBH4
OEt CH30H � OEt
0
0
OH
LiH3A0IH+',.
HOCH2CH3
OEt
OH
Q
0
(biooxed)n" in fworromkisntgarbackwar
ting matedrsiaislstwofosisixxcarcarbbonson or fewer.
ThefragmentproTheducts whigoalhaschiscoultocarsyntdbbeonshjesio,isonzedetthhienelatoagiGrirgcetalgnarcompound
"didscronnect
bond, and Brdouble bonds aro
e madeMgBr
from alcoholsewhiacticohn.arThee thebesprot ductwaystoofmakeGrignarepoxid redactes iiosnsfro. m the double
Methger ...
OH y
O
H30+
VD
H
4
t MCPBA
cro
1 0-43
1)
..
2)
I
(C)
1)
2)
,&
...
1)
...
(e) ('
'\
(f) /
"
2)
r---<
...
1)
+
2)
II
10-44
I
,&
12
I
h
o
+
--
�
H�
225
I
,&
2 S0
v<b
target
steps are reversible.
10-45 All
i
�
}
H
HO H - H20.. ./CHO··I · .. .. H-O H20 :.. :0:
A
A
X
H �
HO><�
� I+
• •
The symbol H-B represents a generic acid, where B-is the conjugate bas0
e.
�
H
H�
OH H- { H :OH
-H
}
:
B
v
f
(a) H H � H -t- C�
H
H lI
T\.
H
.
1 0-46
n
/ + .......
0:
/
�
---
H
(b )
(C )
OH H
H
Q(
1 0-47
I
�
H
\.
..B
r
Q(
6MgBr
H : O -H
_
• •
? , OH
��
H
"
-
I
0:
H
A
B
I)� 1 2) H,o'
OH
�
C -......
H-B
H
•
�
• •
°-...:::
6
c
H2S04 ..
/).
D
+
>-
. .
-...:
::
2
Mg
..
..
°
°
et
h
er
6'
� o:
cr
�
r
crH � + H-B
ao :
:
:
H J H�' H a
+
• •
�
H
H
E
226
H
+ :O-H
.
exH2cesst
�P
2 Br2..
isobutylcyclohexane
F
.
10-v4i8ngThiforscemechani
srmeactisiosinmiislarreltioefclofeavagering softraitnheinepoxithe 4-demiember
n ethyleednecycloxiicdeetbyherGr, whiignarchdisrewhyagentitswi. Thel
driunder
f
o
r
t
h
e
go a Grignard reaction whereas most other ethers wil not.
R "MgX .D . --MgX
�U
10-49 When mixtures of isomers can resCH3ult, only the major product is shown.
CH3
(y O 1) CH3Mg1 OOH HZS04 o- CH3 HPtz ..
2) H3O+
U
t Hz�04
t KM�04
� CH3
(J=O
OH
t 1 )H3CH3OM+ g1
tOH Hz�04
(J-O
Q1 "1 "1" 0
t 2) Mg,cycloetpentheranone
t PhC03H
3) H3BrO+ HEr
U
0
he deskunkion tnhge smiulxfuturr; ealilsofhydrthesogene fupernctiooxinalde.grThioupsolars are eacioxididci.zed
tThe1o0-s5trs0uoctdiTheuurmesmosbihavicart nibmgonatporone,etainttswbasro,eactoricaenough
tnthreine toxygens
can wash them away. sulfenic acitodisonize these acids, smakiulfinnicg acithemds water soluble whersulfeotnihec acisoapds
�O
H
O
H
O
H
H
O
z
z
z
z
z
z
P
SH thiol �OHS � S
S-OH
3-�
methylbutane-lHzOz �S HzOz � S'OH HzOz
�SH
S-OH
OH
(2-butbut-2e-ne-lene-l--thtihoilol)
+
+
...
A '"
A
B
"
"
G
D
2)
F
C
I)
..
E
..
...
I
..
I
...
II
II
°
...
227
I I
°
II
°
...
�°
II
�
I I
°
1 0- 5 1
�0
1
)
Mgether �MgBr H30+ H �
Br
OH
t KOH, H20
S
04
,
+
H2
�OH
21 )
+ Na
03
)
�
Br
2
�
�O- Na+
Br
H G�
+
�O
C
�
Br
�O
O�
KOH, �
H
H �C::C�
plus other alkynes
1
NaNH
,
150°
C
+ H20
2
)
O� I)Sia2BH H -C::C ---"" .......,
7- HBrnon-l-+yne
H
Br
Na+-C =C�
�
Br
�
----..
2)
D
..
E
fj,
A
F
�
B
/' Br
+
fj,
2)
/"'..
..
1
/"'..
I
2
J
CH2CH31
L
o
�
HO- CH3C -C=C�
I
K
228
.......
/"...
.......,
((ab)) botoxihdatreiacton,iooxinsdarateiooxin, dreatductionsion, oxidation
((cd)) onereductcariobn:on C-O
is oxidiiszedreplandacedone carC-Hbon is reduced-no net change
(e) oxineitdhateriooxin (adddiatiotnionnorofrXeduct2) ion-the C stil has two bonds to 0
((hg)) neineitthherer oxioxiddatatiioonn nornor rreeductductiioonn ((aelddiimtiinoatniofon of H20)
(i) oxithedfatirsiot n:reactaddiionngisanoxi0dattoioeachn as cara newbonC-Oof thebonddoublise fbondormed to each carbon of the alkene; the
shasecondonerbondeactiotno ioxygen
s neither oxidation nor reduction, as H20 is added to the epoxide, and each carbon
ion: OHH-Bareisadded,
addedsiontthhere efirisstnoreactnetiooxin, dandatioBn inors replreaductcediobyn.0(Ninottehe
st(hkeatcond) neiin ftruhencteractoxiiioonaln;datovergrioonupsanorl , onlirnevolducty Hand
two cartboonsthelfikuenctalikoenesnal groroupalkiynes,s clasbotifiehdcarasboxionsdathaveion toro bereduction.)
oxidized or reduced before thevingchange
OH
0
0
(a)
H2Cr04
6
6H
6
PCC no reaction
(b) no reaction H2Cr04
o
0
0
OH
(c) cY H2Cr04
PCC
cY
cY
0
(d) no reaction H2Cr04 6
no reaction
(e) no reaction H2Cr04 0 PCC no reaction
0
C
r04
H2
no reaction
H3C -C-OH PCC no reaction
0 H2Cr04 CH3CH2OH PCC H3C -C-H
0
(g) H3C -C-OH
0
0
C
r
0
4
H2
(h) H3C -C-OH
H3C -C-H PCC no reaction
C H A PTER l l-REA CTIONS OF ALCOHOLS
1 1-1
by
(f)
HX)
(j )
sti l l
net
1 1-2
pcc
..
..
�
..
�
pcc
..
..
(f)
..
II
II
�
�
II
�
..
..
�
�
II
�
229
II
troxiansdfoizred.medDMSO
to the allodsehyde,
and a fralomcoholthe sluolsfeusr;oneit is clearly
(hydra) Aogen alincfoholonnilnogseaskettwoone;hydreachogensalcwhen
ohol
i
s
e
s
an
oxygen
ous, oxalto CO;yl chlhowever
oride under, thegnetoes
diefrefduced.
sepctroporon tt(hiIoefnatyoueleiomentngotretactshiosnioeoxaln:twoneyo,l youchlCoidirsidoxidethidseiz"prednootblchange".
oeCO2m corandrectthlye.)otToherbeC irisgroreduced
ction shows that dimethyl sulfide reduces an ozonide. In the process, dimethyl sulfide
(oxib)diTextzed toseDMSO.
(enera) Dehydr
othgenatis reaciontiodoesn, ornotioccur
atmodynami
C-eictaherl y: unfthaervoreaibls ae,hiwighthki[)'nGet>ic barriTheer (alathiergposh actsiibviatlitiyonis
g
y
f
o
r
t
i
s
t
h
er
)
supporC (tseedebytextthseefctaicotnthat theCre)verandsethrereaceftoioren has(cat[)'alGytic O.hydrThiogenats makesion tofheaquescarbtonylion of) kisnpetonticsaneous
at c-a
academi
prwiolceiedmprmusovet wibethusienlcreseaslyinslgow.temperature for virtually all reactions, so both the
(rhydrebact)andoiogenatn(cth)atioKinetcnaandnnoticsdehydr
orgenat
ion.n rTheeactansionswerwilisgothatfatshteerrm. Iodynamics
n this case,wihowever
, tthhies quesreacttiioonn asis howthe to
fteampervor thaetudehydr
o
genat
i
o
n
e
act
i
o
l
f
a
vor
es[)'5ti>matesitnhrceeationesMi>raimolsed.esiculThencee tikeyhseconver
prisotducthetefduketndament
oto tnewo:platuhlserthydrheerfomroeodynami
gen, [)'is5lecs equatstAtablieloontwhan[)'teGmperthe sMiataturtri-engTflSalcoholC. )We, Mi. Alcanso.
all, so [)'negat
G > O.ive.AtFora hitghhe enough
overdomiwnathelesmbecaus
Mi, ande T[)'iGs swio slmbecome
reactionteinmperquesattuioren,, tthheis-T[)'mus5t tbeermthewicasl begie at n to C.
(a)
AND HOPCC
(b) all three reagents give the same ketoneCH3pro(ducCHt2wi)4 Y"­
th a secondary alcohol
(c) - H Na2H2CS04r207 AND PCC
OOH
1 1 -3
10
20
a
is
8- 1 5B
1 1 -4
250
1)
O.
2)
250
<
1 0- 1 1
is
=
0
0
-
T
<
O.
(250
3000
1 1 -5
--.....,.�
°
•
•
(d) all three reagent><s givHe no reaction with a tertiary alcohol
NR
(
n
o
r
e
act
i
o
n
)
V
�
-
-.-
230
Note-PCC
to the st(puyrdentidin: iuForm chlsimoplroichrcityo,matthise)bookto oxiwidlizeuse talhescohole stasndarto alddlehydes;
aboratory methods of oxidation:
-H2
C3,r0H2S04,
4 (chromiacetc acioned)(Jtoonesoxidreizagente )altcooholoxidsitzoecarbaloxylcoholicsacitodkets; ones.
-cr
0
Under
are legi, andtimatCole; floinrsexampl
e,orSwemPCC oxiwildatoxiiodnizwore aks about
welo a lketasone
PCCas welinsltasathndechrprthoeatparmicotahtacieriondchoi.ofIfalcyouesdehydes
r
e
agent
al
c
ohol
t
consult the table in the text befohavere Proablquesem tion about the appropriateness of a reagent you choose,
0
(a) �OH PCC � H
0
C
r
0
4
H2
(b )
� OH
cr03 �
(c) �OH H2SO4
acetone 0
0
PCC
(d )
�H
OH
0
C
r
0
4
H2
(e) OH
� OH
(y
oH cr03 &0
BH3•
H2SO4
H2SO4
oH - H2O 6 H202, HOacetone
esent toitihandl
etenzyme
hanolA, chrmolso roeeniquiculc aleres.csoholmoriec ethashanolinduced"antidmorotee" ADH
moleculenzesymeto actto asbe aprcompet
ve inhie blairtgoer tamount
o "tie up"s thiemextbibreda
OH OH CH3-C-CH
o0
0
o
pyruvaldehyde pyruvicCH3-C-COH
pyr
u
vi
c
aci
d
d is a norof mglalucosmeteabol("biloteod sugar")
in the bracieakdown
1°
1°
2°
2°
as
1 1 -2.
1 1 -6
•
� OH
•
•
�
•
�
(f)
..
THF
1)
Ll
•
•
2)
•
of
1 1 -7
1 1 -8
[0]
..
I I
II
[0]
231
II
I I
"Ts"
�� CH3
Ts � -��
CH3
(a) CH3CH2 - OTs KO - C-CH3
CH3
1 1 -9 From this problem on,
will refer to the "tosyl" or "p-toluenesulfonyl" group:
o
I
+
I
�
CH3
CH3CH20 - CH3
CH3
I
C
(E2 is also possible with this hindered base; the product would be ethylene,
�
+
+
(c)
R
(d)
cS
�
(e)
1 1 - 10
NH
OT
s
..
c5
+
�
+
"
S
excess
..
-
Na+ :C=CH
+
---l.�
6
(c)
(d)
inversion--SN2
+ NH
�C=CH NaOTs
+
OT NaBr
� OH
� OH TsCI � OTs NH3
� OH TsCi �OTs NaOCH2CH3 � O -...../
� OH TsCI � OTs KCN � CN
TsCI
..
pyridine
�
s
•
excess
(b)
s
� I NaOTs
�OTS NaI
H
TsO H
NaCN
�
� NaOTs
+ H3 -0TS
NH2
T,
NH3
3
+ , OTs
(b)
(a)
OT
CH2=CH2)
+ K
-
I
pyridine
pyridine
pyridine
� Br
�
•
•
..
•
•
•
NH2
1 1-1 1
(a)Q- CH20H TsCl
- d-
pyn me
"
o
Q-CH20TS OR Q-CH20-�-o-CH1
232
o
1 1 - 1 1 continued
( b)
0-
CH20Ts
( d)
( c)
Pt
minor
maj or
�
O
O
6� B
Br
H
o
o ro
0
+
0
----
o
t
0
: Br :
nBr
� o�
oo HV
·
1 1-13
CH3 �
H3C -
t - OH
I
n
H - C�
0 0
CH3
� �H2
+
"-- :B�:
o
V
0
�+
CH
I
H 3 C - C - OH2
I
�
V Sr
--=-
0 0
_
+
H3 C - C
CH3
H20
----'l.�
I
:
:Cl
CH 3
I
+-- H3C - C - Cl
I �
CH3
+ H20
_
I
CH3
CH3
carbocation
intermediate
1 1 - 1 4 The two standard qualitati ve tests are:
1 ) chromic acid-distinguishes 3 ° alcohol from either 1 ° or 2°
R
R
+ OH
R
H2Cr04
(orange)
•
R ( or H)
n o reaction
(stays orange)
R
+ OH
H
233
H2Cr04
(orange)
•
R(or H)
I
R-C=O
+
erJ+
blue-green
1 1 - 1 4 continued
2) Lucas test-<listinguishes 1 ° from 2° from 3° alcohol by the rate of reaction
R
R
OH
+
R
+ HCl
soluble
R
R
2°
+ OH
+ HCl
+ OH
+ HCl
H
soluble
H
1°
R
H
soluble
ZnCl2
ZnCl2
R
..
R
Cl
+
R
+ CI
insoluble-" c loudy in
+ H2O
insoluble-"cloudy" in 1 -5 minutes
+ H2O
insoluble-"cloudy" in > 6 mi nutes
( no observable reaction at room temp. )
II
I minute
R
..
R
H
ZnCl2
<
+ H20
H
..
R
+ CI
H
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - . - - - - - - - - - - - - - - _ . .
OH
(a)
Lucas:
H2Cr04:
(b)
Lucas:
H2Cr04:
A
c loudy i n 1 -5 min .
i mmediate blue-green
OH
cloudy in < 1 min .
n o reaction-stays orange
cloudy in 1 - 5 min .
i mmediate blue-green
no reaction
no reaction-stays orange
A
�
(c)
Lucas:
H2Cr04:
Lucas:
H2Cr04:
(e )
Lucas:
H2Cr04:
�
OH
cloudy in 1 -5 min .
no reaction
D OES NOT DISTINGUISH-i mmediate blue-green for both
� OH
(d)
o
� OH
no reaction
cloudy in < 1 min. * *
D OES NOT D ISTINGUISHi mmediate blue-green for both
�
+OH
( * *Remember that allylic catioIlS are reson �nce­
stabilized and are about as stable as 1° cations.
Thus, they will react as fast as 3 u i n the Lucas test,
even though they may be 1 °. Be careful to notice
subtle but important structural features !)
no reaction
cloudy in < 1 min.
D OES NOT D ISTINGUISH-stays orange for both
methyl
shift
..
---1
3° �
jI' ,CH3 : Br :
I
�
C
/ 1 .... C H2
+
• •
-
Br
�
1°
Even though 1 ° , the neopentyl carbon is hindered to backside attack, so SN2 cannot occur easi ly.
Instead, an SN 1 mechanism occurs, with rearrangement.
234
a=
H
This 3 ° c arbocation is planar at the
C+ so that the C l - can approach
from the top or bottom gi ving both
the cis and trans isomers.
CH3
C - CH3
+
approach
: C l : from
above
H
0=
: Cl :
approach
from
below
CH3
;: Cl
CH 3
1 1-17
CH 3
0
"
,
1 1-18
3
1 1 - 19
( a)
o�
" H
• •
�
---
•
OH
+
PBrJ
SOCl2
-
CH3
6
-
CH3
·H
+ C 2°
�
-
hydride
shift
Br
6
+
(from Hel)
CH
I
.�
CH
+C '
: Cl :
"
O
•
•
•
'
CI
O
P(OHh
retention
OT s
OH
( b)
3
"
3°
Cl
OH
6
-�
H3
G
H
ZnCl2
�
TsC I
pyridine
6
Cl
NaCI
SN2-inversion
23 5
o
Another possible answer
would be to use PCl 3 .
(a) OH1 : �\S =O0 +0H1 -1S ,- O· D +O-S=O
�I
I)
.
.
··
·
..
(C
.
Cl
X I
..
+
0
O
O
1 1 -20
Cl
D
I
I
:o - s = o :
·0
S02
j
+
..
:Cl :
+
{
I
.
-
--
al
- . .
• • -
D
I
• •
H
Cl
• •
A·
� tH
allylic!
�
�0 }
- � - S = R:
r.
D "
+
-
l
e
b
0 0+
� - HCI
:Cl :
Cl
O
I
D
te carbocata "firoene icars alblocatylic,iovern".y Thestablnucle andeophirelaltiicvelchly oloring-de lcanived.attaItckcanany
tcar(hber)beTheonforwiekeyesthcposiapeis thttatihveethicharoeninpaitgee,rmedi
r andnot jabecome
ust the one closest. Since two carbons have partial positive charge, two
products result.
(a) OH ZnCIHCl 2 no reaction unles heated, then �CI
HEr �Br
PBr3 �Br
P12
SOCI2
1 1 -2 1
�
...
�
�
-
�
�I
�CI
236
IOH ZnCIHCl 2 IClr
B
(C) �OH ZnClHCl 2 �C
1 1 -2 1 continued
(b)
..
�
�
�I
12
SOCI2 �C
�
HE r
PBr3
-
1SOCl2 2
P
-
-
I
II
I
C
I
HBr
Br
PBr3
�
..
�OH ZHCl 2
nCI
1 °, neopentyl
HB r
•
•
l
P
no reaction unless heated, then
S N I -rearrangement
�Br
�
12 �I
-
Br
..
(poor reaction on 3°)
SN2-minor
(hindered)
PBr3
Br
(poor reaction on 3°)
-
(d)
l
Br
P
-
SOCIl. �Cl
237
+
CI
(poor reaction on }O)
�
Br
�
l
S N l -rearrangemen t
11-21 continued
(e)
carbocat
ionfromi ntermedi
ate
CI
HCI
can
be
att
a
cked
ei
t
h
er
'eY
ZnClz
side by chloride
,Br SN2 with inversion of configuration
"
HEr �
U
� ,Br SN2 with inversion of configuration
U
� " I SN2 with in version of configuration
Iz
U
SOClz �CI retention of confi guration
U
SN 1 ;
+
..
"
p
"
11-22 OH
(a)
�
(b)
(c)
�
OH
�
+
major
major
OH
HzS04 ,
major
HZS04 ,
(cis
+
+
minor
trans)
�
Jx OH
(d)
U
(e)
rrunor
+
(cis + trans)
major
minor
major
minor
mmor
�
238
trace
11-2 3
(y O:
..
V
H
I
°
� II�
P, '
Cl'l Cl
Cl
•
�N
-CI+ II
l
�- f �-.�
V
0
°
H
goo d leavi ng gro up
�
..
lo �
O
C
E2
0
cy clohex ene was fonn ed
withouta car boca tion
in ter media te
""i
..
Cl
°
• •
I
Cl
• •
• •
II
I
N
+
H
!O
: O-P-CI
I+
- -CI
o
H
1+
°
+
H
N
+
o
Cl
this produc tca n reac t with two
morea lcohols ot b eco me leavi ng
groups in the elimi na tion
E2
11 2- 4
B oth mecha nis ms b egin with protona tion of the oxy gen.
H
H
H
�
H+
.
•
H-C-C-O-H
• .
1
1
1
..
1
H
1
H
H
1+
1
H-C-C-O-H
1
H
H
1
• •
H
O ne mecha nis m inv olv es a nother molecule of etha nola ctingas a bas e, giv ing elimi na tion.
H
H
1
H
1
1+
T�
T�
i{q
H- .-A
·
H
-H
H
H
/
\
C=C
\
/
H
H
•
CH2CH3
H20
+
The other mecha nis m inv olv es a nother molec ule of etha nola ctingas a nuc leophi le, givi ngs ubs it tution.
H
1
H
1
H
1+
H
H
�
+ I)
�
U·
H-C-C-O-H
1
H
• •
.. CH3CH2-O-CH2CH3
··
• •
HOCH2CH3
··
ItqCH2CH3
.. CH3CH2 -
O-CH2 CH3
11-2 5 A n equi mola r mi xtur e of metha nola nd etha nol would produc ea ll ht ree possib le ethers . T he
diffi culty ins epa ra ting thes e compounds would pr eclude this method fromb ein ga prac tica l route toa ny
one of them. T his method si prac tica lo nly fors ymmetri c ethers , tha tis , wherebo tha kl y l groups ar e
identica l.
CH3CH20H
+
HOCH3
H+
tl.
..
H20
+
CH3CH20CH3
239
+
CH30CH3
+
CH3CHzOCH2CH3
'
�
H
H
CH3
9
H20 CH3-O-CH3 .H CH3-O-CH3
3CH -O-H H+ CH3 -O-H CH-3-O-H
�
H
HC H
(a) X�H ' H+ X� H
H
0
2
:
#"--.
.
a
&H
H
2
0
VH
H
VH
H-OH
a
:CHl H20:
OCH3
H
H H+
H
H
H
O
O
:
:
d
� OCH3 (cjOCH3
O
H
H
/··
CH
-CH30H HH H+ H
:
aH
aH
H
l H20:
a
11-2 6
• •
,"",1+
�
•
• •
�I+
•
• •
• •
•
• •
• •
11-27
• •
�
�
I+
+
�
1
y---..
• •
•
0.
•
(b)
+
9.CH3
--­
+
• •
1+
..
• •
-
�
H
-
er
CH2 '-"h!'
H
� H+
O O - H2 0
H
H
�
(C� CH2-�H
�
----;.�
+1
1°
+
..
(1: � L) 240
�
H
H(5
hydride
shift_
HC
H,I/.H
)b
�H
� H 20 )
+
H
�+6
(!j
(d) a�H,- Ol+-
11 2- 7 conti nued
�
CH
H3C
w
H
il
- ''""'
Hz O
)
"
7
CH3
'--C+
H3c
(a) n
++0
CH,
H3C
w· H3C
n
-T-rfHO
HO :"OH
"
�
· H3C
CH3
OH
+
1
H
M ethy ls hift
H'C
-H20
H
HO
H�:
CH3
CH3
.
rmg
H20:
contraction
n
-t- 1
-CH3
HO
� � Q-� : QCH
1
;"C
_
___
H2C
11 -2 8
o
CH3
'"
H
R+ CH3
CH3
-
QCHl
0
"- ""
CH3
H20:
A lky ls hift ri ng contraction
H3C
-W---HO
C-CH3
+
Q
""
H,C,+
C
1
H-O:
4
CH3
H'C
·
� -0
'"
H-O+
241
CH3
H20:
H3C
o
CH3
0H
oH
(b) oH.f
+
H
H
2
..
CX;; OH cYf G:O-H O<�C
�
/\ ���;
+OH
�-H
o H2··0 :
/
a
a
a
30 and doubly
benzylic carbocation
1 1-2 8 co ntinued
H
�
1+
-
°�
• •
Ph
Ph
Ph
Ph
\
+
�
Ph
Ph
U
Ph
•
/
f'-.-- /
/
"
+
1
rIng
expa ns oi •n
°
H3C",C'H
+H�
if
II
(a )
(c)
11 -3 1
a( )
(c)
(b)
2
°
Ph
°
(d)
O-H
[c
� CH
+
/
• •
-
1
1
S imila r ot the pina co l rea rra ngement, this mecha nis m
invo lves a ca rbo ca tio n next ot a na lco ho l, with
rea rra ngement ot a p ro ot na ted ca rbo ny l . Reliefof
so meri ngs tra in in the cyclo propa ne si a na dded
a dva nta geof ht e rea rran gement.
11 -30
Ph
Ph
OH
OH0
H+
H
C
H
V,
/
�C-C-H
C==C
H H
H H
�
�
+
.----- -
Ph
�
Ph
11 2- 9
nS oi n
H
�
3
C
[�-CH3
� HS�4-
1
H
c:fCH]
0
(Y"' CH2=0
+
�:
°
°
II
+O-H
ij�
°
II
Cl-C-CH2CH2CH3 + HOCH2CHzCH3 (b) CH3(CH2hOH Cl-CCH2CH3
0
H3C-<, }-OH + CI-CCH(CH3h
[>-OH Cl-C-0
+
°
II
(d)
242
+
II
� !J
11-32
0:
: ..
: 0:
II
: O-S- O CH3
.. II
...
..
__--i..
�
I
-
0==S- O CH3
•• II
....
..f-.
-l..
�
:0 :
:0 :
· 0·
• II .
0==S- O CH3
••
I
..
:0 :
11-33 Proton transfer ( acid- base) reactions are much faster than almost any other reaction. M et ho xide wi ll
act as ab ase and remove a proton fromt he oxyg en much fast er than methoxide will act as a nuci eophile and
displace water.
CH 3CH2 -O H + H+ ---
11-34
( a)
o-
-OH
o-
O CH2CH3
(b) T here are two problems with this att emptedb imolecular dehydr at ion. F irst, all three possible ether
combinations of cyci ohexanol and ethanol would be produced. S econd, hot sulfuric acid aret he co nditions
for dehydrating secondary al cohols like cyci ohexanol, so elimination would comp ete with substituti on.
11-35
(a) W hat the student did:
N a+ 0,
H
X
X
CH3 CH2- O ""
H
+ TsO- CH2CH3
H3 C
CH2CH3
C H2 CH3
H3 C
sodium (S) -2- butoxide
(S)-2- ethoxybutane
T he product also has the S config uration , not the R. W hy? T he sub stitution is indeed anSN2 rea ct ion,
b ut the substitution did not take place at the chiral center, sot he config uration of the starting material si
retained, not inverted.
�
---�
..
,
( b) T here are two ways to make (R) 2- - ethoxybutane. S tart with (R)-2-b utanol, mak e ht e anion, and
substitute on ethyl tosylate simi lar to part ( a), or do anSN2 inversion at the chiral center of (S)-2- butanol .
SN2 works better at 1° carb ons s o the for mer method would b e preferred t o the latter.
( c) T his is nott he optimum methodb ecause it req uiresSN2 at a2° carbon, as discussedi n part b( ).
X
HO
H
CH2 CH3
H3 C
(S)-2-b utanol
•
TsCI
pyri dine
T
S-
O
X
X
H
H
H 3C
CH2CH3
(S)-2- butyl tosylate
no inversion yet
243
low temperature
( high temp.
favors el mi
i natio n)
O CH2 CH3
CH2CH3
H3 C
(R)-2- ethox yb utan e
INVERSI O N!
11-36
OH o-� + CH/�"'0-S03CH3 ---o-� OCH3 :0-S03CH3
CH2
C
H3
SOCI
2
CH3
C
H2MgBr
CH3CH2-COH-CH2CH3
CH3CH2-C-OH • CH3CH2-C-CI H30+
� H2S04
BH3
•
CH2CH3 H202, HO- CH2CH3
CH3CH C-CH2CH3
CH3CHCHCH2CH3
OH
cr03
OH
b( ) CH3CH2CH20H PCC CH3CH2-C-H CH3H30C+ H2MgBr.. � H2S04• �
(alo- 0H H+ HC03H OH
o H30+ U
'
O
H
OH HCI CI
(blo- 0H cr03 O
H3
0
+
Q--\
Q--\
H2PBr3SO: Mg,CH3CH2MgBr'"
CH3CH20H CH3CH2Br
O
H
OH
Br
MgBr
�
�
3
Ho+
g
U
H+
M
�
���
T
H2S04o- Br Mg
OR:o-cr030H �
�
�! ���A
+
H
,
o
)
-- ..
Q-f; 0H
rn
0
) - �gBr O-(;;C
o- 0H � O- Br ��.
0 CH3Br
HJ
Q-f;
CH30H �r3
o�
-
1 1-37
�
-
0: N a +
.. �
°
o
1) 2
II
II
+
-
(a)
"
I
•
2)
I
I
2)
......I-.
---
I
°
----
TIIP
1)
I
=
°
1)
II
acetone
2)
11-38
--f!.
any peroxy acid canb e usedt o fOlm the
epox ide whi ch is c l eaved to the trans­
dia l in aqueous acid
•
"
_
�
U
___
(c)
---
�
/'
ether
---
ether
--­
ether
�
.. ---
from Dart (b )
----'�
�
-
�
(d)
ether
from (c)
see the nex t pag efor an al tern ati ve endi ng
244
_
rom the previous page � H30+ 0--* CH30H
SN
(d) contcontinOHuedinuedf�
I
/"'yBr �
OH H2S04 �OCH3.
ether fmm
are severexamplal possie mayblebecomdibf ienratentionsandofsGrtil igbnare cord rreeactct. ions on aldehydes or ketones. This is
e) There.e Your
(exampl
CH3CH2CH20H �C O
'>
Br etMgher h H)O' �OH cr03
o- 0H PB,)
H2
S
04
�o
�OH H+ Br2 �Br etMgher �MgBr
� CH3SNCIH20H �';­
OCH2CH3" H2S04 �OH
Please refer to solution page of this Solutions Manual.
(a) �OH SOCl2 - �CI (b) Q- 0H SOCl2 o- CI
PBr3 -�Br
PBr3 - o-Br
P12
P12 0-1
1 1- 38
°
/"'y
U
U
____
eel
/"'y
U
one
/'..
_ ____
___
L\
___
1 1- 39
_
_ ____
�
1-2 0,
hv
____
___
H 3 0+
12
1 1- 40
2°
..
1°
---�I
(c) oH HCl - 01
HBr - 0'
-0
..
3°
Cd) (jOH SOCl2 &CI
PBr3 &Rr
P12 &1
..
..
III
-
245
"
(a) �
COOH
Q
11-41
(b)
�
(e)
R
(f ro min version )
R
(f)
(e)
Cl
Q
cJ
(g) �Br
lfH
°
(d)
(h)
CH20MgBr
bCH3CH3
H
(I) C;-H
(i) + OCH] � CH30H 0
(m) �O� �
major �
mInor EtOH
(y 0- Na+ CH)CII,Br (y 0CH2CH3
(y 0H
(a) V
ether synthesis
V
V
Br
OH
Mg
d NaOH. d� V
�H
(( d
,
t
t H30+
H2S04
OH
MgBr H� � cr03
0
H
SO
,
;
p'
O H
�
OH
H
�
(d) VOH �o 2) CH3H30C+ H2MgBr .. U
U)
+
(k)
+
(n)
+
+
11-42
•
( b)
W li li amso n
�
M gB r
+
y
..
(C )
B rM
-E..
ether
t:,.
1)
�
ether
°
°
II
•
2)
�
Sr
1)
246
a ceton e
Major product for each reaction is shown.
(a)�
(c)� (dl O (el O 0
Note that (d), (e), and produce the alkene.
(c) V COOCH2CH3
(a) CH3CH2CH2COOCH3
P-OHCH3
(d) CH3CH20-OCH2
HN CI+
(y 0-�-CH3
Cl-�-CH3 pyridine V 0 + a
chlmetohrianesul
de fonyl
(a) HO", �H SOCl2 C�
S-retention
HO', -.;H pyriTsCdiIne TSO� KEr
R-inversion
(c) HO", �H pyrTsCidiIne TSO�
R-inversion
11-43
cis
+
trans-rearranged
(b)�
cis
+
cis
trans
+
(t)
trans
rearranged
( f) ,
same
11-44
I
h-
o
II
I
o
11-45
II
I
o
II
..
o
11-46
/�
_
S
(b)
.
..
/"-..../ "-..../
_
S
/"-.�
.../
_
S
Alt�rnatlvely,
PBr3 c()uld be
.
.
S
..
S
used,
could use NaOH
if kept cold to
avoid elimination
247
inued ts alcohols to bromides without rearrangement because no carbocation intermediate is
cont3 conver
PBr
produced. OH
Br
� PBr3 �
qU H
q
U
OH �o
U
PBf3•
Bf
OH
V'
V
0 0H 0°- Na CH3.I
d) H OH SOCI2 l
l
ct- •
TspyriCdiIn..e
OR
HCH3
QCH3H �CH3H SN2 CH3H
qCH3' H
OH
HI04 q. oH
(e) ct H2S04 CK. 2) Me203 S
OH
Y
CH2
COOH
H
0
H2
C
r
4
0
a
a
g -t-o- 0H HCr2S0O43 • +0 0
acetone
-t-o- 0H H2SO4 -to
�I TsCI
OH
aI
P,I2
..
a
CH3 cis OH pyridine �
CH3 cis OTs
1 1- 47
(b)
11-48
·
pec.
(a)
(b)
(e)
�
(
•
inversion
retention
�
( f)
(
..
...
..
)
(h)
(i)
1)
I
•
�
..
..
(j)
248
yaH PBr3 )J""Br
yaH SOCl2 yCi
yaH ZnClHCl2 QCI
and
yaH HBr QBr
and
yaH TsCl, pyridine )J""Br
O
H
�
(a) �OH
cloudy in min.
no reaction
O
H
(b) cloudy�
�
immediiante blumie-gnr.een nocloudyreactiinon-stmin.ays orange
(c)
Q-OH
o
climomediudy iante blumie grn.een nono rreeactactiioonn-stays orange
(d)
Q- OH
0
0
iclmmoudyediiante blumie grn.een nono rreeactactiioon-sn tays orange
(e )
clooc
udy in min.
Lucas: no reaction
(a)
(b )
( c)
(d)
(e )
1 1 -49
..
inversion
•
retention
SNi
•
cis
trans
SNi
•
cis
trans
inversion, SN2
1)
•
2) NaBr
1 1- 50
1- 5
Lucas:
OH
1 -5
<
1
<
1
1 -5
1-5
249
1
(b){
o:s
1 1-5 1
\
H
H
...
/
p =c\
(a)
H
-
•
: 0:
0:
�
'
cO
..
...
: 0:
:::::-....
..
�
cO
•
t
: 0:
•
...
. .
-
:00
..
•
..
�
'
c6
�
.
.
. .
: 0:
00
•
: 0:
: 0:
c6
H
: 0:
: 0:
t
H
. .
. .
cO
}
/
\:c-c
/
\\
H
: 0:
...
cO
•
�
�
�
The las t three res onance for ms ar es imi lar to the firs t three; the change si that the electr ons ar e shown in
al tern ate pos itions in the benzeneri ng. To ber igorous ly cor rect, thes e threer es onance for ms sho uld be
included, but mos t chemis st would not wr ite thems ince they do notr eveal extr a char ge del ocalization;
unders tand that they woulds till bes ignificant, even if not wri tten with the others.
(c)
II
: O-S- CH3
•
II
.
�
OH
A
....
...
f---l.�
: 0:
1 1-5 2
I
: 0:
: 0:
: 0:
�
Br
C
I
0 ==S- CH 3
•
.
II
....
..
f---l.�
: 0:
Mg
•
ether
D
�
M gBr
+
�
0
B
250
II
:0 :
0==S- CH3
•
.
I
---
E
O M gBr
0'
�Br M g
__ --i.�
U
ether
X
w
v
1 1- 54
[��+ f)]J,
00
�
M gBr
HO
era
+
CH
00 _-_H_2-i�� �5
10
alkyl
shift-.
.
n ng expansIO n
recall that
c arbocations
probably do not exist; this could
be considered a transition state
N OTE :
O-D
_----;.�
ri ng exp ansion fr om 1 °
car bocation to 2°, resonance­
stabilized c ar bocation
T he mi gration directly above does N OT occur
as the cationpr oduced is notr esonance- stabilized.
An alternative mechanism could be proposed: protonate the ring oxygen, open the ring to a 2°
carbocation followed by a hydride shift to a resonance-stabilized cation, ring closure, and dehydration.
r?\
:96
o
�
t OH j
(
H: 9 II:� b
hydride sh i ft to resonance
stabilized cation
OH
-
2°
car bocation
. .
H+ off o ne 0, H+
on the other °
..
251
�90. H
H-O+
1 1 -5 5
.
( a) ("f OH
V
( b)
.
'----A Z
•
nCl2
-
� o -z nCI 2
V H
r ,.,
-
.
V � OH
• •
H i" QS 0 3H
�
'---'tf
Cl- + H20
9+
Cf(+
�
H -Cl
+ H20
H
�H2
C
----;.�
__
.
OH2
,-- .
..
�
�
_
• •
H
\
�c� \ .
� + H20:
c( ) A ll thr ee of ht e produc ts go thr ough a c omm on carboc ati oni ntermediate.
H ... ... H
�
H�
�
•
•
�
OH
H dS 03H
•
t-
r'0 � � �+ V A.. H V ; C ' H
H
Onc e ht i sc arboc ati on si formed,
removal of adjac ent protons
produces thec ompounds sh own.
Cl
��
+ HO - Z- n Cl 2
I n ht i s presentati on of the mec hanis m, ht e
rearrangementi s sh ownc onc urr ently with
c leavage of ht e C- O bond wi th no 1 0
c arbocati oni nter medi ate.
r'0
oc/ :ci: 0
}\
H
'"
-
OH
OH
-
�
�
..
H dS 03H +
�: O ... H
f
�
i OU+r:(ll-_H
�
CO
)
�J
!
00
H20:
ci?
C
Y
+ 00
�
H20 :
+
'
' 0)
H ·· H
T h;s;s the pc oduc t fc om a p; nacol re arr ange ment.
252
ZnCI2
' .
CH3
(a) 00Hcr03 a CH3MgBr o-CH3
0-°1
H2Saceto0ne4
MgBr
t
PBr
3
CH30H CH3Br
p�HO
CH3OH HOI
OR: alternati"ve end" ("{1
H2
S
4
�
0
V
d
Griaftegrnarhydrd prolyosductis of SN1
OH
(b) 0oH PBr3 0 Br e�
Mg o MgBr .. HO+3 if
r
0
3
PCC
2
4
H
SO
CH3CH20H H �
acetone�
if
MgBr 0- MgBr'
o
CH3
(c) OH H2CrS003 4
I
oE
.
�
.
(
b
f
r
o
m
)
V
acetone
orshouswne theparS t (meta) hod
�OH PBr3..
Br etMgher ! MgBr EtOH, H+
H3
cr03
0
..
.
C
C
..
.
..
..
..
..
..
Et
O
H
OH
H
4
0
� H2S
�
twoerRMgX
cang a addaltcooholan H2Cr04 t
acet
o
ne
OH
est
,
maki
n
1t1) H3Br0M+g Mg PBr3 HO grwiothupstw;oseeequisovlualtieontn Rto CH.,OH
problem
OH
(e) MgBr VOH
fO
rom H30+
11-56
a
a
..
..
Mg,
ether
+
�
I
---
109
...
..
a
..
a
�
°
a
�
I
I
•
II
I
�
10
"
(d)
2
2
�
-
2
�
+
a
II
•
a
3°
-..../
2)
___ �
--..../
10-18
a
l)�
..
(b)
2)
253
..
N
1
a
II
1 1 -56 conti nued
(f) HO�
V
�
OH
(g)
HO"
H
Mg
P B r3
..
ether
P CC..
B rM g�
�o
..
H
�
K CN
TsCI
..
pyri di ne TS O"
H
o
..
B r2
-
hv
OS 04
�
H'
NC
..
Br
6
PCC
Mg
-
ether
CH 3CH2B r
P B r3
t
CH 3CH20H
I I -57 F or a compli cated synthesi s il ke this, begi n by work ing backwar ds. Tr y to fi gur e out where the
carbon framework came from; in thi s problem we are restricted to alcohols containing five or fewer
carbons. The dashed boxes show the fragments that must be assembled. The most practical way of
formi ng car bon- carbon bondsi s by G ri gnard reacti ons. The epoxide must be f ormed from an alkene, and
the alkene must have come fromd ehydration of an alcohol produced in a Gri gnard reaction.
�o�<>,,l,
,H
maj or
i somer
1.
2.
°
�
TsCl, p
KO H,1\
BfM g�.
y
t
)
H30:
!W
0:
254
\
MgB r
L�
2. Mg
avoid c arbocation conditions
to prevent rearrangement
Mg
ether
o-
0H
cr 03 , H2S 04, aceto ne
synthesese, thsufe fNer fcondi
rom tthioensmiofsconcept
iioonntcannot
hat incompat
ibtlhe rtheagent
scondi
or conditionstioofns
can(sao)diBotco-umehmetxiofsth.thoxiIesnedthe.pseudoe fTheirst exampl
i
o
ni
z
at
exi
s
t
wi
e
tert,iawoulryincartdhbegiocatsveecondithoen desiistnetp.hreed(fTiprrhestosductitreopnywoulwiisththoutdatnottthheewaifsodiirtstarustomeundpmetbylohitngoxiselenough
f
o
r
t
h
e
sodit-butnuymtlhbemetrosemicondhoxide dinreemetactto beihoanoln,added
f
,
t
h
e
s
o
l
v
ol
y
s
i
s
of
d
e.
)
e acidtiiconscondiof tthioensseofcondthesfteirp.st sItfebpasinicwhisodichumthemetalchoholoxidies werproteoadded
nated artoethe
Ilfuric aciibdleswioluthtiothn,e tbhasie itcnhscondi
isnucompat
alizat.ion would give methanol, sodium sulfate, and
the starting alcohol. No reacttiaontnaonneousthe alacicohold-baswoule neutdroccur
SN1 solvolysis conditions
war
m
Several synthetic sequences are possible for the second synthesis.
�OH Na �O-Na+ CH31 �O/
OR �OH pyrTsCIidine �OTs NaOCH3 �O/
OR�OH PBr3 �Br NaOCH3 �O/
Compound OH al-muscoholt;becan'a t be° oral2°lylaliccasoholthiswiwoulth andalgikvene;e a posnoirteivacte Lucas
ion witthestLucas leads to a
Compound possi
-musbltebe a cyclic ether, not an alcohol and not an alkene; other isomers of cyclic ethers
6
J5
TS
NaOCH3
Ts
C
I
"-J5 pyridinetworhiskresactfinie,on NOcannotREACTIdo anOSN!N2 reaction
but
wai
t
.
.
.
.
Theare twWiolreiaasmsonsonwhyetherthsyntis tohsylesiasteiscannot
an SN2unerdisplgoacement
ofreacta leioavin. nFig rgrsto, upbacksiby andealatktaoxickdcannot
e ion. Theroccure
an
S
2
N
de of thunder
e bridggehead
bbriecausdgeheade thecarbackbonsicannot
o invercarsionbonbecausis bleockedof thebyconsthetotrahinertsbriofdtthhgehead.
eisbricardbgedonSecond,
rcannot
ing sytsihnteeverm.t
�
whimechanich issrmequired in the
OTS
H�
atbblackstoackedckidise H -§r OTs
\:::J::t'
side view
side view
1 1 -5 8
S
SN2
1
(b)
11-59
�
X:
..
.-
.-
.-
.-
..
1
10
Y:
___
..
---l'-�
SN2
255
11-61 LetThe'saxibegialnalbycoholconsisidoxieridnigzedthetefnacttism. es as fast the equatorial alcohol. (In the olden days, this
observation was used as evidence suggesting the stereochemistry of a ring alcohol.)
as
Second,
i
t
i
s
known
t
h
at
t
h
e
oxi
d
at
i
o
n
occur
s
i
n
t
w
o
s
t
e
ps
:
1
f
o
r
m
at
i
o
n
of
t
h
e
chr
o
mat
e
est
e
r
;
and
)
2) loss of H and chromate to form the C=O. Let's look at each mechanism.
'
S
.
TER
:
FA
�
H
p2
St
e
�
(
(
.
:7
;
�
0
���
0,\.Cr03H
I
A
X
A
OH
H
H
EQU TORI L
�
A
HO
A
H
aboutin ththeeseaxiasyl spostemsiti?onWebecausknowe anythatgrsuobsuptiattuentthesaxiarealmorposeitisotanblhase the
So whatoriadol posweitknow
i
o
n
t
h
an
equat
diequataxiaolriinaltechrractoimatonse. estSoewhatr woulifdStfoerpm fwerasteer tthheanratthee-laxiimiatlinchrg sotematp? eWeestewoulr; sindceexpectthis isthcontat thraery to
whatis ratethliemdatitinag,shweow,woulStepd expect
is not tlhikeelappry toobeachraofte-tlhime basitineg.(pHowrobablabouty watStere)pt2?o the equatthe eloimriainl ation
hydrhydrstericoogengenconges((eaquatxitiaolnochrassoci
riaolmatchraeotematestd wieert)esthwoulsucher).daMorbelarfgaestovere egrrot,hup.tanhe taxiThiheaapprsl iests consioeachr issmoroftenttheewimotbasthietvhatteoertdehleatotaxiivleeavearlatedues ofto
rmechani
eaction sfrmomis exper
immienttin.g.Thus, it is reasonable to conclude that the second step of the
r
a
t
e
l
i
11-(<0
I
C
Cl
/
H
Cl
Cl
H
Cl
0
a) 62 oH01 ° �'
:
I:
H
C
o
1O
,O
O
-p
(y'
oo
p
O O <00-P=CI O QVO-P=O
o Cl - 'Cl O
C
l
(Cl
Ell H H
Cl
Cl
H
(';(
�H
o-p=o
:o-f=o
CI
CI
LiZ
\.finH
�>
continued on next page
in
1 ,3-
1
If
1
�
\ /
0
__
00
00
�
I
0
+
0
�
I")
00
1
1
00
00
" 00
00
�
__
+
1
H
256
00
H
�:N
00
I 1-62
(a) co ntinued
H
CI
G5H J� N)
• •
E2
'0
• •
O( H
•
H
I
C
O-P=O
C
H
N
+1
+
0
1
+
1
l
H3� POC�I3 H3CK:;> NOTH3C �H
h�h
H
H
OH
H
(b)
.....-
•
pyrid ine
Z aitsev
Therem ust be a stereo chemi cal requirem ent in this elimi nation. I f theS ay z
t eff alkene is not p roduced
because them ethy l gro up is trans ot thel eaving group, then the H and the leaving gro upm ust be trans and
the el m
i i nationm ust be anti- the characteri stic stereochem istryo f E2 elimi natio n. This evidence
differentiates between the two po ssibilities inp ar t (a).
H
-CH3
0
..
+
uRCH3�H · U�HQCH3 1 Qtc5� }
H2� �
H
O
H+
O
H2
f)
0
--0
r;
:
O
H
9.) :OH
O
H
H
H
11-63
(a)
HO
..
1+
°
°
: 01
+ -H
t
H
H
--
---
• •
+
•
'---
•
�
• •
H
• •
• •
..
\
---.-
..
•
• •
'---"" H
H
0
H2
H
H..V
0
�
·9t
H
OCH3
� ('l
H .. �-� H H+kO � 0
O
•
°
+
• •
'
H
°
It is equally likely for pro ot nation to occur first on the ringo xy gen, fol ol wed by ri ng opening, ht en
replacemento f
by water.
D r. K anto rowski suggests this altern ative. H e and I will arm wrestle ot determi ne whi ch mec hanism
is co rrec
_
�
2
��+
'
O
�
H
257
H0
°
H
8SZ
H�
\ O�
I)
H
(Il
�qtJ
..
(q)
p;}nUllUO::l £9-11
CHAPTER 12-INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
See p. 270 for some useful web sites with infrared and mass spectra.
12- 1 The tabl e is com pleted by r eco gniz ing that: (V) (A)
v
(cm·l)
A Jl
( m)
=
1 0,000
4000
3300
3003
2198
1700
1640
1600
400
2 .50
3.03
3. 3 3
4.55
5.88
6.10
6.25
2 5 .0
12-2 I n general,o nly bo nds with di pol e mom ents will have an I R abso rptio n .
H-C
C-H
t t
H-C
t
C-C�
yes
no y es
H
I
k ytes
H3C-C-H
�C-C
/
\
CH 3
I
C=C
\
I
H
H
t
y es
t t
no y es
y es
H 3C
C-C�
\
I
C H3
H
t
C=C
\
I
H
H
no
y es
(weak)
12- 3
(a) al kene: C=C at1 640 cm-I , =C-H at 3080 cm-l
(b) al kane: no peaks indicating spo r sp2 car bo ns pr esent
( c) This IR sho wsmo re thano ne gro up. There is atermi nal alky ne sho wn by: C=C at 2 100 cmI
- , =C-H
at 3300 cm!
- . These signals i ndicate an aro matic hy dro carbo n as well: =C- H at 3050 cm!
- , C=C at16 00
cm-! .
12-4
(a) 2° ami ne, R- N H-R : o ne peak at 3300 cm-! indicates an N-H bo nd; this spectr um also sho ws a
C=C at1 640 cm-!
(b) carbo xy lic acid: the extremely bro ad abso rptio n in the 25 00-35 00 cml
- range, w ith a "sho ul der"
aro und 2 500-2 700 cm-! , and a C=O at1 7 10 cm!
- , ar e com pel ling evidence for a carbo xy lic acid
(c) al co hol: stro ng, bro ad O- H at 3330 cm-!
12 -5
( a) co nj ugatedk eto ne: the small p eak at 3030 cm!
- suggests =C-H, and the stro ng peak at 1 685 cm-! IS
co nsistent with a keto ne co nj ugated wi th the alkene. The C=C is indicated by a very sm all peak aro und
1 600 cm!
- .
(b) ester: the C=O absor ptio n at 1 73 8 cm!
- (higher than the keto ne's 1 7 1 0 cm-! ),i n co nj unctio n with the
stro ng C-O at 12 00 cm-! , poi nts ot an ester
(c) ami de: the two peaks at 3 160- 3360 cm-! ar el ikely ot be an NH2 gro up; the stro ngp eak at 1 640 cm-! is
ot o stro ng fo r an al kene, so itm ust be a differ ent yt peo f C=X, in this case a C=O , so ol w because it is part
o f an a mi de
12 -6
(a) The sm al l peak at1 642 cm-! indicates a C=C, co nsistent with the=C- H at 3080 cm-! . This appears
ot be a simpl e al kene.
(b) The stro ng absorp tio n at 1 69 1 cm-! i s unmi stakably a c=o. The smaller p eak at 1 62 6 cm-! indicates a
C=C, pro bably co nj ugated with the C= O . The two peaks at 2 7 12 cm-! and at 2 8 14 cm-! r epresent H-C= O
co nfirm ing that this is an al dehy de.
259
1 2- 6 conti nued
(c) The str ong peak at 1 650 cm-' is C=C, probably conjugated as it is unusually str ong. The 1 703 em'
peak appear s to be a conj ugated C=O, undeni ably a car boxy lic acid because of the str ong , br oad O-H
absor pti on from2 400-3 400 cm-' .
(d) The C=O absor ption at 1 742 cm'
- coupled with C-O at 122 0 cm-' suggest an ester. The smallp eak at
1 604 cm-' , p eaks above3 000 cm-', and peaks in the 600- 800 cm-' regioni ndicate a benz ener ing.
12 -7
(a) TheM an dM +2 p eaks of equal inten sity i dentify the presen ce of bromi ne. The mass ofM ( 1 56 ) mi nus
the weight of the lighter isotope of br omi ne (79) gi ves the mass of the rest of the molecule: 1 56 79 = 77.
The C6HS (p hen yl) group weighs 77; this compound isb romob enz ene, C6HSBr .
(b) The mJz 12 7 peak shows that iodine is presen t. The mol ecular ion mi nus iodine gives ther emainde r of
ht e molec ul e: 15 6 -12 7 = 2 9 . The C2HS (ethyl) group weighs 2 9; this compound is iodoethane, C2HSI.
c( ) The M an d M +2 peaks have relative intensities of about3 : 1 , a sure sign of chlor ine. T he mass of M
mi nus the mass of the lighter i sotope of chl orin e gi ves the mass of ther emainder of the molecul e: 90 35 =
55. A fragmen t of mass 55 is not on e of the com mon al ky l groups (IS, 2 9, 43 , 57, etc.,i ncreasi ng in
ni cremen ts of 1 4 mass units (CH2)), so the presence of an atoml ike oxy gen must be consider ed. I n addit oi n
to the chlorin e atom, mass 5 5 coul d be C4H7 or C3H30 . P ossible molecular formulas ar e C4H7CI or
-
-
C3H3ClO.
(d) The odd-mass molec ul ar ion in dicates the p resenc e of an oddn umber of nitrogen atoms (alway s beg ni
by assumi ng onen itrogen). The rest of the mol ecule must be: 1 1 5 - 1 4 = 1 0 1; this is most likely C7H 17
The formula C7H17N is the correct formula of a molecul e withn o elements of un saturation. The seven
carbon s probably incl ude alky l gr oups like ethyl or p ropy l or isopr opy .l
12- 8 Recall that radical s are not detected ni mass spectr ometry; only positively - char ged ions ar e detected.
CH3
+
I
CH3-CH-CH2 +
12 -9
[
J.
j
r
�
�
CH
CH3- H
57
43
CH3
CH3
I
CH3-CH
-
+
85
mJz
mass 2 9
r adicals
not detected
mJz 57
CH3
CH2 - H
CH2-CH3
+
CH3
I
CH2-CH-CH3
mJz 43
1 00
CH3
I
CH3- <;H
-
mass 57
CH3
I
CH2-CH-CH3
+
+
mass43
-
mJz
CH3
CH3
I
I
CH3-CH-CH2-CH
mJz
260
85
+
+
57
CH3
mass
15
1 2- 1 0 The molecular weig ht of each isomer is 1 16g/ mole, so the molecular ion app ear s at rnJz 1 1 6. The left
h al f of each str ucture is the same; loss of a three car bon radicalg ives a stabiliz ed cation , each with rnJz 73:
iH2C:O� .. H2C";�t iH2C: �'" H2C,,+�}
..
.°
.
•
..
..
rnlz73
·
°.
rnlz73
.
W here the two stru ctures differ is ni the alpha-cleavage on ht e right side of the oxy gen . A lpha-cleavage on
the left structure loses two carbon s, whereas alpha- cleavage on the right structure loses on yl on e carbon.
[ �I o0-/-r CH2CH3
t
alphacleavage
loss of
{�..O.. �H2 ... �O�.. CH2}
C9H200,
[ �o*- r
I alphat cleavag e
J �..,�,
R +
1
loss of
...
.
CH3
�+��,}_
R
rnJz 87
rnJz 1 0 1
1 2- 1 1 2,6-D imethy lheptan -4-01 ,
has molecular weight 144. The hig hest mass peak at 1 26
the molecular ion , but rather is the loss of water ( 1 8) from the molecular ion .
[�r
-
rnlz 1 44
H,o+
(CH3)
[�r
is
not
rnlz 1 26
from thef ragmen t of rnJz 1 26 . This is called ally lic
Th e peak at rnJz III is loss of an other 1 5
cleavage; i t gen erates a 2°, ally lic, reson an ce-stabiliz ed carbocation.
[ � r-i�+- �}+ CH)
rnJz 1 26
rnJz III
The peak at rnlz 87 results fromfrag men tation on on e side of the alcohol :
�:����!C
H}
[�r {JJH )ji + �
rnlz 1 44
_
mlz 87
reson an ce-stabiliz ed
.
mass 57
1 2- 1 2 P lease ref er to solution 1 -20,p age 1 2 of thisS olution s M an ual .
1 2- 1 3 D ivide then umbers ni to 1 0,000 to arrive at the an swer.
(b) 2959 cm1
-
(a) 1 603 cm-1
1 2- 1 4
(a)
H C==C CH2CH3
H t 'H
,
/
/
1 660 cm-1
or
(c) 1 709 cm1
-
(d) 1 739 cm-1
CH2
C
H3
O==C H
t '
(e) 22 1 2 cm-1
/
1 7 10 cm1
-
stronger absorption-larger
261
dipole
(f) 3 300 cm-1
1 2- 14 continued
H
I
\H
t
1660 cm-1
C=C
(b)
H
H
I
\H
t
1 660 cm1
-
"N=C
I
\H
t
1 640 cm1
-
I
stronger absorption-larger
H
CH2CH3
(c)
OCH2CH3
\
C=C
or
I
H
H
CH2CH3
\
CH2CH3
\
H
I
\H
t
1 660 cm-1
C=C
or
/
d ipole
I
stronger absorption-
larg er dipole
H
\
\H
t
1 660 cm1
-
C=C
(d)
H 3C
H
CH3
I
H
I
/
\H
t
1 660 cm1
-
C=C
or
I
CH2CH3
\
stronger absorption
(no dipole moment)
-
1 2- 1 5
H3C
\
H3C
I
30003 100 cm1
and
t 1 660 cm1
I
\
CH3
O�
1 620 cm-l
� .
CH3
\
I
1 660 cm-1
moder ate i ntensity
weak or non- existent
(b)
<'c=cl
H
t CH(CH3h
H
CH3
C=C
(a)
larg er dipole
and
0
�
1 645 cm�1
conj ug ated
not conj ug ated
(c) both car bony ls show str ong absor ptions ar ound 1 7 10 cm-1
o
II
+
o
CH3(CH2)3 - C H
(d)
0\
II
CH3(CH2h - C - CH3
and
2700-2800 c m-1
two small peaks
o y.
H
1200
3300 cm1
br oad, str ong
c m-1
and
a�
262
1 7 1O cm�1
str ong
12 - 1 5c ontinu ed
C 3 300 cm1
CH3(CH2h - C:: C - H
t
(e )
CH3(CH2)6 -C:: N
and
t
22 00-2 300c m1
moder ate tostr ong int ens ti y
2 100-22 00c m-1
weak to moder at e intens ti y
( f) bot hc ar bonyls s hows rt ong abs or pt ions ar ou nd 1 7 1 0c m-1 � 3300c m-1
fH
br oad,str ong
0
'
0
o
1\
1\
I
and
CH3CH2CH2 - C - OH
CH3 - CH - CH2 - C - H
t
2 � 00 cm1
ver y br oad
2 700-2 800 cmi
t wos mall peaks
1 7 1O cm-i � 0
0/ 1 650 cm-1
II
CH3CH2CH2 -C -
(g)
�-�
o/ 3300 cm1
II
CH3CH2 - C -CH2CH3
and
two peaks
o�
12 - 1 6
(a)
II
H
\
(b)
C-OH
t
/
.......-
C=C
2 400-3400c
\
H
CH3
/
o�
1 700 cm1
-
II
1 7 1 5c m-I
CH3-CH-C-CH3
I
m-1
CH3
1 640c m-1
tV 3000- 3 1 00 cm-I
(c )
�
V,
CH2 -C
"-
12 - 1 7
(a)
[
CH3
�71
CH3
�
H
43
j
mlz 86
t
(d)
N
1 600 cmI
-
CH2CH2CH3
a/"
I
22 o cm1
./
H
3000-3 1 00 cmi
3400 cmi
H
N ..... V
\
CH2CH3
,
,
"'=::
�,
T
2 900 -3000 cmI
1 600c m-I
1
+
CH3
•
•
I
+ CH
- CH2CH2CH3
mlz
71
CH3
•
I
CH3-CH
+
263
mlz 43
[
b�� �
1 2- 1 7c ontinued
(b) CH3 -CH=
CH2CH3
9
(c )
cH
H
CH3
b
CH,- H-CH1
45
rnJz 1 02
[
CH3
�
CH3 -CH==CH - H -CH3
rnJz 84
j
�i
----l..
]
rnJz 98- allylic c leavage
[ Jb��j-
r
CH3_CH=
1
t alpha-
c leavage
..
----I�
+
··
i�
H'
rearr ange
..
F c�
•
H'
rnJz 43
G
•
rnJz 1 29
+ OH
CH3
II
I
___ CH - CH2 - CH - CH3
O-
rnJz 85
� =c �
H3
264
rnJz 45
H
C
o
T he tropylium ion si a
ch ar ac teris tic fragment
fr om phen ylalkanes.
m1z 9 1
CH
1O-
CH,
CH3- H
�H3
H3
rnJz 69
b }
:OH
.
��
rnJz 1 44
alpha-c leavage
�
CH3_ H_
alpha­
c leavage
/CH3
HC+
rnJz 77
-
:OH
CH3
I
I
+ CH - CH2 - CH -CH3
benzylc ation
rnJz 1 3 4
<iH,
H3
rnJz 87
(d) [&f-. () �
•
b
H3
}
C
y
r earr an ge
..
H3C, + "CH3
C
I
CH3
rnJz 57
t
1 2- 1 8
(a )
�CH2+ + H2C' CH3
[�r �CH2+ H2C......./
�CH2+ Hi �
H
o
+
CH3
(5 cJ
•
-----
mass 29
mJz 85
rnJz114
+
mass 43
mJz 7 1
+
+
(b)
:
mass 57
mJz 5 7
.
-----
mass1 5
rnIz 83
mJz 98
(d)
CH3
CH3-C==CH-CH2+
I
c( )
+ oCH3
--
+j i :OH +OH }
OH
[ �Hj CH, -CH,CH,CH, �H' gH, +. CH,CH,CH,CH,
.
•
31
.
{ CH2=CH- �H2
CH2=CH -CH2-CH2 + CH3
+
0
55
�
----
mass 57
mJz 31
CH,=CH- c::t C� CH,
mJz 70
.
-
rnIz 88
-
[
mass 15
.
-
! H20
m1z 69
�
mJz 4 1
H2-CH = CH2 } CH2CH3
-----
mJz 1 2 1
c ontinued o n nex t page
mJz 5 5
+
0
mass 29
Oc+
1.0
rnJz 77
+
0
NHCH2CH3
mass 44
265
mass 1 5
12- 1 8 (e) c ontinu ed
[a�::rr
mlz 121
12- 1 9
(a) Thec har act er si t ci fr equ enc ies of t he OH abs or pt ion andt he C=C abs or ption will indic ate the pr es enc e
or abs enc e of t he gr ou ps. A s pectru m with an abs or pt ion ar ou nd 3300c m-i will haves omec yc lohexanpl i n
it; i ft hat s ames pec tru m als o has a peak at 1 645c m-i ,t hent hes ample will als oc ont ains omec yc ol he xene.
Pur es amples will have peaks r epr es ent at ive of only one of thec ompou nds and not the other . N ote that
quantitation of thet woc ompou nds wou ld be ver y diff ci u lt by I R bec aus e thes rt ength of abs orpt ions ar e
ver y diff er ent . Us ually, ot her met hods ar eus ed in pr ef er enc et o I Rf or qu antitat ive measur ements .
0\
o y'H
O
3 300c m- i
hr oad,str ong
--
1 645c m-1
moder ate
12 00c m-1
(b) M ass s pec trometr yc an be mis leading wit h alc ohols . Usu ally, alc ohols dehydr ate in the inlet s ystem of
a mass s pec rt omet er , andt hec har ac ter si t ci peaks obs er ved in the mass s pec tru m ar e thos e of t he alken e,
not of t he par ent alc ohol. F or t his par ticu lar analys si , mass s pectr ometr y wou ld beu nr eliable and per haps
mis leading.
12-2 0
(a) The "stu dent pr ep"c ompou nd mus t be I- br omobu tane. The mos t obvious featur e of the mass spec rt um
si t he pair of peaks atM andM +2 of appr oximat ely equ al heights ,c har act er si tic of a br omine atom. L oss of
br omi ne (79)fr om the molecu lar ion at 1 36 gi ves a mass of 57, C4H9, a butyl gr oup. W hic h of t hef our
possi ble bu tyl gr ou ps? The peaks at 1 07 (loss of 29, C2Hs) and 93 (loss of 43 , C3H7) ar ec ons si tent w ti h a
linear c hai n, not a br anc hedc hain.
(b) The bas e peak at 57 si s os rt ong bec aus e thec ar bon-halogen bond si the weakes t in the molecu le.
Typic ally, loss of halogen si the dominant fr agmentat ion in alkyl halides .
[4
:r
�{�
/' Br
r
rnJz 1 36
+
+
�
rnJz 1 07
mass 29
+
H2C.
Br
rnJz 93
+
mass 43
/'-../
+
rnJz 57
H2C-..../
+
Br
mass 79
•
266
12 -2 1
(a) D euter ui mh as twic e ht e mas s ofh ydr ogen,bu ts imilar s pr ingc ons tant,k . Compar e th e fr equenc y of
C-D vibr ation to C-H vibr ationb ys ettingu p ar atio,ch an ging only ht e mass (su bs titu te 2 m for m).
=
=
�lt2
=
0.707
0.7 07 vH = 0.7 07 (3000 em-I) 2100 cm-1
(b) The func tional gr oup mos t lik ely to bec onfus ed with a C- D str etc h si the alk yne c( ar bon- c ar bon tn ple
bond), whic h appears in ht es amer egion and si of ten ver y weak.
12-22
(
Vo
=
'"
�
m/z 1 14
[+r
r*r
r-
�
+
�
m/z 57
1+
a 1°c ar boc at ion- not favor able
a 2°c ar boc at ion-r eas onable
m/z 57
m/z 1 14
�
�.
a 3°c ar boc ation-t he best
m/z 57
m/z 114
The most likely rf agment ation of 2 ,2, 3,3- tetr amet hylbut ane will give a 3°c ar boc at ion,th e mos tst able of the
c ommon alkylc at ions. The molecu l ar ion s hould bes mall or non- exis tent whi le m/z 57 si lik elyt o be the
bas e peak , wher eas t he molecu lar ion peaks will be mor e pr omi nent for n -oc tane and for 3 ,4- di methylh exane .
12 -2 3
(a) Th e infor mat iont hat this mys teryc ompound si ah ydr oc ar bon mak es inter pr et ni g the mass s pectru m much
eas ier . ( It si r elat ivelys i mple to tel l i f ac ompound has c hlori ne, br omine, or nitr ogen by a mass s pec rt u m, but
oxygen si di ffic ult to determ ine by mass s pec rt ometr y alone. ) A h ydr oc ar bon with molec ular ion of 1 1 0c an
have only 8c ar bons ( 8 x 12 = 96) and 1 4 hydr ogens . The for mula CgHl4 has t wo elements of uns atur at ion.
(b) The IR will be us eful in detenni ning whatt he elements ofu ns atur ation ar e. Cyc loalk an es ar e gener ally
not dist inguish able inth e IR. An alk en esh ouldh ave an abs or ption ar ound 1 600- 1 650 em-I; none si pr es ent in
ht si I R. A n alkynes houldh ave as mall,shar p peak ar ound 22 00 em-I-PRESENT A T 2 12 0 em-I! A s
l o, a
sh ar p peak ar ound 3 300c m-1 indic ates ah ydr ogen on an alk yn e,s o ht e alk yne si at on e end oft he molec ul e.
B ot h elements of uns at ur at ion ar e acc ounted for byt he alk yne.
267
12-2 3
(c ) T he onl y question is how ar e theo ther carb ons arr anged. T hem ass spec rt um sho ws a pr ogr ess oi n of
peak s fr om themo lec ular ion at 1 10 to 95 (lo ss of CH3), to 8 1 l(os so f C2Hs), ot 67 (loss o f C3H7). The
mass spec rt um suggestsi ti s a il near chain. T he extr a ev idence that hydro genation of the mystery com po und
giv es n-o ctanev eri fies that the chain is linear . T he origi nal c ompoundm ustb eo ct- 1- yne.
( d) T he base peak is so str ongb ecause the ion pr oduc ed iss tabil z
i edb yr esonanc e.
�
m/z 95
�
m/z 8 1
�_
m/z 67
+
m/z l lO
+
+
+
'-.....
m/z 39
c:::=:::>
12 -2 4
(a) and (b) T hem as s spec is consistent with the for mul a of the alkyne, CSHJ4' mass 1 1 0. T he IR isno t
c onsistent with the al kyne, howev er . O ften, symmetric all y substituted alk ynes hav e a mi nisc ule C=C peak,
so the fac t that the I R does not sho w this peak does not prov e that the alkyne si ab sent. T he m
i pOlt ant
I
ev idenc e in the I R si the signific ant peak at 1 62 0c m- ; this abs orp tion isc har ac teri stic of a conjugat ed
diene. I nstead of the al kyneb eing form ed, the reac tion must hav eb een a doubl e el m
i ination to the dic nc .
�*+
Br
12 -2 5
(a)
c( )
b( )
II �
Q
-
,
\
OT
0
g
oII
'/
\\
-
"
�
'\
H
I
k
-CH3
I
12 00 cm-J
1 5 80cm-J
268
11--
t
C-C-H
1 600c mI
-
17 60cm-I
O , ...,... 1 72 0 cm-1
27 1 0,
2 82 0c m-1
o
12 2- 6
(a)
� Br
Mg
ether
..
/""-....
/""-....
,,/
........",
- M gB r
A
H30+.. �OH
C7HI60
2m
- eth yl hexan-2-01
m ol. wt. 116
b( ) T he mol ec ul ar i oni s not vi sib le in the spec rt um. Alc ohols typic ally dehydr atei n the hot inlet system of
the mass spec rt om eter , especi all y true for 30 alc ohol s that ar e the easi est type to dehydr ate. T he two
fragmentati ons that produc e a resonanc e-stabiliz edc arb oc ation give the m ajor peak s in the spec rt um at mlz
59 and 1 0 1 .
[ � OH l t {
___
�
mass 1 1 6
;<�-H
_�
..
_
H3C CH3
rnIz 59
..
/<O+-H }
H3C CH3
�
rnIz 1 01
O-H
c(+CH3
. }
12 -2 7 T he unk nownc om pound has peak s in the mass spec at rnIz 1 98, 1 55 , 127 ,7 1 , and 43. I ti s helpf ul
that them asses of two of the fr agm ents, 1 5 5 and 43, sum to 198, as do the other two fragm ent masses,
127 and 7 1 . W ec an say wi thc ertai nty that the mol ec ul ari oni s at rnIz 1 98, and that the unk nowni s a
rel ativel y sim pl em ol ec ul e wit h twom ai n fragm entati ons.
T hi si s a fairl y hi ghm ass for a sim pl ec ompound; som e heavy group mustb e pr esent. W hat is N O T
pr esenti s N b ecause of the even mol ec ul ar ionm ass, nor nor B rb ec ause of thel ack ofi sotope peak s,
nor a phenyl groupb ec ause of no peak at77 . T he progression of alk yl gr oupm asses: 1 5 ,2 9,43 , 57 ,71 ,
85, 99- incl udes two of the peak s, soi t appear s that the unk nownc ontai ns a pr opyl gr oup and a pentyl
group (the pr opyl c oul db e part of the pentyl gr oup). T he 12 7 fragm enti sk ey; the fragm ent
has
thi s m ass,b ut we would expec tm uc hm or e fr agm entati on fr om a ninec arb on pi ec e. T here mustb e som e
o ther expl anati on for thi s 127 pe ak.
CI
C9HI9
A nd ther e is! T her ei s one pi ec e-m ore speci ifc al ly, one atom- that hasm ass 127: i odine! I n all
prob abil ity, the iodi ne atom i s attac hed to a fr agment of mass7 1 whic hi s
' a pentyl group. W e
c annot tell withc ertai nty whati som er i t is, so unl ess ther ei s som e other evidenc e,l et' s pr opose a str ai ght
c hai ni somer, 1 i- odopentane.
CsHl1
127
155
I
T he 155 rf agment probabl y
has thisb ir dged struc ture
b ec ause iodi nei s sobi g:
269
12-2 8
(a)
CH3
CH3 CH3
I
I
I
I
I
HC-C-C-CH3
3
OH OH
s rt ong peak
17 1 5 cm 1
T
broad,s rt ong OH peak
around 3300 cm-I
o
HC-C-C-CH3
3
II I
/ 0 CH3
around
-
OH
str ong peak ar ound
1 690 cm-I
(b)
br oad, strong 011 pe ak
ar ound 3 300 em-'
two peak s at
27 00 and 2 800 cm-I
o
(c)
o
0/
t
H
br oad COOH band
mi ss ing
ver y br oad
2 500- 3 500 cm-1
with a "shoulder "
ar ound 27 00 cm-I
still has a str ong
O H at 3300 cm-I
butd iffer ent fr om
COOH
I f you wish to find IR spectr a and mass spectr a of comm on compounds, ther e ar e two w eb sites
that ar e ver y helpfuL E nter ing a name or molecular form ula will giv e isom er s fr om which t o
choose the desir eds rt uctur e and the IR or MS if av ai lable in their d atabase.
http:// webbook .nisLg ov/
"NIS T " is the U.S . N ational I ns titute ofS tandards and T echnology.
http://www.aist.go.jpIRIODB/SDBS/menu-e.html
T his si fr om the N ational I nstitute of A dvanced I ndustr ialS cience and T echnology of J apan.
270
CHAPTER 13-NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY
A b enzener ing canb e wri tten with thr ee altern ating doub leb ond s or with a cir cle in ther ing. A l l of the
carb ons and hydr ogens in an un sub stitutedb enzener ing ar e equivalent,r egar dless of which sy mb olis m is
used.
o
equi val entto
The Japanese web site listed at the bottom of p. 270 also gives proton and carbon NMR spectra.
Reminder: The word "spectrum" is singular; the word "spectra" is plural.
1 3- 1
( a)
650 Hz
300 x 1 06 Hz
=
2. 17 x 1 0-6
(b) D iffer ence in magnetic field
=
=
2. 17 ppm dow nfield rf om TMS
7 0,459 gauss x (2 . 17 x 1 0-6)
0. 1 5 3 gauss
=
( c) T he chemical shift does not change w ith field str ength: «52 . 17 atb oth 60 MHz and 300 MHz.
( d) 2( . 17 ppm) x ( 60 MHz) = 2( . 17 x lO-6 ) x ( 60 x 1 06 Hz) = 130Hz
*
1 3-2 N umb er s are chemi cal shi ft values, in ppm, derived fr om Table 1 3 -3 and the A ppendix in the text.
Your predictions should be in the given range, or within 0.5 ppm o/the given value.
( a)
b
( b)
.... CH3 �
b
CH3
'
c
CH3 \
/
b
C= C
b
\
/
CH3
H
C....
a CH3 \
......
b
�
( c)
�
c p
a
c
H
H
c
a
b
c
=
""
=
=
=
«5 5-6
«5 0.9
CH3
�
I
h-
b
�
a
b
OC J
b
a
b
CH3
=
=
«57 2.
«52 .3
b
a
( d)
H
b
CH3
H
b
=
=
/)7 2.
/) 3 .6
� H3
c
b
CH3-C-C::C-H
c
em
a
b
c
=
=
=
«52- 5
/)2 . 5
/) 1-2
a
a
c
H
b
CH3
C�
H
�
--uH
c
�
*
H
( e)
......
a
b
a
H
�
( f)
a
CH2-C-O-H
b
b
�H3
CH3 -C-CH2 Br
I
Br
c
/) 10- 12
«5 3 ( between tw o deshielding gr oups)
/)7 2
.
( Th eh ydr ogens l ab el ed "c" ar e not equival ent.
Th ey appear atr oughly the sam e chemi cal shift
b ecause the sub stituent is neither stro ngly electr ond onating nor withdr awing.)
271
a
a
b
=
=
«5 3- 4
«5 1- 2
1 3 -3
(a)
c
b
a
(b )
CH3CH2CH2Cl
1 3 -4
�
(a)
*:
H
3
h
c
a
(c)
CH3CHCH3
I
�H3 b
a
(d)
c
b
H
CH3-C-CH2CH3
I
two types of H
*
H
�
C 3
h
c
a
three types of H
a
�
H
CH3
t
H
d
fiv e types
Br
of H
H
c
H
a b
Cl
three types of H
a
b
(b) T he three types of aromatic hydrogens appear in a relativ elys mall
s pace around () 7.2 . The signal si complexb ecause all the peaks from
the three types of hydrogens ov erlap.
d
four types of H
N ote: N MR spectra drawn in thisS olutions M anual w ill repres ent peaks as single lines. T hese lines
may not look like "real" peaks,b ut this av oids the problem ofv ar iation am ong spectrometers and
printers. I ndiv idual spectra may look different from the ones presented here,b ut all of the i mpor tant
information willb e contained in thes e representational spectra.
1 3 -5
c
CH3
c
I
I
a
a
II
a
II
�
9H
b
H3C - C - O - C - CH 2 - C - CH3
12
3H
CH3
c
�
�
2H
-.J
I
I
10
I
I
8
9
7
I
6
I
I
I
4
5
8 (ppm)
/
I
3
2
TMS
I
I
I
o
1 3 - 6 The three spectra are identified with their str uctures. D ata are giv en as chem ical shiftv alues, with the
integration ratios of each peak giv en in parentheses.
S pectrum (a)
S pectrum (b )
b
a
a
OH
H
H
a
=
cS
2.
4
(
1
)
(
1
H)
C
a
I
CH3 - C - C::C - H b = cS 2. 6 ( 1 ) ( 1 H)
a = cS 6.8 (2 ) (4H)
I
c = I) 1 . 5 (6) (6H)
b = cS 3.7 (3 ) (6H)
c ,o
OC J
CH3
c
�
b
CH3
a
CH3-� - CH2Br
I
Br
H
a
S pectru m (c)
b
*
a =
b =
I) 3 . 9
cS
9
H
�
a
( 1 ) (2 H ) (Thisi s the compound in Problem 1 3-2 ( f). You
1. (3) (6H) may wish to check your answer to that questio n
agains t the spectrum. )
272
1 3- 7 Chem ical shift values ar e appro ximate and mayv ar y sli ghtly fro m yo ur s . T he splitti ng and
integr atio n values sho uldm atch exactly, ho w ever.
(a)
b
b
b
b
Q
H3C ,
/CH3
a HC - O - CH a
/
,
CH3
H3C
12 H
.1!
(To show the pattern
of peaks, this multiplet
is larger th an it would
appear on a real spectrum . )
I
10
(b)
I
9
I
8
I
I
I
6
7
2H
TMS
" III,
I
T
3
4
5
(5 (ppm)
I
9
I
8
�
6
5H
Note : the three types of
aromatic protons above
are accidentally
(5 (ppm)
*�
a
H
-
� /;
H
a
equivalent because the
I
4
5
a
H
a
H
III
I
I
I
7
.1!
(c)
isopropyl group has little
C
2H
I
3
LI
I
9
I
8
I
7
I
6
I
5
(5 (ppm)
273
I
I
o
�
Q
IH
CH3
c
4
I
1
I
6H
"
I
I
c
CH]
(To show the pattern
o f peaks, this m ultiplet
is larger than it would
appear on a real spectrum . )
shifts.
TMS
2
/
b
H
a
effect on their chemical
10
1
I
T
o
3H
.1!
I
I
Q
b
a c 0
II
ClCH2CH2 - C - O CH3
2H
10
T
2
1
3
,
III,
I
2
TMS
I
1
I
o
1 3-7 co ntinued
*
aH
(d)
c b
l!
CH3CH20
4H
9
I
I
5
6
7
8
H
b
a
0
II
b
�
6H
a
8 ( ppm)
I II
I
4
TMS
3
0
b
a
II
1
I
I
I
I
o
2
12
4H
( e)
c
e
OCH2CH3
12
4H
I
I
T
I
a
� Ii
aH
All four aromatic
protons are
equivalent .
10
H
c
�
6H
CH3CH20 - C - CH2CH2 - C - OCH2CH3
l!
N o splitting is ob serv ed when chemically
equivalent hydr ogens ar e adjacent to each
other , as with hydr ogens "b " ab ov e.
4H
TMS
I
I
I
9
10
1 3-8
( aJ
):
8
I
T
T
7
6
5
0
5 2.4 H2
?
8 1 . 8 H2C
\
H3C
'(
8 1 .2
H 8 5.8
H 8 6 .7
CH3 J
T he signals at 1 . 8 and 2.4 ar e
rt ipletsb ecause each seto f
pro tons has two neighbo ring
hydro gens. T he N + 1 r ule
corr ectly pr edicts each ot b e a
triplet.
8 ( ppm)
I II
I
I
3
4
I
2
I
I
T
o
b( )
o
8 1 . 5 H2C
I
8 1 .65 H2C
,
C
H2
8 2. 1
H 8 6. 2
CH3
8 1 .7
allylic
allylic
T he thr ee CH2 gro ups canb e distinguishedb y chemical shift
andb y splitting. T he a l l y l i c CH2 w il l be thefarthest downfield
o f the thr ee, at 2. 1 . T he signal at 1 . 5 is a rt iplet, so that mustb e
the one witho nly two neighbor ing pr oto ns. T he CH2 sho w ing
the multiplet at 1 .65 mustb e theo neb etween theo ther two
CH2 gr oups.
274
1 3 -9
aH
(a)
i!
5H
T he thr ee typeso f
aro matic pro ot ns
ar e accidentally
9
1
J ", 15 Hz
equivalent.
10
IH
IH
7
a
1 5 Hz
5
6
8 (ppm)
(b)
9H
C=C
/
\
C(CH3h
H
b
,......I-..
,......I-..
8
=
�*� }
H
�
12
Q
Ha
4
d
3
3H
c
CH30
CH3
/
C=C
\
/
H
Cl
I
2
12
b
TMS
0
�
3H
\
i!
IH
a
TMS
I
10
c( )
I
9
e
I
I
8
7
o
(CH3hC
I II
6
I
5
8 (ppm)
9
4
I
3
I
I
I
2
I
o
�
9H
Q
3H
�
b
8
I
d
- OCH2CH 3
\
F
C =C
\
/
H
H
II
a
10
c
T
2H
IH
IH
7
6
TMS
5
8 (ppm )
275
4
3
2
1
o
(x
1 3-9 continued
c
H
I
11
The chemical shift
of -COOH is
variable. It is
usually a broad
peak.
I
I
9
10
x>=
Q
d
b
H
The electron-withdrawing
�
Q
2H
c arbonyl group deshields the
2H
I
O
protons adjacent to it, moving
them downfieid from their
usual position at
8 7 .2.
I
I
I
I
7
8
6
.&
�C
c
H
b
H
"' OH
a
I
I
3
4
5
8 (ppm)
'-'::
3H
TMS
I
I
2
1
I
o
The formula C3H2NCI has three elements of unsaturation. The IR peak at 1 650 cm- I indicates an
alkene, while the absorption at 2200 cm- I must be from a ni tri le (not enough carbons left for an alkyne).
These two groups account for the three elements of unsaturation. So far, we have:
1 3- 1 0
\
C
I
C=C
\
/
N
+
2H
+
Cl
The NMR gi ves the coupling constant for the two protons as 1 4 Hz.
This large J value shows the two protons as trans (cis, J = 10 Hz;
geminal , J = 2 Hz). The structure must be the one in the box .
H
C=N
1 \
C=C
\
I
H
Cl
1 3- 1 1
(a)
C3H7CI-no elements of unsaturation;
a b c
Cl - CH2CH2CH3
a
c
=
=
<5 3 . 8
<5 1 .3
3 types of protons in the ratio of 2 : 2 : 3 .
(triplet, 2H) ; b
(triplet, 3H)
=
<5 2. 1
(multiplet, 2H);
The other NMR signals are two 3H singlets,
(b) C9H 1 002-5 elements of unsaturation; four
two CH3 groups . One at <5 3.9 must be a CH30
protons in the aromatic region of the NMR
i ndicate a disubstituted benzene ; the pair of
c=:====�> group. The other at <5 2.4 is most likely a CH3
group on the benzene ring.
doublets with J = 8 Hz indicate the
substituents are on opposite sides of the ring
+ OCH3 + C + 0 +
(para)
CH3
1 element of unsaturation
+ 3C + 6H + 20 +
�
1 e lement of un saturation
< 1>
-Q-
One way to assemble these pieces consistent with the NMR i s :
a
b
H
H
0
a = <5 7.9 (doublet, 2H)
b = <5 7.2 (doublet, 2H)
C 3
- OC 3 c = <5 3 .9 (singlet, 3H)
d = <5 2.4 (singlet, 3H)
H
H
276
a
b
�
*g
�
/
(Another plausible structure is to have the
methoxy group directly on the ring and to put
the carbonyl between the ring and the methyl.
Thi s does not fi t the chemical shi ft " a l ues
quite as well as the above structure, a s the
methyl would appear around <5 2. 1 or 2 . 2
instead o f 2.4.)
�
. . '
J bc =
l .4 Hz
13-13
(a)
c
f
e
d
c=c
\
/
CH3CH2CH2
.
.
1
... .. ..
I
,
1
, .
.
.
I � J bc
,
=
1 .4 Hz
JJ----_
--JU
o
H
,
�
,
.. .. ..
5. 1
II
a
b
c
d
e
f
a
C - OH
/
\
H
b
= 8 1 2. 1 (broad singlet, I H)
= 8 5 . 8 (doublet, I H)
= 8 7 . 1 (multiplet, I H)
= 8 2 .2 (quartet, 2H)
= 8 1 . 5 (sextet, 2H)
= 8 0.9 (triplet, 3H)
(b) The vinyl proton at 8 7 . 1 is He ; it is coupled with Hb and Hd, with two different coupling constants,
and Jed ' respectively. The value of Jbe can be measured most precisely from the signal for Hb a t 8 5 . 8 ;
the two peaks are separated by about 1 5 Hz, corresponding to 0.05 ppm in a 300 MHz spectrum. The
value of Jed appears to be about the standard value 8 Hz, judging from the signal at 8 7 . 1 . The splitting
tree would thus appear:
J be
8 7. 1
1
J be = I 5 Hz
" ..
. ... ,
.
.. ...
,
J ed = 8 Hz
J ed = 8 Hz
�
�
II
' ,, '
Jed = 8 Hz
over� n NMR
U
277
,
Jed = 8 Hz
�
,
'I
L
13-14
(a)
*
-
o
c
d
H
d
H
H
\
C
�
Ii
H
a
II
a
C- H
b
I
C
,
H
�
c
d
=
=
=
=
8 9.7 (doublet, I H)
8 6.7 (doublet of doublets, I H)
8 7.5 (doublet, I H)
8 7.4 (multiple peaks, 5H)
The doublet for He at 8 7.4 OVERLA PS the 5 H peaks of Hd.
b
H
d
d
(b) J ab can be determined most accurately from Ha at 8 9.7: Jab "" 8 Hz, about the same as "normal " alkyl
coupling.
be measured from Hb at 8 6.7, as the distance between either the first and third peaks or the
second and fourth peaks (see diagram below): J bc "" 18 Hz, about double the "normal" alkyl coupling.
J be can
0 6.7
... ..
.. . . . . . . . . . . 1 . . . . .. . . . . . . .
J bc 1 8 Hz
1
1
(c )
=
f"��:� �'� '1
r;.:� �'� '1
1 3- 1 5
J
b
H
a
0
II
d
C - CH3
I
\
c=c
I
,
(a) a
b
c
d
=
""
=
=
0 1 .7
0 6.8
0 5 -6
0 2. 1
�
(b) a
b
c
d
=
=
=
=
H
H3C
(c) using J bc
�
L
doublet
mu l t ip l e t (two overlapping quartets-see part (c»
doublet
singlet
c
=
1 5 Hz and J ab = 7 Hz:
.
6.8
J .1:· ;: · · .
·
·
·
.
.
;
:
.
1
r
. .
.
. . . . 'I;� ·;: :::·1- · · · · · · · · · · · '1';:'-;'
1-· · · · · ·
.
(j:�'= 7 Hz
�
o
-;.
. · ·
=
�.
�..
(
•
• •
�• • • • • ,
'�� : :
� L J �
278
••••
• •,
L
13-16
replace He
�
cis diastereomer
replace Hd
�
trans diastereomer
13-17
(a)
replace Ha
replace Hb
..
non-superimposable mirror i mages =
nantiomers; therefore, Ha and Hb arc
enantiotopic and are not distingui shable by
NMR
�
(b)
replace He on the left
�
enantiomers
replace He on the right
..
(c) The
1 3- 1 8
(a)
Hd protons are also enantiotopic.
Xy
Br
a
CH3
b
H
"
�
(b)
e
e
H
e
H
CH3
-::.,
H H
c
d
H
b
This compound has six types of protons He and
Bct are diastereotopi c , as are � and Hr. a =
8 2-5 ; b = 8 3 . 9 ; c , d = 8 1 .6 ; e,f = 8 1 .3
This compound has fi ve types of protons.
� and Hd are diastereotopi c . a = � 1 . 5;
b = � 3.6; c,d = � 1 . 7; e
8 1.0
=
279
1 3- 1 8 continued
d
(c )
H
b
Br
H
a
H
(d)
\
Cl
/
C=C
/
H
c
\
H
c
b
This compound has three types of
protons. Ha and Hb are diastereotopic.
a,b = & 5-6 ; c = & 7-8
H
H
b
This compound has six types of protons.
He and Hf are diastereotopic. a,b,c = & 7.2;
d
& 5.0; e,f = & 3 .6
==
13-19
.. �
(a)
+ H-B
CH3CH2 - O-H
•
(b)
.
•
+:B
') "
CH 3CH2- O-H
.
""
• •
-
H
I') ". -
--- CH3CH2 - �-H + .B
+
.. �
""
+ H-B
--- CH3CH2-R:
H
1
--- CH3CH2 - �:
H
--- CH3CH2- O:
.
1
.
+
H-8
+ :B
1 3 -20 The protons from the OH i n ethanol exchange with the deuteriums in D20. Thus, the OH in ethanol
is replaced with OD which does not absorb in the NMR. What happens to the H? It becomes H OD which
can usually be seen as a broad singlet around & 5 . 25. (If the sol vent is CDCI3, the i mmiscible H OD will
float on top of the solvent, out of the spectrometer beam , and its signal will be missi ng.)
R OH
CH3CH2 OH
+
D 20 - R OD
+
+
H OD
D20
3H
2H
H OD
Peaks from OH are usually broad
because of rapid proton transfer.
This is especial l y true in water,
or HOD.
I
10
I
9
I
J
8
7
I
6
TMS
I ,
I
5
& (ppm)
280
I
4
II
I
3
I
2
I
I
o
I
1 3-2 1
(a) The fonnula C4HlO02 h as no e lements of unsaturation , so the oxygens must be alcohol or ether
functional groups. The doublet at 8 1 . 2 represents 3H and must be a CH3 next to a CH. The peaks
centered at 8 1 .65 integrating to 2H appear to be an uneven quartet and signify a CH2 between two sets of
non-equi valent protons; apparently the coupling constants between the non-equi valent protons are not
equal , leading to a complicated p attern of overlapping peaks. The remaining five hydrogens appear in four
groups of IH, IH, IH, and 2H, between 8 3.7 and 4.2. The 2H multiplet at 8 3.75 i s a CH2 next to 0, with
complex spli tting (doublet of doublets) due to diastereotopic neighbors . The IH multiplet at 84.0 is a CH
between many neighbors . The two IH signals at 8 4. 1 -4.2 appear to be OH peaks in different
envi ronments, partially split by adj acent CH groups .
( r
H
H3-
I
+
-
-CH2-
Of the two possible structures:
H
-OH
+
y
I
�
+
-CH2
�
12
.
I
1.65
375
CH3-C-CH2CH20H
I
I
OH-- 4.1-42
CH20H
)
H
H
V
::
::
'/
--r===========;::
o
OR
CH3 -C-CH2 OH
-OH
+
HO
H3C
H
::
. %
��
=
OH
The 3-di en onal view
shows clearly that the two Hs
on a CH2 are not equi valent
H
CORRECT!
this one has no chiral centers and would
not display such complex splitting
If the possibility of intramolecular H-bonding is
considered, the diffe rences in the environments
for each H become even more obvious.
�
H
(b) The fonnula C2H7NO has no elements of unsaturation . The N must be an amine and the ° must be an
alcohol or ether. Two triplets, each with 2H, are certain to be -CH2CH2- . Since there are no carbons left,
the Nand 0, with enough h ydrogens to fi l l their valences, must go on the ends of thi s chain. The rapidly
exchanging OH and NH2 protons appear as a broad, 3H hump at 82. 8 .
H
HOCH2CH2NH2
8 3.7
1 3-22
�
'-- 82.9
�
3H
!!
3H
12
2H
I
TMS
I
I
I
9
10
8
°
II
CH3 -C - OCH2CH3
8 2.05
8 4. 1
I
7
I
6
5
(ppm)
I
8
I
4
3
I
I I
2
I
I
1
I
0
The key is what protons are adjacent to the oxygen . The CH30 in methyl
propanoate, above, absorbs at 8 3.9, whereas the CH2 next to the carbonyl
absorbs at 8 2.2. Thi s i s in contrast to ethyl acetate, at the left.
281
1 3 -23
Ha
=
Hb
=
He
=
Hd
=
d
[) 2.4 (singlet, IH)
[) 3.4 (doublet, 2H)
[) 1 .8 (multiplet, 1 H)
[) 0.9 (doublet, 6H)
I
I
CH3
H2
H-O-C -C-H
b
Q
c
1 3 -24
(a) The formula C4H802 has one e le ment of un saturation . The IH singlet at [) 1 2. 1 indicates carboxylic
acid. The IH multiplet and the 6H doublet scream isopropyl group.
CH3•
CH
i
II
°
CH-C-OH
(b) The formula C9HlOO has five e lements of unsaturation . The 5 H pattern between [) 7.2-7 .4 i ndicates
monosubstituted benzene. The peak at [) 9.85 is unmi stakably an aldehyde, trying to be a triplet because it i s
weakly coupled t o a n adj acent CH2. T h e t w o tri plets a t [) 2 . 7 - 3 . 0 are adj acent CH2 groups .
V
I
h-
H2
°
c
,
c
H2
Jl H
(c) The formula CSH802 has two elements of unsaturation. A 3 H singlet at [) 2.3 is probabl y a CH3 next to
carbonyl. The 2H quartet and the 3H triplet are certain to be ethyl; with the CH2 at [) 2 . 7 , this also appears
to be next to carbonyl.
II
II
°
°
CH3CH2 -C-C-CH3
(d) The formula C4H80 has one element of unsaturation, and the signals from [) 5 .0-6.0 i ndicate a vinyl
pattern (CH2=CH-). The complex quartet for IH at [) 4.3 is a CH bonded to an alcohol, next to CH3. The
OH appears as a IH singlet at [) 2 . 5 , and the CH3 next to CH is a doublet at [) 1 . 3. Put together:
OH
I
CH2=CH-CH-CH3
but-3 -en-2-ol
(e) The formula C7H160 i s saturated; the oxygen must be an alcohol or an ether, and the broad IH peak at [)
1 . 2 is probably an OH. Let's analyze the spectrum from left to right. The expansion of the IH multiplet at ()
1 .7 shows seven peaks-an i sopropy l group! Six of the nine H in the pattern at.5 0.9 must be the doublet
from the two methyls from the i sopropyl. The 2H quartet at.5 1 . 5 must be part of an ethy l pattern, from
which the methyl trip let must be the other 3H of the pattern at.5 0.9. This leaves only the 3 H singlet at <5 1.1
which must be a CH3 with no neighboring Hs.
-OH
+ -CH2CH3
�-
H
+
I
CH3-
+ -CH3 +
1 C
CH3
These pieces can be assembled in only one way :
CH3
I
HO-C-CH(CH3}z
I
CH2CH3
282
2,3-dimethylpentan-3-ol
1 3 -25 Chemical shift values are estimates from Figure 1 3 -4 1 and from Appendix lC , except in (c) and (d),
where the values are exac t.
(a) estimated
Q
8 70
l!
8 1 30
a
a
C�C
I
b
C
\
'0/
C
b
TMS
I
I
I
1 60
1 80
200
I
1 40
I
I
I
I
100
1 20
80
Carbon - 1 3 8 (ppm)
60
I
40
I
I
I
0
20
(b) estimated
b
a /C ' c
C
c
8 1 25
I
II
Q
8 30
i!
c, / C
a C
c
s;:.
8 22
b
I
200
I
1 80
TMS
I
1 60
(c) exact values
I
1 40
I
I
I
I
80
1 00
120
Carbon - 1 3 8 (ppm)
60
I
40
Q
8 1 45
C
c
CH2
d
b,..-:
;' '
CH 3 a C /"
s;:.
8 1 26
I
I
o
20
g
8 26
0
I
�
<5 1 7
I
CH3
e
l!
8 200
TMS
I
200
I
1 80
I
1 60
I
1 40
I
I
I
1 20
1 00
80
Carbon - 1 3 8 (ppm)
283
I
60
I
40
I
20
l
I
o
1 3-25 continued
(d) exact values
0
�
8 1 35
12
8 1 36
II
/ a
\
C=C
I c b\
H
H
H
f!
8 192
C-H
TMS
I
I
I
1 60
1 80
200
I
1 40
I
I
I
I
I
80
1 00
1 20
Carbon - 1 3 & (ppm)
40
60
I
I
20
I
o
1 3-26
�
t
(a)
'68--
12
3H
(b) 30 -0- 2 = 15
37 -0- 2 . 7 = 14
8 -0- 1 = 8
'637 0
'630
--
As a general rule,
the 1 5 -20 rule works
fairly well.
'6209
�
3H
f!
2H
0
II
b
I
I
9
10
1 3-27
a
/
b
6
I
5
& (ppm)
I
4
I
II
3
I
2
I
1
I
o
'6 1 30
\
'0/
I
f!
C=C
C
7
8
a
(a)
I
I
c
a
I
TMS
H3C-C-CH2-CH3
C
b
12
'670
TMS
I
200
I
1 80
I
1 60
I
1 40
I
I
I
80
1 00
1 20
Carbon - 1 3 & (ppm)
284
I
60
I
40
I
20
1II1
I
o
1 3-27 continued
(b)
b
a,..,... C, c
C
C
�
I
II
8 125
C, ,..,... C
a C e
b
h
8 30
£
8 22
,II,
TMS
I
I
1 80
200
I
1 60
I
I
1 20
1 40
I
I
I
80
1 00
60
I
40
I
20
o
I
Carbon - 1 3 8 (ppm)
(c)
0
C
c
CH2
d
b/.
;'
CH3 a 'c /'"
h
8 14 5
I
CH3
Q
�
8 26 8 1 7
e
�
8 200
£
8 1 26
JII,
TMS
I
I
200
1 80
I
1 60
(d)
I
I
I
60
I
40
I
20
o
I
0
h
8 1 36
�
I
I
80
1 00
1 20
Carbon - 1 3 8 (ppm)
1 40
J !t I I
II
H
C- H
I a
\
C=C
Ie b\
H
H
8 1 92
� 135
£
I
200
I
1 80
I
1 60
I
1 40
I
TMS
I
I
1 20
1 00
80
Carbon - 1 3 8 (ppm)
285
I
60
I
40
I
20
,11,
I
0
1 3-28 The ful l carbon spectrum of phenyl propanoate is presented belo w .
f
·
·
CH
200
I
1 80
'6 1 2 2
·
·
CH
·
·
·
g
J
1 60
e
_
C
I
O
f
'6 1 5 1
C
I
·
.d
(5 1 73
I
�
·
'6 1 29
�
12
g/'6 1 26/CH
f
I
1 40
I
I
�
'628
89
CH2
CH3
0
g
O- -CH2CH1
c
b a
e
I
80
100
1 20
Carbon - 1 3 8 (ppm)
I
60
I
40
1
TMS
I
20
I
o
The DEPT-90 wi l l show only the methine carbons, i.e . , CH. All other peaks disappear.
f
g/'6 1 26/CH
�
'6 1 29
CH
(5 1 22
CH
DEPT-90
The DEPT- 1 3 5 w i l l show the methyl, CH3, and methine, CH, peaks pointed up, and the methylene, CH2•
peaks pointed down.
f
g/'6 1 26/CH
(5 1 29
CH
�
'6 1 22
CH
DEPT-135
CH,
I
TMS
By analyzing the original spectrum, the DEPT-90, and the
DEPT- 1 35, each peak in the c arbon NMR can be assigned
as to which type of c arbon it represent.
286
-
13 29
S i nce allyl bromide was the s tarting material, it is reasonable to expect the allyl group to be present in the
i mpurity: the triplet at 1 1 5 is a =CH2 ' the doublet at 1 38 is =CH- , and the triplet at 63 is a deshielded
al iphatic CH2 ; assembling the pieces forms an allyl group. The formul a has c h anged from C3HS Br to
C3H60, so OH has replaced the B r.
H2C=CH-CH20H
8 115
tnplet
t
---' 8 I
s
doublet
"-
8 63
triplet
Allyl bromide is easi ly hydrolyzed by water, probably an SN 1 process.
H20
H2C=CH-CH20H + HBr
H2C=CH-CH2Br
°
1 3 -30
II
c
C
/Ca ,
\
d
1 3 -3 1
°
/
C-C
I
C
C
a ' / c
C
b
13 -32
Two elements of un saturation in C 4�02' one of which
i s a carbonyl, and no evidence of a C=C, prove that a
ring must be present.
b
b
C
a / , c
C
C
II
a = 8 180, singlet
b = 8 70, triplet
c = 8 28 , triplet
d = 8 22, triplet
a = 8 1 28 , doublet
b = 8 25, triplet
c = 8 23 , triplet
Two elements of unsaturation in C6HIO must be a C=C
and a ring . Only three peaks i ndicates symmetry.
Using PB r3 instead of H2S04iNaBr would give a
hi gher yield of bromoc yclohexane.
Compound 2
Mass spectrum: the molecular ion at mJz 1 36 shows a peak at 1 3 8 of about equal height, indicating a
bromine atom is present: 1 36 79 = 5 7 . The fragment at mJz 57 is the base peak ; this fragment is most
likely a butyl group, C4�' so a l i ke l y molecular formula is C4H9B r.
Infrared spectrum: Notable for the absence of functional groups : no O-H, no N-H, no =C-H, no
C=C, no C=O =} most l ikel y an alkyl bromide.
NMR spectrum: The 6H doublet at 8 1.0 suggests two CH3's split by an adj acent H-an isopropyl group.
The 2H doub let at 8 3 . 2 i s a CH2 between a CH and the B r.
-
Putting the pieces together gives isobutyl bromide.
1 3 -33
The formula C 9H\\B r i ndicates four elements of unsaturation , just enough for a benzene ring.
Here is the most accurate method for determining the number of protons per signal from integration
values when the total number a/protons is known. Add the integration heights : 4.4 cm + 1 3 .0 cm + 6.7
cm = 24 . 1 cm. Divide by the total number of hydrogens: 24. 1 cm.;. l lH = 2 . 2 cmIH. Each 2.2 cm of
integration height = I H, so the ratio of hydrogens is 2 : 6 : 3.
The 2H singlet at I) 7.1 means that only two hydrogens remain on the benzene ring, that is, it has 4
substi tuents. The 6H singlet at 82 . 3 must be two CH3's on the benzene ring i n identical environments. The
3H si nglet at <') 2 . 2 is another CH3 in a slightly different environment from the first two. Substitution of the
three CH3's and the Br in the most symmetric way leads to the structures on the next page.
287
13-33 continued
a
H
�
C 3
a second structure is also possible although it is
less likely because the Br would probably
deshield the Hs labeled "a" to about 7.3-7.4
b
CH3
*
a = 07 . 1 (singlet, 2H)
b = 0 2 . 3 (singlet, 6H)
c = 0 2 . 2 (singlet, 3H)
B'
CH3
b
H
a
*
b H3C
�
C 3
b H3C
H a
B'
H a
13-34 The numbers in i talics i ndicate the number of peaks in each signal.
(a)
(c)
*
H
10
C H3 -CH - CH 3
I
2
CH3
7
CH 3-CH - OH
1
I
CH3
2
(b)
CH3 - CH2-CCI2 - CH3
4
3
1
(d) H
H
H
7
(e)
C 3
H
1
(assume OH exchanging
rapidly-no splitti ng)
Q-CH1
o
2
(all of these benzene H's
are accidentally equivalent
and do not split each other)
13-35 Consult Appendi x 1 in the text for chemical shift values. Your predictions should be in the given
range, or within 0.5 ppm oflhe given value.
(a)
(c)
<}
(b)
H
al l at 07 . 2
03 .8
(f)
CH3 - CH2 - C - CH3
0 2.0
02 . 5
o
�
g
>=<
-
H
08. 0
H
07 . 5
02 . 5
CH\
o
06-7
II
CH3-CH=CH -C- H
o 1 .7
8 5-6
8 9-10
(h)
-H
09-10
H
08 .0
The C=O has its strongest deshielding
effect at the adjacent H (ortho) and
across the ring (para) . The remaining
H (meta) i s less deshielded.
02.2
04 . 3
<5 2 - 5
CH-O-CH2 - CH 2 -OH
�
03.8
03 .8
CH3
00. 9
II
(j 1.0
CH3- CH2-C- C-H
8 1 .2
03 . 2
o
(g)
(d)
01.6
CH3 - 0-CH2 - CH2 - CH2Cl
03 .4
(e)
all at 0 1.3
288
1 3 -35 continued
CH3
o
/
II
HOOC- CH2- CH2- C-O - CH - 8 4.0
(i)
8 1 0- 1 2
02. 3
U)
"
CH3
02 . 3
i /C,,�-�/F=C\ }
H
H
01 .3
H
01 . 1
/
H
,
, /
H
H
H
I
C-C
I
I'
��
0 1 .7
al lylic
8 1 .3
(k)
H
�
Y:7(
H
� �
H
H
07.2
1 3-36
H
l
f
(I)
H
I
OH
b
�
�
\
H
al lylic and
benzylic
a
=
04.0 (septet, 1 H)
b
=
02 . 5 (broad singlet, 1 H) (rapidly exchanging)
c
=
0 1 . 2 (doublet, 6H)
1 3-37
(a) The chemical shift in
ppm
would not change: 04.00.
(b) Coupling constants do not change with field strength: J = 7 Hz, regardless of field strength.
(c) At 60 MHz, 04.00 = 4 .00 ppm = (4.00 x 1 0-6) x (60 x 1 06 Hz) = 240 Hz
The signal is 240 Hz downfield from TMS in a 60 MHz spectrum.
At 300 MHz, (4.00 x 1 0-6) x (300 x 1 0 6 Hz)
=
1 200 Hz
The signal is 1 200 Hz downfield from TMS in a 300 MHz spectrum.
Necessarily , 1 200 Hz i s exactly 5 times 240 Hz because 300 MHz i s
exactly 5 times 6 0 MHz. They are directly proportional.
1 3-3 8
*
H
H
H
� Ii
H
06 . 0
-H
& l .4
8 2. 5
benzyl i c
c
a
c
CH 3 - CH - CH 3
H
R
a
b
c
d
c
b
d
CH2- CH2-O - C - CH3
H
'-----y--'
a
289
=
=
=
=
04.5
07 . 2-7 . 3 (multiplet, 5H)
04. 3 (tri p let, 2H)
02.9 (triplet, 2H)
02.0 (singlet, 3H)
13-39
(a)
Q
3H
c
a
b
CH3-O-CH2CH3
£
3H
g
2H
1
TMS
III
10
(b)
9
8
7
6
5
8 (ppm)
4
3
o
1
2
Q
3H
0
b
a II
c
(CH3hCH -C-CH3
£
6H
f!
IH
(To show the pattern
o f peaks. this multiplet
is larger than it would
appear on a real spectrum.)
10
(c)
9
8
7
6
5
8 (ppm)
a
a
b
CICH2CH2CH 2Cl
4
j I,
3
TMS
I
o
2
I
a
411
I
Q
2H
I
TMS
II
10
9
8
7
6
5
8 (ppm)
290
4
3
2
..--
o
13-39 continued
(d)
12
2H
I
7
8
9
6
I
I
I
I
2H
I
I
9
10
02N
12
2H
I
-
6
I
I
Hb
� ;)
Hb
O-C
�
IH
II
,I I
I
5
8 (ppm)
�/
I
I
I
I
3
4
*
I
TMS
I
I
a H
7
8
NH2
5
8 (ppm)
a H
(e)
f!
H
b
g
4H
very
broad
and
variable
The NH2 group
shields nearby
protons on benzene.
IH
10
d
NH2
����
�
IH
f!
b
H
o
2
g
6H
d
C H1
C
C H3
d
(To show the pattern
of peaks, this multiplet
is larger than it would
appear on a real spectrum.)
I
4
I
I
I
3
TMS
1
2
I
o
I
(f)
S ignal (a) is split into a quartet because of the adjacent
CH3 with J 7 Hz. Each of those peaks is then split
into a doublet because of the coupling with the trans
H, J 15 Hz. This is called a doublet of quartets, and
it is drawn here as two quartets . In a real spectrum,
these peaks would overlap and would not be a clean
doublet of quartets. See the splitting tree for problem
=
=
13-15.
I
V
�
C
C H3
bH
12
f!
IH
.
d
H3C
,
3 I
�
g
3H
TMS
I III
I
10
I
9
I
8
I
7
6
I
I
5
8 (ppm)
291
I
4
I
3
I
2
I
I
I
o
1 3 40
-
(a) The NMR of I-bromopropane w ould have three sets of signals, whereas the NMR of 2-bromopropane
would have only two sets (a septet and a doublet, the typical isopropyl pattern).
(b) Each spectrum would have a methyl singlet at 8 2. The left structure would show an ethyl pattern (a 2H
quartet and a 3H triplet) , whereas the ri ght structure would exhibit an isopropyl pattern (a IH septet and a 6H
doublet) .
(c ) The most obvious difference i s the c hemical shift of the CH3 singlet. In the compound on the left, the
CH3 singlet would appear at 8 2. 1 , while the compound on the right would show the CH3 singlet at 8 3.8.
Refer to the sol ution of 13-22 for the spectrum of the second compound.
(d) The spl itting and integration for the peaks in these two compounds would be identical, so the chemical
shift must make the difference. As described in text section 1 3-SB, the alkyne is not nearly as deshielding as
a carbon y l , so the protons in pent-2-yne would be farther upfield than the protons in butan-2-one, by about
O.S ppm. For example, the methyl on the carbonyl would appear near 8 2. 1 while the methyl on the alkyne
would appear about 8 1 .7.
13-4 1 The multiplicity in the off-resonance decoupled spectrum is gi ven below each chemical shift: s =
si nglet; d = doublet; t = tri plet; q = q uartet. It is often difficult to predict exact chemical shift values; your
predictions should be in the ri ght vicinity. There should be no question about the multiplicity and DEPT
spectra, however.
(a) estimated
q
t
s
.!!
8 1 70
c
I
200
I
1 80
I
I
1 60
I
1 40
I
I
I
80
1 00
1 20
Carbon - 1 3 8 (ppm)
I
60
I
40
I
20
o
I
The DEPT-90 spectrum for ethyl acetate would have no peaks because there are no C H groups.
DEPT-135
292
1 3-4 1
continued
(b) actual values
c
b a
H2C = CH - CH2Cl
1 1 8 1 34
d
t
12
45
t
�
01 1 8
045
CH2
CH2
(CHJ)4Si
g
CH
I
200
I
1 80
I
I
I
1 40
1 60
DEPT-135
CH
M
Carbons b and c
are accidentally
equivalent.
CH
C
200
1 80
I
I
1 60
DEPT-90 and DEPT- 1 35
Q
1 27
d
o 1 27
0 1 39
I
d
0 1 29
CH
I
1 40
I
I
80
100
60
Carbon - 13 8 (ppm)
DEPT-90
I
r
1 20
CH
(c) actual v&lues
I
TMS
0 1 34
I
1 20
on next page
<
c
b
'\
c-b
1 29
d
1 29
d
20
40
r
o
1 39/s
>/H2 H2
'/
I
I
C -C
e
f
39 33
t
-
Br
f
033
CH2
t
(CHJ)4Si
TMS
I
1 00
Carbon - 1 3 8
293
I
80
(ppm)
I
60
I
40
I
20
I
T
0
1 3-4 1
(c) continued
CH
DEPT-90
CH
DEPT-135
CH
CH2
13-42 The multiplicity of the peaks in this off-resonance decoupled spectrum show two different CR's and
a CH3. There is only one way to assemble these pieces with three chlorines.
Cl
CI
I
I
020-- CH3- CH - CH
q
CI
I � 07 5
060
d
t
d
13-43 There is no evidence for vinyl hydrogens, so the double bond is gone. Integration gives eight
hydrogens, so the formula must be C4HsBr2' and the four carbons must be in a straight chain because the
starting material was but-2-ene. From the integration, the four carbons must be present as one CH3, one
CH, and two CH2 groups. The methyl is split into a doublet, so it must be adjacent to the CH. The two CH2
groups must follow in succession, with two bromine atoms filling the remaining valences. (The spectrum is
complex because the asymmetric carbon atom causes the neighboring protons to be diastereotopic.)
H
I
Br H
I
I
Br
I
H- C- C- C- C-H
I
H
�
d
I
H
a
I
H
c
I
H
�
b
a
b
c
d
=
=
=
=
8 4.3 (sextet, IH)
8 3.6 (triplet, 2H)
8 2.3 (multiplet, 2H)
0l.7 (doublet, 3H)
1 3 -44 There is no evidence for vinyl hydrogens, so the compound must be a small, saturated, oxygen­
containing molecule. Starting upfield (toward TMS), the first signal is a 3H triplet; this must be a CH 3 next
to a CH2. The CH2 could be the signal at 8 l.5 , but it has six peaks: it must have five neighboring
hydrogens, a CH3 on one side and a CH2 on the other side. The third carbon must therefore be a CH2; its
signal is a quartet at 0 3 .6 , split by a CH2 and an OH. To be so far downfield, the final CH2 must be bonded
to oxygen. The remaining I H signal must be from an OH. The compound must be propan-l-ol.
a
b
c
d
=
=
=
=
03 .6 (quartet, 2H)
0 3.2 (triplet, I H)
0 1 . 5 (6 peaks, 2H)
00.9 (triplet, 3H)
294
13-45
CH3
\
/
/
\
C=C
CH3
CH3
+
H
Isomer A
(a)
/
/
\
C=C
\
a
b
c
d
=
=
=
=
d
CH3
b
CH3
C=C
/
H
d
H
CH3
/
H
\
\
CH2CH3
Isomer B
\
/
C=C
/
f
CH3
\
CH2CH3
H
e g
d
Isomer B
d = 84.7 (singlet, 2H)
e = 82.0 (quartet, 2H)
f = 81.7 (singlet, 3H)
g = 8 1.0 (triplet, 3H)
CH3
H
a
c
Isomer A
85.2 (quartet, 1H)
81.7 (singlet, 3H)
81.6 (singlet, 3H)
81.5 (doublet, 3H)
(b) With NaOH as base, the more highly substituted alkene, Isomer A, would be expected to predominate-­
the Zaitsev Rule. With KO-t-Bu as a hindered, bulky base, the less substituted alkene, Isomer B , would
predominate (the Hofmann product).
13-46 "Nuclear waste" is comprised of radioactive products from either nuclear reactions, for example,
from electrical generating stations powered by nuclear reactors, or residue from medical or scientific studies
using radioactive nuclides as therapeutic agents (like iodine for thyroid treatment) or as molecular tracers
(carbon-14, tritium H-3, phosphorus-32, nitrogen-I 5, and many others). The physical technique of nuclear
magnetic resonance neither uses nor generates any radioactive elements, and does not generate "nuclear
waste". (Some people assume that the medical application of NMR, medical resonance imaging or MRl,
purposely dropped the word "nuclear" from the technique to avoid the confusion between "nuclear" and
"radioactive" .)
13-47
The molecular ion of m/z 117 suggests the presence of an odd number of nitrogens.
3
2
Infrared spectrum: No NH or OH appears. Hydrogens bonded to both sp and sp carbon are indicated
around 3000 em-I. The characteristic C=N peak appears at 2250 em-I and aromatic C=C is suggested by
the peak at 1600 cm-I.
NMR spectrum: Five aromatic protons are shown in the NMR at 8 7.3. A CH2 singlet appears at 8 3.7.
Assemble the pieces:
Mass spectrum:
*
H
H
H
H
mass 77
H
+ C H2 +
mass 14
o-�
C:::N
-
mass 26
H
C -C:::N
H
I
mass 117
295
13-48 This is a challenging problem, despite the molecule being relatively small.
Mass spectrum:
or fewer.
The molecular ion at 96 suggests no Cl, Br, or N. The molecule must have seven carbons
The dominant functional group peak is at 1685 cm-I, a carbonyl that is conjugated
with C=C (lower wavenumber than normal, very intense peak). The presence of an oxygen and a molecular
ion of 96 lead to a formula of C6HsO, with three elements of unsaturation, a c=o and one or two c=c.
Carbon NMR spectrum: The six peaks show, by chemical shift, one carbonyl carbon (196), two alkene
carbons (129, 151), and three aliphatic carbons (23, 26, 36). By off-resonance decoupling multiplicity, the
groups are: three CH2 groups, two alkene CH groups, and carbonyl.
Infrared spectrum:
Since the structure has one carbonyl and only two alkene carbons, the third element of unsaturation must be
a ring.
C=O
CH2 + CH2 + CH2 + 1 ring
C=CI
H
I
H
,
Since the structure has no methyl group, and no H2C=, all of the carbons must be included in the ring. The
only way these pieces can fit together is in cyclohex-2-enone. Notice that the proton NMR was
unnecessary to determine the structure, fortunately, since the HNMR was not easily interpreted except for
the two alkene hydrogens; the hydrogen on carbon-2 appears as the doublet at 6.0.
>
The mystery mass spec peak at mlz 68 comes from
a fragmentation that will be discussed later; it is
called a retro-Diels-Alder fragmentation.
13-49 The key to the carbon NMR lies in the symmetry of these structures.
C 3
"
/
b
c
a CH3
c �/
3
a
"
d J" h
c ha CH3
. ,/ c b CH3
b
"
meta-xylene
ortho-xylene
cyclohex-2-enone
cr
er:
,
,
,
,
(a) In each molecule, the methyl carbons are equivalent, giving one signal in the CNMR. Considering
the ring carbons, the symmetry of the structures shows that ortho-xylene would have 3 carbon signals
from the ring (total of 4 peaks), meta-xylene would have 4 carbon signals from the ring (total of 5 peaks),
and para-xylene would have only 2 carbon signals from the ring (total of 3 peaks). These compounds
would be instantly identifiable simply by the number of peaks in the carbon NMR .
(b) The proton NMR would be a completely different problem. Unless the substituent on the benzene
ring is moderately electron-withdrawing or donating, the ring protons absorb at roughly the same
position. A methyl group has essentially no electronic effect on the ring hydrogens, so while the para
isomer would give a clean singlet because all its ring protons are equivalent, the ortho and meta isomers
would have only slightly broadened singlets for their proton signals. (Only a very high field NMR, 500
MHz or higher, would be able to distinguish these isomers in the proton NMR.)
296
13-50 (a), (b) and (c) The six isomers are drawn here. Below each structure is the number of proton
signals and the number of carbon signals. (Note that splitting patterns would give even more clues in the
H
proton NMR.)
H3 C 'rI" CH3
l-.
/'
H
CH3
H
H
CH3
H
y�
>=<
>=<
1
6
H
H
H3C
H
H3C
CH3
H
D
2xH
5xH
3xH*
2xH
2xH
IxH
3xC
4xC
2xC
2xC
3xC
IxC
;L
_
*Rings always present challenges in stereochemistry. When viewed in three
dimensions, it becomes apparent that the two hydrogens on a CHz are not
equivalent: on each CHz, one H is cis to the methyl and one H is trans to the
methyl. These are diastereotopic protons. A more correct answer to part (b)
would be four types of protons; whether all four could be distinguished in the
NMR is a harder question to answer. For the purpose of this problem, whether
it is 3 or 4 types of H does not matter because either one, in combination with
three types of carbon, will distinguish it from the other 5 structures.
(d) Two types of H and three types of C can be only one isomer: 2-methylpropene (isobutylene).
only isomers that would not be distinguished from each other would be cis- and trans-but-2-ene.)
�NMR' �
(The
13-51 For DEPT to be useful in distinguishing isomers, there should be different numbers of CH3, CHz,
CH, and C ,ignal, in
6 C signals
one CH3 peak
four CHz peaks
one CH peak
~
5 C signals
one CH3 peak
two CHz peaks
two CH peaks
5 C signals
one CH3 peak
two CHz peaks
one CH peak
one C peak
A carbon N MR could not distinguish isomers Band C, and if there is any overlap in the signals, it might
have a difficult time distinguishing A. Fortunately, the patterns that would appear in the DEPT-90 would
distinguish B (two CH peaks up) from the other two (l CH peak up), and the DEPT-135 would instantly
distinguish A (4 CHz peaks down) from C (2 CH2 peaks down).
13-52
(a) MS or IR could not easily distinguish these isomers: same molecular weight and same functional
group. They would give dramaticaIIy different proton and carbon NMRs however.
+
OCH]
2 singlets in HNMR
3 singlets in CNMR
�
OCHZCH3
4 signals, all with splitting, in HNMR
4 singlets in CNMR
297
13-52 continued
(b) The only technique that would not readily distinguish these isomers would be MS because they have
the same molecular weight and would have similar, though not identical, fragmentation patterns.
o
)lo�
IR: C=O
HNMR:
CNMR:
about 1 730 cm-i
methyl singlet and ethyl pattern
ester C=O about 8 1 70
IR: C=O about 1 7 1 0
HN MR : 3 singlets
CNMR:
cm-i
ketone C=O about 8 200
(c) The big winner here is MS : they have different molecular weights, plus the Cl has the two isotope
peaks that make a Cl atom easily distinguished. The other techniques would have minor differences and
would require having a detailed table of frequencies or chemical shifts to determine which is which.
F-o-OH
M+
=
mlz 1 1 2
C'-o-OH
M+ = mlz 1 2 8 , 1 30
298
CHAPTER 14-ETHERS. EPOXIDES AND SULFIDES
1 4- 1
The four sol vents decrease i n polarity in this order: water, ethanol , ethyl ether, and dichloromethane. The
three solutes decrease i n polarity in this order: sodium acetate, 2-naphthol, and naphthalene. The guidi ng
principle i n determi ning solubi l i ty is, "Like dissolves like." Compounds of s i milar pol arity w i l l dissolve (in)
each other. Thus , sodium acetate w i l l dissolve in water, will dissolve only s lightly in ethanol , and wi ll be
vi rtual l y insol uble in ethy l ether and dichloromethane. 2-Naphthol w i l l be i n soluble in water, somewhat
soluble i n ethanol , and soluble i n ether and dichloromethane. Naphthalene w i l l be i n soluble i n water,
partially soluble i n ethano l , and soluble in ethy l ether and dichloromethane. (Actual solubi lities are difficult
to predict, but you should be able to predict trends.)
1 4-2
CI
I
I
-
CH2CH3
+
/
Cl-A l-O :
CI
\
Oxygen shares one of its electron pai rs with aluminum; oxygen is the
Lewis base, and aluminum is the Lewis acid. A n oxygen atom with three
bonds and one unshared pair has a positive formal charge . An aluminum
atom with four bonds has a negative formal charge .
CH2CH3
14-3
The crown ether has two effects on KMn04: first, it makes KMn04 much
more soluble in benzene; second, it holds the potassium ion ti ghtly , making
the permanganate more avai l able for reaction. Chemists cal l this a "naked
an ion" because it is not complexed with solvent molecules.
o
'� -/
,0
Mn
/' '\:,
0'
0
benzene
..
IS-crown-6
a " crown ether"
Please see the note on p. 1 3 of this Solutions Manual regarding placement of position numbers.
1 4-4 IUPAC name first; then common name (see Appendi x I in this Solutions Manual for a summary of
IUPAC nomenclature)
(a)
(b)
(c)
(d)
(e)
methoxycyc lopropane ; cyc lopropyl methyl ether
2-ethoxypropane; ethyl i sopropyl ether
l-chloro-2-methoxyethane; 2-c h loroethyl methyl ether
2-ethoxy-2,3-di methylpentane; no common name
2-t-butoxybutane; sec-butyl t-butyl ether
(f) trans-2-methoxycyc lohexan -l-ol; no common name
1 4-5
(a)
+
()
o
+
2 H20
The alcohol is ethane-l,2-di o l ; the common name is ethylene glycol.
299
1 4-5 continued
(b)
HO
I
H
:o�
I
H2C -- CH2
..
H+
:60
IJ
HO
I
o
HO
I
H2C
I
--
:0:
=
H2
H2C -- CH2
----l�
H2C
()
HO
I
H2C
\.
-
�
HO
: OH
I
I
H2C
CH2
-
+
H20:
CH2
I�
)
.. �
H O:
I
H2C-CH2
:O �H
I
I
0
+
I
�
(I
H2C
.
+ H30 + -
�
CH2
I
H O:
0
H2C
CH2
I
CH2
I
H2C-CH2
H OH
\
I
:0:
HO
CH2
CH2
H+
I
The mechanism shows that the acid c atalyst is regenerated at the end of the reaction .
1 4-6
(a) dihydropyran
(b) 2-chloro-l,4-dioxane
(c) 3-isopropylpyran
(d) trans-2,3-diethyloxirane; trans-3,4-epoxyhexane; trans-3-hexene oxide
(e) 3-bromo-2-ethoxyfuran
(f) 3-bromo-2,2-dimethyloxetane
1 4-7
CH3
I
+
+CH
,
CH3
CH3CH2CH2CH2
mJz 57
mJz 43
mJz 73
mJz 1 01
300
1 4-8 SN2 reactions, including the Williamson ether synthesis , work best when the nucleophile attacks a
1 ° or methyl c arbon . I nstead of attempting to form the bond from oxygen to the 2° c arbon on the ring,
form the bond from oxygen to the 1 ° carbon of the butyl group.
The OH must first be transformed into a good leaving group: either a tosyl ate, or one of the hal ides (not
fluoride).
TsCI
� OH
pyridine
� OTs
..
O�
0- Na+
OH
Q �Q
(f
1 4-9
(a)
0H
Na
OH
Na
A
(b)
� OTs
CH3CH2CH2Br
...
..
OH
¢
(c)
CH3I
Na OH
...
..
N0 2
(d)
¢Hl
CH3CH20H
A
CH3
I
CH 3-C- OH
I
CH3
Na
Na
Na
...
�
CH3CH2Br
CH3CH2CH2Br
CH3CH 2CH20CH2CH3
...
CH3
C
CH 3 - - OCH2Ph
I
CH3
PhCH2Br
•
CH3CH2CH20CH2CH3
...
--_..
1 4- 1 0
(a)
(1)
�
or
�
(2 )
0�
N0 2
CH3CH2CH20H
(e)
(f
..
Q
OCH3
CH3I
...
..
OH
�
Hg(OAc)2
CH30H
Na
...
•
CH31
NaBH
�
OCH3
�
OCH3
...
�
30 1
3-butoxy-l,ldi methy Icyclohex:me
14-10 continued
(b) (1)
(2)
(c)
(1)
(2)
(d)
(1)
o
Hg(OAch
..
(yoH
6
Na
--
Hg(OAch
(e)
(1)
CH31
..
NaB H4
..
..
Hg(OAch
NaBH4
CH30 H
(2)
(y
Alkoxymercuration i s not practical here ; the product does not h ave Markovnikov orientation .
------�..
�
y
O
OCH 1
CH 1
..
OH
(2) Wi lliamson ether synthesis would give a poor yield of product as the hal ide i s on a 2° c arbon.
(f)
(1)
O
H
Na
--
==< o-
Hg(OAch
K�
..
OH
NaBH4
..
( }-o+
(2) Wi l liamson ether synthesis is n o t feasible here. SN2 does n o t work on either a benzene or a 3 ° halide.
14-11 An i mportant pri nciple of synthesis is to avoid mixtures of i somers wherever possible; minimizing
separations increases recovery of products. Bimolecular dehydration is a random process . Heating a
mixture of ethanol and methanol w ith acid wi l l produce all possible combinations: dimethyl ether, ethyl
methyl ether, and diethy l ether. Thi s mixture would be troublesome to separate.
302
1 4- 1 2
Ether formation
..r"'..
CH3CH2CH2-�.H
H
nl
'--
CH3CH2CH2 -..O+- H
H+
4
..
H�-CH2CH2CH3
.. �
H20
H
( 1+
CH3CH2CH2 - � - CH2CH2CH3
• •
Dehydration
H
nl+
CH3CHCH2 - ..
° -H
I..)
-
H�
H2 �
Remember I:!.G = M/- TM? Thermodynamic s of a reaction depend on the s i gn and magnitude of AG. As
temperature i ncreases, the entropy term grows i n i mportance. In ether formation, the I:!.S is smal l because
two molecules of alcohol give one molecule of ether plus one molecule of w ater-no net change in the
n umber of molecules. In dehydration , however, one molecule of alcohol generates one molecule of alkene
plus one molecule of water-a l arge i ncrease in entropy. So TI:!.S is more i mportant for dehydration than for
ether formation. As temperature increases, the competition will shift toward more dehydration .
1 4- 1 3
(a) This symmetrical ether a t 1 ° carbons could b e produced in good yield b y bimolecular dehydration.
(b) This unsymmetrical ether could not be produced in high yield by bimolec ular dehydration . Wil l iamson
synthesis would be preferred.
/'-..../ OH
/"'- OH
(c) Even though thi s ether i s symmetrical , both carbons are 2°, so bimolecu l ar dehydration would give low
yields. Unimolecular dehydration to give alkenes would be the dominant pathway. Alkoxymercuration­
demercuration is the preferred route.
or
}
H g ( O Ac ) z
�
OH
�
303
14- 1 4
0
•
-
H- Br
•
R� "-.;
(Q
Br
---­
• •
'4
I
:Br:
£L oH
H-Br
Br
o
(b)
HBr
� O)
HI
_
< }-
OH
�
� OH� I
HBr
(e)
1 4- 1 6
�
o
• •
� �r
/
+
BrCH2CH2 - CH- CH2 - Br
I
CH3
B
Br' 'Br
---
r-\ +
�
Br
1_
:'0- B- Br
I
H3C
1
Br
!
HOBBr 2
2
+
H20
B( OHh
-
�. O+-B, Br +
H3C
+ 2 HBr
•
o xygen, another H+ goes on
the other oxygen; several
d ifferent scenarios of how
this happens could be
proposed
304
.Br.
�
I
J
\
�
H'.. H
O�
..
�
Bu +
Bu
Bu, "' Br
H20:
11_
,
'
,
,
..
..
� : O-B-Br
:·O-B-Br ..
:0-B
.
.
,
I� I
two proton transfers happen
-Sr
Br
H Br
here: one H+ comes off of
H'
0
• •
I BU OHI
JL Br
HI
�
(d)
"-.;
H
• •
+
•
•
• •
Ea
CH3Br
nucleophilic attach on CH3
faster than on I 0 carbon
Begin by transforming the alcohols into good leaving groups like halides or tosylates:
NaSH
NaOH
PBr3
/'....
/'....
/'....
/'....
•
•
.......", 'S- N a+
/'
/'
.......", 'SH
- �Br
�OH
1 4-17
OH
A
TsCI
pyn·d·me
1
A
OTs
I
�S �
14- 1 8 The sulfur at the center of mustard gas is an excellent nucleophile, and chloride is a decent leaving
group. Sulfur can do in internal nucleophilic subsitution to make a reactive sulfonium salt and the sulfur
equivalent of an epoxide.
very reactive
alkylating agent
(a)
�
HN� �CI
S
inactivated enzyme
Cl� �Cl
C
'-- S
•
•
. .
(b) NaOCI is a powerful oxidizing agent. It oxidizes sulfur to a sulfoxide or more likely a sulfone,
either of which is no longer nucleophilic, preventing formation of the cyclic sulfonium salt.
NaOCI
Cl�· �Cl
S
o
Cl� �Cl
S+
+ CI�I�CI
S++
• •
I
I
o
o
sulfoxide
14- 1 9
sulfone
Generally, chemists prefer the peroxyacid method of epoxide formation to the halohydrin method.
Reactions (a) and (b) show the peroxyacid method, but the halohydrin method could also be used.
(a)
==<
HO
(b)
(c)
(d)
(e)
MCPBA
CI
U
CI
D(
..
H2 SO4
'r-
Ph
o
t1
-
BH3 · THF
U
/
Ph
� OH
Cl
H2 02
HO-
..
Hg(OAc}z
H2 O
MCPBA
•
CH2CI2
..
NaBH4
..
NaOH
..
CI
l)
CI
P
Ph
NaOH
..
U
OH
�
305
0
0
NaOH
..
�
1 4-20
(a) 1 ) te rt-B utyl hydroperoxide is the o xidizing agent. The (CH3hCOOH contains the 0-0 bond just
l i ke a peroxyacid. 2) Diethyl tartrate has two asymmetric carbons and is the source of asymmetry; i ts
function is to create a chi ral transition state that is of lowest energy , leading to only one en anti orner of
product. Thi s process is c alled chirality transfer. 3) The function of the titanium (IV) isopropoxide is to
act as the glue that holds all of the reagents together. The titanium holds an oxygen from each reactant­
geraniol , t-B uOOH, and diethyl tartrate-and tethers them so that they react together, rather than j ust
havi ng them in solution and hoping that they w i l l eventuall y col lide.
(b) Al l three reactants are required to make Sharpless epoxidation work, but the key to enantioselective
epoxi dation is the chiral molecule, diethyl tartrate . When it complexes (or chelates) with titanium, it forms
a l arge structure that is also chiral . As the t-B uOOH and geraniol approach the complex , the steric
requirements of the complex allow the approach in one preferred orientation. When the reaction between
the alkene and t-B uOOH occ urs , it occ urs preferentially from one face of the alkene , leading to one major
stereosiomer of the epoxide. Without the chiral diethyl tartrate in the complex, the alkene could approach
from one side j ust as easily as the other, and a racemic mi xture would be formed.
(c) Using the enantiomer of diethyl L-tartrate, cal led diethyl D-tartrate, would give exactly the opposite
stereochemical results.
t-B uOOH
: +�
Ti(Oi-Pr)4
O
O
H
1 4-2 1
R
H
Me
,
" ' 'C
'
').
== '"
C
�
/
HO
�
'
H
trans-but -2-ene
Me
OH
+
�" Me
�
""
"
HM e
""
'"
H
Me
" , Me
H
Ho
0
"7
H
Me
y
trans
CH3
H30 +
---.
H
)
H
I
l---<.
Me �
-:;
....
..'-..
_
'"OH
�
.. .
�
''
'
e
�
'
Me
.
OR : O - H
I
tI
H
OH
+
H
Ho
l
e
K
H
.: �
� Q)
'
I,
+
)
stereochemistry shown i n Newman projections:
1 V ): H
+
:O -
H-O:
IDENTICAL-MESO
CH3
H
OO
diethyl D-tartrate ,
H
COOC zCH3 the enantiomer of diethy l L-tartrate
· A
C=O
t
H2CH J
� �O
I
"
�" " " '-... OH
OH
:O :
rO ' �' H
V O
"
+
•
"'Me
t<::>r
H3C y H
--H
Ii
H
OH
OH
306
C H3
H
'\
\'
'\
+
OH
H
rotate
•
H
H3C
H
H3C
MESO
14-2 1
continued
H"
H
"C==C" "
Me'
'Me
cis-2-butene
I,
>
\'
x
�
Me
'"
HO
OH
'"
H
Me
HO
+
�---------Y
�'"
H Me
H"'/
Me
OH
/
----------
ENANTIOMER S
stereochemistry shown in Newman projections:
OH
CH3
H
H--+---;;;.....-+-- CH3
rotate
..
C1h H 20:
Cli
--
--
3
• •
�
H � CH I
H 3C
H
CH R
I AL­
RACEMIC
MIXTURE
CIS
1 4-22
RC0 3 H
-I..
�
---
CH3 CH20CH2CH20H
Cellosolve®
H2C-CH2
\/
o
1 4-23
Br:
/ : ..
t
anhydrous HBr--only Bc nucleophiles present
H 2C ,,/CH2
..
·0·
H 0r
'-----A
•
•
� ..
H 2C /C 2
"
·0 �
-
•
1+
H
1
: O-H
CH2CH 2Br
:Br:
..
---l"�
H
\
Br
� --...."
CH2CH2Br
�I+
H -O-H
BrCH2CH2Br
+
H20
aqueous HBr-many more H 20 nucleophiles than Bc nucleophiles
H 2C
CH 2
, /
,:0 :
H 0-r
'-----A
_
H2C
0:O-H
.
" Q
:?+
k
H
---
c
307
..
1+
:? H
,
H2C-CH2
I
HO
H
..
H 29.
�
H2C-CH2
I
HO
1
OH
1 4-24 The cyclization of squalene via the epoxide is an excellent (and extraordinary) example o f how
Nature uses organic chemistry to its advantage . In one enzymatic step, Nature forms four rings and eight
chira l centers ! Out of 256 possible stereoisomers, only one is formed !
•
HO
\
�
1 4-25
�
:\.
-
..
� ,,----" :0 - CH3
H
H
•
1 4-26
(a)
(d )
� O�
(JI
�
�
1 4-27
(a )
O- Na+
H
N�
OH
(b)
' '" ,
, ,A
A
" '"
Et M
H
e
CH30 Me
" '
...,,,,
Et , Me
"
Me
H
"
"
CH30H
H+
•
CH30H
�
H
(c)
+
:�-'C H 1
OCH3
� s � O-
o
0 _ Na+
C�
Na+
:N == N == N � _' a
O N +
+
O- K+
X3oH
'>---f/
Ho
H0
•
bonds formed
(f)
N
""- /
.
OCH3
H2N �
(e)
OH
HI 80 �
o
(d)
H
•
(b)
o
(c)
H �CH3
•
=
Me
'
Et " 1
Me
"'- OCH3
�
; �
Me
CH30
H
308
"
OH
Me
H
�
1 4-28 Newly fonned bonds are shown in bold.
(a )
I
� OH
(b )
OH
OH
(c)
1 4-29 Please refer to solution 1 -20, page 1 2 of this Solutions Manual.
1 4-30
�
(a )
/" O
(d)
� O�
(g)
;A
H
(b) � O �
(c)
(e) � O .......
(
(h)
�
''''
�
Br
+
y
Br
(c) and (d) no reaction
(g)
�
HO'"
"
(i)
�
0
H
0
A
H3C
CH3
H
(e) trans-cyclohexene glycol
(f) cyclopentyl methyl ether
(g) propylene o xide
(h) cyclopentene o xide
14-32
(a ) 2-methoxypropan- l -01
(b) etho xybenzene or phenoxyethane
(c) metho xycyclopentane
(d) 2,2-dimethoxycyclopentan- l -01
(e ) trans- l -methoxy-2-methylcyclohexane
(a )
f)
,
HO H
14-3 1
(a) sec-butyl isopropyl ether
(b) t-butyl isobutyl ether
(c) ethyl phenyl ether
(d) chloromethyl n-propyl ether
1 4-33
0 ""-"""
+
H2 O
(e)
(h)
( trans-3-chloro- l ,2-epoxycycloheptane
(g)
f) trans- l -metho xy- l ,2-epoxybutane ; or,
trans-2-ethy 1- 3 -methoxyoxirane
(h) 3-bromooxetane
(i) 1 ,3-dio xane
(b)
< )-OH
�N
�
Br
OH +
HC H3
H
309
CH3CH21
+
� Br
(
f)
(i)
+ H2 O
� OH
OCH3
-+o�
1 4-33 continued
HO
�
o
(j )
(k)
(n)
�
Br
(I)
OH
ry."OCH3
CH3
�1 I 0H
H
1 4-34
(a) On long-term exposure to air , ethers form peroxides. Pero xides are explosive when concentrated or
heated. (For exactly this reason, ethers should never be distilled to dryness. )
(b ) Peroxide formation can be prevented by excluding oxygen. Ethers can be checked for the presence of
peroxides, and peroxides can be destroyed safely by treatment with reducing a gents .
1 4-35
(a ) Beginning with (R)-butan-2-o l and producing the (R) sulfide requires two inversions of configuration.
H Br
HO H
�
�
R
5
An a lternative approach would be to make the tosylate, displace with chloride or bromide (SN2 with
inversion ) , then do a second inversion with Na SCH3.
(b) Synthesis of the (5) isomer directly requires only one inversion .
HO H
�
R
TsCI
Ts O H
�
pyridine
R
1 4-36
(a )
./
.-.
molecular ion mlz
mlz 73
1 02
3 10
))� }
........,.,
�
" C H2
1 4-36 continued
(b)
.
+
3 1 0CH3
H
\+
C-O:
/
molecular ion mJz
+
/
�? ,
mJz 7 1
H
14-37
o: �
�
H+ _
i
0·+
(
�
C
I
(a
�
I
CH2
-:;:::.CH2
MCPBA
�
CH2Cl2
�
MCPBA
•
CH2Cl2
(c �
MCPBA
..
CH2Cl2
(b
1 02
�
H
+1
CH
"
CH3
• •
II
+. .
II
OH
I
CH2
CH2 _
��20 :
• •
I
CH3
�
�
�
311
1)
2)
CH3
CH30H
H+
CH30H
mJz 8 7
OH
I
�
•
�
CH2 0H
• •
• •
PhMgBr
NaOCH3
}
mJz 59
CH2
_
CH2
H
C/ + / H20 :
I -0\"
CH3 H
O
H2O
(after H migration)
H
OCH3
� CH
•
mJz 3 1
• •
CH
"
•
+
\
OCH3
..
O
+
. .
+1
-
H
:OCH3
C=O
/
-----
: OCH3
� CH
O- H
CH 3
1 4-3 8
H
\
H
\
J
OH
+-
� Ph
OH
H3 0+
�OCH3
OH
�OH
OCH3
CH3
o
A
(X
"
0H
"'
G
G
(X
" " '0
0 Na
-
0
Na �
D
+
CH3Br
C
(X
0CH]
�
H
''
'"
0
The student turned in the wrong product ! Three pieces of information are consistent with the
desired product: molecular formula C4HIOO; O-H stretch in the IR at 3300 cm-I (although it should be
strong, not weak); and mass spectrum fragment at rnJz 59 (loss of CH3). The NMR of the product should
have a 9H singlet at 8 1 .0 and a IH singlet between 8 2 and 8 5. Instead, the NMR shows CH3CH2 bonded
to oxygen. The student isolated diethyl ether, the typical solvent used in Grignard reactions.
1 4-40
Predicted product
59 CH3
CH]
1 4-4 1
(a)
(b)
H
Isolated product
- OH
CH3
C4HIOO
O-H at 3 300 cm-I
U OH
/"-.../ OH
\ }-
OH
�OH
Na
-
PBr3
�
NaOH�
U
O- Na+
� Br
\ }-
C4HIOO
O-H at 3300 cm-I due to water contamination
}
O- Na+
TsCI �
�OTs
pyridine
312
O
_ U �
� \ }- o �
1 4-4 1 continued
Hg(OAc)z
(c)
�
--l..
NaBH4
---
t
(d)
1 ) BH3
•
THF
C
HB r, ROOR
..
..
OH
NaO H 3
Br
..
(e)
1 4-42 In the first sequence, no bond i s broken to the chiral center, so the configuration of the product i s the
same as the configuration of the starti ng material .
�
[a]o
==
-
H
�CH2CH
Na
�
[a] o
8 . 24°
(Assume the enantiomer
shown is levorotatory . )
-Q-
==
-
1 5 .6°
In the second reaction sequence , however, bonds to the chiral carbon are broken twice, so the
stereochemistry of each process must be considered.
� 0
o
II
• •
OH
CI - S
H
� II
� !J
- -' ,;; .
,
�
�
0
�
..
"-/ "-/ "-/ \ . H 0
. . . ,
H,;�\-�-Q-
�
) ----
undOUbtedly, some E2
produ cts will als �
.
form In thiS reacHon.
1
s
CH 3 C H2 0 ·•• •
Na+
H,
t :t�
o-s
�
-Q-
0- _
0
RETENTION OF
CONFIGURATION
SN2-INVERSION
The second sequence invol ves retention fol lowed by inversion, thereby producing the enantiomer of the 2ethoxyoctane generated by the first sequence . The optical rotation of the final product w i l l have equal
+ 1 5 .6° .
magni tude but opposite sign, [a]o
==
313
�
14-43
. .'-
HO :
'\
CH3 /""\
CH3
. '. -
I
I
H -- HO - CH2 - CH - �:
H2C
\F
:0:
O
t
O
CH3
I
I
I
.
.- �
HO - CH 2 - CH - O - CH2 - CH - O - CH 2 - CH - O :
+
HO H �
u
I
I
I
• •
C�
C�
C�
HO - CH2 - CH - 0 - CH2 - CH - 0 - CH2 - CH - OH
1 4 -44 The text uses
CH3
I
I
I
..
H -- HO - CH2 - CH - 0 - CH2 - CH - � :
H2C
\F
:0:
CH3
CH3
CH3
�
CH3
CH3
+
I -
H2C - CH
a
HO-
HA to indicate an acid; A- is the conj ugate base.
•
•
.
.
.O - H
+ : ANote that all three carbocation intermediates are 3 ° !
· A­
•
OH
+
..
"-. H " I
'-.... {... C
C+
.
tI
..
OH
H- A
This process resembles the cyclization of squalene oxide to lanostero l . (See the solution to problem L 4-24 . )
I n fact, pharmaceutical synthesis o f steroids uses the same type o f reaction called a "biomimetic cyclization" .
314
(b)
(c)
:" , /0
No bond to the chiral center
was broken. Configuration
is retained; R stays as R.
\CH2
I
"
H
H3 C
Attack of w ater gave inversion of
configuration at the chiral center; R became S.
R
'--HO :
• •
-
R
(d) The difference i n these mechan i s ms lies in where the nuc leophi le attacks. Attack at the chiral c arbon
gi ves i n version; attack at the ach i ral carbon retains the configuration at the chiral carbon. These products
are enantiomers and must necessari l y have optical rotations of opposite sign.
methyl cellosolve
1 4-46
CH30CH2CH20H
To begin, what can be said about methy l cellosolve? Its molecular weight is 76; i ts IR would show C - -O i n
the 1 000- 1 200 cm- I region a n d a strong O-H around 3300 em- I ; and i t s NMR would show four sets o f
signals in the ratio o f 3 : 2 : 2 : 1 .
The unknown has molecular weight 1 34 ; this i s double the weight of methy l cellosolve, minus 1 8 (water) .
The IR shows no OH, only ether C-O. The NMR shows no OH, only H-C-O i n the ratio of 3 : 2 : 2 .
Apparentl y, t w o molecules of methyl ce lloso l ve have combined in a n acid-catalyzed, bimolecular
dehydration .
- ° C H2CH20CH3
________
CH30CH2CH2
"-
-
1-
..
.
(th i s compound i s c a lled "diethylene
glycol dimethy l ether" , or a shortened,
common n ame is "di g l y me" )
..
H2
0:
3 15
CH30CH2CH2 �O
�H
- CH2CH20CH3
I)
1 4-47
The fonnula CgHgO has five elements of unsaturation (enough (4) for a benzene ring). The IR is usefu l for
what is does not show . There i s neither OH nor C=O, so the oxygen must be an ether functional group.
The NMR shows a 5H signal at 8 7 .2, a monosubstituted benzene. No peaks in the 8 4. 5 -6.0 range indicate
the absence of an alkene, so the remaining element of un saturation must be a ring. The three protons are
non-equi valent, with complex splitting.
<>
+ 2C +
0
ether
+ 3 H + ring
These pieces can be assembled in only one manner consistent with the data.
�
H
H
H
(Note that the CH2 hydrogen s are not equi valent (one is cis and one is trans to the phenyl) and therefore
have distinct chemical shifts . )
1 4-48 The key concept is that reagents always go to the less hindered side o f a molecule first. I n this case,
the " underneath" side is less hindered; the " top" side has a CH3 hovering over the double bond and
approach from the top w i l l be much more difficult, and therefore slower, than approach from underneath .
MCPB A
J
less
hindered
side
B r2
B
MCPBA approaches from the
bottom; the epoxide is fanned from
MCPB A w i thout the participation
of any other reagent. The epoxide
fonns on the less hindered side.
I
fonnation of the bromohydrin begins
with fonnation of the bromonium ion
on the less hindered side
2,6-lutidine
o
water must attack
from the side opposite
the bromonium ion ,
from the TOP
c
Br
Epoxides B and C are di astereomers ; they will have different chemical and physical properties.
Nucle� philes can react with C much faster than with B for precisely the same reason that explai ned thei r
fonnatlOn: approach from underneath i s less hindered and is faster than approach from t h e top.
316
CHAPTER I S-CONJUGATED SYSTEMS, ORBITAL SYMMETRY, AND
ULTRAVIOLET SPECTROSCOPY
C=C,
1 5 - 1 Look for: 1 ) the number of double bonds to be hydrogenated-the fewer
the smaller the MI;
2) conj ugation-the more conjugated, the more stable , the lower the � ; 3) degree of substitution of the
alkenes-the more substituted, the more stable, the lower the !1H
.
(a) � <
� <�<
smallest !1H
� < == C � < �
biggest !1H
<
<
smallest !1H
1 5 -2 Reminder:
conj ugate base.
H
H
H H
H-B
H
<
<
bi ggest !1H
i s used to symbolize the general fonn for an acid, that i s , a protonated base;
Hf"B
H H
H
H
H
B-
is the
H
•
H
H
H
H
H
allylic
The key step i s hydride shift from a 2° carbocation
to a 2° allyiic, resonance-stabilized carbocation,
which can subsequently lose a proton to fonn a
conjugated diene.
H
H
3 17
! B:
H
H
H
H
H
1 5 -3 (You may wish to refer to problem 2-6.)
(a)
2
3/
" 1
C == C == C
"
"
H""
H
tsp
orbital picture
H
H
The central carbon atom makes two n bonds with two p orbitals. These p orbitals must necessari l y be
perpendicul ar to each other, thereby forcing the groups on the ends of the allene system perpendicular.
(b)
/
H"
Cl
" " c == C == C
'"
"
Cl
mirror
Cl
H
H
non-superimposable mirror images
i
'"
"
C == C == C '
'
/
=
\
\\
H
CI
enantiomers
1 5-4 Carbocation stabi lity depends on conj ugation (benzylic, allylic), then on degree of carbon carry i ng the
positive charge.
H
H
I
\
C-C +
II
\
H2C
CH3
2°
�
:�� Jl
H
H
1°
more significant
contributor
H
C
I
H
less significant
contributor
H
H
H
H
H
equi valent to the
first resonance form
3 18
H3C
Y'�
allyJic
...... H
H3C ,
�
():-=
/ ==: fv
·
+
two carbon
electron deficient
so nucleophi le
can attack ei ther
�
+
C
�e
H
o •.
I
C H2 C H 3
The most basic
species in the
reaction mixture
will remove this
proton . The oxygen
of ethanol i s more
basic than bromide
Ion .
6� � (i
1 5 -6
I I YI i C
A gC I
+
�
H
:
C H CH'
'----f- OCH2CH3
H
0.
H
••
��
two c arbons are
electron defic ient
so nucleophi le
c an attack either
H - O - CH 2 C 1 3
H3
�
+
H-O:
..
I
CH2CH3
?,H
319
H
• •
'---+-+-
H
0.
CH2CH3
CH,CH3
�I
HJ
'-l(
H
1 5 -7
'"'
� H - Br
..
(
H
H3C
/
\
/C == C
�
H 3C
H
\
("' 1 +
CH 2 - O
0 0- H
-
H20
H
H3C
/
\
+ /C - C
\\
electron deficient
so nucleophile
c an attack either
00 B r° :-0
0
(b)
�
'"'
00
/
\
O0 -/C - C
o
\\
H/
H3C
CH2
same carbocation
as i n (a)
{
H - Br
...
�
-
H 20
..
H3C
CH2
0° B0 �0 o0
_
320
two c arbon s are
electron deficient
so nucleophile
can attack either
�
1 5 - 7 con
/
(c)
H
.....
&{ I �
n
Br - Br
Br
H2C -
? ?
+
-
H
H
.......
H
I II I
Br
allylic
H2C
H2C - C - C == CH2
+
-? ?
+
CH2
==
H
�
..
:Br :
Br
H
I )}
Br
H
:B�:
_
IIII
two carbons are
e lectron deficient
so nucleoph i le
can attack either
Br
Br
H2C - C == C - CH2
H
H
H
II - I I
}
r
f
(
i
�
)
II O II
!
I
I
II
II
While the bromonium ion mechanism i s typical for isolated alkenes, the greater stabi lity of the resonance­
stabilized carbocation w i l l make it the lower energy intermediate for conj ugated systems.
C�
(d)
H3C - C == C - CH2
I
H
H
Ag +
..
AgCl
allylic
+
H3C - C == C - CH2
H
H
�
�
-
H3
. .
•
H2O
.
H20
..
+
H- �H
••
• •
?
H3C - C == C - CH2
H
+
v?
+
-
9.
H3C - C == C - CH2
H
_
H
H20
. .
+
OH
H3C - C - C == CH2
H
321
H
two carbons are
e lectron defic ient
so nucleophile
can attack either
H3C - C - C - CH2
H
OH
H
H
H
H2
CH2
-
H
1 5-7 continued
(e)
+
H3C -
� (? ?
?� A
? ?
:;
== CH2
-
H
_
H
g
two c arbons are
e lectron defic ient
so nucleophile
can attack either
same carbocation as in (d)
allylic
+
H3 C
H2
-
H
R
== CH2
--
H
? ? f
+
H
==
3C -
-
H
I
+
H
}R
H2
,
H2
+
OH
OH
I
I
H3C - C - C == CH2
I
I
H
1 5 -8
(a)
Br
I
H2C - C - C == CH2
I
H
A
( b)
I
.....
H
H
Br
I
H2C - C == C - CH2
+
Br
I
H
+
H
20
H
H
I
I
B
H
�
n
/
B r - Br
I
I
H
I
I
H3C - C == C - CH2
H
Br
Br
+
H
. .
-
--
allylic
• •
:Br :
Br
I
Br
I
I
A
• •
Br
H2C - C - C == CH2
H
I
Br
I
I
+
two carbons are
electron deficient
so nucleophi le
can attack either
H2C - C == C - CH2
B
H
-
:Br:
I
H
I
H
(c) The resonance form A + , which e ventuall y leads to product A , has positi ve charge on a 2° carbon and is
a more significant resonance contributor than structure B+ . With greater positive charge on the 2° c arbon
than on the 1 ° c arbon , we would expect bromide ion attack on the 2° carbon to have lower activation energy.
Therefore, A must be the kinetic p ro d uc t . At higher temperature , however, the last step becomes reversible,
and the stabi lity of the products becomes the dominant factor in determining product ratios. As B has a
disubstituted alkene whereas A i s only monosubstituted, it is reasonable that B is the major, thermodynamic
product at 60° C .
322
1 5 - 8 continued
1
0
}
)
(d) At 60° C, ionization of A would lead to the same allylic carbocation as shown in (b), which would give
the same product ratio as formation of A and B from butadiene.
Br (Br
I
I
- B rH2C - C - C == CH2
Br
I
H2C -
+
I
I
H
A
I
T T
==
allylic
H
_
Br
I
Br
I
I
A
H
10 %
I
H2C - C == C - CH2
+
I
H
Br
I
H2C - C - C == CH2
B
90 %
I
H
I
H
1 5 -9
(a)
nn
(b)
Br - Br
hv
---
�Q)
HB r +
2 Br -
eX 0:
H
H
allylic
H
___
{� n
Br - Br
.
H
H
�---recycles in chain mechan i s m
Br - +
("Pr" i s the abbreviation for n-propyl , used below . )
NB S generates a low concentration of B r2
�
�
a
N - Br
+
H - Br
+
CCl4
-
a
323
I
H
H
Br
15-10
CH2
H
H
:B�:
Br
+
Br - Br
H
two carbons are
electron deficient
so nucleophile
can attack eIther
1 5 - 1 0 continued
initiation
nn
Br - Br
+
hv
propagation
Br ·
2 Br ·
---
pr -
��
-
�ri?
== CH2
HB ,
--.�
I -hexene
�
rec ycles in c hain mechanism
L--__
__________
+{
P, -
�J
��O
Br .
+
H
I
= CH'
�
t---l.
....
..
H
I
Pr - C - C == CH2
I
P,
JJ
�� )
H
I
+
- �H
J
H
I
Pr - C == C - CH2
I
Br
Br
The HBr generated i n the propagation s tep combines with NBS t o produce more B r2 ' continuing the chain
mechanism.
1 5- 1 1
(a)
U
+
�
B,
Br
(b)
<)
(e)
benzylic radicals are
even more stable than
allylic
These are the major products from
abstraction of a 2° al lylic H.
1 5 - 1 2 B oth halides generate the same allylic carbanion.
+
Br
I
H3C - C == C - CH2
� �
�
MgBr
•
�
H3C - C = C - CH2
� �
M
1
1 5- 1 3
(a)
Mg
---l��
ether
MgBr
I
CH3CHCH3
+
CH3CH == CHCH2B r
324
elI,Br
H
+
I
H3C - C - C == CH2
I
H
I
H
1 5 - 1 3 continued
dec - 5 -ene
H2C - CH2CH2 CH,
+
Thi s synthesis could also
be performed sequentially.
F
o
Br - CH2
add one-half equivalent
1 5- 1 4
C O OCH3
o
o
CN
C O OCH3
!
C O OCH3
15-15
(b) CH) O
+
(d )
O
CH30
OMe
(e)
+
O Me
CH ,
:(
o ,(
O
+
NC
+
NC
X
O Et
(c )
CN
(f)
CN
15-16
( a)
(b)
(C)
0
Co
0
6. . .
+
�
eN
<
0
+
0
0
CH3
I
. C O O CH3
.. ... . CHO
CH3
racemic mixture; wedge and
dashed bonds show relative
stereochemistry
325
1 5- 1 7 These structures show the alignment of diene and dienophi le i n the Diels-Alder transition state,
leading to 1 ,4-orientation in (a) and 1 ,2-orientation in (b).
(b)
(a)
:0:
this left structure is a VERY minor resonance contributor;
however, it explains the orientation for the diene as carbon- l
is more negative and carbon-2, a 3°C, is s lightly more
positi ve because of methyl group stab i lization
1 5 - 1 8 For clarity, the bonds formed in the Diels-Alder reaction are shown in bold.
(b)
COOCH3
1 5 - 1 9 For a photochemically allowed process, one molecule must use an excited state in which an elec tron
has been promoted to the first anti bonding orbital . All orbital interactions between the excited molecule's
HOMO* and the other molecule's LUMO must be bonding for the interaction to be al lowed; otherwise, it is a
forbidden process.
HOMO *
excited state o f
diene HOMO
�J
anti bonding interaction
(destructive overlap)
bonding interaction
(constructive overlap
LUMO of
dienophi le
In the Diels-Alder cycloaddition, the LUMO of the dienophile and the excited state of the HOMO of the
diene ( l abeled HOMO * ) produce one bonding interaction and one anti bondin g i nteraction. Thus, this is a
photochemically forbidden process.
326
15-20 For a [4 + 4] cyc\oaddition:
(a)
Photochemical
Thermal
HOMO
HOMO*
)
(
)
(
bondi ng
interaction
anti bondi ng
i n teraction
both interactions are bonding
one interaction is anti bonding
� allowed
� forbidden
(b) A [4 + 4] cycloaddition is not thermal ly al lowed, but a [4 + 2] (Diels-Alder) i s !
HOMO
bonding
bonding
'-interaction �
J� interaction
11:3
LUMO
1 5 -2 1
A
=
EC
l
A
E =
E =
6
x
c
1 0--6 moles
I O mL
1 mg x
l L
X
4
1 0- ) ( 1 )
=
833
I cm
1 000 mg
=
""
=
1 g
1 000 mL
x
0.50
(6
l
l
-
convert mass to moles:
C =
11:3
x
A
1 mole
1 60 g
--
4
6 x l 0-- M
8
327
=
=
0.50
6 x 1 0--6 moles
1 5-22
(a) 353 nm: a conj ugated tetraene-must have highest absorption maximum among these compounds ;
(b) 3 1 3 nm: c losest to the bicyclic conj ugated triene in Table 1 5 -2; the diene i s in a more substituted ri ng,
so it is not surprising for the maximum to be slightly higher than 304 nm;
(c) 232 nm: simi l ar to 3 -methylenecyc lohexene in Table 1 5 -2;
(d) 273 nm: 1 ,3 cyclohexadiene (256 nm) + 2 alkyl substituents (2 x 5 nm) = predicted value of 266 nm;
(e) 237 nm: like 3 -methylenecyclohexene (232 nm) + 1 alkyl group (5 nm) = predicted value of 237 n m
1 5-23 Please refer to solution 1 -20, page 12 of this Solutions Manual .
1 5-24
(a) isolated
(b) conj ugated
b 0
3
4
5
1 5 -25
(a)
(d)
�
�
Br
(f)
Br
(g)
(d) conj ugated
1
W
(b)
7
o-Cl
0
5
+
3
(e) conj ugated
�
(e)
"
OH
Br
�
Br
+
�
Br
+
B r minornot conj ugated
�
Br
(h)
(i)
328
0
(f) cumulated ( 1 ,2) and
conj ugated (2,4)
H
/C �
H 2C = C = C
" CH2
H
(c)
one eq ui valent
of HC l
Br
�
(c) cumulated
Br
+
Br
Br �
OH
15-26 Grignard reactions are performed in ether solvent. For c l arity, the bonds formed are shown in bold.
(a)
(b)
� Br +
BrMg �
I
N'
+
BrMg
r
•
5
4
15-27
(a)
+
�CH2""'''I--·· �?�CH2 ��H2
C H2
CH2
C H2
•
•
.
..1-....
(b)
(c)
·6
:0:
:0:
H
H
H
-........ C .....
:0
(d)
a:
o
H
O+I
H
H
329
:0 :
:0 :
:0:
:0:
H
'iC,
H
H
_
2
H
�
&
15-27 continued
(e)
(f)
{
V
(g)
15-28
(a)
E=
A = Ee l
E
=
A
3.9 x 10-6 moles
100mL
cl
0.00 10 g x
x
0.74
'"
( 3.9 x 10-5 ) ( 1 )
H
__
(C yO:
HC �
l = 1 cm
-
convert mass to moles:
c=
(!=.yO:
H
1000mL
lL
1 mole
=
255 g
= 3.9
A = 0.74
x
3.9 x 10-6 moles
10-5 M
El
(b) Thi s large value of E could only come from a conj ugated system, eliminating the first structure. The
absorption maximum at 235 nm is most l i kely a diene rather than a triene. The most reasonable structure is:
compare with:
co
"max = 235 nm
Solved Problem 15-3
330
"max
=
232 nm
Table 15-2
15-29
�
(a)
�
trans
+
Br
(b)
+
Br
Br
�
cis
CPr" is the abbreviation for n-propy l , used below . )
�
NB S generates a low concentration of Br2
o
+
N- Br
I1-B r
--
o
�
o
N-H
+
Br-Br
o
initiation
nn
Br-Br
hv
2 B r·
---
propagation
first step is abstraction of allylic hydrogen to generate allylic radical
I
I
H H
+
Pr -C
",,--,r
Br·
==
CH2
C
-
HBr
---I"�
+
radical w i l l be a
mixture of cis
and trans
I-hexene
I
I
H H
Pr- C ==C- �H 2
n�)
Br - Br
�--
recycles in
chain mechanism
0
�
Br·
+
I
I
H H
I
Pr - C -C
==
CH2
I
+
Pr -C
Br
CIS +
15-30
(a)
(c )
(b )
COOH
COOCH3
331
I
H H
I
C-CH2
==
Br
trans
,- k
�
COOH
t
15-30 continued
0=
(e)
I
_
"
,
(f)
eN
,
"
""
"
",CN
",
""'eN
eN
'
(Note: Yield of the product in (f) would be small or zero due to severe steric interaction in the s-cis
conformation of the diene. See Figure 1 5- 1 6 . )
15-3 1
(a)
1f
H2C
�+ /
k
�
..
0(
more significant
contributor: 2°
(b)
more significant
contributor: 2°
(c)
(d)
i ..
I
:0
{
{
+
�C
H
C - CH3
:0:
0(
• •
CH3-C-NH2
..
0
�
H
==
:
II
-
:0:
C H3-C=NH2
(e)
H2 C
0(
I
H
H
0(
t
t
more significant contributor-negati ve charge on more electronegatIve
more significant contributorno charge separ ation
0
..
H
.
H
H
:0:
.
�
�
H2C
H
:0:
H
..
..
H
H
H
most sign i ficant contributornegative c h arge on more electronegati ve atom
332
H
atom
(f)
15-31 continued
most significant
contributor---all atoms
have ful l octets
..
(a) The absorption at 1630 c m
migration of the double bond_
15-32
-
i
more significant contributor­
negative charge c an be stabilized
by electronegati ve ch lorines
..
suggests a conj ugated alkene_ The higher temperature al lowed for
UCH,CH,CH] OCH,CH2CH]
(b)
desired
actual
(c )
expected:
+
[OCH,:f-CH2CH]] --{oCH,
doubly allylic
mlz 122
actual :
.
�
�
+
.6
mlz 122
(
15-33
(aj
(dj
(
+
�H
(b)
0
+
o
o
+
¢
o
(e)
o
+
CH2
C
H3
�CH' }
�
�
C'lL
Cl
333
mass 43
mlz 79
mlz 93
+ •
+
mass 29
CI CI CI +Q-CI
::;:,- CI CI CI
CI*
1 5-33 continued
(g)
15-34
(a)
+
Co
0
I
NH
0
S�O
CI CI CI +
d'
:
CN
CI
tb
::;:,CI*CI
I
f
NC
0
(i)
(h)
+
0
H
7
90°
�
exo
H
(b) The endo isomer is usually preferred because of secondary p orbital overlap of
in the transition state.
C, C,
C=O
C
with the diene
(c) The reasoning i n (b) applies to stabi lization of the transition state of the reaction , not the stability of
the product. Arguments based on transition state stabi lity apply to the rate of reaction, i nferring that the
endo product is the kinetic product.
(d) At 25°
the reaction c annot easily reverse, or at least not very rapidly. The endo product i s formed
faster and is the maj or product because its transition state is lower i n energy-the reaction is under
kinetic control. At 90°
the reverse reaction is not as slow and equilibrium is achieved. The exo
product is less crowded and therefore more stable-equilibrium control gives the exo as the maj or
product.
t
,/
,.
,.
.
,
�
,,
."
..
" -'.,
,.
'..
�
...
endo
.
exo transition state-no secondary orbital overlap
.
transition statesecondary orbital overlap
stabilization
,.
.,
.
...... ...
endo product
..
exo product-more
�----reaction -
334
stable, less crowded
(a)
15-35 Nodes are
represented by dashed l ines.
1t *
6
1ts*
1t4*
1t3
1t
(b)
-
-
()
1[6*
1[5*
H ::
_H '"
H
1[
1
()
(C)
()
0
0
'
0
335
2
1t
l
15-35 continued
(d) Whether the triene i s the HOM O and the alkene is the LUMO, or vice versa, the answer wi ll be the same .
Thermal
Photochemical
HOM O*
LUM O
LUMO
one anti bonding interaction
�
two bonding interactions
�
forbidden
(e)
allowed
o
+
o
J H2C== ?- ? == ? - CH2
1 H H H
15-36
(a)
?
?
?
H2C== - - ==CH2 ..
H H H
....
..f--.
�
..
(b) Five atomic orbitals will generate five molecular orbitals.
(c) The lowest energy molecular orbital has no nodes. Each higher molecular orbital will have one more
node , so the fifth molecular orbital will have four nodes .
336
15-36 continued
(d) Nodes are represented by dashed l ines.
11:5*
11:4*
11:3
(non bonding)
11:2
11:1
(e)
__
_1
1L
JL
11: *
5
11:4*
11:3
n,
( f) The HOMO,
11:3,
contains an unpaired electron giving thi s species its radical
character. The HOMO is a non-bonding orbital with lobes only on carbons 1,
i�-�-�}
3 , and 5, consistent with the resonance picture.
n,
337
15-36 conti nued
(g)
--
---
H
_H
(h)
__
1£5*
Agai n , it is
e lectron in
1£4*
1£3
1£3
that determines the character of this species. When the single
of the neutral radical is removed, positive charge appears only in
the posi tion(s) which that electron occ upied. That i s , the posi ti ve charge depends
on the now empty 1£3' with empty lobes (positive charge) on carbons 1, 3, and 5,
1£3
consistent wi th the resonance description .
J�
1+
1£2
___
+
�
___
l
+f
�
1£]
1£5*
Again, it is
1£3
that determines the character of this spec ies. The negative
charge depends on the filled 1£3' with lobes (negative charge) on carbons 1, 3,
i-·�-�-�;t
and 5, consistent with the resonance description.
H
H
_H
a
A
Mg
..
ether
I
MgBr
H3 C-C-OH
CH3
(b) Use Appendix 3 to predict Amax values . Alkyl substituents are circ led .
transoid cyclic diene = 2 17 n m
4 alkyl groups = 20 n m
exocyclic alkene = 5 n m
TOTAL = 242 n m
desired product
cisoid cyclic diene = 253 nm
3 alkyl groups = 15 nm
TOTAL = 268 nm
338
c isoid cyclic diene = 253 11m
4 alkyl groups = 20 nm
TOTAL = 273 nm--AHA!
actual product B
15-37 continued
(c)
hydride
shift
B is produced in preference to the other ci soid diene
above bec ause Bls diene system is more highly
substituted and therefore more stable
�
Q
Hi) :
B
H3C-
? -H
CH3
15-38 It is stunni ngly c le ver reactions like this that earned E. J. Corey his Nobel Prize.
�o oD
Diels-Alder
1
+
COOCH3
CH3
---
endo product
imagine the diene on top
of the dienophile in the
transition state, leading
to the endo adduct
new
0= = 0
retro-Diels-Alder
C
o
�
dienophi le
same as
o
339
o
COOCH3
amazing!
o
CHAPTER 16-AROMATIC COMPOUNDS
Note: The representation of benzene w ith a circle to represent the 1t system is fine for questions of
nomenclature, properties, i somers, and reactions. For questions of mechanism or reacti vity,
however, the representation with three alternating double bonds (the Kekule picture) is more
informative. For c l arity and consi stency , thi s Solutions Manual will use the Kekule form exc lusively.
o
�
equivalent representati ons of henzene
Kekule form used i n
the Solutions M anual
H
16- 1
H
H
H .
C
CH ' ' .. C · H
H '
H
•
•
•
' -C .
C·
•
•
C.
•
•
•
•
•
•
H
•
C·
C
.
•
C,, . H
.C.
' 'C .
C·
•
•
H
H
1 6-2 All values are per mole .
benzene
(a)
-208 kJ
-240 kJ
1 ,4-cyclohexadiene
MI
(b)
-
(c)
-
7.6 kcal
- 49.8 kcal
cyclohexene
- 1 20 kJ
- 28.6 kcal
- 88 kJ
- 2 1 . 2 kcal
f.,}{
=
1 ,3-cyc lohexadiene
-232 kJ
- 55 .4 kcal
cyc lohexene
- 1 20 kJ
- 28.6 kcal
- 1 1 2 kJ
-26 . 8 kcal
f.,}{
.... C,..C.::::-C' H
H
I
C
H
'c-:::' 'c'
I
1----
II
=
H
..... C, -:::.C ....
H
H
C
I
{
+
-208 kJ
H
I
H
32 kJ
- 57.4 kcal
benzene
1 6-3
(a)
+
=
- 49. 8 kcal
II
..... C.::::- ,..C ....
H
H
C
I
H
H
H
"H
"H
"C-C
C=C
---- II II
I
I
C-C
C=C
,
,
H
"-
H
/
/
H
H
H
}
341
c::::::>
©
1 6-3 continued
(b)
16-5 Figure 1 6-8 shows that the fi rst 3 pairs of electrons are in three bonding molecular orbitals of
cyclooctatetraene. Electrons 7 and 8, however, are located in two different nonbonding orbitals . As in
cyclobutadiene, a planar c yc looctatetraene is predicted to be a diradical , a particular l y unstable electron
configuration.
Models show that the angles between p orbitals on adjacent
1l:
bonds approach 90°.
1 6-7
(a) nonaromatic: internal hydrogens prevent planarity
(b) nonaromatic: not al l atoms in the ri ng have a p orbital
(c) aromatic: [l4]annulene
(d) aromatic: also a [ 1 4]annulene in the outer ring: the internal alkene is not part of the aromatic system
16-8 Azulene satisfies all the criteria for aromaticity, and it has a Huckel number of 1l: electrons: 10. Both
heptalene ( 1 2 1[ electrons) and pentalene (8 1l: electrons) are anti aromatic .
1 6-9
(a)
1l:6*
1l:4
1[5
1l:2
(b )
1[7*
nonbonding
�!1�!g.x
1[3
1l:6*
1l:4
1[2
1
1l:s*
H H H
1l:1
1
1l:7*
1l:s
1l:3
This electronic c onfiguration is antiaromatic.
342
16-9 continued
(c )
These are TOP views. Nodes
are shown by dashed li nes.
16- 10
( aj
____
�y.
____
_
______ _
n2 *
�-----iII[
---
(b)
n2 *
n3*
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - non bonding
nI
i s bonding;
n2 *
and
n3*
nI
are antibonding.
(c)
n,
��-------- �------�) \�------Y
Y
nI
/
�------�
cation-aromatic
ani on-antiaromatic
343
16- 1 1
(a)
---
(b)
n2
n5*
antibonding
n3
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _ .
R"
(c)
7l:z
\.
j
H
y
n1
("
R"
n3
7l:z
)
\.
cation-antiaromatic
j�
non bonding
bonding
H
y
n1
rj "
anion-aromatic
16-12
(a) antiaromatic: 8 n electrons, not a Huckel number
(b) aromatic : 10 n electrons, a Huckel number
(c) aromatic: 18 n electrons, a Huckel n umber
(d) antiaromatic: 20 n electrons, not a Huckel number
(e) nonaromatic: no cyclic n system
(f) aromatic : 18 n electrons, a Huckel number
344
n3
)
The reason
r the dipolepicture
can begives
seenonein aring
resonance
distributing
electrons
give each
ring
electrons.
This foresonance
a negativeformcharge
and thetheother
ring a topositive
charge.
1 6- 1 3
61t
tropylium ion
electrons
aromatic
61t
cycIopentadienyl anion
electrons
aromatic
61t
�
�
"\
several
resonance
f
o
rms
several
resonance
f
o
rms
delocaJizing
thearound ___
delocalizing
thethenegative
____ charge
positive
charge
around
thenngseven-membered
membered ring five­
"'---- ----�
(
�---------
�----------
�--------�
�--------�
composite
resonancewithpicture
rings
are aromatic
chargeshows
separationboth
that
1 6- 1 4
+
AgBF4
AgCl
--
+
aromatic cyc\opropenium ion
CI
!KCI
The
crystalline
material
soluble
/
incycIopropenium
polar organic solvents
tetrafluoroborate. Cl-<Jl
carbon.Draw0 resonance forms showing the0carbonyl polarization,leaving0 a positive charge on the carbonyl
is
1 6- 1 5
&
t
:0I :
+C
B
electronsAROMATIC
21t
0
t
..
:0:I
0
electronsAROMATIC
.:0:t.
+CI
The three-and
sevenmembered
rings
aromatic;the fivemembered
and, reactive.
0 antiaromatic
surprisingly,very
electronsANTI AROMATIC
41t
61t
345
are
ring
is
nut
:(:
l6-l6
'I
�oo
o 0
7N
9
H
I
N
I
N
1
0
)
0 0
N
3
0
The structure of purine shows two types of nitrogens. One type (N-l, N-3, and N-7 )
has an electronic structure like the nitrogen in pyridine; the pair of electrons is in an
sp2 orbital planar with the ring. These electrons are available for bonding, and these
three nitrogens are basic. The other type of nitrogen at N-9 h as an electronic
structure l i ke the nitrogen of pyrrole; its electron pair is in a p orbital , perpendicular
to the ring s ystem, and more importantly, an essential part of the aromatic pi system.
With this pair of electrons, the pi system is aromatic and has 10 electrons, a Huckel
number, so the electron pair is not avai lable for bondi ng and N-9 is not a basic
nitrogen .
l6-l7
(a) The proton N MR o f benzene shows a single peak a t 8 7 . 2; alkene hydrogens absorb a t 84.5-6. The
chemical shifts of 2-pyridone are more simi l ar to benzene's absorptions than they are to alkenes. It would be
correct to infer that 2-pyridone is aromatic .
(b)
H
H H�H
J:�
.
H
--..
� x):�
H
N
H
--..
""
0:
+
The lone pair of electrons on the nitrogen in the
first resonance form is part of the cyclic pi
system. The second resonance form shows three
alternating double bonds with six electrons in the
cyclic pi system, consistent with an aromatic
electronic system.
H
H
x�.Hi�. -.... . Hn�.
H H 0:
7
H
N
&7 . 26
0:
0 0
�
...
_
N
These resonance forms
show that the hydrogens at
positions with greater
electron density are
shielded, decreasing
chemical shift.
These resonance forms
show that the hydrogens at
positions with greater
positive charge are
deshielded, increasing
c hemical shift.
87.3l
:0 :
This resonance form of thymine shows a cyclic
pi system w i th 6 pi electrons , consistent with an
aromatic system. Four of these electrons c ame
from the two lone pairs of electrons on nitrogens
in the first resonance form.
346
1 6- 1 8
(a)
/
(..0 ;�:
(b)
(c )
\
:N �..S :
aromatic: 4 n
e lectrons in
double bonds
plus one pair
from sulfur
aromatic: 4 n
electrons in
double bonds
plus one pair
from oxygen
6 6
.
(e)
0.
�
.o.
.
0
+
H
1
H � , B ........ � H
�N
N'
•
1
•
I
�
9
�
10
anthracene
�
I
.
.
N
H
�
....
..f. -I.
_
OCO
�
I
(�)�6
�
h-
0
.o.
.
.
+
aromati c : c ation on carbon4 indicates an empty p
orbital ; two n bonds plus a
pair of electrons from
oxygen makes a 6 n
electron system
sp3
H
1
H +, B ...... + H
..... N'
N'
1 1_
1
, B .;:::. , B ......
H
H
N
+
1
H
, B ....... . , B ......
H
H
N
1
H
1 6-20
(a )
O
not aromatic :
no cyclic n
system
3
because of sp
carbon
aromatic : resonance fonn
shows "push-pul l " of
electrons from one
to
the other, making a cyclic
n system with 6 electrons
1 6- 19
d0
not aromatic :
no cyclic n
system
3
because of sp
carbon
(f)
.-
H
(d) H
� --h-
347
aromatic: resonance form shows
electron pair from N making a cyclic
n system with 6 electrons
Borazole is a non-carbon equivalent of
benzene. Each boron is hybridized in Its
2
2
nonnal sp . Each nitrogen is also sp with Its
pair of electrons in its p orbital. The system
has six
n
electrons in 6 p orbitals--aromatic!
�
�
---
�
�
1 6-20 continued
(b)
anthracene
H
n
B r� B r
�0
__
----,..
©-c00 �cQo� I�
H
..
. r.
.
.B
H
H
H
plus many other
resonance forms
Br
�
�
Br
H
bromide attack at
C-lO leaves
two aromatic rings
cis and trans
H
bromide attack at
C-9 leaves two
aromatic rings
H
(c) A typical addition of bromine occurs with a bromonium ion intermedi ate which can gi ve only anti
addition. Addition of bromine to phenanthrene, however, generates a free c arbocation because the
carbocation is benzylic, stabi lized by resonance over two rings. In the second step of the mechanism,
bromide nucleophi le can attack either side of the carbocation giving a mixture of cis and trans
products.
(d)
C
L1
-----
E1
1 6-2 1
l
6
C
chlorobenzene
�Cl
�Cl
Cl
1 ,2,3-trichlorobenzene
+
H
�'"
,
+
Br
H
HB r
·B r.
•
•
--../
H
resonance-stabilized
�
V
Cl
q
Cl
o
�
Cl
a-dic h lorobenzene
( 1 ,2-dichlorobenzene)
Cl
Cl
p-dichlorobenzenc
( 1 ,4-dichlorobenzene)
Cl
Cl
Cl
Cl
o
�
Cl
1 ,3,5-trichlorobenzene
348
¢
m-dichlorobenzene
( 1 ,3-dichlorobenzene)
Cl
Cl
1 ,2 ,4-tri c hlorobenzene
Cl
(y
Y
CI
CI
CI
1,2,3 ,4-tetrachloro­
benzene
CI
� CI
CI V CI
CI
� CI
16-2 1 continued
CI Y
CI
1 ,2,3 ,5-tetrachloro­
benzene
CI
1 ,2,4,5-tetrachloro­
benzene
16-22
(a) fluorobenzene
(b) 4-phenylbut-I-yne
(c) m-methylphenol , or
3-methylphenol
(common name: m-cresol)
< }- O-CH3
CI
� CI
�
CI
CI
CI Y CI
CI
¥
CI
CI
1 ,2 ,3 ,4,5-pentach loro­
benzene
(d) o-ni trostyrene
(e) p-bromobenzoic acid, or
4-bromobenzoic acid
(f) isopropoxybenzene, or
isopropyl phenyl ether
CH3 --<
CI
}- S03H
1 ,2 ,3,4,5,6-hexachloro­
benzene
(g) 3 ,4-dinitrophenol
(h) benzyl ethyl ether, or
benzoxyethane, or
(ethoxymethyl)benzene, or
a-ethoxytoluene
< }-
02N -{
}- OH
1 6-23 These examples are representati ve. Your examples may be different from these and sti l l be correct.
(a)
(b)
p-toluenesulfonic acid
methyl phenyl ether
or methoxybenzene
or anisole
(e)
�CH-OH
o-
<
(0
I�
.0
diphenylmethanol
16-24
H
H
CH2
+
H
H
H
(c)
phenyllithium
CH3CH20
(g)
f<>
Br
H
+
..
..
H
H
H
..
H
..
H
These are phenols.
Thi s is 4-nitrophenol .
� >- CH,OH
Br
1 ,3-dibromo-2-phenylbenzene
H ...... C
(d)
Li
ii
I � I
I
H
3-ethoxybenzyl alcohol
or 3-ethoxyphenylmethanol
H
H
H
..
..
H
H
H
349
H
0-:O�
1
16-25
Ij_�
CH-CH ==CH2
Amax
H+
<
..
2 20 nm (strong)
258 nm (weak)
CH-CH ==CH2
i O-
CH=CH-
o�
t
1
--
H20 :
OH
250 nm (strong)
290 nm (weak)
<>
�
H20 :
CH ==CH-CH2
Amax
)+¥3� -H20
_
o-�
•
�
�H-CH=CH2
plus resonance fOnTIS with positive
charge on the benzene ring
CH==CH-
?�2
H
r
c
-
electronic systems with extended
conj ugation absorb at longer wavelength
N02
O
�
9 9 6
16-26 Please refer to sol ution 1 -20, page 12 of this Solutions Manual .
1 6- 27
(a)
(0
OH
OCH3
1.0
(b)
(g)
HO
1.0
1.0
CH O
CH30
CH2 OCH3
6
1.0
1
(; )
(h
(m)
Ih
¢
I�
.0
CH3
350
CH3
NH2
(d)
9 �
Br
(I)
C O OH
NH2
HC=CH2
HC=CH2
(k)
(e)
OCH3
HC=CH2
9
�
1.0
OCH3
}
(e)
.0
N02
H
C. Cl-
0
(j)
-
OCH3
CH3
H
(n)
.0
O
I .0
�
CH3 (0)
H
I
c-
0
II
Cl
Na+
1 6-28
(a) 1 ,2-dichlorobenzene (ortho)
(b) 4-nitroanisole (para)
(c) 2,3-dibromobenzoic aci d
(d) 2,7 -dimethoxynaphthalene
16-29
(e) 3-chlorobenzoic aci d (meta)
(0 2 ,4,6-trichlorophenol
(g) 2-sec-butylbenzaldehyde (ortho)
(h ) cycIopropenium tetrafluoroborate
(5
�
q
CH3
llA
¢
o-xylene
toluene
CH l
CH3
CHl
CH3
CH3
1 ,2,4-tri methylbenzene
1 ,2,3-tri methylbenzene
£
.&
CHJ
p-xylene
C H3
1 ,3,5-trimethylbenzene
(common name: mesitylene)
1 6-30 Aromaticity is one of the strongest stabilizing forces in organic molecules. The c ycIopentadienyl
system is stabilized in the anion form where it has 6 'It electrons, a Huckel number. The question then
becomes: which of the four structures can lose a proton to become aromatic?
3
Whi le the first, third, and fourth structures can lose protons from sp carbons to give resonance­
stabilized anions, onl y the second structure can make a cycIopentadienide anion. It w i l l lose a proton most
easily of these four structures which, by definition, means it is the strongest acid.
1 6-31
�
H
H
(a)
H yly H
H VH
H
H
� �
H
AROMATIC
H
H
H
H
H
H
H
H
H
H
H
H
351
H
1 6-3 1 continued
(b)
(c)
6
6
u
o
Br
Br
Jl
U
+
Br
Br
Br
c::::::=>
+
¢
Br
Br
6
+
flZ);
a:;;
+
Br
Br
Cl
�
a L6
Br
Br
Br
+
Br
\.
�
I
B'
+
B'
Y
�
I
J
different positions
of double bonds
+
Br
(ignoring enantiomers)
(d) The only structure consistent with three isomers of di bromobenzene is the prism structure, called
Ladenburg benzene. It also gives no test for alkenes, consi stent with the behavior of benzene. (Kekule
defended his structure by claiming that the "two" structures of ortho-dibromobenzene were rapidly
interconverted, equilibrating so quickly that they could never be separated. )
(e) W e n o w know that three- and four-membered ri ngs are the least stable, but this fact was unknown to
chemists duri ng the mid- 1 800s when the benzene controversy was raging. Ladenburg benzene has two
three-membered rings and three four-membered rings (of which only four of the rings are independent),
which we would predict to be unstable. (In fact, the structure has been s ynthesized. Called p rismane, it is
NOT aromatic, but rather, is very reactive toward addition reactions. )
352
1 6-32
H
H
A
(a)
H
H
nonaromatic
(b)
0
U
(c)
N
H
I
aromatic6 11: electrons
aromatic6 11: elec trons
H
nonaromatic
(d)
N
o
o
0
H
anti aromatic4 11: electrons
o
nonaromatic
1+
H
H
o
o
1-
C
aromatic6 11: e lectrons
[)
0
0
aromatic6 11: electrons
(g)
ill
W
antiaromatic12 11: electrons
(B is Sp 2 but donates
no e lectrons to the 11:
system)
o
+
N
I
H
nonaromatic
H
aromatic6 11: electrons
(00]0
o
(::]
o
o
0
-c
+c
I
I
H
2
if oxygen is sp => anti aromatic (8 11:
e lectrons ) ; if oxygen is sp3 => nonaromatic
=>
(t)
o
.
.
antJaromatlc4 11: electrons
I
o
o
nonaromatic
I
N
aromatic6 11: electrons
H
H
B
o
° °
nonaromatic
aromatic6 11: electrons
anti aromatic4 11: electrons
N
(e)
+"N'
C
N
aromatic6 11: electrons
o
aromatic2 11: electrons
H
H
aromatic6 11: electrons
not aromatic in either case
+\
� oo
H -N/'
N- H
I
aromatic6 11: electrons
nonaromatic
H
this i s a tough call-it has 10 11: electrons s o
could be aromatic , but internal H's might
force it out of planarity
aromatic10 11: electrons
353
it
1 6-32 continued
(h)
H
H
+
0
H
H
C
0
0
nonaromatic
2
I
B
C-
anti aromatic8 1t electrons
aromatic6 1t electrons
(B is sp , but
CH3
I
I
donates no
electrons to the
1t system)
0
aromatic6 1t electrons
1 6-33 The c lue to azulene i s recognition of the five- and seven-membered rings. To attain aromaticity, a
seven-membered c arbon ring must have a positive charge; a five-membered c arbon ring must have a
negati ve charge . Drawing a resonance form of azulene shows thi s :
OJ ---. C»
-..
--..
--�
ri:�
ance
s
} <0@
composite picture
The composite picture shows that the negative charge is concentrated in the five-membered ring, giving rise
to the dipole.
16-34 Whether a n itrogen i s strongly basic or weakly basic depends on the location of i ts electron pair. If
the electron pai r i s needed for an aromatic 1t system, the nitrogen w i l l not be basic (shown here as "weak
2
3
base" ) . If the electron pair i s i n ei ther an sp or sp orbital, i t i s avai lable for bonding, and the nitrogen is a
" strong base" .
H
(a)
I
�
HN:
"
\
weak base
I
N:
I' strong
base
ce·�'\
I
strong base
(c)
(f)
()
I
(e)
0
strong base
strong base
�
\
O
N:
I
strong base
H
weak
�
/N
0
N
strong base
N / base
H
(d)
(b)
/
• •
N
strong
base
g1
weak base
1 6-35 Where a resonance form demonstrates aromaticity, the resonance form is shown.
(al
1"-':
O
+R
/.
aromatic
(b) : 0 :
6
+
�
aromatic
:6
(c)
..
:0:
aromatic
(d)
B AD
+
0
:0 :
:0 :
NOT aromatic-although it can be drawn ,
the resonance form o n the left i s NOT a
significant contributor because the oxygen
does not have a ful l octet; the form on the
right shows the correct polarization of the
carbony l , but it's still not aromatic because
of only 4 1t electrons
354
(e)
0
:0:
aromatic
16-35 continued
�+
�+
+ �,..... not basic
Nitrogens whose e lectrons are needed to complete the aromatic
(D
(g)
o
aromatic ,
not basic
(k)
•
0..
•
• •
-
(I)
(::]
. o.
�H
o
6
C
NOT aromatic
3
because of sp carbon ;
both N are basic ,
although the top one
is conj ugated,
lowering its basicity
O
23
..
..
_
.. ---
-"6
nn
Br- Br
·
hv
-
6
CH 2
CH 2
..
..
..
-
2 B r-
6·:·
-
C H2
..
propagation steps on the next page
355
-
NH2
-
N
H
a, i C
O:
aromatic ; bottom N is
not basic because it
has donated its
electron pair to make
the ri ng aromalic
o
..
,..... basic
tf.:
• •
+
+
-
initiation
ic
+
..
-.. �-
"
NOT aromatic: if 0
is sp 2 , then the pi
system has 8 e-
NOT aromatic : i f
both N and 0 are
sp 2 , then the pi
system has 8 e- ;
N will be basic
aromatic;
not basic
3
sp
aromatic
(ml
(j)
�,..... basic
system w i l l not be basic .
C� �
(i )
iC
-
H
N
B­
H
(b)
•
c;):;
•
aromatic;
N i s not
basic but
the 0 is a
weak base
[j
(a)
9
:0 :
+�
16-36
(h)
1t
(5
1 6-36 continued
HEr
propagation
6
+
resonance-stabilized
� r-·CH2
n!
B r - Br
+
1.&
----l..
�
B r·
+
(c) Both reactions are S N2 on primary carbons , but the one at the benzylic carbon occurs faster. In the
transition state of S N2, as the nuc leophi le is approaching the carbon and the leaving group is departmg, the
electron density resembles that of a p orbital . As such, it can be stabi lized through overlap with the 1'[
8system of the benzene ring.
: Br :
• •
H
�
stabi li zation
through overl ap
1 6-37
(a)
VlCH�
8-
:O-CH3
• •
+
3 isomers
only 1 isomer
(c ) The original compound had to h ave been meta-dibromobenzene as this is the onl y dibromo isomer that
gives three mononitrated products .
Br
Br
Br
�
VlBr
�
�N02 o
Vl
y
+
Br
Br
16-38
(a) The fonnula CgH70C I h as five e lements of unsaturation, probably a benzene ring (4) plus either a
double bond or a ring. The IR suggests a conj ugated carbony l at 1 690 c m-1 and an aromatic ring at 1602
cm-I. The NMR shows a total of five aromatic protons, indicating a monosubstituted benzene.
singlet at 8 4.7 is a deshielded methylene.
<>
o
+
+
I I
C
CI
}
o-�
-
A 2H
C - CH2
o
I
CI
(b) The mass spectral evidence of molecular ion peaks of 1 : 1 i ntensity at 1 84 and 1 86 shows the presence
of a bromine atom. The rnlz 1 84 minus 79 for bromine gi ves a mass of 1 05 for the rest of the molecule,
which i s about a benzene ring plus two carbons and a few hydrogens . The NMR shows four aromatic
hydrogens in a typical para pattern (two doublets), indicati ng a para-disubstituted benzene. The 2H quartet
and 3H triplet are characteristic of an ethyl group.
-0-
1 6-39
(a)
+
Br
}
+
�
�
�
�
like the ends of a
conjugated diene
Br
-o-
CH2 CH3
Diels·Al der
product
o
new sigma bonds shown in bold
16 40
(a) No, biphenyl is not fused. The rings must share two atoms to be labeled "fused".
(b) There are 1 2 n electrons i n biphenyl compared with 10 for naphthalene.
(c) B i phenyl has 6 "double bonds ". An i solated alkene releases 1 20 kJ/mole upon h ydrogenation .
-
predicted: 6
x
1 20 kJ/mole (28 .6 kcaVmole)
""
observed:
720 kJ/mole ( 1 7 2 kcaUmole)
4 1 8 kJ/mole ( 1 00 kcal/mole)
resonance energy:
302 kJ/mole (72 kcaVmole)
(d) On a "per ring" basis, biphenyl is 302 -:- 2 = 1 5 1 kJ/mole, the same as the value for benzene.
Naphthalene's resonance energy is 252 kJ/mole (60 kcaVmole); on a "per ring" basis, naphthalene has only
1 26 kJ/mole of stabilization per ring. This i s consistent with the greater reacti v i ty of naphthalene compared
with benzene. In fact, the more fused rings, the lower the resonance energy per ring, and the more reactive
the compound. (Refer to Problem 1 6-20. )
3
2
16-4 1 Two protons are removed from sp carbons to make sp c arbons and to generate a n system with to n
electrons.
H
H
357
1 6-42
(a)
(c)
�H
l:J"
V- Cl
""--
0
H
CH
aromatic
Ag+
� CH
r>-- +
..
(d)
aromatic
H
(e)
O
�C-H
1-
�:::
C
aromatic
r:}
NaOH
.
-H
aromatic
H
H
O
C
..
C4 H9Li
(f)
comatk
0
Y
o
Ag+
Cl
..
+
()-o
• •
• •
0
H
aromatic
1 6-43 These four bases can be aromatic, partially aromatic, or aromatic in a tautomeric form. In other
words, aromaticity plays an i mportant role in the chemistry of all four structures. (Only electron pairs
involved in the important resonance are shown . ) For part (b) , nitrogens that are basic are denoted by B.
Those that are not basic are shown as NB. Note that some nitrogens change depending on the tautomeric
form.
(a) and (c)
cytosine
aromatic to the extent that this
resonance form contributes
..
:0:
{
� �
�
B H
/
N
NB I
r
7�..:_
l_�
:0:
O.
•
• •
H
uracil
continued on the next page
•
+
I
N
NB
O·
•
• •
H
aromatic to the extent that this
resonance form contributes
358
aromatic
OH
r
��
N
BN
B
OH
tautomer­
aromatic
16-43 continued
�)
H '�
·
•
0
..
aroratic
·
•
N
'
H,NB �BN Jl- N NB
...
.
_
__
2
I
I
�N
BN
B
adenine
16-44
(a) Antiaromatic-on ly 4
n
·.
N
\
H
tautomer­
ful l y aromatic
aromatic to the extent that this
resonance form contributes
guamne
:):.
0
N
\
H
B NH
��
HN�
.
H,NB"/�B Jl- NB
·.
N
B
'\\
.
H
ful l y aromatic
N NB
\/
electrons.
(b) This molecule is electronically equi valent to cyc lobutadiene. Cyclobutadiene is unstable and undergoes
a Diels-Alder reaction with another molecule of itself. The t-butyl groups pre vent dimerization by blocking
approach of any other molecule.
(c) Yes , the n itrogen should be basic. The pair of electrons on the nitrogen i s i n an sp 2 orbital and is not
part of the n system.
a
b'
'
...
I
NMR,
2
Analysis of structure I shows the three t-butyl groups in unique environments i n relation to the nitrogen.
We would expect three different s ignals in the
as is observed at _1 1 00 C. Why do signals coalesce as
the temperature is increased? Two of the t-butyl groups become equivalent-which two? Most likely, they
are a and c that become equi valent as they are symmetric around the n itrogen. But they are not equi valent
in structure I-what i s happening here?
What must happen is an equilibration between struc tures I and 2, very slow at -1100 C, but very fast at
room temperature, faster than the NMR can differentiate. So the signal w h ic h h as coalesced is an average of
'
a and a and c and c'. (Thi s type of low temperature NMR experiment is also used to differentiate axi al and
equatorial hydrogens on a c yclohexane. )
The NMR data prove that I i s not aromatic , and that I and 2 are i somers, not resonance forms. I f I were
aromatic , then a and c would have i dentical NMR signals at all temperatures.
359
is loss of methyl.
Mass spectrum: Molecular ion at base peak at
Infrared spectrum: The broad peak at cm-l is OH; thymol must be an alcohol. The peak at cm-l
suggests an aromatic compound.
spectrum: The singlet at is OH; it disappears upon shaking with D20. The 6H doublet at
and the IH multiplet at are an isopropyl group, apparently on the benzene ring. H singlet at
is a methyl group, also on the benzene ring.
Analysis of the aromatic protons suggests the substitution pattern. The three aromatic hydrogens
confinn that there are three substituents. The singlet at is a proton between two substituents (no
neighboring H's). The doublets at and are ortho hydrogens, splitting each other.
1 6-45
ISO;
135, M
-
15,
3 500
NMR
(') 3 .2
�Y
Y
(') 1 .2
(') 4.8
(') 2.3
A3
(5 6 . 75
X
1620
(5 6.5
(5 7. 1
CH]
+
Several isomeric combinations are consistent with the spectra (although the single H giving suggests
that either or is the OH group-an OH on a benzene ring shields hydrogens ortho to it, moving them
upfield). The structure of thymol is:
Z
8 6.5
Z
thymol
¢r:;�
CH3
The final question is how the molecule fragments in the mass spectrometer:
H
H
CH3 'CI
CH3 I
+ OH
OH
+OH
..
I hI hCH3
�
¢r
CH3
CH3
CH3
mlz
resonance stabilization of this benzylic cation includes fonns with
positive charge on three ring carbons and on oxygen (shown)
'c
..
..
..
135
360
+
•
Mass spectrum: Molecular ion at two prominent peaks are
(as we
(loss of methyl) and
shall see, most likely the loss of acetyl, CH3CO).
Infrared spectrum: The two most significant peaks are at cm-1 (conjugated carbonyl) and cm-I
(aromatic C=C).
spectrum: A 3H singlet at is methyl next to a carbonyl, shifted slightly downfield by an
aromatic ring. The other signals are seven aromatic protons. The IH at d is a deshielded proton next to
a carbonyl. Since there is only one, the carbonyl can have only one neighboring hydrogen.
Conclusions:
rnJz including H mass for carbons
C
The fragment C1oH7 is almost certainly a naphthalene. The correct isomer (box) is indicated by the
This
isomer
would would
II
have
two
deshielded
C-CH3
protons in the
1 6 -46
M
1 70 ;
-
15
M
1680
-
43
1600
8 2.7
NMR
8.7
+
1 27
7
=}
1 20
=}
10
NMR.
o
NMR.
Although all carbons in hexahelicene are sp2, the molecule is not flat. Because of the curvature of
system, one end of the molecule has to sit on top of the other end-the carbons and the hydrogens would
bump into each other if they tried to occupy the same plane. In other words, the molecule is the beginmng
of a spiral. An "upward" spiral is the nonsuperimposable mirror image of a "downward" spiral, so the
molecule is chiral and therefore optically active.
The magnitude of the optical rotation is extraordinary: it is one of the largest rotations ever recorded.
In general, alkanes have small rotations and aromatic compounds have large rotations, so it is reasonable to
expect that it is the interaction of plane-polarized light (electromagnetic radiation) with the electrons in the
twisted pi system (which can also be considered as having wave properties) that causes this enormous
rotation.
16-47
the nng
three-dimensional
picture
oftwist
hexahelicene
showing
the
in the system of six rings
361
The key concept in parts (a)-(c) is that an aromatic product is created.
(a)
+
16-48
5.:\
U
o· ----- 0
I
:O-H
+ H+ �
I
:O-H
..
a
c
+ -. · ·
0
..
:O-H
..
..
The
protonated
gives a resonance-stabilized
Protonation of the singly
bonded
oxygen carbonyl
does not generate
a resonance-stabilizedcation.
product.
A
+
6� 6·
I
:O-H
I h.
+ H+
----
:O-H
o
0
c
..
..
I
:O-H
c
·
..
..
The last
resonance
formnonaromatic
shows that theion.cation
produced
is aromatictheandproducts
therefore
more
stable
than
the
corresponding
In
this
second
reaction,
are
more
than in the first reaction which is interpreted as the reactant being more basic than favored
(b)
+0
B
er
(r
••
CI
".......-A
CI- +
C
Ih••
CI
".......-A
O
o
D
H
O
o
CI- +
+
..
..
+
H
..
..
AROMATIC
H
B
O
+0
O
Ih-
A.
H
The
product
from ionizationbutofis alsois stabilized
by resonance.
The stable.
ionizationreaction
product that
of is not
only
resonance-stabilized
aromatic
and
therefore
more
produces astate
moreleading
stable toproduct
will usually
fasterleading
under milder
conditions
transition
that product
will behappen
stabilized,
to a lower
activationbecause
energy.the
C
362
AROMATIC
A
D
(e)
continued
u
16-48
GH
H+
E
H+
d
( )
catalyst
catalyst
F
HO
o
Dehydration
of that
produces
anstable
aromatic
product
is
more
than
the producta more
fromstable product
reaction
that
produces
will usually
happenbecause
faster under
milder
conditions
the product
transition
state
leading
to
that
will be stabilized,
activation
energy. leading to a lower
o
. .1)- .
:0
+
H20
AROMATIC
� base
---
V
phenol
HO
umbelliferone
(Umbelliferone is one of about
coumarins isolated from plants,
primarily the families of
.
·0
:0
Resonance
stabilization
ofcharge
the phenoxide
anion
shows
the
negative
distributed
over the one para and two ortho carbons
.
base
--1000
Resonance
stabilization
of the
anion
ofsame
umbelliferone
gives
not
only
the
three
forms
as
the
phenoxide
anion,
but informaddition,
gives an extra
resonance
with
(
-) charge on a
carbon,
andcontributor,
the most significant
resonance
another
form
charge
on
the
other
with
the
()
carbonyl
oxygen.
This phenoxide
anion is
much
more
stable
than
which weof the
interpret
as material,
enhanced
acidity
starting
umbelliferone.
In fact,while
the phenol
pKa ofis
umbelliferone
is
about
.
.
:0
Angiosperms: Fabaceae,
Asteraceae, Apiaceae, and Rutaceae.)
10.
:0
7.7
363
F
E. A
Humulon
(sometimes
spelledathumulone),
even
though
highly
resonance
stabilized,
cannot be A
aromatic
because
the
carbon
shown
the
bottom
of
the
ring
is
tetrahedral
and
must
be
sp3
hybridized.
ring is aromatic only when all of the atoms in the ring have a p orbital which the sp3 carbon does not.
Areyouyoulovefamiliar
withownthegood?
conceptThe"tough
love",is that
that sometimes
is, sometimesto demonstrate
you have to one
be stem
with the
someone
for
their
concept
emotion,
behavior
has to atoms
appearlike
exactly
the and
opposite.
Granted,
thisconflicting
is a stretcheffects:
to applythey
it tocanatoms,
but theelectron
point
that
sometimes
oxygen
nitrogen
can
have
withdraw
density bydensity
their strong
electronegativity
(an
inductive
effect) but atwill
the same
time,
they can
donate
electron
through
their
resonance
effect.
This
phenomenon
prove
important
in
the
reactivity
of substituted benzenes described in Chapter
:0 :
N
'"
pyridine N-oxide
pyridine I '"
I
16-49
1 6-50
is
yr
4
""":::3
17.
98.60
2
yr�+
• •
4
9 7.25
98.19
""":::3
2
9 7.40
Nitrogen
is electronegative,
so anit exerts
a deshielding
effect onit is harder
in pyridine.
Thewhyeffect
diminishes
with
distance
as
expected
with
inductive
effect,
although
to
explain
H-4
is deshielded
more than
pyridinethatN-oxide,with
an even
more
electronegative
oxygen
attached
to theweN,can
it would
be reasonable
toonIn expect
the
hydrogens
would
be
deshielded.
This
is
the
case
with
H-3;
infer
that
the effect
H-3 is purely an inductive effect.
Evenmust
morereflect
interesting
is thesideshielding
effects personality,
on andthe donation
these chemical
shiftsdensity
are shifted
upfield.
This
the
other
of
oxygen'
of
electron
through
a
resonance
effect.
Drawing
the
resonance
forms
clearly
shows
that
the
electron
density
at
and
(and presumably
by resonance,
perfectinductive
agreementeffect.
with the
results.
As we will seeincreases
in Chapterthroughin this
mostdonation
cases, resonance
effectintrumps
9 7.32
9 7.64
H-2
H-3.
H-2
H-2
H-6)
17,
:0. . :
'"
:0 :
+ 11
:0 :
C
¢en QtH
I
......:::
9
H-4;
--- I
9
.
......:::
9
.
9
+ II
---
(�J(H
I
_C
9
9
H-4
NMR
:0 :
HC�
�¢:H
--- . I
�
9
9
Note:example,
The practice
ofthatapplying
human
emotions
to itinanimate
objects
iselecctrons.
called "anthropomorphizing".
For
we
say
an
atom
is
"happy"
when
has
a
full
octet
of
In casualfor example,
conversation,
this
gets
a
point
across,
but
it
is
not
appropriate
in
rigorous
scientific
terms,
exams.
conditons,
expected
to use the specific terms of science because they
are
well Under
definedmore
and formal
do not permit
sloppywe orarefuzzy
concepts.
As our equipment technician says: "Don't anthropomorphize computers. They hate that."
on
364
The representation of benzene with a circle to represent the system is fine for questions of
nomenclature, properties, isomers, and reactions. For questions of mechanism or
however,
the representation with three alternating double bonds (the Kekule form
picture) is more
CHAPTER 17-REACTIONS OF AROMATIC COMPOUNDS
1t
clarity and consistency, this Solutions Manual will use the Kekule
reactivity,
informative For
exclusively.
1 7- 1
Oc:�
� C I
O
H
H
�
:!=!H2
I
addition
product­
Sigma
H
NOT AROMATIC
complex
While theis addition
of waterThus,to theit hassigma
complex
can be shown
in a reasonable
mechanism,
the
product
not
aromatic.
lost
the
kllmol
kcal/mole)
of
resonance
stabilization
energy. The addition reaction is not favorable energetically, and substitution prevails.
H
1 7 -2
: CI-Cl:
CI
AI-Cl
CI
I
+
H
+
H
H
H
HCI
H
H
+
H
�
.
CI
H
---
+•
OH
(36
.
+•
•
-----..
:CI:
AICI3
t
CI
:CI-AI-C)
:U I
CI
.
/ "
CI
: CI -CI - Al -CI
CI
.
CII :..CI-CI
v·- Al -Cl
CII
.
H
H
H
)
152
�I
H
H
/ ,,+
•
1
1-
CII:Cl-AI-CI
CII
CI
.
..
+
H
H
+
H
H
..
t
H
H
..
H
H
1-
.
365
Cl
C
+'
H
Cl
H
H
1 7-3
Like most heavy metals, thallium is highly toxic and should not be used on breakfast cereal.
�
�====..
1 7-4
+
Tl(OCOCF3h
:0:
CH3
I
gD
:
CF3COO-
I
.. --.. . V H
C H3
CH3
N+
CH3
+
H
----
--
N02
CH3
Benzene's sigma complex has positive charge on three carbons. The sigma complex above shows
positive charge in one resonance form on a 3° carbon, lending greater stabilization to this sigma complex.
The
morewillstable the intermediate, the lower the activation energy required to reach it, and the faster
reaction
2°
3°
2°
the
be.
1 7 -5
delocalization of the positive charge on the ring
delocalization of the negative charge on the sulfonate group
:0:
:0:
:0:
O ==S-Ar
:O - ISO == S- Ar
I
:0:
:0:
: 0:
•• I
•
.
Ar
.......I--J�-
• • II
.
•• •• II
. .
----
.
366
("Ar"
is the general
abbreviation
for"aryl"
an
"aromatic"
or
group,
benzene;abbreviation
is the for
general
angroup
"ali.)phatic" or "alkyl"
in this case,
"R"
1 7-6
(a) The key to electrophilic aromatic substitution lies in the stabi l ity of the sigma complex. When the
electrophile bonds at ortho or para positions of ethylbenzene, the positive charge is shared by the 3D carbon
with the ethyl group. B onding of the electrophile at the meta position lends no particular advantage because
the positive charge in the sigma complex is never adj acent to, and therefore never stabilized by, the eth yl
group.
�I
ortho
CH 2 CH3
�
;r - Fe B r3
�
B q-
{
-- a:
CH 2 CH3
�
�
CH
+
�
CH 2 CH3
I
+C
Q
Br
H
3 D-good
(b) Electrophilic attack on p-xylene gives an intermediate in which only one of the three resonance forms I S
stabi lized by a substituent (see the solution to Problem 1 7 -4). m-Xylene, however, i s stabil i zed in two of its
three resonance forms. A more stable intermedi ate gives a faster reaction .
1 7-7 For ortho and para attack , the positive charge in the sigma complex can be shared by resonance with
the vinyl group. Thi s c annot happen with meta attack because the positive charge i s never adj acent to the
vinyl group. (Ortho attack is show n ; para attack gi ves an intermediate with positive charge on the same
carbon s . )
�
367
&: }
"extra "
resonance form
1 7-8 Attack at only ortho and para positions (not meta) places the positive charge on the carbon with the
ethoxy group, where the ethoxy group can stabi lize the positive charge by resonance donation of a lone
pair of electrons. (Ortho attack is show n ; para attack gives a simi l ar intermediate . )
"extra "
resonance form
1 7 -9
O IthO
c
0
NH
&
2
:
H
B,
�
H
H
H�
�
�
Br 6
NH2 NH2
Q
Q
Br
mCla
�
�
�
H
Br
�
�
H
17-10
Br
6
o
+
�
Br6
�
H
"extra "
resonance form
H
OCH(CH3 h
+
NH2
'
B
Q
H
+
C
�
Br
-H
"extra "
resonance form
H H
+ 1
H
V
�
H
H
para
+ C1
B'
?'
'
B
�
ll)
+
NH2
8"
Br2 (X'
___
"
Br
+
¢
Br
H
H
HB r (g)
Br
S ubsti tution generates HBr w hereas the addition does not. If the reaction is performed i n an organic
solvent, bubbles of HB r can be observed, and HB r gas escaping into moist air will generate a c loud. If the
reaction is performed in water, then adding moi st li tmus paper to test for acid w i l l differentiate the results of
the two compounds.
368
1 7- 1 1
(a) Nitration i s performed with n itric aci d and a sulfuric acid catalyst. In strong acid, amines in general,
including ani line, are protonated.
+
o
(b) The NH 2 group i s a strongly activating ortho,para-director. In acid, however, i t exists as the protonated
ammonium ion-a strongly deactivating meta-director. The strongly acidic nitrating mixture itself forces
the reaction to be slower.
(c) The acetyl group removes some of the electron density from the nitrogen, making it much less basic;
the nitrogen of this amide is not protonated under the reaction conditions . The N retains enough electron
density to share with the benzene ring, so the NHCOCH3 group is sti l l an activating ortho,para-director,
though weaker than NH2 .
:0:
:0:
+ I
II
..
�
Ph
Ph
C - CH3
- N = C - CH3
• •
I
H
H
1 7- 1 2 Ni tronium ion attack at the ortho and para positions places positive charge on the carbon adjacent to
the bromine, allowing resonance stabilization by an unshared electron pai r from the bromine. Meta attack
does not gi ve a stabilized intermedi ate.
ortho
+ Br:
Br
:Br :
Br
I
N02
N02
NO l
02
+C
a
::::-...
H
CH
+
..
..
a
HC
+
h
H ..
t)
..
..
::::-...
::::-...
G
..
Q .. Q .. ..
meta
Br
Br
H
N 02
N 02
H
para
:Br :
I
+C
Br
H
N0 2
..
H
..
H
•
..
(J
H
N0 2
..
+ Br :
Q
H .
e
Br
(
C
I
h
H
"extra "
resonance form
N0 2
H H
.. ..
H
N 02
"extra "
resonance form
369
•
..
QH
Br
H
N0 2
Cl
H
(b) O; H Ocd a �:- a� : } : �)� d !r
H
H
1 H
(a) O= !r
17- 13
atom can stabilize positive charge by sharing a pair of electrons. Bromine can this
(c ) bromine
cationic
species, whether from electrophilic addition or electrophilic aromatic substi
tution.
resonance-stabilized
do
A
(a)
1 7- 1 4
(b)
<iCOOHCl
�
(c)
(e)
N02
+
Br
(x N02
�
OH
~CH3
�
CH3
CH3
(r
'Q
:::::-.. Cl trace Cl
COOHBr
02 N
+
O2N
+
02 N
N 02
(a)
1 7- 1 5
h ��"g
VCH3
+
I
:::::-..
O
OH
'(\
:::::-.. CH3
I
+
N 02
�
OH
(x
trace
�
o,p-d'''''"'
weak o,p-director
370
CH3
N 02
(d)
COOH
~OCH3
�
N02
in any
1 7- 1 5 continued
(b)
�
Y
<r
02N
+
N0 2
N02
(c)
OH
n
� I
+
N 02
V N02
trace
strong m-director
NO'
Jy
(d)
OCH3
q'I �
I I
N0 2
trace
C - CH3
0
NH -
strong o,p-director
CH3
o
strong o,p-dlrector
V N 02
N0 2
-
(f)
N02
strong o,p-director
C l weak o,p-director
(e)
Jy
Cl weak o,p-director
Cl
Cl
+
weak o,p-director
h
° 2N
•
strong m-director
•
CH3 - C - NH
strong o,p-director
'-----y---'
acti vating
1/
I
\\
\
-
"
0
C - NH 2
�
strong m-director
deacti vating
1 7- 16
(a) Sigma complex of ortho attack-the phenyl substituent stabilizes positive charge by resonance :
H
C
+
I
I
E
E
H
____
H
____
Para attack gives similar stabilization. Meta attack does not permit del ocalization of the positive charge
on the phenyl substituent.
371
H
1 7 - 1 6 continued
(b)
< ) < }- N02
(i)
(ii)
02 N
\>
trace
OH
�)
+
+
\
(v)
f
N02
\ }- N02
� N02 \
> °2N)
>
02
Q-o-
+
N02
OH
OH
+
N
o
(i v)
�>
\> \
02N -{ > < }- C -
(iii )
\>
°2 N
CH3
+
� II
� II
°I I
C - CH3
°2 N
\
f0 � >
2
N
minor
maj or (mi nor amounts of ni tration on the outer rings)
�H
17-17
(a)
f
�
��):
+
Al C I4-
AlCl3
-:?
:¢�
..
+
Cc�-n
•
372
..
}
1 7- 1 7 continued
+
•
•
CH3 - Cl - AlCI3
v·
B
C1I3
l: a:]
��
..;...-.
..,
�
.
.
....
H
para I somer
also formed by
simi l ar mechanism
CH3 :ci : �
AICl3
I
CH 3 - C - CHCH 3
•
I
CH3
I
(c )
OCH3
5�H
3
I�
CH 3 - C - CHCH 3
I + 2°
CH
..
: OCH3
I
+c
CH3
A lCI 4-
CH3
. ·�
....
..----l
O
1
..
H
5
..
•
I
CH]
�
methyl
3° CH3
+
I
shift
---l.� CH3 - C - CHCH3
I
CH 3
CH3
+
I
CH3 - C - CHCH3
I
CH3
+
:CI :
\ ��
�I
CH(CH3h
_
/
CH3
O n l y a small amount of the ortho isomer might be produced as steric
interactions will discourage thi s approach path of the electrophi le.
CH3
1 7- 1 8
(a J
(b)
ex :
C H3
CH3 - � - OH
I
CH3
Ct o
o
H
+
HF
+
H
--
C.
+ H
BF3 �
373
-c
u
CH
CH 3
I
CH3
If
�
-
H
1 7 - 1 8 continued
(c)
(d) HO - CHCH3
I
+
+
B F3
CHCH3
--
I
CH3
CH3
1 7 - 1 9 In (a), (b), and (d), the electrophi le has rearranged.
(a)
-c
U
CH
CH 3
I
CH3
(b)
(c)
(d)
(e)
r; �
_
~
0
0
(excess)
6
C(CH3h
(CH3hCCI
•
AICI3
(excess)
(b)
plus over-alkylation products
gives desired product
gi ves desired product, plus ortho i somer; use excess bromobenzene to avoid overalkylation
gives desired product, plus ortho i somer
gi ves desired product
1 7-2 1
(a)
(c) No reaction : nitrobenzene is
too deactivated for the Fri edel
Crafts reaction to succeed.
(b)
CH3I
•
AICI3
6
C(CH3 h
HN03
•
H 2SO 4
S03
•
H 2SO 4
¢
S03H
374
¢
(separate from ortho, although
steric hindrance would prevent
much ortho substitution)
N0 2
(separate from ortho)
1 7-2 1 continued
(c )
0
•
AlCl3
(excess)
°
1 7-22
(a)
(b)
(e )
(d)
0
+
� I
0
0
Q
o
?'
� I
+
(f)
0
?'
� I
+
II
Cl 2
°
I I
I I
Cl - C
...
II
Cl - C - C(CH3h
•
I I
1 ) AlCl3
•
2) H2 0
H3C0
1 ) AlCl3
-o-
•
2) H20
II
C l - C - C H 2CH2 CH3
C - C(CH3h
2) H2 0
� II
AlCliCuCl
°
� I
..
2) H2O
1 ) AlCl3
-0
CO, HCl
°
1 ) AlCl3
C - CH 2CH(CH)),
II
I I
+
a
°
�
o °
a c 'Q
Cl
C l - C - C(CH3h
0
¢°
(separate from artha)
•
AlCl3
Cl - C - CH 2 CH(CH3h
+
OCH3
(e)
2)
CH3I
CHO
°
a
?'
II
C - C(CH 3 ) 3
� I
°c
a
Zn(Hg) ' HCl
•
II
..
I ) AICl)
2) H 2 0
�
I
375
- CH,Cll,CH)
•
Zn(Hg), HCl
a
� I
CH2c(C H)))
a
� I
CIl 2CH 2CH2CH3
-:� N:
N02
1 7 - 23 Another way of asking this question i s thi s: Why is fluoride ion a good leaving group from A but not
from B (either by S N I or S N2) ?
�� :NU
A
C
B
Formation of the anionic sigma complex A is the rate-determining (slow) step in nucleophilic aromatic
substitution . The loss of fluoride ion occurs in a subsequent fast step where the nature of the leaving group
does not affect the overall reaction rate . In the SN 1 or SN2 mechanisms, however, the carbon-fluorine bond
is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the
rate.
t
t
;>,
�
<!)
f=:
<!)
nucleophilic aromatic
substitution
SN I
SN2
1 7 -24
Cl
H
376
.
.
'
'
0
CH
:
Cl
OCH3
:
]
X;
I
�
;R
N
N
(
.0.... �
: �' � y
y
N02
0.... N+ O0.... N+ 0
O
C
H
R U
02N
-o:N �
1 �C.
y
N02
0...... ..N+ 0
CH
]
�
:
�
�2N
17-25
•
..
_ •
�
..
'¢
I
�I
•
.....
....
....
3
{:
-Cl-
.....
(b)
Cl
O
//C;,-
..
Cl
CH3
...H
H
H
_
CH3
I
.
I
C_
I
.
....
+
• • -
.....
oell,
I
...
_
�
Cl
...
�
- . •
.... + .....
-
A4�:9.n�,,)l�
c�o
CH3
CH3
..
...
OCH3
H
C
4
.....
0N Cl OCH3
-0 ....
DN
:0.... 0:
R t
-o:N 1�
y
0: N 0..:
+/1
I
�
CH3
-o o
�
CH3
'L \CH{JLf� coj �
377
H
o
H
17-25 continued
(c) Br
I
Br NH2CH3
HC.
�
.....
CH3NH2
.
..
N02
°2N
I
..
- ........ N .......
:0 + 0 :
-
(H-�HCH3
CH3NH2
-Br•
• •
_
-
N02
..
-.. ; N
:0
..
+
NH2NH2
..
N02
-Cl-
0
+11
N
- 0"
----..-
_
R
..
"N,
_
0 "+ 0
378
I
•
o
II
+N
-0"
Cl
+
C­ I
..
I
N, ..
�
.0 + 0:
..
N,
0"
..... + 0
_
15NH2NH2
...... N .......
0"""+ 0
+
NH 2NH2
:0:
N,
"+ 0
0"
N HNH2
N02
...... N .......
· 0""" + 0 :
+
NH2NH2
N02
..
+
NH2CH3
C
"'+
..
02 N
..
..
NHCH3
CI
O
Br
+
...... N .......
0 /+ 0
N02
(d)
Br NH2CH3
I .. I
..
..
+
+N
-0"
..
Cl
l
-..", N
:0
..
-
+
NH2NH2
_
....... ..
+ 0:
..
17-26 Assume an acidicClworkup to each of theseCl reactions to produceOHthe phenol, not the phenoxidc ion.
(a) Clz
from additionHN03
NaO
It,
1 elimina�i o n
mechani
sim;somer
0 AICl3 6 HZS04 NOz
onl
y
t
h
i
s
NOz
ort
h
o
(b)
Cl
OH
OH
Bf
3 Brz B'
Clz
NaOH
0 AlCl3 6 3500 C 6
(c)
CI
Cl
Cl Br
Clz
NaOH
0 AlCl3 Cl
Q
OH
vi
a
benzyne
mechani
s
m
OH
ortho
(d)
Clz
NaOH
AlCI3
OH
vi
a
benzyne
mechani
s
m
Cl
OH
+
ort
h
o
CH2CH2CH'
(e)
o 0= CCHzCHzCH3
CH3CHzCHz-C-C� 61 Zn(Hgl
HCl
O AICl3
I Clz
t AI Cl3
CHzCHzCHzCH3 CHzCHzCHzCH3
CHzCHzCHzCH3
Jr
NaOH
+ ortho
V OH
via benzyne mechanism OH
Cl
'7
..
..
¢�
¢
1'1.
+
..
..
2
¢
..
6 ¢
I�
1'1.
..
+
..
1'1.
II
'7
+
�
¢I
�
..
..
q
I�
�
¢
& ¢
6
+
+
�
..
379
1'1.
¢
17-27
from chlorobenzene
17-28
(a) First, theocarboxylic acid proton is neutralized.
C"'OH + NH3
+
NH/
0r
1 �
V
+
NaOH
+
heat
"
--
+ Na+
Q:HOC
IH Li/.
the negatavoiiveds
charge
(b)
�c
r?'
U
+
..
H ·
HH
HpluI s + Li+
resonance
forms
H
plus
resonance
forms
G"
I�
380
II
h
H
o
o
n
� y OCH3 H-O� r(_I(�
t-B� � OCH3
------.. lL�) Li7
tL.)
/
C
�
the carbon
beari
eldonatectrnon­inggthe
methoxy
group
-NH3ry0C"' O
.... C
plus other resonance forms ( piresonance
us
Na. � forms
-HCI
H
H
I .. I
resonance
forms
plus
resonance
forms
17-29
(b)
t
i
o
n
of
chl
o
ri
n
e
morepressure
C12, CI Xtl addi
.. V' ttmiohetxhriteunreg,systofgivsteienmreoig ofa somers
first product
benzyl
ic substformed
itutionfrom CI
(f)
(c )
�,
1t
CI
CI
COOH
17-30 COOH
cr
COOH
(b)
)
c
(
(a) V
I
I
COOH
COOH
17-31
CI-Cl hv 2 CI. initiation
CH3
· C-C1
cis + trans
h-
r-,,,
..
¢
I
CI·
CH3
CI-C-C1
I
+
6
II
h-
h-
6
propagat
i
o
n
step
1
C.
OH
o
CI-CI
stpropagat
ep i o n
r\r
..
2
381
CH3
·C-C1
I
6HC
t
I
6
CH3
C-CI
..-..-
NH2
17-32
Br-Br hv 2 Br- initiation
CH3
CH3
· C-H
�-�-H
r'\ r\
Br-(A
Vpropagat
step 1 ion
+
CH3
C-H
�
�
�
I
I
6
CH3
Br-C-H
6
6HC
t
CH3
·C-H
I
I
Br-Br
BrI propagation
step
17-33 As must
statisbeticalcorrect
mixtuered forwoulthed numbers
give 2 : 3oforeach40%ty: pe60%of hydrogen.
to . To calculate the relative reactivities, the
percent
44%
56%
28
rel
a
t
i
v
e
react
i
v
i
t
y
2H
3H 14.7 relative reactivity
The reactivity of to is 14.287 1.9 to 1
17-34tioReplns, butacement
ofonalatiparomat
hatic hydrogens
wiis tunfavorabl
h bromine ecanbecause
be doneof under
freehigradih energy
cal substofittuhteioaryln
condi
react
i
i
c
carbons
t
h
e
very
radical. Benzyl�H3ic substitution is usual�Br3l y the only product observed.
Br Br
cc)
B
r
�
(a) (1) Br6
- C-CBr3 (b)(l)
-C-CH3 (2) Br6
�
I
Br ic
:--.."'.. I
....:--".. I
alhydrogens
l fourBrbenzylreplaced
I excess Br2
• hv, time
+
a:
=
a
�
'/'"
�
nr
..
2
6
�
a
�:
"'-':
=
(2)
h
382
��,
� Br
Br Br Br
17-35
n
CH2-Br
,
t
�
5
6
6
�
6
6�o
HC
H
+
+
-
-
CH3CH20-CH2
�
H ! CH3CH2-A:
CH3CH - -CH2
6 CH3CH2-0:
I
�
I
..
2 R)
+1
6
�
I
17-36
ic cations are stabipropane.
lized by resonance and are much more stable than regular alkyl cations. The
(a) Benzylis l-bromo-I-phenyl
product
H
H
(b) HC [Hl
HC-CHCH3
CH3
,.,
H-Br
moee stable than
I
benzylic:B. r:
Br H
HC-CHCH3
I-bromo-I-phenylpropane
(]
..
V
• •
1
6
6
+
�
1
.
-
..
1
1
+
6CH
2°
2°
resonancestabilized
383
a) The combi
nateintonatofion.HEr(Recal
withl athfreeradiecveral inspeciitiatoesr generat
es tbromi
nekeneradidetcalesrmiandnesleadsorietnto antatioi­n.) The
(Markovni
k
ovori
at
what
adds
o
an
al
product
wi
l
be
2b
romo-l-phenyl
p
ropane.
(b) Assume the free-radical initiator is a peroxide.
R�rR 'l RO·
RO· H r'Br
ROH Br
Br
HCV
HCH3
C: Br
more stable than
I
17-37
first
6
+
�
2
---
.
+
---
-
6C
·
I
..
2°
2-bromo-l­
(Br nrecycl
phenylpropane
chai
mechanies isnm) Br·
Br
gBr
M
CH2CH20H
(a) HC=CH2
3
CHCH3
H3
Mg
H+
.
L\
--6� ether 6 Br
OCH3
C�2CH3
CH3CH2Br
CH30H
hv
9OCH3 AICl3 �OCH3 ortho Br2NBSor OCH3
OCH1
B'
CN
HN03
NaCN
H2S04
Br2NBhvSor
N02
N02
ortN02ho
•
17-38
I
(b)
C( )
•
6
6ClI
¢
+
•
¢
6C
3
3
¢1
I
H
¢]C
I
,
--­
I
•
+
•
¢
+
384
I
¢
1'1
•
•
0
enzyme
H-OH HO-H
HO-OH catalysts
0
OH
17-40
0
OH
CH] (c) OH Br '(\
(a) OCH2CH3 (b)
�I
�
� I CH3
� I CH3
Q
CH3
CH3
Br
(e) 0
OH
(d)Br OH Br
(CHJhC
� I CBr3
�
CH3
CH
3
¢t
Br
C(CH3h
17-41
(a ) o
(b)
(c) o
17-39 OH
¢
+
..
&
II
*
�
¢
cQ
17-42 OH
o
I
I
�
*
+
(f)
�
,
+
+
°
o
OH
'(y
o
Br
Br2 Br h Br Br2 Br
6
yBr
Br Br
C6H30Br3
C6H20Br4
17-43 Please refer to solution 1-20, page 12 of this Solutions Manual.
17-44
(a) C(CH3h (b) CH H2CHJ (c) C(CH3h (d) Br (e) C(CH3h
6
3
-
3C
o
---
6
rearrangement rearrangement
385
6 6
(08H
rearrangement
-45 Product
i17sopropyl
group.s from substitution at the ortho position wil be minor because of the steric bulk of the
(a)CH Br CHl ( CH3h ( CH3)2 (d) COOH ( CH3h
Q;
Q
O
Br
17-46
I
(a)
71
�
0
Cl
�
AlCl3
°
6
S03H
Q0,.C" CH3
..
o
t:.
...
I
...
°0
...
...
CH(CH3h
Br
NBS �
Zn(H"g) � -�1
~ HCl � 1
CH30H
°
....
Br
from (a)
~
OH
(c)
CI
1) BH3• THF
CH2=CH-CH2 �I
0 AICI3 U 2) H202, HO- �
Br
OH
�
Br2' hv I �
Mg U H,o+ I
or NBS �
OH
(d)
CI
Cl2
NaOH
NaOH
0 AlCI3 6 350°C 6 CH3CH2Br
Cl
Cl
(e)
Cl Cl
HN03
Cl2
Cl2 �I
0 AlCl3 6 SO N0 AICI3 N0
2
2
----
('QCHlh
..
°
...
----
�
...
H ¢
2
62CH3
¢r
..
..
4
..
386
17 -4t6hreecontmetinuedhods
NaBH,
�
Hg(OAch
..
..
Ih
Ih
OR CI
H2O
OH
�
V Mg I
�
---;;° CH2MgBr
..
+
3
H
0
h
et
h
er
V
�
CH3-CH
I : from (c)
from (d)
O
COOH
COOH
(g)
Fe or Sn or Zn
n04
RN03
HCl(aq)
H2SO4
N02
N02
NH2 OH
(h) two possible methods B r
M g :r
new
bond
Mg "
Br2 hv
H
O
2
i
n
bol
d
..
�
or NBS I ether
°
OH
OR
Br
MgBr
Mg
B
r
2
_
-l
_
7,:, I AlCl3 6 ether 6+�
0
(0
from (c)
..
2) 89 ¢
8
2)
KM
..
f1
..
..
'
..
--
�
¢
�
..
major
9Br
major
COOH
COOH
RN03
H2SO4
N0
2
~
Br
Br
¢
..
387
�
17 -46 continued
Zn(
H
g
)
KOH �CI
(I) 0 FeCi;
CI2
-HCI
6 6 AICI3
17-47 (a
J)
0
N
1
;;;
(d) noacylreaacttioniodoesn: Frnotiedeloccur-Crafonts
ri
n
gs
wi
t
h
st
r
o
ng
deact
i
v
at
i
n
g
US03
H
yN02
gr
o
ups
l
i
k
e
N0
2
C(CH3h
OCH3 Br (gl CI (h)¢CH3
( l Br *
(e) CH30q0C
�
'CHl
or
YN02 NH2
CH
Br
CH
Br
2
CH3
2
S03H (koCH] (I) COOH
(i) Ph-C-� -o-� oCCH2CH3 U) �
aCOOH
H�
N02
0 3
(m) HN-C-CH
CH2CH3
� stronger o,p-director than CH3
�
..
.
CH3
CH3 Co
17-48 Major products are shown. Other isomers are possible.
�
I
o
OH
CI
¢
¢H
.
--
( bl
IJ,.
�I
I
--
(el
�
_
OH
OH
I
� I
basic workup
to Isolate
free amine
�I
�I
II
II
as (same
H
388
E)
17-50
product
mollosseofcular weilgohtss of132H20
stmolarteinculg matar weierigahtl 150
IR spectrum: The dominant peak is the carbonyl at 1710 cm-I. No COOH stretch.
NMR
pofuadjacent
l. In themethyl
regioneofnes, 2.CH6-3.2CH2, 2th.ere
ospectsignalricum:regis eachTheon wifromsplthitinint7.eggrat3is-7.compl
i8onhasvaliciunateteeofgratd but2H;ionththofeesein4H,temustgratso itobehne tirihsnehelgtwmust
areThetwaromat
be
di
s
ubst
i
t
u
t
e
d.
our ofngthae diaromat
itcusitegdnalbenzene.
s talAl, insodiicnatdiincatg eC-H;
two are ishort
, with noandattatched
NMR:alsoFshowi
Carbon
hydrogens,
s
ubst
i
d
are
carbons
n
a
carbonyl
wo
meth�
ylenes.
I�S\
o
(F-� CH2CH2 - C -8
or cO
. , , �
molecular weight 132
lose 1 H retain
The product must be the cyclized ketone, formed in an intramolecular Friedel-Crafts acylation.
17-51
(a)
< } CH2CH2COOH
18
=
=
8
8
___
are
c:=:::>
"w.��
H
A
H
+
(jl .H
�
Br H
H
389
..
17(b)-51Assume
continuedthe free-radical initiator is a peroxide.
R�rdR r'J ROinitiation
RO H - Br --- ROH Br- }
�
� lC. BrH .. -.- (9,
� �C BrH
� I H )-Br· ro
H
H
H propagat
i
o
n
step 1
-
r
C'
---
+
2
+
0
---
I
..
Brpropagat
i
o
n
step
17-5�
2 H CI
HO:.. m
�, �
a
subst
i
t
u
t
e
d
benzyne
Attack A
H
HOm
C�n
Ho
m
..HO:..- 2CO
H
0
I
l!
�,
�
�I
product A
Attack B
OH
2
+
+
�
+
____
/'"
'7
H�
product B
lJlJ
----
390
17-53 The electron-w ithdrawing carbonyl group stabilizes the adjacent negative charge. resonance
plus
forms
:0:
H-OEt dJ
��::o:
�
- ... C
��\
OO
\ �
Na+ plus other resonance forms
Na'
0
•
•
+
U
H
'
HcO .
H
(e)
(b) �
17-55
(f)
�
h-
H
:�
�A)
U
H .
(d)
plresonance
us
forms
+
Na·f.
�
H3h
�
�
�
r
H .
°
4
(a)17-5�
°
H
�
< }-Q
C + HS04- H30+
H2S04
0
H
2
6
conjugat
colorless
yellow ed­
ratnatedessulthfeuriwatc acier dto"dehydrat
ees"reverse
the alcohol
, iproduci
ng addi
a hignhlg ymoreconjugat
er,d,however,
colored carbocati
otn,oo
andConcent
prot
o
prevent
t
h
react
o
n.
Upon
wat
e
t
h
ere
are
many water molecules for the acid to protonate, and triphenylmethanol is regenerated.
()
-y
o� 0H
6
+
2
+
---
391
2
+
17-56
o
17-(a) 57
bromination at C-2
Cl is this
(why
tishomer?)
e major
Br
Cl is this
(why
tishomer?)
e major
(
�
H
-i
�'b
-zfH
- HH� i�
""'C�
�<H�
H
�
o
three resonance forms
:Br: +
,- 0 y Br
�H
•
bromination at C-3
H
C:O
�
l;
YCl
H
/
��C Br
H
• _
_
�:�!: j
(,.0yH
� Br
O(H
Honly two resonance forms
Br
opposedy. to only two
at stgiabiveslizaninginattetarmedi
bilizednatbyiotnhreeat C-2resonance
(b) Attackforms
resonance
ck ataC-3.te staBromi
wil occurforms,moreas readil
C-2
392
17-58
abbreviate h phenylalanine as
·0 · F
F
NH2
I
P CH2 - CHCOOH
NH2
I
R
I
II
+
- ../N
:0
- ..;N,..
C
:0
02 N
...
+N
-0/
.
O-/
�
NH2R
�
1. . C
R �
•
...
N, _
0/
/+ a
N02
+
- /
0
-/H
..
..
� 'O-
�
+
I
o-:/�'o-
H2 R
I
..
..
- ../N,.. ;
:0 + O.
I
HN-CHCOOH
N H2 R
N02
° N
2
...
N02
393
1(j
- "'0
CH2Ph
° N
2
:0:.
.
•
•
I
1
.
.
C_
I
+
H2 R
I
/ N,
:0 /+ O·
.
• •
-
)l
17-59
H3C
·. 0·. �
CH3
�
H-C�
i
+ H
H3C
CH3
OH
:Cl :
�
H
>b�: } � 1
H
OH
OH
+OH
CH3 Ho i CH3 Ho i CH3
CH3
CH3
CH3
OH
OH
I
HO XCH,
CH3
)l
/
·. 0
l
I
H
OH
OH
I
I
��[J ["1
CH,
C-CH3
C-CH3
I
CH3
CH3
L (�H3
�
'!?�
-�
H 't
r
Ho i CH3
CH3
..
..
I
:CH
"
..
plhuers
otresonance
forms
OH
OH
Ho
CH3
-O-FQOH
CH3
bisphenol A
� C-CH3
H20: �
�:� CI H3
.. .. HO
394
plus other resonance forms
17-a) 6Thi0 s is an example of kinetic versus thermodynamic control of a reaction. At low temperature, the
(kinet
ic product
predomimatneatlyes:equalin trathisecase,
alC.mostAta10001 : 1 C,mihowever,
xture of ortenough
ho andenergy
para. These
twdoedisfomers
must
be formed
at approxi
s
at
00
i
s
provi
o
r
t
h
e
to occur rapidly; the large excess of the para isomer indicates the para is more stable, even
products from
e 00iConreact
rat(b)ioTheof product
as thethreact
whiciohnwaswil runequiinliitbiratal ye asat i1000t is warmed,
C. and at 1000 C wil produce the same
17-6 1
0
CH3CCH2CH3 H+
Br2
Mg
ether
FeBr3
Br
MgBr
CH3
-C
-CH2CH3
(possible alternative syntheses include acylation followed by Grignard) OH
desulfonation
though it is formed initially at the same rate as the ortho.
6 Q Q
II
•
--
---
17-62
--
I
from
chi
oro­
benzene
reactive sites NaOH at 350°C
As wetiosawn, leaviin Chapt
eris16,olattheed carbons
ofrinthgse oncenttheer ends.
ring ofBenzyne
anthraceneis suchare suscept
iivbeleditoenophi
electrlophie thatlicthe
addi
n
g
t
w
o
benzene
a
react
reluctant anthracene is forced into a Diels-Alder reaction.
17-(a 63 Cl
0- Na+
J
CI NaOH
CI CICH2COO- Na+ ;Y��H
Cl �
CI �
CI
�
CI
Cl
2,4Cl,5-T
cl a, 0 :CC
Cl
�
2 Cl- CI � 0 � CI
TCDD
Two
c substd folitutloiownsthform
a newaddisix-membered
(Thoughsnotm.)
shownnuclhere,eophithilsicreactaromationiwoul
e standard
tion-eliminriatniog.n mechani
+
+
-<r
11
•
-<r
•
+
395
w.
(This is the
compound used to
pois on Ukrainian
political leader,
Boris Yushchenko.)
17(c)-63Tocontmininimuedize fOnTIation of TCDD during synthesis: 1) keep the solutions dilute; 2) avoid high
taddemperat
ure; 3)ofrepltheahalce ochlacetoroacet
an excess
ate. ate with a more reactive molecule like bromoacetate or iodoacetate; 4)
ToT. The
separat2,4e,5TCDD
from
2,v4e,5in-Tanataqueous
the end ofsoltuhteiosyntn ofhaesiweak
s, takebaseadvant
aNaHC03.
ge of the aciThedicTCDD
propertwiiels ofremai2,4,n5T
wi
l
di
s
sol
l
i
k
e
e andpitcanatedbefromfilteaqueous
red or extsolractuteiodninbytoaddian organi
caninsolbeublpreci
ng acicdsol. vent like ether or dichloromethane. The 2,4,S-T
17-64
a( ) �:O:�H+ �:OH
�O
�O
0
plfOnTISus 3 wiresonance
ththposie tive
charge
on
benzene ring
OH
OH
OH
plfOnTISus resonance
wie charge
th
posi
t
i
v
onon tthhee oxygen
ring and
: OH
OH
('i�
�O:
resonance
fOnTIS
onplthuebotsoxygen
h benzene
riphenol
ngs, ,
OH
of
t
h
e
HO
OH and the oxygen in the ring
+
.. =-'\
----.
-
�
o
¢
o
---.J...-J
o
o
th positive charge
onplutshresonance
e ring and fOnTIS
on thewioxygen
o
o
396
..
red dianion
(c)
17-65
°
q
CH
CH3
3
Br2 Br I� Br H2O Br I� Br
AICl3
O
S03H
para posiS03Htion blocked
17-66 A benzyne must have been generated from the Grignard reagent.
((I Br � �S MgBr
0
+ 0
1
1
F
�
6
Hfumi2SO4ng
..
¢
..
A
..
+
�
.
\--�----- �----�)
V
°
397
t
17-67 The intermediate anio�
n forcesHtheOHloss of hydroxide. ax
H OH
CH3
For simplicity, abbreviate I NHCH3 >1 R
OHLi
+
H
H
H
H OH
/OH
s:,
�c c2R
c
.
R
ffR
..
..-;
a
___
I
I
.o
L/ H 'C:Y
HO
H"S:
H �
plus
plresonance
us other
�
resonance
forms
forms tI Li·
I
.0
.0
\
•
r
:·
+
I
..
---.-
H H
� ,C CH3
y
�
NHCH3
H
.
I
�R
.0
J'-R
D
u�
H
'"cH3CH,o - H-
I
plus
resonance
forms
17-68 OH
OH equiv.
OH
lcqui
v
.
NaOH CH3I_ � Cl-C(CH3h � C(CH3h
BHA, butyl.ted hydroxy.niso\c
I
AlCl3
YOCH3
OH
OCH3
hydroquinone
plus other isomer
OH
OH 2equiv.
OH
CH3I A Cl- C(CH3: (H3C)3C : c(CH3h
BIrr, butyl.ted hydroxytoluenc
I
6 AICI; Y AlCl3
¢
¢1
•
.0
CH3
•
V
CH3
398
ve the probl
i17-69
denticalSolmechani
sm.) em by writing the mechanism. (See the solution to problem 17-2S(a) for an
Br �
l.5(OH Br
�
I
I
I
attack C- ;
HCHydroxi
d
e
at
t
a
ck
on
C-l
put
s
t
h
e
negat
ive charge
ongroups,
carbonssoththatesedo
notaniohave
t
h
e
ns are not stabilized.
02 N
J):I
�
�B
r
2
"
Br
•
•
-
HO:
•
•
02N
N02
..
N02
N02
Br
OH
a
only product formed
02 N
N02
..
Hydroxi
dcharge
e attackononcarbons
C-2 putwis tthhe
negat
i
ve
niatrbiolgroups,
stnegat
iizvate charge.
ion bythereby
delThiocalsiniincreasi
ztienrmedi
g tnhge athtee
is formed preferential y.
399
also stabilized by resonance onto
the nitro group
J3t:Cr -�H
Br
�
...... N02
02N
also stabilized by resonance onto
the nitro group
18-1(a) 5-hydroxyhexan-3-one; ethylj3-hydroxypropyl ketone
((bc)) t3-rapns-2-met
henyl buthaoxycycl
nal;j3-pohenyl
butyraldehyde
hexanecarbal
(or nameif you named this enantiomer); no common name
(d) 6,6-dimethylcyclohexa-2,4-dienone;dehyde
no common
18-2
O
5
el
e
ment
s
of
unsat
u
rat
i
o
n
(a) C9Hl
O
IH5H doubl
e
t
(very
smal
l
coupl
i
n
g
const
a
nt
)
at
9.
7
al
d
ehyde
hydrogen,
next
t
o
CH
mUl
t
i
p
l
e
peaks
at
7.
2
-7.
4
monosubst
i
t
u
t
e
d
benzene
IH multiplet at 3.6 and 3H doublet at 1.4 CHCH3
-CH
�
-CH<�
CH3
Thelookssplliikteinagquart
of theet duehydrogen
onsplcarbon-2,
nextthe tadjacent
o the aldCH3.
ehyde,Aisclwortoserhexami
examininatniog.n ofIn tihtse overal
lsho\NS that
t
o
t
h
e
i
t
i
n
g
from
peaks
each
of the quartandet tishesplmetit hinyltohydrogens
two peaks:arethnotis isequiduevtaloetnt,he splso iitt iins gtofrombe expect
the aleddehyde
hydrogen.
The
alconstdehydepeak
hydrogen
t
h
at
t
h
e
coupl
i
n
g
ainntgsconst
wil notantsbe, thequal
. Ifbea hydrogen
is separat
coupledeltyo, justdifferent
neiwoul
ghboring
hydrogens
byitdiinfferent
coupl
ey
must
consi
d
ered
as
you
d
by
drawi
n
g
a
spl
g tree for
each type of adjacent hydrogen.
5
el
e
ment
s
of
un
sat
u
rat
i
o
n
(b) clCgHgO
upeaksteatr of 1974 peaks atcarbonyl
128-145carbon (tmono-or
paras
ubst
i
t
u
t
e
d
benzene
ring
h
e
smal
l
peak
hei
g
ht
suggests
a
ket
o
ne
rat
h
er
t
h
an
an
al
d
ehyde)
peak at 26 methyl next to carbonyl or benzene
< }- C -CH3 or CH3 -o- C- H
more likely
also possible
3 A compound
has to haveto occur.
a hydrogen
or other atom) in order for the
y rearrangement
Butan-2-onoaney carbon
hasy posino (y-hydrogen.
M18-cLaffert
tion-no carbon
CH3-C -CH2 -CH3 t
CHAPTERl�KETONESANDALDEHYDES
(R,R)
=:}
0
0
+
0
=:}
0
=:}
o
II
+
I
=:}
sh:lpe,
=:}
0
0
=:}
=:}
0
=:}
o
o
o
II
a
a
J3
401
it
18-4
]
; Lr
1j :+R:C-CH3 :I�C-CH3
�H2CH2CH2CH2CH2CH3
CH3 It
85
rnIz
43
rnIz
�+CH2CH2CH2CH2CH2CH3
rnIz 85
[
43
t
---
---
128
McLafferty rearrangement
The first value is the to the second value is the n to The values are approximate.
(a) 200 nm; 280 nm; this simple ketone should have values similar to acetone
d systnmembase(210)valpluue)s 2plalusk30yl groups
valueeofbond310 nm310 nm
si(b)mi230lar tonm;Fig310ure nm;18-7:conjugat
ketonee(280
nm for (20)the conj230;ugattehdedoubl
360 nm;ngconjugat
e bondbase(30)valpluuse 4ofal290kyl groups (40)
(280;c) 280simnm;ilar reasoni
for the otedhersysttreansim (210)
tion, stplaurtsin1gextwirtah doubl
an average
(d) 270 nm; 350 nm; same as in (c) except only 3 alkyl groups instead of 4
REMINDERS
ABOUT
PROBLEMS:
SYNTHESIS
There
may
be
more
t
h
an
one
l
e
gi
t
i
m
to a synthesis, especial y as the list of reactions
getsBegilonger.n your analysis by comparing thateetaapproach
rgetoneto tofhe tstheartsmaling lmatnumber
erial. ofIf react
the product
hasformmorecarbon­
carbons
t
h
an
t
h
e
react
a
nt
,
you
wi
l
l
need
t
o
use
i
o
ns
t
h
at
carbonWherebonds.possible, work backwards from the target back to the starting material.
4. KNOW THE REACTIONS.
problems. There is no better test of whether you know the reactions than
18-5
n
n*.
n*;
<
=
=
is
=
1.
2.
3.
attempting synthesis
402
18-6 Alclarbon-bondthree target fmolormieculng ereact
s in ithions.s problSo far,em thave
more
thantysipesx carbons,
soonsaltlhanswers
wil include
carbonh
ere
are
t
h
ree
of
react
i
at
form
carbon-carbon
SN2 substitution by an acetylide ion, and the Friedel-Crafts reactions
ion) ionon,benzene.
(albonds:kylatitohne andGrigacylnardatreact
0
OH
0
f
0 MgB l) H �
0 Bf �
ether
H2C�; �
2) H30+ �
0
(b) Bf
identical sequence as in part (a)
I �
U
~
0
OR
s metheffiodcusiientngasFriit eidels onl-Craft
s acylstepation
Cl �... � ithis more
y
one
o AlCl3 U
(c) a synthesi s as in part (a) could also be used here
OR
0
� Br Mg � MgBr
� �
L-.l
� L-.l
2) H30+ L-.l 6H �C
�4
OR
(:j' Br Na+ -C=CH (:j' C=CH Hg2+, H20
H2SO4 og
18-7
(a) (J BuLi (J BrCH2CH2Ph (J H30+... PhCH2CH2 -CH0
SXS
S C'" S
S X S HgCI2
H H
PhCH2CH2 H
H
(b) (J CH3I (J BuLi
(J Ph H30+ 0 Ph
S ::-C" S
SXS
� HgCI2 �
Ph
CH3 H
H
Br �
from
(a)
..
..
o
1)
..
..
...
• •
..
I
..
::- I
II
..
• •
...
...
(a)
403
�
18-7 continued
H30+ Ph � Ph
(c) rl PhCH2Br SrlS BuLi PhCH2Br SrlS HgCl
2
S' " S
X
X
CH
PhCH2 CH2Ph
PhCH2 H
rl Q- CH2B' rl BuLi PhCH2CH2Br
('j()
S:-C"S
o-:x
S S
� Ph
H
H (a)
('j
�
�
H30+ , HgCl2
�
Ph
18-8 0
0
(c) CH3(CH2h -C0 -CH2CH3
( a)
(b) �
Ih
if
18-9
0
(a) CH3(CH2h -C-CH2CH3 (b) (siPhCH2CN
mple SN2) (c) PhCH2C\l-(J
MgB' CH3CH2C= � H30+
CCH2CH3
(a) I � Br �
O ether O
O
(b) CH3CH2C::N BrMgCH2CH2CH2CH3 --- H30+ �
H30+
(c) CH3(CH2hCOOH 2 CH3CH2Li ----�
CH2C -o
CH2Br
CN MgBr 6
NBS. 6 NaCN.
I�
o
•
•
•
•
.
•
I
(d)
from (a)
�
-
•
I
from
II
II
o
18-10
II
--
h
+
--
+
(
d
6
o
o
'
H,
o
6
6
::::'--1�---
404
o
h
IS-II
(a)
V
CH, H (b)
o
V
CH
o
OH (d)
�
�
o
(C)
II
IS-12dirRevi
ndersstonepsp.i s402.usualThere
more
ect routew tehwie remithPhBrfewer
ly betareter. often more than one correct way to do syntheses, but
( a ) � . Mg
0
OH
� Cr03,H20 �
+
H30
PhMgBr
-..;:
::
H
•
Ih
V V H2S04 V V
PhBrMg
.
H30+
� PhMgBr
o
•
VV
•
•
(c) 0
•
H,o'.
OH
Li
A
I
H
4
NaBH4
OR
�
�
CH30H
VV
VV
(
s
ol
v
ent
)
HC::CCH2CH3
OH
t NaNH2
(d) 0
+ C::CCH2CH3 H30+ � C�CCH2CHJ
�
Na
H
Ih
V
cetoxyborohydride
replIS-13acedThethreetriahydri
des. ion is similar to borohydride, BH4-, where three0acetoxy groups have
(b) .... C...
(a )
C
.... C...0 :0:
" '- 0 0 ' 0
0
'
0
)
0
�O-B-=-H
� O - B + R--1' H
Na+ H- 0� -O -1/\
\. - Na-+ +. )!.::
�
R
H
/
/
O
O H
y
y
y �
0
H3CCOO-H
1O-H
R+ H
..
_
•
•
o
II
o
II
II
-I
o
•
o
I
• .
405
--
1
• •
I
H
a
ylides. TrimCH3ethylphosphine has a-hydrogensCH3that could be removedCH3by butyl ithium, generating undesired
1+
1+
BuLi CH3-P-CHR CH2-P-CH
2R
R
CH3-P-CH
2
1
1
1
CH3
CH3
CH3
wrong ylide
desired ylide
�+
�h PPh3
:
rotation '" I" I I
HMe MeH
:O---! PPh) /
H I I I C - C """ H Ph3PO
1.:1:1 I I I
""""
'
H',
r
Me cis-2-butene Me
Me �Me/ H
Thefashistoen,reochemi
is inpverthosphied. nThee oxinuclde emustophielleimitripnathenyle wipthosphi
ne mustry.attack the epoxide anti
yet the tsrtirpyhenyl
h syn geomet
(b)
:0: rotate and
'H
el
i
m
i
n
at
e
..
PPh3
7
1
fl
o
O
.
trans
+
PhCHO
..
CH2=CHCH-PPh3
PhCH=CH-CH=CH2
(a) CH2=CHCH2Br 1)2) Ph3P
BuLi
1) Ph3P... PhCH-PPh3 O=CH-CH=CH2.. PhCH=CH-CH CH2
OR
2) BuLi
PhCH=CH -CH2Br 1)2) Ph3P
BuLi... PhCH=CH-CH-PPh3
CH2-PPh3 PhCH=CH-CH=O PhCH CH-CH CH2
18-14
1+
�
-
+
• •
..
+
_
....
..
f.
-
\
\\
.
" ,
in
. .
"
o
CIS
�
:
,
..
18-16
�
-
(b)
+
=
+
+
�
406
=
=
an
enes form
can bethesyntphosphoni
hesized ubym tsalwot difromfferentthe Wilesst ihignreact
The18-17onesManyshownalkhere
deredionsalk(asyl halin itdhe.e previous problem).
PhCH=C(CH3)z
(a) PhCHzBr 1) Ph3P
BuLi.. PhCH-PPh3 )l
(b)
BuLi CHz-PPh3
(c) PhCHzBr Ph3P.. PhCH-PPh3 PhCH CH-CH 0.. PhCH CH-CH CHPh
2) BuLi
O �CHz
CHz-PPh3 �
U
U
+
o
+
2)
H3C
CH3
-
+
2)
1)
+
+
18-18
(a)
=
=
=
=
_
+
OH
Cl3C-C-H
OH
"
:O: � H-OH
OH
CH3 -C-CH3
CH3 -C-CH3 HOOH
OH
I
I
I
. .
I
18-19
I
lofeasthydratamounte
I
407
great
est amount
of hydrat
e
+
:0 : H-CN 3 OH
CH CH2 -C-H
CH3CH2-C-H
18-20
. .
I
I
�
CN
(c) t-Bu-C-t
err -Bu
'"""
I
CN
..
I
OH
H-CN
t
B
u-C-t
B
u
t
B
u-C-t
B
u
� :C N CN
CN
0
OH
(a)�
HCN vi'
V
OHI pcc
0
OH
(b)
.
�
�
·
.
H
H HCN �H
H
CN
OH
(c)
CHCOOH
u
smsivofe charnuclgees.ophiThese
lic attspeci
ack atescararebonylverycashorrbont-lifvredequentas eachly incharcludegespeciis quiecsklwiythneutbotrhalized
Nposioteti:veMechani
and
negat
on tr ansfwierl; ishow
n factt,htesehese stepsstearpseasthoccur
e fasterstingofatththeewholsamee mechani
sm.though
In mostyoucahases,ve
tbbyheenisaSolrapiadmoni
udtioprnsoshedtManual
t
i
m
e,
even
sm separstoodatetlhy.atThethesepraracteice offashowi
transfers in one stetop show
is legitimstateepsasofloangmechani
as it is under
st steps.ng these proton
::::..;:::
::
�
::=
:o:� '"""
I
I
I
I
18-21
II
I
all
two
two
408
(a)
18-22
(J<-:NH;
0:
H
H
H
'
)
JW
:
O
h
(J
:0:\
:o-�
d,+NHPh HO+3 . C} NHPh HO+
�20:
(
HI
H20:
two fast prot'---./
on transfers
Ph-C-H
•
•
:o:�
1
��
(twro fast�
proton transfers
1
---•
•
•
Ph-C-H
1
NHCH,
H
�
3
•
:O�
1
��
_
O
-H
H- Ph-C-H
( 1+
NHCH3
1
+ - H20
Ph-C-H .. H20: i Ph-C-H Ph-�-H }
Hplposiusttihvreee charge
resonance
on theforms
benzenewith
II
: NCH
3
409
�U�CH3
II
�
NCH3
I
ring
isomers.a double bond is formed, stereochemistry must be considered. The two compounds are
t18-he 23andWhenever
Z
E
Z
18-25 This mechanism is the reverse of the one shown in 18-22(c) on the previous page.
H-O-H
1+
H20:
Ph-C-H
..
Ph - C - H H30+ Ph-C-H
Ph-C-H
�
+
I
II
I
NHCH3
H
-NCH3
H
NCH3
.NCH3
1 plus th+ree resonance forms with ') t
( Positive charge on the benzene ring
H-O: -CH3NH2 H-O:
H-O:
: II0 :
1
1
I
+
..
.. Ph-C-H
Ph-C-H
Ph-C-H
Ph-C-H ..
I CH3
(I
H
0+
3
H-NHCH3
�
NH
t
H-O+
VII
Ph-C-H
18-26 02N
NH2-NH -O- N02 A
abbreviY. ate "Z"
:o � + H-O-H
+
H30+
Q:
H
1
I
I
30
(
CH3-C-CH3
CH3-C-CH3
--CH
-C-CH
3
CH
-C-CH
3
3
'-- :NH2Z
I
I
�
H2
0
:
+
.�
.1 1
j
�
•
�
• •
..
..
. .
....f----
+
o
..
\
)
:o:�
-­
(k�
two fast proton transfers
NHZ
NHZ
•
•
3
CH3 -�-CH t
H-NZ
I
..
3
NH
N(b) oJ 2
o
I�
.b
(a) PhCHO + H2NNH -C -NH2
(c) 0
N-NHPh
H
2
oJ
NH2 CH3
(e) C(f
18-28
(b)
II
(d)
+
I
N ",NHPh
Ph )l Ph
(d)
ho + H2NOH
0
0
02N
+ NH2-NH � }-N02
0
o
+
·0· + � :O-H
Ph-CH H Ph-CH
:O-H
O-H
+1
�
.
Ph-CH•• }-H�CH3
.-----l Ph- 1 +
.
"---.:.
: OCH
3
)
�k
H.qCH3 �
H-O-H
}
,----- :O-H
:
+1)
�
�
H
:
0
�
2
�
ili
��
Ph-CH ---H+ Ph-CH
j Ph CH ,...-W ---j
OCH3
OCH3al
1 + �CH3 (�H3
hemiacet
18-29
. . �
II
---
II
plus resonance forms
with (+) on benzene ring
----
'"
J
plus resonance forms
(+
ne ring
_
I
...
....
H..2:0CH3
Ph-CH
OCH3
VI
I
OCH3
Ph-CH acetal
OCH3
I
I
41 1
I
CH
I
I
18-30CH3 CH3
1
1
CH31
:000: ""\H 0+
• •
• .
:0,
t
�
18-31
6
(c) 0 0
. .
1
• .
3
..
: O�H
+
•
1
6
1
H20:
.
+ 1"
. .
.
.
..f--....
H2O:
:0:
•
6
(b) 0
CH3 -CH + 2 (CH3hCHOH
(d)
HO
+
O
O HO)
II
+ 2 HO OH
(e)0 HOD
H+
�
+
.
t -CH30H
..
:�
.�H20: ! V
CH3 H
CH3 H
0
0
o
O-H
:O
:
:
H�
0+ 0
f
oo:0 / H
CH31
r\
(f)
HO )(l OH
II
o
412
�';led lX�U uo p�nu!luo:>
WSlUnlj:>�W S!SAI0.1pAlj
t
:9
zH
H
o'0:
o
:0 '0
. .....
J�
H-O
J,
'----1
��
HO
H
-
+
H-O:
• •
J,
+0
-Q�Q
0
0
0:
•
'----1
+
•
H-O�
+
'---1
�:o
H
-1
'---(q)
0- Q
0 '0: H 0: 0:
J,
'----1
+
• •
-
J,
• •
'----1
+
FHQ
Q-Q � :0:
(n)
H
+6 :0·o/�
:0
H
-------
+
contiynsiueds mechanism continued
(b)18-33hydrol
OH
OH
r
+ r
H-O� .... OH
:0, .... OH
C
C,
� rOH
____
� H30+
HO
I\
�6
6-6
(c) The mechanisms of fonnation of an acetal and hydrolysis of an acetal are identical, just the reverse
·0·
((·O �
.. H
+ H
+
order. Thipaths,hasthentothbee reverse
true because
thave
his process
iws antheequiidentlibiricalum:minifimthume forward
stpatepsh. folThiloswis tmihenimum
energy
st
e
ps
t
o
fol
l
o
energy
Principle Microscopic Reversibility, text section 8-4A.
(d)
of
;;o:CO
CO-;
9
o
0 � +
H
RJ
CJ:::'t
/h··
OH
+ O�
18-34
a
( )
;-0-
0
NaBH4
A
CH30H
V CHO
1
equivalent
1\
HO
00-00
o �
··�
�
CCl �CCl
�
'--• •
•
+0
O
• •
famous
:0
I
H
H20:
�+
OH '--- R
W
..
1 equivalent
0
a
RH
�
��
OH
OH
\ �
CO
.
.:
+O
1
H
OH
OH
arecolreduced
fastcreaseer thselaneketctivonesity). (keeping
tAlAlhidtseehydes
react
i
o
n
d
wil
l
i
n
be usedrnatiforvely,thissodireductum itorin;acetseeoxyborohydri
problem 18-13.de could
..
OH
414
18-34 continued equivalent
0
0
o
0
o
0
Mg
(c) CH2B' HO OH
CH
R'
CH
MgB'
2
2
6 H+ a ether a + H�
1
(\
(\
(\
..
..
Theoxygen,last stdehydrat
ep protoesnattheesalthceohol,
and hydrolyzes the acetal.
18-35
(a) oCOO O
HOafter adding H+ : Ag
(d) HO COOH
HO V
0
\..
t
..
)
V
[MgBr]+
t
HC::C -CH2 -CH2 C
(b) DCOOH (c) COO:
y
o
0
after adding H+
H30
+
- - CH3
415
a
II
A gO
j
18-36 :
hydrazone formation
cJ'-
-000
0,
+ H0
..
•
",-- :NH2NH2
�H
V
�
"H"OH
..
�
:OH
"
2
"
NHNH2_
(I
:OH
two ofastn transfers
prot
reduction of the hydrazone
H
• •
•
O
H--./ ..
..
C ....... N
H-N
II
O
C� H --
•
•
-·..N
('i2:�IIN
V
416
H
NJfNH 2
18-37
(a)
H
CO
(b)
H
�
H
H
(c)
(d)
1\
�
V H ' 'H
18-38 Please refer to solution 1-20, page 12 of this Solutions Manual.
18-39 nIgUPplacement
namesoffiposirst; ttihoenn numbers.
common names. Please see the note on p. 136 of this Solutions Manual
regardi
(a)(b) hept
aan-n-24--oone;ne; metdi-nh-propyl
yl n-pentketyloneketone
bromo-p2rop--met2h-eylnalhexanal
; no common
name
(gh)) 4-3-phenyl
(
;
ci
n
namal
d
ehyde
hept
(i) hexa-2,
4-dainalenal; ;nonocommon
commonname
name
anal; no simdiplpehenyl
commonketoname
(j)
3o
xopent
(d)((ec)) buthept
benzophenone;
ne
(
k
)
3o
xocycl
o
pent
a
necarbal
d
ehyde;
no
common
name
a
nal
;
but
y
ral
d
ehyde
)
ci
s
2
,
4
-di
m
et
h
yl
c
ycl
o
pent
a
none;
no
common
name
(I
(18-f) 4propanone;
acet
o
ne
accept
s
"acet
o
ne")
0 I0n order of increasing equil0 ibrium constant for hydrat0ion:
l
2
lofeasthydratamountion
est amount
ofgreathydrat
ion
18-4 1
AC
(lUPAC
II
CH3-C-CH3
C CH
<
o
b2
o
-
II
C-H
}c
I I
H-C-H
<
�
9H
CH3
II
I
H-C-CH -C-CH3 I
f!
CH 3
18-42
o
comparison with similarly substituted molecules shown in the text:
A By1t--t1t
* value (210) plus 3 alkyl groups (30) 240 nm
U CH3 n--t1t* base
300-320 nm
o
=
417
18-43
C6HJ002 indicates two elements of unsaturation.
1
Thetwo elIRement
absorpts ofiounn atsat1708
em·
suggests asiketngloene,ts inorthpossi
biolyoftw2o: ket3 inodinescatsienacehitghhlerey symmet
are two rioxygens
ande.
u
rat
i
o
n.
The
e
rat
c
mol
e
cul
to carbonyl
liThekelysintoglbeet atCH22.15on thise probabl
other siydemetof hthyle next
carbonyl
. , and the singlet at 2.67 integrating to two
H3C -C -CH2 -CH2 -C -CH3
Since the molecular formula is double this fragment, the molecule must be twice the fragment.
Two questofionshydrogens.
arise. WhyWhyis tdon'
he inttethgrate twioonmet: hylandenesnotshow4 : 6?splIint tiengratg? iAdjacent
on provi,des a hydrogens,
not absoluwite th
numbers,
identical chemical shifts, do not split each other; the signals for ethane or cyclohexane appear as singlets.
4dehyde
The formul
ndicatievse 5Tolelelement
suggests
ant18-heal4unknown
ormusta ketabeCone,JOa HIketbut2Oone.ainegat
ns tesstofpreclunsatudesurattihoen.possiA solbiliidty2,of4-DanNPaldderiehyde;vativtheerefore,
Themonosubst
NMR ishows
the typicatal et7.hyl3 pat5H,temulrn attip8le1.t)0. The tsirinpglleet)t andat 3.7.5i s a CHquartet
), andte fara downfield,
t
u
t
e
d
benzene
,
but
qui
2
apparently deshielded by two groups. Assemble the pieces:
0- CH2 C
NMR
(5
(5
>
o
is
o
II
II
2 3
8
(3H,
(
[
o
+
+
18-45
.
H
,CH3
"
0
)
(
CH
(a)
H /C ,�CH
CH2 2
rn/z 86
)H>-. �CH3 .
(b)
� CH
CH3/C,�,CH2
CH
rn/z 114CH3
+
II
I
+
o
II
I
I
ratio,
identical
II
..
..
(5
(5 2
(2H,
H
CHCH3
H J:CH2 CH2
rn/z44
mass 42
0....... H . CH/CH3
CH/3 C',CH
CH2
mass
42
CH3
z72
r
+
II
+
I
rn/
418
I
+
II
18-45 continued
(c) O} (
H
H3C""""
II
C
18-46
CH
. CH3
I
mJz 114
.Jr-.... CH
'c""""
'CH3
.
+
..
H2
[
0
H3C
....... H
mJz 58
I
....... C"
-""':CH2
r
•
HO XO
X
�O
+
mass 56
CHCH3
+
0
II
CHCH3
OH
ATol18-mol4le7nsecultestariniodincofatemJa zket70one.meansThe°a carbonyl
derivativmass
e andfora negat
ive
fairly smal(CO)l molhaseculmasse. A28,solsoid semi
c28arbazone42, enough
70
onl
y
with t.wo elements of unsaturation, we
The molofecula doubl
ar formul
canmoreincarbons.
fer the presence
e bonda isorprobabl
a ring yin addition(mtoassthe70);carbonyl
I, indicative of a ketone in a small ring. No peak in the 1600-1650
Thecm-IIRregishows
a
st
r
ong
peak
at
1790
cmo
n
shows
t
h
e
absence
of
an
al
k
ene.
The
onl
y
possi
b
i
l
i
t
i
e
s
for
a
smal
l
ri
n
g
ket
o
ne
cont
a
i
n
i
n
g
f
o
ur
carbons are these:
0
0
�
4<)'
0
A
3
=
C4H60
0
B
The HNdoes show
can dias4Htinguitripslhetthatese.3.1;Nothmetis sihylgnaldoublcomeset appears
ienttwheo NMR
spect
r(C-um,2rulandingC-out4) adjaThecent to
from
t
h
met
h
yl
e
nes
the carbonyl
by the twoof splhydrogens
roughl
y a qui, nspltetitbecause
it ing by onfourC-3.neigThehborisinggnalprotforons.the methylene at C-3 appears at 2.0,
unknownion ofis cycl
obutanone,at 1790Thecm-symmet
ry inediricsatticedofbysmalthelcarbon
NMR
rulrinegsstoutraistn rstuctrengturehensThe
TheIR absorpt
I
t
h
e
carbonyl
i
s
charact
ri
n
g
ket
o
nes;
carbon-oxygen double bond, increasing its frequency of vibration. (See Section 12-9 in the text.)
18-48The conjugated diene has a maximum at 235 nm (see ved Problem 15-3) and the ketone has a
(a) mum at about 237 nm, so the to transition cannotSolbe used to differentiate the compounds.
maxi
(b) The ketone has an n to transition around 315 nm that the diene cannot have.
NMR
MR
3
CH3
&
A.
n*
n
B.
8
B.
n*
41 9
the
18-49
(a) (�l���etal)
cetalketal)
old: ahemi
(hemi
o
6
II
CH3CHCH2
-C -CH3
2
(b)
(c) acetal
(d) (olacetd:alketal)
18-50
(a)
G.o:
II
,--
3
HC-C-H
+
o
+
CHC
3 Ho
, H
..
0H
1\
:0: �
I
H2
H
O.
H3C-C-H ..
II
:o�
I
H30+
.. H3C-?-H
H
II
HC-C-H
3
l l
"----/' U
CH3-CH
I
I
acetal
OCH3
420
0"
0
........CH
I
",CH
HN
2
+
..
1+
-Q
H-O-H
(
..
I
HC-C-H
3
NHNHPh
---
H3C-
H-
}
NNHPh
� -H
I
• •
O-H
CH3-CH
=(
+
1+ ·
:kV
I
OCH3
H
H
NH2
+NHNHPh 2 •
NHNHPh
(�
J
two fast proton transfers
i
0:
2
: NNHPh
� NNHPh
+
I
H 3C-C-H
H30+
O
OH
(
imine(s)
(h) imine 0=0
OH
+
(f) diether, inert to hydrolysis
(g)
HO
: NH2NHPh
2 CH0
3 H
(X
(e) acetal
O
H CH3
I
:O-H
•
hemiacetal
I
CH3-CH
OCH3
18-(c) 50 continued.
C0
3
h
P
O
P
C
h
)
H
,P
2
cr
�
Ph'
;
'
�
:
U"---- CH2 tPh, �V C�2 V rCH2
/\
/\
/\
0
�-H
'
O
-H
:
O
� O-H
o
�
°
°
o
(d)
:�H30+
0 o
..
:
+
�
�
-
+
----
.
�
:
9
+
�
..
Hi)
�
:
OH
OH <---JOHH
H-+/0 �O r'0"�
"c/ OH
.
0
H
2
+
H
30
o ,. 0 :
l 0' H
H
OH :
HO� o
H1 +
(e) �
'
O
-H
:
o
m
H H,O+ �
H
02
c
N
�
+H
��
H
'
O
H
'
H
0
:�.;-H
r:J2-�';�
+
+
O
.. C,H 4) <;:,H
H H ..H3. EJD2
H
H H
H
.
0.
0
'
;
� O·
C,
C ,H
� 'H H30+ 0
N
H
3
H 'H
H
.
...
•
•
-6 r�:. 6
..
HH H
/N,
�b
1
N/
1
•
+
/ .'
--
..
/N,
'/
•
�/+
o
+
�
I
+
H
'
·
O
/
•
/N
�
N
+
/N �
,
•
•
421
..
� .'
I
�
H
H
18-51 o
OH
OH
H30+
(a) CH3-CH KCN
..
..
HCN CH3-CH-C::N 0=CH3-CH-COOH
,Ph
HC
o
PhCH2Br Ph3P BuLi .. PhCH -PPh3 ..
6
0
0
0
1 equivalent 0
(c)
NaBH4
Al
t
e
rnat
i
v
el
y
,
usi
n
g
sodi
u
m
t
ri
a
cet
o
xyborohydri
d
e
..
sel
e
ct
i
v
el
y
reduces
t
h
e
al
d
ehyde;
see
Probl
e
m
CH30H
CHO
CH20H
o 1I10equi/\val0ent1! �o NaBH4 H OH 30+.
11
(d) 0
� ,H CHPH"
Q
C
C
0'" '0
0'"C'0 Q�1I0
CH3CH2CH2Br PPh3
t
tBuLi CHCH2CH3 5112:113
1
equi
v
al
e
nt
o
o
(e) 0 HO/\OH.. o CII3CH2CH -PPh�
1130+.
H
o
C
C
Co
0'" '0
o,..C,0
o 1 equiH2valent.. roo
ro Pt
OH
o
ro RaneyH2 .. co- H
OH
(h) roo NaBH4
CII,oIl· roll
I
I
II
+
(b)
..
18-13.
'
LJ
'--1
+
+
,II
LJ
LJ
(
(g)
f)
Ni
422
II
18-52 All of these reactions would be acid-catalyzed.
(a) q + H2NOH
(b )
(c)
o
�
� NH
0
+
H2N 'O
�CHO
(d) ro +
0 HO OH
6 + CH30H
O
�
2+
LJ
I
1\
�
,
C
H3
ONH2 O=CCH3
18-53
°2N
0
a( )
(b)
C
NH2
,
O
N
N
}
N
�
H
r
N/
cf
V
(c) NOH
(d)
(e) CH3-CHOCH3
�
OCH}
L
CH2
C
H3
OH
h
)
(g)
(
rl
H -OCH3
CH
< }- CCH2CH3
CH3CH2
(f)
+
\
(c )
o
2
II
1\
I
I
( f)
I
I
N
...
SX S
H
423
18-54
(a) (] BuLi
SX S
H H
..
CH31
(]
S, .. ..... S
_C
H
I
S(]S
CH3XH
C '/ HeH,
C:··S ' H
..
• •
..
+
HS �S-H
..
..
/' CH, H2O:
:··S ' HH
t
"c
..
H ..... C ..'CH3
2+, assists the hydrolysis in two ways. First, mercuric ion is a Lewis acid of moderate
(stcr)engtMercuri
c
i
o
n,
Hg
h, performing the same function as a proton from a protic acid.
Theits charge:
effectiveven
enesscompl
of Hge2xed+ aswia Lewi
s acifur,d tihs epartmercury
ly dueatom
t
h
a
sul
stil has a positive charge, attracting the sulfur's electrons.
S ...... Hg2+
Athisecond
expl
a
nat
i
o
n
for
mercuri
c
i
o
n'
s
effect
i
v
eness
i
n
hydrol
y
si
s
of
lms.s liesThiin sthestacompl
ex formed
betivweleeny removes
the ion andHSCHthe2CHtwo2SH
sulfromofuracetthateaoequi
bl
e
compl
e
x
effect
t
l
i
b
ri
u
m,
shi
f
t
i
n
g
t
h
e
equi
l
i
b
ri
u
m
t
o
product
.
(
A
n
exampl
e
S
of whose principle? His initials are "Le Chatelier".)
Thiols are often referred to as "mercaptans" because of their ability to CAPTure MERcury.
0:
II
to
424
18-55 icTheallykeyaffectto thitheis rproblchemistem irsy.understanding that the relative proximity of the two oxygens can
dramat
1
,
2
d
i
oxane
Thedescribed:
"second"twoisomer
oxygens
connect
e
d
by
a
si
gma bondbondarei s
aeasiperoxi
d
e.
The
0-0
l
y
cl
e
aved
to
gi
v
e
radi
c
al
s
.
In the presenceradiofcalorgani
cions
compounds,
react
can be explosive.
Mechanism of acetal hydrolysis
1,
3
d
i
o
xane
Thetwo oxygens
"third" isbonded
omer descrito thbeed:
3 carbon constitute an
same
sp
acetal whiacichd.isSeehydroltheyzed in
aqueous
mechanism below.
�O - H
C:0
:0 / H
H/C, H
I
•
+
o
CH
+
" '
H
---
•
..
....
..
f-.
-i
�
0 2
l,
4
-di
o
xane
Thean excel"firstle"nti ssolomerventdescri
bed:gh
(al
t
hou
toxic), thapart
ese oxygens
are far
enough
t
o
act
i
n
depen­
dently. It is a simple ether.
. .
..
HO
0
()
}
425
..
..-
"C ' H H2O:
(:'0
•
/
H
" '
H
(a)18-56
(b)
6
(e) N-N-Ph
]
(t)
6H
(i)
0)
0
6
PoH
CH ]
(c) NOH
c5
18-57 Thecomesnewfrombondthteoprotcarbonic solcomes
oxygen
vent. from the or
(a)
NaBD4
OD
(b)
NaBD4
OD
(c)
NaB
0
NaBH4
H2O
II
CH3 - C - CH2CH3
..
..
0
"
CH3 - C - CH2CH3
..
0
H4
"
D 20
..
D 20
(d)
6
(g) no reaction
CH]
(h)
(k) Na+ -(j ON (I)
NaBD4,
OH
I
1\
0
er
0
H
HO
.;; C
/
H
COOH
shown in bold below. The new bond
to
CH3 - C - CH2CH3
I
D
I
CH3 - C - CH2CH3
I
D
I
58 Whito lapproach
e hydridefromis a smal
legroup,
the actsiduealofchemi
cal especi
e, sthsuppl
ythinegsiitd,eAlopposi
H4-, itsefaitherlymetlarge,hyl . so
iThi18-t prefers
t
h
e
l
ss
hi
n
dered
t
h
e
mol
cul
e
at
i
s
,
s forces the oxygen to go to the same side as the methyl, producing the isomer as the major product.
O
�
�fd
Q
lessndered AI·
hiface
majproductor
H
CH3 - C - CH2CH3
..
..
CH3 - C - CH2CH3
I
H
cis
H
H
" /
/
C
H
H
_-,
� H
"- I H·
H
�
..
H3C
0
H
H+
---.
H
H
CIS
426
CH3
18-(a) 59
e bond i sic
BuLi CHz -PPh3O•O O CH2 lAndoubl
essexocycl
steablbond.e tihcandoublan endocycl
methylenecyclohexane
onhexane
from
cycl
ohexanone
withoutoutsusiidentghethrienWig t ig
(b)reactTheion direstffisciulnttyheinstsyntabilhitesiy ofzinthgemetdoublhyleenecycl
bond
i
si
d
e
t
h
e
ri
n
g
(
e
ndocycl
i
c
)
versus
(exocyclic).
HOoCH] :
+.
H20.
+ CH] MgBr
mi
n
or
maj
o
r
Amechani
dehydratsmiopassi
n folnlgowithrough
ng the a
t cold
carbocat
i
o
n
i
n
t
e
rmedi
a
t
e
will
gi
v
e
t
h
e
more assubsttheitmajuted,orendocycl
icThedoublonley Br CH3
bond
product
.
chance
he exocyclis toicdodoublan e o K+ -O-I-Bu
bondeliasmofitnhatmaking
eimajon usior ntproduct
give Hofmann orientgaatibulon k(Cyhaptbaseerto7).
minor major
18-60 CI -CCHzCH2CH3
zn(Hg� �
(a)
•
� HCI V
0
Br H3O+ 1 .0 C-CH2CHl
(b) V C'N CH3CHzMg
ether
V
Cl
OH
OCH3
CI2 • 6
NaOH
Na0
,!;l 6 HCl •
I)
1
0 AICI3 3500 C 6 CH31 .0 AlCuCICI3
CHO
0::5
0
(d)
H
Cr
0
2
7 I:
Z
conc. H2SO:
I �
c6
0:>
AlwittehrnatSOCIivelz,yt,htenhe cyclacidizchledobyridFericouledeld-Cberaftmades
acylation with AICI3.
+
-
6
�
+
El
HB r
E2
E2
°
°
II
•
c5 6
+
°
II
•
(e)
�
co
2)
.0
¢l
°
�
•
427
18-61
(a)
(b)
OHCP ,
(y h H
V
OCH2CH3
CI, H
()OCH2CH3
(c)
H
0
2
()N-N-C-NH
C-H
I
II
0C
.0O
()
Ag
+
(e)
('I
()CH1
c: �
d
+
c
CH3Mg1
H30
r03, H20
� H ether
OH H2S04 �
(b)�C=CH HgS040
H2H2S04 �
+
H
0
3
('I
('I
(
0
SXHS CH31 Sx S CH3 CH2hBr S� HgCI..2 �
H
CH3
CH3 Hc
r03, H20
� H2S04
�
� OH excess +CH�Li
H30 �
+
M
CH3
g1
H30
C=N ether
�
(g)
03
,CH3
O=C
Me2S � + CH3
(d)
I
II
(f)
I
18-62
(a)
II
..
o
�
�
I
..
o
..
(c)
1) BuLi
..
2)
..
1) BuLi
2)
--��
o
/"'.-..
I
I
..
(d)
o
(e)
1)
2)
o
(f) �
o
..
_
�
o
1)
2)
..
o
428
\
18-63
o
�H
PCC
(a) � H
---.-
O
o
�H
(b) �CH2
(c) �C=CH
1)
BuLi
..
�H
HgCl2
o
Cl
(e)
o
� Cl
�H
KOH
H20
---.-
(This reaction needs a solvent like THF to keep all reactants in solution.)
o
II
(f) �
OR
2)
OH H30+
ft
OH
�
o
(b)
..
�
SOCl,
�
18-64
(a)
LiAIH4
1)
--
o
II
�H
OH �
(
Ot-BuhH
ft
" � Cl
�
o
PCC
+
(e)
�
0
+
18-65
(a) ketone: no reaction
(b) aldehyde: positive
(c) enol of an aldehyde-tautomerizes to aldehyde in base: pOSitive
(d) hemiacetal of an aldehyde in equilibrium with the aldehyde in base: positive
(e) acetal-stable in base: no reaction
(f) hemiacetal of an aldehyde in equilibrium with the aldehyde in base: positive
429
(c)
�
o
A
18-66
J
K.
be deduced
from
itsioreact
ion wibethGrigandnard reactWhationsiswicommon
tn-2-one,
o both product
s
stirouctns uisrethofe heptcanan-2-01
ofmustthesebeThereact
part
;
t
h
e
react
ns
must
t
h
hept
a
so
heptan-2-one.
S(l
S
~
X
H H3 BuLi '
B
4
)
H2S04
�H�
HgS04
H
20
S S
�CH3
2
+
Hg
+
MgBr
H30
�
CH3CHO
+
H30 OH N
0
� a2Cr207 �
H2S04 •
MgBr
N�
PhMgBr+
�
6
0�
H30
H3:'
!
OH
�OOH
�Ph
�C=CH
C
1
A
1)
G
D
F
E
B
1)
2)
c
A
1)
1)
2)
J
K
I
o
430
bsorption at nm in tbhe spectrum suggest
I
The
very
st
r
ong
t
o
a
s
a
conj
u
gat
e
d
ket
o
ne
or
onyl
at
emand
smal
l
al
k
ene
at
u
gat
e
d
car
alThedehyde.
ba senceTheof peaksconfiatrms this: strong,m I conj
shows that the unknown is not an aldehyde.
The molecular ion at leads to the molecular formula:
b
C=C-C-C
mass units add car ons and hydrogens
mass
molecular formula C6H80 elements of unsaturation
Two elements of unsaturation are accounted for ibn the enone. The other one is likely a ring.
b
)
hydrogens.
The
dou
l
e
t
at
neiTheghboringshows
carbonstwo(twovinyldoupeaks
one
nei
g
h
ori
n
g
H
blet: neighboring H . says that the two hydrogens are on
H H 0
C-C=C-C-C C H ring
NComo metbinhiylngstarehe piapparent
eces: in the so the group of peaks at is most likely CH2 groups.
18-67
11:
IR
UV
225
11:*
2700-2800
e
16 10 em-I.
1690
-
96
o
,
'(
II
96
-64
I
=>
32
64
=
=
NMR
2
8
3
86.0
=
1
..-I--.
I
I
II
NMR,
+
+
1
+
6
1
82.0-2.4
6H
3
b
b
The mass spectral fragmentation can e explained y a "retro" or reverse Diels-Alder fragmentation:
0
0
H xC
H TH 2
CHz
�CHz
Hz
C
H CHz
H CHz
loss of
j3-unsaturated carbonyl
I3
Ibnecause
the of the resonance
one of the form
vinylthydrogens
appears
at
Thi
s
i
s
t
y
pi
c
al
of
an
a,
hat shows deshielding of the -hydrogen.
H H 0
H H :0:
C-C=C-C-C
C-C-C=C-C
+
�
+
.
.
II
+
--
II
28
rnIz 96
rnIz 68
HNMR ,
87.0.
I
13
I
a
II
87.0 --.....
+1
....
..
f-.
-t
..
�
431
13
I
a
I
(a) Building a model wil help visualize this problem.
H
H " ··H
H H
:O+-"":CH O CH20H :O +CCH:O CH20H
:O-"":CH O CH20H
HO
OH
open-chaiOHn fonn
�
HO y .y CH20H
OH
HO
OH same as HO � OH
OH
cyclic fonn-hemiacetal
(b) Yes, the cyclic fonn of glucose wil give a positive Tollens test. In the basic solution of the Tollens test,
tconcent
he hemiratacetion.al iHowever,
s in equilibitriiusmthewiopen-chai
th the open-chai
n aldtehyde
wisthwiththe cycl
icifonn
inevenmuchthough
largerthere is
n
al
d
ehyde
h
at
react
si
l
v
er
o
n,
so
open-chai
nc fonn
fonn present
at tanyo replgiavceentthime consumed
e, more open-chai
of the open-chai
n fonn
isually all
oxionlof tydhieazedcyclsmalbyiclsiamount
lfonnver iwion,lofmore
cycl
i
wi
l
open
n
fonn.
Event
carboxylate. Chatelbeier'dragged
s Principkilecstkirinkgesandagaiscreami
n! ng through the+open-chain fonn to be oxidized to
oxidized attoe
open-chain fonn NOTAg carboxyl
cyclic fonn """""
latargerequiconcent
at equilerliconcent
brium ration reversible
libriumration smal
18-68
• •
y
I
w
--
I
y
I
I
• •
-
-
y
I
as
the
Le
•
432
18-69
Any
with singroups
gle bondsan
belongscarbonto thwie acetth twalofamioxygens
ly. Ifbonded
one of tthoeitoxygen
groupketatal as iits came
a hemifromacetaal.ket(Tohene.ol) d name for this g;rollp
hemi
(b) Models wil help. Ignore stereochemistry for the mechanism.
is
OR,
then the functional group is a hemiacetal. Thus, the functional
C-2
is
H
OH
OH
1
HO
oo f' H+
o
*'
\
H
OH
00
o
,...
H
HO
HO
1
I
\.
(00 H
00,...
+
HO
OH
OH
HOH2C
OH
OH
OH
OH
---" C-O
HO
OH
H2�
H
OH
HO
OH
HO
0
O�
o
HO
same as
OH
433
H
18-70
Recall that "dilute acid" means an aqueous solution, and aqueous acid wil remove acetals.
XMgJ
1__)) _ _ _+�
�
_
-=�.
2 H30
V U
+
Ht
OH 0
OH
H
C
H3C()'
3
OH �aBH4 � H
)) CH3MgI
+
(2removes
H30)
acetal
equi
v
.
OH
0
0
+
�
cY OH excess
Ag no reaction
PCC
H HO + OH
H
is identical to
j H2Cr04
H20
0
PhNHN NNHPh
OH
�H
&H
I
II
G
H
�
1
(\
--
..
B
(P�
A
o
cr
j HCIZn
E
c
(Hg)
(i'oH
o
F
434
--
D
A
18-7 1
C, H HSfI
fI
fi
H3 .. '
BuLi
SH
/S
S
/S
S
/"
'
,
,
..
-----.
C
I
fu Ph/"C, CH2 Ph HgCI2 Ph/"C CH2Ph
PhCH
2
Ph
H
()
H
H
H
)(a
H w
H
�
'
C+
OH �
d
l...... _� H
0"""" � H
�H /..
RH
0
II
W
�
__
II
2)
18-72
o
0
0+
1)
--- (1-:
�
Cl:
•
R
0=:OR
(b) This is not an ether, but rather an acetal, stable to base but reactive with aqueous acid.
(C)Q- ..
H
O:R
W
U
9-: - ROH. { QtC:o H}
H2� !
O
O-O-HH -Ct.. H H2.. Q
+
('O
H
O
'--· -H
·
/
..
H
/
.
'-...
t
w
/
"
�� +-H .. . 0
H
�- � \L)
l...... _ : H
H2
H Oo
I
I
•
•
o
IIY"\
I :0-
H
+
J
O
• •
435
o
• •
..
:
-
U -H
II
I :0:
o
H
\
0
H
18-73
(a) First, deduce what functional groups are present in A and B. The IR of A shows no alkene and no
carbonyl: the strongest peak is at 1060 cm-1, possible a C-O bond. After acid hydrolysis of A, the IR of B
shows a carbonyl at 17 15 cm-1: a ketone. (If it were an aldehyde, it would have aldehyde C-H around
2700-2800 cm-1, absent in the spectrum of B.) What functional group has C-O bonds and is hydrolyzed to
a ketone? An acetal (ketal)!
°
R'O
OR '
'\..
R
C
/ ,
A
II
/
R-C-R
B
R
mol. wt. 1 16
=>
C6H 1 ZOZ
There is only one ketone of formula
C4HgO:
butan-2-one.
°
II
H3C - C - CHzCH3
t--...
..
B
A must have the same alkyl groups as B. A has one element of unsaturation and is missing only CZH4 from
the partial structure above. The most likely structure is the ethylene ketal. Is this consistent with the NMR?
4H)
03.9 (singlet,
�
HzC-CH2
I
o 1.3 (singlet, 3 H)
i
0
\
O
X�
H3C
A
-
CH3
0 1.6
}
00.9 (triplet, 3H)
(quartet, 2H)
What about the peaks in the
ion at 1 16.
+
'W
H3C
MS at rnJz 87 and
WI? The 87 peak is the loss of 29 from the molecular
.
plus two resonance
forms with positive
charge on the oxygen
atoms
CHz-CH3
rnJz116
.....
_---I..�
The peak at rnJz 10 1
comes from loss of CH3
from the molecular ion at
1 16.
rnJz 101
436
plus two resonance
forms with positive
charge on the oxygen
atoms
18-73
continued
.
"c O
r:"
o
:
H
/
"-
Me
1
:
1
C+
/ "-
Me
:
-H
....
..1--__
• _
II
C
/ "-
Me
Et
p
3VH
:0:
.
H20
• •
Et
}
II
�
C
/ "-
Et
Me
B
Et
18-74 The strong UV absorption at 220 nm indicates a conjugated aldehyde or ketone. The IR shows a
strong carbonyl at 1690 em-I, alkene at 1625 em-I, and two peaks at 2720 em-I and 28 10 em-I -aldehyde!
o
II
C==C- C-H
proton at 89.5 split into
The NMR shows the aldehyde
a doublet, so it has one neighboring H. There are
only two vinyl protons, so there must be an alkyl group coming off the B carbon:
,..-----,.
H
H
0
H
H
0
I
I
II
The only other NMR signal is a 3H doublet:
18-75
(a)
0
II H
P,-
EtO""'" I
......
�
I) R
C
OEt 1
R
. -
·B
--
0
II
R
must be methyl.
......
D
� :O.
R + " II
+
• �
R
II
CH3-C==C-C- H
•
P,EtO""'" I C OEt I
I
I
R-C==C-C- H
.....
R './
C
"' R '
"crotonaldehyde"
·0·
.r-"
EtO· I ...
EtO l.O
" �: ;
"
EtO-P 0
EtO -P � O.
•
.
I� I
.
--
.
R- C-C - R '
I
R
I
1'\'-- 1
-R-C ..!.. C -R'
R'
R
437
(EtOhP02-
+
I
R'
/
,
C==C
'
R'
�
I
R
I
R'
18-75
nued OEt�
(b) conti(X
: -OEt
OEt
(c)
R
+
�
/1
�
+
(i)(EtOhP
Br
�
....
COOMe
.. (Et0}z0P,
� .... COOMe
�
-EtBr
�
� COOMe (CH3}zCHCHO MeO­
_
�
�
I
t
(------...
..
(ii)
°
(EtOhP BrCH2COOMe .. (EtOhOP COOMe
-EtBr "- �
-'
-- �)
�
y
COO M
eO+
+
------
-----
18-76
(a)
�
NM H+
U e2 ..
0
H
('::�( )�.- H
\!!.)C+
1
�
(b) Aminoacetallinkage�� in the dashed bOXCS' 2
/.:
:
:
-olD
.----1 N�lo
N
HO
HO
: N
:
W
W
OH
deoxyadenosine deoxycytidinOHe
'0
I
:
I
-'
�-----
:
'0
.�----
)
I
1
--
438
((c) The first step in the mechanism in
ectron
painucla) ri.seosiprTheodtes,oninatthowever,
roiogensn of ofthearethamie partDnNe'Asofelaromat
ifcor
rithnegs,aromat
and tihcietyelofecttrhone ripaing.rs (areSeerequi
r
ed
the ion
solof tuhteioarn otomatproblicityemof these fornuclaedescript
osidwie l not
bases.
)
Prot
o
nat
i
o
n
of
t
h
e
ni
t
r
o
gen
hlel notacidprotis extonatremele they Nstrandong;
dioccur
ltherefore
ute unlacidethsssewitnucl
eoside wil be stable.
part
16-42
CHAPTER 19-AMlNES
19-1 These compounds satisfy the criteria for aromaticity (planar, cyclic 1t system, and the Huckel number
of 4n + 2 1t electrons): pyrrole, imidazole, indole, pyridine, 2-methylpyridine, pyrimidine, and purine. The
systems with 6 1t electrons are: pyrrole, imidazole, pyridine, 2-methylpyridine, and pyrimidine. Th�
systems with 1 0 1t electrons are: indole and purine. The other nitrogen heterocycles shown are not aromatic
because they do not have cyclic 1t systems.
19-2
(a)
CH3
H3C- �-NH2
CH3
I
(d)
19-3
(a) pentan-2-amine
(c) 3-aminophenol (or meta-)
(e) trans-cyclopentane-l,2-diamine
(b)
NH2
CH3-CHCHO
I
(c)
Q
N
H3C/ ...... CH3
(e)
(f)
(b) N-methylbutan-2-amine
(d) 3-methylpyrrole
(f) cis-3-aminocyclohexanecarbaldehyde
19-4
(a) resolvable: there are two asymmetric carbons; carbon does not invert
(b) not resolvable: the nitrogen is free to invert
(c) not resolvable: it is symmetric
(d) not resolvable: even though the nitrogen is quaternary, one of the groups is a proton which can
exchange rapidly, allowing for inversion
(e) resolvable: the nitrogen is quaternary and cannot invert when bonded to carbons
19-5 In order of increasing boiling point (increasing intermolecular hydrogen bonding):
(a) triethylamine and n-propyl ether have the same b.p. < di-n-propylamine
(b) dimethyl ether < dimethylamine < ethanol
(c) trimethylamine < diethylamine < diisopropylamine
19-6 Listed in order of increasing basicity. (See Appendix 2 for a discussion of acidity and basicity.)
(a) PhNH2 < NH3 < CH3NH2 < NaOH
(b) p-nitroaniline < aniline < p-methylaniline (p-toluidine)
(c) pyrrole < anil ine < pyridine
(d) 3-nitropyrro\e < pyrrole < imidazole
19-7
(a) secondary amine: one peak in the 3200-3400 cm-i region, indicating NH
(b) primary amine: two peaks in the 3200-3400 cm-i region, indicating NH2
(c) alcohol: strong, broad peak around 3400 cm-i
439
19-8
A compound with formula
lypes of
H,
with the
NH2
C4H"N
NMR 8l.15,
NMR
has no elements of unsaturation. The proton
appearing as a broad peak at
of hydrogens on the four carbons. The carbon
shows five
meaning that there are four different groups
also shows four carbons, so there is no symmetry in
this structure; that is, it does not contain a t-butyl group or an isopropyl group.
8l.05, 3H H.
8l.35, CHNH2 81.15, 3HCH CH2
2H
NH2
-C­
HC-C3 H
H
CH,
80.90 H, / 8 .
H3C-C� -CH-CH 8 l.05
8l.35 / NH�
deshielded by the nitrogen; integration shows it to be one
The multiplet farthest downfield is a
There is
a
a
peak at
the broad
multiplet at
80.90. The latter two signals must represent methyl groups next to a
and a
doublet at
triplet at
respectively. So far:
and a
I
The pieces shown above have one carbon too many, so there must be one carbon that is duplicated: the
only possible one is the
and the structure reveals itself.
2
8 (multiplet)
(triplet)
3
I
(doublet)
8l.15
"'---r""'"
(multiplet)
(singlet)
(b)
44.7
25.8
4l.0 CH3
0
CH3CH2-N -CH2CH3 CH3C+ H2-CH
+
H2CH2OH
C
CH3
�12.51.41/
7.t9 20l.t 9 10.t 0 63.t 6
--
(c)
I
(d)
II
(An older printing of the text
used values of 13.8,47.5, and
58.2. The values shown here
are taken from a spectrum.)
19-10
(a)
[H)C+;�2-r��1-CH2CHf iCH-) CH2-t-�H2 - CH)-CH2-�=CH2}
j
CH3CH229 58
CH3
�
mlz
87
.
+
mlz
mass
+
•
mass
15
(c) The fragmentation in (a) occurs more often than the one in (b) because of stability of the radicals
produced along with the iminium ions. Ethyl radical is much more stable than methyl radical, so pathway
.
(a) tS preferred.
440
19- 1 1 Nitration at the 4-position of pyridine is not observed for the same reason that nitration at the 2position is not observed: the intermediate puts some positive character on an electron-deficient nitrogen, and
electronegative nitrogen hates that. (It is important to distinguish this type of positive nitrogen without a
complete octet of electrons, from the quaternary nitrogen, also positively charged but with a full octet. It is
the number of electrons around atoms that is most important; the charge itself is less important.)
GOOD:
1+
1
mechanism
�II
f
� _J
N
+-
D
-N
• •
I
+
�
a�� 0 :�l)
H
o
0
VERYBAD:
- N-
H
NO
--'.
--
·
N
H
NO
+N
VERYBAD­
N does not have
an octet of electrons
N
N02
�
6
N
not produced
because of
unfavorable
in termediate
19- 12 Any e lectrophilic attack, including sulfonation, is preferred at the 3-position of pyridine because the
intermediate is more stable than the intermediate from attack at either the 2-position or the 4-position.
(Resonance forms of the sulfonate group are not shown, but remember that they are important!)
The N of pyridine is basic, and in the strong acid mixture, it will be protonated as shown here. That is part
of the reason that pyridine is so sluggish to react: the ring already has a positive charge, so attack of an
electrophile is slowed.
441
�-o:OoCH3 j;�- o�H �5Hl
c
�'
GOOD
19- 13
Cl
�
I
-Cl
0 0
_
_
__
..
_N
N
N
�
N
N
19- 14
(a)
� : NH�
� _ jlC
/
N
-0 0
GOOD
Br
stabilized by induction
from nitrogen
� NH2
(jib:,
N
NH
- �
:
This is a benzyne-type mechanism. (For simplicity above, two steps of benzyne generation are shown as one
step: first, a proton is abstracted by amide anion, fol lowed by loss of bromide.) Amide ion is a strong
enough base to remove a proton from 3-bromopyridine as it does from a halobenzene. Once a benzyne is
generated (two possibilities), the amide ion reacts quickly, forming a mixture of products.
Why does the 3-bromo follow this extreme mechanism while the 2-bromo reacts smoothly by the addition­
e limination mechanism? Stability of the intermediate! Negative charge on the electronegative nitrogen
makes for a more stable intermediate in the 2-bromo substitution. No such stabilization is possible in the 3bromo case.
442
19-15
H
..--....
Pr- N-H
I
H�
n
n
..--....
CH3 - I
+
HC03CH3 - I
..
.. Pr- N-H � Pr- N -
I),
H
19- 16
(b)
excess NH3
+
(c)
excess NH3
+
Br
�
---
H2N
�
PhCH2Br
19- 17
°
II
(a) CH3C -NHCH2CH3
19- 18
If the amino group were not protected, it would do a nucleophilic substitution on chlorosulfonic acid.
Later in the sequence, this group could not be removed without cleaving the other sulfonamide group.
6�
NH-�
====::> A
I
Y
� J
NH-S03H
/
19-19
/
c.
I
�
both
sulfonamide,
2NH2
sulfathiazole
continued on next page
443
¢
continued
OCHl
19-19
+
NH2
6
S02CI
19-20
(b)
(a)�
(d)
H3C,
9'-'::
NH2
NHCOCH3
(e)
,CH3
CXJ
�
HCI
H20
h-
!1
S02
NH
I
lll
6
t?
9
•
S02
NH
I
6
sulfapyridine
+
,CH3
V \H1
CH3
N
o
a
CH3
H1C-N�
H3C,
,..CH3
N
CH3
I
(c)
,
(I)
+
H2C=CH2
�N-CH]
19-2 1 Orientation of the Cope elimination is similar to Hofmann elimination: the less substituted alkene is
the major product.
HO' �
N
(b) H2C=CH2
(a)�
�
minor
major
(CH3hNOH
+ (CH3CH2)2NOH
+
+
�
+
(c)
0
+
(CH,hNOH
(d)
H2C=CH2
+0
N
OH
+ r0
N�
OH
nunor
19-22 The key to this problem is to understand that Hofmann elimination occurs via an E2 mechanism
requiring anti coplanar stereochemistry, whereas Cope elimination requires syn coplanar stereochemistry.
(a)
H(CH3h
H
CH3
,
.:- ...;
Hofmann orientation loses a hydrogen from the CH3 and the N(CH3h
C
�
/
group
to make the less substituted double bond
�
\
H 2C
H
F
444
continued
(b) Hofmann elimination
19-22
(CH3) 3N+ \�
,)
"C
H
H3C
"I
5=H(CH3h
f�CH3
C
"--\
---
HO:
H�
H
",CH(CH3h
'I,
IIC::::: C"
�
"'
H3C
CH3
..
"
,,\
E
(Saytzeff product-more highly substituted)
Cope elimination
..
,CH3
" "
IIC-C
� - "'
H3C
CH(CH3h
H"
'1
,\\
/
Z
(Saytzeff product-more highly substituted)
19-23
o
+
Aliphatic diazonium ions are very unstable, rapidly decomposing to carbocations.
(b) /' " NO
N
(c)
�
19-24
i
(d)
NO
I
o
+
N2
6
ClAryl diazomum IOns are
relatively stable If kept cold.
The diazonium ion can do aromatic substitution like any other electrophile.
�
� N=NV� N=N
�
--- -03S -o-oN(CH3h
�(CH3h �
HO
.
. ..
plus two other resonance forms
most slgmflcant resonance contnbutor ..
with positive charge on the ring
..
:Cl:
03S
_
-
03S
j
H
I+
_
_
(or some other base)
-< >- N N -< >- N(CH3h
=
methyl orange
445
}
19-25
(a)
(b)
8
HBF4
NaN02
HCI a
CI
N2+ CICuCI
•
a
-
t::..
6
..
from (a)
(c)
N2+ CI-
F
6
°
°
II
�
II
HN-CCH3
2
*
CH3C-;I : I CH31- CH3'¢rI CH3 H30+ CH3 � I CH3
NaN%
2 HCI CH".
CH3
H
CI
2
N
q
CH3 CH3 H3P02 CH3 CH3
�I
�I
'¢r
CH3
CH3
N2 CIB
N2 CICuBr
8
(SCCHl
3
NH
�
t::..
+
9'
•
+
(d)
a
..
(h)
+
N2 CI-
a
+
N2 CI-
+
r
from (a)
(f)
9'
..
CuCN
..
6
CN
(g)
6
a HO� }-OH
+
(e)
a
I
KI
..
+
N2 CI-
H2O
a H2SO4
..
..
t::..
< }-N'Ni }-OH
HO
446
6
OH
6
delinNa(CH3COOhBH
es for choice of reagent
forutioreduct
ivethamie imniatnieoorn: iuseminLiiumAliHon4 when
tihseoliamteid.ne or
oxiAltmernate iGeneral
siviselolya, tcateguid.alUse
i
n
sol
n
when
i
s
not
ytic hydrogenation works in most cases.
0
(a)
(i' H
NOH
0
NH2
H2NOH
Li
A
l
H
4
PhCH2 -C -CH3
PhCH2 - C -CH3 H2O PhCH2 -CH -CH3
(c) H
CH2Ph
0 Na(CH3COOhBH
0 Ph)l H
0
N,Ph
(d) 0
NHPh
LiAlH4
H+
H2O 6
6
6
NOH
(e) 0
H2NOH
LiAlH4
OR
H2O
6 6
H
19-26
(b)
II
if
1)
II
..
2)
I
I
..
+
W
2)
I
0
+
+
1)
-.
..
2)
0
Cl �
CI 'l)
---
a
..
2)
1)
..
19-27
(a) H
6
I
I
+
(b)
..
c5
0-r
c�H
LiAlH4 r"
00 H20 0
HN�
HNJ)
6 I)UAIH20H� 6
:::::...
1)
..
2)
A
---
A
2)
:::::...
447
le alkylations of each NHnitr2ogen.
Use a large excessBrof ammonia to avoid multip�
NH3
� excess
19-28
�N:
19-29
(a)
+
____
C<N- CII2Ph NH2NH2 H2NCH2Ph
Br(CH2hCH3 c<N - (CH2)sCH3 NH2NH2 H2N(CH2)sCH3
o
o
BrCH2Ph
..
o
(b)�
o
v--t
o
-
:::-...
..
!l
o
I
!l
•
(c)� Br(CH2hCOO- �
N: must use anion :::-... I N -(CH2hCOO- NH2NH2
H+
� tthoeavoiphthdalprotimideonataniingon 0
Assume that LiAIH4 or Hz/catalyst can be usedH interchangeably.
PhCH2Br NaN3 PhCH2N3 Pt2 PhCH2NH2
� Br � � C = N LiAlH4 � NH2
H20 • V
V
V
(c) 0 LiAIH4
TsCl
Ts
H20
pyridine �
�
+ NaN3
Li
A
I
H
4
�NH2 H20 �N3
0 C2
NH3 �
�
� Cl
NH2
LiAIH4 + H20
�NH2
OTs��C=N LiH20AIH4 �NH2
�
from (c)
o
o
o
o
' !l
...
1)
-4_
----
..
2)
o
19-30
(a)
+
..
--
(b)
1)
2)
OH
1)
2)
•
�
OH
O
..
1)
•
2)
OR
OH
�
SO l
..
o
o
-
1)
-
1)
2)
448
..
2)
H2N(CHlhCOOH
continued
(e) Br\�H NaN3
H N3 LiAIH4 �
�
�
�
inversion
Br\�H NaCN
�H2NH2
Li
A
I
H
4
�
inversion
(g) 0
HO�
CN
H
�
Li
A
I
H
4
KCN
� HCN
Tortualreduce
nirchangea
troaromatblyic. s,Assume
the reducia workup
ng reagentin base
s (H2toplgiusvea tmethe freeal catamialynste, fiornala metproduct
al pl.us HCI)can be
used
vi
l
y
i
n
t
e
(a)
nHCl
o
Br
Br
Br
(b)
FeHCI
6
o
N0orth2o
nHCI ))
f;
U Br
U Br
from (a)
COOH
COOH
(d)
FeHCI
Jr
6 U NO
CH30 y-:�H :0:
�' 0
R-C- � : R-C=N,: R-C N) Br
hCH T-C � H
H
H
CH3
H
H
(
,
- H
19-30
SN2-
R
(D
SN2-
R
..
1)
..
1)
NH2
•
S
..
1)
..
S
..
19-3 1
S
•
¢ ¢
•
+
S
2)
19-32
P
2-
�
,
II
,
-
. �..
,"-,
2
II
.. -
449
----
..
:0: }
,
B,
U
II
..
'
�
mechan;,m cont;nued on next page : �
!
19-32 continued
:OH R-N=C=O
.. � --- BrII ..R-N=C-OH
R-C=N-Br
--R-C-N-Br
i :0:- -.. � U· i �U
(:0:- "" .. t
t
t
�H OH
II �
-.. II
?� :OH2 R-N:I - CO2 ----l.. R-NI.R-N-C-OH H""OH.. R-..7� CH HJ
H
�
..
I
:0:
�
• •
:0:
'-,1
:0:
..
..
+
-
�
.
H
H
'"
at
a
chi
r
al
carbon
i
s
l
o
st
i
f
t
h
e
carbon
goes
t
h
rough
a
pl
a
nar
Stinteereochemi
s
t
r
y
�
NH2
rmedi
a
t
e
,
ei
t
h
er
a
carbocat
i
o
n
or
a
free
radi
c
al
.
Confi
g
urat
i
o
n
at
a
chi
r
al
carbontcane thebeleiavinvertngegroup.
d durinHowever,
g substitutwhen
ion bythaenuclcarboneophiretleaifrom
thfoure sidpaie rs of
opposi
n
s
al
l
H CH3
electrons, as in this Hofmann rearrangement, it retains its configuration.
1(a)9-34The acyl azide of the Curtius rearrangement is similar to the N-bromoamide of the Hofmann
rearrangement
hat bothehave
an amiydsiesnithtrrough
ogen withethcarbami
a good cleaciavidngtogroup
mimechani
gratiosnms.to theinistocyanat
and hydrol
the amiatntaeched.
are idSubsequent
entical in botalhkyl
(manb) Theor beast
leavi"n, gasgroup
in thtoe say.Curtius rearrangement is N2 gas, one of the best leaving groups known "to
we
used
(c) ,----------------,,
e.g.,
R
.-----------------,
abbreviate as "R"
19-35 Please refer to solution 1-20, page 12 of this Solutions Manual.
6 mary amine; 2,2-dimethylpropan-l-amine, or neopentylamine
((b)19-a)3prisecondary
amierocyclne; iN-met
heylandpropan-2agroup;
mine, or3-niisopropyl
minetehylamine
t
e
rt
i
a
ry
het
c
ami
n
a
ni
t
r
o
t
r
opyri
d
iec oxiammoni
umhylio-n;N-N,mNet-dihylmaetnihliylnepioxiperiddeinium iodide
(d)(e) tquatertiaeryrnaryaromatheteicrocycl
ami
n
d
e;
N-et
ic amiic nammoni
e; N-ethuylm-N-ion;metpyrihyldaininiluimnechloride
(f)(g)tteertrtiiaaryryaromat
het
e
rocycl
(h) secondary amine; N,4-diethylhexan-3-amine
19-37 ShownH in order of increasing basicity. In sets a-c, the aliphatic amine is the strongest base.
(a) Ph-N-Ph < PhNH2 < o NH2 aliphatic amine is the strongest base
H
H
(b) N < N <
aliphatiicciamity woul
ne isdtbehe lstorstongest
base;
pyrrol
e's
aromat
i
f
prot
o
nat
e
d
0
pyrrole 0
(c) CNH < HN "N <
aliphatiicciamity woul
ne isdtbehe lstorstongest
base;
pyrrol
e's
\ I
aromat
i
f
prot
o
nat
e
d
pyrrole
Basi
c
i
t
y
i
s
a
measure
of
t
h
e
abi
l
i
t
y
to
NH2 <
N
2
H
O
(d) O
donat
e
t
h
e
pai
r
of
el
e
ct
r
ons
on
t
h
e
< I
I
amine. lElikeectN0ron-wi
thdrawibasingcity,
H3C
02N
groups
decrease
2
whileinelcrease
ectron-donat
CH3
basicityin. g groups
NH2 cr CH2NH2 aliphatic amine is the strongest base
(e) �
O
NH
2
<
I
I
< I
(amides are not basic)
(c)
I
I
�
�
h
h
°
h
h
like
h
Congrat
ritgyhton! CH3
is onlThey second
slightlystelructectruon-donat
ingbasibycin(byductabout
ion so1 tpKhe first
stbecause
ructureuNputlaHti2soinstsheleifleeyouctastron-donat
elgotectthronisinonedensi
t
h
e
re
i
s
more
g ofbyresonance.
resonance. (See
The ltahste stlarstucttoupireciisnthAppendi
e strongestx Twobaseofforthisa Manual
complet.e)lyThe
diortfferent
reason:
st
e
ri
c
i
n
hi
b
i
t
i
o
n
substconjituuentgatsedonwiNthandtheiaromat
t forcesicthpie NRsyst2em.out ofEssentplanariial tyy, twihitsh the
ribecomes
nhg,osometthanhatylalthiinepthateelrferes
ecticramine.
onwipaithrthonThee itshopropyl
epKbN iofs not2-met
hyly -alN,ipNhat-diiectamihylannielbecause
ine is 6.9th, eabout
2.5icpKrinunig istssomewhat
more basic
telheanctranionlwiine.thdrawi
It is notng asbystinrongductiaobase
as
a
t
e
rt
i
ar
aromat
make the N significantly more basin,cyet. the ortho methyl reduces the pi overlap of the nitrogen's electrons to
unit)
N.
N
451
19-38
(a) not resolvable:
(c) not resolvable:
(e) not
(g) not resolvable in conditions
proton on N can exchange
planar ric
symmet
resolvable: symmetric where the
pKa
he weaker
of the aciditywiandpKbth tbasi
city. )
19-39 The values of
side
reaction
discussion of
(a )
vvablablee:: asymmet
rnicversicarbonon is very slow
(b) resol
resol
ni
t
r
ogen
i
(h)resol
resolvvablablee:: asymmet
asymmetrriiccninittrrogen,
ogen, unabl
unablee to iinnvertvert
be favored at equilibrium. (See Appendix forThea
�
�JII + CH3COOpKa
I[J + CH3COO­
pKa
(d)
(f)
to
of amines or
of the conjugate acids can be obtained from Table 19-3.
acid and base will
2
products are favored
N+
I
H
(b)
I[)
..
--
N
H
(c )
d)
H
.N,
reactants are favored
H
"'- 1
o
H
(
8.75
I
O
pKa
+
NH3
�
+
4.60
19-40
(a) PhCH2CH2NH2
Q
H
o
N
products are favored
+
NH2
pKa
n
OI pKaH N+,H
�
+
....
1 1.27
( )
e
19-4 1
(a) PhCH2CH2CH2NH2 (b) �NH2
(e) (D
CH3
1 1. 12
�N
I
452
e)
(
products are favored
cr(\�
d��-
(h)
NH,
retentgiuraton ofion
confi
CHJ
eN,
(d
)
o
H2after workup
N
OI with base
�
OH
og
19-4 1
(i)
continued
(I)
U)
NHCH1
(m) /"'N�
CH2NH2
I
�
LiAIH4
..
H20
HO CN
CH2CH2NHCHl
- �
HO CH
19-42
(k)
(b) CH1
Q
2NH2
N02
'O
from (a)
(c) CH1
'O
from (a)
(d) CH1
'O
0
CN
+
N2 Cl-
Kl
+
N2 Cl-
H2S04
..
H20
�
(f)
CH3
I
'0I
�
CH1
•
0
+
°
�
CN
I
o
'0I
�
CH3
NHCOCH3 HN03
H2S04
CH3
after workup
with base
°2N
NH2
I
/
CH3 � OH
°
CH3
NH2
CH3C-Cl
II
0
o
---
from (a)
(e) CH3
CH3 � CH2NH2
LiAIH4
..
2) H20
1)
qH
OCH2CH3
+
0
I
(0)
CH3
(a) CH3 � NH2 NaN0 CH3 � N2 Cl2
Cuctt
HCl
o
NHCH3
CH3(CH2)3CHCH2CH3
(n)
�
PhCH2CHCH3
(p)
0
=<J Na(CH3COOhBH453
0
?'
�
I
NH2
°2N
0I
�
+
JJH3O
CH1 �
o
!1
� --o
NHCOCH3
(+ isomer)
19-43 This fragmentation is favorable because the iminium ion produced is stabilized by resonance. Also,
there are three possible cleavages that give the same ion. Both factors combine to make the cleavage facile,
at the expense of the molecular ion.
t?
+
.
CH3
CH3
I
-NH2
CH3
---
CH3
oCH3 +
mass 15
....
..f-.
-l..
�
mlz 58
mlz 73
19-44
(b)
0
O
�I
�I
�-o
Cl-C
..
NH2
+
1)
(e)
(I)
I
CH3
58
(a)
I
CN-CH3
C
N-CH3
H202
H202
..
..
/N�o
C CH3
/o
N�
C CH3
..
LiAlH4
t:.
----
(g) CH3 Y'1( COOH SOCl2
o
..
454
(
oH
N.
"'CH3
+
C = NH2
19-45 The problem restricts the starting materials to six carbons or fewer. Always choose starting materials
with as many of the necessary functional groups as possible.
HO-Q-NH2
(a)
phenacetin
Mg
-< )
HO
NBS
O
..
HO
HO
19-46
O
HO
HO
(c)
CH2CH2NH2 ..
H2, Pt
dopamine
455
HO
methamphetamine
CH2B'
�
continued
19-46
(b)
H
H
2
2
N� 00=7 Pt
�n C
--­
o
H
r'Y:l -­
��: !)
I
H
H :.
l bas :
co
HH
19-47
(a)a0C
'OH SOCl2
� H+
H NH2
TI
1
- H20
..
1
• •
..... N.
co
+
I
H
t
H2
Pt (X)�
�
�
N)
..N base: �
---
1
R:
NH2
Wco
H+O: H
HO:..�
1 }
H
1
H2
CO
) NH,
�
+
o
o
1
H)
,
C ...Cl NH3 o
C
CH2NH2
Li
A
I
H
4
NH
2
r?'Y
r?'Y
2)H20 0
aI
(b) c'
a CH2NH2
Na
H3COOhBH
(C
H
I
aI NH3
(c) C
LiAIH4 N - CH2CH3
NH CI-C-CH3 CN-C-CH3
2)H20 " C
�
"
•
I I
:::::-...
I I
1)
..
o
I I
:::::-...
+
�
o
II
•
::::
::::
: :-...
1)
456
..
continued r-, Na(CH3COOhBH 0- r-,
..
N0
0 o HN0
NH H2N � NH2
(e) HO� OH SOCI2... CI �CI �
o 0
o 0
/ 0
0
LiAl y 2)H20
19-47
(d)
+
_
1)
0
19-48
(a) �Br
CH2CH3
(b)
6
�
(c)
+
-N
HNO] ..
H2SO4
I
�C
0
0
--
NH2NH2 �NH2
CH]
..
L1
Q QNH2
I
�
� OH
Ph
H3
H2N�NH2
N02
SOCI2
...
Sn '"
HCI
NH3 � NH2
0
0
Nao�
� H2
N
�
NH2
NH2NH2
____
� CI
Ph
Ph
Ph
(d) �Br
+
Ph
-�
o
--
L1
...
Ph
(e)�Br KCN � CN 2 iAOH4"' � NH2
)H2
Ph
...
I
Ph
457
1) L
l
Ph
19-49
(a) When guanidine is protonated, the cation is greatly stabilized by resonance, distributing the positive
charge over all atoms (except H):
..
�
w
..
NH
II
H, N-C- NH,
1
�
..
NH2
II
..
..
NH2
�
I
..
+
H,N-C- NH, �H,N- - NH, �H'N=
�
NH2
I
NH,
• •
• •
NH,
I
+
}
H2N-C=NH2
.
(b) The unprotonated molecule has a resonance form shown below that the protonated molecule cannot
have. Therefore, the unprotonated form is stabilized relative to the protonated form. This greater
stabilization of the unprotonated fonn is reflected in weaker basicity.
..-o�
H2N
o·
.
+ //
N
\
-
..
:0:
(c) Anilines are weaker bases than aliphatic amines because the electron pair on the nitrogen is shared with
the ring, stabilizing the system. There is a steric requirement, however: the p orbital on the N must be
parallel with the p orbitals on the benzene ring in order for the electrons on N to be distributed into the TC
system of the ring.
If the orbital on the nitrogen is forced out of this orientation (by substitution on C-2 and C-6, for example),
3
the electrons are no longer shared with the ring. The nitrogen is hybridized sp (no longer any reason to be
2
sp ), and the electron pair is readily available for bonding � increased basicity.
H
\/
H
C ___
As surprising as it sounds, this aniline is about as
basic as a tertiary a liphatic amine, except that the
aromatic ring substituent is electron-withdrawing
by induction, decreasing the basicity slightly. This
phenomenon is called "steric inhibition of
resonance". We will see more examples in future
chapters. A lso, it is the last topic in Appendix 2.
(See another example in the solution to 19-37(f).)
H
458
19-50 In this problem, sodium triacetoxyborohydride
(a)
/'.... /'....
........, -O
/
( b)
TsCI
---
H
pyridine
will be represented as Na(AcOhBH.
KCN
. /'...
/'.......
. /'...
/'.......
...., -. OTs --...., -. CN
..
1) LiAIH4
/
/
2)
H20
/'.... /'....
/
........,
........,
_
H C H2
3
--- �
..
3 �NH2
cr03
H2
�H 2 0..
H20
�
0
CH3-CH
tpcc
CH3CH20H
"
-H
�
�CH20H
(r � a Na0't
V PctV
aCH2NHCH2CH2CH]
�NH2
(rCOCI ac-" NH2
(rCH3 �
I
I
I
PCC
�
OH
(c) �OH
TsCI
O
�OTs
HN
Na(AcOhBH
NH
�
NHCH3
•
pyridine
H S 4
O
./"--./
II
o
Na(AcOhBH /"'--.../
+ �N
I
HN�
•
(ACOhBH
/"'--.../
I
I I
�N�
(d)
CH3
o
C
CH2Br
•
Na(AcOhBH
from (c)
(e)
o
COOH
•
KMn04
�
o
·
---?"
�
NH3
j
�
Br2
NaOH
N02 � NH2
Na
..
0
0
�HNO] �
HCI
(I)
0
�
I
.&
0
�
�OH
AICI3
CI
•
H2Cr04
I
.&
SOCI2
.. ---
0
•
�
.&
H2SO4
°
N02
CI
459
J
zna[g
HCI
Clemmensen
reduces both
y
.&
NH2
NH2
o� y� � � IuY
U
,? 5 Y
Y
�oI
°
19-50 continued
(g)
'
I
AICl3
/
H2C
OH
19-5 1
'
Zn(Hg
°
Cl
II
CI- CCH2CH2CH3
Zn(Hg)
�
b( )
O
Ph
-Q�
19-52
h
HCI
HN03 Fe, HCI
•
�
H 2SO4
h
H2N
after workup with base
°
(a)
(c)
f)I
I
l
S
� '''''
•
'--C
N-CH1
Hofmann elimination
�
N02
HCI
t)
�
HCI
H
Ph
-Q-
NH2
H
I
coniine
460
19-53
unknown X
(a)
-fishy odor
�
amine
-molecular weight
10 1
�
�
odd number of nitrogens
if one nitrogen and no oxygen, the remainder is C6H)S
Mass spectrum:
-fragment at
86
=
M
-
15
=
loss of methyl
�
have this structural piece: CH3
IR spectrum:
-no OH, no NH
�
� a-cleavage
N
the compound is likely to
-
must be a 3° amine
-no C=O or C=C or C=N or N02
NMR spectrum:
--only a triplet and quartet, integration about
3
:
2
�
ethyl group(s) only
assemble the evidence: C6H)SN, 3° amine, only ethyl in the NMR:
(b) React the triethylamine with HCI. The pure salt is solid and odorless.
(CH3CH2hN
+
HCI
+
-.
(CH3CH2hNH
salt
CI -
(c) Washing her clothing in dilute acid like vinegar (dilute acetic acid) or dilute HCI would form a water­
soluble salt as shown in (b). Normal washing will remove the water-soluble salt.
---
yQ
o
H
H
(b) Both answers can be found in the resonance forms of the intermediate, in particular, the resonance
form that shows the positive charge on the N. This is the major resonance contributor; what is special
about it is that every atom has a full octet, the best of all possible conditions. That does not arise in the
benzene intermediate, so it must be easier to form the intermediate from pyrrole than from benzene.
Also, acylation at the 3-position puts positive charge at the 2-position and on the N, but never on the
other side of the ring, so this substitution has only two resonance forms. The intermediate from acylation
at the 3-position is therefore not as stable as the intermediate from acylation at the 2-position.
461
19-55
(a)
OZN
'Q
000
F
�
�
I
oo
N H zR
0 0
:0'"
..
..
+
NH2R
o o F
li
_
Y
'c.
N:"
I
0
�
+
... N....
N02
0'"
y
�
0
0
-
0
: 0'"
0
�..
� 1j
... ..
...
0o-;
1
N+
�
_
0 '"
0
+
N....
(b) Why is fluoride ion a good leaving group from A but not from
F
_
HC
Nu
I
N02
NOz
A
��
A
B
0II
+
N HzR
F
I
-0'"
�
_
N+
+
NH
2R
F
I
00
:0'"
0
C1
N+
I
0 0-
0:
B (either by SNI or SN2)?
:NU
C
Formation of the anionic sigma complex A is the rate-determining (slow) step in nucleophilic aromatic
substitution. The loss of fluoride ion occurs in a subsequent fast step where the nature of the Jeaving group
does not affect the overall reaction rate. In the SN 1 or SN2 mechanisms, however, the carbon-fluorine hond
is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the
rate.
t
t
>-.
lrl�
________________
nucleophilic aromatic
substitution
(c) Amines can act as nucleophiles as long as the electron pair on the N is
available for bonding. The initial reactant, methylamine, CH3NHZ' is a very
reactive nucleophile. However, once the N is bonded to the benzene ring, the
electron pair is delocalized onto the ring, especially with such strong electron­
withdrawing groups like N02 in the ortho and para positions. The electrons
on N are no longer available for bonding so there is no danger of it acting as a
nucleophile in another reaction.
462
delocalized ,
N02
19-56
Compound
A
Mass spectrum:
-molecular ion at 73
=
=
odd mass
odd number of nitrogens;
if one nitrogen and no oxygen present
- -, t
-base peak at 44 is M - 29
N
I
44
�
�
molecular formula C4H11N
this fragment must be present:
CH3-
CH2CH3
�
EITHER
OR
�
H-C-CH2CH3
I
H
a-cleavage
2
IR spectrum:
-two peaks around 3300 cm-I indicate a 1 ° amine; no indication of oxygen
NMR spectrum:
-two exchangeable protons suggest NH2 present
-I H
A
multiplet at 8 2.8 means a CH-NH2
The structure of A must be the same as 1 above:
Compound B
an isomer of A, so its molecular formula must also be C4H11N
IR spectrum:
-only one peak at 3300 cm-I
NMR spectrum:
-one exchangeable proton
-two ethyls present
The structure of B must be:
�
�
2° amine
NH
B
+
-
CH2 = NCH2CH3
I
H
rnJz 58
resonance-stabilized
19-57
(a) The acid-catalyzed condensation of P2P (a controlled substance) with methylamine hydrochloride gives
an imine which can be reduced to methamphetamine. The suspect was probably planning to use zinc in
muriatic acid (dilute HCl) for the reduction.
0Jl
+
-
Cl
methamphetamine
phenyl-2-propanone, P2P
(phenylacetone)
(b) The jury acquitted the defendant on the charge of attempted manufacture of methamphetamine. There
were legal problems with possible entrapment, plus the fact that he had never opened the bottle of the
starting material. The defendant was convicted on several possession charges, however, and was awarded
four years of institutional time to study organic chemistry.
463
1 9- 58
Mass spectrum:
-molecular ion at 87
=
odd mass
-if one nitrogen and no oxygens
-base peak at mJz 30
�
=
�
odd number of nitrogens present
molecular formula CS H13N
R
structure must include this fragment
IR spectrum:
1
-two peaks in the 3300-3400 cm- region
�
t
H
H
I
\
-.. C = N+
\
/
H
H
CH2NH2
30
mJz 3 0
1 ° amine
NMR spectrum:
-singlet at 80.9 for 9H must be a t-butyl group
-2H signal at 8 1 .0 exchanges with D20
�
80.9
must be protons on N or
{
°
8 1.0
CIl3_
�
H3
CH2-
� t
H3
l
,
82.4
Note that the base peak in the MS arises from cleavage to give these two, relatively stable fragments:
CH3
I
CH3 -C·
1
CH3
+
H
H
/
\
C=N+
/
H
\
H
mJz 30
19-59 (a tough problem)
molecular formula CII Hl6N2 has 5 elements of unsaturation, enough for a benzene ring; no oxygt:ns
precludes N02 and amide; if C:::N
:: is present, there are not enough elements of unsaturation left for a
benzene ring, so benzene and C=N are mutually exclusive
IR spectrum:
--one spike around 3300 cm-1 suggests a
2° amine
-no C=N
-CH and C=C regions suggest an aromatic ring
Proton NMR spectrum:
- 5H multiplet at 8 7.3 indicates a monosubstituted benzene ring (the fact that all the peaks are huddled
around 7.3 precludes N being bonded to the ring)
-1 H singlet at 8 2.0 is exchangeable
�
NH of secondary amine
-2H singlet at 83.5 is CH2; the fact that it is so strongly deshielded and unsplit suggests that it is between
a nitrogen and the benzene ring
< >-
fragments so far:
CH, -N
+
NH
continued on next page
+
4C
+
8H
464
+
I clement of unsaturation
19-59 continued
Carbon NMR spectrum:
-four signals around 5 125-138 are the aromatic carbons
-the signal at 5 65 is the CH2 bonded to the benzene
-the other 4 carbons come as two signals at () 46 and () 55; each is a triplet, so there are two sets of two
equivalent CH2 groups, each bonded to N to shift it downfield
fragments so far:
CH2
CH2-N + NH + CH2
CH2 + CH2 + 1 element of unsaturation
<}
There is no evidence for an alkene in any of the spectra, so the remaining element of unsaturation must be
a ring. The simplicity of the
spectra indicates a fairly symmetric compound.
Assemble the pieces:
19-60
(a)
Ph
H
NMR
� N�
Jl
(b)
�H CH3
�a(AcO h BH
0
11
�
V
(
N�
(
HN�
465
1 9-6 1 Not only is substitution at C-2 and C-4 the major products, but substitution occurs under surprisingly
mild conditions.
Begin by drawing the resonance forms of pyridine N-oxide:
..:0:
:0:
:0:
:0:
1+
.. ..
0
11+
11+
. .. ..
..
..
OR (�)
N �
11+
RO
- N
CH
Resonance forms show that the electron density from the oxygen is distributed at C-2 and C-4;
these positions would be the likely places for an electrophile to attack.
-
Resonance forms from electrophilic attack at C-2
..
:0:
a
1++
N
E
.0
H
a .. (:f
:0:
II +
"
N does not have an
octet; not a significant
resonance contributor
1+
E
N
_
·. ·0··. ­
0
N
H "
HC
+
GOOD! All atoms
have octets.
Resonance forms from electrophilic attack at C-3
( \ ..
·. ·0··. ­
1+
N
H
+
C
H
-
E
1+
Q
H�
,,=
�
1+
H "
-
OR
H
E
..
continued on next page
�
N
H "
CH
-
�
E
+CH
H
These two Eresonance forms place twoE
positive charges on adjacent atoms-not good.
-
Resonance forms from electrophilic attack at C-4
....
:0:
1+
N'
c�
1+
E
a
..:0:
·. ·0··.­
N
0
.
·. ·0·....
:0:
1++
-
0
.. ..
H E
N does not have an
octet; not a significant
resonance contributor
466
:0:
11+
..
.-
.
:0:
1+
0 -Q
H E
GOOD! All atoms
have octets.
HC
+
H
E
H
19-61 continued
The resonance forms from electrophilic attack at C-3 are bad; only one of the three is a significant
contributor, which means that there is not much resonance stabilization. When the electrophile attacks at
C-2 or C-4, however, there are two forms that are good plus one great one that has all atoms with full
octets. Clearly, attack at C-2 and C-4 give the most stable intermediates and will be the preferred sites of
(f0: �C-!l
attack.
19-62
(aJ
O
�
<>?: 0/9.:
H
H
I
I
....
..f--l
.
.
�
H� NC I!3
CH3
H-N-H
- Q.H
+ I�
O
1�
B- is the conjugate base B:
of the acid HB
CH3
CH3
I
I
H-N:
H - N:
C-OH
H
C
'-""1+
H
,...... d 9..
�
d··
�
-
�
called an aminal
continued on next page
467
19-62
(b)
cr
.�
continued
O:'�
H -B
+C
..
-
q
B- is the conjugate base
of the acid HB
0
C-OH
• •
11
B:
. -H20
.
/ �H�_ O�:·· +
o�:.
aD
-' �� �
0::0 � (X/O
C
o
C
/'
I
-
H
OH H�B
-
0
C
-
OH
enamme
To this point, everything is the same in the two mechanisms.
But now, there is no H on the N to remove to form the imine.
The only H that can be removed to form a neutral intermediate
is the H onC next to the carbocation.
(c) A secondary amine has only one H to give which it loses in the first half of the mechanism to form the
neutral intermediate called an aminal, equivalent to a hemiacetal. In the second half of the mechanism, the H
on an adjacent carbon is removed to form the neutral product, the enamine. The type of product depends
entirely on whether the amine begins with one or two hydrogen atoms.
468
CHAPTER 20-CARBOXYLIC ACIDS
20-1
CH3
I
(a) CH3CH2 CHCOOH
(d)
(g)
(
)
COOH
Cl
�
'>
H
tH
(c)
OH
cf:�:�
(f)
H
(h)
CI
COOH
N
� COOH
.
(1)
CH3
(j) HOOC � COOH
Cl
(k)
o
� COOH
(I)
N H2
I
COOH
HOOC �
-
0(y 0 H
V a
20-2 IUPAC name first; then common name.
(a) 2-iodo-3-methylpentanoic acid; a-iodo-p-methylvaleric acid
(b) (Z)-3,4-dimethylhex-3-enoic acid
(c) 2,3-dinitrobenzoic acid; no common name
(d) trans-cyclohexane-1,2-dicarboxylic acid; (trans-hexahydrophthalic acid)
(e) 2-chlorobenzene-\ ,4-dicarbox ylic acid; 2-chloroterephthalic acid
(f) 3-mcthylhexanedioic acid; p-methyladipic acid
20-3 Listed in order of increasing acid strength (weakest acid first). (See Appendix 2 for a review of acidity.)
Br
CH3 - CCOOH
(a) CH3CH2COOH < CH3 - CHCOOH <
Br
Br
The greater the number of electron-withdrawing substituents, the greater the stabilization of the carboxylate
ion.
(b) CH3CHCH2CH 2COOH < CH3CH2CCH2 COOH < CH3CH2 CH2 CHCOOH
I
I
I
Br
Br
Br
The closer the electron-withdrawing group, the greater the stabilization of the carboxylate ion.
(c) CH3CH2 COOH < CH3-CHCOOH < CH3 - CHCOOH < CH3-CHCOOH
N02
o
C�N
The stronger the electron-withdrawing effect of the substituent, the greater the stabilization of the
carboxylate ion.
I
I
I
I
I
469
I
�COOH
�CHO
�
CH20H H2S04
"Y
shake with ether and water
ether
water
�COOH
isoln etublhere �CH20H
�CHO
shake with NaOH (aq)
ether
NaOH (aq) ionized form
�
CH2OH
ca�
b
oxyl
i
c
t
a
COON
NaOH �
byunchanged
sol
u
bl
........",- -CHO acidify with HCl (aq) and ether
in e
HCl (aq)
ether
NaCl
�COOH
evaporate ether t
�COOH
20-4
{
/"'-..
./
/"'-..
........",-
of
aCId is
water
/"'-..
470
20-5
The principle used to separate a carboxylic acid (a stronger acid) from a phenol (a weaker acid) is to
neutralize with a weak base (NaHC03), a base strong enough to ionize the stronger acid but not strong
enough to ionize the weaker acid.
�-o- OH
< }- co,
0°
y
shake with ether and NaHC03 (aq)
NaHC03 (aq)
ether
< }- cooNa
A
,-------- -------�\
� OH +
CH3
\..
-Q-
y
!
00
)
1. add HCl
2. filter or extract
� COOH
�
- pure
shake with NaOH (aq)
NaOH (aq)
ether
-Q- oNa
! 2.1. filter
add HCl
or extract
CH3 -Q- 0H
00
!
CH3
evaporate
Q�
pure
471
20-6 The reaction mixture includes the initial reactant, reagent, desired product, and the overoxidation
product-not unusual for an organic reaction mixture.
< :N
(a)
eHO
COOH
er03
H
e
2
� H _0- -----------)
�
�
�
,,
____________
ether
�eOOH
"N
C
y
(
)
shake with ether and water
some compoundS
have appreciable
solubility in both
ether and water
water
o
�
b.p. 137°
eHe 20H
�eHO
b.p. 102° e
(b) Pentan-l-ol cannot be removed from pentanal by acid-base extraction. These two remaining products
can be separated by distillation, the alcohol having the higher boiling point because of hydrogen bonding.
eOOH
20-7 The
has a characteristic IR absorption: a broad peak from 3400-2400 cm-I, with "shoulder"
around 2700 cm-I. The carbonyl stretch at 1695 cm-I is a little lower than the standard 1710 cm-I,
suggesting conjugation. The strong alkene absorption at 1650 cm-I also suggests it is conjugated.
a
472
20-8
(a) The ethyl pattern is obvious: a 3H triplet at b 1.15 and a 2H quartet at b 2.4. The only other peak is the
COOH at b 11. 9 (a 2.1 b uni t offset added t o 9. 8).
o
II
CH3CH2-C-OH
(b)
o
Q
II
£
H-C-CH2CH3
.i!
£
Q
2H
.i!
IH
1 :1
3H
II II
TMS
I
8 7 6 b (p5pm 4 3 2 1 0
)
Theand tmulhe CH2tipletgroup.
betweenThese2 andcoupli3ngis const
drawnantass area pentprobabl
et as ythough
it wer, in ewhisplciht equal
lhyebyactthuealalspldehyde
protteornn
unequal
case
t
i
t
i
n
g
pat
wi(c)l Thebe achemi
complcealx shimulfttiofpletht.e aldehyde proton is between b 9-10, not as far downfield as the carboxylic acid
proton. Also, the aldehyde proton is split into a triplet by the CH2, unlike the COOH proton which always
10
I
9
I
I
<5
I
I
I
I
I
I
I
I
<5
e iins spltheitacibydan. extra proton, so it wil give a multiplet with complex
splappearit insg,asinastsieadnglofet.thFie nquartally,etthshown
CH2
20-9
:OH
+ C1 ........
/ OH
H2C==C
.
:OH
• •
\
t
H
.
I
..
..
..
+OH
H2C==C
..
:OH
II
C .
/ ....... OH
\
H
C ........
h
+
:I"
OH
H2C-C\
H
I
..
..
+
/C�
OH
""'
H2C==C
\
H
473
[ CH3CH2tCH' H0-OH ]
20-10
ig
87 CH3
m/z1l6
McLafferty rearrangement
H3C I1�H� :
�ft'C 'OH
CH3
m/z1l6
t
:OH
CH3masCs H,· H2C==C/C,9,..H
CH3
plasushown
s resonance
im/z n8720-9forms
H,0..:
H3C,CH
CH2 HC/',/.C,OH
mass 42 CH3
29
-
+
+
I
\
t
t
--
II
I
+
I
mlz74
(a) �C C� orconc.1 ) 03,KMn04
2) H20 �COOH
C-:::) conc. KMnO� C:COOH
COOH
H 0+
CH2CH20HH2Cr04 6CH2COOH
(c )
< }- Br etMgher < }- MgBr D H30+ 6
(d) � PB r3 �
Br Mg CO2 H30+ COOH
�
ether
COOH
(e)
conc. KMn04
t"H30+
CH3
COOH
Mgether CO2 H30+ �COOH
OR
KCN �CN H30+ �COOH
20-11
•
(h)
t, ,
3
o
�
-­
�
(f)
--
Q
�I
�I
•
--
---
•
¢
---
---
�
•
t,
•
474
�
I
•
�
I
+
..
..
}
:O-H
:O-H
:O-H
first intennediate RC-�H RC+- OH RC==�H
i
:O-H :O-H iY-H
second intennediat RC(
RC
RC
:OR..
+OR
:OR..
(b) Theomechani
smthofataciared-catalreaadylyzedvernucly famieophiliarlitco acylyou.substitution may seem daunting, but it is simpl)'
successi
n
of
st
e
ps
mechanileavismsng,havewithsiaxlistt eleps:resfouronanceprotstoanbitlriansfzatioersn t(htrwoownon,intwthoatoff)makes, a nuclthe eophilic
atwork.tack,TypiTheandcaalsixlleyavi,sttehpsneseg argroup
e labeled in the mechanism below:
nuclprotprotooennophioffonl(ereatsonance
tacks stabilization)
lprotpreaviotoonnngoffongroup leaves (resonance stabilization)
i :O-H
+
:O-H+
:O-H
1
+
CH)C-OH.. CH)C-OH. CH3C=OH.. fL
1
/
H ........CH2CH3
O-H
CH3C-OH
.... H2CH3 H '" ..... CH2CH
H /R�
3
:O-H
O-H
+
- H20 CH3C:O-H
.
+
CH3C
CH3C
:RCH2CH3 +RCH2CH3 :�CH2CH3
20-12 (a)
·
II
•
i
•
�
1
�
/
\\
�
1
• •
+
\
�
}
a
whole thing
StepA
StepB
Step C
Step D
Step E
Step F
• •
II
•
•
_
•
.
�
• •
.. �
StepA
_
•
1
Step B
o
1
StepC
+
AI
StepD
---I��
Step E
1
l
/
\
�
475
/
\\
""0 .....
�
//'0
\
}
of MicroscopiOEtc Rever
Applyin(cg) thAle lstsetpsepsasaroute reversi
lined onble,thwhie previch isousthepage:reason(atbbrhe ePrviinacitinpgleOCH2CH3
) sibility applies.
-B B is the acid catalyst
nucIprotprotooenophin offonl(eresonance
attacks stabilization)
althoughly removes
in
protleavionngongroup leaves (resonance stabilization) :hydr- oilsysithserconjeactiuogatns,ewatbase,er usual
proton off
:O-H
1+
'OEt
20-12
as
H
StepA
Step B
StepC
StepD
StepE
Step F
O
\ :O-H
:O-H
c1�
H
C
C+ C 'OEt
a a �O:t (f
H+
/
: O-H
+1C
cr �
:O- H
�
I
+
CH
HC
+
I
�
:O-H
C-O:
aA)Et \
� H-Ok
�11
:OHQO� EtOH
:O-H
C-O:
same
seri
s
e
cr
�
+· OEt
1). \H
offorresonance
�
ms as above-- - � I ( H
Oil H1
II :OH, OH
(JC=O
1 + I)' O· ·H2
C-O:
aAEt H
:O - H H �
I
•
.
Step B
• •
St.pC
Step D
-
..
___
_-
StepE
StepF
I
476
+�
I
• •
20-13 For the sake of space in this problem, resonance forms will not be drawn, but remember that they
are critical!
'0'
'0'
Ph
XJ
�· + ....H+H
Ph
5
Ph � OH
H� CH 3
H
O ....
+
�
�
+
Ph1 OH
....0+
;» H ·'CH 3
H� CH 3 +
·0 .... H
Ph10 H
+
;» H .... "CH 3
H� CH 3
H
•
•
0h-t-OH
t
c: +
•
•
•
H
'CH 3
h
H+ �
H ... 0. ..... H
+ ,-
+H
Ph
o
'CH 3
+ -H20
+
Ph / c '" H plus resonance forms
O'CH 3
, HERE'S THE DIFFEREN CE!
CANNOT LOSE H+ TO
HR CH 3+
MAKE
CARBON YL
H "'" CH 3
�O
Ph H
O'CH
3
H� CH 3
..... CH 3
o
p
(+
4
• •
+
Ph
�H
o
•
'CH 3
H+
H ...0. ..... H
Ph
+ H�
·0 .... H
).? plus 6 resonance forms
+
0 plus resonance forms
).?
Ph �H
H�CH 3
H
O ....
+
X='
'!J
'CH 3
477
Ph
+OH
+
,-
o
'CH 3
t -H20
yO� plus 5 resonance forms
.\ CAN LOSE H+
'C � MAKECARBONYL
+H� CH 3
o
o
II
CH
�
Ph O"'" 3
TO
20-14
(a)
°
R-C-OH H+
'-
{R-C-OH
}
:0 :
� R-C-OH
II
II
�
• •
:0:-
++I
I
..
+I
H
.
H
BAD-two adjacent
•
positive charges
(b) Protonation on the gives only two resonance forms, one of which is bad because of adjacent positive
charges. Protonation on the
is good because of three resonance forms distributing the positive charge
over three atoms, with no additional charge separation.
OH C=O
: 0: � H+
RC-OH
i+
O-H+
: O-H
: O-H
:
+ � RC=9 H
RC-RH �RC-RH
.
II
II
�
I
I
• •
..
•
• •
•
t
(c) The carbonyl oxygen is more "basic" because, by definition, it reacts with a proton more readily. It does
so because the intermediate it produces is more stable than the intermediate from protonation of the
20-15
(a)
(b)
(c)
OH
Jv COOH
�
°
CH30H
use CH30H
as solvent
+
HC -OH useCH30H
CH30H
as solvent
II
+
OH
Jv COOCH3
H20
�
remove water with
�
+
molecular sieves
°
..
HC-OCH3
remove by
II
distillation
b.p.32°C
..
�
�
�
478
COOCH2CH3
+ H20
remove water with
molecular sieves or
by distillation
OH.
20-16 The asterisk (" * ") denotes the 18 0 isotope.
(a) and (b)
i
+
:O-H
"
..
•
•
CH3C-OII
,.:.---- O-H
• •
•
: O-H
�
1+
..
j
CH3C-OH
�
.. �
/ 0*
H ' " CH3
•
•
�
•
1
.. f
: O-H
I +
CH3C=OH
O-H
1
..
CH3C-OH
1
0 * CH3
H
1
�
/0*'
"
CH
CH3C-OH
1+
�
O*
/ '
H "
CH3
(c) The 18 0 has two more neutrons, and therefore two more mass units, than 160. The instrument ideally
suited to analyze compounds of different mass is the mass spectrometer.
..
20-17
+ O-Et
: O-Et
: o -Et
(a)
II
I
I
H-C
H-C
first intermediate: H -C+
\\ +
\
\
: O-Et
O-E t
: O-Et
..
I
----
H
: O-
second intermediate: H-C+
\
: O-Et
----
..
I
+ O-H
----
H-C
\\ +
O-Et
II
•
: O-H
----
•
H-C
\
: O-Et
The more resonance forms that can be drawn to represent an intermediate, the more stable the intermediate.
The more stable the intermediate, the more easily it can be formed, that is, under milder conditions. These
intermediates are highly stabilized due to delocalization of the positive charge over the carbon and both
oxygens. A trace of acid is all that is required to initiate this process.
479
20-17
continued
(b) :O :�
II
HC-OEt H+
i
..
�
IF
+
:O - H
II
HC-OEt
r
:O - Et
I
HC-OH
i
. .
:O - H
I
+
HC=OEt
• •
H/R'H t
. .
I
�
. .
:O-H
1+
-O t
OH
H-)5-Et
I)
HC-OH
I
20-18
EtOH
-
..
OH
(a) H 3C
'O
�
I
0
C ....
20-19
----
/
HC \
\
+OH
P-
r:-' : O-H
+O-H
o
H
//�
H-C
HC
\
OH
:OH
----
. .
"
\
OH
o
o
(a)
(b)
I I
:O-H
. .
:O- H
/
HC+
\
: RH
}
ar
?'
I
�
2
o
II
CH -C - Cl LiAl(O-t-BuhH
..
ar
?'
�I
480
2
o
II
CH -C - H
B214 selectively reduces a carboxylic
acid in the presence of a ketone.
Alternatively, protecting the ketone as
an acetal, reducing the COOH, and
removing the protecting group would
also be possible but longer.
�
:0 : Li+
'\
20-20
HO
\
R - C - R'
\
.
\
+ H+
R -C- R'
:6:
.
/
Li+
(b)
O
o:
Y
Ph
1-
Y
+ O-H
I IV
R - C - R'
_
H20
: O-H l
R- � - R' f
+
..
�
RC - OH
20-22
HO
..
··
H20:
-{ � _ H_2-10..
Li
2
H+
a hydrate
\I
R - C - R'
+
+1) H
H-�-
HR:---./
o
20-21
(a)
0
CH3CH2 - C - OH
HO
I
R - C - R'
..
Jt? �Cl
:0
�+ Cl If
0
+ � �.
�
H (-O �
rCl
...."
'0'
•
• •
-
•
Cl
•
1
:o�oj(
I'"
'0'
Ph
0
plus resonance forms
�Cl
I(
..
P�
-:O
'
-{
+ CPh;'0iYCI
Cl
0
tt :0.:
:0 :
HO,
• •
• •
I
.......f--. ..-' H+Oy O � CI
Cl
.
plus three other
resonance forms
with positive charge
on the benzene ring
� R �� �Cl
'0'
IrJ
Cl',/\0
ph
°
481
�
o
.. Ph )l Cl +
• •
(
Ph
O=C=O
+
Cl
_
0
+
20-23
(a) Co
:
II
:0 :
Cl + H-O-CH2
Ph -C�
.. CH3
CIIO:
3
(b)
H
�
CH -C-C1
d'
20-24
(a)
(b)
0
+
I
H-N-CH3
II
C
",1+
-
H-OCH2CH3
..
?j
CH3 -C-C1
:
�� R
t
:0 :
+1 U
H-N-CH3
I
• •
H
oH
� OH
0
Ph
-
o
� Cl
20-25 Please refer to solution 1-20, page 12 of this Solution Manual.
20-26
(a) 3-phenylpropanoic acid
(c) 2-bromo-3-methylbutanoic acid
(e) sodium 2-methylbutanoate
(g) trans-2-methylcyclopentanecarboxylic acid
(i) 7,7-dimethyl-4-oxooctanoic acid
(b)
(d)
(f)
(h)
20-27
(a) f3-phenylpropionic acid
(c) a-bromo-f3-methylbutyric acid,
or a-bromoisovaleric acid
(e) sodium f3-methylbutyrate
(g) o-bromobenzoic acid
(i) 4-methoxyphthalic acid
(b) a-methylbutyric acid
(d) a-methylsuccinic acid
(f) /3-aminobutyric acid
(h) magnesium oxalate
482
2-methylbutanoic acid
2-methylbutanedioic acid
3-methylbut-2-enoic acid
2,4,6-trinitrobenzoic acid
Ph
-
C-�CH2CH3
-8
20 2
0
(a)
CH3 - C - OH
(b)
II
(X
�
(e)
o
COOH
(
(c)
COOH
0
ClCH2 -C - OH
II
0-
)2
(h)
20-29 Weaker base listed first. (Weaker bases come from
H-t O-
< }- C - O- Na+
0
stronger
(f)
)
2
Mg2+
0
CH3-C - Ci
II
0
(i)
FCH2-C-0- Na+
II
conjugate acids.)
(a) ClCH2COO- < CH3COO- < PhO­
(c) PhCOO- Na+
20-30
(a)
(b)
(c)
(X
COOH
7"1
� COOH
CH3
+
2 NaOH
-Q- COOH
(X
�
I
COO- Na+
COO- Na+
no reaction
(d) CH3 - CHCOOH + CH3CH2COO- Na+
CH3- CHCOO- Na+
Br
Br
(e)
Na+ -0
COOH +
COO- Na+ + HO
o
031 0
OH
CH3C-OCH2CH3 <
< CH3CH2CH2 -C - OH
� O�
lowest b.p.
(b.p.
highest b.p.
The ester cannot form hydrogen bonds and will be the lowest boiling. The alcohol can form hydrogen
bonds. The carboxylic acid forms two hydrogen bonds and boils as the dimer, the highest boiling among
these three compounds
Listed in order of increasing acidity (weakest acid first):
(a) ethanol < phenol < acetic acid
EWG electron-withdrawing group
(b) acetic acid < chloroacetic acid < p-toluenesulfonic acid
Acidity increa�es. with:
(c) benzoic acid < m-nitrobenzoic acid < o-nitrobenzoic acid
1. closer proxImIty of EWG
(d) butyric acid < (3-bromobutyric acid < a-bromobutyric acid
great number of EWG
Cl
Br
increasing strength
COOH (electronegativity) of EWG
COOH <
COOH <
(e)
483
.-
I
2-
< }II
-1 �
n-
< }-
-1 �
II
143°C)
(n°C)
.
20-32
..
I
0=
(l62°C)
[yF
} 2.
3.
=
20-33
d derivatives
areareoftcommerci
en used aasl ya avai
test lofablelee),ctroni
c effects
ofarea series
ofmeasured
substituentbys:titthey
are fairSubsti
lAceti
y easitcuents
lyacisynthesi
z
ed
(or
and
pK
val
u
es
easil
y
riats ion.
a
on
carbon-2
of
aceti
c
aci
d
can
express
onl
y
an
inductive
effect
;
no
resonance
effect
3 hybridized and no pi overlap is possible.
possibTwo
le because
is
sp
the
CH2
drawntutedfromacitheds aregivenstronger
pKa valthuanes.acetiFirst,c acialldfour
substituents
are teludeectron­
wieletctron-wi
hdrawintconclusi
ghdrawi
becausenognsaleflcanfectfourbeinsubsti
.
Second,
the
magni
of thoen
creases
in
t
h
e
order
:
OH
Cl
CN
N02
.
(It
i
s
al
w
ays
a
safe
assumpti
that nitro is the strongest electron-withdrawing group of all the common substituents.)
20-34
H�
OH 0 0
c
aci
d
i
s
not
a
carboxyl
i
c
aci
d
.
It
i
s
an
exampl
e
(a)of a Ascorbi
"
structurel y acidi
callecdbecause
an ene-diofothel where
one carbonyl
of the OHgroup.
groups HOCH2
iSees unusual
adjacent
-ene
part
(c).
HO
OH
(b)of acetiAscorbic
aci
d
,
pKa
4.71, is alm ost identical to the aci d i ty
acid, acipKadi4.74.
'(ol
(c)givesThethcemore
c
H
wi
l
l
be
t
h
e
one
t
h
at,
when
removed,
di
more stable conjugate base.
R -<;l(0 _R -}:(O �� -}:(O� R -}:(O_R 1�O
:0 OH :0: OH HO OH
HO :0: HO 0:
t
more
THREE
onl
y
resonance
fOnTIS
t
d
c
i
aci
esonance
r
fOnTIS
e
0
»(
R 0 !. :
(d)present
In the assligthhtle conjugate
y basic pHbase,
of physi
ologiec,alwhose
fluid,structure
ascorbic canacid
is
ascorbat
:R
OH '\..J.
beof part
represented
(c). by any of the three resonance fOnTIS on the left
20-35
(b) 0(a)
(C) � (d) 2�COOH
�
CH2COOH
r; �
- CH20H
�
(e) o- COOH
Ph
(h) COOH
(g)
H �
CH3CH2 -CHCH20H
�
V-- COOH
(i) COOH
(k)
COOH
CHO
<
<
<
_
{
. .
_
•
•
. .
}
start here
�
(f)
.
_
. .
two
��
o
�o C
I
484
}
I
.
Q)
\
{
,
20-36
(a)�
Mg ..
ether
Br
(b)�
0
�H
(c)
conc. KMn04
Ag+ ..
NH3 (aq)
� OH
SOCI2
11) LiAlH4 ..
2) H30+
� COOH
KCN
OR
...
,./"-COOH
2
..
�, H30+
0
(d)
H30+
..
CO2
---
OR H2Cr04 ..
0
� OH
0
�C I
.. � H
0
LiAI(Ot-BuhH
OH
�
J
PCC
CH2N2
CH30H
OR
...
... �COOCH3
H+
OR
SOCI2
C H3 0 H
�COOH
.. �COCI
.. �COOCH3
(e) �
COOH
O'COOH
(f)
(g)
U
1
h
H30+
...
O' CH2OH
( COOH
+
( (X
�
--I
Diels-Alder
OR B2H6
...
QgHN'
CH2COOH SOCI CH3 NH2
�
... I
(h) CI
I
LiAlH4
..
CI
COOH
h
H2
--
PI
0
(X
485
CH3
CI
COOH
Hp+
.�
20-37
� }-
COOH
< }- OH
o- CH,oH
----V
_
0-,
shake with ether and HCl (aq)
('hCOOH, PhOH, PhCH20H)
+
PhNH3 Clshake with NaOH (aq)
and ether
ether
NaOH (aq)
NaCI
PhNH2
V
shake with NaHC03 (aq)
!
PhCOONa
shake with
NaOH (aq)
NaCI
ether
!
< }-pureCH20H
PhCH20H
evaporate
evaporate
shake with
ether and
HCl (aq)
< }-pureNH2
ether
evaporate �
COOH
PhCOOH
�
_
pure
•
NaOH (aq)
PhONa
NaCI
PhOH
486
evaporate..
pure
20-38
H "OH OH
1;(t N
(a)
"",
CH3
S
OH
+
racemic
(R + S)
0
(b) Isomers which are R,S and S,S are diastereomers.
20-39
TsCI KCN PhCH2CH2CN H30+.. PhCH2CH2COOH
(a) PhCH2CH2OH pyridine
�
P B r3
Mg CO2 H30+
..
PhCH2CH2COOH
PhCH2CH2Br
ether
(b)
CH3 Mg
CH2 HB
('-r( CH3
r
COOH
Br ether..
V
!
cr
(f
..
..
---t
..
�
c;cr�
o
(d)
h
()
..
..
(e)
COO
Br
,
W�
:
2
O�
1\
HO
6
Li
H_30_ +...
_ _
o
dD
487
� COOH
yV
o
(l
HO OH
W, �
..
20-40
(a) Mass spectrum:
-mlz 152 ==> molecular ion ==> molecular weight 15 2
-m1z 107 => M 45 => loss of COOH
-m1z77 => monosubstituted benzene ring,
H
-
-Q
K
IR
+
H H
spectrum:
-3400-2400 cm-i , broad ==> O-H stretch of COOH
-1700 cm-i ==> C=O
-1240 cm-i ==> C - O
-1600 cm-i ==> aromatic C=C
NMR spectrum:
� 6.8-7.3, two signals in the ratio of 2H to 3H ==> monosubstituted benzene ring
� 4.6, 2H singlet ==> CH2 , deshielded
Carbon NMR spectrum:
� 170, small peak ==> carbonyl
� 115-157 , four peaks ==> monosubstituted benzene ring; deshielded peak indicates oxygen
substitution on the ring
(b) Fragments indicated in the spectra:
COOH
mlz 14
mlz 45 and from IR
mlz77
This appears deceptively simple. The problem is that the mass of these fragments adds to 136, not
152-we are missing 16 mass units oxygen! Where can the oxygen be? There are only two
possibilities:
O - CH2COOH
CH2-O - COOH
<>
<}
�
<}
How can we differentiate? Mass spectrometry!
( } CHj O - COOH
(
8157
Yat
CH2COOH This structure is consistent
with the peak at 157 in the
91
93 J
carbonNMR.
phenoxyacetic acid
The mlz 93 peak in the MS confirms the structure is phenoxyacetic acid. The CH2 is so far downfield
in the NMR because it is between two electron-withdrawing groups, the 0 and the COOH.
(c) The COOH proton is missing from the proton NMR. Either it is beyond 10 and the NMR was not
scanned (unlikely), or the peak was broadened beyond detection because of hydrogen bonding with DMSO.
488
8
20-41
(a)
6
0
a-valemlactone
(b) : O�\
�
o
r �OH
V
t+ ....
O
.
. .
0:
20-42
CoII:�
Ph
/' C .......
Cl
---
o
6
tt='�:
/
\
O-H
o
Ph-C-O - C-CH3
II
:0:
0+
---
..
:0 :
I
+
Ph-C - O=C-CH3
I
I
Cl
:O - H
• •
�
:0:
.. Ph - CII - O - C - CH3
II
:cl
· · :� H �
-?:
plus two other resonance forms
• •
�
--q CH3
H30+
CH2CO 'l
carboxylic
Compound 1 acid
---
.
:0:
+
I
Ph - C - O - C - CH3
I
I
Cl
:O - H
.
• •
.
(s:
489
CH3
OH alcohol
Compound
2
.
Ph-C-O. - C - CH3
I
II
Ccl
:O+ -H
•
ester (an ester in a ring
is called a lactone)
CH3
acetal
0:
AD:
O=C
II
20-(a) 43
O· o·
CH3
• .
.
H-O :
H-O :
20-(b) 4Compound
3 continued has 8 carbons, and Compound 2 has 6 carbons. Two carbons have been lost: the two
1
carbons of the acetal have been cleaved. (This is the best way to figure out reactions and mechanisms:
find out which atoms of the reactant have become which atoms of the product, then determine what
bonds have been broken and formed.)
(c) Acetals are stable to base, so the acetal must have been cleaved when acid was added.
CH3
H30+
oAo
5 � CH3
4
CH2COOH
1
2
(d) The carbons have been numbered above to help you visualize which atoms in the reactant become
which atoms in the product. The overall process requires cleavage of the acetal to expose two alcohols.
The alcohol at carbon-3 can be found in the product, so it is the primary alcohol at carbon-5 that reacts
with the carboxylic acid to form the lactone.
3°
+ 0-H resonance stabilized;
H3C --'(
H
leaves the reaction
+
..
: OH
:·0 0
0
.
•
.
OH
� :.O
�
'----{
•.
-
.
O� +- H
c50:
_
+0
:O - H
�
H - O:
H
CH3
OH
,
....
..
11-.
-
..
o",-,,-H
CH3
OR 490
..
Q
O-H
CH3 two fast
OR H+ transfers
•
•
..
.Q
+0
:O-H
CR3"
OH
C\
H,+
H-.a �- H
:0
.Q·O
•
!
CH3
OR
H20
-
:O-H
C' +
...
CH 1
OR
20-words44 about
(A more complete discussion of acidity and electronic effects can be found is Appendix 2.) A few
the two types of electronic effects: induction and resonance. Inductive effects are a result of
polarized bonds, usually because of electronegative atom substituents. Resonance effects work through
systems, requiring overlap of p orbitals to delocalize electrons.
All substituents have an inductive effect compared to hydrogen (the reference). Many groups also have a
resonance effect; all that is required to have a resonance effect is that the atom or group have at least one p
orbital for overlap.
The most interesting groups have both inductive and resonance effects. In such groups, how can we tell the
direction of electron movement, that is, whether a group is electron-donating or electron-withdrawing?
And do the resonance and inductive effects reinforce or conflict with each other? We can never "tum off'
an inductive effect from a resonance effect; that is, any time a substituent is expressing its resonance effect,
it is also expressing its inductive effect. We can minimize a group's inductive effect by moving it farther
away; inductive effects decrease with distance. The other side of the coin is more accessible to the
experimenter: we can "tum off' a resonance effect in order
to isolate an inductive effect. We can do this
by interrupting a conjugated system by inserting an sp3 -hybridized atom, or by making resonance overlap
impossible for steric reasons (steric inhibition of resonance).
These three problems are examples of separating inductive effects from resonance effects.
(J
11:
11:
(a) and (b) In electrophilic aromatic substitution, the phenyl substituent is an ortho,para-director because it
can stabilize the intermediate from electrophilic attack at the ortho and para positions. The phenyl
substituent is electron-donating by resonance.
plus other
resonance forms
+
..
..
H
H
H
< }-CH2-C-O-H is a stronger acid than H -CH2 -C -0
+
BUT:
E
o
E
E
o
II
-
H
The greater acidity of phenylacetic acid shows that the phenyl substituent is electron-withdrawing, thereby
stabilizing the product carboxylate's negative charge. Does this contradict what was said above? Yes and
no. What is different is that, since there is no p-orbital overlap between the phenyl group and the carboxyl
group because of the
group in between, the increased acidity must be from a pure inductive effect. This
structure isolates the inductive effect (which can't be "turned off') from the resonance effect of the phenyl
group.
We can conclude three things: (1) phenyl is electron-withdrawing by induction; phenyl is (in this case)
electron-donating by resonance; (3) for phenyl, the resonance effect is stronger than the inductive effect
(since it is an ortho,para-director).
CH2
(2)
491
o
o
20-44 continued
(c) The simpler case first-induction only:
II
II
is a stronger acid than H-CH2 -C-O-H
There is no resonance overlap between the methoxy group and the carboxyl group, so this is a pure inductive
effect. The methoxy substituent increases the acidity, so methoxy must be electron-withdrawing by
induction. This should come as no surprise as oxygen is the second most electronegative element.
The anomaly comes in the decreased acidity of 4-methoxybenzoic acid:
CH30-CH2 -C-O-H
CH30
o
-< >-
C -0 - H
-< >- C
0
is a weaker acid than
H
-
0-H
Through resonance, a pair of electrons from the methoxy oxygen can be donated through the benzene ring to
the carboxyl group-a stabilizing effect. However, this electron donation destabilizes the carboxylate anion
as there is already a negative charge on the carboxyl group; the resonance donation intensifies the negative
charge. Since the product of the equilibrium would be destabilized relative to the starting material, the
proton donation would be less favorable, which we define as a weaker acid.
VC O
. .
CH3O
+
o
:0:
- -H
•
-
•
o
Methoxy is another example of a group which is electron-withdrawing by induction but electron-donating
by resonance.
(d) This problem gives three pieces of data to interpret:
II
II
is a weaker acid than H-CH2-C-O-H
Interpretation: the methyl group is electron-donating by induction.
(1)
(2)
CH3-CH2-C-O-H
<
�
CH3
0
-
H is a weaker aeid than
CH3
< >-
0 -H
Interpretation: the methyl group is electron-donating by induction. This interpretation is
consistent with (1), as expected, since methyl cannot have any resonance effect.
(3)
CH3
� ct is a
\4- O-H
CH3
stronger
acid than
o-� it
-
O-H
(1 )
Interpretation: this is the anomaly. Contradictory to the data in and (2), by putting on two methyl
groups, the substituent seems to have become electron-withdrawing instead of electron-donating. How?
Quick! Turn the page!
492
20-44 continued
Steric inhibition of resonance! In benzoic acid, the phenyl ring and the carboxyl group are all in the same
plane, and benzene is able to donate electrons by resonance overlap through parallel p orbitals. This
stabilizes the starting acid (and destabilizes the carboxylate anion) and makes the acid weaker than it would
be without resonance.
< � C - � -H
:0:
�
+
..
..
C-O-H
..
0:9:
plus other resonance forms
Putting substituents at the 2- and 6-positions prevents the carboxyl or carboxylate from coplanarity with the
ring. Resonance is interrupted, and now the carboxyl group sees a phenyl substituent which cannot stahdizc
the acid through resonance; the stabilization of the acid is lost. At the same time, the electron-withdrawing
inductive effect of the benzene ring stabilizes the carboxylate anion. These two effects work together to
make this acid unusually strong. (Apparently, the slight electron-donating inductive effect of the methyls is
overpowered by the stronger electron-withdrawing inductive effect of the benzene ring.)
•
"
CH3
"
H"
•
�---�!�
o-H
H
""
\
\\
\\
CH3
COOH group is perpendicular
to the plane of the benzene ring­
no resonance interaction.
20-45
(a)
•
0
\'
o
�H
•
- t··o
.
���.-----..---­
"
this three-dimensional view down the C-C bond
between the COOH and the benzene ring shows
that COOH is twisted out of the benzene plane
o
� OH
stock bottle
students' samples
(b) The spectrum of the students' samples shows the carboxylic acid present. Contact with oxygen from the
air oxidized the sensitive aldehyde group to the acid.
(c) Storing the aldehyde in an inert atmosphere like nitrogen or argon prevents oxidation. Freshly prepared
unknowns will avoid the problem.
CH2I2 A
20-46 H
h �COOH
Ph' I'
'
Zn, cuci H""
Ph
H
(Simmons-Smith
reaction, Sec. 8-9A)
IIIIICOOH
H
�AH
H ,,,,,
SOCl2
..
(Hofmann
rearrangement)
Br 2 ' HO-
IIIIINH2"
493
H20
A
H"
Ph
",COCI
H
�A
H"','
II'''CONH2
H
20-47 Products
are boxed.
(a) All steps are reversible in an acid-catalyzed ester hydrolysis. (abbreviating O CH2CH3 as OEt)
Step A p oton on (resonance stabilization)
H B is the acid catalyst
Step B nucleophile attacks
:B- is the conjugate base, although in
Step C proton off
hydrolysis reactions, water usually removes H+
Step 0 proton on
Step E leaving group leaves (resonance stabilization)
Step
proton off
F
-
r
..
:O-H
1+
vC'OEt
!
O/��Et
:O-H
1
..-----....
:O-H
:O-H
1
1
C
ac�a �O�t
� CH
HC+ 0
+
\
(f
HC +
:O-H
C1 'OEt
:O-HH�
:O-H
1
1 +
:OH2
.OH2
C-O:�
C-O·
I H
�
I "
" �:REt
V:REt
/
\
Step B
same series
of resonance .....
forms as above
..E---.�
-;
cr� (
I
"/
,
....""-...
Step
F
494
\
•
•
StepC
\
I
:OA,
V
+k�
� A-a)
�11
StepD
:OH
1- EtOH 1
t;)
+ H
Step E
•
A
..
20-47 continued
(b)
-
:0:
(c)
.·0//H.�.H
·�..0 /
~
•
\\ OH
�
HI'B.
OH
I
:..IO-H
O�:
�
• •
\ OH
s.�
---
: O-H
I"
\
..
HI C)h]
:0��/,...-:H.OH
• •
: OH
OC
c
:0/H..
+
+
•
+
: OH
:0:
+O-H
00:
II .....,
·
c
: B-
•
495
0:
II
00:
c
A cyclic ester is called
lactone. Lactones form when
the nucleophile is just a
few carbons away from the
carbonyl electrophile.
OH
a
20-47 continued
�
(d)
HO
___
/ �,
HO � C /'
C �H
� :OH
II
..
o
..O"·-
II
o
Esters can be fonned only in acid, not in base.
Cyanide substitution is an reaction and requ�res a o or ��rbon with � leaving �roup., The
Grignard reaction is less particular about the type of hahde, but ,IS sensitive to, and IncompatIble with,
acidic functional groups and other reactive groups.
(a) Both methods will work.
Mg CO2 H30+
NaCN..
OR
CH2COOH
CH2Br ether
20-48
SN2
< }0_ ',
1 2°
< }-
---- ---- ---­
0"
_
SN2
(b) Only Grignard will work. The reaction does not work on unactivated benzene rings.
Mg CO2 H30+ 1
1
COOH
\
\
Br ether
(c) Grignard will fail because of the OH group. The cyanide reaction will work, although an excess of
cyanide will need to be added because the first equivalent will deprotonate the phenol.*
HO 1 \ CH2Br NaCN.. H3 0+.. HO
CH2COOH
____ ____ ____
-0-"
-
-o-
tJ.
(d) Grignard will fail because of the OH group. The cyanide reaction will fail because SN2 does not
work on unactivated sp2 carbons. In this case, NEITHER method will work.
Ho
-o-
0-
Br
(e) Both methods will work, although cyanide substitution on 2°C will be accompanied by elimination.
Br Mg CO2 H30+
ether
(f)
---- ---- ---­
Grignard will fail because of the OH group. The cyanide reaction will work. Since alcohols are much
less acidic than phenols, there is no problem with cyanide deprotonating the alcohol.
H0
CH2Br NaCN H�O+ H0
CH2COOH
..
..
-o-
-o-
9.1,
16-18,
*
10.0.
The pKa of HCN is and the pKa of phenol is
Thus cyanide is strong enough to pull off some of
the H from the phenol, although the equilibrium would favor cyanide ion and phenol. The pK;l of
secondary alcohols is
so there is no chance that cyanide would deprotonate an alcohol.
*
CN
+
The side with the
weaker acid is favored.
-{ )
pKa 10.0
HO
� H - CN
pKa
496
9.1
+
-
o
-{ )
° ' pKa11.6
H
H .. H+
H ...... ' H ... 4 ° .... H pKa
Hydrogen
peroxide
isizedfourrelati
pK unitve tso(10hydroxi
times)de. stronger
acid than
water, so the hydrof tohperoxide
anion,ve
HOO,
must
be
stabil
Thi
s
i
s
from
the
e
el
e
ct
r
onegati
oxygen
bonded
to
the
by
i
n
ducti
o
n,
the
negati
v
e
charge
is
distri
b
ut
e
d
over
both
oxygens.
The
oxygen in hydroxide has to support the full negati ve charge with no delocalization.
(b) °
.. H) C Jl-° ... .. H) C �O pKa4.74
H) C )lO .-- H
20-49
(a) H , °
...... 0 ,
+
-
°
-
H+
-0
+
"""""
15.7
_
inductive effect
0-
;
-
°
1
.. 1 H3 C Jl /o-.
w
}
+
O
H) C .--l� / } pKa 8.2
°
H) C )l O ...... O ....... H
reason
that carboxyl
icoacin ofds theare carboxyl
so acidicate(o veranio10n wipKthunitwots equival
more acientdicresonance
than alcoholformss) isinbecause
ofThe
the
resonance
stabil
i
zati
whiall ch
alpossil atobmsle resonance
have octetsworlandds.theThenegatiperoxyacetate
ve charge isanionon,thehowever,
more electcannot
ronegatidelvoecalatoizm-the
best
of
e thea wetnegatnosei veoncharge
ontfrigoidthmorni
e carbonyl
oxygen;
that
negati
v
e
charge
i
s
stuck
out
on
t
h
e
end
oxygen
l
i
ke
al the
n
g.
There
i
s
some
delocal
i
zati
o
n
of
the
el
e
ct
r
on
densi
t
y
onto
the
carbonyl
,
but
wi
t
h
al
charge
separat
iperoxyaceti
on, this second
formis more
is a minor
resonance
contributor
.de.ThiIts resonance
doesclexplai
n,
however,
why
c
acid
aci
d
i
c
than
hydrogen
peroxi
does
not
come
o
se
t
o
acetic acid, though.
(c) � ...... O yCH3
-
o
W
+
. .
. .
•
°
H) C )l O J
Carboxyl
the diymbyer,hydrogen
that is,
tbonding.
wo moleculicTheaciesddiaresmboilerhelisdasantightl
8-membered
ring
widashed
th twolinhydrogen
bonds
as
shown
wi
t
h
escarbonyl
in the diaoxygen
gram. Thihasssiworks
because
the
gnifiis cant
negative
charge,
and
the
H-O
bond
weak
it is a relatively strong acid.
The b.because
p. is 118°C.
Dodoesperoxyaci
ds butboilthere
as tharee dithree
mer? reasons
The autthoor
not
know,
thinatsteadtheyofdohigher
not. Fisuggesti
rst, thenb.gpthat. is ltohwer
(notsuspect
l05°C)
ey do
boil
as
a
team
but
rather
i
n
divi
d
uall
y
.
Second,
dimberleshown
is laikellO-membered
rimembered.
ng-stillthepossi
but
l
e
ss
y thanant, 8-the
Third
and
most
import
elimplectireonid inc nature
of the carbonyl
group,
as places
the
resonance
forms
in
part
(b),
lessHnegative
on the carbonyl
oxygen,
and
thebondi
insg lisessmuch
acicharge
dilce,sssuggesti
n
g
that
the
hydrogren
strong.
497
20-50
1002 5 elements of unsaturation
Spect8ru, mIHA: C9HCOOH
811.
87.
3
,
5H
monosubsti
t
ut
e
d
benzene
ri
n
g
(' )- � H-COOH
83.
8
,
IH
quart
e
t
CHCH3
CH3
81.6, 3H doublet CHCH3
H602 2 elements of unsaturation
Spect1r,umIHB: C4COOH
812.
H COOH
86.
2
,
IHsi
n
gl
e
t
H-C==C
C==C
85.
7
,
IH
si
n
gl
e
t
H
-C==C
81.9, 3H singlet vinyl CH3 with no H neighbors CH 3 -C=C
H CH3
C:
Spect
ru
m
C6H
1
0
02
2
el
e
ments
of
unsaturation
812.
0
,
IH
COOH
87.0, IHmultiplet H-C=C-COOH
H
85.7, IHdoublet C=C-COOH
82.2-0.8 CH2CH2CH3
must be in doubldueettoatla8rge5.7coupling
constant
�
�
�
�
�
�
�
\
�
/
�
�
�
�
�
�
I
�
trans
498
/
\
CHAPTER 21-CARBOXYLIC ACID DERIVATIVES
IUPAC name first; then common name
(a) isobutyl benzoate (both IUPAC and common)
(b) phenyl methanoate; phenyl fonnate
(c) methyl 2-phenylpropanoate; methyl a-phenyl propionate
(d) N-phenyl-3-methylbutanamide; l3-methylbutyranilide
(e) N-benzylethanamide; N-benzylacetamide
(f) 3-hydroxybutanenitrile; j3-hydroxybutyronitrile
(g) 3-methylbutanoyl bromide; isovaleryl bromide
(h) dichloroethanoyl chloride; dichloroacetyl chloride
(i) 2-methylpropanoic methanoic anhydride; isobutyric formic anhydride
(j) cycIopentyl cycIobutanecarboxylate (both IUP AC and common)
(k) 5-hydroxyhexanoic acid lactone; 8-caprolactone
(I) N-cycIopentylbenzamide (both IUPAC and common)
(m) propanedioic anhydride; malonic anhydride
(n) I-hydroxycyclopentanecarbonitrile; cycIopentanone cyanohydrin
(0) cis-4-cyanocyclohexanecarboxylic acid; no common name
(p) 3-bromobenzoyl chloride; m-bromobenzoyl chloride
N-methyl-5-aminoheptanoic acid lactam; no common name
(r) N-ethanoylpiperidine; N-acetylpiperidine
21-2 An aldehyde has a C-H absorption (usually 2 peaks) at 2700-2800 cm-I. A carboxylic acid has a
strong, broad absorption between 2400-3400 cm-I. The spectrum of methyl benzoate has no peaks in this
region.
21-3 The C-O single bond stretch in ethyl octanoate appears at 1170 cm-I, while methyl benzoate shows
this absorption at 1120 and 1280 cm-I.
21-4
(a) acid chloride: single C=O peak at 1800 cm-I; no other carbonyl comes so high
(b) primary amide: C=O at 1650 cm-I and two N-H peaks between 3200-3400 cm-I
(c) anhydride: two C=O absorptions at 1750 and 1820 cm-I
21-1
(q )
21-5
(a) The formula C3HSNO has two elements of unsaturation. The IR spectrum shows two peaks between
3200-3400 cm-I, an NH2 group. The strong peak at 1670 cm-I is a C=O, and the peak at 1 610 cm-I is
C=C. This accounts for all of the atoms.
The HNMR corroborates the assignment. The multiplet at 0 5.8 is the
vinyl next to the carbonyl. The 2H multiplet at 0 6.3 is the vinyl
hydrogen pair on carbon-3. The 2H singlet at 0 4.8 is the amide hydrogens.
The CNMR confinns the structure: two vinyl carbons and a carbonyl.
(b) The formula CSHS02 has two elements of unsaturation. The IR spectrum shows no significant OH, so
this compound is neither an alcohol nor a carboxylic acid. The strong peak at 1 730 cm-I is likely an ester
carbonyl. The C-O appears between 1 050-1250 cm-I. The IR shows no C=C absorption, so the other
element of unsaturation is likely a ring. The carbon NMR spectrum shows the carbonyl carbon at 0 1 71,
the C-O carbon at 0 69, and three more carbons in the aliphatic region, but no carbons in the vinyl
region between 0 100-150, so there can be no C=c. The proton NMR shows multiplets of 2H at 0 4.3 and
2.5, most likely CH2 groups next to oxygen and carbonyl respectively.
The only structure with an ester, four CH2 groups, and a ring, is o-valerolactone:
H
a
IH
499
:0:II
-- PhCH2 -O-C-CH3
H�
ClA : 0 :
II 3
PhCH2 -O-C-CH
.
(b)
·0·
:0:.I . CH3
:0:..I CH3
·
CH3
1�
C
C
C
......
......
.
..
.
..
..
..
:0'"I eCI
:0'" 'Cl
:0II/
:0:II �:0:II)
Ph-C=O:I
Ph-C-O:I
-- Ph-C-O:I
Ph-C-O.I Cl .......C,CH3
H
H
H
H
plus thrdelocal
ee re sonance
wit h posirintigve
charge
ized onforms
the benzene
+I }
• •
.
+
.......
'0
---­
+1
.......
+
---­
the carbonyl oxygen i s more nucleophilic
than the single-bonded oxygen because
the product is resonance stabilized
:0I' ....: ... CH3
:0'"....... C
.. Ph-C=O:
cl-�(A
plus alasl theabove
resonance
forms
.·
(c) :0:
0
·0·
:0:
:0:
I I .. I�
II
�I I )
..
II-CH3
II
Ph-C-O:I Cl ...... C, CH3 ---Ph-C+0-C-CH3
--Ph
-C-0
-C
I}H
H
HI Cl�
�CC\
�
:0:II
II
Ph-C-O-C-CH3
I
.
+
o
1"\
nucleophilic attack by this oxygen
does not generate a resonancestabilized intermediate
+
o
• •
500
0
ued
n
conti
21-6
0
:
:0
(d)
Ph_ �J o A
H
:0 :
II
/+ �
�N
---
H
....
0
Ph
:0 :
� HN
I
....
Ph
II
:O �
/
H�
_
0
0
0 0
p I us
resonance
form
+
leaving group
�
� N � Ph �--...
:0 :
---
---
The
l
e
aving
group
i
s
ethoxi
d
e
i
o
n,
CH3CH20-, a very
strong
never bea base.
a leaving group
in an SN2base.reactiEthoxi
on asdeitwould
is too strong
+
H
/
'"
:0
H�
....
:0 :
-
Ph
�
o .
o .
� + HO �
21-7 onal group ("downhill reactions")
Reactionswiwhill occur
ch goreadi
fromlya. more reactive functional group to a less reacti ve
functi
oride will
NOT occur- i t is an "uphil l" transformati o n
aciamided chlttoooaciester
rided tochlwillamide
wi
l
l
occur rapidly
(b)(c)(a ) amide
NOT occur-another "uphill" transformation
acid chldoerideto amide
to anhydri
e willrapioccurdlyra pidly
(d)(e) anhydri
will doccur
o
21-8
o
HOCH2CH3 ---- CH3CH2 - C - OCH2CH3
HCl
o
HO -{ >
�O �
+ HCl �
o
0
(c)
< � C Cl HOCH2 --< > ---- < � C - OCH2 � >
HCl
-<J
(d)
0
HCl
o- C C l HO
---N
I
H
Figure 2 1 ·9 is critical!
II
+
+
-
....
+
----
+
+
-
+
+
501
21-8
conti
n
ued
CH3
This
on would
haved
toelimbeinreact
keptatioin.cold
to
aVOi
CH3
(e)
0
O
=< �H3
rs of
3 HCl butyl alcoholEstareehard
CH3-C-Cl HO-C-CH3 --- O-C-CH
make.
CH3
CH3
o
o
Cl � Cl � OH --- � 0 � 0� HCl
o
o
21-9 o
o
a) H3C-C-Cl HN(CH3h
H3C -C -N(CH3h HCl
o
(b)
0
H3C -C-Cl H2N -{ .> --- H3C -C -NH -{ .> HCl
o
�C-NH2
HCl
O C-CI NH3 --- �
�
o
0
d)
< }- C-Cl HNJ --- < }- C-NJ HCl
21-10
(a) (i) 0
0
0 -Q
o
-0�
H3C -C -0 -C-CH3 HOCH2 Ij_ � H3C -C -OCH2 � II HO-C-CH3
(ii) 0
0
o
r-�
H3C-C -0-C-CH3 HN'-- � H3C -C-N,-- HO-C-CH3
.. 0
(b) (i) 0.0:
0
:0:
II
II
I'.;) �-CCH3
II
�
H3C-C-0-CCH3
H3C-C
HOCH
2Ph
�. .
I(
H-O-CH2Ph
+
II
I
(f)
(
t­
+
I
I
to
+ 2
II
+
+
II
---
II
+
+
+
(e)
+
+
(
+
+
II
II
+
II
II
+
II
+
r--
+
II
+
• •
• •
"""
• •
+
+
• •
:0:I I
:0:I I
0I I
0II
.
.
H3C-C-0-CH2Ph
+ H -0-CCH3
" H3C-CI
:0-CCH3
..
..
.
·
H.0 o�CH,Ph /
+
502
�
II
2
21-10 (b) continued
(ii)
:0:I� . . 0I I
c�:
R
H3 HNEt2
3
H3C-C-0-CC
�
. . -- H3C-C-O-CCH
l,..
H t2
t
:0:
0
:0:
:0-CCH3
H3C-C-NEt2 H-0-CCH3
.. H3C-CI
..
·
..
+I
+
-
o
\I
21-11
�
·
:0:
O
+
CH3C-0-CH2Ph H-,NCH3
�.
II
..
:0:
CH3C-NHCH3
\I
T.S .+ �
+
+
\I
II
+
.
H - R -CH2Ph
.
+
..
�
--
......
f---
• •
+
503
.
\I
:OJ
CH37 l,.."q -CH2Ph
H-NHCH3
� T.S.+
:0:
C�
H37 + : R -t CH2Ph
H-NHCH3
I leavi n g
group
I
•
•
+
:0:8- 8i
CH3IC----O-CH
2Ph
H-NHCH3
I
NE
�.
\I
+
• •
o
C
:f)
)t ��
21-12
. .�
�' H
OCH2CH3
o
HOCH2CH3
+
�
•
.....--....
:0 :
�
II
H+
Ph - C - OEt
i
-
OH H
I "" I +
Ph - C - O - Et
I
O - C�
�
-
•
H' CH3
�
;--...
_ \
..
: P.CH2CH3
+
:O - H
II ..
Ph - C - .OEt
.
n
+
+
II
HO - CCH3
� ·ti
.. .
O
:O - H
I
..
- .OEt
Ph - C+ )
.
CH3P.H
�
..
:O - H
I
Ph - C = �Et
• •
___
OH
I
H+ Ph - C - OEt ..
Ph - C - OEt
I
"" I +
CH30H
· '-.. H � - CH3
O - C�
---
I
nCH,CH3
-
• •
---
o
�-
•
.
CH3C - 0 - H
: R - CCH3
21-13
o
� O:
•
II
+
II
• •
0
-· I I
O
. - CCH3
}
• .
EtOH
.
:O - H
+ O:O - H
V H�
CH30H
g
..
I
I
_
--- Ph
Ph - C + --- Ph - C
II
I
I
+ R - CH3
: R - CH3
:0. . - CH3
.
•
504
o
II
Ph - C - OCH3
21-14
�
H
1 �.,p H
;9.�
o
\
OCH3
;
-Ii'
rR�
0
•
•
OH
OCH3
505
.0. ,HC6s03H 0 • • ·.0e+ .. H
/' 0 -""i /'H'
21-15
• •
o
II
II
�
II
'-
-<->-
0ll • • ·.0"•c+• H
0 + .,·. •0"c• H }
/'H' �H' 'II
0-• • H
CC�OOH
OH
OH
�H
OH
I ••
••
+
�
t
c
CC
� O• • - CI - O• •,+... CH3 CC
� O-C
I - ·O·' ''' CH3
·
·
HSO;
I - o CH3
cc
·
·
�
�
...
CH3
CH3
CH3
I
I
I
H
SO
4
2
COOH H
COOH H
COOH ·•0 •·
�
/
I
�
:R,
�
t
�
:<],
�
�
- 0:
=C
O
CC
+
CH3
I
COOH
t
H � O·
:0 0C : HS04
• • II
•• +
C
I
• •- ,
CH3
R
aspirin CC
I
CH3
I
CC
COOH
COOH
�
o
:R,
II
two rapid proton transfers
••
H
"'
OH
0:
••
CC
-CH3O• •"+� '" CH3 ----. C + CC •O• •• - C +
� ·O· - C1 \...0(
I
C
H3
I
H3C"
..
..
OH
COOH H
COOH
I
'C ..
II
I
I
�
�
�
�
506
I
�
�
-
....
H ••
••
I
II
�
21-16 The asterisk will denote
(a) 0
:0.)
\ 80.
(*)
:0 :
11
�- . .
H3 C - C - O* R
: OH
�
I
..
H3 C-C V
- .O*R
.
I
:R - H
• •
The alcoholcarbon
productwicontai
s the label
tetrahedral
th (R)nconfigurati
on di, dwinotth none
break,i nsothethecarboxylate.
configuratioThen is bond
retainbetween
ed.
(b) The products are identical regardless of mechanism.
:0: �"'
II
H3 C - C - O*R
,,1 +
CH3 C - O*R
I
O- H
�I -
:O - H
H+
---
H+
--
..
HO*R
I
H3 C - C
I
+
: O- H
I
f----I
....
....
.
I
�
: O- H
.
I
..
• •
: O-H
I
• •
+
CH3 C -O*R � CH3 C - O*R � CH3 C = O*R
I
(+
•
H20 :
..
CH3 C - R *R
• •
:O- H
II
�
f
9H ,
OH H
I
i +
\ 80
1
....
O- H
..
H3 C -C
II
+ O- H
OH
I
•
the
t
CH3 C - O*R
,, 1 +
H20 :
'-..."... H - � - H
H20 :
.I I.V �
+ O- H
:O-H
I
�..
•
• •
\ 80 and
�
H3 C - C
I
: O-H
---
:0 :
II
H3 C-C - OH
acetic acid
its nucleus
thanof mlz 60,Mass
spectratheofmoltheseecularproducts
would show the
hasof2acetimorec acidneutratonsits instandard
(molc) The
e
cular
i
o
n
val
u
e
whereas
i
o
n
of
2-butanol
appear at mlz that
76 instead
of between
mlz 74, proxygen
oving thatand thethe carbonyl
heavy isotope
of oxygen
wentnotwithethbond
the albetween
cohol
demonstrates
the
bond
carbon
is
br
o
ken,
oxygenTo show
and theif thealkylalccarbon.
oholueswaswoulchiral
orve racemic,
measuri
ng(Theits optiheavycal actioxygen
vity inisotope
a polahasrimaeternegliandgible
comparing
t
o
known
val
d
pr
o
i
t
s
confi
g
urati
o
n.
effect on optical rotation. )
\ 80
\ 60.
507
would
Th i S
the
lyst ioslydefsisi,nacied dasisausedchemiincalthespecifirsteands thatfourspeeds
apsreofactthioenmechani
but is notsmconsumed
in the ereactd in itohn.e In
t(a)thheirdaciAandcatdicalhydr
t
h
st
e
but
i
s
regenerat
aiostn.steInps.theAcibasid cishydrnot consumed;
the finthale concent
rdateiothnatofinaciitiadl iys atthteackssametheascarbonyl
the initiails never
concent
r
at
o
l
y
si
s
,
however,
hydr
o
xi
regenerat
, but it quithceklbasey neutralizes tthheecarboxyl
moleculetehofd.e estreAnacter,alioonekn.oximolde leeculavese offromhydrthoexicarbonyl
de is consumed;
reactionibutc acidoesd. Fornotevery
(b) Basian estc hydrer. oAcilysidscatis anotlysireversi
ble. Onceis an anequiestliebrrimolum:eculthee miis xhydrtureowilyzedl aliwnaysbase,contthaeicarboxyl
ateer,cannot
form
s
,
however,
n
some
est
ander
tunthe iyil tehleyd wiarle never
be
as
hi
g
h
as
i
n
basi
c
hydrol
y
si
s
.
Second,
l
o
ng
chai
n
fat
t
y
aci
d
s
ar
e
not
sol
u
bl
e
wat
y as th.eir sodium salts (soap). Basic hydrolysis is preferred
higher yield andionigreatzed;etrheysoluarebilsolity uofblethonle product
21-18
:0:
:O
:OH
·
O-H
O
O
°G
�
21-17
promotes
catalyze
�
21-19
�
. .
..
. .
:O-H
II
i +
:O-H
+
-C-RH
R
i
21-20
(a) :��
H3C- C NMe2 _:OH
:ttr
. .
. .
R -C-NH2
• •
C· � C· · C
:O-H
I
�:
+I
:O-H
R-C-OH
. .
0:
----t��
�'
. .
H
. .
. .
• •
�
---l��
..
�
for
:0 :
:0 :
R -C-NH2
• •
II
•
in
. .
+
�
0J
1 H3C- b ==0
. .
508
+
----
0:
�' H
. .
21-20 continued
(b) :0:I I � H+
H3C - C - NMez
.
.
: OH H
I '" I
MeC - NMez
+
I
H
O-
!
i
--
H+
I
-
..
I
O-H
:O - H
I
H3 C - C +
I
:O - H
----­
..
:O - H
I
MeC - NMez
+
(
H zO :
-----
• •
MeC - NMez
�
� MeC - NMez
- IJN (CH , h
..
:O+ - H
II
�
: OH
I
..
MeC -
..
Meb �ez r
:O - H
-----
NMez
=
'" I +
HzO :
H
� .O. - H
..
:O - H
I
H3C - C
II
+O-H
-----
l ( )
+ OV
-H
:0 :
II
I I �HN C H3 Z
.. H3C - C - OH
H3C - C
I
:O - H
• •
+
21-deprot2 1 onatIn tihoen basiof thcehydrcarbooxyllysisic(21-20(
a»th,ethamie stdeep anithaton.driIvnesthteheacireactdic ihydrol
on to compl
et1-io2n0(isb»th,eprotfinalonatsteiop,ntofhe
aci
d
by
y
si
s
(
2
aminesobythaciat tdheisamiexotneherismnoic landongerit prnucleventeophis thliecreverse
nithetrogen
. reaction by tying up the pair of electrons on the
21-22
:- H - OH
� "
n
Ph - C N : - Ph - C = N ·
I
\�: O
. .H
:O. . - H
.. i h .. -
:0 :
II
HO':l H �
P -C-N-H
Ph - C - N - H
II
:
)
.?l :
O. .H
(
h
.
Ph - C - NHz
IU
:O - H
:0:
---l..�
-
:0:
II
�
• •
-
..
+
Ph - C
:NHz
J
I Y'\
:O - H �
509
..
+
HzN(C H3 h
Ph - C = N - H
I Y'\
:O - Hj
!
-
:0:T
: OH
- Ph - = N - H
..
i.. h
r
:
0
:
r
- --- h
:0 :
II
P -C O:
I
P -C O
• •
• •
+
NH3
• •
=
• •
21-23
Hi) :�
H �
H
Ph-C-N-H
Ph- - H2
Ph -C- N-H
.
.
H
:
O
-H
:
: -H
.
Note: species with positive charge on
U
-H
Ph-C=N
H ��+- H
I
+
I
�
..E---l..
....
I IJr\
<./
0
I
2
II
C
----
0
N
carbon
adjacent
to(notbenzene
alswio thave
resonance
shown)
forms
positive charge distributed over theh thering.
21-24
er reductism owhere
n, a newthe C-H
fi(a)rstRsteducti
ep andoninoccurs
the thiwhen
rd step.a newThisC-Hcan bond
also beis formed.
seen in thiIns estmechani
stepsbond
are simformed
ilarly l ainbeltehd.e
(b)
Step 1-on
reducti
:
o
C
IS
alStkeoxip 4-workup;
de protonation �
2 1-25
(a)
� NH
«2 (5CH2CH'
C(
OH
HH
H
I
O
H
Streducti
ep 3-on
O� Q.N ,
H
(d) (' N · H (e)
510
3C
(t) ex>
CH2CH2CH3
.
C H2 NH2
21-26
+
MgI
Ph "MgI MgI
C=N. H+ C=N+ ..
H3C H
"""----""' H Ph H
0
I
H2 : : O1 "\� C-N
+ I :H
1
I
�
H CH3
PhI
Ph
PhI (':H1 - NH3
PhI H
+
+O =C
O-C-N-H
O-C+
:
:
: O-C-N
:
(I I
H CH3
CH'1
HI CH3 H\--... H+ H CH3 H
H) �
Note:
speciadjacent
es with tposi
tive Charge
!
onhavecarbon
o
benzene
al
s
o
forms
(notdisshown)
:0 :
wioverth-thetheresonance
posi
t
i
v
e
charge
tributed
+ Ph '- C-I I - CH3
ring,
NH4
+
:?j
MgBr
CH3CH2Q:
-C - Cl Ph -MgBr .. CH3CH2 -C-Cl
�
PhI � MgBrCl
+
OH H+
- :0 : MgBr
:0 :
G
II
CH3CH2 -CI -Ph ..workup CH3CH2-C-Ph
..
CH3CH2-C-Ph
MgBr
-./
Ph
I
-:::.
Ph
Ph
�
Ph
\
/
H3C
/
•
"--."..
•
....
\
I
/
\
-
I
\
/
• •
\
I
..
I
I
•
I
{
I
I
I
..
•
H
+
21-27
• •
+
I
V
----I�
�
JI'
• •
-
I
511
21-28 0
OH
(a)
H30+
MgBr -.. �
Ph )l OR
Ph
these alcohols can also be synthesized from ketones:
OR
OH
H30+
.. �
� Ph � MgBr
Ph
OR 0
OH
H30+
.. �
� PhMgBr
Ph
OH
(b) 0
H30+
HAOR � MgBr __ .. �
H
0
(c) CH3C N � MgBr __ H30+.. �
ORN ::-....
0
H30+
�C
CH3MgBr -- .. �
21-29
·0+ L «f0CHJ
: 0I I :
: 0I I :
Et-C �AlCI3 --- AlCI4- i Et-C' .. .. Et � � r
+
+
2 �
0
+
--
+
+
---
2
+
�
+
l
+
AICI4-(0
o
o
II
II
C-Et
H C-Et
------
------
:OCH3
..
� CCH2CH3
_
CH30 -o-
+OCH3
..
o
II
+
HCl AlCl3
+
512
o
II
H
I
6-�
C
:OCH3
..
:OCH3
..
21-30
Ij-�
0
(a )
'
+
21-31
(a ) (i)
o
0\\ C - Cl
_
\
o
II
II
� Cl
0
0
HC - 0 - CCH3
+
H2N
AlCl �
-
AlCl3
•
-\ .J
< '}-c0l l -1 �
o
r
�
0
HC - NH
Zn(Hg)
HCl
-\ .J
+
o
II
HO - CCH3
HO - CCH3
+
(b)
C�:
ft
H2.N. Ph
- 0 - CCH 3
H - C�
(i)
+
�
..
0
:0:
I::,) . . I I
H-CO. - CCH3
I l,�
+
H - NHPh
�
:0:
II
H-C
I
H
0
.
0
· 01 1 ·
II
H - C - 0 - CCH3
·· 0·.
0
II
1::,)
H - C (",,9· · - CCH3
1
H -+O - CH2Ph
.
+
HOCH2Ph
�..
• •
�
• •
513
0
.. II
:O. . - CCH3
�
:0:
II
H - C - NHPh
(ii)
o
II
�
+
+
0
.. II
H - R- CCH�
:0:
0
II
II
H-C +
:0CCH3
··
,, 1 +
l{0Q� CH?Ph )
- • •
�
�
�
O",C 'H
�OH + HC-O-CCH3
HC Cl does not exist, so acetic fonnic anhydride is the most practical way to fonnylate the alcohol.
(b)
CH3C -O-CCH3
O",C 'CH3
OH
+
Acetic anhydride is more convenient and less expensive than acetyl chloride.
(c) 0=$
(Vl NH2
+ NH 3
Vy OH
21- 32
(a)
o
0
o
�
I I
I I
o
II
-
�
o
I I
I I
o
0
�
II
II
o
(d)
c$
o
The
chloride would tend to react at both carbonyls instead of just one; only the anhydride wi l l
give acithids product.
o
o
�OCH
3
H
�O
o
o
The
acid chloride would tend to react at both carbonyls instead of just one; only the anhydride will give
this product.
514
SIS
o
...... J�
.
H
I
:0
+/
: 0.
.
.
HO
,
I
:0
• •
------
+0
..
+
II���
.�
• •
H
HO
.
HO
+H
..
HO
HO
• •
�
r
Q:oH
21-34
(a)
I
0
CH3C-O-CCH3
a
II
+
II
COOH
Generally, acetic anhydride is the optimum reagent for the preparation of acetate esters. Acetyl chloride
would also react with the carboxylic acid to form a mixed anhydride.
(b)
0H
0H
CH30H
I
I
COOH
COOCH3
Fischer esterification works well to prepare simple carboxylic esters. The diazomethane method would also
react with the phenol, making the phenyl ether.
(c)
rh CH2N2 .. rh
�COOH
�COOCH3
Fischer esterification would make the ester, but in the process, the acidic conditions would risk migrating
the double bond into conjugation with the carbonyl group. The diazomethane reaction is run under neutral
conditions where double bond migration will not occur.
21-35 Syntheses may have more than one correct approach.
(a) 0
OH
H30+
ether
Ph-C-OCH3 2 PhMgBr
Ph -C-Ph
Ph
(b) 0
OH
H30+
ether
H-C-OCH2CH3 2 PhCH2MgBr
..
H-C-CH2Ph
CH2Ph
(c) o
o
Ph -C-OCH3 H2NCH2CH3
Ph -C- NHCH2CH3
h
Q:
h
II
+
(e)
(f)
..
II
0
Ph -C-OCH3
II
+
+
I
I
II
�
ether.. H30+
2 PhMgBr
---
OH
H-C- Ph
Ph
H30+
LiAIH4 ether..
PhCH20H
Ph-C-OCH3 H30+..
a
II
I
---
+
0
H -C-OCH2CH3
I
---
+
II
+
h
+
II
(d)
Q:
---
Ph -C-OH
a
II
+
CH30H
516
I
I
continued
pyridine PhCHzC:: N
PhCH20H 2)1) TsCl,
KCN
from (e)
21-35
(g)
•
(h) a
PhCHz-C-OCH(CH3h 2 CH3CHzMgBr ether
from (g)
a
(i) How to make an 8-carbon diol
0
from an ester that has no more
than 8 carbons? Make the ester a
1
lactone!
II
OH
PhCH2-C-CHzCH3
CHzCH3
I
•
+
3
0
4
8
I
+
7
H O+
LiAIH4 ether. �
There may be additional correct approaches to these problems.
21-36
(b)
H
I
6
a
a
HC -0-CCH3
II
II
•
H
C=O
I
I
6
5
6
LiAIH4
•
3
vlOH
8
4
5
6
7
OH
CH3
I
�6
Prepare the amide, then dehydrate.
(a)
a
SOClz
� NH3
� POCI3
� OH
�CI
�NHz .�C::N
21-37
•
•
(b)
a
PhC-OH
II
a
SOCl2
----l.�
NH3
o
PhC-CI
II
o
•
a
PhC-NHz
II
21-38
(a)
517
POCI3..
a
PhCH2 -C::N
continued
(b)
2l-38
Fe
HCI
a)
�
(c)
NaN02 CuCN
HCl
..
-�..
�
•
2l-39
(
OH
Cl
�
OH
C ... CH3
II
..
o
•
Ar=
• •
----l.�
¢
Cl
�Brg
M
ether
!
0
.. CH3-N=C-O. :
CH3-N=C=O
Ar-OH i
�
H-O.. Ar
�
21-40
NaCN PhCH2CH 2CN
C
pyridine
CN
Ts I
.. .
1+
-
..- .. }
---- CH3-N- C=O
I
Af
((\J,'two rapid
OH proton
Ar
H
Ar +9 I)
transfers
.
• •
• •
0
518
1
• •
Sevin
(carbaryl)
• •
21-41
(a)
(X0>= 0
(b)
(c)
("y 0H
o
� OH
carbonate ester
(iii) not aromatic
(i)
ao
(i) thiolactone
(iii) not aromatic
(l
SyS
o
(i) thiocarbonate ester
(iii) not aromatic
+
CO2
(lSH
SH
imine
enediamine
(i) a substituted urea
(iii) AROMATIC-more easily seen in the resonance form shown
(The enediamine product would not be stable in aqueous acid. It would probably tautomerize to an imine,
hydrolyze to ammonia and 2-aminoethanal, then polymerize.)
(e) At first glance, this AROMATIC compound does not appear to be an acid derivative. Like any enol,
however, its tautomer must be considered.
H 3 O +.. �
H30+
H 0 COOH
C'
COOH
[['
COOH
U
0N 0
N
NH
N OH
enamine
H
H
H
NH3
imine
lactam
..
...
I
I
o
---
2
I
HO NH2
0)( NH
\=..!
\=..!
(i) a carbamate Dr urethane
(iii) AROMATIC-more easily seen in the resonance form
�
...
f----'l
..
-
519
+
polymer
see part (d)
---
21-42
between carbamic acid and phosphoric acid. It would react
(a) Carbamoyl phosphate is a mixed anhydride
a urea), with phosphate as the leaving group.
(technically,
bond
amide
an
form
to
amine
an
with
easily
two nitrogens on either side; this group is a urea derivative.
(b) N-Carbamoylaspartate has a carbonylthewith
(c) The NH2 group on one end replaces OH of a carboxylic acid on the other end; this reaction is
nucleophilic acyl substitution.
(d) Orotate is aromatic as can be seen readily in the tautomer. It is called a "pyrimidine base" hecause of its
structural similarity to the pyrimidine ring.
OH
o
H oN
N��� pyrimidine
N�
N
N COOO� N COO-��h HO �� HJl
H
21-43 Please refer to solution 1-20, page 12 of this Solutions Manual.
21-44
(g) benzonitrile
(a) 3-methylpentanoyl chloride
(b) benzoic formic anhydride
(h) 4-phenylbutane nitrile; y-phenylbutyronitrile
(c) acetanilide; N-phenylethanamide (i) dimethyl isophthalate, or dimethyl benzene-l,3-dicarboxylale
(d) N-methylbenzamide
(e) phenyl acetate; phenyl ethanoate (j)(k) N,N-diethyl-3-methylbenzamide
4-hydroxypentanoic acid lactone; y-valerolactone
(f) methyl benzoate
(I) 3-aminopentanoic acid lactam; �-valerolactam
21-45
(c) a -./\
(b) 0 0
(a) 0
PhC-� �
PhC-O-CCH3
PhC-OCH2CH3
H
(e) OH
(f) a
Ph -C- Ph
PhC-H
Ph
21-46 When a carboxylic acid is treated with a basic reagent, the base removes the acidic proton rather than
attacking at the carbonyl (proton transfers are much faster than formation or cleavage of other types of
bonds). Once the carboxylate anion is formed, the carbonyl is no longer susceptible to nucleophilic attack:
nucleophiles do not attack sites of negative charge. By contrast, in acidic conditions, the protonated
carbonyl has a positive charge and is activated to nucleophilic attack.
basic conditions
a
\
_ _
I
II
II
I
II
II
II
I
o
R-C-OH
II
acidic conditions
R -C- OH
o
II
+
+
-OR'
H+
a
R-C-OHOR'
anion-not susceptible
to nucleophilic attack
---
II
+
OH
R -C-OH
+
rapidly attacked
by R'OH nucleophile
I
520
_
c(
o
21-47
(h) (
(a)o- o
)- O- oC- H (C) I
NH -{
r; � O-C-CH3
>
OH
h
_
anhydrides react
o only once
OH
o
CH3
OH
CH3
..
.
c
..
.
..
.
(e)
(d)
C
.... C=O
O
I
(f)
o
0
h
H3CO
amines are more nucleophilic
than alcohols
21-48
o 0
o
(a) oo
r; � NH2
C "'CH
NH- C - H
... C ... O .... C ... CH 3
)H
HO
....
(
3
_
_
(b)
0
o
o 0
C
C"'OH SOCl2
C ... O .... C ... CH3
'CI Na + -OOCCH3
II
�
if'
II
(}
(c)
(tH oH
II
I
�
h
H
0
�
OH
(}
CI
0
h
(e) 0
0
C
(f) �
(g)
OH
OH
-H2O
/:!,.
Ag>
OH NH 3 (aq)
0
�
A HolloH.
h
0
C
•
0
0
H+
/:!,.
OH -H20
X
1\
�
6
\...
II
o
� OCH(CH3h
HOCH(CH3h
o
II
h
H:: o Ao
•
� w.d;o
(d)
(}"-'::
a�°Y'°
II
II
+
---�
Cl xO
+
II
II
II
+
)C
H
�
r
�
I
�OH
H+
/:!,.
H20
-
...
0
o
OH
0
NaBH4 �H
)
� COOH�H30H �COOH
Y
(f)
0
)
LiAlH,
A Ac,o
A
X 2) H30+ Y
H+,-H20
Y
MeOOC COOMe MeOOC COOMe
HOH2C CH20H AcOH2CYCH20Ac
I)
•
any ester where ethylene glycol displaces
methanol will be reduced with LiAIH4
521
aqueous acid workup
removes ketal
protecting group
•
21-48 continued
(h)
MCPBA
0
..
0
OR
Bf,
H20
..
d
OH
6 ...
• •
...
:0:-
H+, - H2O
:0.)
I ..
Ph-C-OEt
I
:O-H
+
:O :�
:O-H
II
II ..
H+ PhCPh-C-OEt
REt
<
--..
�
------
1
.
+
�
--
I
:O-H
1
II
+ O-H
Note: species with positive charge
on carbon adjacent to benzene also
have resonance forms (not shown)
with the positive charge distributed
over the ring.
2. CO2
..
:O-H
1
PhC-OEt
(+ ..
•
.
:O-H
:O-H
I
Ph-C+ ------ Ph - C
...
f\.
\=�
H20 : !
OH H
OH
I
I
.
1
H+
PhC-O-Et
PhC-OEt ..
PhC-OEt
1
1
1+
H20: H-O-H
O-H
O-H
..
'-......."..
� - EtOH
""
0
HCl
0 �
-0
" 1 0 :ci: I "-'::
h
-
··
00]
:0:
Ph -CII
Ph -: fI�J
H-O..
v
i
f\.
...
..
(c)
...
0
CtD]
H2SO4
-
OH
6 ,,"COOH Q- COOH
6·,,·CN KOH
H2SO4
H2O,/).
.
o
,
H
Q- Bf HO OH 0- Br d
1. Mg
KCN
H2O
Bf
21-49
(a) :0
I
Ph-CI�R �
\
OH
""
+ O-H�
H20 :
Ph-CII�
...
1
:O-H
522
.
�
+
0
(g)
II
:0:
II
X;
'.
H
0
�
H3 C-C-O
-CCH
'--' '--' H3
/
�
'-'
+
T
R
(a)
'
I
0
'
" " ,
t
H
--
configuration
No bond to the chiral center
is broken, so the
configuration is retained.
21-50
0
11
Ph-C-O
-0
• •
(b)
I::> II
H3 C-C-O-CCH3
I l.,,�'
+
. .Xv
... ...
H-°
0
(g)
C-NHCH3
/j
C
'
OCH3
OH
:0:
0
• •
il
O
-
:0:
523
(c)
(h)
'H
I"'J
Ph-C- N0
0
II
H
>:::
� OH
(d)
H-o
C-N
0
II
C
(i) 0-
I
COOH
0
11
C-CH3
continued
+
C
21-50
(j)
o
II
(k)
a
'
O- N
NHz
(!�
(m)
CH3
(I)
I
PhCHzCHCHzNHz
OH
OH
Products after adding dilute acid in the workup:
0
(b)
(a) 0
21-51
II
HC-OH
+
�
(c)
HO-Ph
(
(d)
COOH
� OH
(a)
�
H
HO
X
0
0
II
CH2 - 0-C-(CH2)12CH3
CH-OH
CH-O-C-(CHz)12CH3
glycerol
CH2-0 -C-(CHz)12CH3
I
I
I
I
o
(b) CHz-O-C-(CHz),zCH3
I
I
I) LiAIH4
..
2) H20
�
6
CO,HCI
..
AICl3/CuCI
Gatterman-Koch
�
OH
"-':
HN03
h
HZS04
...
I
I
0
&"
I
"-':
C,
H
H2Cr04
...
h
+
para
OH
¢ ¢
Fe
..
HCl
+
tetradecan-1-ol
+
CH-OH
3 CH3(CHzhzCHzOH
CHz-OH
OH
OH
OH
trimyristin
glycerol
CHz-0 -C-(CHZ)I2CH3
6
�
CHz-OH
�
21-53
�
II
CH - 0 -C - (CHZ)I2CH3
,
+
O
CHz-OH
CHz-OH
(b)
HOCHzCH3
HO
OH
21-52
(a)
+
OH
ortho
NOz
NHz
:
II
OH
0
&"
I
524
CH3-C
C'
1 equivalent
AczO
&
0
..
3
OH
AczO
OH
¢
CH -C-NH
II
0
..
I
"-':
h
more
NHz is
nucleophilic
\i
C,
0
H
than OH
(a)
21-54
I
� OH
o 0
HC- O-CCH3
II
+
II
H30+
..
2 CH3CH2MgBr -o
OH
�
(e) See the solution to 21.36(b) for one method. Here is another:
+
H Jl H
(f) 0- NH2
(g) 0- Mg
(h)
CH3O
Br
O
CH30X0
o- MgBr
...
ether
+
two equivalents
NH2
C NH2
+
..
CH30H � + ...
large
excess
+
0
II
H -C-OEt
H
O
I
reductive alkylation.
9H -o
0+
H,o
-- -CH
N X0
(
(y CONH2
I
H
o
C
� "'OCH3
t POCI3
II
'--1
525
+
Ho D
(y C::N
21-55 Diethyl carbonate has two leaving groups on the carbonyl. It can undergo two nucleophilic acyl
substitutions, followed by one nucleophilic addition.
o PhMgBr H30+ OH
(a)
0
PhMgBr
<i?
PhMgBr
Ph-C-OEt
Ph-C-Ph
EtO-C-OEt subst. 1
subst. 2 Ph- C- Ph addition
Ph
Mg CH3CH2MgBr
ether
o
OH
EtO-C-OEt
CH3CH2 -C -CH2CH3
3 CH3CH2MgBr
CH2CH3
21-56 Triethylamine is nucleophilic, but it has no H on nitrogen to lose, so it forms a salt instead of a
stable amide.
.
:0:
:�j
CH3C-CI :NEt3
CH3C-CI
�
IV
+NEt3
When ethanol is added, it attacks the carbonyl of the salt, with triethylamine as the leaving group.
II
----�.�
�
---'-I..
----�.�
II
+
G:
1
[
II
I
I
I
.
+
---
+
C�:
CH3C�
-NEt3 HO-CH2CH3
..
+
:0:
---
An alternate mechanism explains the same products, and is more likely with hindered bases:
oI I � like an E2
� ..
H�-CH2CH3
1\ -I
H2C)"" C-CI
.. Et3NH CIH2C=C=O
a ketene
H... .---. NEt3
+
I
I
..
•
+
•
526
�
21-57
H
(a)
H2C � OH
o
(b)
Br2
CH30-o�
_
---
03
2) Me2S
1)
6
•
..
w
!'i
H
O� OH
! Ag+ , NH3 (aq)
0O� OH
Mg
CH30-O-- Br ether
---I"�
CO2 H30+
___
�
CHP -O-- CONH2
•
NHJ
CH30-O-- COOH
*1
*
*1
CH20H
COOH
COOH
1) NaOH
1) LiAIH!
7 1
excess ..
CH3I CH 30 ::::-"" OCH3 2) H20 CH30 ::::-"" OCH3
HO ::::-"" OH 2)excess
OH
OCH3
OCH3
TsCl
pyridine
7
*1
CH2CH2NH2
7
*1
CH2CN
!
*1
CH20Ts
LiAIH4
KCN
..,
::::-"
"
CH30 ::::-.... OCH3 2) H20 CH30
OCH3 CH30 ::::-.... OCH3
OCH3
OCH3
OCH3
..,1)
527
+
CH30 -O-- COCI
(c) � CH2Br KCN � CH2CN 1) LiAIH4 � CH2CH2NH2
..
2) H20 " 0
0
o
(d)
SOCI2
21-58
o
(a) CH3CH2 0-C-OCH2CH3
(d)
(c)
(b)
II
(e)
0
0'" C ..... 0
II
'---1
CH3
CH3-� -OH
I
CH3
0
II
Cl-C-Cl
o
II
+
�
�
_
CH3 0-C- �
H
�H3
CH3 -�- 0
CH3
�
- C -Cl
21-59 (a) The rate of these nucleophilic acyl substitution reactions is controlled by two factors: stability of
the starting material as detennined by the amount of resonance donation from the leaving group into the
carbonyl, and the leaving group ability which is determined by the basicity of the leaving group, the least basic
being the best leaving group.
LEAST STABLE: no
significant sharing of
electrons from chlorine
electrons from oxygen are
also distributed through the
ring and nitro; very little
resonance stabilization
o
..)l
..
:0
leaving group ability and basicity:
Cl-
weakest base;
best leaving group
>
02N
-o-
MOST STABLE: most
significant resonance donation
of electrons from oxygen
electrons from oxygen
are also distributed
through the ring; small
resonance stabilization
0:
do
0-
>
� }-
O-
>
CH3 O-
negative charge
negative charge
strongest base;
delocalized through delocalized through worst leaving group
ring and nitro
ring
The least stable starting material with the best leaving group will be fastest to react. The most stable starting
material with the poorest leaving group will be slowest to react.
(b)
( 0:
:0:
Nuc
··0··
·
..� II � NUC
:O � O:
/
Cl-C
\
C-Cl
/
I 'Cl Cl I
Cl
Cl
C
:o
..
Cl-C '
I
/
Cl
X-o:
C-Cl
Cl Cl / I \
�\
"
..)l
:0
Cl-C'
I Cl
Cl
/
Nuc
+
Cl-
triphosgene
The first step is the standard attack of nucleophile like CH3 0H at the carbonyl carbon to make the
tetrahedral intennediate. The second step is key: the collapse of the tetrahedral intennediate produces
one equivalent of phosgene. Attack of a second nucleophile of the other side of triphosgene would
release a latent (hidden or trapped) equivalent of phosgene from that side too. Thus, the equivalent of
three molecules of phosgene are locked into the triphasgene molecule. Eventually, all six positions
would be substituted with methanol producing three molecules of dimethyl carbonate for each molecule
of triphosgene.
Cl;'
a
528
(a)
II
o
21-60
H
H3C-N-C-OH
methyl isocyanate
a carbamic acid-unstable
of
gaseous products causes a large pressure increase, risking an explosion.
CH3-N=C=O
Both
H20
+
�
CH3NH2(g)
�
+
CO2 (g)
these reactions are exothermic. In a closed vessel like an industrial reactor, the production of
.
(b)
0
�
CH3- N=C=O
�
H-OH
•
•
•
1·· ··-
CH3-N=C-O:
1+ "
---
H-O-H
..
}
( \!
3
CH _�r=-c=o
+
I)
H-RH
•
•
!
I
• •
'-H
H - OH
•
•
two rapid
proton transfers
.. �
II
..
3 .,.../
decompositionV
could be proposed
either acid or base catalyzed
H-OH
H
.. .. H C-N-C- O-H
(c)
o
I I
OH
o
II
phosgene
CI-C-C1
+
as
CI-C-O
�-�
�
VJ
(a) (i) The repeating functional group is an ester, so the polymer is a polyester (named Kodel®).
(ii) hydrolysis products:
21-61
o
HO-
C
-o- C
0
-OH
+
HOCH2
-o-
CH20H
(iii) The monomers could be the same as the hydrolysis products, or else some reactive derivative of the
dicarboxylic acid, like an acid chloride or an ester derivative.
6).
(b)(i) The repeating functional group is an amide, so the polymer is a polyamide (named
0
(ii) hydrolysis product:
Nylon
H2N
II
(iii) The monomer could be the same as the hydrolysis product, but in the polymer
industry, the actual monomer used is the lactam shown at the right.
� OH
529
continued
(c) (i) The repeating functional group is a carbonate, so the polymer is a polycarbonate (named Lexan®).
(ii) hydrolysis products:
-o- CH3�
HO r;_� � � OH
CH3
(iii) The phenol monomer would be the same as the hydrolysis product; phosgene or a carbonate ester
would be the other monomer.
(d) (i) The repeating functional group is an amide, so the polymer is a polyamide.
(ii) hydrolysis product:
H2N -o- COOH p-aminobenzoic acid, PABA, used in sunscreens
21-61
(iii) The monomer could be the same as the hydrolysis product; a reactive derivative of the acid like an
ester could also be used.
(e) (i) The repeating functional group is a urethane, so the polymer is a polyurethane.
(ii) hydrolysis products:
NH2
H2N
HOCH2CH20H CO2
I h
CH3
(iii) monomers:
� N=C=O
HOCH2CH20H O=C=N �
I
� CH3
(f) (i) The repeating functional group is a urea, so the polymer is a polyurea.
(ii) hydrolysis products:
CO2 NH2(CH2)9NH2
(iii) monomers:
°
)l or
or an equivalent carbonate ester
CI CI
+
+
'Ct
+
+
(a) Both structures are 13-lactam antibiotics, a penicillin and a cephalosporin.
(b) "Cephalosporin N" has a 5-membered, sulfur-containing ring. This belongs in the penicillin class of
antibiotics.
21-62
530
21-63 The rate of a reaction depends on its activation energy, that is, the difference in energy between
starting material and the transition state. The transition state in saponification is similar in structure, and
therefore in energy, to the tetrahedral intermediate:
·0·
� � � OCH3 tetrahedral intermediate
�
_ OH
I
The tetrahedral carbon has no resonance overlap with the benzene ring, so any resonance effect of a
substituent on the ring will have very little influence on the energy of the transition state.
What will have a big influence on the activation energy is whether a substituent stabilizes or destabilizes
the starting material. Anything that stabilizes the starting material will therefore increase the activation
energy, slowing the reaction; anything that destabilizes the starting material will decrease the activation
energy, speeding the reaction.
energy
__::---
smaller Ea because
of destabilized
starting material
similar transition state energy
larger Ea because of stabilized
starting material
(a) One of the resonance forms of methyl p-nitrobenzoate has a positive charge on the benzene carbon
adjacent to the positive carbonyl carbon. This resonance form destabilizes the starting material, lowering
the activation energy, speeding the reaction.
:0: /8+ poor resonance contributor,
-'0·
11/
. ,.
+
destabilizing the starting
+
C-C-OCH3
material; no effect in the
r
-transition state
:0:
_. .
(b) One of the resonance forms of methyl p-methoxybenzoate has all atoms with full octets, and negative
charge on the most electronegative atom. This resonance form stabilizes the starting material, increasing
the activation energy, slowing the reaction.
good resonance contributor,
stabilizing the starting
material; no effect in the
transition state
'---- reaction
=C
21-64
�COOH
+
NH3
___
�COO-
A
531
21-65 A singlet at 2.15 is on carbon next to carbonyl, the only type of proton in the compound. The IR
spectrum shows no and shows two carbonyl absorptions at high frequency, characteristic of an
anhydride. Mass of the molecular ion at 102 proves that the anhydride must be acetic anhydride, a reagent
commonly used in aspirin synthesis.
°
°
cS
H
OH,
II
""'"
C
II
' /
0
C
'
Acetic anhydride can be disposed of by hydrolyzing (carefully! exothermic!) and neutralizing in aqueous
base.
21-66
NMR spectrum:
IR spectrum:
-sharp spike at 2250 cm-I
-triplet and quartet
-1750 cm-I
-this quartet at 8 4.3
° }
maybe
an
ester
-2 singlet at 8 3.5 isolated
-1200 cm-i
°
H 3C
===> C
===> C
===> C
CH3
===> CH3CH2
N
===> CH3CH20
=
===>
H
-
+
°
II
+
+
CH2
sum of the masses is 113, consistent with the MS
The fragments can be combined in only two possible ways:
CH3CH20
°
II
CH3CH20 - C - CH2 C
C
CH2
C
N
°
II
N
CH3CH20CH2 - C - C
N
The
proves the structure to be A. If the structure were the between oxygen and the carbonyl
would come farther downfield than the of the ethyl (deshielded by oxygen and carbonyl instead of by
oxygen alone). As this is not the case, the structure cannot be B.
The peak in the mass spectrum at mlz 68 is due to a-cleavage of the ester:
B
A
B,
NMR
CH2
CH2
[
CH3CH20
tR
- CH2 C
mlz 113
68
N
]t
..
1J :R:
mlz 68
+ CCH2CN "
532
..
:
?t
I
CCH2CN
Lr
The formula C5�NO has 2 elements of unsaturation.
IR spectrum: The strongest peak at 1670 cm-1 comes low in the carbonyl region; in the absence of
conjugation (no alkene peak observed), a carbonyl this low is almost certainly an amide. There is one hroad
NHlOH
s on m d .
o
H
II
-C - NI HNMR spectrum: The broad peak at 8 7.55 is exchangeable with D20; this is an amide proton. A broad,
2H peak at 8 3.3 is a CH2 next to nitrogen. A broad, 2H peak at 8 2.4 is a CH2 next to carbonyl. The 4H
peak at 8 1.8 is probably two more CH2 groups. There appears to be coupling among these protons bUl it is
not resolved enough to be useful for interpretation. This is often the case when the compound is cyclic, with
restricted rotation around carbon-carbon bonds, giving non-equivalent (axial and equatorial) hydrogens
the same carbon.
o
H
I
II
CH2 CH2
CH2 - C - N-CH2
The peak at 8 175 is the C=O of the amide. All of the peaks between 8 25 and 8 50 are
CNMR spectrum:
aliphatic sp3 carbons, no sp2 carbons, so the remaining element of unsaturation cannot be a C=C; it must be
a ring. The carbon peak farthest downfield is the carbon adjacent to N.
The most consistent structure:
o
II
C, H
N/
21-67
peak in the
region, hinting at the likelihood of a ec
da ry a
i e
on
+
+
O
8-valerolactam
533
21-68
IR spectrum: A strong carbonyl peak at 1720 cm-I, in conjunction withI the C-O peak at 1200 cm-I,
suggests the presence of an ester. An alkene peak appears at 1660 cm- .
o
C - O-C
C=C
HNMR spectrum: The typical ethyl pattern stands out: 3H triplet at � 1.25 and 2H quartet at � 4.2. The
chemical shift of the CH2 suggests it is bonded to an oxygen. The other groups are: a 3H doublet at () 1.8,
likely to be a CH3 next to one a IH doublet at 05.8, a vinyl hydrogen with one neighboring H; and IH
multiplet at 06.9, another vinyl H with many neighbors, the far downfield chemical shift suggesting that is
beta to the carbonyl. The large coupling constant in the doublet at 05.8 shows that the two vinyl hydlTlgens
are trans.
0
/H
II
C=C
CH3-C
OCH2CH3
C-O-C
1
If
H
There is only one possible way to assemble these pieces:
o 1.8, d H3C
H 05.8, d
\
C=C\
/
06.9, m H
C-OCH2CH3 01.25, t
II
+
o
04.2, q
CNMR spectrum: The six unique carbons are unmistakeable: the c=o of the ester at 0 166; the two vinyl
carbons at 0144 (beta to C=O) and at 0123 (alpha to C=O); the CH2-O of the ester at 060; and the two
methyls at 018 and at 0 14.
Mass spectrum: This structure has mass 114, consistent with the molecular ion.
Major fragmentations:
H3C
H
.
\
CH3CH20.
C=C
/
\
mass 45
C+
H
II
:0:
plus two other
resonance forms
+ H
rnIz 69
H3C
\
C=C\
/
+
C-O=CH2
H
plus one other resonance form
rnIz 99 g ..
II
a
H;
/
+
+
+
/
+
+
---J"�
/
534
/
It
If you solved this problem, put a gold star on your forehead.
The formula C6Hg03 indicates 3 elements of unsaturation.
IR spectrum: The absence of strong OH peaks shows that the compound is neither an alcohol nor a
carboxylic acid. There are two carbonyl absorptions: the one about 1770 cm I is likely a strained cyclic
1150 em-I),
1720 e m
(An anhydride also has two peaks, but they are of higher frequency than the ones in this spectrum.)
o
o
II
II
C-O-C
C-C - C
Proton NMR spctrum: The NMR shows four types of protons. The 2H triplet at 84.3 is a CH2 group
next to an oxygen on one side, with a CH2 on the other. The IH multiplet at 83.7 is also strongly
deshielded (probably by two carbonyls), a CH next to a CH2. The 3H singlet at 82.45 is a CH3 on one of
the carbonyls. The remaining two hydrogens are highly coupled, a CH2 where the two hydrogens
equivalent. There are no vinyl hydrogens (and no alkene carbon in the carbon NMR), so the remaining
element of unsaturation must be a ring.
Assemble the pieces:
o
o
II
II
\: 0-CH2CH2CH
C - C - CH3 1 rinV
21-69
-
ester (reinforced with the C--O peak around
while the one at
-
I is probably
a
ketone.
arc not
-
+
y
o
+
0
C...... CH3
o A�
K
Carbon NMR:
On each carbon, the "up" hydrogen is cis to the acetyl group,
H H H H 1r while
the "down" hydrogen is trans. Thus, the two
hydrogens on each of these carbons are not equivalent,
leading to complex splitting.
o 853 0 8200
8173 --g � g /
0/ 'C/ 'C
830
/
\
868 ---C - C
---
•
824
535
CHAPTER 22CONDENSATIONS AND ALPHA SUBSTITUTIONS OF CARBONYL COMPOUNDS
22-1
(a)
O�
-
OH
CH2 -�==CH2
2
(b) Enol 1 will predominate at equilibrium as its double bond is conjugated with the benzene ring, making it
more stable than 2.
(c)
basic conditions fonning enol 1
H :0:
H 0
_ I II
I II
Ph - C - C - CH3 ...;:::=:=:::: Ph - �-C-CH3
(I
-:··OH
H
::::
�
1
acidic conditions forming enol 1
H :O :� n
H : �- H
I II
I II
H-B..
Ph-C-C-CH3
Ph-C-C-CH3
I
I
H
H
1
basic conditions forming enol 2
o
H
:0:
.
II
\I (I �
:�I-!.
PhCH2 - C-CH2 ...;:::
=:
::: ::: PhCH2-C-�.H2
::::
1
----
H :O-H
I I
Ph-C-C-CH
3
IJ
H
+
�
}
ri'
H :O-H
I I
Ph-C==C-CH 3
1
..t----l.�
....
: O-H
I
PhCH2 - C CH2
==
2
537
(c) continued
acidic conditions forming enol 2
22-1
-H
II
�
+
:O
�
H
B.. PhCH2-C-CH2
PhCH2-C-?H2
I
H
:O:
II
:O - H
I
PhCH2-CCH2
+ .."-I
H
. .
---­
H
�
B: ,
I
PhCH2-C CH2
=
�
2
0"
22-2
:O:
. .
:O-H
'CH3 H.... I')..
B:
planar enol
This planar enol intermediate has lost all chirality. Protonation can occur with equal probability at either
face of the pi bond leading to racemic product.
H
:0/
o
R
�" "CH3
• •
(plus one other resonance form for each)
22-3
o
...
racemicU
mixture
:0:
. .
..
unaffected "
by base
H"
_ ,
cis
mixture of diastereomers
CIS
0
CH3
CH3
11
H20
+
538
0
H
CH3
trans
H
(a)
.
:0:
:0:
I
..- II
.. .. H2C=C- CH3
H2C-C-CH3
..
:0:
:0:
H
tH
22-4
.
(b)
(c)
H3C
C
:0:
...
:0:
(y
..
)l C. )l CH3 - H3C
I
.
..
H
22-5
&'''�H
-
..
22-6
H�
- ..
:OH..
Ph-C�C. - H
I
H
II
H
rl
•
•
.
i
:0:
.
..
:0:
V
0
H
CH3
o
.:0.:- t
:0:
II
I
Ph-C - -S: H2" - Ph-C = CH2
.
�
C
0
+
CI-
CI
I
Ph-C - C-H
I
CI 01
H)
r
-:RH
----
o
II
..:0:- !CI }
CI
:R: j:l
I
j
e
l
... .. Ph-� = �H
Ph - Ci CI ... �
Ph-C-CH
..
l
C: 1 CI 1
!l
..
:OH
�
..
o
_
1
CH"
3 -H 3 C
:0:
H
&
Cf_
"'"'--:OH
•
:0:
V
�
:0:
o
o
..
:0:
....----.
�
_____
'"--"
I
-
:0: CI
:..0: CI
II _ I
I
I
Ph-C-C-Cl .. - Ph-C==C-Cl
• •
---
CI
I
Ph-C-C-Cl
I
Cl \...CI
.(
CI
}
-
539
o
II
22-7
o H �-,
:OH
II rl
Cy-C�C-H
I
H
i
. •
i
Q4
= Cy
}
o
:0:
For this problem, the cyC\ohexyl group is abbreviated "Cy".
:0:
"
-
•
•
---
I
Cy-C-CH2 ..
1
·.0· -Br
o Br
"
Cy-C" -�- Br .. - B Cy-C-�I H
I)
b �
..
•
}
(b)
•
HCBr3
bromoform
. .
(a) � COO- Na+
CHCl3
U
.
II
I
I)
Cy-C-C-H
(H
-..
:OH lI
:0: Br
I
I
Cy-C=CH
na· Bf
�II· I
+
----
rH
(;:1 r
:0:
,,
�
Cy-C-O
-H
+
-
'--../
o Br
" I
(c) Ph-C-C
-CH3
I
Br
:CBr3
Coo- Na +
CHI3
(precipitate)
22-9 Methyl ketones, and alcohols which are oxidized to methyl ketones, will give a positive iodoform test.
All of the compounds in this problem except pentan-3-one (part (d)) will give a positive iodoform test.
+
}
·0·
CY-C�
- C-Br_--::- Cy - C - CBr3
---A Br
r
I
:RH
0
:0: Br
I I
Cy - C=C-Br
o
"
Cy - CR:
22-8
Br0r
•
.. Cy-C=CH2
• •
�J
:0: Br
" -I
Cy - C - C - Br'"
Br
..
�
U
540
+
H
:0""'"
22-10
:0
+
.......
II
�cy"
H
o
22-11
(a)
� COOH
22-13
(a) 0
0
Cl2
CH3COOH
from Solved Problem 22-2
6
Br
22-12
0
0ly
Br
�
•
6
•
�H
�c
y
Br
II
:Br:
0 0-
0
H
�j ci
I
.......
BGBr
..
0
Cl KOH
•
L1
6
E2 elimination
(b) no reaction:
no a-hydrogen
(b) 0
H
I
· �?V
\ I
AcO :
H�
o 0
03
(c)
0
HO 'Jl'I I( OH
Br 0
3
+
541
.......
.
(d) no reaction:
no a-hydrogen
..O
V
�
,
..
J..
.....
VI
�
II
""0
A
::s-
rQ zj
�
n (�+
\l
.
.....
...
.
::c
\
0. .
o
I
N
::c
t :s:
(
\
Q={J
+'iJ
.�+.
n
o =�
t
..�
G
,
J..
\
Q. t t
O
Z
Z
+
Q
QO
t
n+-. :
0
�
:
�
�
r
Q�
q<J
�
5
ln
<
���
C
C
Q
�
�
Z
0
0
G
�;Y
::c
�
Ul
�
N
n
N
N
::c
+n-
::c
�
o
\
I
\.
+
t
�
y
::c
::c
IJ
\
o
::c
::c
::c
::c
J
"
�:q:
�
n
\
0
::c
::c
::c
)
y
N
\'"
·o-::c
n
::c
::c
o
::c
I
::c
z
t��
O.
.::c
C�
0
z
•
::c
:
N ... CH3
Ph-C-CH3
22-16
(a)
II
22-17
(a)
(b)
(b)
I
Any 2° aliphatic amines can be used for this problem.
0
6
+
0
H
(c) 0"
PhC-CH 3
H + ..
- H2O
I
6
�
22-18
Cd) o-iJ
H3C' N ... CH3
Ph-C=CH2
H
I
+
0
n
N
6
H+
- H2O
•
1
)
2)
� Br
H30+
..
0
~
� Ph
o
0
PhC=CH2
I
0"
1) PhC-CI
2) H30+
•
0"
0"
PhC- CH2 -CPh
HO-H
...
Y\
In general, the equilibrium in aldol condensations of ketones favors reactants rather than products. There is
significant steric hindrance at both carbons with new bonds, so it is reasonable to conclude that this reaction
of cyclohexanone would also favor reactants at equilibrium.
543
OH
a
22-19
( )
o
b
( )
H
Ph
22-20 All the steps in the aldol condensation are reversible. Adding base to diacetone alcohol promoted the
reverse aldol reaction. The equilibrium greatly favors acetone.
o
o
CH3 H � 2
CH3
() I ••II
I ("I CO 3 ..II
CH3 -C-CH2 -C-O· :
CH3 -C-CH2-C-O•• :
I
IU
CH3
CH3
+
H
II
I
CH3 -C-CH2
o
i
22-21
+
:O :�
:O-H
II
H - B CH3-g-CH2
CH3 -C-CH2 B
HI
HI
CH3 H B:II
I
I
CH3-C-CH2-?-O
CH3
a
---
�
....
..t---i
.
..
}
:O-H
•·O-H
I
CH3 -�+ -CH2
CH3-C==CH2
"-I
H�B:
�
. .
__
O-H
:I
H
H3
:R�
?
CH3 -C-CH2- C -O�
I
CH3
CH3-C-CH3
+
H3
-H
:
?
?
�:
CH3 -C-CH2C-O
+
I··
CH3
carbons of the eiectrophi Ie
are shown in bold just to
keep track of which carbons
••
+
....
If-..
.
-il
..
�
come from which molecule
544
(a) acidic conditions
22-22
CH3 H
I I
H C-C-CH -- C-O : �W
I
I ·.
H CH3
3
II
o
(b) basic conditions
o
CH3 H
II
I I
H3 C - C - CH-C - O
(I
I
H�- :..OH
H O
I II
H3C-C-C-H ..
- ..
(I
H�:OH
•
{
i
CH3 H
:0:
CH3 H
:0:
I
I I
II - � 1,,1
H3C-C==CH-C-0 ......'---;
. .�H3C - C - CH-C-O:
I
I
CH3
CH3
o
CH]
/
II
H3C-C-CH=C
\
CH3 :RH
. .
----l--<
22-23
•
• •
• .
H :0:
_I II
H3 C-C-C-H ...
�
\
CH]CH2 � fQ
•
• •
j
�
• •
..- }
H :0:
I I
H3C - C == C - H
+
HO CH3 0
n �:O: CH3 0
II
I I
... - H CH3CH2 - TI - TI - CI I - H
CH3CH2 - C - C- C - H HO
I I)
H H
t R0
-:
H-O) CH3 :0:
HO CH3
I I
I I II
. .�CH3CH2 -C - C
CH3CH2 -C-C - C-H ......'---;
IU
I
H ·
H
.:0:
I
C-H
==
545
.
}-
CH3 0
I II
CH3CH2 - C == C - C -- H
I
H
22-24
(a)
(b)
Ph
o
H
22-25
(a)
Step 1: carbon skeletons
�
H3C
�
,0
-&-11
C
C-H
�_I
•
•
--H
H
H
H
I
·
I
:0:
i(2,2-Dimethylpropanal has no a-hydrogen.) t
H-C�C-H
·
•
•
:0:
1
++1
H
H:0:
.
II
•
H
H -C-C-H
.
H-O
H :0:
I I !
33C-C-C�C-H...
(CH)
I
a
\""­
H-OH
C-H
(CH3hC
Step 4: conversion to final product
i
+
H
_I II
Step 3: nucleophilic attack
H
HC��
3
�
CH 3 q .
:0:
H-�-C-H
-
-:OH
H�··
CH3
H3 C
H
0
H
H
CH3
Step 2: nucleophile generation
H 0
H
�0
>
HH
H
H3
M Ph
CH3
H
(c)
H
H-O: H
� II
•
H
0
C-H
(CH3hC
�
H H\
I-:.OH
,
H : 0:
O
H-:
.
}
(I I II
(CH)
33C-C-C-C-H
I U·
H
j
H
I
0
"
(CH3hC-C�C-C-H
Step 5: combine Steps 2, 3, and 4 to complete the mechanism
546
�
22-25
continued
b
Step 1: carbon skeletons
( )
Ph
�
H
O
H3C
H
Step 2: nucleophile generation
>
Ph -{
H
+
o
Step 3: nucleophilic attack
Step 4: conversion to final product
CH3 0
I
II
Ph-C==C - C-H
I
H
Step 5: combine Steps 2, 3, and 4 to complete the mechanism
547
This solution presents the sequence of reactions leading to the product, following the fonnat of the
Problem-Solving feature. This is not a complete mechanism.
Step 2: generation of the nucleophile
H :0:
H :0:
H 0
_I II
I II
H-b = b-c�
H-C-C-CH3 - H-C-C-CH3 ..
(I �:OH
··
H
Step 3: nucleophilic attack
H- O : H 0
�:
-� :�:
1 I II
__ H-OH
Ph-C-C-C-CH3
3
Ph-C-H
H-C-C-CH
� ..
1 I
H H
Step 4: dehydration
H - O) H 0
H 0
1 II
I 1 II
Ph-C=C-C-CH3
Ph-C-C-C-CH3
1
I '--- I
:OH
H
H H� ··
22-26
i
•
• •
.. �
•
+
•
The same sequence of steps occurs on the other side.
- ..
H 0
H 0 H
o
:RH
I II
II
I " I
Ph-C=C-C-CH3 Ph-C-H
.. Ph-C=C-C-C=C-Ph
I
I
I
H
H
H
final product
+
548
:0:
:0:
. .
:��
A
l'�
HC-Ph
EtO-�
CH3
..
There are three problems with the reaction as shown:
1. Hydrogen on a 3° carbon (structure A) is less acidic than hydrogen on a 2° carbon. The 3° hydrogen
will be removed at a slower rate than the 2° hydrogen.
2. Nucleophilic attack by the 3° carbon will be more hindered, and therefore slower, than attack by the
2° carbon. Structure B is quite hindered.
3. Once a normal aldol product is formed, dehydration gives a conjugated system which has great
stability. The aldol product C cannot dehydrate because no a-hydrogen remains. Some C will form, but
eventually the reverse-aldol process will return C to starting materials which, in tum, will react at the other
a-carbon to produce the conjugated system. (This reason is the Kiss of Death for C.)
B
(a)
22-28
H
22-29
Ph
yY
H
0
22-30
0
H
==�>
I
C
Ph
H
(b)
i1n
H
0
>Vy
H
==�>
I
C
o
Ph
o
Ph
Y
0
H
o
- --
The formation of a seven-membered ring is unfavorable for entropy reasons: the farther apart the
nucleophile and the electrophile, the harder time they will have finding each other. If the molecule has
possibility of forming a 5- or a 7-membered ring, it will almost always prefer to form the 5-membered ring.
a
549
:0:
: 0:
22-32
(a)
�)
(c)
II �
a
PhC - CH2CH3
a
H
+
II
a
not feasible: requires condensation of two different
aldehydes, each with a-hydrogens
a
CH3CH2C Ph feasible: self-condensation
-
feasible: only one reactant has a-hydrogen; cannot use excess
benzaldehyde as acetone has two reactive sites
feasible; however,
the cyclization from
the carbon a to the
aldehyde to the
ketone carbonyl is
also possible
550
a
CH3
feasible: symmetric reagent will
give the same product in either
direction of cyclization
22-33 (a) and (b)
- ..
0 :RH
starting ( CkJ
diketone
o
0
o
o
22-34
(a) The side reaction with sodium methoxide is transesterification. The starting material, and therefore the
product, would be a mixture of methyl and ethyl esters.
0
o
II
II
H3C -C-OCH2CH3
NaOCH3
H3C-C-OCH 3 NaOCH2CH3
(b) Sodium hydroxide would irreversibly saponify the ester, completely stopping the Claisen condensation
as the carbonyl no longer has a leaving group attached to it.
o
II
H3C -C-OCH2CH3 + NaOH
22-35 CH3 0
+
:::...;:::=="�
I
II
H3 C-C-C-OEt - ..
:OEt
"( I
H '----.-/
• •
CH 3 0
3 2CH - C - C - C - OEt
(CH)
1
CH3
II I II
o
explanation on next page
551
+
continued
There are two reasons why this reaction gives a poor yield. The nucleophilic carbon in the enolate is 3°
and attack is hindered. More important, the final product has no hydrogen on the a-carbon, so the
deprotonation by base which is the driving force in other Claisen condensations cannot occur here. What is
produced is an equilibrium mixture of product and starting materials; the conversion to product is low.
22-35
22-36
(a)
� OCHJ
(b) Ph Jl Jl
......." '( 'OCHzCH3
Ph
o
0
o
0
_
22-37
i
H O
H :0:
PhCHz �-g-OMe 00 PhCHz 00�-=-g-oMe
:OMe
I)
H
�
,-----/ ..
H 0
I
PhCH2CHz-C" -C-C" -OMe
1
CHzPh
Ph
Ph
- MeO......1-
--
III
�o:OMe
o0
o
o0
:0:
OMe
o
III
!
OMe
1
�
PhCHz � == b-OMe
o0
(?: f R
:0:
Ph
�
W
(workup)
====
:0:
}
PhCHzCHz-C-C-C-OMe
I) 1
MeO CH2Ph
OMe
:0:
Ph
�:D
PhCHzCHz-C - OMe
o
o
III
H
o0
:0:
:0:
:0:
OMe
H 0
" 1 "
PhCHzCHz-C-C-C-OMe
1
CH2Ph
552
o
o
II
o
II
(c) CH3CHCH2 -C -OCH2CH 3
I
CH3
This one would be difficult
because the alphiJ-carnon
hindered.
(b) PhCH2C - OCH3
IS
H�
II
CH - C -OEt _:9_.E_t ._
�
C-OEt
((I
22-39
. .
g
I
I
C
.
This is the final
product, after
removal of the
hydrogen by
ethoxide, followed
by reprotonation
during the workup
a-
I) I
C
0
:g�
- EtO-
0(
0
-
•
.
II
o
a­
G{C-OEt
CH - C -OEt
C "":-...
H�
I
: RMe
CH - C-OMe ---I
I.�
C-OMe
This is the final
product, after
removal of the
hydrogen by
ethoxide, followed
by reprotonation
during the workup.
II
-�.H - C-O
�
:0:
C
0-
C
0
Cg
o
II
CH - C - OEt
I r'4
C-OEt
II
.
:0:
�
:0
yH - C-OMe
C ""::--...
CH=C-OEt
C-OEt
:�)
..CH - C-OMe
{ C-OMe
o
II
j
I
. .
:0:
- MeO-
..
I
CH=C - OMe
C -OMe
Cg
j
I
o
CH - C - OMe
I"
C-OMe
C6)
:
553
.
:0:
(c) C(COOCH3
COOCH3
H 0
I "
H-C-C-OEt -..
I)
H'---.-/ :OEt
22-41
i
Ph
)
H
o
#
�:OEt.
�
o
0
H :0:
_I II_OEt
H-C-C
H
r :
0
�g-OEt
II I II
Ph-C-C-C-OEt
I
o
OEt
Pho/OEt
H
�
0
H
:0:
3
..
�
H :0:
I I
H-C=C-OEt
\
t
(:0: H 0
�I I II
-EtO- Ph-C-C-C-OEt
I) I
EtO H
: 0:
:0:
:0:
. .
:0:
:0:
Ph- �'OEt Ph-� Jl'OEt
� Jl OEt�Jl
T
T
H
H
H
---
-
wj
H
II I II
Ph-C-C-C-OEt
I
o
+
+
NaOCH3
+
�COOCH3
�COOCH3
NaOCH (mixture of products results)
COOCH2CH3
O�
NaOCH2CH3
(d) C
COOCH2CH3
o
The protecting group is necessary to
prevent aldol condensation. Aqueous acid
workup removes the protecting group.
(b)
(a) not possible by Dieckmann-not a !3-keto ester
22-40
0
H
554
(a)
22-42
�
0
Ph
(c)
Ph
OCH3
�
0
EtO
(b)
0
(d)
0
OEt
�
0
V
0
EtO
II
o
'---.--/
Ph -C-OEt
(c)
+
EtO-C-CH2Ph
22-44
(a)
0
(f
(a) two ways:
22-45
(b)
II
0
H
--------
0
if
CH2CH3
II
o
0
II
CH3CH2CH2CH2"C-OMe
(c)
(b) �
0
V'
0
�
VyPh
o
OCHl
OR
o
II
+ CH3CH20-C-C-OCH2CH3
(d) two ways:
0
0�
0
�
II
o 0
�
PhCH
H
0
/'��II
0
- C - OMe + MeO-C-C-OMe
II
o
HO
�CII - Ph
CH3CH2-C-CH2CH3
(c)
0
actually present in the enol form:
6
OCH3
�
EtO-C-OEt
�
o
�
OEt
o
(d)
II
(CH3hC - C-OMe +
II
o
o
II
Ph
0
(b)
+ CH3CH2-C-OEt
II
OCH3
CH3
22-43
0
+
0
o
Ph
plus 2 self-condensation products-a poor choice
because both esters have a-hydrogens
CH3
o
(a)
0
II
0
C-OCH2CHl
555
�
o
OR
0
OCH2CHl
OCH2CH3
:0:
22-46
(a)
:0:
:0:
:0:
¥
¥
· .
:0:
:0:
y
y
· .
H3C
OEt
H3C
OEt
H3C Jl C)l OEt
I
H
H
H
:0: :0:
:0 : :0:
(b) :0: :0:
H3C
CHl
H3C
CH3
H3C Jl C Jl CH3
I
H
H
H
:0:
:0:
:0:
(c) ..
N
:N.�
N�
�C h�
"c,. ;"C
)l
OEt
OEt
OEt
C
I
H
H
H
:0: :0:
:0 : :0:
(d) :0: :0:
lI
- N" 'c-)l
hN
N
:0/
:0 /
CH
:0/
CH3
CH3
I
H
H
H
(other resonance forms of the nitro group are not shown)
22-47 In the products, the wavy lines cross the bonds that must be made by alkylation, before hydrolysis
and decarboxylation produce the substituted acetic acid.
0 0
(a)
0 0
H30+..
.Jl.Jl
1 ) NaOEt
EtO � OEt 2) PhCH-,B; EtO' I' �OEt
�OH CO2 2 EtOH
CH2Ph
CH2Ph
0 0
(b) �
0
1) NaOEt.. I) NaOEt 0 0 H30+
..
OEt 2) CH) 2) CH3I .. EtO
EtO
OEt
OH CO2 2 EtOH
-----
-----
· .
· .
-----
.
-----
-
+1
-----
· .
0
-----
· .
· .
+
-----
•
•
-
• •
• •
yl
yl
0
+
-----
-
• •
• •
3
o
+
II
y
II
�
+
+
+
0 0
(c) �
0 0
0
H30+
I) NaOEt
Ph
� OH CO2
EtO
OEt 2) Ph(CH2);Br EtO V OEt
CH2CH2Ph
0 0
(d) �
0 0 H30 +
I) 2 NaOEt
COOH
"
O2
EtO
OEt 2) Br(CH2)4Br EtO
OEt
..
�
556
II
..
II
6
+
+ C
+
''' 2
2
EtOH
EtOH
22-4 8
(a) Only two substituent groups plus a hydrogen atom can appear on the alpha carbon after decarboxylation
at the end of the malonic ester synthesis. The product shown has three alkyl groups, so it cannot be made hy
malonic ester synthesis.
desired product
0
0 0
0 0
H 0
EtO�OEI ==> EIOYOEI ==> H 00H � � C �OH
R\ R2
R \ R2
H3C CH3
()
Li+ :0: =,i+ :0:
'>-- '>-bH
+
:
' C AOCH3 --- CAOCH
Jl' OCH3 + :N\-HN\-:
x
Li+ /
/
H3C CH3
H3C CH3
H3C CH3
�
•
�
pKa 24
pKa 40
1
'"
1
'"
(to
}
With such a large difference in pKa values, products are favored 99%.
:0:
(c) 0
Be
r
r�
H Jl OCH3 LDA -' C A OCH3
H30+�
'
X
OH
VOCH3 /).
H3C CH3
H3C CH3
SN2
H3C CH3
H3C CH3
plus resonance form
as shown in part (b)
22-49
0
(b)
(c)
(a) PhCH2CH2 - C - CH3
co,
EtOH
CO2 EtOH
CO2 EtOH
22-50 In the products, the wavy lines indicate the bonds that must be made by alkylation, before hydrolysis
and decarboxylation produce the substituted acetone.
0 0
0 0 1) NaOEt
(a)
+ 0
CO2 EtOH
EtO� 2) PhCH2Br EtO� H�O . �
CH2Ph
CH2Ph
0 0
0
(b) 0 0 1) 2 NaOEt
H30+�
CO2 EtOH
EtO� 2) Br(CH2)4Br EtO ()' /).
LDA
»
•
o
'"
"
+
cr
+
..
+
o6
+
+
•
•
o
0
(c) 0 0 1) NaOEt
EtO� 2) PhCH2Br EtO�
CH2Ph
..
\3!'
+
+
+
o
0
NaOEt
2) Br� EIO
�Ph
) A �:'�+
0
1)
+
..
CO2 EtOH
+
557
+
+
�
(a) There are two problems with attempting to make this compound by
acetoacetic ester synthesis. The acetone "core" of the product is shown in the
box. This product would require alkylation at BOTH carbons of the acetone
"core" of acetoacetic ester; in reality, only one carbon undergoes alkylation
the acetoacetic ester synthesis. Second, it is not possible to do SN2 type
reaction on an unsubstituted benzene ring, so neither benzene could be attached
by acetoacetic ester synthesis.
(b)
o
o
� Br
0
LDA
II � . Ph
Ph
Ph �C
Ph � Ph
'
�
H
22-5 1
in
an
..
plus resonance fonns showing
delocalization of e- into ring and C=O
(c)
n
N
0
Ph � Ph
� Br H30+
Ph
Ph � Ph
..
�
-l..
---
0"
Ph - C1H l- �H2 - C - Ph came from Ph - CH
>
II 0
Ph -C-C - CH3
Ph - CH - C - CH3
Michael acceptor
forward direction
o
�CH O
Ph - C - CH2 NaOEt Ph - C� - CH
II) �
( COOEt)
(COOEt) Ph -C - C - CH3
resonance­
temporary
stabilized
ester group
22-52
�
y
"
"
a
"
+
CH2 - C - Ph
Michael
donor
- . ..
o
II
( OOEt�
I
Ph - CH - CH - C - Ph
Ph - CH t- CH2 - C - Ph
I
�
II
o
CO2 EtOH
+
o
I
Ph - CH - C -CH3
Ph - CH - C - CH3
22-53 First, you might wonder why this sequence does not make the desired product:
O
0
0
0
H
.�
� ft
CP H�H 0
..
- +�
..
poor yield
MVK
resonance­
stabilized
The poor yield in this conjugate addition is due primarily to the numerous competing reactions the
enolate can self-condense (aldol), can condense with the ketone of MVK (aldol), or can deprotonate the
methyl of MVK to generate a new nucleophile. The complex mixture of products makes this route
practically useless. (continued on next page)
+
+
�
�
---
ketone
558
continued
What pennits enamines (or other stabilized enolates) to work are: a) the certainty of which atom is the
nucleophile, and b) the lack of self-condensation. Enamines can also do conjugate addition:
22-53
H
MVK.
22-54
o
o
I
o
high yield
The enolate of acetoacetic ester can be used in a Michael addition to an a,l3-unsaturated ketone l i k�
0
0
0 0
�
�
NaOEt.
EtO
EtO
��
•
/::
�
o
CO2 EtOH
+
O
o
+
8
y
�
a
: N -' C - CH = CH2 ----- : N = C - CH= CH2 ----- : N C CH - +CH2 lr
acrylonitrile
Nuc :
22-55 J
+
1
-
- ..
•
•
==
! (
==
.
n �
Nuc
N C - CH2 - CH2 .. - H : N - C-CH-?H2 ----- : N C CH ?H1
I
Nuc
Nuc
Nuc
i
: o:
}
-.
==
==
-
:0:
:0:
I
II
2
: O - N - CH = CH2 ----- : O - N = CH - CH
: O - N+ = CH - CH
�
I
Nuc :
Nuc
+
nitroethylene
+
(some resonance fOnTIS of
the nitro group are not
:0:
hown)
II
:0 - N - CH-CH2
:0 - N CH2 CH2 ..
I
I
Nuc
Nuc
i
-
-
• •
• •
+
• •
• •
+
+
- ..
•
s
•
•
+
-
-
•
559
•
+
-
2
}
.
22-56
o
(a)
II
PhCH CH - C - OEt
�
==
EtO
o
(c) two ways
H2 = CH - C N
:
OR
0t
�
(d)
(f)
0
OEt
followed by hydrolysis
and decarboxylation
(b) CH2 = CH - C - �
�
EtO
followed by hydrolysis
and decarboxylation
0 (1\
�
II
OEt
o
0
followed by hydrolysis
and decarboxylation
o
CH2 = CH - C N
� �aOEt
PhCH CH - C - OEt
�
0
followed by hydrolysis
o
EtO
o
o
o
H3C.... /CH3
0
�
II
� CH3
CH2 == CH - C - Ph
U
followed by hydrolysis
==
OEt
OEt + �
o
(could also be synthesized by the
Stork enamine reactions)
560
22-57
Step 1: carbon skeleton
CH3
comes from
>
o
Step 2: nucleophile generation
. .
0:
o
These hydrogen atoms more
acidic than the other alpha
hydrogens because the anion
can be stabilized by the
benzene ring.
H
plus resonance forms
with negative charge
on the benzene ring
Step 3: nucleophilic attack (Michael addition)
H
continued on next page
. . -
0
561
22-57 continued
Step 4: conversion to final product
(nucleophile
formation)
plus one other (enolate)
resonance form
•
(nucleophilic
attack)
o
I
t
: 0:
o
(base-catalyzed dehydration)
o·
0
..
plus one other (enolate)
resonance form
Step 5: The complete mechanism is the combination of Steps 2, 3, and 4. Notice that this mechanism is
simply described by:
1) Enolate formation, followed by Michael addition;
2) Aldol condensation, followed by dehydration.
22-58
o'"
U
Step 1: carbon skeleton
"
o
II
CH=CH-C-OH
�
Step 2: n leophile generation
II
CH2 - C-OCOCH3
D
H
o
0
II
II
CH3-C-O-C-CH3
comes from
=====�
>
•
CH3COO-
i
-
: 0:
II
: CH2 - C-OCOCH3
562
-.. }
·
....
..I----!.�
CH2
•
0
·
� � OCOCH3
OH
22-58 continued
0
�
g
: O: �H -OAC
'-.."
I
Step 3: nucleophilic attack
II
·0:
II
�
Ph-C -H
: CH2-C-OCOCH3 ---- Ph-CH-CH2-C-OCOCH3
+
Step 4: conversion to final product
0
OH
I
II
Ph-CH-CH-C-OCOCH3
I)
Ph- H -CH2 - -OCOCH3
-
0
_
0
CH3COO-
(OH
I
: 0:
II
Ph -CH - CH-C-OCOCH3
U.
plus one other (enola!e)
resonance fonn
0
..
_
�
H
II
Ph-CH = CH - C-OCOCH3
�
hydrolysis
: O:
II
II
H+
..
PhCH==CH-C-0-C-CH3
�H+
OH
:0:
I
II
PhCH==CH-C-0-C-CH3 ..
••
I
H20:
'"" +
/ H -0 -H
..(
H20:
I two fast
t proton transfers
••
Q
+
OH
.O-H
I
II
PhCH==CH-C-0-C-CH3
IV
OH
plus two other
resonance fonns
i
:
O -H
II
t
0
II
PhCH==CH - C-0-C-CH3
�
t
• •
:O -H 0
I
II
PhCH==CH-C-0-C-CH3
+
.
plus one other resonance fOlm
with positive on the other oxygen
I
O -H
... O==C
I
+
C�
o
II
PhCH==CH - C-OH
i
PhCH==CH -
..
+
t
: OH
� 0�
-H -O.
•• /
H20:
?
··
H -O:
PhCH==CH -
II
?
:OH
..
}
plus one other resonance fonn
with positive on the other oxygen
Step 5: The complete mechanism is the combination of Steps 2, 3, and 4.
563
22-59 The Robinson annulation consists of a Michael addition followed by aldol cyclization with
dehydration. In the retrosynthetic direction, disconnect the alkene formed in the aldoVdehydration, then
disconnect the Michael addition to discover the reactants.
(a) aldol and dehydration forms the a,� double bond:
o
>
o
Michael addition forms a bond to the �' carbon:
;>
o
(b) aldol and dehydration forms the a,� double bond:
o
o
o
;>
o
Michael addition forms a bond to the W carbon:
o
o
22-60 Please refer to solution 1-20, page 12 of this Solutions Manual.
S64
22-61 The most acidic hydrogens are shown in boldface. (See Appendix 2 for a review of acidity.) (Braces
around resonance forms omitted here.)
: 0:
:0:
o
(a)
o
/H
--
·9
.
. .
R:
.
H
H
I
y f::y o: :0
�V--
H
o
o
I
H
:O
. .
--
. .
:0:
:0:
: 0:
H
0:
0:
--
(d)
. .
. .
. .
:0
. .
H
H
. .
. .
. .
. .
0:
0:
. .
same enolate as in (b)
�"
U
(e) 0
0
:0:
�
COCH
(f)
: 0:
: 0:
Because of the stereochemistry of
the double bond, the H atoms on
the left side of the ring are not
equivalent to those on the right
side. However, the two enolates
will be equivalent except for the
double bond geometry.
565
:0:
22-61 continued
a
H
"
I
CH
H ' ...... C�
....
..
"
c
c
I \
I
H H
H
H
(g)
'iC ...... 1 ...... CH
H2C
C
I
I H
"
I
- ........ C� ...... CH
H 2C
"C
....--
I
H
a
H
(h)
---
H
: 0:
H
"
---
"
I
- ........ C� ...... CH
H2C
"C
I
H
: 0:
"
'iC ....... ;-...... CH ..
H2C
C
H
I
....--
I
H
same enolate as in (g)
H
"
I
'iC .....
.. ;-...... CH ..
H2C
C
: 0:
I
H
..
H
: 0:
..
:0:
I
I
'iC ........... CH
H 2C
C/'
I
H
H
..
:0:
I
I
'iC ........... CH
..
H 2C
C /'
H
I
11011 0°11
22-62 In order of increasing acidity. The most acidic protons are shown in boldface. (The approximate
pKa values are shown for comparison.) See Appendix 2 for a review of acidity.
a
H
uCOOCH3
<
a
.6COOCH3
<
(a)
pKa 13
.......... .... .. ....
<
ex:
<
(c)
pKa 11
_
<
(f)
pKa 17-18
a
,.----
,
H
(b)
pKa20
(g)
pKa25
least acidic
1_
H
0COOII
(d)
pKa5
........ ............ ....
fully deprotonated
by ethoxide ion
566
a
<
CN
H
coaCH3
(e)
pKa2
most acidic
22-63
o
0
H.
O� '0
�
The
enolgationfonnandisbecause
stable because
of the
conj
u
of
i
n
tramolecular
hydrogen-bonding in a six-membered nng.
CH
....
CH;�Jl
M
T
H
enolH
keto
0
Inthedielectron-donati
carbonyl compounds
in
general
,
the
weaker
ngtheabienollity fonn:
of the group
G, the
more
i
t
wi
l
l
exist
in
aldehydes
G
N
(ketG ones, areestealmost
completely
enolized,
t
h
en
rs, andcontent.
finally amides which have
virtually no enol
keto
cH
cH3
---
3
H
o
�
)
=
H)
22-64
H
enol
TheO wavy0 line lies across the bond fonned in0the aldol condensation.
(b<i)
0
-H20
H
..
H
OH
0
(c) OH 0 O H20
Ph�Ph .. Ph�Ph
H
(e)
22-65
H
H
o
_
2
o
The wavy 0line lies across the bond formed in the Claisen condensati0on.
0
O (c) A�
(b
�
)
OCH3 �
D
o
.....
,.,
o
Cxj
o
567
o
c6
89S
H
t
.. ..
H
D - H
1 ..
)
H O:
H
o
:.Q:
..
H -OH
�
u
H
H
(e)t9-ll w�lqOJd U! uO!1esu�puo::lloPle)o ws!ue4::l�w (e)
99-ll
69S
OZH-
-----
: 0:
)
t�:O-H
0
\Jy"z:
:a
0
O-H
----
: 0:
H
-----...
H
t
:O-H
+
H-O
H-O
w�
H-O·
H-OI:H
o
+
•
a
a
a-H
V� H
:O:
(q)W-ll W�IqOld U! u0!lusu�puo::> 10PIU)0 ws!ump�w (q)
p�nu!luo::> 99-ll
22-66
continued
(c) mechanism of Claisen condensation in problem 22-65(a)
�o
H
..
..
OMe -:OMe
H�
OMe
I I
A\
_
•
OMc
4
.
�Me
1 J:
OMe
/T"
-MeO-
product wiluponbe deprotonated
but(thisregenerated
acidic workupby) methoxide,
(d) mechanism of Ciaisen condensation in problem 22-65(b)
o
o OMe
H
'---..-- ..
OMe
:0:
..
..
'" -: OMe
:0:
o
•
~
-MeO-
product willuponbe deprotonated
but(thisregenerated
acidic workupby) methoxide,
570
:..
0:
o
All products shown are after acidic workup.
C
(a) aldol self-condensati
o
n
\YT
22-67
CHO
Ha
H
a
Claisen self-condensation
Oll
(c) aldol cyclization
(b)
aEt+
a
a
�
Ha- ..
a
(d) mixed Claisen
lf
OR lfCH3
+
�
caaEt
t
Ea
-
a
OC
H3C �
Eta-..-
+
E1O
Eta-..
�
a
a
a
a
�
a
a
(e) mixed aldol
COaEt
Ha-
-
a
a
OEl
�
+
�
� Ph
lfCH3 H Jl ph
(f) enamine acylation-attempt at aldol would give self-condensation
Q
Ph
CI .Jl ph
a
a
6
Ha-
+
..
�
a
a
+
a
_--1"-_
HP·.
571
a
o
II
(d) 0
o
(c) CH3CH2CH2 hCH3
(b)A
CO2
+
� + CH3CH20H
Ph
0
(a) CH3CH2 -C-CH2CH3
+ CO2 + CH30H
22-68
0
.....
(e) Ph�
U
o
(f)
V
....
CH3
0 H2CH2CH2CH3
CrY
OCH3
o
(g) 6CH2CH2CH2CH)
22-69
(c) reagents: excess (or Br2 or Cl;»,
(b) reagents:
(a) reagents: Br2' H+
NaOH
followed byBr2'H20PBr3,
(d) 0
PhCH=CHCH3 Ph3P=O
Ph-C-H Ph3P-CHCH3
Ph
O -Ph NaOH
Oyo + <
CH3
H
H
In the products,
the wavythe lsubstituted
ines indicatacet
e thiecbonds
and decarboxyl
ation produce
acid. that must be made by alkylation, before hydrolysis
0 0
(a) �
NaOEt .. V0 H30+ �0
EtO
OEt
EtO
OEt
OH CO2 EtOH
CH2Ph
CH2Ph
(b) 0 0
0
NaOEt .. Y0
NaOEt
EtO�OEt
EIO
OEI �
OEt
\.-J Br
�
O
u(
CO2 + EtOH +
0H
12
+
II
-
+
..
(e)
..
o
22-70
1)
II
II
1)
o
W
+
o
!1
o
"
+
1)
•
2
2
572
o
+ 2
continued
NaOEt
COOH + cO2 + EtOH
(c) a a Br(CH2
)sBr
EtO�OEt
In the products,
the wavy
lines indicateacetone.
the bonds that must be made by alkylation, before hydrolysis
and decarboxyl
ation produce
the substituted
(a) a a NaOEt
a
NaOEt
EtO � EtB: EtO
() 2Br Eta
+ CO2 + EtOH
Ih
22-70
II
1) 2
II
•
2)
22-71
�
0
1)
2)
(b) a a
EtO �
1) 2
NaOEt
0
EtO
1)
2)
•
CH
�
()'
a a
2
6
o
o
~
+ cO2 + EtOH
(c) The acetoacetic ester synthesis makes substituted acetone, so where is the acetone in this product?
substituted acetone
The
substituted
acetone canon!be made by the acetoacetic ester synthesis. How can we
makesingle
the bonddoublto ethisbond?
Aldol condensati
a
a,13
make
makeizatibyoalnlddolehydration
additionby conjugate :
cycl
a a
a a
a NaOH a
NaOEt
..
EtO � + �
Eta
A
---
o
.�
573
22-72 These compounds are made by aldol condensations followed by other reactions. The key is to find
the skeleton make by the aldol.
(a) Where is the possible a,\3-unsaturated carbonyl in this skeleton?
reverse
0
0
OH
I
II
L
aldol
II
PhCH2CH2-CHPh ==> PhCHjCH-C-Ph ===> Ph-C-H
o
forward synthesis
II
Ph-C-H
II
+
o
..
NaOH
CH3-C-Ph
II
o
+
0
II
CH3-C-Ph
PhCH = CH - C - Ph
(b) The aldol skeleton is not immediately apparent in this formidable product. What can we see from it?
Most obvious is the \3-dicarbonyl (I3-ketoester) which we know to be a good nucleophile, capable of
substitution or Michael addition. In this case, Michael addition is most likely as the site of attack is \3 to
anotho. carbonyl. 0
O
Ph
Ph '.
+
!�"""'\.
O �
>Q
I
j
,
\ �-ketoester /
OCH3
\
forward synthesis
0
¢o
..
'
Q
�
+
0
0
�
OCH3
0
AHA! The aldol product reveals itself.
(See the solution to 22-71 (c).)
......... ",/
0
NaOH
\.
0
Ph
OCH3
..
Ph
1 ) NaOCH3
2) H+
0
OCH3
���
R
+
(c) The key in this product is the a-nitroketone, the equivalent of a \3-dicarbonyl system, capable of doing
Michael
,-. addition to the \3-carbon of the other carbonyl.
(¥�> Jly-AJl
o
AJl
+A
\'"
".
NO ; '.,
forward synthesis
�o
N 02
aldol
product
NaO �
574
reverse
aldol
==�>
0
SLS
+
H
..
H
y
o o
4d-:)
II
:..
0:
0R:;
4d -
??[(
HO:
..
H�H
0
t
d_Hi?
:R:
4d
: 0:
WJ
H
(I
HO :
-
(q)
:)
/ ..
·. 0·.
0
-1(--
:..
0:
: 0:
H£;J
O:� H::J
H..
I(
o
(n)
(L-ZZ
22-73 continued
(c) Robinson annulations are explained most easily by remembering that the first step is a Michael
addition, followed by aldol cyclization with dehydration.
OMe
OMe
j;�
.c-
:0
..
o
plus other
resonance forms
·1
H
MiChael addition)
two rapid
�
OMe
�
o
.l'mton transrc,
O
J
r- H �� C.I·
HJ
- ..
: RH
n
0
n
H -OH
�
H
plus one other
resonance form
OMe
H -OH
•
OMe
OMe
..
(dehydration)
o
576
plus one other
resonance form
cr-�HP'
22-73 continued
the negative charge in this structure
is stabilized by resonance with the
carbonyl
H
H
(d)
�
N
�O
+
H
H
H
N�\
�;�")
+ OH
H20:
�
H 30+
H
°
H
I
H
•
�.
two rapid
proton transfers
H20:
H30+
two rapid
proton transfers
plus one other
resonance fonn
R)
R:�
+
H
plus one other
resonance fonn
-
/'
�o:
2
H
o
+ -H20
H30+
•
o
+
H20:
..
two rapid
proton transfers
� c;r.
�H
H20:
577
'OH
+
o
•
l
N-O
�
�I
o
H
°
H
H
I
o
�
��
22-74
(a)
H 30+
..
-
o
o
o
H 0+
3
aldol, then
dehydration
•
•
22-75
o
NaOCH3
---
Br
..
Dieckmann
�
�
alkylation
o
o
578
o
•
hydrolysis,
.
decarboxylatIOn
H
22-75 continuued
(b) 0
NaO
o
+
Robinson
reduction
HO
o
Claisen
doing this reaction first
blocks this side of the ketone
Ph
Ph
22-76
(a)
:o:�
rlJ �
'"
H-B
� OH
anion formed in Claisen used
without isolation in the next step
:O- H
I
C
..
---I....
H-B"
..
o
o
9
+ .....
C .....
..
·
plus one oth r
resonance form
plus one other
resonance form
o
O-H
H
�H
plus one other
resonance form
579
B:! -
.
•
(O-H
..
o
22-76 continued
(;) (\
...
·
H B
--.R
OH
I
..
:O -H
H
rb�
)
•
01
-
! c: �<>
B:-
..
c
-
:OH
plus other
resonance fonns
: 0:
0
•
H
1+
: OH
"'\
1
: O -H
H
1+
OH
i
plus one other
resonance fonn
:O -H
..
'1
: OH
�
. .
: O -H
+
1+
H
c
OH
plus one other
resonance fonn
t
�
-
I
--
"
H-B
:��
!
H
G
Q. H20:)
o
0
580
22-76 continued
o
(c) 0
n
,-
o
.. H
C_
I
•
C_N.
•
____
!
HO -.: H
H ""
l
+
lr
-C
1
_
+
Ph
(e
)
0
Uo
+
�
Ao
(b)
o
581
&
..
:0:
____
� CN
22-77 All of these Robinson annulations are catalyzed by NaOH.
(a) CH'
H
(l
�o
+
�
�o
CH 3
H
22-78
a
a
E'O � OE' -:1iE'· E'O
H H�
:0: :0:
OE
Y
H
:0:
;---
.
.
:0:
:0: t :0:
C_� OEt
H
Ph, C /H
EtOOC./ C 'COOEt
+
• •
I
II
plus two other
resonance forms
Ph, C /H
CO2
H./ C 'COOH
+
II
---
[
Ph, C /H
HOOC./ C, COOH
II
22-79
1
CH2OPO32CH20P032CH20P032(plus one other
C==O
C== resonance
C==O
form)
.. HO{CH
HO-CH
HO-CH
HC-O�-H .OH HCi 0:
HC==O
HC-OH
HC-OH
HC-OH
CH20P032CH2OPO32- CH OPO32- CHglycer­
0P0322
To identify the retro­
2
aldehyde
aldol, we must first
C==O
3-phosphate
locate the HO that is
HO-CH2
dihydroxyacetone
I
I
I
I
I
_I
I
..
I
• •
I
•
• •
�
I
-
• •-
I "\......I e.
I
I
I
I
beta to the C=O.
a
I
phosphate
582
22-80
..
This is an aldol condensation. P stands for a protein chain in this problem.
H :O-H
1 1
P-CH2CH2 C J CH
H :O:�
1 II
H+
P-CH2CH2 C - CH --1
..
H
!
H :O-H
1
1
+
� P-CH2CH2 C-CH
1 +
H
this is another molecule
of protonated aldehyde
..;).
1 ..7
H
+
H2 0.
plus one other
resonance form
..
H :0-H
1 1.....1
P-CH2CH2 C=CH
H H
I
+1
:OH C==O
I) 1
P-CH2CH2CH2-CH· CHCH2CH2-P
H
+
I
two fast
:OH C==O-H
proton transfers
..
1
1
P-CH2CH2CH2-CH· CHCH2CH2-P
plus one other
resonance form
• •
CHO
1
P-CH2CH2CH2 . CH = CCH2CH2-P
enol serving as a nucleophile
!-
CHO
1
+
P-CH2CH2CH2 - CH � CCH7.CHr-P
.
..
�-----H
...... 1
22-81
(a) aldol followed by Michael
o
EtOOC
CH�
2
NaOEt
�
+ CH2(COOEth -----l...
+ CH2(COOEth
)l
Y
aldol
H
EtOOC
NaOEI M;chael
�
EtOOC
yy COOEt
EtOOC
iS
..
COOEt
!
(b) Michael followed by Claisen condensation; hydrolysis and decarboxylation
o
o
0
EtOOC
OEt
OEt
NaOEt Etooc
NaOEt EtOOC
...
Michae i
Claisen
I
I
o
�O
J
�
CO2
+
E tO H +
H2 0
p
A �
�
o
583
+
0
0
22-81 continued
(c) aldol, Michael, aldol cyclization, decarboxylation
o
EtOOe
EtOOe ....... A
NaOEt
NaOEt
•
•
d
Michael
� �
(0
�
IT �
\.�
H 0l. H
eOOEt
h
o
Etooe
+
EtOH
+
eOOH
584
�
!
�
EtOOe
NaOEt aldol
EtOOC
e 02
r.
a
eOOEt
CHAPTER 23-CARBOHYDRA TES AND NUCLEIC ACIDS
Reminder about Fischer projections, first introduced in Chapter 5, section 5- 10: vertical bonds are
equivalent to dashed bonds, going behind the plane of the paper, and horizontal bonds are equivalent
to wedge bonds, coming toward the viewer.
o�
H
CHO
"C,..
HO
Fischer
projection:
HO
+
+
H
HO-C-H
H
HO-C-H
I
CH2 0H
X
�'.'T
A'
�
B_
"4
view
from
left
side
X
X
Y
view
from
right
side
y
X
Y
B
y
CHO
H-t--OH
>
'A
"'r.'�
�
.B
CH2 0H
H
HO
H
H
HO
A
X
CHO
23-1
CH2 0H
:
°
OH
HO
H
H
OH
HO
H
H
OH
H0
H
OH
HO
H
H
OH
HO
CH2 0H
glucose
CH2 0H
mirror
image
CH2 0H
i
=t=
OH
H
H
CH2 0H
mirror
image
CH2 0H
fructose
All four of these compounds are chiral and optically active.
23-2
(a)
+
+
H
H
OH
HO
OH
HO
CH2 0H
two asymmetric carbons
+
+
CHO
CHO
=>
+
+
H
H
H
HO
OH
H
+
+
CHO
CHO
HO
H
H
OH
CH2 0H
CH2 0H
CH2 0H
four stereoisomers (two pairs of enantiomers) if none are meso
585
23-2 continued
(b)
one chiral center
�
two stereoisomers
(enantiomers)
(c) An aldohexose has four chiral carbons and sixteen stereoisomers. A ketohexose has three chiral carbons
and eight stereoisomers.
��H
23-3
(a)
23-4
HO
H
HO
HO
-
CH20H
CHO
H
OH
H
CH20H
H
--
L-( )-glucose
23-5
2
(b)
CHO
H OH
HO H
HO H
CH20H
CHO
4H-C-OH
CH20H
:CH1
i
H
L-( +)-arabinose
1
1
3
D =R
23-6
HO
23-7
(a)
H
H
--
t
CHO
Ho H
HO H
CH20H
Ph
CH3
L-( -)-gJyceraldehyde
2
CHO
HO-C-H 4
CH20H
L =S
�H2CHJ
i OH
3
CH20H
Ph
2
CHO
HO+H
CH20H
L-(+)-erythrose
+ OHNHCH3 H3CHNHo t HH
H
CH20H
1
Ho
H
CHO
HO+H
CH20H
CHO
H+OH
CH20H
CH3
H
H3CHN
3
586
+ HOH
Ph
CH3
Ph
HO+H
H+NHCH3
CH3
4
23-7 (a) continued
H
H
Ph
+
=
HO
OH
H3CHN
NHCHl
2
1
Ph
+
=
H
H
H
H3CHN
3
Ph
+
=
OH
H
HO
H
4
Ph
+
=
H
NHCH)
(b) Ephedrine is the erythro diastereomer, represented by structures 1 and 2. Structures 3 and 4 represent
pseudoephedrine, the threo diastereomer.
(c) The Fischer-Rosanoff convention assigns D and L to the configuration of the asymmetric carbon at the
bottom of the Fischer projection. Structure 1 is D-ephedrine; structure 2 is L-ephedrine; structure 3 is L·
pseudoephedrine; and structure 4 is D-pseudoephedrine.
(d) It is not possible to determine which is the (+) or (-) isomer of any compound just by looking at the
structure. The only compound that has a direct correlation between the direction of optical rotation and its D
or L designation is glyceraldehyde, about which the D and L system was designed. For all other compounds,
optical rotation can only be determined by measurement in a polarimeter.
23-8
(a)
H
H
H
H
CHO
(b)
OH
HO
OH
3
HO
H
OH
CH20H
H
H
OH
H
4
H
OH
HO
H
OH
H
CH20H
23-9 D-mannose
HO
HO
H
3
OH
H
OH
CH20H
H�t--OH
r:=::�> HO
H
� inversion at bottom chiral center
HO --t=;;... H
CH20H
CHO
1
5
L-series sugar
L-arabinose
---,3,+-_: \
2
rotare
H ----'i4t-- 0H
H
CHO
D-idose
CHO
CH20H
D-xylose
HO
H
D-talose
CHO
HO
(c)
H
HO
OH
D-allose
(d)
CHO
2
OH
6CH20H
C:===:::::>
587
�
OH
23-1 0 D-allose
HO
1
OH
OH
H
OH
OH at C-3 is axial
23-11 D-talopyranose
OH groups at C-2 and C-4 are axial
H
23-1 2
5
(a) HOH2C
(b)
OH
4
5
HOH2C
H
H
OH
OH
4
H
OH
OH
D-ribofuranose
D-arabinofuranose
6
23-1 3 HOH2C
OH
5
CH20H
1
H
H
23-1 4
(a) ex-D-mannopyranose
OH
(b) �-D-galactopyranose
OH
(c) �-D-allopyranose
OH
equatorial =�
HO
H
H
OH
OH
H
axial =ex
(d) ex-D-arabinofuranose
(e) �-D-ribofuranose
H
HOH2C
equatorial =�
HO
OH cis to CH20H
=�
HOH2C
trans to
OH
OH
H
OH
CHzOH =ex
588
OH
a
23-15
=
-
fraction of galactose as the a anomer; b
a (+ 150.7°) + b (+ 52.8°)
a + b
b
1;
=
=
1
=
=
fraction of galactose as the � anomer
+ 80.2°
a
a (+ 150.7°) + (1 - a) (+ 52.8°)
=
solve for "a"
c::====�> a
+ 80.2°
0.28; b
=
..
=
o.n
The equilibrium mixture contains 28% of the a anomer and 72% of the � anomer
23-16
H,
H�'c ....... ..
0:
0
C/
.......
.
. �+
� 1
H-C-OH
• • -
HO:
H
•
II
C-OH
•
H
OH
erythrose
H
.......
, C/
0
1
H-C-OH
H
+
H,
H
: + :�
+
H
OH
CHzOH
I
0
H
OH
+
OH
CHzOH
threose
CHzOH
1
C=O
CHzOH
1
C=O
I Y"\
.......
C/
HO-C-H
+
CHzOH
erythrose
23-17
OH
CHzOH
CHzOH
The planar enolate can reprotonate
from either side, producing a mixture
of erythrose and threose.
+
:OH
_I
+
+
HO-C:
H
H
OH
OH
CHzOH
fructose
CH,oH
I
C=O
1
HO-C-H
H
H
+
+
OH
OH
CHzOH
589
�
H r'OH
CHzOH
1
C=O
+
:+�:
I
H-C-OH
CHzOH
23-18
H
1
CH20H
1
2C==O
31Y\
: + :�
+
H
:OH
--
CH20H
CH20H
1
I
C==O
C-O:
•
•
II
-1
---- HO-C
HO-C:
H
H
OH
+
+
OH
H
OH
H
CH20H
6CH20H
fructose
+
+
�
•
.
-
OH
OH
CH20H
H
CH20H
1
C-OH
CH20H
1-
:C-OH
� + OH
1
O==C
H
+
----
OH
H"OH
1
CH20H
21
H-C-OH
31
O==C
H
H
+
+
+
� + OH
CH20H
CH20H
II
:O-C
!
+
OH
CH20H
1
HO-C-H
1
O==C
OH
H
OH
H
6CH20H
I
�H C-OH
HO:
I) II
...
:O-C
H
CH20H
CH20H
+
+
OH
OH
CH20H
590
� + OH
H
+
OH
CHO
23-1 9
X
H
HO
CH20H
H
OH
H
----_ ......
..
H
OH
i
CH20H
CHO
HO
H
H
1=
CH20H
OH
H
H
H
H
NaBH4 HO
..
H
OH
H
OH
OH
H
OH
H
OH
COOH
(b)
H
HO
H
HO
H
H
OH
HO
H
H
OH
H
23-22
(a)
COOH
OH
CH20H
D-galactonic acid
(b)
COOH
HO
H
H
HO
H
HO
H
OH
HO
H
H
H
i OH
COOH
mannaric acid
H
H
HO
(c) no reactionD-fructose is a
ketohexose; only
aldoses react
H -+- OH
HO
D-mannonic acid
HO
-
CHO
L-gulose
COOH
CH20H
HO
H
..--
CH20H
D-glucitol
CHO
OH
0H
CH20H
D-glucose
23-21
(a)
CH20H
OH
NaBH4 HO
plane of symmetry
The reduction product of D-galactose still
has four chiral centers but also has a plane
of symmetry; it is a meso compound and
is therefore optically inactive.
H
CH20H
D-galactose
H
........ _------ .. -
HO
OH
23-20
H
HO
NaBH4
HO-+---H
H
OH
OH
OH
COOH
galactaric acid (meso)
591
X
1=
H
H
0H
H
CH20H
L-gulose
COOH
CHO
23-23
H
OH
H
H
HO
HN03
CHO
OH
HO
H
H
HO
H
HO
H
H
OH
OH
HO
..
H
A
H
OH
H
HN03 HO
..
H
H
OH
H
OH
H
OH
H
COOH
mesooptically
inactive
CH20H
galactose
COOH
CH20H
glucose
OH
OH
COOH
optically
active
B
23-24
(a)
(b)
(c)
(d)
(e)
(f)
not reducing: an acetal ending in "oside"
reducing: a hemiacetal ending in "ose"
reducing: a hemiacetal ending in "ose"
not reducing: an acetal ending in "oside"
reducing: one of the rings has a hemiacetal
not reducing: all anomeric carbons are in acetal form
23-25
(c)
OCH2CH3
cis to
CH20H= B
(d) HOH2C
HO
�
H
axial
23-26
=
�
a
OH
H
�W
C�OH
\ ..
It
HO
H
I
�\
H
••
�O
HO
HO
I
H
H HO
I
H
9• •
�
H
\ � I+
It
HO
C_•O• -H
I
�\ 5A
�O
HO
H
I
H
HO
592
..
- H20
H
�
CH3 H
. HO
_OCH
OH
�O
HO
HO
H
OH
H
..
OH
OH
H
�O
HO
HO
OH
axial = a
I
H
�H3
H
9• •
CH3 H
.
·
HO
H
H
23-27
OH
OH
HO
2
H30+
..
HO
OH
H
CN
CN
I
./ HO-CHPh
/ a cyanohydrin
I
O-CHPh
°
H
H
+8
II
Ph-CH
+
HCN is released from amygdalin. HCN is a potent cytotoxic (cell-killing) agent, particularly toxic to
nerve cells.
23-28
HOH2C
OH
CH3CH20H
CH20H
OH
_
W
H20
HOH2C
..
H
H
OH
ethyl B-D-fructofuranoside
Ct and B-D-fructofuranose
+
CH20H
HOH2C
The aglycone in each product is circled.
H
OH
ethyl Ct-D-fructofuranoside
23-29
OH
HO
..
H
�
.0 : I
•
H
OH
OH
�.
H
HO
..
CH'-0
V SO,eH,
H
H
+
593
-OS03CH3
23-30
(a) CH)OH2C
CH20CH)
OCH)
(b)
OCH)
O CH) H
H
23-31
(a)
OAc
(b)
OAc
AcOH2C
H
AcO
OAc
H
23-32
(a)
CHO
r
H
-J.- OH
--------
l
-----
CHO
HO
- -- - '
HO --l-- H
i oH
H -t- OH
r--- .. ..
,
:
--- -
HO
,
CH20H
... H.. .
--
----
r--------
: HO
H
,
,
°
.. -_ ..
-----
H
,
H
H
OH
H
OH
CH20H
,
,
CH20H
D-g\ucose
PhNHNH2
W
~
H
OH
H
OH
CH20H
l
D-mannose
H
I
C=NNHPh
I
C=NNHPh
iH
H -t- OH
r--------·---------
HO
H--i--OH
CH20H
594
/
D-fructose
OAc
23-32 continued
(b)
CHO
H---+-O H
r--------
H
..
I
W
I
I
W
C=NNHPh
HO
HO
HO
H
CHO
--
r--------
HO---+-H
iH
H + OH
......1--
C=NNHPh PhNHNH2
PhNHNH2
HO
H
HO
H
HO
OH
H
CH20H
CH20H
I
H
r--------
H
H
OH
CH20H
D-talose
D-galactose
D-Talose must be the C-2 epimer of D-galactose.
23-33 Reagents for the Ruff degradation are: 1 . Br2, H20; 2. H202, Fe2(S04h .
CHO
CHO
r-
.. H---+.......... OH
--
=t=
HO---+-H
H
H
..
-
CHO
=t=
HO
r--------
Ruff
HO---+-H
•
0H
H
OH
H
-
•
Ruff
HO
0H
H
OH
H
- -
CH20H
CH20H
_ ... - .. - -_ ... --- -_ ...._ .
---------
----
...
H
H
0H
OH
CH20H
--
--------- _
D-arabinose
D-glucose
=t=
r--------
------_ .
D-mannose
t
23-34 Reagents for the Ruff degradation are: 1 . Br2' H20; 2. H202, Fe2(S04h .
CHO
H
HO
I
OH
H
HO�H
t
CHO
Ruff
HO
�
HO
H
•
H
H�OH
H�OH
CH20H
CH20H
D-galactose
CHO
Ruff
Ho
H
H
OH
CH20H
D-Iyxose
D-threose
595
23-35 Reagents for the Ruff degradation are: 1 . Br2' H20; 2. H202, FeiS04h .
CHO
CHO
H
OH
H
OH
H
OH
H
OH
..
Ruff
H
OH
H
OH
H
OH
D-allose
HO
CHO
H
H
OH
H
OH
...
Ruff
H
OH
H
OH
H
OH
CH20H
CH20H
CH20H
23-36
H
HO
CHO
Br2
H2O
..
..
H202
FeiS04h
t
D-altrose
CHO
H
H
CRO
CHO
OH
l) HCN
2) H3O+
H
HO
..
3) Na(Hg)
OH
H
OH
H
OH
+ H
OH
H
OH
H
OH
CH20H
CH20H
CH20H
CHzOH
D-arabinose
D-erythrose
D-arabinose
D--ribose
23-37
H'
CHO
HO
H
H
OH
H
OH
CH20H
D-arabinose
..
H2NOH· HCI
C=NOH
HO
C
H
H
OH
H
OH
..
AC20
CH20H
HO
N
H
H
OH
H
OH
CH20H
oxime
cyanohydrin
596
t
CN- +
..
HOH2O
CHO
H
H
OH
OH
CH2 0H
D-erythrose
*:
23-38 Solve this problem by working backward from (+)-glyceraldehyde.
+
�
H
CHO
H
OH
<�:
==::1
CH20H
H ---If-- OH
H--if--OH
..
HO
H
H ---tl-- ° H
H---If-O
- H
OR
CH20H
cannot be C
COOH
not optically
active
(
HO ---If-- H
HN03
..
COOH
CHO
HO ---tl-- H
H
HO
HO---tl-- H
HN03
�
H�f--OH
CH20H
must be C
D-Iyxose
COOH
optically active
HO -I-- H AND HO ---II-- H
H�I--OH
H---tl-O
- H
COOH
not optically
active
CH20H
A
D-galactose
HO�I--H
H
COOH
/
597
HO_f--H
HN03
•
HO---tl-- H
H---If--OH
CH20H
COOH
optically active
D-talose
r-----",..
HO---tl--H
OH
B
'"
HO---lf--H
H ---tl-- ° H
HO ---tt- H
HO�f-- H
OH
optically active
tartaric acid
(not meso)
/
H ---tl-- OH
H---tO
t- H
HO---lH
l--
CHO
H
COOH
��--------�
CHO \
�------�
COOH
H
CH20H
CHO
COOH
HN03
HO
H
u
+
+
COOH
D-threose
D
D-(+)-glyceraldehyde
HO--if--H
CHO
:=t=::
23-39 Solve this problem by working backward from (+)-glyceraldehyde.
CHO
+
CHO
H
OH
<¢:: ===::J
: r:
CH20H
CH20H
CH20H
F
D-erythrose
D-(+)-glyceraldehyde
optically inactive
erythritol
CHO
CH20H
H-t-OH
H-t--OH
H
OH
H
OH
H
OH
H
OH
CH20H
D-ribose
CH20H
optically inactive
ribitol
E
23-40
(a)
HOH2C
(b)
OH
H
OH
D-fructose
iCH2OCH3
2
CH30H
+
CH30
H
3
4
H
°
H
OCH3
OH
6
CH2OCH3
(c) Determining that the open chain form of fructose is a ketone at C-2 with a free OR at C-5 show", that
fructose exists as a furanose hemiacetal.
598
23-41
(a)
HC=O
I
C=O
CH20H
I
CH20H
+
H
OH
methyll3-D-fructofuranoside
(b)
HO
�
H
D-glyceraldehyde
2H5IO�
OH
O
H
H
OH
H30+
___
OCH3
CH20H
HC=O
I
C=O
I
CH20H
+
methyl I3-D-fructopyranoside
Production of one equivalent of formic acid, and two fragments containing two and three carbons
respectively, proves that the glycoside was in a six-membered ring.
(c) In periodic acid oxidation of an aldohexose glycoside, glyceraldehyde is generated from carbons 4,5,
and 6. If configuration at the middle carbon is D, that means that carbon-5 of the aldohexose must have had
the D configuration. On the other hand, if the isolated glyceraldehyde had the L configuration, then the
original aldohexose must have been an L sugar.
23-42
OH
OH
HO
HO
a anomer of maltose
H
13 anomer of maltose
H
599
H
H
OH
H
�O
OH
H
0
H
OH
B anomer of maltose
H
OH
o
�i
n
_
open chain fonn of maltose
H
-
HO
H
OH
oH
H
�
OH
H
0
o
+
H
AgO
(mirror)
23-44 Lactose is a hemiacetal. Therefore, it can mutarotate and is a reducing sugar.
OH
OH
OH
o
o
H
H
H
a anomer of lactose
H
OH
H
OH
H
B anomer of lactose
OH
H
H
23-45 Gentiobiose is a hemiacetal; in water, the hemiacetal is in equilibrium with the open-chain fonn and
can react as an aldehyde. Gentiobiose can mutarotate and is a reducing sugar.
23-46 Trehalose must be two glucose molecules connected by an
OH
a-I, I '-glycoside.
a-D-glucopyranosyl-a D-glucopyranoside
-
H
H
H
600
0
melibiose = 6 -0-(o.-galactopyranosyl)OH
D-glucopyranose
raffinose
OH
23-47
galactos
H
H
0.-1,6'
invertase
..
glucose
HO
H
H
+
HOH2C
OH
H
The lower glycoside linkage is a-I ,2'
from the glucose point of view, but f}2,1' from the fructose point of view.
�
O� /0
"C
NH2
I
I
0
H
CA
I
H
N,
N
I
"
0
II
0-C-CH3
0
0
H
I H
CCH3
.y
0
eta 2::�
I
CH3
23-49 cytosine:
H
OH
fructose
(D-fructofuranose)
0
II
0-C-CH3
23-48 cellulose acetate
uracil:
OH
HOH2C
fructose
N
o
0
I H
C
.y , CH3
0
II
0-C-CH3
o�
0
I
I H
H
H
H
�C '
C
CH3 // "
0/
CH3 //C"
0
CH3
0
OH
..
O
H
OH
guanine:
N�
�� JL N I NH2
_ _
601
H
23-50 Aminoglycosides, including nucleosides, are similar to acetals: stable to base, cleaved by acid.
(a)
�RA)
HHH
I
R
H
R,I,R
OH
OH
I
0
•
H
1 R
N
OH OH
3° aliphatic
amine­
strong base
CH2
HO-CH2
+
-
OH OH
OH OH
H2 :
I
hen:iacetal form
of nbose
CH2
0
••
H
H
•• H
:0'
H
/ iU
H2 :
..
H�
CH2
H
H
H
O
OH OH
NH2
NH2
H
0
0
N
.
CH2
0
N+
� yl
:0'
H
H
� -t )
C1 ? Cl.. � � I
tH2
OH
+
0:::-...
��\) ��
0 !
�
R
� 0 �
OH
(b)
R2NH2
•
+
• • '---"H
"
----I.�
•
+
•
�
hH2 ..�
OH
R:
H
H
.
N
N
H
N
N
H
site of
OH OH
OH OH
OH OH
protonation; weak
cytidine
base
adenosine
NucJeosides are less rapidly hydrolyzed in aqueous acid because the site of protonation (the N in adenosme,
and in cytidine, the oxygen shown with the negative charge in the second resonance form) is much less
basic than the aliphatic amine in an aminoglycoside. Nucleosides require stronger acid, or longer time and
higher temperature, to be hydrolyzed.
This is important in living systems as it would cause genetic damage or even death of an organism if
its DNA or RNA were too easily decomposed. Organisms go to great length and expend considerable
energy to maintain the structural integrity of their DNA.
23-5\
-- - -:N�
t�N-H
+
+
:O: --- -H-N
N
I ribos�1
H
.
••
N-H-- -:O
H
={
)=
N-H ----:
· O: �
�
guanine Ii
N
_
.
. .
N
.
.
��
t
N
nbose
f
'\
y\
CH3
I
....
.
N ..-: N
N: ----H
I -r--,
N =.!
.
.
+
�
.• 0.•
.
.
�
adenine
thymine
cytosine
The polar resonance forms show how the hydrogen bonds are particularly strong. Each oxygen has
significant negative charge, and in each pair, one H -N is polarized more strongly because the N has
positive charge.
602
23-52 Please refer to solution 1-20, page 12 of this Solutions Manual.
23-53
(a)
CHO
H
HO
OH
(c)
(b)
CH20H
HO
H
H
OH
H
OH
OH
H
H
CH20H
23-54
OH
OH
(b)
(a)
HO
HO
H
OH
H
H
(c)
(d)
OH
HO
OH
OH
H
H
23-55
(a)
(b)
(c)
(d)
(e)
(f)
(g)
O-aldohexose (D configuration, aldehyde, 6 carbons)
O-aldopentose (0 configuration, aldehyde, 5 carbons)
L-ketohexose (L configuration, ketone, 6 carbons)
L-aldohexose (L configuration, aldehyde, 6 carbons)
O-ketopentose (D configuration, ketone, 5 carbons)
L-aldotetrose (L configuration, aldehyde, 4 carbons)
2-acetarnido O-aldohexose (0 configuration, aldehyde, 6 carbons in chain, with acetarnido group at C-2)
603
23-56
H
(a)
0
, C'l
I
tr\
0.0HR :
HO-C - H
HO
H
H
,.-::0:
,c/
I-
HO-C:
--
II
HO-C
..
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH20H
CH20H
CH20H
D-mannose
CH20H
I
I
o==c
HO
H
H
t1
HO-H
H
HO
H
HO
H
..
"D
0
,c/o•'o
H
i)
- OH
,oC /
OH
,C/
.
•. I
o==C
HO-H
..
H0 -j H
H
H
OH
CH20H
• HO'..
.. ..
II
---
: o-c
HO
H ----+- OH
OH
Cl
H
OH
H
H
'C/
�'o0.. C
H
.
HO
OH
H
H
OH
H
CH20H
I
II
H
CH20H
OH
H
OH
OH
CH20H
D-fructose
enediol
(b) The a isomer has the anomeric OH trans to the CH20H off of C-5. The �-anomer has these groups cis.
c
HOH2C
CH20H
�
HOH2C
A
OH
OH
H
�-0- fructofuranose
a-D-fructofuranose
604
23-57
(a)
HO
+
CHO
H
+
CN
OH
HCN
�
�
H
OH
H
H
CN
+
�
OH
B
CH20H
H
H
Ba(OHh
H2O
+
..
H
H
H
HO
OH
+
H
OR
COOH
(-)-tartaric acid
COOH
OH
H
HN03
---
OH
D
+
H
OH
Oll
COOH
meso-tartaric acid
CH20H
COOH
COOH
(S,S)-(-)-tartaric acid
HO
COOH
OH
+
COOH
OH
CH20H
+
H
+
H
C
COOH
HO
H2O
HO
..
CH20H
H
(b)
+
Ba(OHh
H
A
CH20H
+
COOH
COOH
OH
H
H
H
COOH
(R,R)-(+)-tartaric acid
OH
OH
COOH
(R,S)-meso-tartaric acid
23-58
(a) D-(-)-ribose
(b) D-(+ )-altrose
(c) L-( +)-erythrose
(d) L-(-)-galactose
(e) L-(+)-idose
23-59
(a)
(b)
OCH3
OH
H
H
(c) CH30H2C
OCH3
(d)
OH
CH20CH3
OH
H
OCH3
H
H
605
23-60
(a)
(b)
OH
HO
OH
H
H
CH20H
OH
OH
OH
HOH2C
OH
H
HO
°
H
H
H
H
°
I
H
HO
HO
�
OH
H
H
23-61
(a)
(b)
(c)
(d)
methyll3-D-fructofuranoside
3,6-di-O-methyl-I3-D-mannopyranose
4-0-( CI.- D-fructofuranosy I)-I3-D-galactopyranose
/3-D-N-acetylgalactopyranosamine, or 2-acetamido-2-deoxy-/3-D-galactopyranose
23-62 These are reducing sugars and would undergo mutarotation:
-in problem 23-59: (b) and (c);
-in problem 23-60: (a) and (c);
-in problem 23-61: (b), (c), and (d)
23-63
(a)
H
HO
HO
H
COOH
OH
H
H
OH
CH20H
(b)
CH20H
CHO
HO
HO
HO
H
+H
H
H
OH
CH20H
+
(c)
6: 0
HO
H
HO
H
H
OH
CH20H
606
H
+
others
-\4-i \
. -OH
OB
-
H
,.OCH3
H
23-63 continued
COO(d)
(e)
CHzOH
H-t-0H
H
OH
HO--f--H
HO
HO--f--H
HO
H
i oH
H
+
(i)
CHzOH
HO
H
I
H
AcO
H
H
�
H
H
OH
�
CH
,0
H
CHOIO
OAC
OCH
", 3
H
(j)
CHO
H
H
H
OH
H
HO
H
H
OH
H
H
OH
OH
HO
H
CHzOH
HO
H
HO
H
+
HO
H
OH
H
H
HO
OH
H
OH
H
CHzOH
±
CHO
CHO
HO
H--+-OH
HO
�
19}
OAc
OAc
CHzOH
CHzOH
(h)
(f)
CHzOH
CHzOH
CHO
(k)
H
� OH
HO--f--H
excess
HI04
..
5
HO----t=-H
H
II
°
� OH
II
°
H-C-OH
from C-l
through C-5
+
1
H-C-H
from C-6
CHzOH
23-64 Use the milder reagent, CH3VAgzO, when the sugar is in the hemiacetal form; the mild conditions
prevent isomerization. When the carbohydrate is present as an acetal (a glycoside), use the more basic
reagent, NaOHl(CH3)zS04; an acetal is stable to basic conditions.
(a)
CH20CH3
CH3°
H
H
I
(b)
(c)
CHO
°
H
H
CH30
OCH3
CHO
H
H
CH30
OCH3
H
CHzOCH3
bo
+
CH3O-+--H
OCH3
H
OCH3
H
OCH3
H
OH
H
OH
H
OH
H
CH20CH3
using CH3I, AgzO
CHzOCH3
using (CH3)2S04' NaOH
607
I
oCH3
OH
CHzOCH3
CH2OCH3
using (CH3hS04' NaOH
23-64 continued
H
CH30
CH30
H
� OCH3
�
(e)
CHO
CHO
(d)
+
H
OCH3
H
H
CH30
H
CH30
H
H
OH
H
H
OH
H
i oH
CH2OCH3
CH2OCH3
using CH3I, Ag20
(f)
=C
CHO
CHO
r
+
OCH3
H
CH30
OCH]
H
OH
H
CH2OCH3
NHCOCH3
CH30
using (CH3hS04' NaOH
H
H
OH
H
OH
CH2OCH3
23-65
(a) These D-aldopentoses will give optically active aldaric acids.
CHO
D-arabinose
CHO
H
HO
H
H
OH
HO
H
H
OH
H
HO
CH20H
+
+
D-Iyxose
OH
CH20H
(b) Only D-threose (of the aldotetroses) will give an optically active aldaric acid.
CHO
D-threose
HO
H
H
OH
CH20H
608
X
+
OCH3
H
OCH]
OH
CH20H
using CH31, Ag20
CHO
H
H
23-65 continued
(c) X is D-galactose.
COOH
CHO
H
X:
HO
HO
H
I
!
H
OH
H
HN03
HO
H
HO
H
..
H
H
OH
COOH
HO
H
H
H
OH
Ruff
CHO
HO
optically inactive
COOH
CHzOH
HO
OH
--!- H
HO -f-- H
HN03
•
H
OH
i OH
optically active
COOH
CHzOH
The other aldohexose that gives an optically inactive aldaric acid is D-allose, with all OH groups on the
right side of the Fischer projection. Ruff degradation followed by nitric acid gives an optically inactive
aldaric acid, however, so X cannot be D-allose.
t
t
(d) The optically active, five-carbon aldaric acid comes from the optically active pentose, not from the
optically inactive, six-carbon aldaric acid. The principle is not violated.
CHO
(e)
HO
--!- H
HO-f--H
H-f--OH
CHzOH
CHO
Ruff
---
Ho
H
H
COOH
HN03
•
Ho
H
OH
CHzOH
D-threose
H
OH
COOH
(S,S)-tartaric acid
optically active
609
23-66
(a)
+
Ho
CHO
H
t
:=+=::
CN
CN
HCN
OH
..
H
H
+
OH
CH20H
CH20H
CH20H
(b) The products are diastereomers with different physical properties. They could be separated by
crystallization, distillation, or chromatography.
(c) Both products are optically active. Each has two chiral centers and no plane of symmetry.
23-67
�
fc]\
(a) The Tollens reaction is run in aqueous base which isomerizes carbohydrates.
H
HO
'¢f=9 \
u
H
H
-
(II
C-OH
..
HO:
H--II--OH
HO-I--H
H--II--OH
CH20H
H-t-OH
H-It-OH
H-t-OH
H--I--
CH20H
plus one other
resonance fonn
H
HO-f--H
H--II--OH
-H
....... /··
C
+
0:
....... /
C/
1:C
OH
r
HO'l H
HO-t--H
HO
..
-
H
H---1r--OH
H--ll--OH
H--II--OH
H---1r--OH
H
CH20H
D-glucose
CH20H
D-mannose
OH
CH20H
plus one other
resonance fonn
(b) Bromine water is acidic, not basic like the Tollens reagent. Carbohydrates isomerize quickly in base,
but only very slowly in acid, so bromine water can oxidize without isomerization.
610
23-68 Tagatose is a monosaccharide, a ketohexose, that is found in the pyranose form.
61 1
Z19
o
HO
-o
--- -'d 'i°
\
H
0
H
Z
H:J
/
H
o
I
Z
H:J-O-d-OII
o
(q)
HO
0-
I
0
I
0-
I
H :J-O-d- O-d-O-d-O
Z
\I
0
II
0
\1
o
23-72 Bonds from C to
H
are omitted for simplicity.
NH2
V\
o
A
� �
N
5'
?
N
o
\
0
\
P
-0/
�o
c
O
NH2
,)
t
?
�
N
I
" N
A
N
{
o
\
�O
\
:.--'
- /P
0
o
O
H
C J I
?
t
H
T
o
0
\
...,0
p"-0""'"
\
o
?
�
N
:c
O
I
N
,H
A NH2
N
G
°
3'
23-73
�
(a) No, there is no relation between the amount of G and A.
(b) Yes, this must be true mathematically.
(c) Chargaffs rule must apply only to double-stranded DNA. For each G in one strand, there is a
complementary C in the opposing strand, but there is no correlation between G and C in the same strand.
61 3
23-74 Bonds from C to H are omitted for simplicity.
�;:
H
�;:
H
a
CH3
HO
5'
a
OH
l N�
0
2'-deoxythymidine
(abbreviated dT)
a
CH3
HO
5'
a
N
N+
l N�
0
3'-azido-2',3'-dideoxythymidine
(AZT)
II
�'"
No phosphate can attach to the azide group, so
synthesis of the DNA chain is terminated.
�;:
a
CH3
a
a
a
I
II
I
O -P -O-P-O-p -O
I
I
I
a
a
a
-
-
5'
a
-
AZT 5'-triphosphate
N+
N
N
II
II
61 4
l N�
H
0
H-O-N=O
H+
H0
N=O
23-75 Recall from section 19-17 that nitrous acid is unstable, generating nitrosonium ion.
(a)
+
+
2
-
+
AI:�"
A
l
+
0
H
N
N=R
cytosine (C)
+
two rapid proton transfers
N
�'
+
: N=N
tA tA
I
-""'::::N
H
N
H 0:
2
�
o
I
H
N
H
H2O
..
o
H�
\
':tO�
•
•
I
-""':N
:::
H
H+
I
N�O
H
..
I
-""'::::N
O
H 0:
2 �
OH
NH
...
0
===::
enol tautomer
of uracil
(b) In base pairing, cytosine pairs with guanine. If cytosine is converted to uracil, however, each
replication will not carry the complement of cytosine (guanine) but instead will carry the complement n1'
uracil (adenine). This is the definition of a mutation, where the wrong base is inserted in a nucleic acid
chain.
(c) In RNA, the transformation of cytosine (C) to uracil (U) is not detected as a problem because U is a
base normally found in RNA so it goes unrepaired. In DNA, however, thymine (with an extra methyl
group) is used instead of uraci l. If cytosine is diazotized to uracil, the DNA repair enzymes detect it as a
mutation and correct it.
615
H
I
N�O
H
f� r��
NH
if this species exists, its
lifetime is very short as
water will attack quickly
-
!
Q
6q
23-76
I
(a)
< }-
h
C-O
se
(b) Trityl groups are specific for 1 ° alcohols for steric
reasons: the trityl group is so big that even a 2° alcohol is
too crowded to react at the central carbon. It is possible for a
trityl to go on a 2° alcohol, but the reaction is exceedingly
slow and in the presence of a 1 ° alcohol, the reaction is done
at the 1 ° alcohol long before the 2° alcohol gets started.
OH
OH
(c) Reactions happen faster when the product or intermediate is stabilized. Each OCH3 group stabilizes
the carbocation by resonance as shown here; two OCH3 groups stabilize more than just one, increasing
the rate of removal. The color comes from the extended conjugation through all three rings and out onto
the OCH3 groups. Compare the DMT structure with that of phenolphthalein, the most common acid
base indicator, that turns pink in its ring open form shown here. Phenolphthalein is simply another trityl
group with different substituents.
+ H
+ CH3
:o�
:0'"
one of the major
resonance contributors
showing the delocalization
of the positive charge on
the oxygen
¢
I I
0-0�
Ij_
C
�-!J
<
OCH3
¢}
�c�
o
23-77
o
�
(a)
acetal
-\H
Ph
phenolphthalein
H
6
(b)
O
o
3
OH
OH
OH
>
Ph
�
O
benzaldehyde
OH
OH
�-D-glucose
(c)
The 4- and 6-0H groups of glucose reacted with
benzaldehyde to form the acetal.
(c) and (d) The chiral center is marked with (*). This stereoisomer
is a diastereomer since only one chiral center is inverted. This
OH diastereomer puts the phenyl group in an axial position definitely
less stable than the structure shown in part (a). Only the product
shown in (a) will isolated from this reaction.
,
OH
Only the 2'- and 3'-OH groups are close
enough to form this cyclic acetal (ketal)
from acetone, called an acetonide.
616
CHAPTER 24-AMINO ACIDS, PEPTIDES, AND PROTEINS
24-1
(a)
COOH
H2N
I "'''CH2Ph
�
H
���
(d)
H2N
COOH
(c)
H2N
COOH
(b)
-
NH
NH2
COOH
H2N
,oH
I "CH2CH2CH2NH- C
�
H
II
I '''''CH2
�
H
24-2
(a) The configurations around the asymmetric carbons of (R)-cysteine and (5)-alanine are the same. The
designation of configuration changes because sulfur changes the priorities of the side chain and the COOH.
2
(5)-alanine with
group priorities shown
1
H2N
COOH
�
�
3
3
' CH
3
H
4
" "
1
H2N
COOH
2
'"''CH2SH
4H
(R)-cysteine with group priorities
shown; note that the COOH and
the CH2SH priorities are
reversed compared with (5)­
alanine
(b) Fischer projections show that both (5)-alanine and (R)-cysteine are L-amino acids.
+
H2N
H
CH3
(5)-alanine
L-alanine
+
COOH
COOH
H2N
H
CH2SH
( R)-cysteine
L-cysteine
24-3 In their evolution, plants have needed to be more resourceful than animals in developing biochemical
mechanisms for survival. Thus, plants make more of their own required compounds than animals do. The
amino acid phenylalanine is produced by plants but required in the diet of mammals. To interfere with a
plant's production of phenylalanine is fatal to the plant, but since humans do not produce phenylalanine,
glyphosate is virtually non-toxic to us.
24-4 Here is a simple way of determining if a group will be protonated: at solution pH below the group's
pKa value, the group will be protonated; at pH higher than the group's pKa value, it will not be protonated.
H
(a)
r
H
?
1
(b)
H2N - -COOCH(CH3h
(d)
(c)
617
H-
1+
\
C -COOH
�
24-4 continued
(e)
+
lysine
H
alanine
H
I
H3N-C-COO-
(i) pH 6
I
CH3
+
I
?
H2N- -COO-
H2N-C-COO-
(CH2)4NH2
CH2COO-
H
I
H3N-C-COOH
(iii) pH 2
I
?
I
H3N - -COOH
I
I
H
I
. .
H -N -H
I
I
....
.
..
t--I.�
H
H -N -C -N -H
I
H
t
1....
. . --I"�
..
I
+
I
CH2COOH
H -N -H
I
H
H3N-C-COOH
+
H -N-H
+
+
(CH2)4NH3
. .
H -N==C-N -H
I
H
+
CH3
24-5
I
I
CH3
+
H
H
H2N-C-COO-
I
CH2COO-
+
I
?
H3N- -COO-
H3N -C-COOI
(CH2)4NH3
H
(ii) pH 11
+
I
aspartic acid
H
H
+
H -N-C==N--H
I
I
H
H
+
H -N -H
II
H-N -C-N -H
I
I
H
H
Protonation of the guanidino group gives a resonance-stabilized cation with all octets filled and the positive
charge delocalized over three nitrogen atoms. Arginine's strongly basic isoelectric point reflects the unusual
basicity of the guanidino group due to this resonance stabilization in the protonated form. (See Problems 139 and 19-49(a).)
24-6
tryptophan
histidine
H
H
H2N-
?
I
-COOH
?
I
H2N- -COOH
CH2
ro
not bask (like pyrrole)
I
H
CH2
G�
basIc --.. N
(like pyridine)
• •
not basic
(like pyrrole)
The basicity of any nitrogen depends on its electron pair's availability for bonding with a proton. In
tryptophan, the nitrogen's electron pair is part of the aromatic 1t system; without this electron pair in the 1t
system, the molecule would not be aromatic. Using this electron pair for bonding to a proton would
therefore destroy the aromaticity-not a favorable process.
In the imidazole ring of histidine, the electron pair of one nitrogen is also part of the aromatic 1t system
and is unavailable for bonding; this nitrogen is not basic. The electron pair on the other nitrogen, however,
2
is in an sp orbital available for bonding, and is about as basic as pyridine.
618
24-7 At pH 9.7, alanine (isoelectric point (IEP) 6.0) has a charge of -1 and will migrate to the anode.
Lysine (IEP 9.7) is at its isoelectric point and will not move. Aspartic acid (IEP 2.8) has a charge of -2 and
will also migrate to the anode, faster than alanine.
caili
�
� � ��
nOde
t
t
lysine
o
t
alanine
-1
aspartic acid
-2
24-8 At pH 6.0, tryptophan (IEP 5.9) has a charge of zero and will not migrate. Cysteine (lEP S.O) has a
partial negative charge and will move toward the anode. Histidine (IEP 7.6) has a partial positive charge
and will move toward the cathode.
�
catho
�
�
t
nooc
t
histidine
H
� �
t
tryptophan
+
cysteine
?
H
+
I
H3N - -COO-
\N �
I
H
tryptophan
I
CH2Scysteine (partially
deprotonated sulfur)
(The SH is more acidic
than the NH3+ group.)
24-9
(a)
O=C -COOH
I
CH3
(b)
O=C-COOH
H2N -CH-COOH
I
CH2CH(CH3h
I
CH2CH(CH3h
(c)
O=C-COOH
H2N -CH-COOH
I
I
CH20H
(d)
O=C-COOH
CH20H
NH3
H2, Pd
CH2CH2CONH2
I
..
I
H3N -C-COO-
�
histidine
(partially
protonated
imidazole
ring)
H
H2N -CH-COOH
I
CH2CH2CONH2
619
24-10 All of these reactions use: first arrow: (1 ) Br2IPBr3' followed by H20 workup; second arrow: (2)
excess NH3, followed by neutralizing workup.
(a) H2C -COOH
(1)
I
..
Br - CH-COOH
I
.. H2N - CH-COOH
I
H
(2)
H
H
(b) H2C -COOH
..
(1 )
I
CH2CH(CH3h
(c) H2C-COOH
..
(1 )
I
Br -CH-COOH
I
CH2CH(CH3h
(2)
..
I
H2N-CH-COOH
CH2CH(CH3h
I
Br - CH-COOH
I
.. H2N - CH-COOH
(2)
CH(CH3h
CH(CH3h
CH(CH3h
(d) H2C -COOH
(1)
I
..
Br -CH-COOH
I
(2)
..
H2N-CH-COOH
I
CH2CH2COOH
CH2CH2COOH
CH2CH2COOH
In part (d), care must be taken to avoid reaction a to the other COOH. In practice, this would be
accomplished by using less than one-half mole of bromine per mole of the diacid.
o
I
1 ) NaOEt
I
2) BrCH(CH3h
COOEt
N - CH
I
I
I
o
abbreviate
COOEt
G�
?
(}- H
C ?
COOEt
1 ) NaOEt
2) BrCH2Ph-
COOEt
(c)
OOEt
G
COOEt
I
N - C-CH2Ph
COOEt
N-
H
1 ) NaOEt
COOEt
�
!
H30+
H2N - CH-COOH
I
CH(CH3h
N-CH
(b)
COOEt
N - C-CH(CH3h
..
COOEt
�
o
o
..
2) BrCH2CH2COO\
COOEt
.,.
salt, not
acid-why?
I
COOEt
H30+
�
-
H30+
�
- H2N -CH-COOH
I
CH2Ph
H2N - CH-COOH
I
CH2CH2COOH
620
24-1 1 continued
0�
(d)
COOEt
- H
1) NaOEt
H30+
H2N - CH-COOH
tJ.
----,t.�
COOEt
I
CH2CH(CH3h
24-12
COOEt
I
COOEt
1) NaOEt
AcNH -CH
I
COOEt
2) BrCH2Ph
I
H30+
I
AcNH - C - CH2Ph
•
tJ.
..
----,t
�
COOEt
H2N -CH - COOH
I
CH2Ph
acetamidomalonic
ester
o
24-1 3
(a)
II
PhCH2-CH
NH3, HCN
NH2
I
..
H2N -CH-COOH
PhCH2-CH
I
C
I
CH2Ph
N
(b) While the solvent for the Strecker synthesis is water, the proton acceptor is ammonia and the proton
donor is ammonium ion.
�
�
:0:
�+
H - NH3
+NH2
I)
H
+
I
PhCH2-CH
t
:NH2
-H20
....
..f---
'--../
0NH3
°
H-O - H
+Jr\
+ 1)
H3N -H
PhCH2-CH ..
I
NH2
II
I
this protonated
imine is rapidly
attacked by
cyanide nucIeophile
N
PhCH2-CH
I
+ NH2
NH2
(abbreviate as
R -C
� :O - H
I
I
PhCH2 - CH
NH2
C
PhCH2-CH
l
two fast pMon tmnSf
PhCH ,- H
N
on the next page)
621
24-1 3 (b) continued
mechanism of acid hydrolysis of the nitrile
+
J
.�H+
.. l R-C==N -H ..
R-C==N.
_
H
1+
==N - H
1 .�.
R-
T
+
JI'
H
I
..
I
20:
two rapid
OH H
proton transfers
I �I +
.. R - C -N-H
� H+
R - C -N -H
I
HO
•
•
+O. -H
�
I
I
I
.
I
HO
I
-
i
NH3
..
H
•
R - C==N-H
H -O
\.
.. -H
-
H
II
H20:
H -OH H
•
H20:
• •
R -C-N - H
:O - H
+..
l} H
�
R - C==N -H
R -C -N - H '"
:O - H
VI
+
..
_
..
}
�I+
H20:
+
.
H+"'--R - C==N - H
�
I
:O
. -H
:O -H
I
• •
• •
R-C - OH
+
• •
---
:O-H
I
+
R - C==OH
+
H2N - CH - COOH
I
H20:
..1-­
....
.
CH2Ph
24-1 4
(a)
R
I
C-H
�
NH3, HC
H20
CH2CH(CH3h
(b) 0
II
C-H
I
II
C -H
I
CH(CH3h
I
CH2CH(CH3h
NH3, HCN
..
H2O
H
(c) 0
H2N - CH-CN
H2N -CH-CN
I
H
NH3, HCN
..
H2O
T
H2N - H-CN
CH(CH3h
622
H30+
..
!l
H2N-CH -COOH
H30+
..
!l
H2N-CH-COOH
H30+
!l
..
I
CH2CH(CH3h
I
H
H2N - CH - COOH
I
CH(CH3h
:O�
II
• •
R - C -OH
24-1 5 In acid solution, the free amino acid will be protonated, with a positive charge, and probably soluble
in water as are other organic ions. The acylated amino acid, however, is not basic since the nitrogen is
present as an amide. In acid solution, the acylated amino acid is neutral and not soluble in water. Water
extraction or ion-exchange chromatography (Figure 24-1 1 ) would be practical techniques to separate these
compounds.
24-1 6
o
0
+
II
abbreviate H3N -C H - COEt
I
:O :
II
CH2Ph
�
R-C -OEt
II
R-COEt
as
'
' OH �
I I' . .
H20 :
•
RC-OEt
H+
-
+
•
?
o
+ I '"
II
•
?
CH2CH2CONH2
RC -OEt
•
t:J
H
II
OH H
EtOH
�.�---
RC
I
I r\1+
RC-O -Et
l
:O -H
(protonated form
in acid solution)
+
•
�
CH2Ph
H3N- H -COO-
I
..
H20 :
RC-OEt
.
H20 :
H3N- H - COH
24-1 7
OH
I
H - � - H�
•
plus two other
resonance forms
+
OH
O-H
plus two other
resonance forms
PhCH20H
+
.. H3N -CH - COOCH2Ph
--i�
---
H+
I
H2, Pd
--
+
H3N -CH-COO-
I
CH2CH2CONH2
CH2CH2CONH2
+
°
24-1 8
II
+
H3N-CH-COO-
I
PhCH20 - C -Cl
..
o
II
H
I
PhCH20C -N-CH-COOH
!
I
CH2CH2SCH3
CH2CH2SCH3
PhCH3
+
CO2
+
+
?
H2, Pd
H3N- H-COOCH2CH2SCH3
623
CH3Ph
24-1 9
..
N
:0 :
:0 :
1
N
:0:
i
N
:0:
:0:
�
:0 :
:0 :
-
-
:0:
..
:0 :
:0 :
:0 :
:0 :
:0
:
..
1
"
:0:
N�
:0 :
:0:
These are the most significant resonance contributors in which the electronegative oxygens carry the
negative charge. There are also two other forms in which the negative charge is on the carbons bonded to
the nitrogen, plus the usual resonance forms involving the alternate Kekule structures of the benzene rings.
24-20
(a)
(b)
o:
II :
0:
0
II :
iI
H3N - CH·C�NH·CH·C�NH·CH·C-O:
I
:
I
I
CH2Ph :
HO-CHCH3:
CH2CH2SCH3
Phe
'
Met
Thr
+
,
,
o:
0:
0:
0
I I:
II :
II
I I :.
H3N -CH . C � NH . CH . C -+- NH . CH . C -7- NH . CH . C - 0+
I
:
CH20H:
seryl
:
I
:
I
:
(CH2h :
I
H
:
:
HN - C - NH2 glycyl
II
:
I
CH2Ph
phenylalanine
arginyl NH
(c)
0:
0
I I:
II
H3N -CH . C --f NH . CH . C -:- NH . CH . C -7" NH . CH . C -:- NH . CH . C-0:
I
:
I
I
:
I
:
I
:
(CH2h :
(CH2h :
CH2
(CH2)4
CH3CH2CHCH3 :
.
I
:
I
:
I
:
I
I (isoleucine)
SCH3
CONH2
COOH
NH2
M (methionine) Q (glutamine) D (aspartic acid) K (lysine)
+
0:
0:
0:
I
I
I
I
0:
0:
I
I
II:
0:
(d)
+
II:
II :
0:
II :
II :
II :
II :
-
0
II
H3N - CH . C --f NH . CH . C -:- NH . CH . C -7" NH . CH . C � NH CH - C - 0I
:
I
:
I
:
I
:
I
(CH2h :
:
CH2 :,
CHMe
2'
CHCH3:
CH2
.
'
.
I
Try spelling your
''
I
''
'
I
.'
I
COOH
name in peptides!
CHMe2
CH2CH3
OH
L (leucine)
E (glutamic acid)
V (valine)
I (isoleucine)
S (serine)
624
'
I
•
(a)
II
II
(b)
/C,
HN
NPh
\
I
HC -C
�0
I
CH3
S
S
S
S
24-21
II
/ C,
(c)
HN
NPh
\
I
HC-C=O
I
I
II
NPh
I
\
C -C
I
H
(CH2)4NH2
24-22
/C,
C
HN
NPh
\
I
HC-C=O
CH(CH3h
II
(d)
/C,
�0
S
1 ) PhNCS
H2N-Ile--7Gln--7peptide
..
/C,
+
HN
NPh
\
I
HC-C=O
I
CH3-CHCH2CH3
H2N-Gln--7peptide
II
S
Step 4
/C,
HN
NPh
+ H2N-peptide
\
I
HC-C=O
I
CH2CH2CONH 2
1) PhNCS
H2N-Gln--7peptide
Fy
.
I�
24-23 Abbreviate the N-terrninus of the peptide chain as NH2R .
F
(a) This is a nucleophilic aromatic substitution by the addition-elimination mechanism. The presence of
two nitro groups makes this reaction feasible under mild conditions.
'Q
F
02N
·
;-...... . .
NH2R
�
I
.
"
:0
-
�
•
•
..
0
·
".
�-C .
.
�
+
NH2R
I
·
•
N
: 0"
-
• •
�..
F
+
NH 2R
I
�
,. N+
+
,. N
_
"
'0
0
N02
0·
0"
0
- "N
0
�
+
NH2R
D
11+
I
..
C-
I
1+
'0
_
,.
:0"
N'
• •-
0:
\
I
�
N02
�
I
�
+:r::
--
N02
(b) The main drawback of the Sanger method is that only one amino acid is analyzed per sample of protein.
The Edman degradation can usually analyze more than 20 amino acids per sample of protein.
625
I
24-24
I
trypsin
� �
N-tenninus
l
1
r
Tyr-Ile----Gln-Arg-Leu----Gly-Phe-Lys-Asn-Trp--Phe----Gly-Ala-Lys�ly-GIn-GlnoNHz
I
t
C tcnrunu,
(amide fonn)
chymotrypsin
Phe-Gln-Asn
24-25
Pro-Arg-Gly·NH2
Cys-Tyr-Phe
Asn-Cys-Pro-Arg
\
Tyr-Phe-Gln-Asn
�----------
_/
)
Y
---------
Cys- Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly·NH2
24-26 abbreviations used in this problem:
o
I
H2N f-CH - COOH
Rl
II
°
II
PhCH20C - NH - CH COH
I
CH3 R2
CH3
mechanism of fonnation of Z-Ala
:
c��
+
PhCH2O-C-CI
°
II
2
R - COH
o.
H N-R i
�
°
¢:::=l
II
PhCH20 - C-NH-R i
---
626
o
:Cl :
• •
II
H
1+
PhCH2O - C-N-R i
(
�k
+
24-26 continued
two possible mechanisms of ethyl chlorofonnate activation
mechanism 1
R
00,,--/:��
2
R -C -OH
CI -C -OEt
+
�:?j
II
2
R -C -0 -C -OEt
+1)
H"",\
CI-C -OEt
1
+
:O -H
-
00
o
II
2 I
R -C=0
0
� :9!:
o
II
2
R -C-0 -C-OEt
II
�:?j
I+
C)-C-OEt
Cl-
___
-
2 II
R -C -OH
:0
J
1
II
2 II
R - C -0-C -OEt
G0
0
+ II
2
R - C=0 -C-OEt
I
0
H-O :
00
I
H-O :
00-�
i
�
oo
il
II
2
R -C -0 -C-OEt
:Cl
:
. .
..
o
0
0
�200
II
1
0
II
+
3
H N -R
0
----
II
0
PhCH20C - NH - CH - C-NH - CH - C-OH
I
CH3
I
I+ 3
2 "
R -C -N -R
:0 : H
---
CH(CH3h
(I
+
CO2
��
H
627
0
H-O :
0
+
11
II
. .
2
R - C-0 -C -OEt
mechanism of the coupling with valine
o
t
II
2
R - C -0 -C -OEt
+
plus two other
resonance fonns
o
0
-
: Et
1 \���
24-27
o
°
0
0
II
II
II
II
PhCH20C -NH -CH - C -NH -CH - C-NH-CH-C-OH
o
CH(CH3h
0
II
II
I
I
I
CH3
!
CH2Ph
o
II
Cl-C-OEt
0
0
II
II
Z-N H -CH -C -N H -C H-C-N H -CH-C -0 -C-OEt
I
I
CH(CH3h
CH3
o
I
CH2Ph
t
°
0
0
II
II
glycine
H2NCH2COOH
II
II
Z-NH-CH - C -NH-CH -C-NH -CH -C-NH-CH -C-OH
I
I
CH(CH3h
CH3
°
0
II
I
I
CH2Ph
H
j
1
II
H2N-CH-COOH
0
b
leucine
H2CH(CH3h
°
II
II
Z -NH -CH-C-NH -CH - C-NH -CH -C-NH CH-C-NH -CH - COOH
I
I
CH3
CH(CH3h
o
0
/I
-
I
CH2Ph
.
II
I
I
H
CH2CH(CH3h
H2, Pd
°
0
II
II
H N -CH - C-NH-CH -C-NH -CH-C -NH -CH -C-NH -CH - COOH
2
I
CH3
I
CH(CH3h
I
CH2Ph
I
H
628
I
CH2CH(CH3h
24-28
II
II
°
°
PhCH20C - Cl
+
I
H2N -CH -COOH
-­
I
Z - NH-CH -C-OH
CH3 - CHCH2CH3
CH3 -CHCH2CH3
isoleucine
II
o
I
I
II
°
Cl-C-OEt
II
°
Z -NH-CH - C-0-C -OEt
CH3 - CHCH2CH3
II
o
I
t
glycine
H2NCH2COOH
II
°
I
Z -NH-CH-C - NH-CH -C-OH
CH3 - CHCH2CH3
II
0
I
1
H
II
°
CI-C-OEt
II
0
I
II
0
Z -NH CH - C-NH-CH - C-0 -C -OEt
-
1
CH3 - CHCH2CH3
II
o
I
H
?
H2N - H -COOH
asparagine
CH2CONH2
II
°
I
I
Z-NH-CH - C-NH-CH - C-NH-CH -COOH
CH3 - CHCH2CH3
II
o
I
t
H
CH2CONH2
H2' Pd
II
°
I
I
H2N-CH -C-NH-CH -C-NH-CH -COOH
CH3 - CHCH2CH3
629
H
CH2CONH2
24-29 In this problem, "Cy" stands for "cyc1ohexyl".
o-
N=C =N
mechanism
-Q
c=:::::::>
Cy-N =C=N -Cy
DCC
a
\
II
CH3C-� :
a
II
1
CH3C-O-C
N --Cy
"
Cy-N=C
a
I
II
:N --Cy
....
..f--.
..
�
-I
+
H3N -Ph
J-Cy
II
N-Cy
CH3C-O-C
,
N-Cy
• • _
plus other resonance forms
a
II
a
+
(I
CH3C-NHPh
H
�
�
+
+
II
Cy-N-C
- NH-Cy
..
•
•
resonance­
stabilized
II
a
Cy - NH-C - NH - Cy
DCU
630
����
24-30
0
0
II
0
II
0
" -0
II
P
Me3COC - NH - CH - C-NH-CH - C-NH-CH - C -0
I
I
CH3
+
CH(CH3h
I
CH2Ph
CF3COOH
0
0
II
+
0
" -0
II
p
H3N - CH - C-NH-CH - C-NH - CH - C -O
I
I
CH3
DCC
o
0
II
•I
CH(CH3h
I
CH2Ph
�
Boc-glycine
Me3COC - NHCH2COOH
0
II
0
0
II
II
"
Me3COC-NH - CH - C-NH-CH-C - NH - CH - C - NH - CH - C -0
I
I
I
H
CH3
+
0
0
" -0
0
0
II
II
H3N -CH - C-NH-CH - C-NH . CH - C - NH - CH - C -0
I
I
H
1
I
CH3
DCC
o
0
II
II
Mc,c
ci1
CH(CH3h
I
er
Boc-leuc;nc
CH2CHMe2
" -0
0
0
II
II
Boc NH-CH-C- NH - CH - C - NH-CH - C-NH-CH - C-NH - CH - C -0
I
CH2CHMe2
I
H
o
I
CH3
0
II
+
I
II
tHF
CH(CH3h
0
I
CH2Ph
0
II
0
II
II
H N - CH - C - NH - CH -C - NH - CH - C - NH-CH-C - NH-CH - C -OH
3
I
CH2CHMe2
I
H
I
CH3
I
CH(CH3h
631
P
CH2Ph
-NH . H . COOH
0
P
CH2Ph
CF3COOH
II
+
CH(CH3h
I
-0
I
CH2Ph
P
24-31
a
CH2Cl
-
II
Me3COC - NH-CH - COO
I
+
�
CH2CONH2
Boc-asparagine
a
a
11
II
--- Me3COC-NH-CH-C-O
I
1
CF3CO H
11
+
H3N-CH -C-O
a
II
Me3COC - NHCH2COOH
a
a
II
II
I
I
I
l
H
a
11
I
CF3COOH
a
11
H3N - CH -C - NH-CH - C-O
I
H
a
?
II
Boc-isoleucine Me3COC - NH - H - COOH
CH3 - CHCH2CH3
a
a
II
II
II
1
I
CH3 - CHCH2CH3
I
H
a
--0
P
DCC
a
11
I
--0
CH2CONH2
.HF
a
II
+
P
CH2CONH2
Me3COC -NH-CH - C-NH CH -C - NH-CH-C-a
-
--0
CH2CONH2
II
I
P
DCC
a
+
--0
CH2CONH2
Me3COC-NH-CH - C-NH-CH -C-O
a
P
CH2CONH2
o
Boc-glycine
:
--0
a
II
II
H3N-CH -C-NH-CH -C - NH-CH-C-OH
I
CH3 - CHCH2CH3
I
H
632
I
CH2CONH2
P
24-32 Please refer to solution 1-20, page 12 of this Solutions Manual.
24-34
:0 :
:0 :
(a)
:0 :
:0 :
II
0
(c)
CO2
+
N
(d)
?
CH3C -NH- H - COOH
I
� CHO
+
C'x,
'
I
N
(CH2)4NHCOCH3
+
H
H
the enzyme does not
recognize D amino acids
I
(f) H2N-CH - COOH
H2N-CH - CN
I
H3C-CHCH2CH3
,
N
I
I
O==C
H
L-proline
(e)
y:)
HOOC
COOH
N-acetylD-proline
o
II
Br-CH -C-Br
(g)
H3C-CHCH2CH3
CH3
I
CH2CH(CH3h
with a water workup, the acid
bromide would become COOH
isoleucine
o
I
H2N-CH - COOH
(h)
OR
I
II
H2N-CH - C -NH2
CH2CH(CH3h
CH2CH(CH3h
another possible answer, depending on
whether the acid bromide is hydrolyzed
before adding ammonia
leucine
o
24-35
II
(a)
C-COOH
NH3
I
..
I
H2N-CH- COOH
CH(CH3h
(b)
H2C-COOH
I
H3C-CHCH2CH3
valine
CH(CH3h
H20
---i
...
_
PBr3
---
I
Br-CH-COOH
H3C-CHCH2CH3
633
excess
NH3
---
I
H2N-CH - COOH
H3C-CHCH2CH3
isoleucine
24-35 continued
(c)
o
II
NH3, HCN
..
H20
C-H
I
H2N -CH -COOH
I
CH2CH(CH3h
leucine
CH2CH(CH3h
o
o
COOEt
I
..
2) B�H2Ph
N-CH
I
o
COOEt
COOEt
H
H2N - -COOH
?I
1 ) NaOEt
T
N- -CH2Ph
CH2Ph
COOEt
o
phenylalanine
24-36
(a) H2N-CH -COOH
HOCH(CH3h
+
I
CH3
o
II
(b) H2N-CH -COOH
PhC-CI
+
I
CH3
(pyridine)
o
II
.. PhC- NH - CH -COOH
----1�
I
o
(c) H2N-CH -COOH
I
o
II
PhCH20C-CI
+
CH3
II
--­
PhCH20C - NH - CH -COOH
I
CH3
CH3
o
(d) H2N-CH -COOH
I
o
0
II
II
--
Me3COC - 0 - COCMe3
+
II
Me3COC - NH - CH-COOH
I
CH3
24-37
CH3
COOH
-
HO-C-H
-
CH3
COOH
TsCI
-
...
pyridine
TsO-C-H
excess
NH3
...
-
COOH
-
H-<;:-NH2
-
CH3
CH3
D-alanine
634
o
24-38
COOEt
I
N-CH
o
1) NaOEt
N-
I
COOEt
o
o
?
COOEt
-CH,
COOEt
L\!
,-1H
N
H30+
l
H2N -CH - COOH
I
racemic
histidine
'
N
24-39
0
II
NH3, HCN
C-H
H2O
I
H2C
to
0
N
H
0
II
0
II
I
I
CH2
CHCH3
CH2SCH3
OH
II
H2N-CH - C-NH-CH - C-OH
CH2
OH
CH2SCH3
0
racemic
0
tryptophan
0
II
II
II
H2N-CH - C-NH-CH . C-NH - CH - C-OH
II
N
H
I
0
(CH2h
I
HN-CNH2
to
neutral
I
CHCH3
I
I
H2C
0
II
I
H
H2N -C-COOH
I
o
I
..
neutral
H2N-CH - C-NH-CH -C-OH
I
(c)
to
N
H
o
24-40
(b)
L\
I
H2C
(a)
H30+
H
H2N -C -CN
..
==.i
NH
I
CH2
I
I
(CH2)4NH2
COOH
NH
635
basic: two basic side chains and one
acidic side chain
24-40 continued
acidic: carboxylic acid side chain, and the
SH is weakly acidic
(d)
24-41
(a), (b), (c)
(
isoleucine
�A
CH3
�
glutamine
��
\
0
II
II
*
CH3CH2-CH - C H - NH - C-CH - CH2CH2- C-NH2
I
I *
NH - CO - CH2NH2 -- N-terminus
C-terminus --- CONH2
\(0
____
I
________�
\
'(
glycine
Peptide bonds are denoted with asterisks ( * ) .
(d) glycylglutamylisoleucinamide; Gly-GIn-He
NH2
0
o
24-42
•
II
II
H2N -CH . C - NH - CH . C - OCH3
Aspartame:
I
CH2COOH
I
CH2Ph
Ic,-_� ,--___) �'-_--.. ,--_�I
Y
aspartic acid
(from Edman
degradation)
J
'(
phenylalanine
}
methyl ester
no free COOH =>
no reaction with
carboxypeptidase
Aspartame is aspartylphenylalanine methyl ester.
24-43
0
0
0
0
o
II
II
II
II
II
H2N -CH . C - N H - CH . C - N H - CH . C - NH - CH . C - NH - CH . C-OH
I
I
I
I
I
CH2Ph
CH3
H
CH2
CH3
I
CH2SCH3
'-----r--' '-----r--''-----r--' '----r-" '-----r--'
phenylalanine
�
alanine
y
glycine
)
methionine
from Edman degradation
alanine
�
from carboxypeptidase
636
24-44
(a)
°
protect the N-tenninus of the first amino acid 0
II
PhCH20C - Cl
+
I
H2N -CH -COOH
�
II
I
Z-NH - CH -C-OH
CH2CHMe2
CH2CHMe2
leucine
o
I
II
1
°
II
activate the
C-terrninus
CI-C -OEt
°
II
Z -NH - CH -C-0-C-OEt
1
CH2CHMe2
add the next amino acid
o
Z
-
I
II
?
H2N - H -COOH
alanine
CH3
°
I
II
NH-CH - C-NH - CH -C -OH
!
CH2CHMe2
activate the C-terrninus
II
0
I
CH3
�
Cl-C-OEt
II
0
I
II
0
Z -NH-CH-C-NH - CH -C-0 -C -OEt
j
CH2CHMe2
add the next amino acid
o
I
II
CH3
?
H2N - H - COOH
phenylalanine
CH2Ph
°
I
II
I
Z -NH-CH -C-NH-CH - C-NH - CH -COOH
CH2CHMe2
deprotect the N-tenninus
o
I
II
t
CH3
CH2Ph
H2, Pd
°
I
II
I
H2N-CH -C-NH CH - C - NH -CH - COOH
CH2CHMe2
637
-
CH3
CH2Ph
r:
24-44 continued
(b)
°
II
Me3COC - NH - CH - COOI
CH2Ph
�
U
II
---
V
+
°
0
CH2Cl
1
Boc-phenylalanine
attach C-terminus of
N-protected amino acid
to polymer support
CH2Ph
-0
deprotect
CF3COOH N-terminus
1\ -0
o
+
P
H3N -CH - C-O
o
Boc-alanine
T
1\
Me3COC-NH- H - C-0
I
II
Me3COC - NH -CH - COOH
I
CH3
0
0
II
II
I
CH2Ph
.
next ammo
DCC a dd
.
aCId and couple
0
" -0
P
Me 3COC - NH- CH-C -NH-CH-C-O
I
CH3
I
CH2Ph
I
o
t
CF3 COOH
0
II
+
deprotect
.
N-termmus
H3N-CH-C -NH-CH - C-O
I
CH3
o
Boc-leucine
?
II
Me3COC-NH- H - COOH
CH 2CHM e2
0
0
\I
0
II
II
1
I
"
CH2Ph
CH2CHMe2
I
°
1\
0
0
II
+
I
CH3
+
P
add next amino
DCC acid and couple
Me3COC - NH-CH - C-NH· CH . C-NH . CH . C - 0
I
-0
CH2Ph
HF
-0
deprotect and remove from polymer
0
II
II
H3N -CH . C-NH· CH . C-NH . CH . C-OH
I
CH2CHMe2
I
CH3
638
I
P
CH2Ph
24-45
o
o
protect N-terminus
II
PhCH20C - Cl
H2N -CH-COOH
+
I
II
---
Z - NH-CH-C-OH
I
CH3
CH3
alanine
o
II
activate the
C-tenninus
0
1
II
Z -NH - CH - C-0-C-OEt
I
CH3
o
II
T
H2N - H - COOH
valine
add the next amino acid
CH(CH3h
0
II
Z-NH - CH - C -NH-CH-C-OH
I
I
CH3
CH(CH3h
o
II
!
H2• Pd
dcprolect thc N-terminus
0
II
react the N-terminus of the dipeptide
at the left with the N-protected,
C-activated tripeptide below
H2N -CH - C-NH-CH - C -OH
I
I
CH3
CH(CH3h
o
II
1
o
0
II
0
II
0
II
II
Z -NH-CH - C-NH - CH - C -NH-CH - C -0 -COEt
I
H3C-CHCH2CH3
0
I
CH2CHMe2
0
II
I
CH2Ph
0
II
II
Z - NH -CH-C-NH-CH-C - NH CH-C -NH-CH-C -NH . CH-COOH
I
H3C-CHCH2CH3
o
�
-
I
CH2CHMe2
H2, Pd
I
CH2Ph
I
CH(CH3h
deprotect the N-terminus
0
0
II
II
I
CH3
0
II
II
H2N -CH - C -NH . CH . C -NH . CH - C -NH . CH - C-NH . CH-COOH
I
H3C -CHCH2CH3
lie
I
CH2CHMe2
Leu
I
I
I
CH2Ph
CH3
CH(CH3h
Phe
Ala
Val
639
24-46
(a) There are two possible sources of ammonia in the hydrolysate. The C-terminus could have been
present as the amide instead of the carboxyl, or the glutamic acid could have been present as its amide,
glutamine.
(b) The C-terminus is present as the amide. The N-terminus is present as the lactam (cyclic amide)
combining the amino group with the carboxyl group of the glutamic acid side chain.
(c) The fact that hydrolysis does not release ammonia implies that the C-terminus is not an amide. Yet,
carboxypeptidase treatment gives no reaction, showing that the C-terminus is not a free carboxyl group.
Also, treatment with phenyl isothiocyanate gives no reaction, suggesting no free amine at the N-terminus.
The most plausible explanation is that the N-terminus has reacted with the C-terminus to produce a cyclic
amide, a lactam. (These large rings, called macrocycIes, are often found in nature as hormones or
antibiotics.)
24-47
(a) Lipoic acid is a mild oxidizing agent. In the process of oxidizing another reactant, lipoic acid is reduced.
COOH
S-S
SH
oxidized form
�
(b)
O==C
I
I
I
�
H20
0
II
+
�
�R
RC-H
+
� (CH2)4COOH
\ /
HC-CH2S
I
�
HN
H2N -CH-COOH
I
CH2
f--.J
SH
SCH2-CH
I
�R
NH
\not basic
640
RC-OH
+
�
NH
SH
II
---l.�
24-48
basic --- N
I
--
o
S-S
(a) histidine:
�
C==O
I
---l.�
NH
S-S
reduced form
O==C
I
HN
SH
�
C==O
HC -CH2SH HSCH2-CH
(c)
COOH
N
SH
SH
(CHZ)4COOH
24-48 continued
(b)
t
N -H
�
..
..
I
N -H
�
..
�
l
� -H
Jl
+N
:N
:N
H
H
H
I
I
I
H
H
H
In the protonated imidazole, the two N's are similar in structure, and both NH groups are acidic.
(c)
tj
-H
H+
---
tj
-H
-H+
----
(J
+N
I
I
H
H
resonancestabilized
We usually think of protonation-deprotonation reactions occurring in solution where protons can move with
solvent molecules. In an enzyme active-site, there is no "solvent", so there must be another mechanism for
movement of protons. Often, conformational changes in the protein will move atoms closer or farther.
Histidine serves the function of moving a proton toward or away from a particular site by using its different
nitrogens in concert as a proton acceptor and a proton donor.
N
24-49 The high isoelectric point suggests a strongly basic side chain as in lysine. The N-CH2 bond in the
side chain of arginine is likely to have remained intact during the metabolism. (Can you propose a likely
mechanism for this reaction?)
H2N-CH - COOH
I
(CH2h
I
HN
C-NH2
I
NH
arginine
�
H2N -CH - COOH
I
(CH2h
I
+
NH2
ornithine
641
o
II
H2N--C-NH2
urea
0
24-50
0
0
II
II
II
H2N - CH -C -NH -CH -C -NH -CH - C-OH
(a) glutathione:
,.-__-,,1
gluta c acid
(from Edman
degradation)
(b) reaction: 2 glutathione
+
'-----y---"
CH2SH
�
-..
""....__
.
H
---
J
. ,--_-,,'
.
\....._----.
gl cine
(from carboxypeptidase)
cysteine
H202
glutathione disulfide
II
structure of glutathione disulfide:
I
I
I
HOOCCH2CH2
0
+
2 H20
0
0
II
II
H2N-CH - C-NH -CH -C - NH -CH - C-OH
I
I
HOOCCH2CH2
HOOCCH2CH2
I
I
I<
CH2S
H
I
CH2S
II
I
new S-S bond
H
I
II
II
H2N -CH - C -NH -CH -C -NH -CH -C-OH
0
0
0
24-51
end groups: N-tenninus
Ala ------- lie
chymotrypsin fragments
B
A
C-tenninus
Glu-Gly-Tyr (middle)
Ala-Lys-Phe
c
N-terminus
�-----y
Arg-(Ser? Leu?)-lIe
y
c-tenninu
Ala-Lys-Phe-Glu-Gly-Tyr-Arg-(Ser? Leu?)-Ile
trypsin fragments
D
�
Ala-Lys
E
Phe-Glu-Gly-Tyr-Arg
F
y
l
}
' - - - - -'
se r -Leu lIe
Ala-Lys-Phe-Glu-Gly-Tyr-Arg-Ser-Leu-Ile
642
)
I
(a) One important factor in ester reactivity is the ability of the alkoxy group to leave , and the main factor in
determining leaving group ability is the stabilization of the anion . An NHS ester is more reactive than an
alkyl ester because the anion R 2NO- has an electron-withdrawing group on the 0-, t hereby distributing the
negative charge over two atoms instead of just one. A simple alkyl ester has the full negative charg� on the
24-52
oxygen with nowhere to go, RO-.
o
(b»
)l
R
OH
+
- F3C
\
o
abbre\iate
O-Succ
)
�
:0
O=C""'"
I
R
tetrahedral
intermediate
J
:0:
-SUCC
----
R
reactive
anhydride
0
o
t.�
)l
CP3
:O-Succ
. .
o
R
�
)
:0
�
0
)l
o
CF]
o
I
NHS ester
Succ
o
tetrahedral
intermediate
NHS trifluoroacetate is an amazing reagent. Not only does is activate the carboxylic acid through a mixed
an hydride to form an ester under mild conditions , but the NHS leaving group is also the nucleophile that
forms an ester that is both stable enough to work with, yet easily reactive w hen it needs to be. Perfect!
+
(c)
:�
Succ
H2NR
R )l
�
./
+
: 0:
R
----
+
)l N "R
H
/
+
�
HJ
:0:
R
643
-:O - Succ
)l N "R
H
+
HO-Succ
24-53
(a)
COOH
H 2N
�
L
'
R
H
,
NaN0 2 + HC I
•
Y
HONO
)
COO H
COOH
..
nitrous acid
makes diazonium
ions, Sec. 19-17
N
+�
N
"
H
R
intermediate 1
L
NaN )
azide as
nucleophile,
Sec. 19-21B
•
"
R"
N)
H
intermediate 2
D
COO H
H2
-
Pd
R"
"
H
D
(b) The product has the opposite configuration from the starting because of the stereochemistry of the
reactions. Diazotization does not break a bond to the chiral center so the L configuration is retained in
Reaction 1. It is well documented that azide substitution is an SN2 process that proceeds with inversion
of configuration; this is why this process works. The third reaction does not break or form a bond to the
chiral center, so the D configuration is retained.
644
NH 2
CHAPTER 25-LIPIDS
0
25-1
II
CH2 - 0 - C - (CH2 )12CH3
trimyristin
25-2
I
�
�
I
CH2 - 0 - C - (CH2 h2CH3
CH - 0 - C - (CH2 )12CH3
0
o
all c i s
II
II
CH2 - 0 -C -(CH2hCH == CH(CH2hCH3
I
CH
I
CH2
-
-
�
0- C -(CH2hCH
�
0 -C -(CH2hCH
==
CH(CH2hCH3
==
CH(CH2hCH3
triolein, m.p. -40 C
(liquid at room temperature)
25-3
(a)
0
II
2+
2 CH3(CH2 )16 -C -0- Na+ + C a --
0
(b)
II
2 CH3(CH2 )16 -C
(c)
-
0- Na+ + Mg 2+--
0
II
3+
3 CH3(CH2h6-C-0 - Na+ + Fe
--
excess
H2
Ni
[
[
[
CH2 - 0 - C - (CH2 )16CHl
�
I
�
I
CH2 - 0 - C - (CH2 )16CH 3
CH - 0 - C - (CH2 )16CH3
tristearin, m.p. 72° C
(solid at room temperature)
a
II
CH3(CH2 )16-C - 0
a
II
CH3 (CH2 )16 - C - 0
a
II
CH3(CH2 )16-C-0
1
1
1
2
Ca
+ 2 Na+
2
Mg
+ 2 Na+
3
Fe
+
3 Na+
25-4
(a) Both sodium carbonate (its old name is "washing soda") and sodium phosphate will increase lhe pH
above 6, so that the carboxyl group of the soap molecule will remain ionized, thus preventing precipitation.
(b) In the presence of calcium, magnesium, and ferric ions, the carboxylate group of soap will form
precipitates called " h ard-water scum", or as scientists l abel it, "bathtub ring". B oth c arbonate and
phosphate ions will form complexes or precipitates with these cations, thereby preventing precipitation of
the soap from solution.
645
25-5
S03
abbreviate
-
o- S03-
as
.lVV\l\l\l\lVVV
-03S
S03-
S03-
25-6 In each structure, the hydrophilic portion is circled. The uncircled part is hydrophobic.
N '
0(' �
(�?_���O�O-../'O�O-../'O�O�?-�(��I
I
,
� ',
,
h-
,'+
,
"
'
I
-',
I
,
-- ','
,
'
benzalkonium chloride
,
l-
C
.....
..
.........
_
_
---
_
...
__
...
__
...
__
...
__
-
...
_---------- --
....
...
----
- --- --- ------- ---_
....
_--
-
-
--
-
-
---
---_ ....
_
_
----
...
---
... -
-
............
---�----
Nonoxynol-9
-O�N
,
"
"
'
:" Na+
,
II
'."
I
0
"
...
..
_
.....
.....
,
,
0"
,
,
,
,
,
Gardol®
_
...
646
--
electrophile
pentamer
..
�
NaOH
25-8
o
(a)
II
CH 2-0-C-(CH 2)nCH 3
------
I
CH
I
-
�
0 -C - (CH 2)nCH 3
:?:
CH 2 -O - P=O
I
OH
(b)
CH 2-0
I
-
o
"
C - (CH2)nCH 3
�
CH -O-C-(CH 2)"CH]
I
••
...
.
:�:
CH 2 -O - P - O:
I
: .0:
.
• •
: .0:
.
-
:0 :
25-9 Estradiol i s a phenol and can be ionized with aqueous N aOH. Testosterone does not have any
hydrogens acidic enough to react with NaOH. Treatment of a solution of estradiol and testosterone in
organic solvent with aqueous base will extract the phenoxide form of estradiol into the aqueous layer,
leaving testosterone in the organic layer. Acidification of the aqueous base will precipitate estradiol which
can be filtered. Evaporation of the organic solvent will leave testosterone.
647
25- 1 0 Models may help. Abbreviations: "ax" = axial; "eq"
ring junctures are axial to one ring and equatorial to another.
=
equatorial . Note that substituents at cis-fused
ax to B
eq to A
C H3
ax
CH3
(a)
(b)
ax to A
eq to B
eq
HO
C H3
H
ax
H
ax
CH3
(c)
ax
OH
o
H
H
OH
ax
ax
ax
ax
HO
25- 1 1
-
---""
- --
H
0
.......
.
.... - .....
...
"
-'
"
"
,
,
,
,
,
,
.
.
.
,
,
.
.
.
.
,
,
.
,
H:
,
,
OR':
.
,
,
,
:
,
.
geranial
,
'
.
.
OR
menthol
,
.
.
,
,
.
.
,
,
.
,
.
,
.
,
,
.
,
,
,
,
.
,
,
.
,
.........
,
,
.
,
.
..
...
,
.'
.
,
..... - ...........
,
.
.
o
COOR :
.
.
,
.
.
>-':-ii-+--
,
,
.
.
...
,
-
camphor
-- ........
-
,
.
,
.
.....
-
,
,
.
,
,
...
,
,
.
,
.
.
.
.
.
,
,
,
.
,
.
,
OR ..
,
,
... ......... ",
.
camphor shown
in top view
,
648
"
,
,
.
,
,
.
,
,
.
o
.
,
,
,
.
,
,
.
- ..... - ..........
,
o
.
.
abietic acid
,
25- 1 2 B -carotene
.. - .. _- ... _ -.....
,
,�
-
,
,
,
,
.
I
•
�
,
,
--- .. _
..
_
.. --
�--"--"-- ..
.
,
,
..............
25-13
"'''''''' .....
,
,
,
.
.
.
.
.
,
I
.
.
,
,
,
.
.....
... -
- .....
.. � .. ......
,
.
'--
-
.
.
,
.
.
,
,
,
.. .............
.
.
,
,
.
.
-
"- ..
...
........
...... ..
_
---'
..... '
limonene
monoterpene
a-famesene
sesqui terpene
a-pinene
monoterpene
(Limonene can also be circled
in the other direction around
the ring-see menthol in 25- 1 1.)
..
,
"
\..-----Y
,
,
'
.
.
.
.
.
OR
.
.
V
''
--.:.
- -...- '
--,:...::...:;
-
.
,
.
.
�
-.... - .. -'
,
.
.
--
zingiberene
sesquiterpene
25- 1 4 Please refer to problem 1 -20, page 1 2 of this Solutions Manual.
25- 1 5
(a) triglyceride
(b) synthetic detergent
(c) wax
649
(d) sesquiterpene
(e) steroId
CH3(CH2hCH =CH(CH2hCOO-
25- 1 6
3
(a)
soap
(b)
CH2
I
I
o
-0
-g
0
-
)U(- - -
Br
(c)
H, H
CH3(CH2h
'
0
Br
--
tristeann
.
CH3(CH2hCHO
3
o
)l--1<
II
I
I
CH2 -0
+
0
II
C (CH2h
Bf
_
(e)
R
I
R
I
CH2-0 -C - (CH2hCHO
CH3(CH2h
(mixture of diastereomers)
(g)
3
'"
"
H H
_
�
"
_
3
Bf
(CH2)7CH ]
_
Br
(CH2hCH3
CH3(CH2hCOOH
o
+
I
R
R
I
CH2 -0-C - (CH2hCOOH
CH - 0-C - (CH2hCOOH
0
II
CH2 -0-C - (CH2h'
I
H
"'"
R
I
CH2 -0-C - (CH2h
(CH2h C -0 - CH
II
�
II
CH - 0 -C - (CH2hCHO
o
H
CH2 -0 - C - (CH2hCOOH
CH2 -0-C - (CH2hCHO
H
II
)l--1<
"
II
(f)
'"
0
"
II
CH2-0-C - (CH2h
Bf
(CH2);C-O - CH
(mixture 0f diastereomers)
(d)
glycerol
(CH2)16CH3
CH - O-C-(CH2116 CH3
0
II
CH2-0-C-(CH2h6CH3
II
HO CH(CH20Hh
Na+ +
CH3(CH2hCH =CH(CH2hCH20H
+
H
,,
"�'
HOCH (CH20Hh
glycerol
650
C
H
/
(CH2hCH3
(CH2hCH3
H2
(a) CH3(CH2hCH =CH(CH2)7COO H ----:-Nl
LiAIH4
2) H30+
25-17
H2
CH3(CH2hCH = CH(CH2hCOOH � CH3(CH2)16COOH
(b)
(c) CH3(CH2)16COOH
from (b)
+
2)
o
Me2S
excess
Br2
PBr3
•
II
o
II
R
6-
I
H2C-OH
+
o
H2C-0 -C-(CH2)14CH3
I
II
NaOH
+
0I
t1
..
H2C -OH
I
HC -OH
I
H2C -OH
glycerol
651
+
k
CH3(CH2h
Na+
(CH2)6
b:COOH
o
0-C - (CH2)16CH3
II
Na+ oII
0 -C - (CH2hCH = CH (CH2hCH3
Na+
glycerol
R
HC-0 -C - (CH2)14CH3
I
R
H2C -0- -OCH2CH2N(CH3h
P
•
HC-OH
H2C-0- P- OCH2CH2NH3
(b)
H20
0 =CH(CH2hCOOH
+
HOOC(CH2hCOOH
nonanedioic acid
B H H Br
I
t1
HC -0-C - (CH2hCH = CH(CH2hCH3
I
CH3(CH2hCH = 0
nonanal
NaOH H2C-OH
..
H2C-0 -C - (CH2)16CH3
I
t1
KMn04
CH3(CH2)7COOH
H20, t1-
CH3(CH2hCH =CH(CH2hCOOH
25-18
CH3(CH2)16COOCH2(CH2)16CH3
----
1 ) 03
(e) CH3(CH2hCH = CH(CH2hCOOH
(f)
H+
HOCH2(CH2)16CH3
from (a)
(d) CH3(CH2hCH =CH(CH2hCOOH
(a)
CH3(CH2)16CH20H
1)
2
Na+
o
0- P-OCH2CH2NH2
I
Na+ 0
II
o
Na+
0 -C - (CH2)14CH3
II
o
0 - P- OCH2CH2N(CH3h
.
I
Na+ - 0
II
+
3
25-20 Reagents in parts (a), (b), and (d) would react with alkenes. If both samples contained alkenes,
these reagents could not distinguish the samples. Saponification (part (c)), however, is a reaction of an
ester, so only the vegetable oil would react, not the hydrocarbon oil mi xture.
25-2 1 (a) Add an aqueous solution of calcium ion or magnesium ion. Sodium stearate will produce a
precipitate, while the sulfonate w i l l not precipitate.
(b) Beeswax , an ester, can be saponified with NaOH. Paraffin wax is a solid mixture of alkanes and will not
react.
(c) Myristic acid will dissolve (or be emulsified) in di l ute aqueous base . Trimyristin will remain unaffected.
(d) Triolein (an unsaturated oil) will decolorize bromine in CCI4, but tri myristi n (a saturated fat) will not.
o
25-22
II
(a)
CH2 - 0 - C - (CH2 )12CH3
asymmetric
carbon
*
I
I
CH2 -
�
�
- C - (CH2hCH == CH(CH2hCH3
CH - 0 - C - (CH2hCH - CH(CH2hCH3
_
0
(b)
optically acti ve
o
II
CH2 - 0 - C - (CH2hCH == CH(CH2 )7CH3
I
�
I
�
CH2 - 0 - C - (CH2hCH == CH(CH2hCH3
CH - 0 - C - (CH2 )12CH3
not optically active ; symmetric
0
25-23
II
CH2 - 0 - C - (CH2 )16CH3
asymmetric
carbon
*
I
I
CH2 -
�
�
- C - (CH2hCH == CH(CH2hCH3
CH - 0 - C - (C H 2 h C H - C H (C H 2 h CH 3
_
0
652
optically active
25-23 continued
o
CH2-0-g - (CH2)16CH3
(a)
I
CH-O
_ g (CH2)16CH3
0
ICH2-O-g-(CH2)16CH3
0
not optical ly active
_
(b)
'--1"",.",".,
"'
�
Br
/
H
Br
CH,(CH2)7
0 --'
II
(CH2);C-O CH
, l l y acti ve
optlca
(mixture 0 f d'la stereomers
(c) products are not optically acti ve
HOCH(CH20Hh
+
Na+
glycerol
o
(d)
0
II
C1 li2 - O-C(CH2)16CH3
.L.J/
I
/
CH2),CH,
) CH2 0 C _ (CH2),
�
_
-OOC(CH2)16CH3
JCH - -� - (CH2hCHO
I
CH2- -� - (CH2)7CHO
+
+
C
0
C
0
•
• ••
••
•
H
_
II
CH2-0-C-(CH2)16CH3
Br.
...
2
2
Na+
�
Bf
'\
-OOC(CH2hCH CH(CH2hCH3
==
0 == CH(CH2hCH3
not optically active
(b) , (c) and (f) are mixtures of stereoisomers,
optically active
25-24 The products in
(e)
(c)
(b)
(a)
Br
Br
H
Br
Br
Br
Br
(f)
HO
653
+
CH20 0
+
CO2
25-25 Is it any wonder that Olestra cannot be digested !
oleic
0
0
�
L
o
0
#"
#
654
l inolenic
(a)
COOH
H
OH
eq
NHCH 2COOH
I
(b)
HO "
Like a soap or detergent, this cholic aci d-glycine
combination has a very polar " head" and a slightly polar
"tail ". The "tai l " can dissolve non-polar molecules and
the polar " head" can carry the complex into polar medi a.
",
25-27
........... ..
- ....
... .,. ...
... ... ---- ..
...
(a) sesquiterpene
,
OR
\
.
,
.. - .. - ......
,
,
,
,
\
.
,
,
,
,
"
,
,
,
.
.
.
.......... _
.
---_ ........
-
,
,
,
,
,
\
I
,
\,
--
---
:'
-
-
o
(b) monoterpene
-
.
.
..... -- ..
,
.
"
,
\
\
\
\
\
.
,
.
,
.
,
-
,
,
OR
.
,
,
,
.............
.
,
,
.
.
\
.
\
.
\
,
,
,
,
,
--
(c) sesquiterpene
,
,
.. ... __ ...
............
... - ..... - .....
\
,
.
.
,
.
,
,
,
,
,
OR
,
,
.
,
,
,
,
""
--H
-
.
--
.
,
,
,
,
,
,
H
.
\
,
,
,
,
,
--.
..........
655
25-27
(d) sesquiterpene
_ .......
,
,
_
·
, .. -
·
.... - ...........
·
·
,
·
,
,
CH3
,
J OR
,
·
,
,
,
.
:
"
.... --- .............
CH3
·
·
,
·
·
,
,
·
,
CH3
--'
.
.
,
:
'
CH3 :
... :
.
,
\
,
.
"
,
,
·
CH3
"
........................... '
I
.............. -
25-28 The fonnula ClsH340 2 has two elements of unsaturation; one is the carbony l , so the other must be
an alkene or a ring. Catalytic hydrogenation gi ves stearic acid, so the c arbon cannot include a ring; it must
contain an alkene. The products from KMn04 oxidation detennine the location of the alkene:
HOOC(CH 2)4COOH
+
HOOC(CH 2)IOCH3 :::::> HOOC(CH2)4CH
==
CH(CH 2)IOCH3
If the alkene were trans, the coupling constant for the vinyl protons would be about 1 5 Hz; a 10 Hz coupling
constant indicates a cis alkene. The 7 Hz coupling is from the vinyl H's to the neighbori ng CH2·s.
25-29
COOH
linolenic acid
COOH
linoleic acid-known
COOH NEW
COOH NEW
COOH NEW
COOH NEW
COOH
oleic acid-known
In addition to the new isomers with different positions of the double bonds , there has been growing concern
over the trans fatty acids created by isomerizing the naturally occuring cis double bonds. More
manufacturers are now listing the percent of "trans fat" on their food labels.
25-30 The cetyl glycoside would be a good emulsifying agent. It has a polar end (glucose) and a non­
polar end (the hydrocarbon chain), so it can dissolve non-polar molecules, then carry them through pol ar
medi a in micelles. This is found in Nature where non-polar molecules such as steroid honnones,
antibiotics, and other physiologically-active compounds are c arried through the bloodstream (aqueous) by
attaching saccharides (usually mono, di , or tri), making the non-polar group water soluble.
656
25-3 1 (a) Four of these components in Vicks Vapo-Rub® are terpenes . The only one that is not is decane;
it is not comprised of i soprene units despite having the correct number of carbons for a monterpene.
Three of the four terpenes have had their isoprene units indicated in previous problems; in each of those
three, there are two possible w ay s to assign the isoprene units , so those pictures will not be duplicated here.
a-pinene;
cineole;
eucal yptol
see 25- 1 3
.
.
.
.
(
.
.
.
menthol;
see 25- 1 1
.
camphor;
see 25- 1 1
.
.
o
OH
.
.....
..
.. __ ........
camphor
decane
(b) Vicks Vapo-Rub® must be optically active as it contains four optically active terpenes.
........
25-32
"
, .............. _-
,
·
(a) Of the two, only nepetalactone is a terpene.
The other has only 8 c arbons and terpenes must
be in multiples of 5 c arbons.
·
·
·
"
.
CI:?0
.
.
(b) Of the two, only the second is aromatic as can
be readily seen in one of the resonance forms.
.
,
o
......... --_ ....
"
::::-...
o
'1 .. ,
,
-
.
,,
.
0
_ _.
... __
__
:-.
o+�o
°
::::-...
I
00/
//
o
(c) Each compound i s c leaved with NaOH (aq) to give an enolate that tautomerizes to the more
stable keto form.
NaOH
-
NaOH
o
a(cooo
657
0:
.0
25-33
(a) High temperature and the diradical O 2 molecule strongly suggest a radical mechanism.
,---H
(b) Radical stability follows the order: benzylic> allylic
doubly allylic making it a prime site for radical reaction.
(c)
R R
••
·
••
-
·
�
•
initiation
>
3°> 2° > 1°. A radical at C-ll would be
}
COOH
Ll
j'\ .. . .
H
�C;
propagation step 1
..J
•
R-R
COOH
..
....
.
f--l��
plus two allylic
resonance forms
.
�� }
H
••
•
.
0
I
�
••
-
0
••
·
T
H
.. ..
_
_
!
_
COOH
COOH
propagation step 2
:O-OH
��
�
t
I
COOH
••
T-""'=�
COOH
H
recycles to begin
propagation step J
foul-smelling aldehydes, ketones,
and c arboxylic acids = " rancidity"
(d) Antioxidants are molecules that stop the free radical chain mechanism. In each of the two c ases here,
abstraction of the phenolic H makes an oxygen radical that is highly stabi lized by resonance, so stable that
it does not continue the free radical chain process. It only takes a small amount of antioxidant to prevent
the chain mechanism. Interestingly, in breakfast cereals, the antioxidant is usually put in the pl astic bag
that the cereal is packaged in , rather than in the food itself. BRA and BHT can also be used directly in
food as there is no evidence that they are h armful to humans.
:0 :
:0 :
•
(H3C h C
�
Y
OCH 3
658
B RA radical
•
(CH3hC
�c(CH3h
Y
CH 3
BHT radical
CHAPTER 26-SYNTHETIC POLYMERS
Note: In this chapter, the " wavy bond" symbol means the continuation of a polymer chain .
.JVVVVVVV"I..
H
I
H
I
H
I
H
I
IVVVV'C -C - C - C·
I
H
I
Ph
I
Ph
I
H
10 radical, and not resonance-stabilized­
this orientation is not observed
Orientation of addition always generates the more stable intermediate; the energy difference between a 10
radical (shown above) and a benzylic radical is huge. Moreover, thi s energy difference accumulates with
each repetition of the propagation step. The phenyl substituents must necessarily be on alternating carbons
because the orientation of attack is always the same-not a random process.
26-2
PhCOO - OOCPh
Ph·
+
2 Ph·
----I
..
�
2 CO 2
,�
T!)� \H
+
CH3
H
/
\
C=C
H
I
CH3H
I
I
I
I
CH3H
I
I
I
I
CH3
I
Ph-C-C - C-C - C - C.
I
H
H
H
H
H
659
I
H
etc.
26-3 The benzylic hydrogen will be abstracted in preference to a 2° hydrogen because the benzylic radical
is both 3° and resonance-stabilized, and the 2° radical is neither.
benzylic
H
I'V'VVV'
\)H1
H
1
H
1 (I
1
1
1
Ph H
1
Ph H
.JVVVV
(k
(
C-C-C-C ,JVV\J\A
1
Ph
I
H
new benzylic
radical
H
I'V'VVV'
I
1
I
I
1
H
+
H
I
I
H
I
I
H-C-C-C-CJVVVV\
1
Ph
1
1
H
1
H
Ph
tenninated chain
Ph
Ph
Ph H
1
1
I
I
C-C-C- C ,JVV\J\A
I
Ph H
1
H
H
\ rU
c==c
/
\
Ph
H
1
1
H
H
Ph H
H
1
H
1
growing polystyrene chain
middle of a polystyrene chain
H
H
1
oC-C-C-CJVVVV\
I
I
1
1
Ph H
Ph H
+
c-C-C-C ,JVV\J\A
H
1
H2C,
>
H
?
Ph
Ph
/H
Ph
Ph
o
Ph
26-4 Addition occurs with the orientation giving the more stable intennediate. In the case of isobutylene,
the growing chain will bond at the less substituted carbon to generate the more highl y substitited carbocation.
H
I'V'VVV'
I
Me
1+
+
C-C
I
H
1
Me
H
Me
/
\
C==C
�
..
----I
\
/
Me
H
H
.JVVVV
1
Me H
I
1
I
1
1+
C-C-C-C
1
H
Me H
1
Me
3° carbocation-favored
26-5
H1
H
1
1
(a) chlorine can stabilize a c arbocation intennediate by resonance
.JVVVV
1+
C- C
H :Cl:
..
..
H
1
H
I
IVVV\I'C-c
1
II+
H :Cl
660
H
Me
OR
I
Me Me H
1
I
1+
1
1
1
IVVV\I'C-C-C-C
1
H
Me Me H
1 ° c arbocation---disfavored
(also more steric hindrance)
26-5 continued
(b) CH3 c an stabilize the c arbocation intermediate by induction
H
H
1
1+
2° (not the best case imaginable, but still possible)
C -C
IVVVV"
1
1
H
CH3
(c) terrible for cationic polymerization : both substituents are electron-withdrawing and would destabilize
the carbocation intermediate
H
CaaCH3
1
1+
1
1
H
26-6
C
N
benzylic
H
IVVVV"
destabilized carbocation
C-C
.JVVVV
1
\.
H
H
1
1
H
1
C -C - C - C.JVVVV\
1
1
1
1
Ph H
Ph H
+
+
H
1
H
1
hydride
transfer
---
H
1
1
C-C-C-C.JVVVV\
1
1
Ph
1
H
1
Ph
1
H
H
1
1
H
1
H-C - C-C-C.JVVVV\
1
Ph
H
growing polystyrene chain
middle of a polystyrene chain
H
1
1
H
Ph
1
H
terminated chain
+
H
IVVVV"
1
Ph
Ph
1
1
1
I
1
H
1
1
..
.JVVVV
H
H
I
\
C==C
I
\
Ph
H
H
H 2C, / H
C
/
Ph H
I
1
C - C - C -C.JVVVV\
H
H
C -C-C -C.JVVVV\
Ph
Ph
Ph
H
1+
1
1
Ph
H
1
1 +
H
1
H
new henzylic
cation
Ph
Ph
Ph
Polystyrene is particularly susceptible to branching because the 3° benzylic c ation produced by a hydride
transfer is so stable. In poly(isobutylene), there is no hydrogen on the carbon with the stabi lizing
substituents; any hydride transfer would generate a 2° c arbocation at the expense of a 3° c arbocation at the
end of a growing chain-this is an increase in energy and therefore unfavorable.
2°
H
JVVVV
1
"
Me H
1
1
1
C - C - C -C.JVVVV\
1
H
1
1
Me H
Me H
Me
1
+
Me
middle of a poly(isobuty lene) chain
+1
30
1
Me H
1
1
C -C -C -C.JVVVV\
1
1
Me H
1
1
no hydride
transfer
Me H
growing poly(isobutylene) chain
661
26-7
. .
:0:
:0 :
II
H
I\IVVV'
C-OCH3
I
_I
C -C :
I
I
H H
�
..
If-....
-l..
.
JVVVV
C-OCH3
I
II
C-C
I
I
H
0
26-8
I
H
II
H
H
COCH3
\ nl
---C=C
HO:
'---I-"
\
H
C
N
I
I
H
:0 :
:0 :
II
C-OCH3
_I
HO-C-C:
• •
H
I
C
I
H
�---l"'�
I
HO-C -C
I
N:
H
C-OCH3
II
H
�
....
.
-l...
..!--
I
I
:0 :
II
C-OCH,
I
HO-C-C
I
C -N:
H
o
II
C=N:
. .
II
COCH3
\ nl
C=C
I
\
H
C_N
. .-
:0:
:0 :
H
I
COOMe I I
H C-OCH3
I
I
_I
HO-C-C-C-C:
I
I
I
I
H CN H C
_
•
H
...
I
COOMe I
H
C-OCH3
I
I
"
HO-C-C-C-C
!
N:
I
H
I
I
CN H
I
C
H
I
: 0:
COOMe "
H
C-OCH3
I
I
I
I
II
�HO-C-C-C-C
....
.!-.
-l...
N:
I
H
I
CN H
C = N:
etc.
COOMe COOMe COOMe COOMe COOMe
CN
CN
CN
CN
CN
This polymerization goes so quickly because the anionic intermediate is highl y resonance stabilized by the
c arbonyl and the cyano groups . A stable intermediate suggests a low activation energy which translates to a
fast reaction.
662
26-9
H
(a)
I
rvvVV'
H
H
I (I
H
H
I
- I
C - C -C - C .JVVVV'\
I
I
CN H
I
..JVVVV
I
H
I
I
I
CN H
I
C - C - C - C .JVVVV'\
I
I
I
H
I
I
I
I
CN H
CN H
H
H
+
I
H
I
I
H
I
H-C-C-C-C .JVVVV'\
I
H
CN H
I
growing poly(acrylonitri le) chain
!
CN H
_I
H
I
:C - C -C-C .JVVVV'\
middle of a poly(acrylonitrile) chain
H
H
I
I
CN H
I
CN H
terminated chain
H
H
\ nl
C=C
I
\
H
CN
H
I
rvvVV'
H
I
CN
CN
CN H
I
I
>
C - C - C - C .JVVVV'\
I
I
I
H
CN H
I
H2C
CN
CN
CN
CN
"" H
, /
C
1-
CN
(b) The chain-branching hydride transfer (from a cationic mechanism) or proton transfer (from an aniontc
mechanism) ends a less-highly-substituted end of a chain and generates an intermediate on l more-highly­
s ubstituted middle of a chain (a 3° carbon in these mechanisms). This stabilizes a carbocation, but greater
substitution destabilizes a carbanion. B ranching can and does happen in anionic mechanisms, but it is less
like ly than in cationic mechanisms.
26- 1 0
isotactic poly(acrylonitrile)
NC
H
NC
H NC
syndiotactic polystyrene
663
H NC
H
26- 1 1
(a)
all trans
(b) The trans double bonds in gutta-percha allow for more ordered packing of the chains, that is, a higher
degree of crystallinity. (Recall how cis double bonds in fats and oils lower the melting points because the
cis orientation disrupts the ordering of the packing of the chains.) The more crystalline a polymer is, the
less elastic it is.
26-12 Whether the alkene i s cis or trans is not specified.
I
Me
I
H
H
Me
H
H
I
I
I
I
I
Me
I
I
H
H
I
� C -- C--C -- C==C -- C �
\
Y
isobutylene
I
H
\.
Y
)
isoprene
26-13 The repeating unit in each polymer is boxed.
(a) Nomex®
o
IVV'g
0
0
0
--o- --o- --o- --o- �.NV'
�
�
g- �
H
�
�
�H -g
�
�
g - �H
26- 1 4 Kodel® polyester (only one repeating unit shown)
.NV' 0 - CH 2
-0
� !J
�-o-�
CH 2 - 0 - C
664
C.NV'
�
�
H
26-15
o
II
O-C
I
�
\\
-0'/
CHOJVV\
I
CH20
I
'\
_
o
11
H2C -O -C
0
II
I
c-o -rn 0
I
11
H 2C-O -C
-0-0\\
\.
!J
� h
CH,o-"'"
I
0II
CHO
I
""""
C-O -CH2
0
C H,oII
C-O -CH
I
CH20
�
,
Glycerol is a trifunctional molecule, so not only does it grow in two directions to make a chain, it grows
in three directions. All of i ts chains are cross-linked, forming a three-di mensional l attice with very little
motion possible. The more c ross-linked the polymer is, the more rigid it is.
26-16 For simplicity in this problem, bisphenol A will be abbreviated as a substituted phenol.
HO
-o-�
b�
I � OH
--
CH3
represented as
__
bisphenol A
aC :
mechanism
..
C I > 'C I
R .. �OH
� �
·
• •
--
:?)
0C
Cl 1'CI
o+,
H
..
R
-oI�
:R: \.OH
-C
l--
R
HCI(g)
�D
H /C, ·CI·
/C.'CI
CI
O
�9+
,
•
•
•
:
•
R
R
H
:0:
1
R
10
o
II
C
O/ 'O
I
O U
�
R
-HCI
-
.---CI-
(?¢CI
C
o/ ' o·
R
665
+1 u......;
H
as above
+
I
�
R
R
26-17 B isphenol A i s made by condensing two molecules of phenol with one molecule of acetone, with
loss of a molecule of w ater. This is an e\ectrophi lic aromatic subsitution (more specifically, a Friedel-Crafts
alkylation), and would require an acid catalyst to generate the carbocation. While a Lewis acid could be
used, the mechanism below shows a protic acid.
from acetone
OH
HO
from phenol
from phenol
...
...
B 0O
/
_
---0-
H CH3
�
// C� I
HO
'/
-
C-OH
I
CH3
plus three resonance forms
plus four resonance forms
666
Q R
�
26-18
Ph-N = C
i -+k cf.
:0 :
..
I
Ph-N = C-O-Et �
�
-Et
-
:0 :
II
Ph -N - C - O -Et
..
..
f
H -O - Et
• •
• •
+ 1)
;i?
H-O-Et
H
I
• •
}
• •
:0 :
II
Ph-N-C-O-Et
two rapid
proton transfers
H
26-19 Glycerol is a trifunctional alcohol . It uses two of its OH groups in a growing chain. The third
group cross-links with another chain. The more cross-linked a polymer, the more rigid it is.
26-20
urethan linkage
I
-O-I�
� /;
� I
r
0
Me
.1\1\1'0
OH
I
T
I
O-C-N
Me
\...
,
H
H
�
) \.
V
0
� N-CI I
I
Me
)
y
bisphenol A
.N\r
toluene diisocyanate
26-21 Please refer to solution 1-20, page 12 of this Solutions Manual.
26-22
H
(a)
.rvvvv
I
Me
I
I
H
Me
I
H
I
Me
I
C-C-C-C-C-C .I\I\I'
I
H
I
Me
I
I
H
Me
I
H
I
Me
(b) Polyisobutylene is an addi tion polymer. No small molecule is lost, so this c annot be a condensation
polymer.
(c) Either cationic poly merization or free-radical polymerization would be appropriate. The carhocation or
free-radical intermediate would be 3° and therefore rel atively stable. Anionic polymerization would be
inappropriate as there is no electron-withdrawing group to stabilize the anion .
26-23
(a) It is
a polyurethane.
(b) As with all polyurethanes, it is a condensation polymer.
o
(c)
II
.N\r CH2CH2CH2 - N-C- 0 .1\1\1'
I
HOCH2CH2CH2NH2
H
667
+
CO2
26-24
(a) It is a polyester.
(b) As with all polyesters, it is a condensation polymer.
(c)
Using the dicarboxylic acid instead of the ester would produce water as the small neutral molecule lost in
this condensation.
26-25
(a) Ury lon® is a polyurea.
(b) A pol yurea is a condensation polymer.
(c)
.JVV'-
(CH2 )
9
-N
I
°
-
II
C
H
-
N
I
JVV'
H20
�
H
26-26
(a) Pol yethylene glycol, abbreviated PEG, is a polyether.
(b) PEG is usually made from ethylene oxide (first reaction shown). In theory, PEG could also be made by
intermolecular dehydration of ethylene glycol (second reaction shown), but the yields are low and the chains
are short.
°
n
U
+
HO-
ethylene
oxide
HO
",OH
�
"'"
�
ethylene
glycol
(c) B asic c atalysts are most l ikely as they open the epoxide to generate a new nuc\eophile. Acid catalysts
are possible but they risk dehydration and ether cleavage.
668
26-26 continued
(d) Mechanism of ethylene oxide polymerization (showing hydroxide as the base):
-0 0
/"""-....
.., 0 ....
/"""-....
'
�
�
0
HO'
.., 0 ....
...
/"""-....
/"""-.... 0 0 /""'0....
.., 0 ....
' '/"""-.�
'0 :
�
�
'v/
0
f))"
etc.
HO'
:
26-27
(a) Polychloroprene (Neoprene®) is an addition polymer.
(b) Polychloroprene comes from the diene, chloroprene, j ust as natural rubber comes from isoprene:
H
\
Cl
I
C -C
II
H2C
chloroprene
\\
CH2
26-28
(a)
H
H
1
1
H
H
1
1
oJV\r C- 0 -C - 0 -C - O-C - OoJV\r
1
1
H
H
1
Delrin® (polyformaldehyde)
1
H
H
(b) All of these intermediates are resonance-stabilized.
'---/
H+
H
/
O=C---
\
H
H
•• I ,
H
1+
H-O-C",---,
H
1
00
H
H
1
/
H
O=C
\
H
�
1+
H-O- C-O-C
•
1
H
'\0
H
/
O=C
0 0
etc.
4
H
I
\
H
H
I
H
1+
I
H
I
H
H-O-C-O-C-O-C
I
H
trimer
(c) Delrin® is an addition polymer; i nstead of adding across the double bond of an alkene, addition occurs
across the double bond of a c arbonyl group.
669
(a)
26-29
cis
trans
(b) Each structure has a f ully conjugated chain. It is reasonable to expect electrons to be able to be
transferred through the system, just as resonance effects can work over long distances through conjugated
systems.
(c) It is not surprising that the conductivity is directional. Electrons must flow along the 'It system of the
chain, so if the chains were aligned, conductivity would be greater in the direction parallel to the polymer
chains. (It is possible, though less likely, that electrons could pass from the 'It system of one chain to the 'It
system of another, that is, perpendicular to the direction of the chain; we would expect reduced conductivity
in that direction.)
1t
(a) A Nylon is a polyamide. Amides can be hydrolyzed in aqueous acid, cleaving the polymer chain in the
process.
0 0
0
0
H30+
II
II
II
II
+
.JVV' N -C
C - N .JVV'
.JVV' NH3 + HO -C .JVV'C - OH + H3N .JVV'
I
I
H
H
(b) A polyester can be saponified in aqueous base, cleaving the polymer chain in the process .
o
0
0 0
II
II
II
II
NaOH
.JVV'O -C'VVV'C-O.JVV'
.. .JVV' OH + 0-C.JVV' C - 0 + HO.JV\r
26-31
0
0
0
II
II
II
(a)
CCH3 CCH3 CCH3
I
I
I
H 0 H 0 H 0
H OH H OH H OH
H2O
I
I
I
I
I
I
I
I
I
I
I
I
.JV\rC -C -C-C -C-C.JV\r
.JVV'C -C-C -C-C-C.JV\r
H+ or
I
I
I
I
I
I
I
I
I
I
I
I
H H H H H H
H H H H H H
HOpoly(vinyl acetate)
poly(vinyl alcohol)
(b) A polyester is a condensation polymer in which monomer units are linked through ester groups as part of
the polymer chain . Poly(vinyl acetate) is rea lly a substituted polyethylene, an addition polymer, with only
carbons in the chain ; the ester groups are in the side chains, not in the polymer backbone .
26-30
+
JV'V\.
---I'-�
_
.-
(c) Hydrolysis of the esters in poly(vinyJ acetate) does not affect the chain because the ester groups do not
occur in the chain as they do in Dacron® .
(d) Vinyl a lcohol cannot be polymerized because it is unstable, tautomerizing to acetaldehyde.
o
OH
I
II
H 2C==C H
CH3-CH
--
670
(a)
OCOC H3
26-32
�
cellulose acetate
o
I
COC H3
(b) Cellulose has three O H groups per gl ucose monomer, which form hydrogen bonds with other polar
groups. Transforming these O H groups into acetates makes the polymer much less polar and therefore more
soluble in organic solvents .
(c) The acetone dissolved t he cellulose acetate in the fibers . As the acetone evaporated, t he cellulose acetate
remained but no longer had the fibrous, woven structure of cloth. It recrystallized as white fluff .
(d) Any article of clothing made from synthetic fibers is susceptible to the ravages of organic solvents.
Solvent splashes leave dimples or blotches on Corfam shoes . (Yet, Corfam shoes co uld still provide
protection for the toenail polish!)
26-33
OH
OH
OH
OH
Bakelite is highly cross-linked through the
ortho and para positions of phenol; each
phenol can form a chain at two ring positions,
then form a branch at the third position.
mechanism
..
H°
l-w
H
plus four resonance forms
-o- i -oH
°H
further coupling at ortho positions leads to
cross-linked Bakelite
671
·
.0.
26-34 ·
H2N
II
....... C, . . ....
�
\..
N
H
I
H.J
..NH3..
H2N
7,)
....... c, .. -
N
..
..
H2N
6
..
N:...J
I
-HO..
I
H-C-H
H2N
....... C,
eN:
this leads to
cross-linking
/"""
w
....... C,.........
HN
H2N""'"
I
)
N
A
H2N
......
�
0
II
....C
... ,
C
g
+
N
I
H2N
II
.......C,
N
••
H
NH2
C:--,
"0
672
�
.... H
-
N
••
�
a
N
I
H
II
....... C,
NH2
plus another resonance form
�
H""'" 'H
II
�
:0:
N
II
....C
... , .. ....
....... C ,
H2N
+
gA-"
:0:
....... C,..
H
:0:
H""'" 'H
II
l
....
from above
II
I
N
+
H2N
:0 :
'N:
N
H-?-H
U"
� :o:
..
H-C-H
OH
H2N
�
cg
H3N-H
OH
....... C,
II
....... C�
:0:
I
C H
o
H2N
..
'-__--_
:0 :
H2N
H
H' /H
c
H-C-H
I
....... C,.:....
�
H3N:
....... C,...... H
:0:
II
i
H2N
• •
II
;}
..O .
· ·
.0 .
63
2
� � � � � � -y
O
O
O
glycolic
acid
lactic
acid
o
O
O
glycolic
acid
lactic
acid
0
O
�
0
glycolic
acid
lactic
acid
'-----T---" '-----T---" '-----T---" '----r-' '-----T---" '-----T---"
26-36
�
OH
OH
OH
OH
OH
OH
cellulose
cotton
O�
=
polypropylene
As we have seen repeatedly through this presentation of organic chemistry, physical and
chemical behavior depend on structure. The structure of cotton, i.e. cellulose, has multiple
oxygen atoms that form hydrogen bonds with water. When cotton gets wet, it holds onto the
water tightly, as you have seen if you have put cotton clothes in a clothes dryer-it takes a long
time to dry. Polypropylene is a hydrocarbon with no hydrogen bonding groups; the fiber feels
dry because it cannot hold the water the way cotton can. Athletic garments are increasingly
using polypropylene because they allow evaporation and cooling during periods of exertion ;
cotton is just the opposite .
26-37
(a) This addition polymer is called polyvinylidene chloride,
trade name Saran®. It could be made by any of the three
mechanism types : radical, cationic, or anionic .
(b) When the substituent is on every fourth carbon, and one
double bond in the chain in every 4-carbon unit, the polymer
must come from addition across a diene, probably under
cationic conditions.
(c) This polyester is a condensation polymer of two monomers
a diol and a derivative of phthalic acid, either the anhydride, an O
ester, the acid chloride, or the acid itself. Heating the
monomers will make the polymer; no catalyst is required if
done at high temperature.
HO
---<>-monomer
OH
673
Cl
monomer
>===
Cl
H3CO
h
0
one of these is the
other monomer
monomer
0
X
X
X
= OH orCl or
OR
0
26-3 7 continued
(d) This polyamide (Nylon) is a condensation polymer made
from two monomers, a diamine and a derivative of succinic
acid, either the anhydride, an ester, the acid chloride, or the
acid itself. Heating the monomers will make the polymer; no
catalyst is required if done at high temperature.
H2N� NH2
monomer
(a)
26-3 8
COOH
===<CH3 +
o
�
o
one of these is the
other monomer
X=OH or Cl or
OR
o
HO .... /""....
........, ' OH
}- OCH2CH20H
hydroxyethyl methacrylate
====CH3
=\
j
polymerize using radical or
anionic conditions
polymer
(b) This polymer has a few properties that make it useful as the material in soft, extended-wear contact
lenses. First, carboxylic acids usually are crystalline solids with high melting points, but esters and
alcohols are low melting, often liquids, so the polymer with this ester is softer than the carboxylic acid
or even the methyl ester. (The methyl ester, polymethyl methacrylate or Plexiglas, was the first
material used in the original hard contact lenses.) Second, the ability of the free OH to form hydrogen
bonds with water makes the contact lens more fluid and less irritating to the cornea. Third, a hidden
advantage but very important for ocular health: the fluidity of the contact lens also permits oxygen to
go through the lens. Because the cornea does not have a large blood flow, it needs to absorb oxygen
from the air to maintain its health, and this enhanced gas permeability permits the contact lens to be
worn for days at a time without compromising the health of the cornea. Thanks, polymers!
Note to the student: BO N VOYAGE!
I hope you have enjoyed your travels
through organic chemistry.
Jan Simek
674
Appendix I-Summary of IUPAC Nomenclature ofOq:anic Compounds
The purpose of the IUPAC system of nomenclature is to establish an international standard of
naming compounds to facilitate communication. The goal of the system is to give each
structure a unique and unambiguous name, and to correlate each name with a unique and
unambiguous structure.
Introduction
IUPAC nomenclature is based on naming a molecule's longest chain of carbons connected by single
bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or atoms other
than carbon and hydrogen, are indicated by prefixes or suffixes according to a specific set of priorities.
I. Fundamental Principle
II. Alkanes and Cycloalkanes (also called "aliphatic" compounds)
Alkanes are the family of saturated hydrocarbons, that is, molecules containing carbon and hydrogen
connected by single bonds only. These molecules can be in continuous chains (called linear or
acyclic), or in rings (called cyclic or alicyclic). The names of alkanes and cycloalkanes are the root
names of organic compounds. Beginning with the five-carbon alkane, the number of carbons in the
chain is indicated by the Greek or Latin prefix. Rings are designated by the prefix "cyclo". (In the
geometrical symbols for rings, each apex represents a carbon with the number of hydrogens required
to fill its valence.)
CI
C2
C3
C4
Cs
C6
C7
Cs
C9
C IO
Cll
CH4
CH3CH 3
CH3CH2CH3
CH3[CH2hCH3
CH3(CH2hCH3
CH3(CH2]4CH3
CH3(CH2ls CH3
CH3[CH2]6CH3
CH3[CH2hCH3
CH3[CH2lsCH3
CH3[CH2]9CH 3
methane
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
undecane
C12
C13
C I4
C2 0
C21
C22
C23
C3 0
C3 1
C40
C so
H
D
cyclopropane
0
cyclohexane
>
H
,
C
,
/ \
H
'C-C'"
I
H
CH3(CH2]IOCH3
CH3(CH2]IICH3
CH3(CH2]12CH3
CH3(CH2]ISCH3
CH3[CH2]19CH 3
CH3(CH2hoCH3
CH3[CH2hICH 3
CH3[CH2hsCH3
CH3(CH2h9CH3
CH3(CH2hsCH3
CH3[CH2]4SCH3
\
H
H
D
dodecane
tridecane
tetradecane
icosane
henicosane
docosane
tricosane
triacontane
hentriacontane
tetracontane
pentacontane
0
cyclobutane
cyclopentane
0
0
cycloheptane
cyclooctane
The IUPAC system of nomenclature is undergoing many changes, most notably in the placement
of position numbers. The new system places the position number close to the functional group
designation; however, you should be able to use and recognize names in either the old or the new
style. Ask your instructor which system to use.
675
Appendix 1, Summary of IUPAC Nomenclature, continued
III. Nomenclature of Molecules Containina: Substituents and Functional Groups
A. Priorities of Substituents and Functional Groups
LISTED HERE FROM HIGHEST TO LOWEST PRIORITY, except that the substituents within
Group C have equivalent priority .
Prefix
Structure
0
II
R - C-OH
0
\I
R - C-H
0
\I
R - C-R
Functional Group
Suffix
Group A-Functional Groups Named By Prefix Or Suffix
Carboxylic Acid
Aldehyde
Ketone
carboxyoxo(formyl)
oxo-
hydroxy-
R-O - H
Alcohol
Amine
R-N
Functional Group
Structure
amino-
/
"
-oic acid
(-carboxylic acid)
-al
(carbaldehyde)
-one
-01
-amine
Group B-Functional Groups Named By Suffix Only
-ene
/
\
C=C
\
/
-C:::C-
Alkene
Alkyne
Substituent
Alkyl (see next page)
-yne
Structure
Group C--Substituent Groups Named By Prefix Only
alkylR -0 Alkoxy
alkoxy(alkoxy groups take the name of the alkyl group, like methyl or ethyl, drop the "yl", and add
"oxy"; CH 30 is methoxy; CH3CH 20 is ethoxy)
Halogen
F­
fluoro­
CI­
chloro­
Br bromo­
1iodoR-
Miscellaneous substituents and their prefixes
-N0 2
nitro
-
CH=CH2
vinyl
-CH 2CH=CH2
allyl
676
<>
phenyl
Appendix 1, Summary of IUPAC Nomenclature, continued
Common alkyl groups-replace "ane" ending of alkane name with "yl". A lternate names for
complex substituents are given in brackets.
-CH3
methyl
CH3
-CH
CH3
I
\
isopropyl
[ 1 -methylethyl]
-CH2CH3
ethyl
-CH2CH2CH3
propy l (n-propyl)
\
sec-butyl
[ I-methylpropyl]
CH3
-C-CH
CH3
I
isobutyl
[2-methylpropyl]
-CH2CH2CH2CH3
butyl ( -butyl)
n
B.
CH3
-CH
CH2CH3
I
I
3
t-butyl or
tert-butyl
[ 1,I-dimethylethyl]
Naming Substit uted Alkanes and Cycloalkanes-Group C Substituents Only
Organic compounds containing substituents from Group are named following this sequence of
steps, as indicated on the examples below :
C
·Step l. Find the longest continuous carbon chain . Determine the root name for this parent
chain. In cyclic compounds, the ring is usually considered the parent chain, unless it is attached to
a longer chain of carbons ; indicate a ring with the prefix "cyclo" before the root name. (When
there are two longest chains of equal length, use the chain with the greater number of substituents .)
·Step 2. Number the chain in the direction such that the position number of the first substituent
is the smaller number. If the first substituents have the same num ber, then number so that the
second substituent has the smaller number, etc.
·Step 3. Determine the name and position number of each substituent. (A substituent on a
nitrogen is designated with an "N" instead of a number ; see Section 111.0. 1 . below .)
·Step 4. Indicate the n umber of identical groups by the prefixes di, tri, tetra, etc.
·Step 5. Place the position numbers and names of the substituent groups, in alphabetical order,
before the root name. In alphabetizing, ignore prefixes like sec -, tert-, di, tri, etc., b ut include iso
and cyclo. Always include a position number for each substit uent, regardless of redundancies.
CH3
CH2CH2CH3
CH2CH2CH3
1
1
1
1
I CH3 -CH
-CH
4C
-sCH
CH2CH3
CH
H
H
T
T
CH3
-TH
H
TF
2
3
�
_
1
_ _ _ _3-,
1
1
CI
CH3
CH3
CH3
3-bromo-2-chloro-5-ethyl-4,4-dimethyloctane
3-fluor o-4-isopropyl-2-methylheptane
H3C - CHCH2CH3
l 2 I-sec-butyl-3-nitrocyclohexane
(numbering determined by t he
alphabetical order of s ubstituents,
3 N02 "b" comes before "n")
I
.l-__
2
3
__
__
-I
-"
__
Br
6
5
a
F
4
677
Appendix 1 , Summary of IUPAC N omenclature, continued
C. Naming Molecules Containing Functional Groups from Group B-Su ffix Only
1 . Alkenes-Follow the same steps as for alkanes , except :
a. Number the chain of carbons that includes the C= C so that the C=C has the lower position
number, since it has a higher priority than any substituents;
b. Change "ane" to "ene" and assign a position number to the first carbon of the C=C ; place the
position number just before the name of functional group(s);
c . Designate geometrical isomers with a cis,trans or E, Z prefix.
F
2
I
4 C H 3C - C = C H2
H
/
1
F
CH3
\
F
H
I
C ==C - C = CH2
2
\
I
-
4,4-difluoro-3-methylb ut - l -ene
0,
F
I
3
4
H
1 , I-di fluoro-2-methyl­
buta- l ,3 -diene
CH3
5-methylcyclopenta1 ,3-diene
2
3
Special case : When the chain cannot include an alkene , a substituent name is used. See
Section V.A.2.a.
2
6
5
4
C = CH2
3 H
3 -vmy Icyc I 0hex- 1-ene
.
Numbering must be on EITHER
a ring OR a chain, but not both.
2. Alkynes-Follow the same steps as for alkanes , except :
a . Number the chain of carbons that includes the C-C so that the alkyne has the lower position
number ;
b. Change "ane" to "yne" and assign a position number to the first carbon of the C=C ; place the
position number just before the name of functional group(s).
Note : The Group B f unctional groups (alkene and alkyne) are considered to have equal priority: in
a molecule with both an ene and an yne , whichever is closer to the end of the chain determines the
direction of numbering. In the case where each would have the same position n umber, the alkene
takes the lower number. In the name , "ene" comes before "yne" because of alphabetization .
F
2
I
H2
\
H
-C =CH2
CH - C - C == CH
HC = C - C = C - CH 3
HC
C
-C
3/
4;
H H
H
F
C H3
pent-3-en- l -yne
pent- l-en-4-yne
4,4-difluoro-3-methylbut- l-yne
( "yne" closer to end of
("ene " and "yne" have equal priority un less
chain)
they have the same position n umber, when
"ene" takes the lower number)
(Notes : 1 . An "e" is dropped if the letter following it is a vowel: "pent-3-en- l-yne" , not "pent-3ene- l-yne". 2 . An "a" is added if inclusion of di , tri, etc. , would put two consonants together:
"buta- l ,3 -diene" , not "but- l ,3 -diene".)
D. Naming Molecules Containing Functional Groups from Group A-Prefix or Suffix
In naming molecules containing one or more of the functional groups in Group A, the group o f
highest priority is indicated by suffix; the others are indicated by prefix, with priority equivalent to
any other substituents . The table in Section lIlA. defines the priorities ; they a re discussed on the
following pages in order of increasing priority.
678
Appendix 1 , Summary of IUPAC Nomenclature, continued
Now that the functional groups and substituents from Groups A , B , and C have been described, a
modified set of steps for naming organic compounds can be appli ed to all simple structures :
-Step 1 . Find the highest priority functional group. Detennine and name the longest
continuous carbon chain that includes this group.
-Step 2 . Number the c hain so that the highest priority functional grou p is assigned the lower
number . (The number " I " is often omitted when there is no confusion about where the group
must be. Aldehydes and carboxylic acids must be at the first carbon of a chain, so a " I " is rarely
used with those functional groups.)
-Step 3 . If the carbon chain includes multiple bonds (Group B ) , rep lace "ane" with "ene" for
an alkene or "yne" for an alkyne. Designate the position of the multiple bond with the number
of the first carbon of the multiple bond .
-Step 4. If the molecule includes Group A functional groups, replace the last "e" with the
suffix of the highest priority functional group , and include its position number just before the
name of the highest priority functional group .
-Step 5 . Indicate all Group C substituents, and Group A functional groups of lower priority,
with a prefix . Place the prefixes , with appropriate position numbers, in alphabetical order before
the root name.
1. Amines : prefix : a mino-; suffix : -amine-substituents on nitrogen denoted by "N'
NH2
CH P
CH3CH, , /CH 2CH J
propan- l -amine
yY
V
�
3-methoxycyclohexan- l -amine
(" 1 " is optional in this case)
2. Alcohols : prefix : hydroxy- ; suffix : -01
OH
I
H3C-C -C =CH 2
H H
but-3-en-2-01
ethanol
3. Ketones: prefix : oxo- ; suffix : -one (pronounced "own ")
o
CH3 -CH -C -CH3
I
OH
3-hydroxybutan-2 -one
II
o
6
H2C =C -CHCH3
H
N,N-diethylbut-3-en-2-amine
0H
c(
NH 2
2-aminocyclobutan- l-01
( " 1 " is optional in this case)
H3C , /CH3
N
o
II
I
CH3 -C - CH2 - C = CH 2
4-(N,N-di methylamino)pent A-co-2-one
cyclohex-3-en- l-one
(" 1" is optional in this case)
4. Aldehydes : prefix : oxo-, or formyl- (O=CH ); suffix : -al (abbreviation : --CHO)
An aldehyde can only be on carbon I , so the " 1"- is generally omitted from the name.
o
OH
0
o
0
o
"
I
"
"
II
II
C H3 -CH
H 2C- � = �-CH
CH3CCH2CH2CH
HCH
ethanal;
4-oxopentanal
methanal;
4-hydroxybut -2-enal
acetaldehyde
formaldehyde
679
Appendix 1, Summary of IUPA C Nomenclature , continued
Special case : When the chain cannot include the carbon of the aldehyde, the suffix
2
3
"carbaldehyde " is used:
a
4
0-1 1
5
I
6
cyc lohexanecarba ldehyde
CH
5 . Carbox ylic Acids : prefix : carboxy-; suffix : -oic acid (abbreviation : -COOH )
A carboxy lic acid can only be on carbon 1 , so the "1" is genera lly omitted from the name.
(Note : Chemists traditionally use , and IUPAC accepts , the names "formic acid" and "acetic
acid" in p lace of "methanoic acid" and "ethanoic acid".)
a
a
II
methanoic acid;
formic acid
HC - OH
O
�
II
ethanoic acid;
acetic acid
CH3C - OH
_
a
a
g
II
a
CH
II
I
3
HC - C - C - COOH
I
CH3
CH 2 - CH - OH
I
NH2
2-amino-3-phenylpropanoic acid
2 ,2-dimethyl-3 ,4dioxobutanoic acid
Specia l case : When the chain numbering cannot inclu de the carbon of the carboxylic acid, the
suffix "carboxy lic acid" is used:
CHO
2-formyl-4-oxocyclohexanecarboxylic acid
("formyl" is used to indicate an aldehyde as
O
COOH a substituent when its carbon cannot h e in
the chain numbering)
K,
�
5
6
E . Naming Carboxylic Acid Derivatives
The six common groups derived from carboxylic acids are , in decreasing priority after carbox ylic
acids : s alts , anhydrides, esters , acyl ha lides , amides , and nitriles .
1. Sa lts of Carbox ylic Acids
Sa lts are named with cation first , followed by the anion name of the carboxylic acid, where "ic
acid" is rep laced by "ate" :
acet ic acid becomes acetate
butanoic acid becomes butanoate
cyc lohexanecarboxylic acid becomes cyc lohexanecarboxy late
NH2
Li +
lithium 2-aminopropanoate
Na+
sodium chloroacetate
I
CH3 - CHCOO-
CICH2COO-
2. Anhydrides : "oic acid" is replaced b y "oic anhydride "
a
II
alkanoic acid
R - C - OH
a
>
a
II
II
a lkanoic anhydride
R-C-O-C-R
680
CH30
U
COO­
ammonium 2-methox y­
cyclobutanecarboxylate
��
a
a
O
benzoic anhydride
Appendix 1 , Summary of IUPAC Nomenclature, continued
3 . Esters
Esters are named as "organic salts" that is, the alkyl name comes first, followed by the
name of the carboxylate anion . (common abbreviation : -COOR)
CH3 a
carbJ,xylate
(
�
I
II
a
\
a
II
R-C-O�R
"alkanoate" . "alky l"
"alkyl alkanoate"
II
a
II
H3C - C - O + CH2CH3
ethyl acetate
HO
.
a
II
V
a
II
alkanoic acid
R - C - OH
5.
>
Amides : "oic acid" is replaced by "a mide"
a
II
alkanoic acid
R - C - OH
< }a
�
butanamide
II
a lkanamide
R - C - NH2
Nitriles : "oic acid" is replaced by "enitrile"
a
II
alkanoic acid
R - C - OH
I
NH2
�C = N
a
benzoyl
chloride
�
(i.:
I
CI
0
benzamide
�
NH 2
V:� :
butanenitrile
alkanenitrile
+-0
�
l
C
Amides are notable for their role in biochemistry, i .e . , the special amide bond
between two amino acids is called a peptide bond.
6.
CH2COO
cyclohexyl 2-pheny lacetatc
a
a
>
CH3
butanoyl
chloride
alkanoyl chloride
:
I
�
II
R - C - Cl
I
isopropyl 2,2-dimethylpropanoa te
C - �H 3
a
CH3
H3 C - C - C - O � CHCH 3
methyl 3-hydroxycyclo­
pentanecarbox ylate
4. Acyl Halides : "oic acid" is replaced by "oyl halide"
vinyl prop-2-enoate
H2C = C - C - 0 7" C = CH2
. H
H
:
z nil'iIC
(common spelling di ffers
from IUPAC)
"Aromatic" compounds are those derived from benzene and similar ring systems . As with
aliphatic nomenclature described above, the process is : determining the root name of the parent
ring ; determining priority, name, and position number of substituents ; a nd assembling the name in
alphabetical order . Functional group priorities are the same in aliphatic and aromatic
nomenclature. See p . 676 for the list of priorities .
A . Common Parent Ring Systems
(a)
IV. Nomenclature of A romatic Compounds
7
benzene
6
0)
8
I
'-'::
�
1
'-'::
�
naphthalene
5
4
2 (t})
3
681
�
8
9
1
5
10
4
6� 3
7
a nthracene
2
Appendix 1 , Summary of IUP AC Nomenclature , continued
B. Monosubstituted Benzenes
1 . Most substituents keep their designation , followed by the word "benzene ":
CH3
8
6
6
H CHO OH
3
6 6 6 6 6 6 6
C.
�Y �
¢
COOH
H0'Q
{ }OCH3
Ql
6 NH 2
CHO
CH]
CH3
�OCH2CH3
r(YCI 02 A 02
H
O
Y
Y
CH3
N0 2
NH2
nitrobenzene
ethylbenzene
chlorobenzene
2 . Some common substituents change the root name of the ring . IUPAC accepts these as
root names , listed here in decreasing p riority (same as Group A, p . 676):
COOH
benzene- benzaldehyde phenol
benzoic
sulfonic acid
acid
Disubstituted Benzenes
1. Designation of substitution--only three possibilities :
X
U
ortho1 ,2-
common:
IUPAC:
aniline
X
meta1,3 -
toluene
X
�y
(0-)
anisole
(m-)
para- (p-)
1 ,42 . Naming disubstituted benzenes-Priorities from Group A, p . 676 , determine root name and
substituents
Br
Br
p-dibromobenzene
1 ,4-dibromobenzene
y
I �
�
�
o-methoxybenzaldehyde
2-methoxybenzaldehyde
m-aminobenzoic acid
3 -aminobenzoic acid
D. Polysubstituted Benzenes-must use numbers to indicate substituent position
HN
I
VCl
3,4-dichloro-N-methylaniline
N
m-methylphenol
3 -methylphenol
N
2 ,4,6 -trinitrotoluene
(TNT)
682
ethyI 4-amino-3-hydroxybenzoate
Appendix 1, Summary of IUPAC Nomenclature, continued
E. Aromatic Ketones
A special group of aromatic compounds are ketones where the carbonyl is attached to at least one
benzene ring. Such compounds are named as "phenones ", the prefix depending on the size and
nature of the group on the other side of the carbonyl. These are the common examples :
o
o
C
< }-C-CH3
< }- CH2CH3
acetophenone
propiophenone
-
benzophenone
V. Nomenclature of Bicyclic Compounds
"Bicyclic" compounds are those that contain two rings. There are four possible arrangements of t wo
rings that depend on how many atoms are shared by the two rings . The first arrangement in which
the rings do not share any atoms does not use any special nomenclature, but the other t ypes require
method to designate how the rings are put together. Once the ring system is named, then functional
groups and substituents follow the standard rules described above.
Type 1 . Two rings with no common atoms
These follow the standard r ules of choosing one parent ring system and describing the other ring as
a substituent . 5 4
ketone is the highest priority functional group, phenyl is substituent
6
c::==�> 3-pheny Icyclohexan- l-one (" 1" could be omitted here)
o
benzene is the parent ring system as it is larger than
cyclopentane and it has three substituents
c::=�> l-cyclopentyl-2,3-dinitrobenzene
a
y,D
Type 2. T wo rings with one common atom-spiro ring system
The ring s ystem in s piro compounds is indicated by the word "spiro" (instead of "cyelo"), followed
by brackets indicating how many atoms are contained in each path around the rings , ending with
the alkane name describing how many carbons are in the ring systems including the spiro carbon.
(If any atoms are not carbons, see section VI.) Numbering follows the s maller path first, passmg
through the sprio carbon a nd around the second ring.
S
1
5
3
:00,
4
spiro[3.4 ]octane
spiro[4.5]decane
683
6
7
Appendix 1, Summary of IUPAC Nomenclature, continued
Substituents and functional groups are indicated in the usual ways . Spiro ring systems are a lways
numbered smaller before larger, and numbered in such a way as to give the highest priority
functional group the lower position number.
I
10 9
o
8
7 r\A 2
6
5 3
l0V
'---+- CH3
CH 3
7 ,7 -dimethylspiro [4.5]decan-2-one
spiro [3.4 ]oct -5-ene
Type 3 . Two rings with t wo common atom-fused ring system
Two rings that share two common atoms are called fused rings. This ring system and the next type
called bridged rings share the same designation of ring system. Each of the two common atoms i s
called a bridgehead atom, and there are three paths between the two bridgehead atoms . In contrast
with naming the spiro rings, the longer path is counted first, then the shorter, then the shortest. in
fused rings, the shortest path is al ways a zero, meaning zero atoms between the two bridgehead
atoms . Numbering starts at a bridgehead, continues around the largest ring, through the other
bridgehead and around the shorter ring . (In these structures, bridgeheads are marked with dark
circle for clarity .)
2
2
LO
4
bicyclo [2 . i .O]pentane
(path of 2 atoms and a
path of I atom)
CO
1 3
8 6 4
9
4
bicyclo [4.4.0]decane
7
(path of
5
atoms in
a
9
5
4
6
bicyclo [5.3.0]decane
(path of
atoms and
Substituents and functional groups are indicated in the usual ways . Fused rings systems are always
numbered larger before smaller, and numbered in such a way as to give the highest priority
functional group the lower position number.
Type 4. Two rings with more than two common atom-bridged ring system
Two rings that share more than t wo common atoms are called bridged rings . Bridged rings "hare
the same designation of ring system as Type 3 in which there are three paths between the t w'.)
bridgehead atoms. The longer path is counted first, then the medium, then the shortest . Numbering
starts at a bridgehead, continues around the largest ring, through the other bridgehead and around
the medium path, ending with the shortest path numbered from the original bridgehead atom . (In
these structures, bridgeheads are marked with a dark circle for clarity .)
2
64
�
31 5
bicyclo [2. 1 . 1 ]hexane
(paths of 2 atoms, I
atom, and 1 ato m )
each d irection)
6
�84
51 3
bicyc 10[2 . 2. 2]octane
2
(three paths of 2 atoms)
684
path of 3 atoms)
8
7
3
3
bicyclo [3.2. 1 ]octane
(paths of
atoms, 2 atoms,
and 1 atom)
Appendix 1 , Summary of IUPAC Nomenclature, continued
6
� B,
�Br
3
2
5
5,5-dibromo­
bicyclo [2 . 1 . 1 ]hexane
3
3
6
bicyclo [2.2.2]oct-5-en -2-one
o
8,8-dimethyl­
bicyclo [3 .2. 1 ]octan - l-ol
The teon "heteroatom" a pplies to any atom other than carbon or hydrogen . It is common for
heteroatoms to appear in location s that are inconvenient to name following ba sic rules, so a simple
system called "replacement nomenclature" has been devised . T he fundamental principle is to na me a
compound as if it contained only carbons in the skeleton , plus any functional groups or substituents,
and then indicate which carbon s are "replaced" by heteroatoms. The prefixes u sed to indicate these
substitions are li sted here in decreasing priority and listed in this order in the name:
OH
Prefix
Example
Element
o
oxa
nonan- l -01
thia
S
aza
P
phospha
H2
H
H
sila
Si
B '-.,./ Si � S
/"
B
bora
'-.,./OH
2-thia-8-aza-4-sila-6 -boranonan- l -ol
VI. Replacement Nomenclature of Heteroatoms
N
N ......."..
In the above example , note that the (imaginary) compound no longer has nine carbons, even though
the name still includes "nonan" . The heteroatom s have replaced carbon s, but the compound is
named as if it still had those carbon s.
Where the replacement system i s particularly useful i s in polycyclic compounds. Shown below are
three examples of commercially available and synthetically useful reagents that use this system .
parent hydrocarbon
reagent
abbreviation
6
8
7�4
�
S
3
6
bicyclo [2.2.2 ]octane
4
�
8
6
I
1 ,4-DiAzaBiCyclo [2 .2.2]Octane
2
��
i(5 3
DABCO
2
( upper case added to explain abbrev iation)
O
b icyclo[5.4.0]undec-7 -ene
4
:)
r� �
�N8
6
O
1 ,8-DiazaBicyclo [5.4.0]Undec-7-ene
( upper case added to explain abbreviation)
685
DBU
Appendix
I,
Summary of IUPAC Nomenclature , continued
3�5
9
9
3
B ....
H
9-BBN
6
2
I
8
7
9-BoraBicyclo [3 . 1. 1]Nonane
bicyclo [3 . I. l ]nonane
7
( upper case added to explain abbreviation)
How can we distinguish
this structure from its
mirror image ?
Is this alkene cis or trans ?
I
F
VII. Designation of Stereochemistry; Cahn-Ingold-Prelog system
F
,I
"1"
I \\'
CI
Br
Br
Cl
Compounds that exhibit stereoisomerism , whether geometric isomers around double bonds ,
substituent groups on rings , or molecules with asymmetric tetrahedral atoms (which are almost
always carbons) , require a system to designate re lative and absolute orientation of the groups. The
terms cis/trans , DIL in carbohydrates and amino acids , and dlffor optically active compounds, are
li mited and cannot be used generally , although each still is used in appropriate situations . For
example , cis/trans st ill is used to indicate relative positions of substituents around a ring .
A system developed by chemists Cahn , Ingold , and Prelog , uses a series of steps to determine
group priorities , and a definition of position based on the relative arrangement of the groups. In
alkenes , the system is relatively simple :
high
high
low
high
>=<
>===<
>===<
low
low
low
high
this arrangement is defined
this arrangement is defined as
as Z =zusammen , together ,
E = entgegen , opposite , from
from both high priority groups
the two high priority groups
on the same side o f the C=C
on opposite sides of the C=C
As with alkenes , the orientation around an asymmetic carbon can be only one of two choices . In
three dimensions , clockwise and counterclockwise are the only two directions that are definite ,
and even that description requires a f ixed reference point. To designate configuration , the lowest
(fourth) priority group is always placed farthest away from the v iewer ( indicated by a dashed
line) , and the group priorities will follow 1 to 2 to 3 in either a clockwise or a counterclockwise
direction.
to 2 to 3 is
as the
configuration
1
clockwise, defi ned
R = rectus
1
3
)<
1
2
2
686
)<
1 to 2 to 3 is
countercIock
defined as the S
sinister configuration
WIse,
3
=
Appendix 1 , Summar y of IUPAC Nomenc lature, continued
The only step remaining is t o determine the priority of groups , for which t here is a carefully
defined set of rules .
Rule 1. Consider the first atom of the group , the point of attachment. Atoms with higher atomic
number receive higher priority. Heavier isotopes have higher priority t han lighter isotopes.
I
Br
>
>
CI
>
F
Rule 2. If the first atoms of two or more groups are the same , go out to the next atoms to break
the tie . One high priority atom takes priority over any number of lower-priority atoms .
H
CH 3
H
H
I
I
I
I
-C - OH
> -C -H
> -C -H
> -C - H
I
I
I
I
H
H
CH 3
CH 3
Rule 3. Treat multiple bonds as if they were all single bonds ; one wil l be to the real atom, the
others will be to imaginary atoms.
This is the hardest rule to put into practice . This example of an alkyne shows stepwise how to
accomplish t his. In the pictures , imaginary atoms (ones that did not start out in the structure,
had to invent them) are ind icated by italics.
(
I
C
-- C C - H
all atoms are real
>
- C -C -H
I
halfway there-had to
add two fake carbons
O-C
r:::=
==�>
C
....
.
C
.....
R ....... H
H
R .......
added a fake 0 to
the real C , and a
fake C to the real 0
II
R -C :: N
0, /
Examples applying the Cahn -Ingold-Prelog sytem
i
compari n g
F and C l
low
F
high
CI
'-
I
J
t==\
__
E
I
Br
high
low
}
comparing
I and B r
687
I
I
C
C
-- C - C - H
>
I
I
final-had to add four
fake carbons
C
becomes
o
started with 3 bonds to carbon,
so have to rep l ace them
C
C
becomes
we
>
N
C
R -C - N
I
I
I
I
added two fake Ns to the
C , and two Cs to the N
N
C
Appendix 1 , Summary of IUPAC Nomenclature , continued
4 F
,I
"1"
'
I"
Br
2
Often, the hard part of applying the RlS system to asymmetric carbons
is orienting the molecule so that the fourth priority group is farthest
away. Sometimes it is easier to put the fourth priority group toward
you, then take the opposite of what the direction of 1 to 2 to 3 says.
CI
1
I
3
�
'\) ,
flip
Br up so
F goes bac k
I"F
�CI
I
2
Br
4
1
R
3
\1"
R
=
,\\\
Br"
�',/1
"
CI
3
up
�
Putting the F (4th priority) pointing up toward you,
the groups 1 to 2 to 3 appear counterclockwise, but
we have to take the opposite, so instead uf S, it is R.
2
688
F
�
Appendix 2 : Summary of Acidity and Basicity
Imagine that you are at a family reunion w here you can observe t he competition for ice cream cones a mong
your nieces and nephews . Pretty soon , you formulate a generalization : t he older kids can hold onto their ice
cream cones more strongly t han the younger ones. Another way of saying it is t hat t he older ones are less
equilibria between the child with ice cream , and the free ice cream plus the hungry child . The differences in
strength could also be quantitated : the larger the hunger factor , pKH ' the less likely the child will give up the
ice cream.
likely to give up their cones. You could even represent this information i n a table showing a series of
p KH
Approximate pKH Values of Children
---
12
1 2 -year-old with ice cream
---
10
1 O-year-old w ith ice cream
---
8
8 -year-old w ith ice cream
---
6
6-year-old w ith ice cream
---
4
4-year-old w ith ice cream
---
---
---
---
---
Ice cream
+
hungry 1 2-year-old
Ice cream
+
hungry l O- year-old
ice cream
+
hungry 8 -year-old
Ice cream
+
hungry 6-year-old
Ice cream
+
hungry 4-year-old
What good is this table ? It allows anyone to make predictions about what will happen when kids with ice
cream are mixed with hungry kids .
hungry 1 O-year-old
+
4-year-old with ice cream
}?1
---
---
1 O-year-Old with ice cream
+
hungry 4-year-old
If the hungry lO-year -old was left unattended in a room with the 4-year-old with ice cream, which side of this
equilibrium would be favored w hen you came back in a few minutes: "reactants" or "products "? More likely
than not , there would be an ice cream transfer in your absence ; "products" would be fa vored . You could
ha ve predicted this from t he table, and you could generalize : the hungry lO- year-old will be strong enough
to rip the ice cream awa y from an y kid lower on the table than the lO-year-old hi mlherself .
Let's predict the results of another equilibrium :
hungry 6-year-old
+
}?i
l 2-year-old with ice cream
---
---
6-year-old with ice cream
+
hungry 1 2-year-old
Is the hungry 6-year-old stron g enough to pull the ice cream away from the 1 2 -year-old ? Not in most
families . The table shows that the only chance for the hungry 6-year-old is to find a 4-year-old with ice
cream . The 12-year -old with ice cream is pretty safe as long as the hunger table doesn't go to hi gher ages.
If you understand this analogy and can make predictions of ice cream transfer using the table, then you can
understand how to predict the direction of equilibrium in acid-base reactions. Turn the page .
689
Appendix 2 continued, Summary of Acidity and B asicity
A few generalizations:
To set the stage ....
A) This Appendix deals with pro tic acids and bases, called Bronsted-Lowry acids. Similar statements can
be made about Lewis aci ds but they are not the focus of this discussion.
B ) Values of pKa are measures of equilibrium constants, described further below. Values between 0 and
15 .7 are measured by titration in water solution and are known accuratel y , to within 0. 1 pK unit and
sometimes better. Values outside of thi s range cannot be measured in water because of the leveling effect of
water, and there is no universal ly accepted method for measuring these pKa values . This lack of a single
standard of measurement means that the values below 0 and above 15 .7 should be considered relative, not
absolute . If your instructor says the pKa of methane is 46 and this book says it is 50, those should be
considered the same value within experimental variation .
C) Acidi ty is a thermodynamic property, and the acid equi libri um constant, K a, is a measure of the relative
concentrations of species in the protonated and unprotonated form. As most organic acids are weak acids,
meaning they are present mostly in the protonated form at equi librium, the Ka < l. Since the pKa = -log Ka,
the pKa values will be greater than 0 , with the larger pKa representi ng a weaker acid. If this is not clear,
review text section 1 - 13 .
D) Acids do not spontaneously spit out a proton ! Despite our way of writing ionization equilibri a as shown
on the next page, aci ds do not give up a proton unless a base comes by to take the proton away . The
reactions as drawn in the table should be consi dered half-reactions, just as the reactions in the electromotive
series were half-reactions for balancing oxidation-reduction reactions in general chemistry.
Look at the table on p. 6 9 1 and notice how it looks j ust like the "chi ldren-with-ice-cream" table. We c an use
this table to make predictions about equi libri um position in acid-base reactions j ust as we did for the
children with ice cream.
1) A base will deprotonate any acid stronger than its conjugate acid. This is the most important principle
of predicting acid-base reactions. On the table, this means that any base, hydroxide for example, can react
with any acid more acidic than the conj ugate acid of itself, water in our example. So hydroxide is a strong
enough base to pull the proton from any of these : bicarbonate ion, a phenol, c arbonic acid, a carboxylic
acid, or a sulfonic acid. We can also predict that hydroxide is NOT a strong enough base to react with any
acid above water on the table; for example , a mi xture of hydroxide with an alkyne will favor the reactants at
equilibrium, with onl y a small amount of products.
I. Predicting equilibrium position
reactants favored
HO
-
+
RC _ C - H
pKa 25
weaker
base
.. ___
weaker
acid
H20
+
RC - C :
pKa 15.7
stronger
acid
stronger
base
2) Another way of predicting the position of an equilibrium is to assign "stronger" and " weaker" to the acid
and base on each side of the equation, using the table to determine which is stronger and which i s weaker.
Equilibrium will always favor the weaker acid and base. This method will always give the same answer as
the principle in # 1 above.
To lead into the next section, look again at the table on p. 691 and notice two things: a) with only a couple
of exceptions, all the aci dic protons are on either oxygen or c arbon; and b) generalizations can be made
about the acidity of functional groups. Learning to correlate acidity with functional group is i mportant in
predicting reactivity of the functional group.
690
Appendix 2 continued, Summary of Acidity and Basicity
A��roximate �Ka Values of Oq�anic Com�o un ds
p Ka
I
I
"" 50
alkane
- H
H+ +
weaker acid
RC
I
a lkene
"" 45
a mine
3 5 -40
alkyne
ketone and ester
"" 3 5
"" 25
==C
-
/
H
N -H
..
I
R
R -C -O - H
I
l
t
alcohol
canno! be
meas ured in
water solution
I
1 5 .7
1 0.3
HC03-
I
"" 1 7
I
I
"" 1 6
measured in
water solution
+
-N :
H+
R
H
R -C -O -H
R
H
R -C - O - H
H
H2O
H+
== � .
H+
I
"" 1 8
+
H+
RC ::C - H
0 I
-C -C - H
..
..
I
I
H+
H+
H -H
"
20-25
R-C:
-
+
H:
stronge r base
/
I
+
RC ::C :
0 I
-C - C :
+
R -C - O :
+
H+
+
H+
+
H+
+
H+
+
"
R
I
I
R
I
H
R -C - O :
R
H
R -C - O :
H
HOI
I
I
I
- - - � - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
phenol
",, 1 0
6.4
carboxylic acid
sulfonic acid
stronger acid
Ar - O -H
H2C03
0
R -C - O -H
I I
4-5
<0
RS02 - O - H
691
..
..
H+
H+
H+
H+
+
+
+
+
col-
Ar - O :
HC030
R -C - O :
I I
RSO 2 - 0 :
• •
weaker base
Appendix
2 continued, Summary of Acidity and Basicity
II. Correlation of Acidity with Functional Group
A. Ox�gen Acids
0
R
"
-
S
I I
0
-
O-
H
pKa < O
sulfonic acid
>
0
I I
R
-
C
-
O-
pKa 4- 5
H
>
( }O-H
pKa
c arboxylic acid
10.0
>
HO-H
pKa
phenol
1 5 .7
water
>
R
I
R
-
C
I
R
-
O-
pKa
H
16 - 1 9
alcohol
Sulfonic acids are the strongest of the oxygen acids but are not common in organic chemistry.
Carboxylic acids, however, are everywhere and are considered the strongest of the common organic
oxygen acids. (Note that "strong" and " weak" are relative terms: acetic acid, pKa 4.7 4, was a " weak"
acid in general c hemistry in comparison to sulfuric and hydrochloric acids , but acetic acid is a "strong"
acid in organic chemistry relati ve to the other oxygen acids. ) Phenols having OR groups on benzene or
other aromatic rings are sti l l stronger acids than water.
Why are phenols, carboxylic acids, and sulfonic acids stronger acids than water? B ecause their anions
are stabi lized by resonance. (Refer to text sections 1 - 13 , 10-6 , and 20-4, especially Figure 20- 1 .) Let's
look at that statement in more detai l .
Remember that acidity i s a thermodyn amic property ; that i s , acidity equi librium depends o n the
difference in energy between the reactants and products. The more the anion is stabilized by resonance,
the lower in energy it is, and the less positi ve the 1\G, as shown on the reaction energy diagram:
...
,
,
,
,
- alcohol
water
Energy
phenol
,
,
,
-
- - - . carboxylic acid
oxygen anions
stabilized by resonance
sulfonic acid
reaction -.
Why are alcohols weaker acids than water? There are two effects that contribute, both of which are
consi stent with the trend that 1 ° alcohols (pKa 16) are slightly stronger than 2° alcohols (pKa 1 7 ) which
are slightly stronger than 3° alcohols (pKa 1 8). Alkyl groups are mildly electron-donating in their
inductive effect (more about thi s later) and destabilize the anion , as shown in the energy diagram above.
S econd, the more crowded the anion is, the less it can be stabilized by hydrogen bondin g with the solvent.
692
Appendix 2 continued, Summary of Acidity and Basicity
B. Carbon Acids
When we think of "acids", we do not usually think of protons on carbon, yet carbon acids and the
carbanions that come from them are of tremendous importance in organic chemistry.
"Unstabilized" carbon acids are those that do not have any substituent to stabilize the anion. Alkanes,
with only sp 3 carbons are the weakest acids with pKa around 50. The vinyl carbon in a carbon-carbon
2
double bond is sp hybridized with the electrons of the anion slightly closer to the positive nucleus,
leading to some s tabilization of the anion. This type of stabilization is particularly important in alkynes
with sp hybridized carbons.
I
I
R -C:
R -C-H
=C
I
I
3
s p hybridized
2
pKa 50
/
,
=
H
c:
/
RC::C-H
.
sp hybridized
sp hybridized
pKa 45
pKa 25
RC::C:
..
C. Carbon Acids Alpha to Carbonyl (This topic is described in detail in text section 22-2B.)
Look at these huge differences in pKa when the acidic group is next to a carbonyl.
o
O- H
H2 O
e=:>
RC
II
-
pKa 16- 18
H
R -C-O-H
0
N-H
H H
e=:>
RC 2 N
-
H
H
II
pKa 35-40
R-C- N-H
0
C- H
H2
RCH2 C
e=:>
H
pKa 50
pKa 16
H2
II
-
pKa 4-5
R-C-C -H
pKa 20
Hydrogens alpha to carbonyl are unusually acidic because of resonance stabilization of the anionic
conjugate base.
D. Acidities of Acyl Functional Groups
In addition to the significant variation in the acidity of alpha hydrogens depending on which atom the H
is bonded to, what is on the other side of the carbonyl also has a dramatic influence. In this case,the
stabilization is more important on the starting material,not on the conjugate bas e. See the energy
diagram on p. 694.
H
I
O
II
H2C -C-H
pKa
17
no stabilization of
starting material
H
I
amide
es ter
ketone
aldehyde
O
H
II
I
H2C-C-CH3
H
O
II
H2C- C OCH3
-
pKa 20
••
pKa 25
mild stabilization of
starting material by
weak electron donation
from
CH3
H
I
H2C
0
I
+
H2C-C=�CH3
••
II
-
C- N(CH3}z
t
pKa 30
significant resonance
stabilization of
starting material
693
O
I
H 0
I
+
I
H2C-C N(CH3}z
=
s trongest resonance
stabilization of
starting material
Appendix 2 continued, Summary of Acidity and Basicity
,,
,
,
.-
(
least
tabilization
Energy
aldehyde
I
,
,
,,
,
I
,
'
...
,
, " , ...... \
.
" , "
'" \
'
'�
,
,I'
' ,,/
,
"/
,
ketone
ester
amide
greatest stabilization of starting material, largest LlG
__
reaction
--
E. Carbon Acids Between Two Carbonyls (This topic is described in detail in text section 22-15.)
While a hydrogen alpha to one carbonyl moves into the pKa 20-25 range for ketones and esters
respectively, a hydrogen between two carbonyls (cyano and nitro are similar to carbonyl electronically)
is more acidic than water. The increased resonance stabilization of the conjugate base is largely
responsible, but there are subtle variations depending on the type of functional group as noted at the
bottom of p.
693.
0
0
EtO
Look at the enormous influence of the nitro group.
¥
OEt
N:--...
�
C
13.5
0
II
H2C'
I
+
N
pKa
pKa
10.2
EtO
-
H3CO
o
CH3
H
pKa
+
�"
N
10.2
0
-
'0
H3C
0
V
CH3
H
pKa 9,0
+" , "+
0
0
-O,N
H
pKa
0
¥
1 1.2
0
'0
H
C
H
H
pKa
y
0
/N
�
N, O
H
5.8
pKa 3.6
III. Correlation of Basicity with Functional Group
The bulk of this Appendix is on acidity because many more functional groups are acidic than are basic.
Basically (oooh, sorry), only one functional group is basic: amines. There is variation among aliphatic,
aromatic, and heteroaromatic amines; these are covered thoroughly in text sections 19-5 and 19-6. One
point in the text, just before Table 19-3, deserves emphasis: for any conjugate acid-base pair:
pKa + pKb
=
694
14
Appendix 2 continued, Summary of Acidity and Basicity
This simple algebraic relationship is very useful:
Sample problem. Is triethylamine (pKb 3.24) a strong enough base to deprotonate phenol (pKa
1O.0)?
We need to calculate either the pKa of the conjugate acid of triethylamine or the pKb of the conjugate
base of phenol to see which is stronger and weaker. Then we can say with certainty which side of the
equilibrium will be favored.
(}
pKa
10.0
OH
+
( }o-
Et-N-Et
1
Et
pKb 3. 24
pKb
=
4.0
(14 - 10.0)
stronger
acid
stronger
base
weaker
base
H
+1
+
Et - N -Et
1
Et
pKa = 10.76
(14 - 3. 24)
weaker
acid
Aha! Products are favored at equilibrium, so the correct answer to the question is "Yes,triethylamine is a
strong enough base to deprotonate phenol."
Try this for fun: How weak must a base be before it does NOT deprotonate phenol? What algebraic rule
can you formulate to predict whether any combination of acid and base will favor products or reactants?
IV. Substituent Effects on Acidity
So far, we have focused on acidities of different functional groups. Let's tum to more minor,more subtle,
structural changes to see what effect substituents will have on the acidity of a group. Primarily, we imply
electronic effects as opposed to steric effects, but this Appendix will conclude with a discussion of how
steric and electronic effects can work together.
A. Classification of Substituents-Induction and Resonance
Substituent groups can exert an electronic effect on an acidic functional group in two different ways:
through sigma bonds, where this is called an inductive effect, or through p orbitals and pi bonds which is
called a resonance effect. Groups can also be electron-donating or electron-withdrawing by either of the
mechanisms, so there are four possible categories for groups. Note that a group can appear in more than
one category, even in conflicting groups!
a) Electron-donating by induction: only alkyl groups (abbreviated R) have electrons to share by induction;
b) Electron-withdrawing by induction: every group that has a more electronegative atom than carbon is in
this category; some examples: F, Cl,Br,I, OH,OR,NH2,NHR, NR2, N02, C=O,CN,S03H, CX3 where
X is halogen;
c) Electron-donating by resonance: groups that have electron pairs to share: F, Cl,Br,I, OH,OR,NH2,
NHR,NR2;
d) Electron-withdrawing by resonance: N02,C=O,CN,S03H.
695
Appendix 2 continued, Summary of Acidity and Basicity
B. Generalizations on Electronic Effects on Acidity (refer to text section 20-4B)
Electric charge is the key to understanding substituent effects. An acid is always more positive than its
conjugate base; in other words, the conjugate base is always more negative than the acid. Electron­
donating and electron-withdrawing groups will have opposite effects on the acid-base conjugate pair.
Electron-donating groups stabilize the more positive acid form and destabilize the more negative
conjugate base. From the diagram, it is apparent that electron-donating groups widen the energy gap
between reactants and products, making L\G more positive, favoring reactants more than products. In
essence, this weakens the acid strength.
conjugate
base
- "", �
...
- electron-donating substituent
no substituent effect
Energy
acid
electron-withdrawing substituent
reaction
-
Electron-withdrawing groups destabilize the more positive acid form and stabilize the more negative
conjugate base, narrowing the energy gap between reactants and products, making L\G less positive.
Products are increased in concentration at equilibrium which we define as a stronger acid.
Electron-withdrawing groups increase acid strength; electron-donating groups decrease acid strength.
C. Inductive Effects on Acidity
Text section 20-4B gives a thorough explanation of the inductive effect of electron-withdrawing groups
on simple carboxylic acids, from which three generalizations arise:
A) Acidity increases with stronger electron-withdrawing groups. (See problem 20-33.)
B) Acidity increases with greater number of electron-withdrawing groups.
C) Acidity increases with closer proximity of the electron-withdrawing group to the acidic group.
We don't usually look to aromatic systems for examples of inductive effects, because the pi system of
electrons is ripe for resonance effects. However, in analyzing the resonance forms of phenoxide on the
next page, it becomes apparent that the negative charge is never distributed on the meta carbons. Meta
substituents cannot exert any resonance stabilization or destabilization; at the meta position, substituents
can exert only an inductive effect. The series of phenols demonstrates this phenomenon, consistent with
aliphatic carboxylic acids.
696
Appendix 2 continued, Summary of Acidity and Basicity
: 0:
:0:
:0:
6 �-6
'
OH
O
CH3
----
OH
0
�
H
pKa 10.00
pKa 10.09
donating
(]
C
meta
OH
0
----
:0:
the delocalized
negative charge does
not increase electron
density at meta position
6-
OH
OH
OCH3
pKa 9.65
withdrawing
0
0
CI
N02
pKa 9.02
pKa 8.39
withdrawing
withdrawing
D. Resonance Effects on Acidity-benzoic acids and phenols (review the solution to problem 20-45)
Resonance effects can be expressed with placement of substituents at ortho or para positions, but ortho has
the complication of steric effects, so just para substitution is shown here.
�
Electron-withdrawing substituents increase the acidity of benzoic acids and phenols:
I
/
-...;:::
�
H
OH
OH
hO
I
COOH
COOH
COOH
-...;:::
�
'
para
pKa 10.00
OH
¢ ¢ ¢ ¢ ¢
N02
CN
pKa 8.05
pKa 7.95
pKa 7.15
H
pKa 4.20
CN
pKa 3.55
N02
pKa 3.42
Electron-donating by resonance but electron-withdrawing by induction:
There is a group of substituents that donate by resonance but withdraw by induction: alkoxy groups and
halogens are the most notable examples, and the acidity data provide an insight into which effect is stronger.
COOH
¢
COOH
0
�
OCH3
H
pKa4.20
COOH
¢
OCH3
pKa4. 09
pKa4. 47
meta-Methoxybenzoic acid is stronger than benzoic acid,
consistent with electron-withdrawing by induction which
is expressed at the meta position. But the para isomer is
weaker than benzoic acid; electron donation by resonance
has not only compensated for the inductive effect (which
is still operative at the para position) but has decreased the
acidity even further. Thus, the donating effect by
resonance must be stronger than the withdrawing effect
by induction for the methoxy group.
See another example on the next page.
697
Appendix 2 continued, Summary of Acidity and Basicity
OH
I
"'"
h
61
¢
"'"
I
¢
OH
OH
h
"'"
1
h
F
F
H
pKa 10.00
pKa 9.81
pKa 9.28
meta-Fluorophenol is stronger than phenol, consistent
with electron-withdrawing by induction which is
expressed at the meta position. The para isomer is still
stronger than phenol; electron donation by resonance has
not compensated for the inductive effect (which is still
operative at the para position). Thus, the donating effect
by resonance must be weaker than the withdrawing
effect by induction for the fluoro group.
Studying substituent effects on acidity is the standard method of determining whether a group is
donating or withdrawing by induction and resonance.
E. Proximity Effects of Substituents
pKa 10.0
pKa 8.05
pKa 9.19
pKa 9. 90
OH
OH
OH
OH
6
9r
cr
o
B
A
0
D
Three effects influence the pKa values of these substituted phenols. In C, the acetyl group at the meta
position is electron-withdrawing by induction only. In B, the acetyl group at the para position exerts both
resonance and inductive effects, both of which are electron-withdrawing, making the acid stronger. In
theory, substituents at the ortho position should be like para, exerting both resonance and inductive
effects; in fact, the inductive effect should be stronger because of closer proximity to the acidic group. So
we would predict D to be a stronger acid than B, yet it is not. What other effect is operating?
Structure E shows that because of the proximity of the acetyl group to the OH, intramolecular hydrogen
bonding is possible. Hydrogen bonding stabilizes the starting material, lowering the energy of the
starting material and making �G more positive. Intuitively, it should be apparent that a hydrogen held
between two oxygens will be more difficult to remove by a base. Also, after the proton has left as shown
in structure F, the negative charge on the phenolic oxygen is close to the partial negative charge on the
oxygen of the carbonyl, destabilizing product F, raising its energy, also making AG more positive. The
proximity of the acetyl group influences both sides of the equation to make the acid weaker.
intramolecular
hydrogen bond
cr
�
I
E
"'"
h
0 8-
o
..
F
cr
Here are two more examples where intramolecular hydrogen-bonding influences acidity.
)-
OH
Ho
-o-
eOOH
pK\ 4.6; pK2 9.3
(
Hooe
�
COOH
eOOH
pK\ 2.75; pK2 13.4
pK\ 3.02; pK2 4.38
698
Hooe
eOOH
\::=./
Appendix 2 continued, Summary of Acidity and Basicity
F. Steric Inhibition of Resonance
Another type of proximity effect arises when the placement of a substituent interferes with the orbital
overlap required for resonance stabilization. This can be seen clearly in the acidity of substituted benzoic
acids and in the basicity of substituted anilines.
Let's analyze this series of carboxylic acids.
COOH
COOH
COOH
OH
COOH
COOH
H3
C
CHJ
H 6
�
�
0
CH3
¢CH3
I
pKa 3.75
,
pKa 4.27
pKa 4.20
"
�
pKa 3. 46
,
"
h
pKa 3. 21
pKa 4. 34
Formic acid, pKa 3. 75, serves as the reference carboxylic acid. Benzoic acid is weaker because the phenyl
group is electron-donating by resonance, stabilizing the protonated form. Methyl substituents are known to
be electron-donating by induction, strengthening the electron-donating effect, making the meta- and para­
substituted acids weaker than benzoic acid.
Then come the anomalies. Alkyl groups are electron-donating by induction and should weaken the acids,
but the ortho-t-butyl and the 2,6-dimethylbenzoic acids are not only stronger than benzoic acid, they are
stronger than formic acid! Something has happened to tum the phenyl group into an electron-withdrawing
group.
Phenyl is electron-donating by resonance but electron-withdrawing by induction, so what has happened is
that the ortho substituents have forced the
out of the plane of the benzene ring so that there is no
COOH COOH
resonance overlap between the benzene ring and the
orbitals. The
COOH
"feels" the benzene ring as
simply an inductive substituent. Resonance has been "inhibited" because of the steric effect of the
substituent.
•
"CH3
<----.l�CH3<O-H
COOH
\\
\\
\\
•
•
0
•
group is not parallel with
this three-dimensional view down the
bond
between the
and the benzene ring shows
that
is twisted out of the benzene plane
the plane of the benzene ring­
no resonance interaction.
The same phenomenon is observed in substituted anilines.
Anilines are usually much weaker bases than aliphatic
amines because of resonance overlap of the nitrogen's lone
pair of electrons with the pi sy stem of benzene. When that
resonance is disrupted, the aniline becomes closer in
basicity to an aliphatic amine. (See problem 19-49(c). )
CC
COOH
COOH
CH3
0-'\ N.�H3 � N:�H3
- CH3 CH3 CH3
pKb 8.94
�-
pKb
More examples on the next page.
=::
6-7
(estimated)
699
Appendix 2 continued, Summary of Acidity and Basicity
Examples of steric inhibition of resonance:
o
o
strong base similar to aliphatic amine;
geometry of bridged ring prevents
overlap of N lone pair with carbonyl
amide-not basic because of
resonance sharing of N lone
pair with carbonyl
�ND
pKb 3.4
pKb 8. 9
3° aromatic amine
�
pKb 6.2
3° aliphatic amine
pKb
-
2.3
3° amine, and the N
is
bonded to a benzene, but the
bridged ring system prevents
overlap of N lone pair with
benzene
Not only is steric inhibition of
resonance important in this example,
but so is intramolecular hydrogen­
bonding in the protonated form. Draw
a picture.
(yes, negative!)
700