Chapter 21 Homework Treatment of a carboxylic acid with a strong base yields

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subject Authors David R. Klein

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Chapter 21
Carboxylic Acids and Their Derivatives
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 21. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
Carboxylic acid derivatives exhibit the same ________ state as carboxylic acids.
Carboxylic acid derivatives differ in reactivity, with _____________ being the
most reactive and __________ the least reactive.
When drawing a mechanism, avoid formation of ___________ charges in acidic
conditions, and avoid formation of ____________ charges in alkaline conditions.
In a process called the Fischer esterification, carboxylic acids are converted into
esters when treated with an ____________ in the presence of ________________.
Esters can be hydrolyzed to yield carboxylic acids by treatment with either
aqueous base or aqueous _______. Hydrolysis under basic conditions is also
called ________________.
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514
CHAPTER 21
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 21. The answers appear in the section entitled
SkillBuilder Review.
21.1 Drawing the Mechanism of a Nucleophilic Acyl Substitution Reaction
IDENTIFY THE TWO CORE STEPS OF ANY NUCLEOPHILIC ACYL SUBSTITUTION REACTION
21.2 Interconverting Functional Groups
OH
O
IDENTIFY THE REAGENTS NECESSARY TO ACHIEVE EACH OF THE FOLLOWING TRANSFORMATIONS
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CHAPTER 21
515
21.3 Choosing the Most Efficient C-C Bond-Forming Reaction
C-C Bond Forming Reactions
for which the Functional Group
Remains in the Same Location
C-C Bond Forming Reactions
Involving a Change in the
Location of the Functional Group
Z
O1) Xs RMgBr
2) H
2
O
NaCN
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 21. The
answers appear in the section entitled Review of Reactions.
Preparation of Carboxylic acids Reactions of Carboxylic Acids
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516
CHAPTER 21
Preparation and Reactions
of Acid Chlorides
Preparation and Reactions
of Acid Anhydrides
OR
O
NH
2
O
N
O
R
O
O
O
R
Preparation of Esters Reactions of Esters
R OH
O
R
O
O
CH
3
R OR
O
R OH
OROH
+
Preparation of Amides Reactions of Amides
O
O
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CHAPTER 21
517
Preparation of Nitriles Reactions of Nitriles
R Br R C
N
R C N ROH
O
Solutions
21.1.
a) IUPAC name = pentanedioic acid
Common name = glutaric acid
b) IUPAC name = butanoic acid
Common name = butyric acid
21.2.
O
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518
CHAPTER 21
21.3.
21.4. The compound below is more acidic because its conjugate base is resonance
stabilized. The conjugate base of the other compound is not resonance stabilized.
H
3
COH
O
21.5.
O
O
O
O
21.6. meta-Hydroxyacetophenone should be less acidic than para-hydroxyacetophenone,
because in the conjugate base of the former, the negative charge is spread over only one
21.7.
O
O
21.8. The conjugate base predominates under these conditions:
= 10 = 10
(5.76 - 4.76)
= 10
1
= 10
(pH - pK
a
)
[conjugate base]
[acid]
21.9.
a) 2,3-dichlorobutyric acid is the most acidic and 3,4-dimethylbutyric acid is the least
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CHAPTER 21
519
21.10.
a) Na
2
Cr
2
O
7
, H
2
SO
4
, H
2
O
21.11.
a)
Br OH
1) Mg
2) CO
2
3) H
3
O
+
4) LAH
5) H
2
O
21.12.
a) propionic anhydride
b) N,N-diphenyl-propionamide
21.13.
a)
OO
O
O
b)
O
O
c)
N
H
O
d)
C
l
O
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520
CHAPTER 21
21.14.
a)
Cl
O
NH
N
O
Cl H
Cl
O
H O Me
O
O
Cl
H
Me
Cl O
O
H
Me
H O Me
-
d)
NH
3
Cl
OO
N
Cl
H
H
Cl
N
O
H
H
NH
2
O
-
H
H
NH
3
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CHAPTER 21
521
e)
NH4OH
O
NH3
OH
OH
OH
OH
NH3
H
O
H H
- NH3
+
O
O
MeOH
+
g)
OH
O
H O
Me
H
OH
OH
HOMe
OH
OH
O
Me H
HOMe
OH
OH
O
Me
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522
CHAPTER 21
21.15.
Cl
O
Cl
O
O
Cl
H
O
O
H
OH
-
21.16.
HO
O
OH
H O
H
H
HO
O
OH
H
O
HO OH
H
21.17.
N
O H H O
Me
H
N
O H
H
MeOH
N
H
HO
O
Me H
MeOH N
H
HO
OMe
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CHAPTER 21
523
21.18.
a)
Cl
O1) xs LAH
2) H
2
O
OH
d)
Cl
O1) Et
2
CuLi
2) LAH
3) H
2
O
OH
21.19.
OH O
Cl
1) Na
2
Cr
2
O
7
, H
2
SO
4
, H
2
O
2) SOCl
2
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524
CHAPTER 21
21.20.
CCH
H
H
H
H
O
HO
H
H
OH
HO
H
H
O
H H
21.21.
a)
O
O
OH
O
O
O
HO
O
+
b)
N
H
O OO HO OO N
+
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CHAPTER 21
525
21.22.
O
O
1) NaOH
2) CH
3
I
21.23.
a)
OH
O
O
1) NaOH
2) CH
3
I
b)
O
O
1) NaOH
2) CH
3
I
21.24.
a)
OMe
O
OH MeOH
1) xs LAH
2) H
2
O+
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526
CHAPTER 21
d)
OEt
O
H
3
O
+
OH
O
+ EtOH
21.25.
O
OH O
H
H
O
O
H
HOH
OHO
O
H H
HOH OHO
OH
21.26.
a)
NH
2
O1) excess LAH
2) H
2
ONH
2
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CHAPTER 21
527
21.27.
21.28.
a)
N
O H H O
H
H
N
O H
H
HOH
N
H
HO
O
H H
HOH N
H
HO
OH
b)
N
O H
N O
O
OH N
H
O
OH
H H
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528
CHAPTER 21
21.29.
a)
CN 1) xs LAH
2) H
2
ONH
2
c)
CN 1) EtMgBr
O
2) H
2
O
1) LAH
2) H
2
O
OH
21.30.
a)
O
Cl CN
1) excess NH
3
2) SOCl
2
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CHAPTER 21
529
21.31.
H O
H
H
HOH
C N C N H O
NH
H
H
HOH
O
N
H H
21.32.
a)
OMe
O
Cl
O
1) H
3
O
+
2) SOCl
2
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530
CHAPTER 21
d)
OEt
O
NH
2
2) SOCl
2
3) excess NH
3
4) LAH
5) H
2
O
1) H
3
O
+
f)
O
O O
OH
O
H
2
O
i)
OH
CN 1) H
3
O
+
2) excess LAH
3) H
2
O
21.33. Four steps: 1) oxidize to a carboxylic acid, 2) convert into an acid halide, 3)
convert into an amide, and 4) reduce to give an amine.
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CHAPTER 21
531
21.35.
a)
OH
OOH
1) SOCl
2
2) Et
2
CuLi
3) LAH
4) H
2
O
c)
CN O
O
1) H
3
O
+
2) SOCl
2
3) excess MeMgBr
4)
C
l
O
21.36.
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532
CHAPTER 21
21.37.
a)
2) LiAl(OR)
3
H
1) SOCl
2
3) H
2
O
b)
Br
PBr
3
OH
Mg
1) xs LAH
OH
O
H
3
O
+
CH
3
CN
2) H
2
O
1) Mg
2) CO
2
3) H
3
O
+
c)
PBr
3
1) xs LAH
O
H
3
O
+
2) LiAl(OR)
3
H
1) SOCl
2
3) H
2
O
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CHAPTER 21
533
d)
Br
PBr
3
OH
1) xs LAH
OH
O
H
3
O
+
CH
3
CN 2) H
2
O
21.38. The signal at 1740 cm
-1
indicates that the carbonyl group is not conjugated with
the aromatic ring (it would be at a lower wavenumber if it was conjugated),
2) H
2
O
1) LAH
OH
OH
O
O
21.39.
a)
Increasing acidity
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534
CHAPTER 21
21.40.
a) The second carboxylic acid moiety is electron withdrawing, and stabilizes the
conjugate base that is formed when the first proton is removed.
21.41.
a) cyclopentanecarboxylic acid
b) cyclopentanecarboxamide
21.42.
a) acetic anhydride
21.43.
OH
O
OH
O
OH
O
hexanoic acid
2-methylpentanoic acid
3-methylpentanoic acid
chirality centers
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CHAPTER 21
535
21.45.
a)
OH
O1) excess LAH
2) H
2
OOH
c)
OH
O
1) excess LAH
2) H
2
O
3) PBr
3
4) NaCN
5) H
3
O
+
OH
O
21.46.
b)
Br 1) NaCN
2) H
3
O
+
OH
O
21.47. As discussed in Chapter 19, the methoxy group is electron donating via
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536
CHAPTER 21
21.48.
a)
N
O
H
d)
O
H
O
f)
NH
2
O
21.49.
a)
Cl
O
b)
OH
c)
ONa
O
d)
O
O
21.50.
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CHAPTER 21
537
c)
NH
2
OSOCl
2
CN
f)
OH O
O O
O
OHO
O
+
g)
NCl
O
pyridine
H
N
O
j)
O
OH
3
O
+
OH
O
O
H
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538
CHAPTER 21
21.51.
a)
HO
O
H
O
+
21.52.
OH
2) H
2
O
1) LiAl(OR)
3
H
O
Cl
O
H
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
21.53.
a) NaOH, followed by Na
2
Cr
2
O
7
, H
2
SO
4
, H
2
O
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CHAPTER 21
539
b)
N
O
1) Br
2
, AlBr
3
2) Mg
3) CO
2
4) H
3
O
+
5) SOCl
2
6) excess (CH
3
)
2
NH
d)
C
l
O
N
O
H
3)
1) Cl
2
, AlCl
3
2) NaNH
2
, NH
3
, pyridine
21.55.
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540
CHAPTER 21
c)
Br N
O
1) Mg
2) CO
2
3) H
3
O
+
4) SOCl
2
5) excess CH
3
NH
2
1) LAH
2) H
2
O
3)
Cl
O
N
H
O
21.56. A methoxy group is electron donating, thereby decreasing the electrophilicity of
the ester moiety. A nitro group is electron withdrawing, thereby increasing the
electrophilicity of the ester group.
21.57.
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CHAPTER 21
541
21.58.
O
1) Mg
21.59.
O O 1) excess MeMgBr
2) H
3
O
+
O
OH OH
+
21.60.
OH
O
HO
H
H
OH
OH
HOH
OH
OH
O
H H
HOH
OH
OH
O
H
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542
CHAPTER 21
21.61.
a)
Cl Cl
OH O R
O
O
Cl
Cl
H
R
Cl
Cl O
O
H
R
-
b)
OO
O
c)
Ph Ph
P
OH
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CHAPTER 21
543
21.63.
1) Br
2
, AlBr
3
21.64.
21.65.
a)
Ph Cl
O
H O Ph
O
O
Cl
Ph
H
Ph
Cl Ph O
O
H
Ph
Ph O
O
-
Ph
N
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544
CHAPTER 21
c)
O
O
OH
OOH
O
O
O
O
H
d)
Cl
O
Cl
O
H
2
N NH
2
Cl
Cl
O
O
NN
H H
H
H
Cl
O
NN
HH
H
H
O
-Cl
N
O
N
H
H
O
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CHAPTER 21
545
e)
O
OC C H
H
H
H
HO
O
O
O
21.66. The three chlorine atoms withdraw electron density via induction. This effect
renders the carbonyl group more electrophilic.
21.68
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546
CHAPTER 21
21.69.
21.70.
O
O
O
O
O
O
O
O
21.72.
Cl
OO
O
Cl H
HO Cl
O
HO HO
Cl
O
page-pf23
21.73.
a)
OH
OOO
1) SOCl
2
2) Et
2
CuLi
3) HOCH
2
CH
2
OH,
[H
+
], -H
2
O
c)
O
N
O
1) MCPBA
2) H
3
O
+
3) SOCl
2
4) (CH
3
)
2
NH
e)
COOH
O
O
1) SOCl
2
2) LiAl(OR)
3
H
3) HOCH
2
CH
2
OH,
[H
+
], -H
2
O
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548
CHAPTER 21
21.75.
OH
OO
O
H O
H
H
OH
OOH
HO
H O
H
H
OH
OO
O
H
OH
OOH
HO
H
O
HO
HO OH
H
OH
OOH
O
HOH
HO
-
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CHAPTER 21
549
21.76.
Cl
O
xs NH
3
NH
2
O
Compound A
21.79. The
1
H NMR spectrum of para-chlorobenzaldehyde should have a signal at
approximately 10 ppm corresponding to the aldehydic proton. The
1
H NMR
spectrum of benzoyl chloride should not have a signal near 10 ppm.
S
O
O O H
O
HO
R
S
O
O
OOH H
O
HO
R
H
O
O
O
O
R
OH
H
S
O
O
O O H
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550
CHAPTER 21
21.82. The lone pair of the nitrogen atom in this case is participating in resonance and is
less available to donate electron density to the carbonyl group. As a result, the carbonyl
21.83.
a) DMF, like most amides, exhibits restricted rotation about the bond between the
carbonyl group and the nitrogen atom. This restricted rotation causes the methyl groups

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