Chapter 7
Substitution Reactions
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 7. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• Substitution reactions exchange one _____________ for another.
• Evidence for the concerted mechanism, called S
N
2, includes the observation of a
__________-order rate equation. The reaction proceeds with ____________ of
Review of Skills
Follow the instructions below. To verify that your answers are correct, look in your
textbook at the end of Chapter 7. The answers appear in the section entitled SkillBuilder
Review.
SkillBuilder 7.1 Drawing the Curved Arrows of a Substitution Reaction
CHAPTER 7
115
SkillBuilder 7.3 Drawing the Transition State of an S
N
2 Process
SH
NaSH
Cl
DRAW THE TRANSITION STATE OF THE FOLLOWING REACTION
TRANSITION STATE
SkillBuilder 7.6 Drawing the Complete Mechanism of an S
N
1 Process
IDENTIFY THE TWO CORE STEPS AND THREE POSSIBLE ADDITIONAL STEPS OF AN S
N
1 PROCESS
TWO CORE STEPS
THREE POSSIBLE ADDITIONAL STEPS
116
CHAPTER 7
SkillBuilder 7.8 Determining whether a Reaction Proceeds via an S
N
1 Mechanism or an S
N
2
Mechanism
S
N
2
NUC
SUBSTRATE
S
N
1
LG
SOLVENT
FILL IN THE TABLE BELOW, SHOWING THE FEATURES
THAT FAVOR S
N
2 OR S
N
1 REACTIONS
Review of Reactions
Follow the instructions below. To verify that your answers are correct, look in your
textbook at the end of Chapter 7. The answers appear in the section entitled Review of
Reactions.
H
H
HLG Nuc H
HH
Nuc LG
S
N
2
+
DRAW THE CURVED ARROWS THAT SHOW
THE FLOW OF ELECTRON DENSITY DURING
THE FOLLOWING S
N
2 REACTION
CHAPTER 7
117
Solutions
7.1.
a) 4-chloro-4-ethylheptane
7.2.
a)
Br
SH
SH
+Br
7.3.
Br O
O
O
O
+Br
7.4.
OBr O
Br
+
118
CHAPTER 7
7.5.
Br
Cl
Br
Cl
+
7.6.
7.7.
a)
SH
b)
C
l
c)
OH
7.8.
2
CHAPTER 7
119
7.9.
7.10.
7.11.
7.12.
a)
N
N
H
H
RSR
CH
3
N
N
H
CH
3
HN
N
H
CH
3
− H+
N
i
c
o
t
i
n
e
120
CHAPTER 7
b)
H
3
C
N
H
3
C OH RSR
CH
3
CH
3
N
CH
3
OH
H
3
C
c
h
o
l
i
n
e
7.13. a) The rate of the reaction will be doubled, because the change in concentration
7.14. Draw the carbocation intermediate generated by each of the following substrates in
an S
N
1 reaction:
(a)
(b)
(c)
(d)
7.15.
7.16.
a)
+ NaΙ
Cl
CHAPTER 7
121
7.17.
SH
+
HS
D
i
a
s
t
e
r
e
o
m
e
r
s
7.19.
7.20.
a) No b) Yes c) Yes d) No e) No f) No
7.21.
a)
O H
H Br
O
H
HBr
Br
+
122
CHAPTER 7
d)
ΙEtOH
O
H
Et OEt
EtOH
f)
MeOH
OH
H
O
H H
H
OH
h)
Ι
EtOH
O
E
t
H
EtOH
OEt
CHAPTER 7
123
7.22.
c)
LG Nuc Attack H
+
––
h)
Problem 7.20c and 7.20h exhibit the same pattern. Both problems are characterized by
three mechanistic steps: 1) loss of a leaving group, 2) nucleophilic attack, and 3) proton
transfer.
7.23.
H
OHH
O
H H
H
OH
124
CHAPTER 7
7.24.
a)
Cl
MeOH
O
H
Me MeOH
OMe
d)
Br
OH
O
H
OH O
7.25.
N
H
HH
Me Ι
N
H
H
H
Me
NH
3
N
H
MeH
Me Ι
NH
N
H
H
Me
Me
CHAPTER 7
125
N
N
N
N
N
7.28.
TsO
Cl
I
OMe
Cl
7.29.
a) S
N
1 b) S
N
2 c) S
N
1 d) S
N
2
e) S
N
1 f) S
N
2 g) S
N
2 h) S
N
1
7.30.
Acetone is a polar aprotic solvent and will favor S
N
2 by raising the energy of the
nucleophile, giving a smaller E
a
.
7.31.
a)
I
MeOH
OMe
S
N
1
Br Cl
Cl
126
CHAPTER 7
IH
2
OOH S
N
1
R
a
c
e
m
i
c
7.32.
No. Preparation of this amine via the Gabriel synthesis would require the use of a tertiary
alkyl halide, which will not undergo an S
N
2 process.
7.33.
(a)
I
NaOH
O
H
(b)
Br
OH
HBr
(c)
I
O
H
HI
(d)
Br NaSH
D
M
S
O
SH
7.34.
OH SH
(R)-
2
–
b
u
t
a
n
o
l
(R)-
2
–
b
u
t
a
n
e
t
h
i
o
l
1) TsCl, pyridine
2) NaI, DMSO
3) NaSH, DMSO
CHAPTER 7
127
7.35.
N
Cl
O
HO
NH
2
N
Cl
ClO
HO
NH
2
melphalan
Nuc
7.36.
a) Systematic Name = 2-chloropropane
Common Name = isopropyl chloride
b) Systematic Name = 2-bromo-2-methylpropane
Common Name = tert–butyl bromide
128
CHAPTER 7
7.37.
Ι
Ι
ΙΙ
Increasing reactivity (S
N
2)
7.38.
7.39.
No. Preparation of this compound via the process above would require the use of a
tertiary alkyl halide, which will not undergo an S
N
2 process.
7.40.
7.41.
a)
Cl
Cl
tertiary primary
b)
Br
Br
primary tertiary
CHAPTER 7
129
7.43. a) The rate of the reaction is doubled
b) The rate of the reaction will remain the same.
7.44. a) aprotic
b) protic
7.45.
a)
Br
O
2
1
3
4
H
R
7.46.
Ι
O
OMe
H H
δ− δ−
7.47.
Iodide functions as a nucleophile and attacks (S)-2-iodopentane, displacing iodide as a
130
CHAPTER 7
7.48.
O H
H
Br
7.49.
O
HH O S
O
O
OH
O
H H
7.50.
Increasing stability
CHAPTER 7
131
7.52.
OH H Cl OH
H
H
7.54.
a)
Cl MeOH O
H
Me MeOH OMe
132
CHAPTER 7
d)
H
OTs
7.55.
a)
Br EtOH OEt
7.56.
O
O
7.57.
CHAPTER 7
133
7.58.
a)
Br
HOHO
H
HOH
HOH
N
7.59.
a)
O
T
s
OH
NaOH
b)
OH CN
1) TsCl, py
2) NaCN
c)
O
H
Br
HBr
134
CHAPTER 7
7.60.
a)
Ι
+OH
b)
Ι
O
O
+
f)
7.61.
a)
SH
b)
SEt
c)
CN
7.62.
N
7.63.
a)
OH 1) TsCl, pyridine
2) NaBr
Br
CHAPTER 7
135
7.64.
a)
Br
S H
SH Br
+
7.65.
a) S
N
1 (tertiary substrate)
b)
O H
H Br
O H
H
Br Br
136
CHAPTER 7
7.66.
a) S
N
2
b)
Br
CN CN
Br
+
c)
Br
NaCN
Rate = k
7.67.
H
I
CHAPTER 7
137
7.68.
a)
O
O
I
O
O
I
+
b) This reaction occurs via an S
N
2 process. As such, the rate of the reaction is highly
sensitive to the nature of the substrate. The reaction will be faster in this case, because
the methyl ester is less sterically hindered than the ethyl ester.
7.71.
When the leaving group leaves, the carbocation formed is resonance stabilized:
O OTs OO
R
e
s
o
n
a
n
c
e
s
t
a
b
i
l
i
z
e
d
7.72.
Iodide is a very good nucleophile (because it is polarizable), and it is also a very good
138
CHAPTER 7
7.73.
OH
HOH
H
OH
H