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 SN2, includes the observation of a
__________-order rate equation. The reaction proceeds with ____________ of
configuration.
SN2 reactions are said to be _________________ because the configuration of the
product is determined by the configuration of the substrate.
Evidence for the stepwise mechanism, called SN1, includes the observation of a
__________-order rate equation.
The __________ step of an SN1 process is the rate-determining step.
An SN1 reaction is a stepwise process with a first-order rate equation.
There are four factors that impact the competition between the SN2 mechanism
and SN1: 1) the _____________, 2) the ________________, 3) the ___________
_______________, and 4) the _______________.
______________________ solvents favor SN2.
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
A CONCERTED MECHANISM
A STEPWISE MECHANISM
DRAW CURVED ARROWS, SHOWING NUCLEOPHILIC
ATTACK ACCOMPANIED BY SIMULTANEOUS LOSS OF
A LEAVING GROUP
DRAW A CURVED ARROW SHOWING THE LOSS OF THE LEAVING GROUP
TO FORM A CARBOCATION INTERMEDIATE, FOLLOWED BY ANOTHER
CURVED ARROW SHOWING THE NUCLEOPHILIC ATTACK
Nuc
LG
- LG
Nuc
- LG
Nuc
LG
SkillBuilder 7.2 Drawing the Product of an SN2 Process
DRAW THE MAJOR PRODUCT OF THE FOLLOWING REACTION
Br
+
OH
+
Br
Nuc
CHAPTER 7
SkillBuilder 7.3 Drawing the Transition State of an SN2 Process
DRAW THE TRANSITION STATE OF THE FOLLOWING REACTION
NaSH
SH
Cl
TRANSITION STATE
SkillBuilder 7.4 Drawing the Carbocation Intermediate of an SN1 Process
DRAW THE CARBOCATION THAT WOULD BE FORMED IF A CHLORIDE ION IS EXPELLED
FROM THE FOLLOWING COMPOUND
- Cl
Cl
SkillBuilder 7.5 Drawing the Products of an SN1 Process
PREDICT THE PRODUCTS OF THE FOLLOWING SN1 REACTION
Br
NaCN
+
SkillBuilder 7.6 Drawing the Complete Mechanism of an SN1 Process
IDENTIFY THE TWO CORE STEPS AND THREE POSSIBLE ADDITIONAL STEPS OF AN SN1 PROCESS
TWO CORE STEPS
THREE POSSIBLE ADDITIONAL STEPS
SkillBuilder 7.7 Drawing the Complete Mechanism of an SN2 Process
IDENTIFY THE ONE CORE STEP (CONCERTED) AND TWO POSSIBLE ADDITIONAL STEPS OF AN SN2 PROCESS
CORE STEP
TWO POSSIBLE ADDITIONAL STEPS
115
116
CHAPTER 7
SkillBuilder 7.8 Determining whether a Reaction Proceeds via an SN1 Mechanism or an SN2
Mechanism
FILL IN THE TABLE BELOW, SHOWING THE FEATURES
THAT FAVOR SN2 OR SN1 REACTIONS
SN2
SN1
SUBSTRATE
NUC
LG
SOLVENT
SkillBuilder 7.9 Identifying the Reagents Necessary for a Substitution Reaction
IDENTIFY THE REAGENTS NECESSARY TO ACHIEVE THE FOLLOWING TRANSFORMATION
OH
1)
CN
2)
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.
SN2
DRAW THE CURVED ARROWS THAT SHOW
THE FLOW OF ELECTRON DENSITY DURING
THE FOLLOWING SN2 REACTION
H
H
H
LG
Nuc
H
Nuc
H
H
+
LG
SN1
DRAW THE CURVED ARROWS THAT SHOW
THE FLOW OF ELECTRON DENSITY DURING
THE FOLLOWING SN1 REACTION
LG
- LG
Nuc
Nuc
CHAPTER 7
Solutions
7.1.
a)
b)
c)
d)
4-chloro-4-ethylheptane
1-bromo-1-methylcyclohexane
4,4-dibromo-1-chloropentane
(S)-5-fluoro-2,2-dimethylhexane
7.2.
SH
Br
+
SH
Br
a)
I
O
OMe
Me
b)
7.3.
O
O
Br
O
Br
+
O
a)
I
+
Cl
Ι
Cl
b)
7.4.
O
Br
O
+
Br
+
I
117
118
CHAPTER 7
7.5.
Br
+
Br
Cl
Cl
7.6.
a)
b)
c)
the rate of the reaction is tripled.
the rate of the reaction is doubled.
the rate of the reaction will be six times faster.
7.7.
OH
SH
a)
b)
Cl
c)
7.8.
2
4
H
F
1
1
Br
H3C
OMe
2
F
MeO
CH3
3
S
H4
3
S
The reaction does proceed with inversion of configuration. However, the Cahn-IngoldPrelog system for assigning a stereodescriptor (R or S) is based on a prioritization
scheme. Specifically, the four groups connected to a chirality center are ranked (one
through four). In the reactant (above left), the highest priority group is the leaving group
(bromide) which is then replaced by a group that does not receive the highest priority. In
the product, the fluorine atom has been promoted to the highest priority as a result of the
reaction, and as such, the prioritization scheme has changed. In this way, the
stereodescriptor (S) remains unchanged, despite the fact that chirality center undergoes
inversion.
119
CHAPTER 7
7.9.
a)
b)
c)
d)
7.10.
7.11.
O
H δ+
Br
δ+
H
Being
formed
Being
broken
This step is favorable (downhill in energy) because ring strain is alleviated when the
three-membered ring is opened.
7.12.
H
N
N
a)
H
CH3
S
R
R
H
H
− H+
N
N
H
CH3
N
N
CH3
Nicotine
120
CHAPTER 7
H3C
CH3
S
R
R
OH
N
CH3
OH
H3C N
H3C
CH3
choline
b)
7.13. a) The rate of the reaction will be doubled, because the change in concentration
of sodium chloride will not affect the rate.
b) The rate of the reaction will remain the same, because the change in concentration of
sodium chloride will not affect the rate.
7.14. Draw the carbocation intermediate generated by each of the following substrates in
an SN1 reaction:
(b)
(a)
(c)
(d)
7.15.
Br
The first compound will generate a tertiary carbocation, while the second compound will
generate a tertiary benzylic carbocation that is resonance stabilized. The second
compound leads to a more stable carbocation, so that compound will lose its leaving
group more rapidly than the first compound.
7.16.
Cl
+ NaΙ
a)
SH
HS
+
+
Br
b)
+
+
O
O
c)
Cl
d)
O
O
+
Ι
Cl
CHAPTER 7
121
7.17.
SH
HS
+
Diastereomers
7.18.
a) No
b) Yes
c) No
7.19.
a) No
g) No
b) Yes
h) Yes
c) Yes
i) No
d) Yes
j) No
e) No f) No
k) Yes l) No
7.20.
a) No
b) Yes
c) Yes
d) No
e) No
d) Yes
e) Yes
f) No
f) No
7.21.
a)
H
O
O H
H Br
+
Br
H
Br
b)
H
H
O
O
H
H
H Br
Br
Br
+ Br
c)
H
Br
O
H
O
H
H
H
O
H
OH
122
CHAPTER 7
d)
H
O Et EtOH
EtOH
Ι
OEt
e)
H
H
O
MeOH
H
OH
O
H
O H
H
Me
MeOH
OMe
f)
H
H
O
H
OH
O
H
H
MeOH
H
Me O
MeO
MeOH
g)
S H
Br
SH
h)
EtOH
Ι
EtOH
O
Et
H
OEt
CHAPTER 7
123
7.22.
- LG
Nuc Attack
- LG
C+
rearr.
Nuc Attack
- H+
+ H+
- LG
Nuc Attack
- H+
+ H+
- LG
- H+
c)
d)
e)
C+
rearr.
- H+
Nuc Attack
f)
g)
- LG
Nuc Attack
- 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.
OH
H
H
O
H
O
H
H
H
H
OH
O
H
H
O
H
O
H
H
The chirality center at C2 is lost when the leaving group leaves to form a carbocation
with trigonal planar geometry. The chirality center at C3 is lost during the hydride shift
in the following step. Once again, the chirality center is converted into a trigonal planar
sp2 hybridized center (which is no longer a chirality center).
124
CHAPTER 7
7.24.
a)
MeOH
Cl
O
H
Me
MeOH
Et
EtOH
OMe
b)
EtOH
Br
O
H
O
c)
Ι
H
O
H
O
H
H
H
O
OH
H
d)
OH
OH
Br
O
O
H
7.25.
H
H
N
Me
Ι
H
H
H N Me
H
NH3
H
H
N
Me
Ι
Me
H
H N Me
Me
NH3
Me
Me N Me
Ι Me
Me
7.26.
a) SN1
e) Both
b) SN2
f) Neither
Me
H3N
N
Me
Me
c) Neither
g) Both
H
Me N Me
Me
d) SN1
Ι Me
Me
H
N
Me
CHAPTER 7
7.27.
a) SN1
b) SN2
c) SN2
d) SN2
e) SN2
7.28.
Cl
Cl
I
TsO
OMe
Br
a)
NH2 F
Cl
Cl
I
TsO
OMe
Br
b)
7.29.
a) SN1
e) SN1
b) SN2
f) SN2
c) SN1
g) SN2
NH2 F
d) SN2
h) SN1
7.30.
Acetone is a polar aprotic solvent and will favor SN2 by raising the energy of the
nucleophile, giving a smaller Ea.
7.31.
I
OMe
MeOH
SN1
a)
Br
Cl
Cl
SN2
HMPA
b)
O H
H Br
Br
SN1
Racemic
c)
NaCN
d)
OTs
DMF
CN
SN2
125
126
CHAPTER 7
H2O
I
SN1
OH
Racemic
e)
NaCN
Br
CN
SN2
DMSO
f)
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 SN2 process.
7.33.
OH
(a)
HBr
(b)
NaOH
I
OH
Br
(c)
(d)
HI
OH
1) TsCl, pyridine
2) NaBr, DMSO
(f)
O
O
Br
O
DMSO
OH
Br
(g)
I
OH
(i)
O
1) TsCl, pyridine
(h)
O
CN
2) NaCN
7.34.
OH
(R)2-butanol
1) TsCl, pyridine
2) NaI, DMSO
3) NaSH, DMSO
SH
DMSO
O
Br
NaSH
I
OH
(e)
Br
SH
(R)2-butanethiol
H2O
CHAPTER 7
7.35.
O
NH2
O
Cl
HO
HO
N
melphalan
NH2
N
Cl
Cl
Nuc
O
Nuc
NH2
NH2
HO
N
HO
O
Nuc
N
Cl
Nuc
O
HO
NH2
Nuc
N
Nuc
7.36.
a) Systematic Name = 2-chloropropane
Common Name = isopropyl chloride
b) Systematic Name = 2-bromo-2-methylpropane
Common Name = tert-butyl bromide
c) Systematic Name = 1-iodopropane
Common Name = propyl iodide
d) Systematic Name = 2-chlorobutane
Common Name = propyl iodide
d) Systematic Name = (R)-2-bromobutane
Common Name = (R)-sec-butyl bromide
e) Systematic Name = 1-chloro-2,2-dimethylpropane
Common Name = neopentyl chloride
f) Systematic Name = chlorocyclohexane
Common Name = cyclohexyl chloride
127
128
CHAPTER 7
7.37.
Increasing reactivity (SN2)
Ι
Ι
Ι
Ι
7.38.
Br
Cl
Cl
a) secondary
Cl
primary
b)
primary
more sterically
hindered
Br
primary
less sterically
hindered
Cl
Br
c)
secondary
tertiary
I
better
leaving group
d)
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 SN2 process.
7.40.
a) NaSH
b) sodium hydroxide
c) methoxide dissolved in DMSO
7.41.
Cl
Br
Cl
a) tertiary
Cl
c)
primary
Br
b) primary
Cl
Cl
allylic
d)
tertiary
OTs
better
leaving group
CHAPTER 7
7.42.
a) The rate of the reaction is doubled.
b) The rate of the reaction is doubled.
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
c) aprotic
d) protic
e) protic
129
7.45.
4
H Br
1
R
O
2
3
a)
2
NC H
4
R
O
3
b)
1
c) The reaction is an SN2 process, and it does proceed with inversion of configuration.
However, the prioritization scheme changes when bromide (#1) is replaced with a cyano
group (#2). As a result, the Cahn-Ingold-Prelog system assigns the same configuration to
the reactant and the product.
7.46.
O
δ−
O
Me
δ−
Ι
H H
7.47.
Iodide functions as a nucleophile and attacks (S)-2-iodopentane, displacing iodide as a
leaving group. The reaction is an SN2 process, and therefore proceeds via inversion of
configuration. The product is (R)-2-iodopentane. The reaction continues until a racemic
mixture is obtained.
130
CHAPTER 7
7.48.
H
O
O H
H Br
Br
H
Br
− H2O
Racemic
The chirality center is lost when the leaving group leaves to form a carbocation with
trigonal planar geometry. The nucleophile can then attack either face of the planar
carbocation, leading to a racemic mixture.
7.49.
H
O
H O S OH
O
O
H
O
H
H
OH
O
HO S O
O
O
H
H
O
Racemic
E
Reaction coordinate
7.50.
Increasing stability
7.51.
secondary
tertiary
primary
secondary
H
131
CHAPTER 7
7.52.
OH
H
O H
H Cl
H
Cl
Cl
7.53.
O
O
+
Br
O
Br
O
7.54.
a)
Cl
MeOH
H
O Me
MeOH
OMe
3 Steps
b)
Cl
SH
SH
2 Steps
c)
OH
H Ι
H
O H
− H2O
Ι
Ι
3 Steps
132
CHAPTER 7
d)
H
OTs
EtOH
OEt
EtOH
H
O Et
4 Steps
7.55.
EtOH
Br
OEt
a)
NaBr
Br
OTs
b)
OH
HCl
Cl
c)
NaCN
DMSO
I
d)
CN
7.56.
O
O
7.57.
O H
Br
OH
Although the substrate is primary, it is still sterically hindered. As a result, SN2 reactions
at neopentyl halides do not occur at an appreciable rate.
CHAPTER 7
133
7.58.
a)
H
O
H
Br
O
H
H
H
O
H
OH
b) The substrate is primary, and therefore, the reaction must proceed via an SN2 process.
SN2 reactions are highly sensitive to the strength of the nucleophile, and the nucleophile
(water) is a weak nucleophile. As a result, the reaction occurs slowly.
O H
OH
Br
c)
+
Br
Hydroxide is a strong nucleophile, which favors the SN2 process.
7.59.
NaOH
OH
OTs
a)
1) TsCl, py
CN
OH
2) NaCN
b)
HBr
Br
OH
c)
Cl
SH
NaSH
d)
O
O
Br
e)
NaO
O
134
CHAPTER 7
7.60.
+
Ι
a)
Ι
OH
O
+
O
b)
Ι
CN
+
c)
SH
+
Ι
d)
Ι
+ H2O
e)
Ι
+ H2S
f)
7.61.
S
SH
a)
Et
CN
b)
c)
7.62.
The second method is more efficient because the alkyl halide (methyl iodide) is not
sterically hindered. The first method is not efficient because it employs a tertiary alkyl
halide, and SN2 reactions do not occur at tertiary substrates.
7.63.
OH
1) TsCl, pyridine
Br
2) NaBr
a)
OH
HCl
Cl
b)
NaOH
c)
Cl
OH
CHAPTER 7
135
7.64.
S H
SH
Br
a)
Rate = k
Br
b)
+
Br
NaSH
c) The rate would be slower.
E
Reaction coordinate
d)
Et
δ−
HS
δ−
Br
H H
e)
7.65.
a) SN1 (tertiary substrate)
Br
H Br
O H
H
O H
b)
Rate = k
c)
OH
d) No. The rate is not dependent on the concentration or strength of the nucleophile.
E
e)
Reaction coordinate
Br
136
CHAPTER 7
7.66.
a) SN2
Br
CN
CN
+
b)
Br
Rate = k
NaCN
c)
d) Yes. The reaction rate would double.
E
Reaction coordinate
e)
7.67.
H
I
H
O
H
H
O
H
H
O
OH
H
Br
CHAPTER 7
137
7.68.
O
O
I
O
O
+
I
a)
b) This reaction occurs via an SN2 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.
7.69.
O
H
O
Base
O
Br
Br
7.70.
Ι
EtOH
Br
Ι
O
H
Et
EtOH
OEt
7.71.
When the leaving group leaves, the carbocation formed is resonance stabilized:
O
OTs
O
O
Resonance stabilized
7.72.
Iodide is a very good nucleophile (because it is polarizable), and it is also a very good
leaving group (because it can stabilize the negative charge by spreading the charge over a
large volume of space). As such, iodide will function as a nucleophile to displace the
chloride ion. Once installed, the iodide group is a better leaving group than chloride,
thereby increasing the rate of the reaction.
138
CHAPTER 7
7.73.
H
OH
H
H
H
O
O
H
O
O
H
H
H
H
OH
H
O
H
OH