Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
Chapter 8
Alkenes: Structure and
Preparation via Elimination 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 8. Each of the sentences below appears in the section entitled Review
of Concepts and Vocabulary.
• Alkene stability increases with increasing degree of ________________.
• E2 reactions are said to be regioselective, because the more substituted alkene,
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 8. The answers appear in the section entitled
SkillBuilder Review.
8.1 Assembling the Systematic Name of an Alkene
PROVIDE A SYSTEMATIC NAME FOR THE FOLLOWING COMPOUND
1) IDENTIFY THE PARENT
2) IDENTIFY AND NAME SUBSTITUENTS
3) ASSIGN LOCANTS TO EACH SUBSTITUENT
4) ALPHABETIZE
140
CHAPTER 8
8.3 Comparing the Stability of Isomeric Alkenes
CIRCLE THE MOST STABLE ALKENE BELOW
8.6 Predicting the Stereochemical Outcome of an E2 Reaction
Cl
PREDICT THE STEREOCHEMICAL OUTCOME OF THE FOLLOWING REACTION, AND DRAW THE PRODUCT.
Strong Base
8.7 Drawing the Products of an E2 Reaction
CHAPTER 8
141
8.9 Drawing the Complete Mechanism of an E1 Reaction
IDENTIFY THE TWO CORE STEPS AND TWO POSSIBLE ADDITIONAL STEPS OF AN E1 PROCESS
TWO CORE STEPS
TWO POSSIBLE ADDITIONAL STEPS
8.11 Identifying the Expected Mechanism(s)
MINOR
MINOR
ONLY
STRONG
NUC
1
2
3
STRONG
BASE
IDENTIFY THE MECHANISM(S) THAT
OPERATE IN EACH OF THE CASES BELOW:
+
+
MAJOR
MAJOR
142
CHAPTER 8
Review of Synthetically Useful Elimination Reactions
Identify reagents that will achieve each of the transformations below. To verify that your
answers are correct, look in your textbook at the end of Chapter 8. The answers appear in
the section entitled Review of Synthetically Useful Elimination Reactions.
OH
Br
Br
Solutions
8.1.
a) 2,3,5-trimethyl-2-heptene
8.2.
a) b) c)
CHAPTER 8
143
8.6. When using cis-trans terminology, we look for two identical groups. In this case,
there are two ethyl groups that are in the trans configuration:
Cl
t
r
a
n
s
8.7.
a)
increasing stability
144
CHAPTER 8
8.8. In the first compound, all of the carbon atoms of the ring are sp
3
hybridized and
tetrahedral. As a result, they are supposed to have bond angles of approximately 109.5º,
but their bond angles are compressed due to the ring (and are almost 90º). In other
8.9.
a)
OH
H
Cl
H
8.10.
Cl Cl H
2
O
-H
CHAPTER 8
145
8.11. This mechanism is concerted:
EtO
OTs
H
8.14.
a)
Br
Br Br
secondary
s
u
b
s
t
r
a
t
e
tertiary
s
u
b
s
t
r
a
t
e
primary
s
u
b
s
t
r
a
t
e
increasing reactivity towards E2
146
CHAPTER 8
b)
Cl
O
major
(less substituted)
minor
(more substituted)
+
I
NaOH
+
8.16.
8.17.
a)
O
HO
Br
Br
CHAPTER 8
147
b)
O
Br
8.18.
a)
major minor
+
b)
the only E2 product
8.19.
Br
Strong Base
148
CHAPTER 8
8.20. The leaving group in menthyl chloride can only achieve antiperiplanarity with one
beta proton, so only one elimination product is observed. In contrast, the leaving group
8.21. Because of the bulky tert-butyl group, the first compound is essentially locked in a
chair conformation in which the chlorine occupies an equatorial position. This
conformation cannot undergo an E2 reaction because the leaving group is not
antiperiplanar to a proton. However, the second compound is locked in a chair
8.22.
a)
Cl
NaOEt
+
major minor
Br
CHAPTER 8
149
d)
NaOEt
I
minormajor
+
8.23.
strong base
RR
R R
X
RR
R R
all R groups are identical
8.24.
strong base
X
+
8.26.
a) Only the concentration of tert-butyl iodide affects the rate, so the rate will double.
b) Only the concentration of tert-butyl iodide affects the rate, so the rate will remain
the same.
150
CHAPTER 8
8.28.
a) b) c) d)
8.29.
a)
H
2
SO
4
heat
OH
+
minormajor
d)
heat
Br
+
minormajor
EtOH
8.30. Both alcohols below can be used to form the product. The tertiary alcohol below
will react more rapidly because the rate determining step involves formation of a tertiary
CHAPTER 8
151
8.31.
H
2
SO
4
OH
8.32.
a) No, the leaving group is not OH
b) Yes, the leaving group is OH
8.33.
a) No. Loss of the leaving group forms a tertiary carbocation, which will not
rearrange.
b) Yes. Loss of the leaving group forms a secondary carbocation, which can
152
CHAPTER 8
8.34.
a)
OH
H O S
O
O
HO
O
H
HH
- H
2
O
HOH
d)
OH
H O S
O
O
HO
HOH
CHAPTER 8
153
8.35. Problem 8.34b and 8.34c exhibit the same pattern because both have leaving
groups that can leave without being protonated, and both do not exhibit a carbocation
rearrangement. As a result both mechanisms involve only two steps: 1) loss of a leaving
8.36. The first method is more efficient because it employs a strong base to promote an
E2 process for a secondary substrate bearing a good leaving group. The second method
relies on an E1 process occurring at a secondary substrate, which will be slow and will
involve a carbocation rearrangement to produce a different product.
8.37.
HMeO
8.38
NH
2
R
Cl Cl
R
R
H H
R
R
Cl
R
H
NH
2
154
CHAPTER 8
8.39.
a) weak nucleophile, weak base
b) strong nucleophile, weak base
8.40. Aluminum is a larger atom and is polarizable. Therefore, the entire complex can
8.41.
a) NaOH is a strong nucleophile and strong base. The substrate in this case is
primary. Therefore, we expect S
N
2 (giving the major product) and E2 (giving the
8.42.
a) NaOEt is a strong nucleophile and strong base. The substrate in this case is
secondary. Therefore, we expect E2 (giving the major product) and S
N
2 (giving
8.43.
a) EtOH is a weak nucleophile and weak base. The substrate in this case is
tertiary. Therefore, we expect both S
N
1 and E1.
CHAPTER 8
155
8.44.
a) An E2 reaction does not readily occur because the base is weak.
8.45. The substrate is tertiary, so S
N
2 cannot occur at a reasonable rate. There are no
beta protons, so E2 also cannot occur.
8.46.
a)
OTs NaCl
D
M
S
O
Cl
b)
I
NaOH
+
156
CHAPTER 8
f)
NaSH
I
SH
g)
Br NaOH
+
major minor
OH
j)
Br NaOH
+
major minor
OH
m
i
n
o
r
m
i
n
o
r
+
CHAPTER 8
157
l)
Br NaOMe OMe
+
major minor minor
+
8.47. There are only two constitutional isomers with molecular formula C
3
H
7
Cl:
a primary
alkyl halide
a secondary
alkyl halide
Cl
Cl
Sodium methoxide is both a strong nucleophile and a strong base. When compound A is
158
CHAPTER 8
8.48.
a)
Cl
b)
Cl
Cl
8.49.
NaOEt
Cl
8.50.
a) trans-3,4,5,5-tetramethyl-3-heptene
CHAPTER 8
159
8.51.
O
Br Cl
Br
E
Z
8.52. Because of the bulky tert-butyl group, the trans isomer is essentially locked in a
chair conformation in which the chlorine occupies an equatorial position. This
8.53.
Increasing Stability
8.54.
a)
Cl
Cl
tertiary primary
b)
Cl Cl
secondary secondary
allylic
8.55.
8.56.
Br
Me
Me
H
Br
P
h
H
Me
Me H
P
h
observer
strong base
160
CHAPTER 8
8.57.
a) The rate of an E2 process is dependent on the concentrations of the substrate
8.58.
a) The rate of an E1 process is dependent only on the concentration of the
8.59.
Increasing Stability
major product
8.60. There are only two beta protons to abstract: one at C2 and the other at C4.
8.61.
a)
B
r
NaOH
major
CHAPTER 8
161
8.62.
H
2
O
heat
8.63.
NaOH
8.64. The reagent is a strong nucleophile and a strong base, so we expect a bimolecular
reaction. The substrate is tertiary so only E2 can operate (S
N
2 is too sterically hindered to
occur). There is only one possible regiochemical outcome for the E2 process, because
the other beta positions lack protons.
Br
NaOEt
162
CHAPTER 8
8.66.
a) one
8.67.
a)
E2
X
8.68.
a)
H O S
O
O
HO
- H
2
O
OH
OH
H H
HOH
CHAPTER 8
163
c)
Proton
Loss of
LG Proton
Transfer
8.69. The primary substrate will not undergo an E1 reaction because primary
carbocations are too high in energy to form readily.
a)
secondary
b)
tertiary
c)
primary
d)
secondary
8.70.
C
H
HO
H
H
δ
δδ
δ-
8.71. The first compound produces a trisubstituted alkene, while the second compound
produces a monosubstituted alkene. As such, the transition state for the reaction
of the first compound will be lower in energy than the transition state for the
reaction of the second compound.
8.72. The first reaction is very slow, because the tert-butyl group effectively locks the
164
CHAPTER 8
8.73. Pi bonds cannot be formed at the bridgehead of a bicyclic compound, unless one of
the rings is large (at least eight carbon atoms). This rule is known as Bredt’s rule.
8.74.
a) The first compound will react more rapidly because it is tertiary
8.75.
a) The Zaitsev product is desired, so sodium hydroxide should be used.
8.76. There is only one beta proton that can be abstracted so as to form the Zaitsev
8.77.
a)
OH
H O S
O
O
HO
HOH
CHAPTER 8
165
c)
OH
H O S
O
O
HO
- H
2
O
OH
H
H
HOH
8.78.
a)
Cl t-BuOK
major minor minor
++
166
CHAPTER 8
d)
O
T
s
NaCl
DMSO Cl
e)
Br NaOEt OEt
+ + +
major minor minor minor
g)
minor
+
h)
Br
NaOEt
EtOH
OEt
+
major minor
CHAPTER 8
167
j)
Br KOC(CH
3
)
3
+
major minor
8.79.
a)
EtO
Br
H
b) For an E2 process, the rate is dependent on the concentrations of the substrate
and the nucleophile: Rate = k[substrate][base]
168
CHAPTER 8
e)
δ
δδ
δ-
Br
8.80.
a)
Cl NaSH SH
e)
OTs
OK
f)
I
OK
CHAPTER 8
169
h)
OTs
NaOH
H
2
O
8.81.
Br
Br
Br
Br
Increasing reactivity towards E2
8.82.
OH
H O S
O
O
HO
HOH
170
CHAPTER 8
8.83.
a)
NaOEt
Br
H
Ph
Br
H
Ph
H
observer
rotate C-C bond H
H
Br
H
Ph
Ph
NaOEt
8.84.
OH
CHAPTER 8
171
8.85.
OTs
EtOH
O H
Et
EtOH
OEt
OTs
OEt
- TsO
8.86. The stereoisomer shown below does not readily undergo E2 elimination because
none of the chlorine atoms can be antiperiplanar to a beta proton in a chair conformation.
C
l
8.87. The first compound is a tertiary substrate. The second compound is a tertiary
allylic substrate. The latter will undergo E1 more rapidly because a tertiary allylic
