CHAPTER 5
STRUCTURE AND PREPARATION OF ALKENES:
ELIMINATION REACTIONS
SOLUTIONS TO TEXT PROBLEMS
5.1 (b) Writing the structure in more detail, we see that the longest continuous chain contains four
carbon atoms.
The double bond is located at the end of the chain, and so the alkene is named as a derivative
of 1-butene. Two methyl groups are substituents at C-3. The correct IUPAC name is 3,3-
dimethyl-1-butene.
(c) Expanding the structural formula reveals the molecule to be a methyl-substituted derivative of
hexene.
(d) In compounds containing a double bond and a halogen, the double bond takes precedence in
numbering the longest carbon chain.
Cl
CHCH
2
CHCH
3
CH
2
1 2 3 4 5
4-Chloro-1-pentene
CH
3
CH
3
CHCH
2
CH
2
CH
3
C
123 4 56
2-Methyl-2-hexene
CH
3
CH
CH
3
CH
3
CH
2
C
4 32 1
90
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(e) When a hydroxyl group is present in a compound containing a double bond, the hydroxyl
takes precedence over the double bond in numbering the longest carbon chain.
5.2 There are three sets of nonequivalent positions on a cyclopentene ring, identified as a, b, and c on
the cyclopentene structure shown:
Thus, there are three different monochloro-substituted derivatives of cyclopentene. The carbons that
bear the double bond are numbered C-1 and C-2 in each isomer, and the other positions are num-
bered in sequence in the direction that gives the chlorine-bearing carbon its lower locant.
5.3 (b) The alkene is a derivative of 3-hexene regardless of whether the chain is numbered from
left to right or from right to left. Number it in the direction that gives the lower number to the
substituent.
(c) There are only two sp
2
-hybridized carbons, the two connected by the double bond. All other
carbons (six) are sp
3
-hybridized.
(d) There are three sp
2
–sp
3
H9268 bonds and three sp
3
–sp
3
H9268 bonds.
5.4 Consider first the C
5
H
10
alkenes that have an unbranched carbon chain:
There are three additional isomers. These have a four-carbon chain with a methyl substituent.
5.5 First, identify the constitution of 9-tricosene. Referring back to Table 2.4 in Section 2.8 of the text,
we see that tricosane is the unbranched alkane containing 23 carbon atoms. 9-Tricosene, therefore,
contains an unbranched chain of 23 carbons with a double bond between C-9 and C-10. Since the
2-Methyl-1-butene 2-Methyl-2-butene 3-Methyl-1-butene
1-Pentene cis-2-Pentene trans-2-Pentene
5
4
3
2
1
6
3-Ethyl-3-hexene
Cl
1
2
3
4
5
Cl
2
1
5
4
3
Cl
1
2
3
4
5
1-Chlorocyclopentene 3-Chlorocyclopentene 4-Chlorocyclopentene
a
a
b
c
b
OH
CHCH
2
CHCH
3
CH
2
5 4 3 2 1
4-Penten-2-ol
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 91
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92 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
problem specifies that the pheromone has the cis configuration, the first 8 carbons and the last 13
must be on the same side of the C-9–C-10 double bond.
5.6 (b) One of the carbons of the double bond bears a methyl group and a hydrogen; methyl is of
higher rank than hydrogen. The other doubly bonded carbon bears the groups @CH
2
CH
2
F
and @CH
2
CH
2
CH
2
CH
3
. At the first point of difference between these two, fluorine is of
higher atomic number than carbon, and so @CH
2
CH
2
F is of higher precedence.
Higher ranked substituents are on the same side of the double bond; the alkene has the Z con-
figuration.
(c) One of the carbons of the double bond bears a methyl group and a hydrogen; as we have seen,
methyl is of higher rank. The other doubly bonded carbon bears @CH
2
CH
2
OH and @C(CH
3
)
3
.
Let’s analyze these two groups to determine their order of precedence.
We examine the atoms one by one at the point of attachment before proceeding down the
chain. Therefore, @C(CH
3
)
3
outranks @CH
2
CH
2
OH.
Higher ranked groups are on opposite sides; the configuration of the alkene is E.
(d) The cyclopropyl ring is attached to the double bond by a carbon that bears the atoms (C, C, H)
and is therefore of higher precedence than an ethyl group @C(C, H, H).
Higher ranked groups are on opposite sides; the configuration of the alkene is E.
5.7 A trisubstituted alkene has three carbons directly attached to the doubly bonded carbons. There are
three trisubstituted C
6
H
12
isomers, two of which are stereoisomers.
C C
CH
3
CH
3
CH
2
CH
3
H
C C
CH
3
H
CH
3
CH
2
CH
3
C C
CH
3
H
CH
2
CH
3
CH
3
(Z)-3-Methyl-2-pentene(E)-3-Methyl-2-pentene2-Methyl-2-pentene
C C
CH
3
CH
2
H
CH
3
Higher
Lower
Lower
Higher
CC
H
CH
3
CH
2
CH
2
OH
C(CH
3
)
3
Higher
Lower Higher
Lower
CH
2
CH
2
OH
C(C,H,H)
Lower priority
C(CH
3
)
3
C(C,C,C)
Higher priority
CC
H
CH
3
CH
2
CH
2
F
CH
2
CH
2
CH
2
CH
3
Higher
Lower
Higher
Lower
CC
H
CH
3
(CH
2
)
7
H
(CH
2
)
12
CH
3
cis-9-Tricosene
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5.8 The most stable C
6
H
12
alkene has a tetrasubstituted double bond:
5.9 Apply the two general rules for alkene stability to rank these compounds. First, more highly substi-
tuted double bonds are more stable than less substituted ones. Second, when two double bonds are
similarly constituted, the trans stereoisomer is more stable than the cis. The predicted order of de-
creasing stability is therefore:
5.10 Begin by writing the structural formula corresponding to the IUPAC name given in the problem. A
bond-line depiction is useful here.
The alkene is extremely crowded and destabilized by van der Waals strain. Bulky tert-butyl groups
are cis to one another on each side of the double bond. Highly strained compounds are often quite
difficult to synthesize, and this alkene is a good example.
5.11 Use the zigzag arrangement of bonds in the parent skeleton figure to place E and Z bonds as appro-
priate for each part of the problem. From the sample solution to parts (a) and (b), the ring carbons
have the higher priorities. Thus, an E double bond will have ring carbons arranged and a
Z double bond .
5.12 Write out the structure of the alcohol, recognizing that the alkene is formed by loss of a hydrogen
and a hydroxyl group from adjacent carbons.
H
H
CH
3
CH
3
1
2
H
H
2
35
4
1
3
(Z)-3-Methylcyclodecene
(E)-3-Methylcyclodecene (E)-5-Methylcyclodecene
(Z)-5-Methylcyclodecene
H
1
2
3
4
5
H
CH
3
H
HCH
3
2
3
1
(c)
(d)
(e)
( f )
3,4-Di-tert-butyl-2,2,5,5-tetramethyl-3-hexene
CC
CH
3
CH
3
CH
3
H
2-Methyl-2-butene
(trisubstituted):
most stable
CC
CH
3
CH
2
CH
3
H
H
(E)-2-Pentene
(disubstituted)
CC
CH
3
CH
2
CH
3
HH
(Z)-2-Pentene
(disubstituted)
CC
CH
2
CH
2
CH
3
HH
H
1-Pentene
(monosubstituted):
least stable
CC
CH
3
CH
3
CH
3
CH
3
2,3-Dimethyl-2-butene
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 93
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(b, c) Both 1-propanol and 2-propanol give propene on acid-catalyzed dehydration.
(d) Carbon-3 has no hydrogens in 2,3,3-trimethyl-2-butanol. Elimination can involve only the
hydroxyl group at C-2 and a hydrogen at C-1.
5.13 (b) Elimination can involve loss of a hydrogen from the methyl group or from C-2 of the ring in
1-methylcyclohexanol.
According to the Zaitsev rule, the major alkene is the one corresponding to loss of a hydrogen
from the alkyl group that has the smaller number of hydrogens. Thus hydrogen is removed
from the methylene group in the ring rather than from the methyl group, and 1-methylcyclo-
hexene is formed in greater amounts than methylenecyclohexane.
(c) The two alkenes formed are as shown in the equation.
The more highly substituted alkene is formed in greater amounts, as predicted by Zaitsev’s
rule.
5.14 2-Pentanol can undergo dehydration in two different directions, giving either 1-pentene or
2-pentene. 2-Pentene is formed as a mixture of the cis and trans stereoisomers.
H11001
1-Pentene cis-2-Pentene
CH
2
CHCH
2
CH
2
CH
3
CC
CH
3
H
CH
2
CH
3
H
trans-2-Pentene
CC
CH
3
H
H
CH
2
CH
3
H11001
H
H11001
heat
2-Pentanol
OH
CH
3
CHCH
2
CH
2
CH
3
OH
HH
Compound has a
trisubstituted
double bond
Compound has a
tetrasubstituted
double bond;
more stable
H11001
H11002H
2
O
H11002H
2
O
H11001
H
3
COH
1-Methylcyclohexanol
CH
2
Methylenecyclohexane
(a disubstituted alkene;
minor product)
CH
3
1-Methylcyclohexene
(a trisubstituted alkene;
major product)
H
H11001
heat
2,3,3-Trimethyl-2-butanol
No hydrogens on this H9252 carbon
2,3,3-Trimethyl-1-butene
C
HO
CH
3
CH
3
H
3
C C CH
3
H9252
H9252
H9251
H9252
C CH
3
C
CH
3
CH
3
CH
3
CH
3
H
2
C
H
H11001
heat
H
H11001
heat
1-Propanol Propene
CH
3
CH CH
2
CH
3
CH
2
CH
2
OH
H9251
H9252
2-Propanol
CH
3
CHCH
3
H9252H9252
H9251
OH
94 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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5.15 (b) The site of positive charge in the carbocation is the carbon atom that bears the hydroxyl group
in the starting alcohol.
Water may remove a proton from the methyl group, as shown in the following equation:
Loss of a proton from the ring gives the major product 1-methylcyclohexene.
(c) Loss of the hydroxyl group under conditions of acid catalysis yields a tertiary carbocation.
Water may remove a proton from an adjacent methylene group to give a trisubstituted alkene.
Removal of the methine proton gives a tetrasubstituted alkene.
5.16 In writing mechanisms for acid-catalyzed dehydration of alcohols, begin with formation of the
carbocation intermediate:
CH
3
CH
3
CH
3
H
H11001
CH
3
OH
H
2,2-Dimethylcyclohexanol 2,2-Dimethylcyclohexyl cation
H
H11001
H11002H
2
O
H11001
H
H
3
O
H11001
OH
2
H11001
H
H11001
H
H
HH
H
3
O
H11001
H
2
O
H11001
OH
HH
H11001
H
2
SO
4
H11002H
2
O
CH
3
H
H
H11001
1-Methylcyclohexene
H
3
O
H11001
CH
3
H
H
2
O
H11001
Methylenecyclohexane
H
3
O
H11001
CH
2
CH
2
H
2
OH
H11001
H11001
1-Methylcyclohexanol
H
H11001
CH
3
H
2
OH11001
H11001
H
3
COH
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 95
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This secondary carbocation can rearrange to a more stable tertiary carbocation by a methyl group
shift.
Loss of a proton from the 1,2-dimethylcyclohexyl cation intermediate yields 1,2-dimethylcyclo-
hexene.
5.17 (b) All the hydrogens of tert-butyl chloride are equivalent. Loss of any of these hydrogens along
with the chlorine substituent yields 2-methylpropene as the only alkene.
(c) All the H9252 hydrogens of 3-bromo-3-ethylpentane are equivalent, so that H9252-elimination can give
only 3-ethyl-2-pentene.
(d) There are two possible modes of H9252-elimination from 2-bromo-3-methylbutane. Elimination in
one direction provides 3-methyl-1-butene; elimination in the other gives 2-methyl-2-butene.
The major product is the more highly substituted alkene, 2-methyl-2-butene. It is the more
stable alkene and corresponds to removal of a hydrogen from the carbon that has the fewer
hydrogens.
(e) Regioselectivity is not an issue here, because 3-methyl-1-butene is the only alkene that can be
formed by H9252-elimination from 1-bromo-3-methylbutane.
1-Bromo-3-methylbutene 3-Methyl-1-butene
BrCH
2
CH
2
CH(CH
3
)
2
H9252
CH
2
CHCH(CH
3
)
2
2-Bromo-3-methylbutane 3-Methyl-1-butene
(monosubstituted)
2-Methyl-2-butene
(trisubstituted)
CH
3
CHCH(CH
3
)
2
H9252H9252
Br
CH
2
CHCH(CH
3
)
2
CH
3
CH C(CH
3
)
2
H11001
3-Bromo-3-ethylpentane 3-Ethyl-2-pentene
CH
3
CH
2
CH
2
CH
3
CH
2
CH
3
BrC
H9252
H9252
H9252
CH
3
CH
CH
2
CH
3
CH
2
CH
3
C
C
H
3
C
H
3
C
CH
2
CH
3
CCl
CH
3
CH
3
tert-Butyl chloride 2-Methylpropene
H
3
O
H11001
OH
2
H11002H
H11001
H11001
CH
3
CH
3
CH
3
CH
3
H11001
H
1,2-Dimethylcyclohexyl cation 1,2-Dimethylcyclohexene
CH
3
CH
3
CH
3
H
H11001
H11001
CH
3
H
2,2-Dimethylcyclohexyl cation
(secondary)
1,2-Dimethylcyclohexyl cation
(tertiary)
96 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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( f ) Two alkenes may be formed here. The more highly substituted one is 1-methylcyclohexene,
and this is predicted to be the major product in accordance with Zaitsev’s rule.
5.18 Elimination in 2-bromobutane can take place between C-1 and C-2 or between C-2 and C-3. There
are three alkenes capable of being formed: 1-butene and the stereoisomers cis-2-butene and trans-
2-butene.
As predicted by Zaitsev’s rule, the most stable alkene predominates. The major product is trans-
2-butene.
5.19 An unshared electron pair of the base methoxide (CH
3
O
H11002
) abstracts a proton from carbon. The pair
of electrons in this C@H bond becomes the H9266 component of the double bond of the alkene. The pair
of electrons in the C@Cl bond becomes an unshared electron pair of chloride ion.
5.20 The most stable conformation of cis-4-tert-butylcyclohexyl bromide has the bromine substituent in
an axial orientation. The hydrogen that is removed by the base is an axial proton at C-2. This
hydrogen and the bromine are anti periplanar to each other in the most stable conformation.
5.21 (a) 1-Heptene is .
(b) 3-Ethyl-2-pentene is .
(c) cis-3-Octene is
(d) trans-1,4-Dichloro-2-butene is
C
ClCH
2
H
C
H
CH
2
Cl
C
CH
3
CH
2
H
C
CH
2
CH
2
CH
2
CH
3
H
CH
3
CH C(CH
2
CH
3
)
2
CH
2
CH(CH
2
)
4
CH
3
Br
H
(CH
3
)
3
C
(CH
3
)
3
C O
H11002
H
H
CC
CH
3
H CH
3
Cl
ClCH
3
OH H11001 H
2
C C(CH
3
)
2
H11001
H11002
CH
3
O
H11002
H11001
1-Butene cis-2-Butene
CH
2
CHCH
2
CH
3
CC
H
3
C
H
CH
3
H
trans-2-Butene
CC
H
3
C
H
H
CH
3
H11001
2-Bromobutane
Br
CH
3
CHCH
2
CH
3
H9252H9252
H
3
CI
1-Iodo-1-methylcyclohexane Methylenecyclohexane
(disubstituted)
CH
2
CH
3
1-Methylcyclohexene
(trisubstituted;
major product)
H11001
H9252H9252
H9252
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 97
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(e)(Z)-3-Methyl-2-hexene is
( f )(E)-3-Chloro-2-hexene is
(g) 1-Bromo-3-methylcyclohexene is
(h) 1-Bromo-6-methylcyclohexene is
(i) 4-Methyl-4-penten-2-ol is
( j) Vinylcycloheptane is
(k) An allyl group is @CH
2
CH?CH
2
. 1,1-Diallylcyclopropane is
(l) An isopropenyl substituent is . trans-1-Isopropenyl-3-methylcyclohexane is
therefore
5.22 Alkenes with tetrasubstituted double bonds have four alkyl groups attached to the doubly bonded
carbons. There is only one alkene of molecular formula C
7
H
14
that has a tetrasubstituted double
bond, 2,3-dimethyl-2-pentene.
5.23 (a) The longest chain that includes the double bond in (CH
3
CH
2
)
2
C?CHCH
3
contains five
carbon atoms, and so the parent alkene is a pentene. The numbering scheme that gives the
double bond the lowest number is
The compound is 3-ethyl-2-pentene.
H
54 1
32
C
H
3
C
H
3
C
C
CH
3
CH
2
CH
3
2,3-Dimethyl-2-pentene
CH
3
CH
2
CH
3
CCH
2
CH
3
H
CH
3
CHCH
2
C CH
2
CH
3
OH
Br
CH
3
Br
H
3
C
C
H
3
C
H
C
CH
2
CH
2
CH
3
Cl
C
H
3
C
H
C
CH
2
CH
2
CH
3
CH
3
98 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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(b) Write out the structure in detail, and identify the longest continuous chain that includes the
double bond.
The longest chain contains six carbon atoms, and the double bond is between C-3 and C-4.
The compound is named as a derivative of 3-hexene. There are ethyl substituents at C-3 and
C-4. The complete name is 3,4-diethyl-3-hexene.
(c) Write out the structure completely.
The longest carbon chain contains four carbons. Number the chain so as to give the lowest
numbers to the doubly bonded carbons, and list the substituents in alphabetical order. This
compound is 1,1-dichloro-3,3-dimethyl-1-butene.
(d ) The longest chain has five carbon atoms, the double bond is at C-1, and there are two methyl
substituents. The compound is 4,4-dimethyl-1-pentene.
(e) We number this trimethylcyclobutene derivative so as to provide the lowest number for the
substituent at the first point of difference. We therefore number
The correct IUPAC name is 1,4,4-trimethylcyclobutene, not 2,3,3-trimethylcyclobutene.
( f ) The cyclohexane ring has a 1,2-cis arrangement of vinyl substituents. The compound is
cis-1,2-divinylcyclohexane.
(g) Name this compound as a derivative of cyclohexene. It is 1,2-divinylcyclohexene.
5.24 (a) Go to the end of the name, because this tells you how many carbon atoms are present in the
longest chain. In the hydrocarbon name 2,6,10,14-tetramethyl-2-pentadecene, the suffix
“2-pentadecene” reveals that the longest continuous chain has 15 carbon atoms and that there
H
H
H
3
C
H
3
C
H
3
C
1
4
2
3
H
3
C
H
3
C
H
3
C
2
3
1
4
rather than
1
2
3
4
5
C
H
3
C
H
3
C
CH
3
C
H
C
Cl
Cl
3
4 2 1
C
CH
3
CH
2
CH
2
CH
3
CH
3
CH
2
CH
2
CH
3
C
1 2 56
34
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 99
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is a double bond between C-2 and C-3. The rest of the name provides the information that
there are four methyl groups and that they are located at C-2, C-6, C-10, and C-14.
(b) An allyl group is CH
2
?CHCH
2
@. Allyl isothiocyanate is therefore CH
2
?CHCH
2
N?C?S.
5.25 (a) The E configuration means that the higher priority groups are on opposite sides of the double
bond.
(b) Geraniol has two double bonds, but only one of them, the one between C-2 and C-3, is capa-
ble of stereochemical variation. Of the groups at C-2, CH
2
OH is of higher priority than H. At
C-3, CH
2
CH
2
outranks CH
3
. Higher priority groups are on opposite sides of the double bond
in the E isomer; hence geraniol has the structure shown.
(c) Since nerol is a stereoisomer of geraniol, it has the same constitution and differs from gera-
niol only in having the Z configuration of the double bond.
(d) Beginning at the C-6, C-7 double bond, we see that the propyl group is of higher priority than
the methyl group at C-7. Since the C-6, C-7 double bond is E, the propyl group must be on the
opposite side of the higher priority group at C-6, where the CH
2
fragment has a higher prior-
ity than hydrogen. We therefore write for the stereochemistry of the C-6, C-7 double bond as:
At C-2, CH
2
OH is of higher priority than H; and at C-3, CH
2
CH
2
C@ is of higher priority than
CH
2
CH
3
. The double-bond configuration at C-2 is Z. Therefore
CC
CH
3
CH
2
CCH
2
CH
2
H
CH
2
OHHigher
Lower
Higher
Lower
32
456 1
E
CC
H
3
C
CH
3
CH
2
CH
2
H
CH
2
Higher
Lower
Higher
Lower
76
8910
5
CC
H
3
C
CHCH
2
CH
2
(CH
3
)
2
C
H
CH
2
OH
Nerol
CC
H
3
C
CHCH
2
CH
2
(CH
3
)
2
C H
CH
2
OH
Geraniol
CC
H
CH
3
CH
2
H
CH
2
CH
2
CH
2
CH
2
CH
2
OH
Higher
Lower Higher
Lower
(E)-6-Nonen-1-ol
2,6,10,14-Tetramethyl-2-pentadecene
100 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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Combining the two partial structures, we obtain for the full structure of the codling moth’s sex
pheromone
The compound is (2Z,6E)-3-ethyl-7-methyl-2,6-decadien-1-ol.
(e) The sex pheromone of the honeybee is (E)-9-oxo-2-decenoic acid, with the structure
( f ) Looking first at the C-2, C-3 double bond of the cecropia moth’s growth hormone
we find that its configuration is E, since the higher priority groups are on opposite sides of the
double bond.
The configuration of the C-6, C-7 double bond is also E.
5.26 We haven’t covered, and won’t cover, how to calculate the size of a dipole moment, but we can de-
cide whether a compound has a dipole moment or not. Only compound B has dipole moment. The
individual bond dipoles in A, C, and D cancel; therefore, none of these three has a dipole moment.
5.27 The alkenes are listed as follows in order of decreasing heat of combustion:
(e) 2,4,4-Trimethyl-2-pentene; 5293 kJ/mol (1264.9 kcal/mol). Highest
heat of combustion because it is C
8
H
16
; all others are C
7
H
14
.
(a) 1-Heptene; 4658 kJ/mol (1113.4 kcal/mol). Monosubstituted double
bond; therefore least stable C
7
H
14
isomer.
H
3
C
H
3
C
CH
3
CH
3
CC
A (H9262 H11005 0 D)
Cl
H
3
C
CH
3
Cl
CC
C (H9262 H11005 0 D)
Cl
Cl Cl
Cl
CC
D (H9262 H11005 0 D)
H
3
C
Cl
CH
3
Cl
CC
B (has a dipole moment)
CC
CH
2
CH
2
C
CH
3
CH
2
H
CH
2
CH
2
Higher
Lower Higher
Lower
76
8910
54
CC
CH
2
CH
2
H
3
C
H
CO
2
CH
3
Higher
Lower Higher
Lower
32
45
1
H
3
C
CH
3
CH
2
CH
3
CH
2
CO
2
CH
3
CH
3
HH
O
6
3
2
7
CC
H
CH
3
C(CH
2
)
4
CH
2
H
CO
2
H
Higher
Lower
Higher
Lower
O
CC
CH
3
CH
2
CH
2
CH
2
H
CH
2
OH
CC
H
3
C
CH
3
CH
2
CH
2
H
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 101
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(d)(Z)-4,4-Dimethyl-2-pentene; 4650 kJ/mol (1111.4 kcal/mol). Disubsti-
tuted double bond, but destabilized by van der Waals strain.
(b) 2,4-Dimethyl-1-pentene; 4638 kJ/mol (1108.6 kcal/mol). Disubstituted
double bond.
(c) 2,4-Dimethyl-2-pentene; 4632 kJ/mol (1107.1 kcal/mol). Trisubstituted
double bond.
5.28 (a) 1-Methylcyclohexene is more stable; it contains a trisubstituted double bond, whereas
3-methylcyclohexene has only a disubstituted double bond.
(b) Both isopropenyl and allyl are three-carbon alkenyl groups: isopropenyl is CH
2
?CCH
3
, allyl
is CH
2
?CHCH
2
@.
g
Isopropenylcyclopentane has a disubstituted double bond and so is predicted to be more sta-
ble than allylcyclopentane, in which the double bond is monosubstituted.
(c) A double bond in a six-membered ring is less strained than a double bond in a four-membered
ring; therefore bicyclo[4.2.0]oct-3-ene is more stable.
(d) Cis double bonds are more stable than trans double bonds when the ring is smaller than
11-membered. (Z)-Cyclononene has a cis double bond in a 9-membered ring, and is thus more
stable than (E)-cyclononene.
(e) Trans double bonds are more stable than cis when the ring is large. Here the rings are 18-mem-
bered, so that (E)-cyclooctadecene is more stable than (Z)-cyclooctadecene.
H
HH
H
(E)-Cyclooctadecene (Z)-Cyclooctadecene
more stable than
(Z)-Cyclononene (E)-Cyclononene
more stable than
more stable than
Bicyclo[4.2.0]oct-3-ene Bicyclo[4.2.0]oct-7-ene
C
CH
2
CH
3
CH
2
CH CH
2
Isopropenylcyclopentane Allylcyclopentane
CH
3
CH
3
1-Methylcyclohexene 3-Methylcyclohexene
more stable than
H H
102 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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5.29 (a) Carbon atoms that are involved in double bonds are sp
2
-hybridized, with ideal bond angles of
120°. Incorporating an sp
2
-hybridized carbon into a three-membered ring leads to more angle
strain than incorporation of an sp
3
-hybridized carbon. 1-Methylcyclopropene has two sp
2
-
hybridized carbons in a three-membered ring and so has substantially more angle strain than
methylenecyclopropane.
The higher degree of substitution at the double bond in 1-methylcyclopropene is not sufficient
to offset the increased angle strain, and so 1-methylcyclopropene is less stable than methyl-
enecyclopropane.
(b) 3-Methylcyclopropene has a disubstituted double bond and two sp
2
-hybridized carbons in its
three-membered ring. It is the least stable of the isomers.
5.30 In all parts of this exercise, write the structure of the alkyl halide in sufficient detail to identify the
carbon that bears the halogen and the H9252-carbon atoms that bear at least one hydrogen. These are the
carbons that become doubly bonded in the alkene product.
(a) 1-Bromohexane can give only 1-hexene under conditions of E2 elimination.
(b) 2-Bromohexane can give both 1-hexene and 2-hexene on dehydrobromination. The 2-hexene
fraction is a mixture of cis and trans stereoisomers.
(c) Both a cis–trans pair of 2-hexenes and a cis–trans pair of 3-hexenes are capable of being
formed from 3-bromohexane.
base
E2
base
E2
3-Bromohexane
cis-2-Hexene trans-2-Hexene
H11001CC
H H
CH
2
CH
2
CH
3
H
3
C
CC
H
H
CH
2
CH
2
CH
3
H
3
C
cis-3-Hexene trans-3-Hexene
H11001CC
H H
CH
2
CH
3
CH
3
CH
2
CC
H
H
CH
2
CH
3
CH
3
CH
2
CH
3
CH
2
CHCH
2
CH
2
CH
3
Br
H9252H9252
H11001
1-Hexene cis-2-Hexene
CH
2
CHCH
2
CH
2
CH
2
CH
3
CC
H
3
C
H
CH
2
CH
2
CH
3
H
trans-2-Hexene
CC
H
3
C
H
H
CH
2
CH
2
CH
3
H11001
base
E2
2-Bromohexane
Br
CH
3
CHCH
2
CH
2
CH
2
CH
3
H9252H9252
base
E2
1-Bromohexane 1-Hexene (only alkene)
CH
2
CHCH
2
CH
2
CH
2
CH
3
BrCH
2
CH
2
CH
2
CH
2
CH
2
CH
3
H9252
CH
3
3-Methylcyclopropene
CH
3
CH
2
1-Methylcyclopropene Methylenecyclopropane
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 103
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(d) Dehydrobromination of 2-bromo-2-methylpentane can involve one of the hydrogens of either
a methyl group (C-1) or a methylene group (C-3).
Neither alkene is capable of existing in stereoisomeric forms, and so these two are the only
products of E2 elimination.
(e) 2-Bromo-3-methylpentane can undergo dehydrohalogenation by loss of a proton from either
C-1 or C-3. Loss of a proton from C-1 gives 3-methyl-1-pentene.
Loss of a proton from C-3 gives a mixture of (E)- and (Z)-3-methyl-2-pentene.
( f ) Three alkenes are possible from 3-bromo-2-methylpentane. Loss of the C-2 proton gives
2-methyl-2-pentene.
Abstraction of a proton from C-4 can yield either (E)- or (Z)-4-methyl-2-pentene.
(g) Proton abstraction from the C-3 methyl group of 3-bromo-3-methylpentane yields 2-ethyl-
1-butene.
E2
2-Ethyl-1-butene3-Bromo-3-methylpentane
CH
3
CH
2
CCH
2
CH
3
Br
CH
2
H
C
CH
3
CH
2
CH
2
CH
3
CH
2
Base:
E2
3-Bromo-2-methylpentane (E)-4-Methyl-2-pentene
:Base
(CH
3
)
2
CHCH
Br
H
CHCH
3
H11001CC
H
(CH
3
)
2
CH
CH
3
H
(Z)-4-Methyl-2-pentene
CC
H
(CH
3
)
2
CH CH
3
H
2-Methyl-2-pentene
E2
3-Bromo-2-methylpentane
Base:
CH
3
C
BrCH
3
H
CHCH
2
CH
3
C CHCH
2
CH
3
CH
3
CH
3
(E)-3-Methyl-2-pentene
CC
H
3
C
H
CH
3
CH
2
CH
3
(Z)-3-Methyl-2-pentene
CC
H
3
C
H
CH
2
CH
3
CH
3
H11001
base
E2
2-Bromo-3-methylpentane
Br
CH
3
CHCHCH
2
CH
3
CH
3
3-Methyl-1-pentene
base
E2
2-Bromo-3-methylpentane
CH
2
CHCHCH
2
CH
3
CH
3
Base:
H
CH
2
CHCHCH
2
CH
3
Br
CH
3
2-Bromo-2-methylpentane 2-Methyl-1-pentene 2-Methyl-2-pentene
CH
2
CH
3
CH
2
CH
2
CH
3
C
base
E2
Br
CH
3
CH
3
CCH
2
CH
2
CH
3
H11001 (CH
3
)
2
C CHCH
2
CH
3
104 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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Stereoisomeric 3-methyl-2-pentenes are formed by proton abstraction from C-2.
(h) Only 3,3-dimethyl-1-butene may be formed under conditions of E2 elimination from
3-bromo-2,2-dimethylbutane.
5.31 (a) The reaction that takes place with 1-bromo-3,3-dimethylbutane is an E2 elimination involving
loss of the bromine at C-1 and abstraction of the proton at C-2 by the strong base potassium
tert-butoxide, yielding a single alkene.
(b) Two alkenes are capable of being formed in this H9252-elimination reaction.
The more highly substituted alkene is 1-methylcyclopentene; it is the major product of this
reaction. According to Zaitsev’s rule, the major alkene is formed by proton removal from the
H9252 carbon that has the fewest hydrogens.
(c) Acid-catalyzed dehydration of 3-methyl-3-pentanol can lead either to 2-ethyl-1-butene or to a
mixture of (E)- and (Z)-3-methyl-2-pentene.
The major product is a mixture of the trisubstituted alkenes, (E)- and (Z)-3-methyl-2-
pentene. Of these two stereoisomers the E isomer is slightly more stable and is expected to
predominate.
(E)-3-Methyl-2-pentene
CC
H
CH
3
CH
2
CH
3
H
3
C
(Z)-3-Methyl-2-pentene
CC
H
CH
2
CH
3
CH
3
H
3
C
2-Ethyl-1-butene
C
CH
3
CH
2
CH
2
CH
3
CH
2
3-Methyl-3-pentanol
CH
3
CH
2
CCH
2
CH
3
CH
3
OH
H
2
SO
4
80H11034C
H11001H11001
H9252H9252
H9252
CH
2
CH
3
H11001
H
H
H9252
H
H
H9252
ClCH
3
H9252
NaOCH
2
CH
3
ethanol
70°C
1-Methylcyclopentyl chloride Methylenecyclopentane 1-Methylcyclopentene
3,3-Dimethyl-1-butene
E2
1-Bromo-3,3-dimethylbutane
(CH
3
)
3
CCH CH
2
(CH
3
)
3
CO
H
(CH
3
)
3
CCH CH
2
Br
H11002
E2
3-Bromo-2,2-dimethylbutane
C
BrCH
3
CH
2 H
CH
3
CH CH
2
3,3-Dimethyl-1-butene
C
CH
3
CH
3
CH
3
CH CH
2
Base
E2
3-Bromo-3-methylpentane (E)-3-Methyl-2-pentene
CH
Br
CH
3
H
CH
3
CCH
2
CH
3
CC
H
3
C
H
CH
3
CH
2
CH
3
(Z)-3-Methyl-2-pentene
CC
H
3
C
H
CH
2
CH
3
CH
3
H11001
Base:
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 105
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(d) Acid-catalyzed dehydration of 2,3-dimethyl-2-butanol can proceed in either of two directions.
The major alkene is the one with the more highly substituted double bond, 2,3-dimethyl-2-
butene. Its formation corresponds to Zaitsev’s rule in that a proton is lost from the H9252 carbon
that has the fewest hydrogens.
(e) Only a single alkene is capable of being formed on E2 elimination from this alkyl iodide.
Stereoisomeric alkenes are not possible, and because all the H9252 hydrogens are equivalent,
regioisomers cannot be formed either.
( f ) Despite the structural similarity of this alcohol to the alkyl halide in the preceding part of this
problem, its dehydration is more complicated. The initially formed carbocation is secondary
and can rearrange to a more stable tertiary carbocation by a hydride shift.
The tertiary carbocation, once formed, can give either 2,4-dimethyl-1-pentene or
2,4-dimethyl-2-pentene by loss of a proton.
The proton is lost from the methylene group in preference to the methyl group. The major
alkene is the more highly substituted one, 2,4-dimethyl-2-pentene.
5.32 In all parts of this problem you need to reason backward from an alkene to an alkyl bromide of mo-
lecular formula C
7
H
13
Br that gives only the desired alkene under E2 elimination conditions. Recall
that the carbon–carbon double bond is formed by loss of a proton from one of the carbons that be-
comes doubly bonded and a bromine from the other.
(a) Cycloheptene is the only alkene formed by an E2 elimination reaction of cycloheptyl bromide.
Br
Cycloheptyl bromide Cycloheptene
base
E2
2,4-Dimethyl-1-pentene
(disubstituted)
CH
3
CH
3
CH
3
CCH
2
CHCH
3
H11001
H11001
CH
3
CCH
2
CH(CH
3
)
2
H
2
C
2,4-Dimethyl-2-pentene
(trisubstituted)
CHCH(CH
3
)
2
(CH
3
)
2
C
H
H11001
hydride
H11002H
2
O shift
2,4-Dimethyl-3-pentanol Secondary carbocation
(less stable)
Tertiary carbocation
(more stable)
CH
3
CHCHCHCH
3
CH
3
CH
3
OH
CH
3
CH
3
H
CH
3
C CHCHCH
3
H11001
CH
3
CH
3
CH
3
CCH
2
CHCH
3
H11001
NaOCH
2
CH
3
ethanol
70°C
3-Iodo-2,4-dimethylpentane 2,4-Dimethyl-2-pentene
CH
3
CHCHCHCH
3
CH
3
CH
3
I
H9252 H9252
(CH
3
)
2
C CHCH(CH
3
)
2
H
3
PO
4
120°C
H11002H
2
O
2,3-Dimethyl-2-butanol 2,3-Dimethyl-1-butene
(disubstituted)
2,3-Dimethyl-2-butene
(tetrasubstituted)
CH
3
OH
CH
3
CH
3
CHCH
3
C
H9252
H9252
CH
2
CH
3
CH(CH
3
)
2
C (CH
3
)
2
C C(CH
3
)
2
H11001
106 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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(b) (Bromomethyl)cyclohexane is the correct answer. It gives methylenecyclohexane as the only
alkene under E2 conditions.
1-Bromo-1-methylcyclohexane is not correct. It gives a mixture of 1-methylcyclohexene and
methylenecyclohexane on elimination.
(c) In order for 4-methylcyclohexene to be the only alkene, the starting alkyl bromide must be
1-bromo-4-methylcyclohexane. Either the cis or the trans isomer may be used, although the
cis will react more readily, as the more stable conformation (equatorial methyl) has an axial
bromine.
1-Bromo-3-methylcyclohexane is incorrect; its dehydrobromination yields a mixture of 3-
methylcyclohexene and 4-methylcyclohexene.
(d) The bromine must be at C-2 in the starting alkyl bromide for a single alkene to be formed on
E2 elimination.
If the bromine substituent were at C-3, a mixture of 3,3-dimethyl- and 4,4-dimethylcyclopen-
tene would be formed.
Br
CH
3
CH
3
CH
3
CH
3
CH
3
CH
3
base
E2
H11001
3-Bromo-1,1-dimethylcyclopentane 3,3-Dimethylcyclopentene 4,4-Dimethylcyclopentene
2-Bromo-1,1-dimethylcyclopentane 3,3-Dimethylcyclopentene
base
E2
H
H
Br
CH
3
CH
3
CH
3
CH
3
Base:
Br
H
CH
3
(H)
CH
3
H(CH
3
)
H
1-Bromo-3-methylcyclohexene 3-Methylcyclohexene
CH
3
4-Methylcyclohexene
base
E2
H11001
H
H
CH
3
(H)
CH
3
H(CH
3
)
Br
cis- or trans-1-Bromo-4-methylcyclohexane 4-Methylcyclohexene
base
E2
Br
CH
3
CH
2
base
E2
1-Bromo-1-methylcyclohexane Methylenecyclohexane
CH
3
1-Methylcyclohexene
H11001
H
CH
2
CH
2
Br
base
E2
(Bromomethyl)cyclohexane Methylenecyclohexane
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 107
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(e) The alkyl bromide must be primary in order for the desired alkene to be the only product of
E2 elimination.
If 1-cyclopentylethyl bromide were used, a mixture of regioisomeric alkenes would be
formed, with the desired vinylcyclopentane being the minor component of the mixture.
( f ) Either cis- or trans-1-bromo-3-isopropylcyclobutane would be appropriate here.
(g) The desired alkene is the exclusive product formed on E2 elimination from 1-bromo-1-tert-
butylcyclopropane.
5.33 (a) Both 1-bromopropane and 2-bromopropane yield propene as the exclusive product of E2
elimination.
(b) Isobutene is formed on dehydrobromination of either tert-butyl bromide or isobutyl bromide.
(c) A tetrabromoalkane is required as the starting material to form a tribromoalkene under
E2 elimination conditions. Either 1,1,2,2-tetrabromoethane or 1,1,1,2-tetrabromoethane is
satisfactory.
Br
2
CHCHBr
2
1,1,2,2-Tetrabromoethane 1,1,1,2-Tetrabromoethane 1,1,2-Tribromoethene
BrCH
2
CBr
3
BrCH CBr
2
or
E2
base
(CH
3
)
3
CBr
tert-Butyl bromide Isobutyl bromide 2-Methylpropene
(CH
3
)
2
CHCH
2
Br (CH
3
)
2
C CH
2
or
E2
base
base
E2
CH
3
CH
2
CH
2
Br or CH
3
CHCH
3
Br
1-Bromopropane 2-Bromopropane Propene
CH
3
CH CH
2
base
E2
H
H
H
Br
H
C(CH
3
)
3
C(CH
3
)
3
1-Bromo-1-tert-butylcyclopropane 1-tert-Butylcyclopropene
base
E2
H
H
H
Br
H
CH(CH
3
)
2
CH(CH
3
)
2
cis- or trans-1-Bromo-3-isopropylcyclobutane 3-Isopropylcyclobutene
base
E2
CHCH
3
CHCH
3
CH CH
2
H11001
Br
1-Cyclopentylethyl bromide Ethylidenecyclopentane
(major product)
Vinylcyclopentane
(minor product)
base
E2
CH
2
CH
2
Br CH
2-Cyclopentylethyl bromide Vinylcyclopentane
CH
2
108 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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(d) The bromine substituent may be at either C-2 or C-3.
5.34 (a) The isomeric alkyl bromides having the molecular formula C
5
H
11
Br are:
(b) The order of reactivity toward E1 elimination parallels carbocation stability and is tertiary H11022
secondary H11022 primary. The tertiary bromide 2-bromo-2-methylbutane will undergo E1 elimi-
nation at the fastest rate.
(c) 1-Bromo-2,2-dimethylpropane has no hydrogens on the H9252 carbon and so cannot form an
alkene by an E2 process.
The only available pathway is E1 with rearrangement.
(d) Only the primary bromides will give a single alkene on E2 elimination.
base
E2
1-Bromo-3-methylbutane 3-Methyl-1-butene
CH
3
CHCH
2
CH
2
Br
CH
3
CH
3
CHCH CH
2
CH
3
base
E2
1-Bromo-2-methylbutane 2-Methyl-1-butene
CH
3
CH
2
CHCH
2
Br
CH
3
CH
3
CH
2
C CH
2
CH
3
CH
3
CH
2
CH
2
CHCH
3
CH
2
CH
2
CH
2
CH
2
Br
base
E2
CH
2
1-Bromopentane 1-Pentene
H9252 carbon has no hydrogens
CH
3
C
CH
3
CH
3
CH
2
Br
2-Bromo-2-methylbutane
CH
3
CCH
2
CH
3
CH
3
Br
1-Bromo-2,2-dimethylpropane
CH
3
CCH
2
Br
CH
3
CH
3
1-Bromo-2-methylbutane 1-Bromo-3-methylbutane 2-Bromo-3-methylbutane
CH
3
CHCH
2
CH
2
Br
CH
3
CH
3
CH
2
CHCH
2
Br
CH
3
CH
3
CH
BrCH
3
CHCH
3
CH
3
CH
2
CH
2
CH
2
CH
2
Br
1-Bromopentane 2-Bromopentane 3-Bromopentane
CH
3
CH
2
CHCH
2
CH
3
Br
CH
3
CH
2
CH
2
CHCH
3
Br
HH
Br
CH
3
CH
3
CH
3
CH
3
or
H
Br
CH
3
CH
3
H
2-Bromo-1,1-dimethylcyclobutane 3-Bromo-1,1-dimethylcyclobutane 3,3-Dimethylcyclobutene
base
E2
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 109
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(e) Elimination in 3-bromopentane will give the stereoisomers (E)- and (Z)-2-pentene.
( f ) Three alkenes can be formed from 2-bromopentane.
5.35 (a) The isomeric C
5
H
12
O alcohols are:
(b) The order of reactivity in alcohol dehydration parallels carbocation stability and is tertiary H11022
secondary H11022 primary. The only tertiary alcohol in the group is 2-methyl-2-butanol. It will
dehydrate fastest.
(c) The most stable C
5
H
11
carbocation is the tertiary carbocation.
(d) A proton may be lost from C-1 or C-3:
CH
3
CH
3
CCH
2
CH
3
H11001
CH
3
H
2
C CCH
2
CH
3
CH
3
CH
3
C CHCH
3
H11001
1,1-Dimethylpropyl
cation
2-Methyl-1-butene
(minor alkene)
2-Methyl-2-butene
(major alkene)
1,1-Dimethylpropyl cation
CH
3
CCH
2
CH
3
CH
3
H11001
2-Methyl-2-butanol
CH
3
CCH
2
CH
3
CH
3
OH
2,2-Dimethyl-1-propanol
CH
3
CCH
2
OH
CH
3
CH
3
2-Methyl-1-butanol 3-Methyl-1-butanol 3-Methyl-2-butanol
CH
3
CHCH
2
CH
2
OH
CH
3
CH
3
CH
2
CHCH
2
OH
CH
3
CH
3
CH
OHCH
3
CHCH
3
CH
3
CH
2
CH
2
CH
2
CH
2
OH
1-Pentanol 2-Pentanol 3-Pentanol
CH
3
CH
2
CHCH
2
CH
3
OH
CH
3
CH
2
CH
2
CHCH
3
OH
base
E2
CH
3
CH
3
CH
3
CH
2
CH
3
CH
2
H11001H11001CH
3
CH
2
CH
2
CHCH
3
Br
2-Bromopentane 1-Pentene (E)-2-Pentene(Z)-2-Pentene
CH
3
CH
2
CH
2
CH CH
2
base
E2
CH
3
CH
3
CH
2
CH
3
CH
2
CH
3
CH
3
CH
2
CHCH
2
CH
3
H11001
Br
3-Bromopentane (E)-2-Pentene (Z)-2-Pentene
110 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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(e) For the 1,1-dimethylpropyl cation to be formed by a process involving a hydride shift, the
starting alcohol must have the same carbon skeleton as the 1,1-dimethylpropyl cation.
Although the same carbon skeleton is necessary, it alone is not sufficient; the alcohol must
also have its hydroxyl group on the carbon atom adjacent to the carbon that bears the migrat-
ing hydrogen. Thus, 3-methyl-1-butanol cannot form a tertiary carbocation by a single hy-
dride shift. It requires two sequential hydride shifts.
( f ) 2,2-Dimethyl-1-propanol can yield a tertiary carbocation by a process involving a methyl
shift.
5.36 (a) Heating an alcohol in the presence of an acid catalyst (KHSO
4
) leads to dehydration with
formation of an alkene. In this alcohol, elimination can occur in only one direction to give a
mixture of cis and trans alkenes.
Br
CHCH
2
CH
3
CHCH
3
Br
CH
OH
KHSO
4
heat
Cis–trans mixture
CH
3
CH
3
H
H11001
CH
3
C
CH
3
H11001H11001
CH
3
C
2,2-Dimethyl-1-propanol
CH
2
CH
2
CH
3
O
H
H
CH
3
CH
3
CH
3
C CH
2
OH
CH
3
H
H
H11001
CH
3
CHCH
CH
3
CH
3
CHCH
2
CH
2
OH
CH
3
H
CH
3
C
H11001 H11001H11001
3-Methyl-1-butanol
CH
2
CHCH
3
O
H
H
CH
3
CH
3
CCH
2
CH
3
CH
3
C
H
H
H11001
H11002H
2
O
CH
3
CH
3
OH
CHCH
3
H
CH
3
C
CH
3
C
H11001H11001
CH
3
3-Methyl-2-butanol
CHCH
3
CH
2
CH
3
H11001
CH
3
HOCH
2
CCH
2
CH
3
H
CH
3
CCH
2
CH
3
H
H
H11001
HO
H
CH
2
CH
3
CCH
2
CH
3
H11001
CH
3
2-Methyl-1-butanol
H
CH
3
C CHCH
3
CH
3
CH
3
CCH
2
CH
3
H11001
CH
3
CH
3
OH
HOCH
2
CCH
2
CH
3
has same carbon skeleton as
H
and
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 111
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(b) Alkyl halides undergo E2 elimination on being heated with potassium tert-butoxide.
(c) The exclusive product of this reaction is 1,2-dimethylcyclohexene.
(d) Elimination can occur only in one direction, to give the alkene shown.
(e) The reaction is a conventional one of alcohol dehydration and proceeds as written in 76–78%
yield.
( f ) Dehydration of citric acid occurs, giving aconitic acid.
(g) Sequential double dehydrohalogenation gives the diene.
CH
3
Cl
Cl
H
3
C CH
3
KOC(CH
3
)
3
DMSO
KOC(CH
3
)
3
DMSO
Bornylene (83%)
CH
3
Cl
H
3
C CH
3
CH
3
H
3
C CH
3
OH
CO
2
H
CH
2
CO
2
HHO
2
CCH
2
C
H
2
SO
4
140–145°C
CO
2
H
HO
2
CCH CCH
2
CO
2
H
Citric acid Aconitic acid
HO
CH
3
O
CN
CH
3
O
CN
KHSO
4
130–150°C
(H11002H
2
O)
(CH
3
)
2
CCl
KOC(CH
3
)
3
(CH
3
)
3
COH
heat
CH
2
C
CH
3
NaOCH
2
CH
3
CH
3
CH
2
OH
heat
H
CH
3
CH
3
CH
3
Br
CH
3
1-Bromo-trans-1,2-
dimethylcyclohexane
1,2-Dimethylcyclohexene
(100%)
KOC(CH
3
)
3
(CH
3
)
3
COH
heat
H
2
C C(OCH
2
CH
3
)
2
ICH
2
CH(OCH
2
CH
3
)
2
112 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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(h) This example has been reported in the chemical literature, and in spite of the complexity of the
starting material, elimination proceeds in the usual way.
(i) Again, we have a fairly complicated substrate, but notice that it is well disposed toward E2
elimination of the axial bromine.
( j) In the most stable conformation of this compound, chlorine occupies an axial site, and so it is
ideally situated to undergo an E2 elimination reaction by way of an anti arrangement in the
transition state.
The minor product is the less highly substituted isomer, in which the double bond is exocyclic
to the ring.
5.37 First identify the base as the amide ion (H
2
N
H11002
) portion of potassium amide (KNH
2
). Amide ion is a
strong base and uses an unshared electron pair to abstract a proton from H9252 carbon of the alkyl halide.
The pair of electrons in the C@H bond becomes the H9266 component of the double bond as the C—Br
bond breaks. The electrons in the C@Br bond become an unshared electron pair of bromide ion.
5.38 The problem states that the reaction is first order in (CH
3
)
3
CCl (tert-butyl chloride) and first order in
NaSCH
2
CH
3
(sodium ethanethiolate). It therefore exhibits the kinetic behavior (overall second
order) of a reaction that proceeds by the E2 mechanism. The base that abstracts the proton from car-
bon is the anion CH
3
CH
2
S
H11002
.
H
CH H11001H
CH
3
CH
2
S
CH
3
CH
2
S C(CH
3
)
2
H11002
Cl
ClH
2
C H11001
H11002
C
H
CH
3
CH
3
H
Br
CH
2
(CH
3
CH
2
)
2
C H
2
N H11001 H11001
H11002
H
H
2
N
H11002
(CH
3
CH
2
)
2
C CH
2
Br
CH
3
CH
3
CH
2
(CH
3
)
3
C
C(CH
3
)
3
C(CH
3
)
3
Cl
NaOCH
3
CH
3
OH
4-tert-Butyl-
1-methylcyclohexene (95%)
4-tert-Butyl-
(methylene)cyclohexane (5%)
H11001
O
CH
3
O
CH
3
OCH
2
CH
3
OCH
2
CH
3
O
CH
3
O
CH
3
O
CH
3
O OCH
3
KOH
heat
O
Br
O
O
O
O
O
O
O
O
Br
Br
KOC(CH
3
)
3
DMSO
(84%)
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 113
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5.39 The two starting materials are stereoisomers of each other, and so it is reasonable to begin by ex-
amining each one in more stereochemical detail. First, write the most stable conformation of each
isomer, keeping in mind that isopropyl is the bulkiest of the three substituents and has the greatest
preference for an equatorial orientation.
The anti periplanar relationship of halide and proton can be achieved only when the chlorine is axial;
this corresponds to the most stable conformation of neomenthyl chloride. Menthyl chloride, on the
other hand, must undergo appreciable distortion of its ring to achieve an anti periplanar
Cl@C@C@H geometry. Strain increases substantially in going to the transition state for E2 elim-
ination in menthyl chloride but not in neomenthyl chloride. Neomenthyl chloride undergoes E2
elimination at the faster rate.
5.40 The proton that is removed by the base must be anti to bromine. Thus, the alkyl groups must be
gauche to one another in the conformation that leads to cis-4-nonene and anti to one another in the
one that leads to trans-4-nonene.
Br
H
H
H
CH
2
CH
2
CH
3
CH
3
CH
2
CH
2
CH
2
Anti conformation of 5-bromononane
E2 transition state trans-4-Nonene
CH
3
CH
2
CH
2
CH
2
Br
H
H
H
OCH
3
CH
2
CH
2
CH
2
CH
3
H9254H11002
H9254H11002
H
CH
2
CH
2
CH
3
CH
3
CH
2
CH
2
CH
2
H
Br
H
H
H
CH
2
CH
2
CH
3
CH
2
CH
2
CH
2
CH
3
Gauche conformation of 5-bromononane
Br
H
H
H
OCH
3
CH
2
CH
2
CH
2
CH
3
CH
2
CH
2
CH
2
CH
3
E2 transition state cis-4-Nonene
H9254H11002
H9254H11002
H
CH
2
CH
2
CH
3
CH
2
CH
2
CH
2
CH
3
H
CH(CH
3
)
2
CH(CH
3
)
2
ClH
3
C
H
H
H
Cl
Menthyl chloride Most stable conformation of menthyl chloride:
none of the three H9252 protons is anti to chlorine
H
3
C
CH(CH
3
)
2
CH(CH
3
)
2
ClH
3
C
H
H
Cl
Neomenthyl chloride Most stable conformation of neomenthyl chloride:
each H9252 carbon has a proton that is anti to chlorine
H
3
C
114 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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The alkyl groups move closer together (van der Waals strain increases) as the transition state for for-
mation of cis-4-nonene is approached. No comparable increase in strain is involved in going to the
transition state for formation of the trans isomer.
5.41 Begin by writing chemical equations for the processes specified in the problem. First consider re-
arrangement by way of a hydride shift:
Rearrangement by way of a methyl group shift is as follows:
A hydride shift gives a tertiary carbocation; a methyl migration gives a secondary carbocation. It is
reasonable to expect that rearrangement will occur so as to produce the more stable of these two car-
bocations because the transition state has carbocation character at the carbon that bears the migrat-
ing group. We predict that rearrangement proceeds by a hydride shift rather than a methyl shift, since
the group that remains behind in this process stabilizes the carbocation better.
5.42 Rearrangement proceeds by migration of a hydrogen or an alkyl group from the carbon atom adja-
cent to the positively charged carbon.
(a) A propyl cation is primary and rearranges to an isopropyl cation, which is secondary, by mi-
gration of a hydrogen with its pair of electrons.
(b) A hydride shift transforms the secondary carbocation to a tertiary one.
This hydride shift occurs in preference to methyl migration, which would produce the same
secondary carbocation. (Verify this by writing appropriate structural formulas.)
(c) Migration of a methyl group converts this secondary carbocation to a tertiary one.
1,2,2-Trimethylpropyl cation
(secondary, less stable)
1,1,2-Trimethylpropyl cation
(tertiary, more stable)
CH
3
CH
3
CHCH
3
CH
3
C
H11001
CH
3
CH
3
CH
3
C
H11001
CHCH
3
1,2-Dimethylpropyl cation
(secondary, less stable)
1,1-Dimethylpropyl cation
(tertiary, more stable)
H
CH
3
CHCH
3
CH
3
C
H11001
CH
3
CH
3
CCH
2
CH
3
H11001
Propyl cation
(primary, less stable)
Isopropyl cation
(secondary, more stable)
CH
3
CHCH
3
H11001
H
CH
2
CH
3
CH
H11001
H11001
Isobutyloxonium ion Secondary cation Water
O
H
H
H
CH
3
CH
2
CH
3
C O
H
H
H11001
CH
3
CCH
2
CH
3
H
H11001
H11001
Isobutyloxonium ion Tertiary cation Water
O
H
H
CH
3
CCH
3
CH
3
H11001
O
H
CH
3
CH
2
CH
3
C
H
H
H11001
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 115
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(d) The group that shifts in this case is the entire ethyl group.
(e) Migration of a hydride from the ring carbon that bears the methyl group produces a tertiary
carbocation.
5.43 (a) Note that the starting material is an alcohol and that it is treated with an acid. The product is
an alkene but its carbon skeleton is different from that of the starting alcohol. The reaction is
one of alcohol dehydration accompanied by rearrangement at the carbocation stage. Begin by
writing the step in which the alcohol is converted to a carbocation.
The carbocation is tertiary and relatively stable. Migration of a methyl group from the tert-butyl
substituent, however, converts it to an isomeric carbocation, which is also tertiary.
Loss of a proton from this carbocation gives the observed product.
(b) Here also we have an alcohol dehydration reaction accompanied by rearrangement. The
initially formed carbocation is secondary.
H
OH
H
H
H11001
H11002H
2
O
H11001
C
CH
3
CH
2
CH
3
C
CH
3
CH
2
H
H11002H
H11001
CH
3
H11001
C
C
CH
3
CH
3
CH
3
H11001
H11001
CH
3
CH
3
CH
3
C(CH
3
)
3
C(CH
3
)
3OH
H
H11001
H11002H
2
O
H11001
HH11001
H11001
CH
3
CH
3
2-Methylcyclopentyl cation
(secondary, less stable)
1-Methylcyclopentyl cation
(tertiary, more stable)
2,2-Diethylbutyl cation
(primary, less stable)
1,1-Diethylbutyl cation
(tertiary, more stable)
CH
3
CH
2
CH
3
CH
2
CH
2
CH
3
CH
2
C
H11001
CH
3
CH
2
CH
3
CH
2
C
H11001
CH
2
CH
2
CH
3
116 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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This cation can rearrange to a tertiary carbocation by an alkyl group shift.
Loss of a proton from the tertiary carbocation gives the observed alkene.
(c) The reaction begins as a normal alcohol dehydration in which the hydroxyl group is proto-
nated by the acid catalyst and then loses water from the oxonium ion to give a carbocation.
We see that the final product, 1-methylsantene, has a rearranged carbon skeleton corre-
sponding to a methyl shift, and so we consider the rearrangement of the initially formed sec-
ondary carbocation to a tertiary ion.
Deprotonation of the tertiary carbocation yields 1-methylsantene.
5.44 The secondary carbocation can, as we have seen, rearrange by a methyl shift (Problem 5.16). It can
also rearrange by migration of one of the ring bonds.
The tertiary carbocation formed by this rearrangement can lose a proton to give the observed
byproduct.
Isopropylidenecyclopentane
H
C
H11001
CH
3
CH
3
C
CH
3
CH
3
H11002H
H11001
H
HH11001
CH
3 C
H11001
CH
3
CH
3
CH
3
Tertiary carbocationSecondary carbocation
H
CH
3
CH
3
CH
3
Tertiary carbocation
H
H11001
H11001
CH
3
CH
3
CH
3
methyl shift
CH
3
CH
3
CH
3
1-Methylsantene
H11002H
H11001
OH
H
CH
3
CH
3
CH
3
4-Methylcamphenilol
OH
2
H
H11001
CH
3
CH
3
CH
3
KHSO
4
H
H11001
CH
3
CH
3
CH
3
Secondary carbocation
H11002H
2
O
H
H11001
H11002H
H11001
H
H11001
H
H11001
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 117
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5.45 Let’s do both part (a) and part (b) together by reasoning mechanistically. The first step in any acid-
catalyzed alcohol dehydration is proton transfer to the OH group.
But notice that because this alcohol does not have any hydrogens on its H9252 carbon, it cannot dehy-
drate directly. Any alkenes that are formed must arise by rearrangement processes. Consider, for
example, migration of either of the two equivalent methyl groups at C-2.
The resulting carbocation can lose a proton in three different directions.
The alkene mixture shown in the preceding equation constitutes part of the answer to part (b). None
of the alkenes arising from methyl migration is 2-methyl-2-heptene, the answer to part (a), however.
What other group can migrate? The other group attached to the H9252 carbon is a butyl group. Con-
sider its migration.
Loss of a proton from the carbocation gives the alkene in part (a).
A proton can also be lost from one of the methyl groups to give 2-methyl-1-heptene. This is the last
alkene constituting the answer to part (b).
H11001
CCH
2
CH
2
CH
2
CH
2
CH
3
H11002H
H11001
2-Methyl-1-heptene
CCH
2
CH
2
CH
2
CH
2
CH
3
H
2
C
CH
3
CH
3
CH
3
H11001
CCH
2
CH
2
CH
2
CH
2
CH
3
H11002H
H11001
2-Methyl-2-heptene
C
CH
3
CH
3
CH
3
CHCH
2
CH
2
CH
2
CH
3
CH
3
O
H
H
CH
3
CH
3
CH
2
CH
3
CH
3
CH
3
CH
2
CH
2
CH
2
C
H11001
CCH
2
CH
2
CH
2
CH
2
CH
3
H11001
H11001
O
H
H
CH
3
CH
2
CH
2
CH
2
CCH
2
CH
3
CH
3
H11002H
H11001
CH
3
CH
3
CH
2
CH
2
CH CCH
2
CH
3
H11001 H11001CH
3
CH
2
CH
2
CH
2
CCH
2
CH
3
CH
2
CH
3
CH
2
CH
2
CH
2
C
CH
3
CHCH
3
3-Methyl-3-heptene
(mixture of E and Z)
2-Ethyl-1-hexene 3-Methyl-2-heptene
(mixture of E and Z)
H11001
CH
3
CH
2
CH
2
CH
2
C O
H
H
CH
3
CH
3
CH
3
CH
2
CH
2
CH
2
CCH
2
CH
3
CH
3
CH
2
O
H
H
H11001
H11001
H11001
CH
3
CH
2
CH
2
CH
2
CCH
2
H
2
SO
4
OH
CH
3
CH
3
CH
3
CH
2
CH
2
CH
2
CCH
2
O
H
H
CH
3
CH
3
H11001
118 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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5.46 Only two alkanes have the molecular formula C
4
H
10
: butane and isobutane (2-methylpropane)—
both of which give two monochlorides on free-radical chlorination. However, dehydrochlorination
of one of the monochlorides derived from butane yields a mixture of alkenes.
Both monochlorides derived from 2-methylpropane yield only 2-methylpropene under conditions of
E2 elimination.
Compound A is therefore 2-methylpropane, the two alkyl chlorides are tert-butyl chloride and
isobutyl chloride, and alkene B is 2-methylpropene.
5.47 The key to this problem is the fact that one of the alkyl chlorides of molecular formula C
6
H
13
Cl
does not undergo E2 elimination. It must therefore have a structure in which the carbon atom that
is H9252 to the chlorine bears no hydrogens. This C
6
H
13
Cl isomer is 1-chloro-2,2-dimethylbutane.
Identifying this monochloride derivative gives us the carbon skeleton. The starting alkane
(compound A) must be 2,2-dimethylbutane. Its free-radical halogenation gives three different
monochlorides:
Both 3-chloro-2,2-dimethylbutane and 1-chloro-3,3-dimethylbutane give only 3,3-dimethyl-1-
butene on E2 elimination.
3-Chloro-2,2-
dimethylbutane
KOC(CH
3
)
3
(CH
3
)
3
COH
CH
3
CH
3
ClCH
2
CH
2
CCH
3
1-Chloro-3,3-
dimethylbutane
CH
3
CH
3
CH
3
CHCCH
3
Cl
3,3-Dimethyl-1-butene
(alkene B)
or
CH
3
CH
3
CHCCH
3
CH
2
CH
3
CH
3
CH
3
CH
2
CCH
3
2,2-Dimethylbutane
(compound A)
CH
3
CH
3
CH
3
CH
2
CCH
2
Cl
1-Chloro-2,2-
dimethylbutane
3-Chloro-2,2-
dimethylbutane
Cl
2
light
H11001H11001
CH
3
CH
3
ClCH
2
CH
2
CCH
3
1-Chloro-3,3-
dimethylbutane
CH
3
CH
3
Cl
CH
3
CHCCH
3
CH
3
CH
3
CH
3
CH
2
CCH
2
Cl
1-Chloro-2,2-dimethylbutane
(cannot form an alkene)
KOC(CH
3
)
3
dimethyl
sulfoxide
Isobutyl chloride 2-Methylpropenetert-Butyl chloride
(CH
3
)
3
CCl (CH
3
)
2
CHCH
2
Cl (CH
3
)
2
Cor CH
2
KOC(CH
3
)
3
dimethyl
sulfoxide
2-Butene (cis H11001 trans)
1-Butene2-Chlorobutane
H
2
C CHCH
2
CH
3
H11001 CHCH
3
CH
3
CHCH
3
CHCH
2
CH
3
Cl
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 119
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SELF-TEST
PART A
A-1. Write the correct IUPAC name for each of the following:
A-2. Each of the following is an incorrect name for an alkene. Write the structure and give the
correct name for each.
(a) 2-Ethyl-3-methyl-2-butene (c) 2,3-Dimethylcyclohexene
(b) 2-Chloro-5-methyl-5-hexene (d) 2-Methyl-1-penten-4-o1
A-3. (a) Write the structures of all the alkenes of molecular formula C
5
H
10
.
(b) Which isomer is the most stable?
(c) Which isomers are the least stable?
(d) Which isomers can exist as a pair of stereoisomers?
A-4. How many carbon atoms are sp
2
-hybridized in 2-methyl-2-pentene? How many are sp
3
-
hybridized? How many H9268 bonds are of the sp
2
–sp
3
type?
A-5. Write the structure, clearly indicating the stereochemistry, of each of the following:
(a)(Z)-4-Ethyl-3-methyl-3-heptene
(b)(E)-1,2-Dichloro-3-methyl-2-hexene
(c)(E)-3-Methyl-3-penten-1-ol
A-6. Write structural formulas for two alkenes of molecular formula C
7
H
14
that are stereoisomers
of each other and have a trisubstituted double bond. Give systematic names for each.
A-7. Write structural formulas for the reactant or product(s) omitted from each of the following.
If more than one product is formed, indicate the major one.
A-8. Write the structure of the C
6
H
13
Br isomer that is not capable of undergoing E2 elimination.
A-9. Write a stepwise mechanism for the formation of 2-methyl-2-butene from the dehydration
of 2-methyl-2-butanol is sulfuric acid.
A-10. Draw the structures of all the alkenes, including stereoisomers, that can be formed from the
E2 elimination of 3-bromo-2,3-dimethylpentane with sodium ethoxide (NaOCH
2
CH
3
) in
ethanol. Which of these would you expect to be the major product?
(CH
3
)
2
COH
C(CH
3
)
3
H
3
PO
4
?(d)
KOCH
2
CH
3
CH
3
CH
2
OH
?
(only alkene formed)(c)
Cl CH
3
NaOCH
3
CH
3
OH
?
(b)
(CH
3
)
2
CCH
2
CH
2
CH
3
OH
H
2
SO
4
heat
?
(a)
(CH
3
)
3
CCH C(CH
3
)
2
(a)(c)
(b)(d)
Br
Br
OH
120 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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A-11. Using curved arrows and perspective drawings (of chair cyclohexanes), explain the forma-
tion of the indicated product from the following reaction:
A-12. Compare the relative rate of reaction of cis- and trans-1-chloro-3-isopropylcyclohexane
with sodium methoxide in methanol by the E2 mechanism.
A-13. Outline a mechanism for the following reaction:
A-14. Compound A, on reaction with bromine in the presence of light, gave as the major product
compound B (C
9
H
19
Br). Reaction of B with sodium ethoxide in ethanol gave 3-ethyl-4,4-
dimethyl-2-pentene as the only alkene. Identify compounds A and B.
PART B
B-1. Which one of the alkenes shown below has the Z configuration of its double bond?
B-2. Carbon–carbon double bonds do not undergo rotation as do carbon–carbon single bonds.
The reason is that
(a) The double bond is much stronger and thus more difficult to rotate
(b) Overlap of the sp
2
orbitals of the carbon–carbon H9268 bond would be lost
(c) Overlap of the p orbitals of the carbon–carbon H9266 bond would be lost
(d) The shorter bond length of the double bond makes it more difficult for the attached
groups to pass one another
(e) The statement is incorrect—rotation around double bonds does occur.
B-3. Rank the following substituent groups in order of decreasing priority according to the
Cahn–Ingold–Prelog system:
(a)2H11022 3 H11022 1(b)1H11022 3 H11022 2(c)3H11022 1 H11022 2(d)2H11022 1 H11022 3
B-4. The heats of combustion for the four C
6
H
12
isomers shown are (not necessarily in order):
955.3, 953.6, 950.6, and 949.7 (all in kilocalories per mole). Which of these values is most
likely the heat of combustion of isomer 1?
(a) 955.3 kcal/mol (c) 950.6 kcal/mol
(b) 953.6 kcal/mol (d) 949.7 kcal/mol
12 3 4
CH(CH
3
)
2
CH
2
Br CH
2
CH
2
Br
12 3
(a) (b) (c)(d)
OH
H
3
PO
4
heat
H11001 other alkenes
CH
3
CH
3
Br
NaOCH
3
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 121
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B-5. Referring to the structures in the previous question, what can be said about isomers 3 and 4?
(a) 3 is more stable by 1.7 kcal/mol.
(b) 4 is more stable by 1.7 kcal/mol.
(c) 3 is more stable by 3.0 kcal/mol.
(d) 3 is more stable by 0.9 kcal/mol.
B-6. The structure of (E)-1-chloro-3-methyl-3-hexene is
B-7. In the dehydrohalogenation of 2-bromobutane, which conformation leads to the formation of
cis-2-butene?
(e) None of these is the correct conformation.
B-8. Which of the following alcohols would be most likely to undergo dehydration with re-
arrangement by a process involving a methyl migration (methyl shift)?
B-9. Rank the following alcohols in order of decreasing reactivity (fastest A slowest) toward de-
hydration with 85% H
3
PO
4
:
(a)2 H11022 3 H11022 1(b)1 H11022 3 H11022 2(c)2 H11022 1 H11022 3(d)1 H11022 2 H11022 3
B-10. Consider the following reaction:
Which response contains all the correct statements about this process and no incorrect ones?
OH
H
3
PO
4
heat
H11001
(CH
3
)
2
CHCH
2
CH
2
OH (CH
3
)
2
CCH
2
CH
3
(CH
3
)
2
CHCHCH
3
1 2 3
OH OH
OH
OH
OH
OH
OH
(a)(c)(e)
(d)(b)
H
Br
H
H
CH
3
CH
3
(a)
H
H
H
Br
CH
3
CH
3
(b)
Br
H
H
CH
3
CH
3
H
(c)
H
Br
H
H
CH
3
H
3
C
(d)
Cl
Cl Cl
(a)(c)
(b)(d) None of these
122 STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS
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1. Dehydration
2. E2 mechanism
3. Carbon skeleton migration
4. Most stable carbocation forms
5. Single-step reaction
(a) 1, 3 (b) 1, 2, 3 (c) 1, 2, 5 (d) 1, 3, 4
B-11. Select the formula or statement representing the major product(s) of the following reaction:
B-12. Which one of the following statements concerning E2 reactions of alkyl halides is true?
(a) The rate of an E2 reaction depends only on the concentration of the alkyl halide.
(b) The rate of an E2 reaction depends only on the concentration of the base.
(c) The C@H bond and the C@X (X H11005 halogen) bond are broken in the same step.
(d) Alkyl chlorides generally react faster than alkyl bromides.
B-13. Which alkyl halide undergoes E2 elimination at the fastest rate?
B-14. What is the relationship between the pair of compounds shown?
(a) Identical: superimposable without bond rotations
(b) Conformations
(c) Stereoisomers
(d) Constitutional isomers
B-15. Which one of the following will give 2-methyl-1-butene as the only alkene on treatment with
KOC(CH
3
)
3
in dimethyl sulfoxide?
(a) 1-Bromo-2-methylbutane (c) 2-Bromo-2-methylbutane
(b) 2-Methyl-1-butanol (d) 2-Methyl-2-butanol
H
3
C
H
3
C
H
H H
H
Br
C(CH
3
)
3
Br
C(CH
3
)
3
Cl
C(CH
3
)
3
Cl
C(CH
3
)
3
(a)
(b)
(c)
(d)
HCH
3
CH
2
CH
3
CH
2
CHCHCH
2
CH
3
CH
3
HCH
2
CH
3
CH
3
CH
3
(a)(c)
(b)(d) Both (a) and (b) form in approximately
equal amounts.
CH
3
CC
H
Br H
H
3
C CH
2
CH
3
CH
3
KOCH
2
CH
3
CH
3
CH
2
OH
?
STRUCTURE AND PREPARATION OF ALKENES: ELIMINATION REACTIONS 123
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