1
CHEM 2423
Unit 7 Homework Answers
1. Name the following by IUPAC rules. Designate as E,Z when necessary.
(a)
hp
CH3
1
2
lpH
hp
4 5 6
hp
CH2CH(CH3)2
3
(b)
hp
Br
Cl
2
1
lp
CHCH2CH2CH3
3 4 5 6
lp H
lp CH3
CH3
Z-3,5-dimethylhex-2-ene
(c)
lp
Cl
hp
Br
ηπ I
lp CH2CH3
Z-2-bromo-1-chloro-3-methylhex-1-ene
(d)
lp
BrCH2CH2CH2
(CH3)3C
E-2-bromo-1-chloro-1-iodobut-1-ene
H lp
hp
E-1,7-dibromo-4-t-butyloct-4-ene
5
1
(e)
(f)
4
6
1
2
3
1,4-dimethylcyclopentene
(g)
hp
CH2CHBrCH3
4
3
1
2
1,4,4-trimethylcyclobutene
5
2
3
4
1-methylcyclohexa-1,4-diene
(h)
(CH3)3CCH=CCl2
1,1-dichloro-3,3-dimethylbut-1-ene
2
Br
Br
(j)
(i)
CH2CH(CH3)2
4,4-dibromocycloheptene
1-isopropylcyclooctene
2. Write the structure for the following.
H
OH
H
OH
allyl alcohol
H
Br
H
H
F
vinyl alcohol
allyl bromide
H
vinyl fluoride
3. List in order of relative stability.
(a)
least
=
stable
= most
stable
<
<
<
(b)
H
CH3
H
least
=
stable
CH3
CH3 H
<
<
H
H
CH3
H
H
CH3
CH3
= most
stable
<
H CH3
CH3
4. The following are examples of dehydration reactions. Show the product(s). If more than one
product is possible, label each as major or minor.
Br
Br
there are no H's on this beta-carbon
α
(a)
β
CH CH2CH3
OH
KHSO4
Δ
CH CHCH3
3
HO
CN
CN
α
(b)
KHSO4
β
Δ
CH3O
CH3O
β
CH2CO 2H
CH2CO 2H
α
(c)
HO
C
CH2CO 2H
H2SO 4
β
C
Δ
CO 2H
CHCO2H
CO 2H
the two beta positions will lead to the same product
5. Show the mechanism and products for the following elimination reactions. Label major and
minor products, as necessary. Label the rate-determining step.
(a)
Cl
KOEt, EtOH
heat
This is a secondary alkyl halide, with a fairly strong base. E2 elimination is expected.
Cl
+
-OEt
(RDS)
H
+ EtOH + Cl-
(b)
H2SO 4
+
heat
OH
major
minor
This is a tertiary alcohol. Reaction is expected to occur by E1 elimination.
4
+
H2SO 4
+ HSO4-
OH
OH2+
+ H2O
(RDS)
OH2+
α
β
β
+
+ H2O
H
H
major
minor
+ H3O+
(c)
H3PO 4
CH3CH2CH2CH2OH
heat
CH3CH2CH=CH2
This is a primary alcohol. This would react by E2 elimination.
CH3CH2CH2CH2OH
+
+
CH3CH2CH2CH2OH2
H3PO 4
H
CH3CH2
CH CH2 OH2+
+
H2O
CH3CH2CH=CH2
(RDS)
5
CH3
(d)
Br
CH2
CH3
EtOH
heat
+
minor
major
Here we have a tertiary alkyl halide, treated with a very mild base and heat. E1 elimination
should predominate.
CH3
+
CH3
(RDS)
Br
a
H
a
+
b
CH2
H
+
a
EtOH
CH3
major
b
b
CH2
minor
6
6. Show the products and their stereochemistry of E2 elimination in the following reactions.
(a)
H
Ph
CH3
Ph
Br
Ph
Ph
CH3
Br
Ph
Ph
Br
rotate back C
H
CH3
H
H H
H
EtO-
Ph
(b)
CH3
Ph
Ph
Ph
H
Ph
Br
Ph
CH3
Ph
H
Ph
CH3
H
CH3
H
Br
Ph
Br
Ph
H
H CH3
H
H
- OEt
CH3
Ph
Ph
Ph
H
H
CH3
Ph
7. Explain the following experimental results. Specifically, explain why Zaitsev’s rule is followed
in the first example, but is not in the second example. You will need to draw the chair
conformations to complete your answers.
Both reactions involve dehydrohalogenation of secondary alkyl halides with strong bases. This
means the reactions will go by E2 elimination which is a concerted reaction that requires antiperiplanar configuration between the leaving group and the β-hydrogen.
7
The cis isomer reacts easily by this mechanism yielding a product that obeys Zaitsev’s rule.
When the large tert-butyl is equatorial, the leaving group and hydrogen on the more substituted
β-carbon are anti-periplanar.
EtO-
H
H
C(CH3)3
C(CH3)3
Cl
cis
C(CH3)3
In the case of the trans isomer:
H
Cl
Cl
C(CH3)3
H
β
EtO-
H
β
α
H
H
H
C(CH3)3
trans
Cl must be axial to achieve
an anti-periplanar configuration
to a beta hydrogen. Hence,
ring inversion occurs, forming
the less stable configuration
this is the only beta hydrogen that
is anti-periplanar to Cl, hence elimination
occurs between these two carbons
C(CH3)3
C(CH3)3
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8. Explain the following experimental results. Specifically, why is Zaitsev’s rule followed in the
dehydration, but not in deydrohalogenation?
In the case of the alkyl halide, we have already shown how this occurs in problem 7. The main
thing is that 2o alkyl halides undergo E2 elimination and require an anti-periplanar configuration.
On the other hand, 2o alcohols undergo E1 elimination. There are no stereochemical
requirements placed on the transition state, so the preferred regiochemistry is achieved – that
which follows Zaitsev’s rule. Because there are no stereochemical restrictions, we don’t even
need to worry about chair conformations.
C(CH3)3
C(CH3)3
+
H2SO 4
OH2+
OH
C(CH3)3
H
C(CH3)3
+
H2O
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9. Show the mechanisms that best explains the following results.
(a)
OH
CH3CH
CHCH3
H3PO4
heat
CH3
CH3
OH2+
CH3CH
CHCH3
CH CH2
CH3CH
CH3
CH3C
CH3
1
C
CHCH3 + CH2
CH2CH3
CH3
2
3
H
H
CH3C
+
1
+ H2O
CH CH2
+
2
2
+
CH3
4
H
CH3C
H
CH CH3
2o cation
CH3
4
~H
CH2
H
C
CH3
CHCH3
3o cation
+ H2O
3
10
OH2+
OH
(b)
H2SO 4
+
10
8
1
2
4
10
H
methylene
shift
7
6
3
9
9
8
7
1
5
6
2
3
+
+ H2O
+
5
4
H3O+
10. Elimination and substitution are always competing reactions when a base/nucleophile is
added to an alkyl halide. Predict the products and mechanism of these competing reactions.
Explain which reaction is likely to yield the major product.
major
(a)
CH3CH2Br
+
CH3CH2ONa
EtOH
minor
CH3CH2OCH2CH3
SN2
major
CH3
(b) CH3
C
CH3
EtOH
CH3
C
E2
minor
CH3
CH2
Br + CH3CH2ONa
CH2=CH2
+
+ CH3
C
CH3
CH3
E2
SN1
OCH2CH3
11
minor
major
Br
(c) CH3
EtOH
CH CH3 + CH3CH2ONa
OCH2CH3
CH3CH=CH2 + CH
3
E2
CH CH3
SN2
minor
major
(d)
CH3CH2Br + t-BuOK
t-BuOH
CH2=CH2 +
CH3
C
CH3CH2O
CH3
E2
SN2
CH3
(e) CH3
C
Br
CH3OH
CH3
CH3
CH CH3
minor
CH3
CH2
C
OCH3
+
NaI
acetone
CH3
CH3
C
CH3
CH3
SN1
E1
major
I
Br
(f) CH3
major
CH CH3 +
SN2
minor
CH3CH=CH2
E2
CH3
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11. You are working in a lab, and your boss wants you to synthesize the molecule (CH3)3COCH3
from an alkyl halide (RX) and a sodium alkoxide (RONa). Show two methods by which you might
achieve this goal, and explain which is the better method, and why.
method 1
CH3
CH3
CH3
C
Br
CH3ONa
+
CH3
o
CH3 3
C
CH2
OCH3 + CH3
CH3
CH3
probably not formed
method 2
C
CH3
probably formed
exclusively
CH3
CH3
CH3
C
ONa
+
CH3Br
CH3
C
OCH3
CH3
Of the two methods, the second one is superior. Alkoxides, as a general rule, are fairly strong
bases. The synthesis, shown in method 1, where reaction of the alkoxide is with a tertiary alkyl
halide, will most likely result in elimination only. In method 2, the methyl bromide easily reacts
via substitution, and elimination is not possible.
12. You have available 2,2-dimethylcyclpentanol (A) and 2-bromo-1,1-dimethylcyclopentane (B)
available and wish to prepare 3,3-dimethylcyclopentene (C). Which would you choose as the
more suitable reactant, A or B, and with what would you treat it?
The alcohol is not the best choice in this case. To dehydrate any alcohol, you need a nonnucleophilic acid (e.g., sulfuric acid). The alcohol is secondary, so E1 elimination is favored.
Because a carbocation is formed, rearrangement is always a possibility. In this case, it is a likely
reaction.
13
H3C
CH3
H3C
OH
H3C
CH3
OH2+
H2SO 4
CH3
CH3
+
CH3
CH3
+
CH3
H
+ H2O
H2O
The alkyl bromide is more likely to react by E2 elimination, with a relatively strong base.
Rearrangement is not a possibility, so the desired product can be obtained.
H3C
CH3
H3C
Br
H
KOEt
EtOH
CH3