2010 CHE 230 Chapter 9 SOLUTIONS - Illinois State University

ILLINOIS STATE UNIVERSITY
Organic Chemistry 1, CHE 230, spring semester 2010, Homework #7
Chapter 9: Elimination Reactions
1. Complete the following reactions and identify all products. Determine if the dominant product is
that of substitution or elimination.
Cl
CH 3 ONa
E2
High temperature favors
elimination.
Br
CH 3 ONa
25 o C
S N 2
OCH 3
I
CH 3 ONa
25 o C
S N 2/E2 mixture
OCH 3
Cl
CH 3 ONa
E2
H 3 C
H 3 C
Br
Br
CH 3 ONa
25 o C
E2
CH 3 ONa
CH 3 CH 2
Tertiary haloalkanes
are very sensitive
substrates and will
undergo elimination
CH 3
with basic
CH 2 nucleophiles no matter
what the temperature.
E2
O
S
O
H 3 C
O
CH 3 ONa
25 o C
OCH 3
Primary haloalkane = S N 2 substitution
1

2. Complete the following reactions and identify all products. Determine if the dominant product is
that of substitution (S N 2 or S N 1) or elimination.
Cl
H
CH 3
CH 2 CH 3
NaN 3
dmf
S N 2
H 3 C
H
N 3
CH 2 CH 3
Br
CH 3
CH 3
CH 3
E1 is minor
KCN
dmf
S N 1
NC
CH 3
CH 3
CH 3
H 3 C
CH 3
CH 3
CN
CH 2
CH 3
NaOCH 3
OH
Br
CH 3
CH 3
CH 3
E2
H 3 C
O
O
CH 3
O
I
CH 3
CH 3
CH 3
S N 1
E1 is minor
Ph
O
CH 3
CH 3
CH 3
H 3 C
CH 3
O
Ph
H
H
Br
CH 3
CH 2 CH 3
LiSCH 2 CH 3
dmf
S N 2
H 3 C
CH 2 CH 3
SCH 2 CH 3
Cl
CH 3
CH 2 CH 3
CH 3 CH 2 OH
S N 1-racemization
E1 is minor
OCH 2 CH OCH 3
2 CH 3
CH 3 H 3 C
CH 2 CH 3 CH 2 CH 3
Br
H
CH 3
CH 2 CH 3
NaOCH 3
100 o C
E2
H
H 3 C
C
CH 3
C
H
Br
H
CH 3
CH 2 CH 3
KOC(CH 3 ) 3
25 o C
H
C
H 3 C
CH 3
C
H
E2-For secondary haloalkanes large bulky bases will
trigger the E2 reaction no matter what the temperature is.
2

3. Provide the products of the following transformation. Also state if the reaction is S N 2, S N 1, E1, or
E2.
Br
Na OH
S 25 o C
N 2 + E2
O
O K
Cl
dmf
S N 2 + E2
CH 3 CH 2 S Li
OMs nitromethane
S
solvent
N 1 + E1
MS = mesylate, a good leaving group
OH
H
CH 3
O
O
SCH 2 CH 3
CH 2
CH 2
CH 3
4. Provide the products of the following transformation. If there is no reaction then write NR. If
there is stereochemistry, you must write out the proper answer involving stereochemistry.
CH 3
Br
H
CH 2 CH 3
Na CN
S N 2
CH 3
H
CN
CH 2 CH 3
H
Na OCH 3
CH 3 CH 2 CH 3
There is no reaction. There is no leaving
group present.
H
T his is also wr itten as NaOCH 3.
Br
OH
OCH(CH 3 ) 2
CH 2 CH 3
CH 3
CH 2 CH 3
S N 1 + E1
CH 3
CH 3
OCH(CH 3 ) 2
CH 2 CH 3
H 2 C C CH 2 CH 3 CH 3 C CHCH 3
CH 3 C CH 2 CH 3
3

5. Complete the following reactions. You must indicate stereochemistry. If there is no reaction, then
write NR.
CH 3
Br
H
CH 2 CH 3
K
S N 2
C
N
CH 3
H
CN
CH 2 CH 3
O
OH
O
O
H
Cl
S N 1 + E1
H
O
Ph
Ph
O CH 3
H
I
CH 3
CH 2 CH 3
H
S
Li
CH 3
CH 2 CH 3
CH 2
CH 3
H
H
S N 2
H
S
6. Complete the following reactions. If there is no reaction, then write NR. If there is any
stereochemistry, then you must address it.
H
Br
CH 3
CH 2 CH 3
Br
CH 3 CH 2 OH
H
S N 1-E1
Et = -CH 2 CH 3 (ethyl)
OEt
CH 3
CH 2 CH 3
OCH 3
CH 2 CH 3
OEt
H
CH 2
H CH 3 CH 2 CH 3
H
CH 3
CHCH 3
NaOCH 3
25 o C
S N 2-E2 mix
Cl
NaOCH 3
100 o C
E2 is dominant
Br
KSCH 2 CH 3
dmf
S N 2 is dominant
SCH 2 CH 3
4

7. Complete the following reactions by drawing the major and minor products. If there is no
reaction, then write NR. If there any stereochemistry, then you must address it.
Br
CH 3
CH 3
CH 2
KOH
H 2 O
E2 elimination due to a
basic nucleophile
Et
Et
Cl
NaN 3
dmf
N 3
S N 2-inversion of configuration
KOt-Bu
Br
E2 elimination due to a
basic nucleophile
t-BuOH
100 o C
Cl
CH 3
CH 3
CH 2
NaOCH 2 CH 3
E2 elimination due to a
basic nucleophile
CH 3 CH 2 OH
100 o C
5

8. Complete the following reactions. If there is no reaction, then write NR. If there is any
stereochemistry, then you must address it.
Cl
H
CH 3
CH 2 CH 3
NaN 3
dmf
S N 2
H 3 C
H
N 3
CH 2 CH 3
OH
CH 2 CH 3
Br CH 3
H
F CH 3
CH 2 CH 3
CH 3
H CH 2 CH 3
OH
CH 3
CH 3 N
Cl
CH 3
Br
CH 3
NaCl
dmso
H 2 O
S N 1 + E1
K O CH 3
KCN
dmf
O
1. CH 3 SO 2 Cl, base
2. KSCH 2 CH(CH 3 ) 2
S N 2
HO
O
H
S N 1 + E1
NaOH, cold temp.
E2
No Reaction. Alcohols are not
reasonable leaviing groups.
HO
CH 2 CH 3
CH 3
H 3 C
CH 2 CH 3
OH
CHCH 3
CH 2 CH 3
CH 3
CH 2
CH 2 CH 3
CH 3
No Reaction. There is no leaving group.
No reaction. There is no reasoable leaving group.
The carbon-fluorine bond is extraordinarily strong.
CH 3
H
N
CH 3
CH 2 CH 3
OSO 2 CH 3
O
CH 2
C
CH 3
H
SCH 2 CH(CH 3 ) 2
CH 3
CH 2 CH 3
CH 3 CH 3 N
CH 3
O
O
O
H CH 3 N CH 3
H
CH 3 N CH 2
Cl
I
one equivalent of
NaOAc
Cl
sodium acetate
S N 2-iodo system is more reactive.
O
O
CH 3
Cl
C C
S N 2
Li
C C
6

9. Rank the following alkenes in terms of their relative stabilities.
A B C D
B > A > D ≈ C. B is the most stable because it is tetra-substituted. D and C are nearly equivalent because they
are both di-substituted alkenes.
10. Provide a mechanistic explanation for the following sequence of reaction.
O
D
O
S
O
CF 3
D
O
H
O
S
O
CF 3
CH 3 OH
CH 3 O
The triflate group is a powerful leaving group that is considered
to be superior to the halogens in terms of leaving group ability.
7

11. Provide a mechanistic explanation for the following sequence of reaction.
CH 3
O
S
O
The Hofmann base, potassium tert -butoxide is a large base that
removes the most accessible proton to induce the elimination.
D
O
C
H
H
D
C
C
H
CH 3
+ (CH 3 ) 3 CO-H
H
O
CH 3
CH 3
CH 3
H
O
S
O
O
8

12. Complete the following reactions. Indicate which product is the major product and which is the minor. State the
pathway of elimination, either Zaitsev or Hoffmann. What is the dominant pathway in each case
Br CH 3
NaOCH 3 in CH 3 OH
CH 3
CH 2
100 o C
Zaitsev
major
minor
Br
KOC(CH 3 ) 3
(CH 3 ) 3 COH
100 o C
Hofmann
major
cis + trans (minor)
CH 2 CH 3
CH 2 CH 3 CH2 CH 3
Cl
LiOCH 3 in CH 3 OH
100 o C
Zaitsev
major
minor
CH 2 CH 2 I
KOEt in EtOH
EtOH = CH 3 CH 2 OH
100 o C
This reaction would not be
classified as either Zaitsev
or Hofmann as the alkene
generated is the only possible
alkene.
OH
LiOCH 2 CH 3
CH 3 CH 2 OH
100 o C
No reaction. There is no leaving group to
undergo the elimination process. Hydroxide
is not a suitable leaving group for the E2
process. The alcohol would only deprotonate.
NaOCH 3 in CH 3 OH
OTs
100 o C
TsO = tosylate leaving group
Zaitsev
major
minor
Br
KOC(CH 3 ) 3
(CH 3 ) 3 COH
100 o C
major
CH 2
minor
CH 3
Hofmann
9

13. Provide the major and minor alkenes from the following reactions and circle the major alkene. Indicate the level of
substitution for this alkene. What is the dominant reaction mechanism for all of these reactions
The most stable alkene is always the dominant product under conditions involving strong
acid. The elimination of secondary and tertiary alcohols involves carbocations! The
unwritten by-products for each reaction is water.
H 3 PO 4
heat
No reaction. There is no leaving group. Cyclopentane will
not react under these conditions.
OH
Et Et
Et
trisub.
disub.
CH 3
OH
H 3 PO 4
heat
+
major
minor
OH
H 2 SO 4
heat
major
trans and cis-isomers
+
OH
disub.
H 3 PO 4
heat
H 2 SO 4
heat
trans and cis-isomers
OH
disub.
H 2 SO 4
heat
OH
H 3 PO 4
heat
There is no reaction in terms of forming a new alkene.
There is no suitable β-hydrogen for straightforward
removal.
CH 3
This is the exclusive alkene product
for this reaction.
tetrasub.
major
+
minor
monosub.
monosub.
CH2
disub.
minor
10

14. Provide a mechanistic explanation for the following transformation. Identify nucleophiles and electrophiles. What is
the name given to key process that leads into the alkene product
electrophile
H
O
H
nucleophile
CH 3
CH 3
CH 3
CH 3
H 2 SO 4
heat
H
O
H
CH 3
CH 3
H
O
H
E1 elimination pathway
CH 3
CH 3
H
H
H
CH 3
CH 3
H
H
H
This hydrogen was always there. It did not appear
from nowhere.
The major transformation is a carbocation.
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