Organic Chemistry Ⅰ for Department of Agricultural Chemistry
2nd Midterm Examination
Dec. 11, 2001 10:10 a.m.-12:00 noon
Ⅰ. Each of the following problems has five statements. Each problem is worth
four points. Any wrong guess will cost you one extra point. : (40 pts)
1. Of the following statements, which is not true for the nucleophilic substitution
occurring by SN2 mechanism?
(A) The absolute configuration of the product is opposite to that of the reactant
when an optically active substrate is used.
(B) The probable mechanism involves a carbocation intermediate.
(C) The rate of reaction depends markedly on the nucleophilicity of the attacking
nucleophlie.
(D) The rate of reaction is proportional to the concentration of the attacking
nucleophile.
(E) The rate of reaction depends on the nature of the leaving group.
2. Of the following statements, which is not true for nucleophilic substitutions
occurring by SN1 mechanism?
(A)
(B)
(C)
Tertiary alkyl halides react faster than the secondary alkyl halides.
The reaction shows first-order kinetics.
The absolute configuration of the product remains the same as that of the
reactant when an optically active substrate is used.
(D)
(E)
The rate of reaction is independent on the nature of the nucleophile.
The rate of reaction depends on the nature of the leaving group.
3. Which of the following is the most reactive nucleophile in protic solvent?
(A) F- (B) Cl-
(C) Br-
(D)
I-
(E) H2O
4. 1-Iodobicyclo[2.2.2]octane is very unreactive towards nucleophilic displacement
reaction.. Which of the following statement is ture?
I
(A) Since it’s a tertiary iodide, SN2 reaction does not occur.
(B) In aprotic solvent, SN1 reaction occurs as fast as t-butyl iodide.
(C) Since E1 reaction involves a carbocation intermediate, bridgehead alkene will
be the major product.
(D) In polar protic solvent, 1-iodobicyclo[2.2.2]octane will undergo E2 reaction
rather than SN2 reaction.
(E) The carbocation prefer non-planar geometry.
5. Which of the following alkyl chlorides should undergo dehydrochlorination upon
treatment with potassium tert-butoxide most rapidly?
t-Bu
t-Bu
t-Bu
(A)
(C)
(B)
Cl
Cl
t-Bu
t-Bu
(E)
(D)
Cl
Cl
Cl
6. Which of the following statements is false?
(A) The reaction of ethyl iodide with potassium tert-butoxide gives
2-ethoxy-2-methylpropane as the major product.
(B) The conjugate base of an acid tends to be a better nucleophile than acid itself in
the SN2 reaction.
(C) Secondary alkyl halides react with sodium methoxide to give more substitution
product than they do with potassium tert-butoxide.
(D) The rate equation for the elimination reaction of tert-butyl chloride and sodium
methoxide is second-order.
(E) The major product from the elimination reaction of tert-butyl chloride and
sodium methoxide is isobutene.
7. Rank the following alkyl chloride in order of increasing reactivity (least reactivity
→ most reactivity) in an SN2 reaction.
NaI
R Cl
R I
+
NaCl
acetone
Cl
Cl
Cl
¢¹
(A) Ⅰ,
(B) Ⅱ,
(C) Ⅰ,
(D) Ⅱ,
(E) Ⅲ,
Ⅲ,
Ⅲ,
Ⅱ,
Ⅰ,
Ⅱ,
Ⅱ,
Ⅰ
Ⅲ
Ⅲ,
Ⅰ
¢º
¢»
8. What is the major organic product from this reaction?
CH3
H3C
CCH
H2O
CH2
H2SO4
CH3
(A)
(B)
CH3
H3C
H3C
CH3
H3C
OH
(C)
CH3
H3C
H3C
HO
CH3
CH3
OH
(D)
(E)
CH3
OH
H3C
H3C
OH
H3C
CH3
CH3
CH3
9. What is the major organic product from this sequence of reactions?
BH3.THF
(A)
H2O2, OH
(C)
(B)
OH
OH
(E)
(D)
OH
product
OH
mixture of isomers
OH
10. Optically pure (S)-2-iodobutane is treated with 10 % NaI in acetone until the
equilibrium is reached. Predict the product(s).
CH3
H2C
I
10 % NaI
H
acetone
CH3
optically pure (S)-form
(A) 50 % (R)-form & 50 % (S)- form
(B) 10 % (R)-form & 90 % (S)- form
(C) 90 % (R)-form & 10 % (S)- form
(D) 100 % (R)-form
(E) 100 % (S)-form
(?)
Ⅱ. Answer the following team problems: (70 pts)
1. Compound A drawn in Fischer projection formular is optically active. It undergos
the following sequence of reactions. (15 pts)
(A) What is the absolute configuration of the asymmetric centers in A, if any.
(B) What are the structures of B-E. Indicate their pertinent stereochemistry, if any.
H
H3C
CH3
OTs
H
C2H5
NaCN
NaI
acetone
B
C
DMF
A
CH3O-Na+ / DMSO
t-BuO-K+ / t-BuOH
E
D
( C6H12)
( C7H14O )
Answer
S
CH3
OTs
H
H
H3C
C2 H5
S
NaI
acetone
SN2
A
t-BuO-K+ / t-BuOH
S
R CH3
H
I
H
H3C
H
C
2 5
S
NaCN
DMF
SN2
B
C
CH3O-Na+ / DMSO
E2
C2H5
H
CH3
H3C
D
( C6H12)
CH3
CN
H
H
H3C
C2H5
S
E
( C7H14O )
2. Draw the major product formed when the optically active compound shown below
is treated with H2O in the present of a catalytic amount of H2SO4. How many
product(s) will be obtained from this reaction? Is(are) it(they) optically active?
Give your reasoning.(15 pts)
(optically acive)
Answer
H
H
H
path A
H
H
H
H3C
H2O
H3C
H
O
H
CH3
H
H
path B
H
O
H
H
-H
H3C
HO
OH
CH3
+
path A
major
path B
minor
Both of two products are optically inactive.
3. Compound F is treated with a catalytic amount of HCl to give the cyclic pyrane
derivative G. Write a plausible mechanism for the transformation. Is the product
optically active? Give your reasoning. (10 pts)
O
HO
HCl (cat.)
F
G
Answer
path A
HO
H
F
H
HO
H
O
path B
CH3
-H
O
CH3
racemic mixture
optically inactive
4. Upon treatment with potassium tert-butoxide, H gives I as the only product. On
the other hand, J yields a mixture of K and L. Rationalize the above observation.
(15 pts)
CH3
CH3
Base
Cl
(only)
I
H
CH3
CH3
CH3
Base
+
Cl
K
J
L
CH3
H3C
H
H
CH3
H3C
Cl
H
H
Cl
HH
H
CH3
H
H
CH3
more stable form, but
no H-Cl trans co-planar
H
Cl
less stable form, but has
H-Cl trans co-planar
CH3
H3C
base
CH3
CH3
H
H
H
CH3
H
CH3
Cl
I
CH3
(only)
K
base
base
path A
H
H
CH3
H3C
Cl
HH
J
CH3
H
CH3
Cl
most stable form has
two H-Cl trans co-planar
path B
L
5. Consider the hypothetical two-step reaction that is described by following energy
profile.
(A)
Is the overall reaction A→C exothermic or endothermic?(2 pts)
(B)
(C)
Label the transition state. Which transition state is rate-determining? (3 pts)
What is the rate law of the reaction? (use steady-state approach) (10 pts)
R
Q
M
E
k1
N
k2
N
M
esothermic
P
reaction coordinate
Answer
1. Reaction is exothermic.
2. Both Q & R are transition state, and R is rate-determining.
3. For the steady-state approach:
d[N]
dt = k1[M] - k2[N] - k3[N] = 0
So, [N] =
k1[M]
k2+k3
Rate law =
d[P]
dt
= k3[N] =
k1k3[M]
k2+k3
k3
P