Massachusetts Institute of Technology
Dr. Kimberly L. Berkowski
5.13, Fall 2006
Organic Chemistry II
EXAM #3 EXTRA PROBLEMS
KEY
What to expect on Exam #3:
1. ~1 Labeling experiment
2. ~2 Mechanisms
3. ~2 Syntheses
4. ~5 transformations – supply missing product
5. ~5 transformations – supply missing reagents
6. ~3 General questions
1. (4 points each, 8 points total) In the boxes, please provide the reagents for the illustrated
transformations. More than one step may be required.
(a)
1. KCN
2. H3O
i-Pr
or
+
1. Mg ,ether
2. CO2
+
3. H3O /workup
Br
O
i-Pr
OH
(b)
1. SOCl2
2. LiAl(OtBu)3H
3. workup
O
i-Pr
or
1. LiAlH4 (XS)
2. H2O
3. PCC
O
i-Pr
OH
H
Figure by MIT OCW.
2. (2 points each, 8 points total) Please provide the products of the following reactions. If no
reaction is expected, write “NR”.
(a)
O
Et
(b)
1. Excess Na BH4
Cl
O
Et
(c)
O
(d)
O
Et
NR
or
O
2. Workup
1. Excess MeLi
OMe
Et
2. Workup
O
HO
Me
Et
Me
2. Workup
1. Excess LiAlH4
NMe2
OH
2. Workup
1. Excess MeMgBr
OH
Et
Et
Et
NMe2
Figure by MIT OCW.
NAME____________________________
1
3. (2 points each, 16 points total) Please provide the requested products or reagents. If no
reaction is expected, write “NR”.
(a)
Br2,
NaOH
O
Me
NH2
Me
Me
NH2
H2O2, ∆
Me
(b)
O
Me
H2N
Me
OH
N-OH
1. LiAlH4
cat. H+
NH2
2. workup
(c)
POCl3
O
n-Bu
n-Bu
NH2
CN
H3O+
H2O
O
nBu
OH
(d)
NH2
NaNO2
HCl
N2
CuBr
Br
Figure by MIT OCW.
Name_________________________
2
4. (11 points) Please provide a detailed mechanism for the illustrated conversion of acetic acid
(A) to acetyl chloride (B).
O
Me
OH
+
Cl
O
S
O
Cl
Me
O
S
Me
+
HCl
Cl
Cl
OH
O
S
Cl
O
O
Me
+ SO2
B
A
O
Cl
S Cl
O
OH
Cl
OH
+
Cl
Me
Cl
O H
Me
+ SO2
Cl
Figure by MIT OCW.
Name_______________________
3
5. (11 points each, 22 points total) Please provide syntheses for only two of the three indicated
compounds. All the carbon atoms should be derived from the allowed starting materials. You
may use any common reagents.
Allowed Starting Materials:
Me
Me
CO2
Me
Me
OH
Me
A
Pick Two:
N
C
Me
B
C
HO Me
H
N
Synthesis # 1:
D
A
O
OH
2
O
Me
1.O3
2.MeS
3.Na2Cr2O7
H2SO4
SOCl2
Cl
NH3
B
Me
Me
Me
KMnO4
Me OH
1.BH3, THF
2. H2O2,−OH
or
H3O+/H2O
O
NH2
OH
POCl3
C
1.PBr3
N
MgBr
O
1.CO2
OH
2. H +
2.Mg
SOCl2
OH
O
(excess)
1.MeMgBr
2.H2O
Cl
Figure by MIT OCW.
Name__________________________
4
5. (Continued)
Allowed Starting Materials:
Me
Me
Me
CO2
Me
A
Pick Two:
B
Me
C
N
OH
Me
Me-OH
C
HO Me
Me
Me
H
N
Me
Me
O
1.
Synthesis #2:
C
MeOH
OH
O
2. H2NNH2
PBr3
or
TsCl, Pyridine
Br
O
NH2
PCC
H
H
N K
NH2
or 1. NaN3
2. LiAlH4
3. H3O+
or XS NH3
Cat H
N
1. LiAlH4
2. H2O
H
N
Figure by MIT OCW.
Name________________________________
5
6. (11 points) Provide a synthesis that will selectively convert A to B. Show all the key
intermediates and furnish all the important reagents. This is not a one-step process.
O
Me
A
O
Me
HNO3
H2SO4
O 2N
O
HO
Me
B
O
Me
H2, Pd
H2N
N2
O
Me
NaNO2, HCl
O
Me
H2SO4, H2O
HO
O
Me
Figure by MIT OCW.
Name________________________________
6
7. Methyl acetimidate (A) is hydrolyzed in aqueous sodium hydroxide to give mainly acetamide
and methanol (eq 1). In aqueous acid, A hydrolyzes to give primarily methyl acetate and
ammonium ion (eq 2).
a) Provide a detailed mechanism for the illustrated process. Please show all arrow
pushing.
HO-
NH
Me
N-H
Me
OMe
A
O
H2O
OMe
Me
NH2
PT
Me
OH
NH2
O
OMe
Me
O
MeOH
+
H OH
+ MeO
(1)
MeOH + OH
NH2
Figure by MIT OCW.
b) Provide a detailed mechanism for the illustrated process. Please show all arrow
pushing.
NH
Me
H
Me
N
A
H
Me
OMe
OMe
NH2
OMe
excess H
H 2O
PT
OH2
O
+
Me
Me
OMe
NH4
+
NH3
OMe
OH
H
O
Me
(2)
OMe + NH3
H2O
O
Me
OMe +
NH4
Figure by MIT OCW.
c) Briefly explain why the two reactions provide different products.
Basic conditions:
NH2 worse L.G. than
OMe . Elimination favors amide
Acidic conditions: Acid/base equilibrium favors protonation of
ntrogen making it a good
L.G. Also, NH4 is not nucleophilic + formation of ester is reversible
Figure by MIT OCW.
7
Massachusetts Institute of Technology
5.13: Organic Chemistry II
8. Synthesize the indicated compounds from the allowed starting materials shown below. All of
the carbons of the target compounds should be derived from the allowed starting materials.
8.
b)
TsCl
MeOH
NaCN
MeOTs
MeCN
N
1) LAH
NH2
2) H2O
O
O
HBr
EtOH
EtBr
Mg
1)
H, O
2) H3O+
EtMgBr
O
Na2Cr2O7
OH
EtNH2, H+
PCC
O
N
T
H
O
O
c)
Me
OH
Me
OH
Me
OH
KMnO4, H+
PBr3
2
Br
Me
Cl
1
Br
2
O
Me
Me
OH
Me
PCC
CN
SOCl2
Me
Me
3
H
C
N
LiAlH4
O
Me
4
H
NH2
Me
Me
Me
NH
Me
H
3
LiAlH4
O
1
Cl
Me
Me
Me
N
T
O
Figure by MIT OCW.
8
Massachusetts Institute of Technology
5.13: Organic Chemistry II
1 from part c
O
Cl
(d)
Me
CN
OH
O
Me
H-CN
Me
AlCl3
LiAlH4
OH
NH2
Me
NO2
H2SO4
(e)
NH2
H2/Ni cat
Br
FeBr3
HNO3
NH2
Br2(excess)
Br
Br
NaNO2
HCl
Br
Br
O
H2O, H
OH
Br
Br
Br
Br
C N
N N
CaCN
Br
Br
Br
SOCl2
Br
Br
O
Cl
Br
Br
+
H 2N
O
N
H
Me
4 from part c
Br
Me
Br
Figure by MIT OCW.
9
Massachusetts Institute of Technology
5.13: Organic Chemistry II
9. Provide the best stepwise mechanism for the illustrated process. Please show all arrow
pushing.
(9)
O
O
H
NH2
O
H+
H
O H
H
H
NH2
O
O
O
H
H
N
H
H 2O
H
H
H
N
H
+
H
H
OH
N
H
H
H
OH
NaBH3CN
"H "
O
O H
H
H
H
N
H
H
N
H
HO
N
H
H
H
H
H
NaBH3CN
N
H
N
"H
"
H
N
H2O
H
O
H
N
H
Target
Figure by MIT OCW.
10
Massachusetts Institute of Technology
5.13: Organic Chemistry II
10. (a) Provide the best mechanism. Please show all arrow pushing.
(a)
O N
benzylic cation
H
H
C Me
O
OH
CH3
N
C CH3
C
N
OH
OH
OH
OH
etc
N C
CH3
O H
or workup
H
Me
N
O
Target
Figure by MIT OCW.
11
Massachusetts Institute of Technology
5.13: Organic Chemistry II
(b) Provide the best mechanism. Please show all arrow pushing.
Me
N
H
N
H
O
OH2
O
H 2O
H H
Me
N
O H
O
Me
H
N
H
Me
O H
O
H
H
NH2
O
NH3
H
O
O
O
Me
Me
+
H
NH3
O H
Me
O
H O
H
NH3
O
O
H
Me
OH2
H
NH3
O H
Me
O
OH
H
OH2
NH3
OH
O H
HO
Me
workup
NH2
OH
Figure by MIT OCW.
12
Massachusetts Institute of Technology
5.13: Organic Chemistry II
11. Consider the labeling experiments outlined below:
11.Start with the mechanisms:
O H
NHMe
H2O
HO
k1
PT
O H2
OH
H 2O
O
k1
NHMe
NHMe
NHMe
PT
PT
HO
OH
k2
O H
OH
OH
NH2Me
Proton transfer are very fast.
O H
Acid/base equilibria favor the protonated
N (not O). Therefore, once the tetrahedral
intermediate forms, loss of NH2Me (k2) is
faster than loss of OH (k1).Very little O is
incorporated into the unreacted starting
material.
H
O
O
OH
OH
O H
OMe
H2O
k1
HO
O H2
PT
H2O
OH
O
k1
OMe
OMe
OMe
PT
PT
HO
All of the oxygens in the tetrahedral
intermediate are roughly equally basic.
Therefore, each protonated form 13
present in the same concentration ,and
k2 ~= k1. as a result, you would expect
more O incorporation into the ester starting
material than you would into the corresponding
amide.
OH
OMe
H
k2
O H
O H
OH
OH
H
O
O
OH
OH
Figure by MIT OCW.
Name_______________
13