exam 3 probs key
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7/17/2019 Exam 3 Probs Key
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Massachusetts Institute of Technology 5.13, Fall 2006
Dr. Kimberly L. Berkowski Organic Chemistry II
EXAM #3 EXTRA PROBLEMS
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
KEY
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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.
2.
(2 points each, 8 points total) Please provide the products of the following reactions. If no
reaction is expected, write “NR”.
NAME____________________________
1
Figure by MIT OCW.
Figure by MIT OCW.
O
OHi-Pr
O
Hi-Pr
O
OHi-Pr
i-Pr Br
(a)
(b)
1. KCN
2. H3O+
or
or
1. Mg ,ether
2. CO2
3. H3O
+/workup
1. SOCl22. LiAl(OtBu)3H
3. workup
1. LiAlH4 (XS)
2. H2O
3. PCC
Et
O
Et Cl
O
OEt OH
O
Et OMe
O
(a)
(b)
(c)
(d)
Et NMe2
O
1. Excess Na BH4
2. Workup
2. Workup
1. Excess MeMgBr
2. Workup
1. Excess MeLi
2. Workup
1. Excess LiAlH4
OH
NR
or
HO Me
MeEt
Et NMe2
Et
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3. (2 points each, 16 points total) Please provide the requested products or reagents. If no
reaction is expected, write “NR”.
Name_________________________
2
Figure by MIT OCW.
NH2 N2 Br
(d)
(c)
NH2
O
O
n-BuH2O
H3O
+
Me Me
H2 N
cat. H+
(b)
1. LiAlH4
2. workup
Me
MeMe
Me NH2 NH2
H2O2, ∆
(a)
Br 2,
NaOH
N-OH
POCl3
NaNO2
HCl
CuBr
O
n-Bu
NH2
O
OHnBuCN
OH
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4. (11 points) Please provide a detailed mechanism for the illustrated conversion of acetic acid
(A) to acetyl chloride (B).
Name_______________________
3
Figure by MIT O
OHMe
O
OHMe Me
O
O
OHMe
O
Me
O
ClMe
OO
+S
Cl Cl
OSCl
Cl
O
S Cl
OS C l
C l
+ +
+
+
SO2
SO2
HCl
BA
Cl
Cl
H
Cl
OH
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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.
Name__________________________
4
Figure by MIT OC
Allowed Starting Materials:
Me
MeMe Me
Me
Me CO2 OH OH
Me
Me
C N
C N
Me
Me
HO
MeMe N
H
A
A
B
B C
Pick Two:
Synthesis # 1:1.O3
2.MeS
3.Na2Cr 2O7
H2SO4
OH
O
2KMnO4
D
Cl
O
O
NH2 NH3
SOCl2
POCl3
1.BH3, THF
2. H2O2,−OH
or
H3O+/H2O
OH 1.PBr 3
2.Mg
MgBr 1.CO2
2. H
O
OH
O
Cl
SOCl2
1.MeMgBr
2.H2O
(excess)OH
+
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5. (Continued)
Name________________________________
5
Figure by MIT OCW.
Allowed Starting Materials: Me
Me
Me
Me Me
Me
Me Me
CO2 OH
Me
Me-OH
NH2Br OH
PBr 3
or
or
or TsCl, Pyridine
Me
Me
C N
HO
NH
Pick Two:
Synthesis #2:
O
O
N K
2. H2 NNH2
1. NaN3
2. LiAlH4
3. H3O+
1.
NH2
N N
H
H
H
O
MeOHPCC
Cat H
1. LiAlH4
2. H2O
XS NH3
A B C
C
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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.
Me
O
Me
O
Me
O
A
HO
Me
O
B
H2 N
HNO3
H2SO4
Me
O
Me
O
H2, Pd
NaNO2, HCl
O2 N
N2
Me
O
H2SO4, H2O
HO
Name________________________________
6
Figure by MIT OCW.
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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.
b)
Provide a detailed mechanism for the illustrated process. Please show all arrow
pushing.
c) Briefly explain why the two reactions provide different products.
7
Figure by MIT OCW.
Figure by MIT OCW.
Figure by MIT OCW.
OMe NH2
OMe
Me
Me
A
H2O
NH
Me
NH2
O
O
NH2Me
O
+
OMeMe
N-H
OH
PT
(1) MeOH
MeOH + OH+ MeO H
HO-
OH
Basic conditions: NH2 worse L.G. than OMe . Elimination favor s amide
Acidic conditions:
L.G. Also, NH4 is not nucleophilic + formation of ester is reversible
Acid/base equilibrium favors protonation of ntrogen making it a good
OMe OMe
OMe
Me
Me
A
H2O
NH
OMeMe
N
Me
NH2
OH2
O
OMeMe
NH4+
OMeMe
NH3
OMeMe
O
NH4+
OH
PT O
(2)
H H
H2O
NH3
excess H+
+
H
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Massachusetts Institute of Technology
5.13: Organic Chemistry II
8. Synthesize the indicated compounds from the allowed starting materials shown below. All ofthe carbons of the target compounds should be derived from the allowed starting materials.
8
Figure by MIT OCW.
Me OH
Me OH
Me OH
Me
Me OH
KMnO4, H+
O
H
H
O
Me H
O
Me
O
SOCl2Cl
Me
O
Cl
Me
O
PBr 3Br
Me Br Me Me
Me
Me
Me
PCC
C N NH2
O
N
Me
Me NH
H
HLiAlH4
LiAlH4
CN
T
3
1
2
3
2
4
1
T
N
MeOH MeOTs MeCN1) LAH
2) H2O
NaCN NH28.
EtOHEtBr EtMgBr
EtNH2, H+PCC
OHOMgHBr
TsCl
H O
1)
2)
O O
c)
b)
O
,
3+
Na2Cr 2O7
N
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Massachusetts Institute of Technology
5.13: Organic Chemistry II
9
Figure by MIT OCW.
Cl Me
Me
AlCl3
O
Me
O
H-CN Me
CNOH
OH
LiAlH4
NH2
NH2
Br Br
Br NH2 NO2
Br Br
Br
N N
Br Br
Br
C N
H2SO4
HNO3
H2/Ni cat Br 2(excess)
FeBr 3
NaNO2
HCl
CaCNH2O, H
Br Br
Br O
OH
(d)
(e)
1 from part c
Br Br
Br O
Cl + H2 N Me
Br Br
Br O
N Me
H
4 from part c
SOCl2
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Massachusetts Institute of Technology
5.13: Organic Chemistry II
9. Provide the best stepwise mechanism for the illustrated process. Please show all arrow
pushing.
10
Figure by MIT OCW.
O O O OH H
H H
H
H
NH2 NH2
H2O
H2O
+
H
H
+H
O
H
H HHH N OH
H
H N
HO
O
O
H
HH
H N
O
H
H
N
H
O
H
H
H
N
H
H
H
H
H NOH
N NaBH3CN
HH N " "
O
H
HH N
H
NaBH3CN
" "
NTarget
(9)
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Massachusetts Institute of Technology
5.13: Organic Chemistry II
10. (a) Provide the best mechanism. Please show all arrow pushing.
O O
OH
OH
OH
OH
N C
C C
C
O H
H
HH
CH3
N
O
CH3
CH3
N
N
N Me
etc
OH
benzylic cation
Me
Target
or workup
(a)
11
Figure by MIT OCW.
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Massachusetts Institute of Technology
5.13: Organic Chemistry II
(b) Provide the best mechanism. Please show all arrow pushing.
N
O
Me N
O
Me
H2O
H
H
H
OH2
NO
O
MeH
H
H N
OO
O
Me
Me
NH2
O
H H
H
HO
Me
NH3
O
H
O
Me
NH3
O
H NH3
O
O
OOH2
OH2
OH
H
H
H
NH3
O
O
H
H
H
H
Me
Me
Me
O H NH3
OH
NH2
O
O
H
H
workup
+
12
Figure by MIT OCW.
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Massachusetts Institute of Technology
5.13: Organic Chemistry II
11. Consider the labeling experiments outlined below:
Name_______________
Figure by MIT OCW.
HO
NHMek 1
k 1
k 2
NHMeHO
NHMe
H2O
NHMe
PT H
OMe
OMeOMe OMe OMe
O
OH
NH2Me
H
OH
HO
O
HO
PT PT
k 2HO
PT PT
H
OH
OH
H
H
O
H
Proton transfer are very fast.Acid/base equilibria favor the protonated
N (not O). Therefore, once the tetrahedral
intermediate forms, loss of NH2Me (k 2) is
faster than loss of OH (k 1 .Very little is
incorporated into the unreacted starting
material.
All of the oxygens in the tetrahedral
intermediate are roughly equally basic.
Therefore, each protonated form 13
present in the same concentration ,and
k 2 =~ k 1. as a result, you would expect
more
incorporation into the ester startingmaterial than you would into the corresponding
amide.
11.Start with the mechanisms:
H2O H2O
O
O
O
O
HO
O
HO
k 1
k 1HOH2OPT
HH2O H2O
O
O
O
O
HO
HO
HO
HO
HO
HO
)