Chem 331 Final exam December 13, 2002 Prof. Fox

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Chem 331
Final exam
December 13, 2002
Prof. Fox
180 minutes
The exam is open book,
Open notes. Models are permitted
Show your work in detail
300 points
WRITE YOUR NAME ON EVERY PAGE
NAME
Chem 331, Final Exam. December 13, 2002
1. Identify the following functional groups
NAME
(Total 10 points)
e.g.
O
O
C N
OH
O
H
alkene
nitrile
2. Provide IUPAC names for the following
a
Br
O
Carboxylic acid
aldehyde
(15 points each)
(2R)-2-methylbutyl-(3R)-3-bromobutanoate
O
b
CH3
H 3C
H
HO
H
Br
(1S)-1-bromo-2-methyl-1-propanol
epoxide
Chem 331, Final Exam. December 13, 2002
NAME
3. Draw the structures of A and B. You do not need to provide mechanisms. Stereochemistry is important!
(20 points)
1) Mg, ether
SOCl2
OH
2)
Cl
A
O
OH
B
4. Circle the compound that is more stable (0 points). Draw clear 3-D chair representations of the two
compounds and describe how your drawings support your answer (20 points).
iPr
iPr
H
Me
H
Me
H
Me
this conformation
is most stable
both the iPr and Me
are equatorial
H
axial Me costs
1.70 kcal
H
iPr
Me
H
axial iPr costs
2.20 kcal
H
iPr
H
Chem 331, Final Exam. December 13, 2002
NAME
5. Compound A would undergo E2 elimination if treated with NaOtBu. Explain which one of the four
possible products below would be formed. To receive full credit, you must:
• circle the correct product (5 points)
• draw a clear 3-D representation of the trans-decalin framework 1 (12 points)
• provide a mechanism and description of why only only one compound is formed (13 points)
D
H
NaOtBu
H
Cl
Br
H
H
H
D
Br
H
H
D
Cl
A
H
H
H
H
(total 30 points)
Br
H
H
Cl
–
O
H
D
H
Br
H
H
H
DO
H
Br
Cl–
Cl
Anti-periplanar E2 elimination. The axial substituents (D and Cl) are eliminated. Transdecalin cannot ring flip, and therefore a conformation cannot be adopted in which
the Br or H is axial.
Chem 331, Final Exam. December 13, 2002
NAME
6. The reactions below would not proceed as shown. Explain why (be concise), and indicate which product
would be formed instead.
a.
MCPBA
O
X
O
b.
Et
D
H
Et
D
Cl
the more highly substituted (and more electron rich)
double bond is the most reactive towards epoxidation
NaOtBu
D
X
Et
H
Me
you must show a mechanism
with attention to stereochemistry
Me
(20 points)
Cl
Et
H
H
Me
H
(20 points)
D
Et
Cl
H
Me
major product
H
Me
tBuO–
antiperiplanar elimination will
predominantly occur from the more
populated conformer.
Newman projections
Cl
Cl
D
H
H
Et
Et
Me
H
H
Me
D
this gauche interaction cost ca. 0.7 kcal/mol
Chem 331, Final Exam. December 13, 2002
NAME
6( continued) The reactions below would not proceed as shown. Explain why (be concise), and indicate which produc
would be formed instead.
c.
O
1) CH3MgBr
CN
2) H+
CH3
X
(20 points)
O
O
reduction of the nitrile is not selective in
the presence of the ketone. Either of the answers below were accepted
O
CN
CH3
CH3
HO
d.
HO
+
–
NMe3 OH
CH3
∆
CH3
(20 points)
X
Hoffman Eliminations give the least substituted double bond product via reaction of the most
sterically accessible hydrogen
+
NMe3
H
–
OH
Chem 331, Final Exam. December 13, 2002
NAME
(40 points)
7. Propose a multistep synthesis of 2 using 1 and any other materials.
OMe
O
NC
1
OMe
2
NaCN
1) O3
2) Me 2S
OMe
O
O
TsCl
O
NaBH4
O
pyridine
HO
OMe
TsO
OMe
Chem 331, Final Exam. December 13, 2002
NAME
(35 points)
8. Propose a detailed mechanism for the interconversion of 3 and 4.
O
Ph
C*
Ph
O
H+, H2O
Ph
Ph
Ph
C*
Ph
Ph
Ph
3
4
13
C* = C
all other carbons are C12
H+
O
Ph
Ph
+ H
O
:
Ph
C*
Ph
Ph
C*
Ph
Ph
Ph
O
+
H
:
:
:O
C*
Ph
Ph
+
Ph
Ph
Ph
Ph
C*
Ph
Ph
H
:
: OH2
H
H
H
+O
Ph
*C
Ph
Ph
: O:
Ph
+
Ph
O
*C
Ph
*C
Ph
Ph
Ph
Ph
Ph
Ph
The following was NOT accepted:
:
:O
H
H
C*
Ph
Ph
Ph
+
Ph
Ph
+
C*
Ph
O
Ph
Ph
Ph
Ph
O
+*
Ph
Ph
H
+O
Ph
Ph
C* Ph
Ph
Chem 331, Final Exam. December 13, 2002
NAME
9. Propose a detailed mechanism
(35 points)
OH
OH
Br
Bu3SnH
O
O
AIBN
Initiation
NC
N N
CN
NC
NC
+
H
•
+
•
H
SnBu3
+
CN
N N
• SnBu3
Propagation
Bu3SnBr
Bu3Sn •
OH
OH
atom extraction
Br
O
multiple bond
addition
•
OH
O
O
•
multiple bond
addition
H SnBu3
OH
Bu3Sn •
atom extraction
+
O
OH
OH
•
fragmentation to
form a double bond
O
•O
The solution uses the three reactions that you know for radical reactions. The following was not accepted:
single bond additions are not part of your synthetic toolkit!
OH
•
OH
•
O
X
O
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