Name:
Student ID number:
12/12/07
152/252:
Synthetic Methods in Organic Chemistry
Final Exams
Instructions:
1.
Please write your name at the top of every page. Please write your Student ID
number at the top of the first page, right below your name.
2.
The exams are from 11:30-2:00 pm.
3.
This exam is composed of 8 pages total. Page 8 is blank.
4.
This is an open book exam, and you are free to consult your notes anytime
during the exam. However, the more time you spend looking at your notes, the less
time you may have to finish the exams....
5.
Answer the questions concisely, within the provided space and erase any
irrelevant structure/reagent or unnecessary information. If you need additional space
please use the back side of any page.
6.
Your grades, exams and related material will be available for pickup from 12/15-
12/30 from my assistant's office at PacHall 5224B.
7.
The grades will be submitted to the department on 12/15/07.
GOOD LUCK !!! and HAPPY HOLIDAYS!!!
Problem
Points
Score
1.
60
________
2.
60
________
3.
60
________
4.
60
________
5.
60
________
TOTAL
300
_______
1
Name:
PROBLEM 1: (3 parts, 60 points)
1. The Diels Alder reaction between diene A and dienophile B gives a mixture of products
C and D in a ratio: C/D = 1/6.
Me
CO2Me
CO2Me
Me
+
Me
+
CO2Me
A
B
C (10%)
D (60%)
a. Explain the regioselectivity of the reaction using a pneumonic device and 1-2 sentences.
pneumonic device showing charges generated at transition state:
Me
Me
CO2Me
A1
Formation of D is
favored because the
methyl substituent of
the diene stabilizes
the partial charges as
shown in structure A1.
CO2Me
D
Me
CO2Me
Me
CO2Me
A2
20 points
C
What product ratio would you expect (approximately) for the following reactions, if
performed under identical conditions as above and why (answer with 1-2 sentences).
MeO
CO2Me
CO2Me
MeO
+
b.
MeO
+
CO2Me
E
F
G
H
The methoxy substituent at the 2-position of the diene stabilizes better than the
methyl group the partial charges formed at the DA transition state (shown as A1
above).
Based on this, I expect the ratio of G/H to be smaller than that of the ratio C/D (which
is 1/6). Compound H should be formed as the major product in better than 6/1 ratio.
20 points
I predict that the ratio of G/H= 1/9
MeO
SO2Ph
SO2Ph
MeO
+
c.
MeO
+
SO2Ph
F
G
F
H
The sulfonyl group of the dienophile is a better electron withdrawing group than the ester
(structure F). Thus the DA reaction would proceed more efficiently and with higher
selectivity in favor of product H. The pneumonic device A1 applies better here than in the
cases above.
Based on this, I predict that the ratio of G/H= 1/20
20 points
2
Name:
PROBLEM 2 (60 points):
a. Provide all reagents and conditions that you can use to achieve the following
transformations. Multiple steps are needed. Possible sources of deuterium are: CDCl3,
LiAlD4, NaBD4, NaCNBD3, Bu3SnD, D2O, D2, C6D6, CD3CO2D, Et3SiD).
OH
D
O
H
A
B
CrO3•pyr
20 points for correct answer
-5 points per wrong reaction
or other
oxidants
BF3•Et2O
Et3SH
O
Cl
O
O
OH
O
O
DIBAL-D
O
OH
D
or other
reductants
mCPBA
(Baeyer-Villiger)
D
OH
C
CrO3
H2SO4
(Jones [O]
or other [O]
N
O
OH
D
20 points for correct answer
-5 points per wrong reaction
S
OH
h!
Bu3Sn-D
O
O
N
S
DCC
3
Name:
b. Give the structure of products or reagents A-H
O
G
O3, then
NaBH4 (excess)
B
E
O
NaIO4
OsO4
A
DMSO, (COCl)2
-78 °C
then Et3N
NMO
OH
D
CH3-PPh3 I
1. TBSCl, Et3N
(excess)
2. mCPBA
NaHMDS
H2,
Pd/C
(excess)
H
C
O
OH
OH
F
O
O
OTBS
O
O
OH
OTBS
H
A
OH
C
OH
B
Li(tBuO)3AlH
O
OTBS
or
CH2
NaCNBH3
D
H
O
mild reducing
reagents selective
for aldehydes
E
OH
OH
H
CH2
O
20 points total
-3 points per wrong answer
OH
G
F
H
OH
4
Name:
PROBLEM 3 (2 parts 60 points):
Using the starting materials A and C and any reagents of your choice suggest an
enantioselective synthesis of compounds B and D respectively. (Possible sources of
deuterium are: LiAlD4, DIBAL-D, NaCNBD3, Bu3SnD, D2O, D2, Et3SiD).
O
OH
enantioselective
synthesis
D
A
B
– DET
[O]
D
D
30 points total
+/-10 points for
enantioselective
step
OH
NaIO4
OH
OH
O
TBAF
SAE
(-) DET
Ti(OiPr)4/
tBuOOH
SAE gives access
to one enantiomer
TBSCl
imid
O
D
LiAlD4
O
OTBS
OH
OH
OTBS
OH
OH
enantioselective
synthesis
Me
D
C
CrO3
H2SO4
(Jones [O]
O
OH
O
LiAlH4
O
Ph
O
O
Me
30 points total
+/-10 points for
enantioselective
step
PhCH2O-Li+
N
O
Me
SOCl2
Ph
O
H N
O
O
O
Ph
Cl
N
Et3N
Ph
5
LDA
MeI
O
O
Evans oxazolidinone
forms one enantiomer
Name:
PROBLEM 4 (60 points)
Prostanglandins are a family of extremely potent hormone-like compounds with many biological
functions, including muscle stimulation, lowering of blood pressure and generation of
inflammation. A synthetic scheme towards prostanglandins as developed by E. J. Corey is shown
below. (Corey, E. J. et al. J. Am. Chem. Soc. 1969, 91, 5675; 1971, 93, 1490).
Complete the synthetic scheme toward prostanglandin PGF3a by adding the appropriate reagents
and/or providing the structure of the products.
(More than one reagents/reactions may be correct for each step. Full credit will be given even if
you have performed the desired transformation(s) using less/more steps than the ones indicated).
NC
Cl
MeO
1. KOH, H2O
MeO
OMe
Cu(BF4)2
NC
2.
Cl
1. NaOH, then H+
2. KI3, NaHCO3
6 pointd/box
no partial
O
O
O
1.
O
O
O
O
BBr3
1. Ac2O, pyrid
I
2. Bu3SnH,
AIBN, 80 °C
2.
PCC
AcO
AcO
HO
OMe
OMe
H
O
Ac:
O
Me
O
then H+
1. Ph3P
2.
O
O
mCPBA
O
THP:
O
K2CO3, MeOH
THPO
O
3.
OTHP
, TsOH
1.
2.
HO
CO2H
AcOH, H2O
DIBAL-H
PPh3
then
HO
H+
CO2H
HO
PGF3a
OH
THPO
6
OTHP
CO2
Name:
PROBLEM 5 (60 points)
Complete the synthetic schemes, by adding the appropriate reagents and/or providing
the structure of the products. More than one answers may be correct. Full credit will be
given if more/less steps are used than the ones indicated.
H
O
O
H
1. TBSCl, imidazole
H
O
Ph
O
H
H
2. 9-BBN, then
H2O2, NaOH
OH
Ph
O
O
H
H
H
OH
OTBS
TBSCl : tBuMe2Si-Cl
1. PCC
12 points/box
2. Ph3P=CHCO2Et
H
O
Ph
O
H
H
O
O
H
OTBS
OH
O
1. DIBAL-H
2. (–)DET, Ti(iPrO)4
tBuOOH
Ph
O
O
H
H
OTBS
CO2Et
1. PCC
2. [ Ph3P=CH2 ]
1.
H
O
Ph
O
TBAF
2. PPTS, CH2Cl2
or other acids, LA
H
H
O
O
O
H
Ph
OTBS
7
O
H
O
H
H
OH
O