Name________KEY____________
Page 1
I. (42 points)
Provide structures of the products or reactants as indicated by the boxes. If there is more than one
possible outcome, provide the major product as implied by the conditions and any given information.
When a racemic mixture forms, show the stereochemistry of one enantiomer and write "and
enantiomer".
(a)
CH3
O
CH3
O
CH 3CH2 C
C
H
C
CH3
CH3
O
3 points structure
2 points stereo
CH2CH 3
(CH3)3C-O-OH
(an initiator)
C 11H12O3
C
C
6
3 points structure
3 points stereo
O
2. HCl, H 2O, heat
H
Tetrahedron Lett. 2001, 42, 4281
O
S CH2CH 2CN
(1 pt)
H
OOH
1. Ph2 CuLi
OCH3
H
C H
C
and enantiomer
R configuration at new s tereocenter
O
CH3
(d)
6
CH3
CH 3CH2 C
O
C
3 points structure
3 points stereo
H O
O2
J. Org. Chem. 2001, 66, 4087
O
O
O
(b)
O
CH3
+
J. Org. Chem. 2001, 66, 7129
(c)
H O cis only
O
DON'T HAVE TO ASSIGN
R/S CONF IGURATION, BUT
M UST BE DRAWN CORRECTLY
CH2NH2
N
J. Am. Chem. Soc . 2001, 123, 8053
H
CH2OH
6
-1
no partial
CH2
N
C
S
strong IR abs orptions at 3550 and 1770 cm
no partial
N
heat
R
H
S CH2CH 2CN
6
NaBH4
NH CH2
ethanol
CH2
N
S CH2CH 2CN
6
Name________KEY____________
Page 1
(e)
CH3
C
C C
C
(CH3O)2CH CH2 CH2
4 points structure
2 points stereo
Cl
O
H
CH2OH
1. Cl
CH3
O
CH3 S CH3
O ,
C
C
(CH3O)2CH CH2 CH2
C
6
(1 pt)
CH3
O
H
1. LiCH2Cl
2. NH4Cl, H 2O
(protonation)
2. (CH3CH2 )3 N
and enantiomer
J. Am. Chem. Soc . 2001, 123, 8210
H
H
C
(CH3O)2CH CH2 CH2
3 points structure
2 points stereo
C
H
C
OH
CH2 C l
6
Name________KEY____________
Page 2
II. (40 points)
A. Complete hydrolysis of an octapeptide gave the following amino acids in alphabetical order:
Ala, Asp, Gly, Lys (2), Phe, Ser, Tyr
Treatment of the peptide with 2,4-dinitrofluorobenzene (DNFB, Sanger's method) identified Ala as the
N-terminal amino acid.
Trypsin digestion of the octapeptide gave three smaller fragments containing the following amino
acids.
Fragment A: Asp, Tyr
Lys, Phe
Fragment B: Ala, Lys, Ser
Fragment C:
Gly,
Chymotrypsin digestion gave three smaller fragments containing the following amino acids.
Fragment D: Ala, Lys, Phe, Ser
Fragment E: Gly, Lys, Tyr
Fragment F: Asp
Write out the full sequence of the amino acids in this octapeptide. Use the standard convention for
writing peptides (i.e., start with the N-terminal residue on the left).
no partial except +1 pt for correct N-terminal residue
and +1 pt for correct C-terminal residue
Ala-Ser-Lys-Phe-Gly-Lys-Tyr-Asp
9
B. Draw the full structure (include stereochemistry) of the product formed when Ala reacts with
DNFB.
CH3
H
NO2
O
C
+
C
O2N
NO2
F
O2N
O
H2N
Ala
CH3 H
NH
C
C
O
4 points structure
1 point stereo
5
OH
DNFB
C. Draw the full structure (include stereochemistry) for the tripeptide Cys-Pro-Arg in the predominant
form it would exist at its isoelectric point. pKa's for underlined hydrogens are as indicated.
O
OH 2.05
O
C
H
8.00 HSCH2
C
OH
C
NH3
Cys
10.25
N
H2
10.60
H
Pro
2.00
H2N
C
OH 2.01
O
12.48
NH2
C
H
NHCH2CH 2CH2
Arg
C
NH3
9.04
Name________________________
Page 2
Structure at the is oelectric point:
H
SCH2
H
N
C
C
C
O
H3N
The predominant form of this tripeptide
at pH 1 has a net charge of: (circle one)
+2
H
N
O
O
C
0
-1
-2
no partial
NH2
C
+2
8 pts correct structure and stereo
8 pts correct charges
5
The predominant form of this tripeptide
at pH 11 has a net charge of: (circle one)
O
C
NH2
CH2CH 2CH2NH
H
+1
+1
0
16
-1
-2
no partial
5
III. (38 points)
A. When 1-phenyl-1-propene is heated with N-bromosuccinimide (NBS) in carbon tetrachloride
solvent, a substitution reaction occurs to give predominantly one regioisomer.
O
CH
CH CH3
Br
N O
(NBS)
CCl4
Br
CH CH CH2Br
CH CH
+
MAJOR
heat
CH2
MINOR
(a) Using the curved-arrow convention, provide the detailed step-wise mechanism for this free-radical
chain reaction as directed in the boxes below.
Reactants , Products and Mechanism for the Initiation Step
O
O
N
N
Br
1 error = 3 pts
>1 error = 0 pts
+
Br
5
O
O
Reactants , Products and Mechanism for the First Propagation Step
CH CH CH2
H
1 error = 3 pts
>1 error = 0 pts
CH
Br
CH CH2
+ H
Br
5
(b) The regioselectivity occurs in the second propagation step where the radical generated in the first
propagation step reacts with Br2. Draw all five important resonance forms for this particular radical
and briefly explain the observed regiochemical outcome.
CH
CH CH2
CH
CH CH2
CH CH CH2
CH
CH CH2
CH
CH CH2
1 pt for each correct resonance form
Addition of bromine to the aromatic ring would destroy aroma ticity thus not favore d. Of the other two
sites of e lectron deficiency in this de loca lized radical, bromine adds to the site that gives the more sta ble product
(double bond conjugated to benzene ring and also more highly substituted)
5 pts for explanation of regioselectivity
B. Complete the following reaction scheme as indicated. Clearly show stereochemistry.
10
Name________KEY____________
Page 3
O
CH3 C O CH2
O
O
CH3 C O
CH3 C O
O
O
CH3 C O CH2
O
O
CH3 C O
OH
O
O C CH3
O C CH3
O
BrCH2 CH CH2Br
BF 3 . O(CH 2CH 3)2
(a Lewis acid catalyst)
J. Org . Chem. 200 1, 66 , 3783
no partial excep t
O
transcriptio n errors
CH3 C OCH2
CH2NH2
O
O
CH3 C O
O CH CH2NH2
CH3 C O
C 17 H 28 N 2 O10 O
O C CH3
O
6
H2
PtO2
CH3 C O
-an omer O
O
CH3 C O CH2
O
O
CH3 C O
CH3 C O
O
3 p ts structure
3 p ts stereo
CH2Br
O CH CH2Br
O C CH3
O
6
NaN 3 (exces s)
no partial excep t
transcriptio n errors
CH2N3
O CH CH2N3
O C CH3
O
6
Name________KEY____________
Page 4
IV. (40 points)
Provide structures of the expected reactants or major products as indicated by the boxes and any
given information. Show stereochemistry where known.
(a)
O
CH 3
H2N
H
C
CH3 C O
CH2 CH
O
CH 3
CH2
no partial
O
C
(CH3)3CO C NH
2
C
H
CH2 CH
CH2
Tetrahedron Lett. 2001, 42, 4301
4
O
O
no partial
(CH3)3CO C NH
C
H
Ph3P
H
CH
(CH3)3CO C NH
C
I
I
C
CH2 CH
H
O
1. O3
CH2 CH
2. (CH3) 2S 2 pts
or Zn, H 2O
4
4
(b)
O
HO
CH3
CH3
Tetrahedron Lett. 2001, 42, 4001
(c)
O
Br
1.
O
, AlCl3
O
Br
2. H 3O+
(work-up)
CH3
O
O
C
C
O
O
CH3 C
CH3
2 pts
no partial
O
O
CH3
CH3
pyridine
4
O
O
1.
C CH2CH 2 C OH
2 pts if ortho isomer
O
O
C CH2CH 2 C OCH2CH3
N C N
no partial
2. CH3CH2OH
4
Br
4
J. Org. Chem. 2001, 66, 7283
(d) For full credit you must clearly indicate where the isotopically labeled atoms (15N and 18O) end up
in these products.
Name________KEY____________
Page 4
Box A
OCH2CH3
O
O
1. BrCH2 CH OCH2CH3
15 N
K
O
2. HCl, H 2O18
short time
O
Product in NaBH4
Box A
methanol
18
Box B
sto ich iometry not
mand atory
O
15 N CH2
C
H
+
2 CH3 CH2OH
-2 p ts if lab eled
C 10 H 7 NO3 O
J. Am. Chem. Soc. 200 1, 123, 9246
C 10 H 7 NO3
- 2 pts if u nlabeled
18
15 N CH2
O
- 2 pts if u nlabeled or
inconsistently lab eled
4
4
draw all organic pro ducts
OH
NH2-NH2
(1 eq.)
CH2
15
H2N CH2 CH2
O
18
OH
2pts
NH
+
NH
heat
4
- 2 pts if u nlabeled or
inconsistently lab eled
2pts
O
4
Name________KEY____________
Page 5
V. (38 points)
A. Compound A and ethyl acetate in the presence of lithium hexamethyldisilazane (LHMDS, a strong
base) react to form Compound B (Tetrahedron Lett. 2001, 42, 6907).
O
ROCH2
O
RO
C OCH2CH3
O
CH3 C OCH2CH3
O
O
ROCH2
RO
C
OR
RO
RO
Compound A
+ LiOH
H
LiN[Si(CH3)3]2
OR
Compound B
(R = PhCH 2)
Use the curved-arrow notation to provide a detailed step-wise mechanism for this transformation. Do
not add any other reagents. Draw only one resonance form for intermediates where applicable. Ignore
stereochemistry. You may use B- to represent LHMDS and HB for its conjugate acid.
5 pts per step (1 error = 3 pts, >1 error = 0 pts)
OR
O
CH3CH2O
C
OR
O
OR
O
O
CH3CH2O C
O
Step 1
CH2
H
CH3CH2O C
+ H
B
ROCH2
CH2
OR
RO
CH2
OR
O
O
CH2OR
Step 2
B
H
B
Step 3
O
CH3CH2O C C
OH +
H
OR
O
ROCH2
O
O
Step 5 CH3CH2O C
CH2
O
OR
OR
OH
OR
OR
OR
B
CH3CH2O C
B
CH
OH
OR
H O
ROCH2
ROCH2
+ H
Step 4
OR
OR
20
B. Provide structures of the expected major products as indicated by the boxes and any given
information. Show stereochemistry where known.
Name________KEY____________
Page 5
O
CH2 C Cl
O
C
N
HO
O
CH2 C O
NO2
(1 eq.)
N
K2CO3
OCH3
PhCH2O C O
N
O
C
OCH3
N
H
C 13 H 17 N3 O 3
O
C
OCH3
PhCH2O C O
O
CH2 C NH
no partial
CH2 C NH
H2
Pd/C
6
N
NO2
no partial
N
N
O
C OCH3
PhCH2O C O O
no partial
but don't penalize twice
for the same mistake
6
6
NH2
(1 eq.)
Name_______KEY_____________
Page 6
VI. (42 points)
A. (a) Provide the structures of the missing reagents in the following scheme. You do not need to
specify solvents or reaction temperatures. Abbreviations for reagents are not acceptable.
or Fe/HCl or Sn/HCl
OCH3
H2
OCH3
HNO 3
or (CH 3CO) 2O
Ni, Pd or Pt
3
O
Step 2
NH2
OCH3
OCH3
H2SO4
3
Step 1
NO2
CH3COCl
OCH3
Step 3
NH C CH3
OCH3
OCH3
3
C
3 O2N
O2N
N
Step 6
N
H2O, - OH
or 1. H 3O+, H 2O
2. OH 3
HCl, HBr or H2SO4
O2N
O
O2N
NH C CH3
NaNO2
CuCN
3
NH2
Step 5
Step 4
N
(b) One of the steps in the above synthesis would give a mixture of regioisomeric products. Identify
which step this would be and draw the structure of the other possible regioisomer.
A mixture of regioisomers is formed
in Step: (circle one)
1
2
3
4
5
other poss ible regioisomer
no partial
NH C CH3
CH3O
6
1
O
no partial
O2N
4
B. Provide structures of the expected major products as indicated by the boxes and any given
information. Show stereochemistry where known.
(a)
H
CH3CH2O
C
O
H
C
H
C
C
H
H
C N
HOCH 2CH2
C
1. LiAlH4 (excess)
OH
4 pts
2. H2O
H
CH2CH 2NH2
OH
+
CH3CH2OH
1 pt
5
J. Org. Chem. 2001, 66, 5352
(b)
O
C H
H
HO
OH
H
H
OH
CH2OH
KNO3
H2O
HO
This product would be formed as:
(circle one)
3 pts structure
2 pts relative stereochem
O
a single enantiomer
a single dias tereomer
OH
a racemic mixture
a mixture of dias tereomers
HO
cyclic product
OH
5
must be consistent with your answer
no partial
4
Name_______KEY_____________
Page 6
(c)
CH3
O
O
CH3
O
1. excess CH2 =CHMgBr
2. NH4Cl, H 2O
OH
J. Org. Chem. 2001, 66, 6490
CH3
O
CH2OH
CH3
O
CH CH=CH2
OH
no partial except
transcription errors
5