Chemistry 233
Organic I
Due April 29, 2010
George A. O’Doherty
Exam #3
Name:
ID#
(Please also put your name and ID# on the last page)
There is a total of 200 points for this exam.
[I]
(55 points)
i) When (5S)-5-tert-butyl-1-methylcyclohex-1-ene reacts with Br2 in water (Br2/H2O) it forms two
diastereomeric bromohydrins A and B. Where A is the (1R, 2R, 5S)-isomer and B is the (1S, 2S, 5S)-isomer.
ii) When bromohydrin A is treated with NaH it rapidly closes to form product C. Similarly, when B is treated
with NaH it closes to form product D (see reaction below).
iii) An identically mixture of products C and D is also formed when (5S)-5-tert-butyl-1-methylcyclohex-1-ene
reacts with mCPBA.
iv) Use these details to help you answer the questions below (A - E).
C
A
OH
(1R)
(1R)
NaH
(5S)
(5S)
(2R)
Br2/H2O
O
(2S)
Br
+
(1S)
OH
(5S)
(2S)
B
(1S)
NaH
(5S)
Br
O
(2R)
D
A) In the boxes above, draw the structure of the products A, B, C and D (please include stereochemistry).
(20 points)
B) Below, please draw both A and B in their two chair forms and label their substituents as either equatorial
(eq) or axial (ax). Thus the conformer of A will be A' and the conformer of B will be B'. (20 points)
OH (ax)
(eq)
(eq)
(eq)
OH
Br (ax)
(ax)
More Stable
A
(ax)
A'
(ax)
(eq)
Br
(eq)
(ax)
Br
(eq)
OH
(eq)
Br
(eq)
OH (ax)
(ax)
B
More Stable
B'
C) Indicate which of the two conformers A/A' and B/B' is the more stable by writing "more stable" in the
appropriate box with the more stable structure. (5 points)
D) If the starting material and/or any of the products (A, B, C and D) are chiral indicate it by circling the
appropriate labels below? (1 points/each)
starting material
A
B
C
D
E) If so, please then label (on the structures on page 1) any stereocenters with an R or S. (1 point/structure)
[II]
(1 points/1 points per arrow)
Please clearly draw mechanistic arrows to show the mechanism for the conversion of the structure on
the left to the right. An example is drawn in the box below.
Example (3 arrows)
Li
O
O
O
H
+
Li
OH
+
O
H2O
a) (hint 2 arrows)
C
CH2
H3C
b) (hint 2 arrows)
c) (hint 1 arrows)
+
Br
Br
d) (hint 1 arrows)
+
O Me
H
O Me
H
e) (hint 1 arrows)
O Me
O Me
+
H
H
f) (hint 3 arrows)
Br
+
H
O Me
+
O Me
H
[III]
(20 points/4 points each)
Please rank the following compounds in order of their relative stability (1 most stable and 3 least
stable).
H
C
H
H
H
3
1
a)
H
C
H
H
H
3
1
b)
H
3
1
c)
O
OH
OH
H
3
1
d)
H
H
H
1
e)
3
[IV] (40 points)
Please provide reagents for the following conversions. Some transformation may require more than one step.
(HINT: < 4 steps) (10 points each).
HOW?
K/NH3
A)
HOW?
H2 Lindlar's (Pd on BaSO4/quinoline)
or
BH3 then H+
B)
H
HOW?
HOW?
O
1) KNH2 in NH3
2) BH3 then NaOOH
1) KNH2 in NH3
2) H3O+, Hg(II)
O
C)
OMe
OMe
HOW?
(+/–)
mCPBA
KH then MeI
O
H+
in MeOH
(+/–)
OMe
OH
(+/–)
D)
OMe
OMe
HOW?
(+/–)
KH then MeI
(2 equiv)
OsO4/NMO
OH
OH
(+/–)
[V] (75 points/3 points each)
In the boxes provided on the next page draw the products (A-Y) for the reactions drawn below. Please indicate
relative stereochemistry when appropriate. If the product is formed as part of a racemic mixture only draw one
enantiomer. Also, if the product is formed as part of a racemic mixture this will be indicated by a (+/–) next to
the letter. (Hint: some products are formed more than once)
O
O
+
H
OH
NaBH4
H
H
P
O
OH
H3O+
O3 then
DMS
+
H
P
O
O
O
OH
+
OH
H3O+
+
O3 then
DMS
(+/–) N
A
H2
Pd/C
OH
OH
Na
in NH3
OH
H2
Lindlar's
Catalyst
B
H3O+
OsO4
NMO
NaH
then
EtBr
H
NaH
then
EtBr
H
M
D
C
NaH
then
MeBr
H
(+/–) E
OH
J
L
BH3 then
NaOOH
O
H2
Lindlar's
Catalyst
mCPBA
K
H
Na
in NH3
O
OH
H3O+
OsO4
NMO
H
F
OH
H3O+
OH
G
(+/–) E
NaBH4
(+/–) Q
1% HCl
in MeOH
OCH3
NaH
then
MeBr
(+/–) W
Hg(OAc)2/H2O
PCC
OH
OH
OH
NaBH4
O2
+
OH
HgOAc
OH
(+/–) E
NaBH4
O
OH
D
NaBH4
(+/–) I
H3O+
HgSO4
TsCl/Et3N
U
Ph3P=CH2
OTs
H
(Et)2CuLi
H2
Pd/C
J
H
T
S
BH3 then
NaOOH
NaNH2/NH3
Na
O
H2
Pd/C
EtMgBr
MeMgBr
HO
H3O+
CH3
(+/–) V
OH
R
(+/–) Q
H2O
(+/–) X
Mg
KOt-Bu
(+/–) Y
LiBr
PBr3
MgBr
OTs
TsCl/Et3N
+
G
H2
Pd/C
KOt-Bu
Br
LiAlH4
(+/–)
R
OH
A=
B=
C5H12O
C5H8
OH
C=
D=
OH
C6H14O2
C5H10
OH
E=
O
F=
OH
(+/–)
H
C6H14O2
C4H8O
O
G=
H=
C6H12
C6H12O
HgOAc
I=
J=
OH
(+/–)
C8H16HgO3
K=
C6H10
H
L=
C4H6
C6H12
OH
M=
H
N=
H
C2H2
(+/–)
C5H12O
O
O
O=
P=
H
H
C3H6O
C2H4O
OH
Q=
R=
(+/–)
C6H14O
OTs
S=
C6H14
T=
H
C6H10
C11H16O3S
O
U=
OCH3
(+/–)
C7H16O
CH3
(+/–)
C6H12O
W=
HO
V=
C7H16O
MgBr
X=
(+/–)
C6H13BrMg
OTs
Y=
(+/–)
C13H20O3S
Name
Problems
Points Possible
1
55
2
10
3
20
4
40
5
75
200
Total
Points Awarded