Additional Problems for practice:
1. Show how to accomplish the following synthetic conversions:
OH
a
O
CH3
PCC
CH2Cl2
PCC
CH2Cl2
OH
O
H
1. CH3MgBr
CH3
2. H3O+
OH
OH
b
PBr3
Br
PCC, CH2Cl2
Mg
MgBr
Ether
O
+
2. H3O+
O
Br
PCC
CH2Cl2
c
Mg
MgBr
OH
H
MgBr
1. CH2=O
2. H3O+
PCC
CH2Cl2
O
1.
2. H3O+
HO
d
CN
BH3•THF
H2O2, HO-
SN2
NaCN, DMF or DMSO
TsCl
OH
OTs
Pyridine
e
CN
OH
PBr3
NaCN, DMF or DMSO
Br
2. Propose a mechanism for the following transformation:
Br
Br
P
Br
O
O
OH
PBr3
NH
N
H
O
Br
O
NH
Br-
O
H
P
N
Br
HO
-
Br
P
Br
3. Design a preparation of the following molecules from the indicated starting
materials:
a
OCH3
BH3, H2O2
HOOH
1. NaH,
DMF
OCH3
2. CH3I
H
OH
b
H
OH
O
HgSO4
NaBH4
H2SO4
H2O
CH3O
c
OH
PBr3
Br
OH
KOtBu
E2
KOtBu
E2
or:
TosCl
OH
OTos
Pyridine
O
d
Br
OH
Br
Mg
MgBr
CH2OH
CH2=O
Na2Cr2O3
H2SO4,
H2O
O
OH
e
O
starting material
1. Hg(OAc)2,
H2O
PCC
2. NaBH4
CH2Cl2
HO
O
Additional Problems for practice:
4. Acid-catalyzed dehydration of 2,2-dimethylcyclohexanol yields a mixture of 1,2dimethylcyclohexane and isopropylidenecyclopentane. Propose a mechanism to
account for the formation of both products.
H+
HO
H3C
CH3
H2SO4
CH3
CH3
CH3
+
CH3
H
O
H2O
H
H
H
CH3
CH3
methyl shift
CH3
CH3
CH3
CH3
alkyl shift
CH2
H
CH3
CH3
CH3
H2O
5. When the following diene is treated with MCPBA in CH2Cl2, an intermediate is
produced with formula C9H16O3. Treatment of this intermediate with sodium
hydride leads to the formation of the observed products. Draw the structure of the
intermediate, indicate the type of reaction which is taking place to produce the
observed products, and rationalize the stereochemical outcome of the reaction.
C9H16O3
MCPBA
H
OH
CH2Cl2
H
H
+ enantimer
CH2
CH2
OH
H
H
H
O
O
syn bis-epoxide
+
CH2
NaH
OH
H
H
H
O
+ enantimer
O
anti-bis epoxide
SN2
CH2
O-
H
H
H
O
O
backside
opening of
epoxide
SN2
O
O-
H
H
H
backside
opening of
epoxide
O
syn bis-epoxide
H2O
O
H
H
O
O-
H
O
O
H
H
H
OH
+ enantiomer
SN2
CH2
O-
H
O
H
H
O
backside
opening of
epoxide
SN2
O
H
H
O-
H
O
backside
opening of
epoxide
anti-bis epoxide
H2O
O
H
H
O
H
O-
O
H
H
O
H
OH
+ enantiomer
because a bis-epoxide is produced, two different relative stereochemistries are possible : syn and anti.
the syn bis-epoxide leads to one set of racemic products via double SN2 reaction, while the anti bis-epoxide
leads to another set of racemic products via double SN2 reaction
6. Design a preparation of the following molecules from the indicated starting
materials:
O
a
O
CH3
CH3
starting material
1. CH3MgBr
O
2. H3
OH
BH3•THF
OH
H2O2, HO-
OH2
H
O
OH
O+
H
CH3
H2SO4
H
O
OH
OH
CH3
O
CH3
CH3
OCH3
b
OCH3
starting material
O
CH3COOH
H3O+
OH
1. xs NaH
OCH3
OH
2. CH3I
OCH3
O
c
Br
O
HO
Mg
MgBr H3C
Br
CH3
ether
H2SO4
E1
H
HO
H
OH
HO
H2SO4
O
OsO4
H2O2
HO
d
OH2
O
OH
OH
CH3
starting material
O
CH3
OH
O
PCC
1.CH3MgBr
OH
H2SO4
E1
CH3
2. H3O+
CH2Cl2
PCC
O
H
CH3
HBr
ROOR
heat
H
CH3
1.
O
CH2Cl2
OH
2. H3O+
Mg
CH3
ether
MgBr
Br
7. Treatment of the following diol with sulfuric acid leads to two products. Propose
a mechanism which rationalizes the formation of both observed products
H+
a
O
OH
HO
O
H2SO4
+
a
H+
OH2
H
b
H
HO
OH
H
O
O
O
alkyl shift
SN2
-H3O+
OH2
H
H
H
O
OH
OH
O
O
-H3O+