CHEM 12A
EXAM II
Name KEY
Fall 2003
1.(16 pts)
Draw the structure of the major product expected from each of the following sets of
reactants. Specify stereochemistry where appropriate.
CH3
CH3
Pt
H2
+
CH3
H
CH3
H
Br
CH3
CH3
Br2
+
CH2Cl2
CH3
CH3
Br
HO
CH3
+
enantiomer
CH3
1) Hg(OAc)2, H2O, THF
2) NaBH4
CH3
CH3
+
tBuO OtBu
HBr
Br
CH3
CH3
1) B2H6/THF
OH
-
2) OH + H2O2
CH3
HC CCH2CHCH3
+
NaNH2
CH3
CH3CH2C CCH2CHCH3
CH3
HC CCH2CHCH3
NH3
Na
CH3
C CCH2CHCH3
Na
NH3/-35°C
Cl
+
HCl
(excess)
CH3
CH3
CCH2CHCH3
Cl
2.(8 pts)
Is the compound below optically active? Explain your answer, being as concise and
complete as possible, and include a definition of "optically active" in your explanation.
H
OH
H
OH
Br
Br
H
H
An optically active compound will rotate the plane of plane-polarized light, due to a
lack of symmetry. This compound has no elements (points, lines or planes) of symmetry,
so it will rotate the plane of plane-polarized light. The absolute configuration of each
chiral carbon is R, so a sample of this compound that is pure (that is, with none of its
enantiomer present) will not have any mirror-image conformations present, which are
necessary to cause a lack of optical activity.
3.(12 pts)
Write the IUPAC name for each of the following compounds.
H
HO C CH CH
2
3
CH3
(S)-2-butanol
Br
Cl
CH3
CH2CH3
(R)-2-bromo-2-chlorobutane
CH3
H
CH3
OH
F
CH2CH3
(2S,3R)-3-fluoro-3-methyl-2-pentanol
CH3CH2CHC
CH3
CCH2CHCH3
CH2CH3
3,7-dimethyl-4-nonyne
4.(8 pts)
Show each step in the reaction mechanism for the reaction below. Include only
reactants, intermediates and products - no transition states!
H2SO4
CH3
CH3
OH
O
CH3
CH3
O
H
OH
+
OSO3H
OSO3H
5.(12 pts)
Addition of D2O to Z-3-methyl-2-pentene in the presence of D2SO4 can occur by
syn or by anti addition mechanisms. Draw the products of both types of addition, clearly
indicating the stereochemistry around any chiral carbons that may result, and clearly
labelling which is the syn product and which is the anti product.
+
D2O
D2SO4
H
H
CH3 OD
+
D
OD
syn product
D
CH3
anti product
(there are more than one of each of these possible)
6.(12 pts)
A high yield of 2-bromo-3-methylbutane is desired. Explain, using both words and
a mechanism, why the reaction below would not be successful in forming a high yield of
2-bromo-3-methylbutane, and draw the structure of the main side product.
CH3
CH3
CH2
CHCHCH3
+
CH C
CH3
Br
CHCHCH3
Br
Br
Br
CH3
CH3
H
CH3
CH3
CH3
CH2 C
H
CH3
Br
Once the initial carbocation (2˚) is formed, bromide ion could immediately bond to form
the desired product. However, a more stable 3˚ carbocation can be formed by a 1,2-hydride shift.
This leads to substantial amounts of 2-bromo-2-methylbutane being formed.
7.(12 pts)
For each of the following pairs of structural formulas, tell whether they represent
identical species, conformers of the same species, constitutional isomers, enantiomers
or diastereomers.
CH3
F C H
CH3CH2
CH3
H
C
F
CH3
identical
H
CH2CH3
HO
CH2CH3
CH3CH2
OH
CH3
H
enantiomers
Cl
F
H
H
Cl
H
F
H
H
diastereomers
H
Cl
Br
Br
Cl
H
Br
Cl
H
Cl
Br
diastereomers
8.(8 pts)
The enzyme aconitase catalyzes the hydration of aconitic acid to two products, citric
acid and isocitric acid. Isocitric acid is optically active; citric acid is not. What are the
respective structures (use perspective formulas or Fischer projections) of citric acid and
isocitric acid?
HO2CCH2
CO2H
C
C
HO2C
H
aconitic acid
CH2CO2H
HO2C
CH2CO2H
OH
H
HO2C
H
H
H
CO2H
OH
CO2H
citric acid
isocitric acid
9.(12 pts)
The alkane formed by total hydrogenation of (S)-4-methyl-1-hexyne is optically
active, while the one formed by total hydrogenation of (S)-3-methyl-1-pentyne is not.
a)
Show these products and explain this difference. Structures of the alkynes
obviously (I hope!) do not have their stereochemistry specified.
(S)-4-methyl-1-hexyne
(S)-3-methyl-1-pentyne
CH2CH3
H C CH C
2
CH3
CH2CH3
H C
CH3
CH
H2/Pd
C
H2/Pd
CH2CH3
H C
CH3
CH
CH2CH3
CH2CH2CH3
H C
CH3
CH2CH3
The product of hydrogenation of (S)-4-methyl-1-hexyne is chiral: (S)-3-methylhexane.
The product of hydrogenation of (S)-3-methyl-1-pentyne has two ethyl groups bonded to the
"chiral" carbon, rendering it achiral, and the molecule achiral.
b)
Would each of the products of hydrogenation of these two compounds by a
poisoned catalyst (e.g., Lindlar catalyst) be expected to be optically active? YES !
CH2CH3
H C
CH3
CH2CH
CH2CH3
CH2
H C
CH3
CH
CH2