Chem 343–Spring 2007
Problem Set 2 – Key
1. Indicate the relationship between the two structures.
(a)
CH3
S H
H
H
S
Cl R
S
(b)
H
S H
OH
Cl
S H
H3C S
S
CH3
H
OH
CH3
H3C
Br
H
R
R
H
Br
H
Br
CH3
Br
meso
CH3
H
(c)
CH3
H
C C C
Cl
H
C
C
C
C
mirror
images C
C
H
H3C
Cl
H
H
Cl
o
o
o
o
o
enantiomers
o
o
o
o
o
enantiomers
diastereomers
o
o
o
o
o
enantiomers
o
o
o
o
o
enantiomers
diastereomers
conformers
same
structural isomers
conformers
same
structural isomers
diastereomers
conformers
same
structural isomers
-90 ° rotation
H
H
(d)
OH
H
HO
O
OH H
H
OH
R
OH
H
HO
OH
O
OHHO
H
H
S OH
H
H
diastereomers
conformers
same
structural isomers
only one configuration is changed
(e)
O
S Cl
CH3
RH
R R
Br H
Br H
H
H
Cl R
S
S
H
S
CH3
1
O
o
o
o
o
o
enantiomers
diastereomers
conformers
same
structural isomers
2. (a) Label each of the stereogenic centers below as R or S. Is the molecule chiral?
If it is chiral, draw the enantiomer of the compound. If it is not chiral, indicate any
symmetry elements present.
CH3
CH3
CH3
R H
H3C
S HS
H
H
S
CH3
R H R
CH3
H
chiral
enantiomer
(b) Label each of the stereogenic centers below as R or S. Is the molecule chiral? If it
is chiral, draw the enantiomer of the compound. If it is not chiral, indicate any symmetry
elements present.
H
H3C
CH3
HS
CH3
R
H
R
H
CH3
S
plane of symmetry
achiral
3. Write the major organic reactant(s) or product(s) of the following reactions. Clearly
show the stereochemistry of the reactant(s) or product(s) where isomers are possible.
Indicate whether the starting material is chiral or achiral by circling the appropriate
designation under the starting material. When more than one product is formed, clearly
state whether equal or unequal amounts of the compounds are formed. Also indicate
whether the product(s) is(are) optically active or optically inactive.
(SN1 – racemization)
(a)
Br
CH3OH
OMe
OMe
+
25 °C
racemate
chiral
achiral
optically acitve
(b)
CH3CH2CH2
C
H
H3C
chiral
Br
achiral
NaCN
optically inacitve
CH3CH2CH2
NC
C
DMF, 25 °C
H
CH3
(SN2 – inversion)
optically acitve
2
optically inacitve
4. (a) When the optically active tosylate (A) shown below was reacted with CH3SNa in
CH3OH, the reaction was observed to be second order and the substitution product
shown was formed almost exclusively. Show the mechanism for this reaction and
predict the stereochemistry of the product.
C OTs
H
CH3 (A)
+ CH3S
SCH3
CH CH3
CH3OH
+ Na
‡
!H3CS
C
!OTs
H3CS C
H
CH3
inversion of configuration
H CH3
(b) If the solvent in part (a) was changed from methanol to dimethyl formamide [DMF,
HCON(CH3)2], the same product was formed but at a much faster rate. Explain.
Polar aprotic solvent like DMF solvates Na cation, which will make –CN
available for nucleophilic attack on alkyl tosylate substrate. Therefore, this
reaction will be faster in DMF.
5. (b) Would you expect the following reaction is to occur faster in DMF or in ethanol?
Explain your answer.
Br
+
NaCN
CN
+
NaBr
This reaction should follow a SN2 reaction mechanism (primary bromide and
strong nucleophile). In this reaction, the nucleophile’s capacity to attack the
substrate is dependent upon the availability of free anionic form of
nucleophile (-CN). Polar aprotic solvent like DMF solvates Na cation, which will
make –CN available for nucleophilic attack on alkyl bromide. In ethanol both
Na cation and –CN will be solvated. Therefore, this reaction will be faster in
DMF.
3
6. The reaction of B with NaCN in ethanol is much faster than the reaction of A with
NaCN in ethanol. The rate of reaction of A with 0.2 M NaCN is twice as fast as with
0.1 M NaCN. The rate of reaction of B with either 0.1 M or 0.2 M NaCN is the same.
A
NaCN
CH3CH2OH
Br
C
CN
H
B
Br
H
H
NaCN
CH3CH2OH
H
+
D
CN
E
H
H
CN
(a) A reacts with NaCN by a___ SN2____ mechanism.
(b) The reaction of the R isomer of A would give:
____ the R isomer of C, __ X __ the S isomer of C, ____ a racemic mixture of C.
(c) B reacts with NaCN by a___ SN1____ mechanism.
(d) The reaction of the R isomer of B would give:
____ the R isomer of D, ____ the S isomer of D, __ X __ a racemic mixture of D.
(e) Explain why B reacts faster than A.
A reacts via a SN2 mechanism but the substrate is a secondary bromide,
thus reaction will be slow due to the steric hindrance. Although
secondary bromide B reacts via SN1 B is a special type of substrate
because the secondary carbocation intermediate will be stabilized by
the double bond through resonance.
(f) Explain why two products are formed from B.
H
Br
H
NaCN
CH3CH2OH
H
H
H
H
CN
CN
H
H
D
CN
H
4
E
H
CN
7. How the following syntheses can be efficiently accomplished. More than one step
may be required. Show any intermediate products that would be isolated in the
course of the synthesis.
H2, P-2 catalyst
H
or Lindlar catalyst
NaNH2
SN2
Br
8. Write an electron pushing reaction mechanism for the acid catalyzed equilibration of
3-ethyl-1-pentene and 3-ethyl-2-pentene. Which isomer is more stable?
H
H+
H
OH2
H2O
more stable: trisubstituted
less stable: monosubstituted
9. Depending on the reaction conditions and reagents employed, 3,3-dimethyl-1-butene
(t-butyl ethylene) can be converted into three different alcohols. Reactions of 3,3dimethyl-1-butene with H2SO4 and H2O produces 2,3-dimethyl-2-butanol. Draw a
detailed mechanism for its formation. Use electron pushing arrows (curved arrows) to
show how each intermediate is converted to the next intermediate in the sequence.
H2SO4
H2O
OH
H+
OH2
H
H
H
OH2
methyl migration
2 ° carbocation
H
O
H
3 ° carbocation
5
-H3O+
10. trans-2-Methylcyclohexanol is subjected to acid-catalyzed dehydration. Provide the
structure of the major product with a reasonably detailed reaction mechanism to explain the
product formation.
OH2
CH3
OH
CH3 H
O H
20 %
H2SO4
Ha
CH3
80 °C
Ha
-Hb
CH3
Hb
CH3
-Ha
favored : formation of more stable
trisubstituted double bond
11. trans-1-Bromo-2-methylcyclohexane is subjected to dehydrohalogenation reaction. Provide
the major product with a reasonably detailed reaction mechanism to explain the product
formation.
CH3
CH3
Br
Br
!OEt
CH3
H
anti periplanar
H
Br
NaOHEt
H
H
EtOH
60°C
CH3
CH3
CH3
H
Br !-
12. For each pair of reactions, circle the reaction that is faster. Convincingly explain
your choice.
HO
(a)
Br2/H2O
Br
faster due to the invlovement of 3° carbocationic
character compared to the other path, which
should involve a 1° carbocationic character.
!+
versus
Br
!+
vs.
OH
H2O
6
Br
Br
H2O
HCl in ether
(b)
vs.
H
H
versus
faster due to the invlovement of 3° carbocationic
character compared to the other path, which
should involve a 2° carbocationic character.
HCl in ether
1M CH3CO2H
in H2O
(c)
OH
versus
Stronger acid: more free H+ available to
react with the double bond
1M H2SO4
in H2O
13. Write an electron pushing reaction mechanism for the following acid catalyzed
rearrangement. Comment on the change in ring strain energy for steps involving
change in ring size.
Br
H3C
CH2
H3C
H–Br
BrCH2
CH3
CH3
CH3
CH3
CH3
CH3
CH3
+
CH3
Br
CH3
CH3
Although both are 3° carbocation, due to the ring-strain of the fourmembered ring the initially formed carbocation will rearrange itself to
a more stable carbocation.
CH3
equal amount: racemate
14. Write the two major organic products of the bromination of (S)-3-methyl-1methylenecyclohexane, clearly showing their stereochemistry. Indicate whether each
product is optically active or optically inactive. Will the amounts of the two products be
equal or unequal? What is the stereochemical relationship between the two different
products (i.e. structural isomers, diastereomers, enantiomers, etc)? Draw the most
stable chair form of one of the products.
!+
Br
Br
!+
H
H3C
H
H3C
Br2
Br-
optically active
Br
Br
more stable
diastereomers: unequla amount
CCl4
H
H3C
H3C
Br
Br
!+
H
H3C
H3C
Br !+
H
H3C
these two intermediates are diastereomers:
formed in unequal amount
7
Br
Br
Br
Both products are optically active
because the original stereogenic carbon
was not affected by the reaction.
15. Write an electron pushing mechanism for the following reaction. Show all
intermediates and use the curved arrow notation to indicate how one intermediate is
converted to the next. Draw the most stable chair conformation of the product.
H
Br
a
+
Br
Br
H
OCH3
Br
H3CO R
R
H
Br
racemate
b
a
H
OCH3
H
CH3OH
b
Br
H
H3CO
H
S S Br
H
16. The following scheme involves a strategy for a stereospecific synthesis of Z and E
alkenes. Provide appropriate reaction conditions and structures A and B.
NaNH2,
Li/EtNH2
Br
H2, Lindlar catalyst or P-2 (Ni2B)
H
H
H
H
B (C6H12)
A (C6H12)
17. The following scheme involves a strategy for a stereospecific synthesis of vicinal
dibromides
(a) Provide appropriate reaction conditions and structures A–D in the boxes. (The
reaction conditions and correct structures of A and B can be deduced from the
reaction conditions and reagents as well as from the structures of C and D
respectively)
8
NaNH2,
Br
H2, Lindlar catalyst
Li/EtNH2
H
H
H
H
B (C6H12)
A (C6H12)
Br2
Br2
racemic
meso
Br
Br
H
H
Br
+
H
H
Br
C (C6H12Br2)
Br
H
H
Br
D (C6H12Br2)
(b) Provide the mechanistic rationale for the above bromination reactions to generate C
and D.
meso
+
Br
a
H
Br2
A
a
c
Br
H
Br-
b
S
Br2
B
d
d
H
+
mixture of enatiomers:
racemate
c
H
S
R
H
Br
Br
H
H
a
Br
H
a
+
H
b
H
Br
H
meso
Br
R
Br-
meso
9
H
S
racemate
b
Br
b
Br
Br
S
H
R
R
H
Br
18. Write the appropriate intermediate(s) and product(s) for the following reactions.
Clearly show the stereochemistry of the intermediate(s) and product(s) where isomers
are possible.
(a)
Hg(OAc)2
NaBH4
THF, H2O
(b)
AcOHg
BH3:THF
3
-OH
H
OsO4
O
O
O
Os
HO
H
H2O
HO
OH
O
Br2
H2O
C6H10Br
O3
C6H11OBr
O
Zn/AcOH
O
O
O
O
OH
Br
+ Br
(e)
OH
NaHSO3
pyridine
(d)
-OH
H2O2
B
(c)
OH
O O
O
C6H10O2
ozonide
10
19. Spodoptol is the sex attractant produced by the female fall armyworm moth to attract
male moths. The formula of spodoptol is C14H28O. Hydrogenation of spodoptol with H2
and Pt produced the straight chain primary alcohol 1-tetradecanol, C14H30O, H3C–
(CH2)12–CH2OH. Ozonolysis of spodoptol followed by work-up with Zn and acid produced
a mixture of two aldehydes A, C5H10O and B, C9H18O2.
H2, Pt
OH
Spodoptol
C 14H28O
1) O3
2) Zn, H+
A
C 5H10O
B
C 9H18O 2
+
(a) What are the structures of A, C5H10O and B, C9H18O2?
O
A=
B=
OH
O
(b) Draw two possible structures for spodoptol. Both of these compounds were
synthesized and only one of them attracted male fall armyworm moths.
OH
trans-isomer
or
OH
cis-isomer
20. Starting from 1-methyl-1-cyclopentene, a tertiary alcohol, a secondary alcohol, and a
diol can be synthesized. Indicate specific reagents and the reaction mechanism involved
in each of the following transformation. Specify unambiguously the stereochemistry of the
intermediates and the products if relevant (Don’t use acidic hydration conditions for the
first transformation).
OH2
1. Hg(OAc)2, H2O-THF
2. NaBH4, HO-
OH
via
OH
HgOAc
Oxymercuration-Demercuration
+
1. BH3:THF
2. H2O2,
via
OH-
-OOH
H
H
BR2
OH
HgOAc
Hydroboration-Oxidation
BR2
O
OH
1. OsO4, pyridine
OH
2. NaHSO3, H2O
OH
Dihydroxylation
11
via
O
Os
O
O
O
21. What reagents would you use to accomplish the following transformations?
Cl
H C Cl
(a)
H
Cl
Cl
Cl
KOtBu
H
H
KMnO4, cold H2O, -OH
OH
(b)
OH
or OsO4 then NaHSO3
(c)
(d)
H
O3 then
CHO
Zn/AcOH
CHO
KMnO4, hot H2O, -OH
CO2H
CO2H
22. Methylcyclopentene can be transformed to two different ethers as shown. Provide
appropriate reagents and conditions for each reaction. Use ethyl alcohol in the first
reaction and ethyl chloride in the second.
1. Hg(OAc)2, EtOH
2. NaBH4, OHO
or H2SO4, EtOH
1. BH3-THF
2. H2O2, OH3. NaH, EtCl
O
racemic
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