Download a) b) c) d) e) f) g) h) i

Chemistry& 241
Clark College
Fall 2008 Exam 3 Review SOLUTION
1. Complete the following reactions. Be sure to specify stereochemistry, when appropriate.
OH
a)
1) Hg(OAc)2,
H2O
2) NaBH4
Br2
b)
Br
CH2Cl2
Br
O
H 2O 2
Na2WO4
c)
O
OH
OH
H2
Pt
d)
e)
H
H
Cl
H-Cl
OH
1) BH3
f)
2) NaOH, H2O
g)
O
O
1) O3
2) (CH3)2S
h)
OsO4
OH
HOO
i)
Exam 3 Review Sheet
Cl2
H 2O
OH
Cl
HO
H
Fall 2008
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Chemistry& 241
Clark College
OsO4
Syn addition of a 1,2-diol
OH OH
HOO |
O
6
5
H 1
1) O3
2) (CH3)2S
1 2
3
4
2
3
4
5
6
O
1) BH3
2) H2O2
NaOH
H
Syn, anti-Markovnikov
addition of an alcohol
OH
Catalytic hydrogenation,
Syn addition of H2
H2
Pt
Cl2
H2O
Halohydrin reaction,
Anti addition
Cl
OH
O
H2O2
Na2WO4
O
O
CO2
HO
OH
O
O
O
O
OH
OH
Exam 3 Review Sheet
Fall 2008
Page 2 of 8
Chemistry& 241
Clark College
2. Give all products for the following reaction. Indicate which products would be formed as a racemic
mixture. [10 pts]
NCS, !
Products?
NCS (N-chlorosuccinamide) selectively chlorinates the allylic positions. There are three allylic positions
on this molecule, and each one has a resonance structure. Racemic mixtures are formed when the
product has a chiral center at the Cl group, since the radical intermediate formed is planar, and can
react on either side. The resonance structures, and all chloro products, are shown below.
Cl
Cl
Cl
racemic
Cl
racemic
Cl
racemic
Cl
racemic
3. Mechanisms. Provide the mechanisms for the following reactions.
Br2
H2O
a.
OH
Br
OH2
Br Br
O
Br
Br
H
H
OH2 (or Br-)
OH
Br
(+ H3O+ + Br-)
Exam 3 Review Sheet
Fall 2008
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Chemistry& 241
Clark College
Br2
CH2Cl2
b.
Br
Br
Br Br
Br
Br
Br
Br-
c.
1) Hg(OAc)2,
CH3OH
2) NaBH4
OCH3
AcO
Hg
OAc
Remember that chemically, alcohols and water
work in the same way! The product does not show
stereochemistry, as a racemic mixture forms.
HOCH3
HgOAc
HgOAc
OAc-
O
H
NaBH4
CH3
HgOAc
O
OCH3
CH3
d.
Initiation Br Br h! or !
2 Br•
Propagation
Br•
+ HBr
H
Br Br
Br
Termination 2 Br•
+ Br•
Br Br
Br•
Br
H
Exam 3 Review Sheet
Fall 2008
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Chemistry& 241
Clark College
4. Give the mechanism for the reaction between 2-methylpropane and Cl2, in the presence of light.
Include all steps. [12 pts]
Cl2
HCl
Cl
Initiation
Cl Cl
Propagation
H
2 Cl•
•Cl
HCl
•Cl
Cl Cl
Cl
Termination
Cl2
2 Cl•
•Cl
Cl
5. The reaction shown below is used to make a synthetic precursor to the antibiotic monensin,
Provide the mechanism for its formation. [8]
O
Br2
OH
Br
Br
H
Br
Br
O
OH
Br
O
OH
Br
Br
6. Using both words and figures, explain why the hydroboration (addition of BH3) of propene is an antiMarkovnikov reaction that results in 1-propanol. [8 pts]
1) BH3
2) H2O2, NaOH
H2O
OH
In BH3, boron is less electronegative, and has an empty p-orbital. Therefore, boron is the Lewis acid in
the reaction, and the hydrogen is a hydride – H-, which is the Lewis base that attacks the carbocation.
The carbocation that is formed is still the most substituted carbocation.
Exam 3 Review Sheet
Fall 2008
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Chemistry& 241
Clark College
The partial mechanism:
H
H
H
H
B
H
BH2
H
This is a concerted mechanism- the pi bond attacks the boron and the hydride attacks
the carbocation in one step. This accounts for the syn addition of the two groups.
7. Briefly explain why a 3° radical is more stable than a 2° radical. Use a figure to illustrate your
explanation. [8 pts]
A tertiary radical is more stable than a 2° radical since there are more groups that surround the radical
to stabilize it through hyperconjugation. Hyperconjugation is the process by which electron density is
shared through space, in a “π-bond” fashion, into partially-filled (radical) or empty (carbocation) p
orbitals. A 3° radical has three groups surrounding it, while a 2° radical has two.
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
A 3° radical
A 2° radical
8. Consider the reaction between 3-methylheptane and bromine in the presence of light. [8]
a. What is the product of this reaction?
Br
b. The product is formed as a racemic mixture of two enantiomers. Briefly explain (using
words and figures) how/why this racemic mixture forms.
H3CH2C
H3C
C4H9
Exam 3 Review Sheet
Br The carbon radical intermediate becomes sp2 hybridized, with the
radical in the unhybridized p orbital. Since this is a planar
intermediate, the bromine radical can react with either lobe of the p
orbital equally, so both enantiomers form = racemic mixture!
Fall 2008
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Chemistry& 241
Clark College
9. For each of the following reactions, explain why the addition of HBr is regioselective in these cases.
Use both word and figures in your explanations. [8]
Br
OCH3
HBr
OCH3
H
OCH3
H Br
OCH3
H
or
OCH3
When the formed carbocation is adjacent to
the oxygen, the oxygen can donate lone
pairs to form a resonance-delocalized
carbocation. This more-stable carbocation
leads to the given product.
OCH3
H
O
Br
OCH3
O
HBr
OCH3
O
H
OCH3
O
O
or
H Br
OCH3
OCH3
H
When the carbocation is adjacent to the carbonyl carbon, the
partial-positive charge on the carbonyl carbon destabilizes the
formed carbocation. Therefore, the more stable carbocation
forms away from the ester group, and results in the given product.
10. Which of the following alkenes cannot be used to prepare 3-chloro-3-methylhexane? [3]
11. What type of intermediate is formed in the reaction of an alkene with HBr to give a bromoalkane?
a. A carbocation
b. A carbanion
c. A radical (single-electron species)
d. A bromonium ion
12. Which of the following reactions is not stereospecific?
a. Bromination (Br2/CH2Cl2)
b. Hydrogenation (H2/Pt)
c. Acid-catalyzed hydration (H2O, H2SO4)
d. Bromohydrin formation (Br2, H2O)
Exam 3 Review Sheet
Fall 2008
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Chemistry& 241
Clark College
13. Which of the following concepts explains Markovnikov’s rule as applied to the addition of HBr to
propene?
a. The relative stability of carbocations.
b. The nucleophilicity of the bromide ion.
c. The acidity of HBr.
14. What is the expected product of the following reaction?
O
1) Hg(OAc)2,
CH3OH
2) NaBH4
?
OCH3
CH3
OH
OCH3
O
O
O
O
The most stable carbocation is next to the oxygen, which can help with lone pairs!
15. Radical bromination occurs selectively at which position on this molecule?
A
B
D
C
A
Exam 3 Review Sheet
B
C
Position B is a benzylic position!
Fall 2008
D
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