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Exam 3 SOLUTION
Please work and place your answers in the spaces provided. Show your work for maximum credit!
The last page may be torn off and used for scratch paper. Do not write on the IR/NMR sheet. Bring all
material up when you are finished.
1. Provide the reagents for the following 2-step transformation. [6]
Br2
h!
H2
Pt, Pd or Ni
Br
This could actually just be done in one step with HBr – gave credit for both.
2. 4-Methylpent-3-en-1-ol is heated in the presence of a non-aqueous acid catalyst to form a
compound with the formula C6H12O. Determine the product of the reaction, and provide the
mechanism for its formation. [10]
H2SO4
OH
OH
H+
5
4
3
OH
2 1
?
O H
O
The first step of the reaction is the straightforward process of the pi bond (Lewis Base) going
after a proton (Lewis Acid) from sulfuric acid, to create the 3° carbocation. At this point the
available Lewis bases to go after the carbocation are the alcohol and the hydrogen sulfate ion.
The alcohol is a stronger base, and an intramolecular reaction is often faster than a bimolecular
reaction, since the nucleophile (the alcohol) is tethered to the electrophile (the carbocation).
From there, there is the finishing step of losing the proton off of the positively charged oxygen,
which can be picked up by the hydrogen sulfate ion (which makes the reaction catalytic with
respect to acid).
3. When 1-butene reacts with HBr, racemic 2-bromobutene forms. Explain, using both words and
pictures, why a racemic mixture forms in this reaction. [8]
H
Exam 3
Br-
When the carbocation forms, it is an empty p orbital in a planar geometry.
The bromide ion can then react with either lobe of the p orbital, causing
either enantiomer depending on reaction from the "top" or "bottom".
Fall 2006
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Chemistry& 241
Clark College
4. Predict the product, give the starting material, or give the reagents needed for the following
reactions. Indicate stereochemistry, where appropriate. Do 8 of 9. You may do the 9th for extra
credit. Indicate which six you would like graded for credit by placing a ‘star’ (*) next to them. If
there are no indications, the first six will be graded for credit. [24+2 pts]
OH
H
1) BH3
2) H2O2
NaOH
2
1
3
4
5
Anti-Markovnikov, syn addition
7
6
O
H2O2
Na2WO4
HO 1
O
3
2
HO
Cl2
H2O
HO
6
7
H
CH3
H3C
OsO4
5
4
Peroxide/sodium tungstate
gives carboxylic acids and
ketones- and H's on the
original double bond
become OH's.
Cl
OH
You can also use OsO4, followed
by NaHSO3 (numbered steps).
HOO
1) Hg(OAc)2, H2O
2) NaBH4
Using H+/H2O will result in a
rearrangement of the carbocation.
OH
1) O3
2) (CH3)2S
O
Br
O
Br
HBr
H2O2
The reducing conditions
of dimethylsuflide
O
O
results in aldehydes
H
H and ketones. The H's
H
on the original double
bonds remain as H's.
This is the "Greener" Bromination
of cis-Stilbene. The two Br's
undergo anti addition.
cis-Stilbene
*
NCS
h!
Br2
h!
Exam 3
Halogenation always favors the
allylic spot, yielding 2 products
(the original allylic position and
Cl the resonance product) and
retention of the double bond.
Cl
Br
Fall 2006
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Chemistry& 241
Clark College
5. Using curved-arrow notation, give the complete mechanism for the following reaction. [12]
Br2
h!
+
HBr
Br
Initiation
Br
Propagation
2 Br•
Br
H
Br
•Br
HBr
Br
•Br
Br
Termination
Br2
2 Br•
•Br
Br
An extra point was given
for including this
disproportionation step.
H
6. Give the structures for all possible products for the following reaction. How many unique products
are there? [12]
*
*
NBS, !
*
Products?
OO
Every allylic position generates two radical
intermediates, for a potential of 6 products. Only 5 are
unique.
Br
Br
Br
Br
The same
compound!
Br
Exam 3
Fall 2006
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Chemistry& 241
Clark College
7. Use both words and figures to explain why an allylic radical is more stable than a 3° radical.
(Remember, this would hold true for carbocations as well as radical species!) [8]
H
H2C
H
A 3° radical can be stabilized by hyperconjugation from neighboring
CH2 groups - the small sharing of e- density through space.
CH2
H
When electron density is shared through resonance, there is a greater
extent of electron "sharing" than hyperconjugation. Since the electrons are
shared by moving through like orbitals, the allylic radical is more stable.
8. When R-3-tert-butyl-1-methylcyclohexene reacts with hydrochloric acid, to different diastereomers
are formed. These products are NOT formed in equal amounts – one isomer is favored over the
other. Which isomer forms in a greater yield? Why does this happen? Explain your response. [8]
HCl
Cl
Cl
The first molecule is formed in a larger amount, for two reasons. First, the steric bulk of the t-butyl
group can block the Cl- from reacting with the “top” side of the carbocation, forcing the Cl- to add anti to
the t-butyl group more often than syn. Second, the two (sterically) larger groups – the t-butyl and the
methyl, are both equatorial in the first compound, making a slightly more stable product.
Exam 3
Fall 2006
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Chemistry& 241
Clark College
Multiple Choice. Select and clearly circle the choice that best answers the question. [3 ea. 12 total]
9. Which of the following alkenes undergoes allylic bromination with NBS to form a single
monobrominated product?
10. Which alkene will generate a product that is a meso compound when treated with osmium tetroxide
and t-butylhydroperoxide?
Osmium tetroxide adds to alcohols in a syn fashion.
11. The product(s) of this reaction is/are:
Br2
CH2Cl2
a. a pair of enatiomers.
b. achiral. Both C’s of the alkene are achiral, since they have two identical groups attached.
c. a pair of diastermers.
d. two completely different compounds.
12. Which of the following constitutional isomers of C5H12 will be most reactive to Br2 in the presence of
light?
It has a 3° center.
Exam 3
Fall 2006
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