Average Score = 55
CHEM 3311 Spring 2006
High Score = 98
Exam 2
March 16, 2006
Professor Rebecca Hoenigman
Low Score = 7
I pledge to uphold the CU Honor Code:
Signature________________________________________________________
Name (printed)____________________________________________________
Last four digits of your student ID number_______________________________
Recitation TA_____________________________________________________
Recitation number, day, and time______________________________________
You have 1 hour and 15 minutes to complete this exam.
No model kits or calculators allowed; periodic table and scratch paper are
attached.
DO NOT TURN PAGE UNTIL INSTRUCTED TO DO SO.
Put your name on ALL pages of the exam
Score:
Recitation Sections:
Number
121
131
141
151
153
152
171
Day
Tuesday
Tuesday
Wednesday
Wednesday
Wednesday
Wednesday
Thursday
Time
8 am
1 pm
8 am
12 pm
12 pm
5 pm
12 pm
TA
Andrew
Heather
Chris
Andrew
Nicole
Chris
Heather
Page 1 ___________/15
Page 2 ___________/20
Page 3 ___________/21
Page 4 ___________/8
Page 5 ___________/24
Page 6 ___________/12
Total ______________/100
Name: _________________________________
1. (3 pts) Give the molecular formula for three stable inorganic radicals.
(1 point each)
O2
NO
NO2
2. (12 pts) You have available 2,2-dimethylcyclopentanol (A) and 2-bromo-1,1dimethylcyclopentane (B) and wish to prepare 3,3-dimethylcyclopentene (C).
Which would you choose as the more suitable reactant, A or B, and with what
would you treat it? Give an explanation for your choice.
OH
A
Br
B
C
Book Problem 5.43
(2 points for choosing correct reagent, 4 points for giving an example of a strong
base, 6 points for the explanation)
B is the more suitable reactant because it can undergo an E2 reaction with
a strong base (such as CH3O-) to give only compound C. Compound A will
undergo an E1 reaction, and since the cation intermediate will rearrange,
compound C will be a minor product.
1
Name: _________________________________
3. (20 pts) Write a complete stepwise mechanism for the bromination of
methylcyclobutane. Label each step.
(5 points, -1 arrows, -1 no step label)
Initiation
Br
hν
Br
Propagation
2Br
(-1 arrows, -1 no step label)
(5 points each for the two propagation steps)
H
+
Br
CH3
+
HBr
CH3
Br
CH3
+
Br
+
Br
Br
CH3
Termination (5 points, -1 arrows, -1 no step label)
(only one termination step is needed)
Br
+
Br2
Br
or
CH3
+
H3C
or
Br
CH3
+
Br
CH3
2
Name: _________________________________
4. (15 pts) Circle the more stable compound in each of the following pairs and
give the reason for your choice in the adjacent box.
(2 points
each circle,
3 points each
explanation)
A. 1-methylcyclohexene
or
3-methylcyclohexene
or
(E)-cycloheptene
Book Problem 5.30a
B. (Z)-cycloheptene
1-methylcyclohexene has a
more substituted double bond.
Cis (Z) double bonds are more
stable than trans (E) double
bonds in cycloalkenes with
fewer than 10 or 11 carbons.
C.
The first carbocation is
stabilized by resonance.
or
5. (6 pts) There are 11 chiral centers in cholic acid. Give the stereochemical label
for the indicated atoms.
(2 points each)
H
S
CH3
OH
CH3
CH3
HO
R
H
H
CH2CH2CO2 H
H
OH
H
R
3
Name: _________________________________
6. (8 pts) Circle the relationship between the following pairs of compounds.
(2 points each)
A. Book Problem 7.33e
CH2 OH
H
CH2OH
OH
HO
H
CH2 OH
Identical
constitutional isomers
CH2OH
enantiomers
diastereomers
B. Book Problem 7.33j
HO
HO
Identical
CH2OH
constitutional isomers
CH2 OH
enantiomers
diastereomers
C.
CH3
CH3
Cl
H
H
Cl
H
Br
H
Br
CH3
Identical
CH3
constitutional isomers
D.
enantiomers
CH3
Br
C
Cl
H
Br
Identical
diastereomers
constitutional isomers
C
H3 C
Cl
H
enantiomers
diastereomers
4
Name: _________________________________
7. (24 pts) Give the organic products for the following reactions. Be sure to
clearly label the stereochemistry of the products. If possible, label the major and
minor products.
A. (3 points) Book Problem 6.34c
2-tert-butyl-3,3-dimethyl-1-butene
1) B2H6 in diglyme
2) HOOH, HOHO
B. (3 points each enantiomer, no stereochemistry = no credit)
Br
Br2
Br
+
CH3CH2OH
OCH2 CH3
OCH2CH3
C. (3 points for major product, 3 points for minor product, 1 point for
enantiomer, 2 points for major and minor product labels)
I
HI
+
I
+
I
Minor
Major
D. (3 points)
OH
OsO4
HOOH
OH
E. (3 points, no deuterium = no points)
CH3
CH3
NaOCH2CH3
CH3CH2OH, heat
Br
H
D
D
5
Name: _________________________________
8. (12 pts) Each of the following transformations can be carried out in two or
three steps. For each transformation show above and/or below the arrows the
necessary reagents and between the arrows show the organic intermediate that
is formed in the first reaction and serves as the starting material for the second
reaction.
(1 point each reagent, 1 point each intermediate)
A. Book Problem 6.36e
O
2-propanol
O
PhCOOH
H2SO4
Δ
or
O
CH3COOH
B. Book Problem 6.36f
isobutyl alcohol
tert-butyl alcohol
dil. H2SO4
H2SO4
Δ
(no credit if not dilute
or 50% or with H2O)
C.
CH3
CH3
Br2
Br
hν
OCH3
(or any
strong base)
(Must have both Br2 and light or heat.
No credit if Cl2.)
CH3
D
D
CH3
D2
Pt
(1 point for each reagent)
6