Exam #1
Chemistry 334
Principles of Organic Chemistry II
Thursday October 5, 2006
Name:
KEY
.
The exam is worth a total of 100 points; there are six questions.
Please show all work to receive full credit for an answer.
By putting your name on this exam, you agree to abide by
California State University, Northridge policies of academic
honesty and integrity
Molecular models are allowed for this exam. Calculators are
not needed.
Good Luck!
1. Predict the products of the following reactions. Remember to indicate
stereochemistry where relevant. (20 pts)
CH3
CH3
HBr
a.
40°C
H3C
CH3
Br
CH3
CH3
H3C
CH3
H3 C
1. Br2, H2O
b
O
2. NaOH
H
c
H
1. MCPBA
OH
2. CH3OH,
H+
CH3
H3CO
CH3
OH
d.
H3 C
O
1. Hg(OAc)2
CH3
+
2. NaBH4, HO-
H3C
e.
1. SO3, H2SO4
2. Br2, FeBr3
3. H2SO4, H2O, D
H3 C
Br
2. Indicate reagents to accomplish the following transformations. More than one step
may be required. (20 pts)
OH
O
1. NaH
a.
2.
Br
OH
CH2
1. MCPBA
b.
OCH2CH3
2. NaOEt
O
CH3
CH3
1. Zn(Hg)/
Aq. HCl
c.
2. HNO3, H2SO4
O2N
CH3
d
OH
H3 C
O
H3C
CH3
H3 C
O
e.
CH3MgBr
OH
O
1. Na,NH3, EtOH
2. H2, Pd-C
OH
3. Draw the structure of the product of the following reaction, being sure to indicate
both regiochemistry and stereochemistry. Provide a brief rationalization for your
regiochemical and stereochemical assignments (15 pts).
H
N
C
CN
heat
+
CH3
+ enantiomer
H3 C
H
CN
N
C
H3C
+
CH3
s-cis: reactive form
of diene
regiochemistry: 1,4 rule
stereochemistry: endo approach of dienophile:
CH3
secondary
orbital overlap
H
methyl group and CN trans
N
C
H
4.
Treatment of epoxide A with a catalytic amount of HCl in diethyl ether leads to the formation of
alcohol B. Show a detailed, step-by step mechanism for the transformation of A to B using the
curved arrow notation. Be sure to include all intermediates and resonance forms where
appropriate. (15 pts)
+ HCl
+ HCl
CH3
HO
O
CH3
H3C
A
B
O
CH3
O
H
CH3
H
CH3
CH3
HO
H
CH3
H3C CH
3
3° carbocation
Cl-
HO
HO
H
H3C CH
3
H
H3C CH
3
5. a. Drawing (resonance) structures to support your answer, explain why an unusually stable
cation is formed when cycloheptatrienone is treated with acid (H+), and why
cyclopentadienone is so unstable that it undergoes spontaneous Diels-Alder reaction with
itself at low temperatures when formed. (10 pts)
O
OH
O-
H+
aromatic cation-stable
6 pi electrons
aromatic resonance form
1
-
O
O
4 pi electrons
antiaromatic resonance form
unstable
b. Consider the heterocyclic base guanine, found in DNA. It has an easily formed tautomeric
(isomeric) form as shown below. Is the tautomer aromatic? Draw out all lone pairs on
heteroatoms, and then diagram the three-dimensional orbital structure of the tautomer to explain
your answer. Include all p orbitals and any hybrid orbitals bearing lone pairs (10 pts).
O
OH
H
N
N
H2 N
N
N
N
N
H2 N
N
H
H
tautomer
guanine
sp2
N
N
N
sp2
N
sp2
N
H
aromatic: 10 pi electrons
6. The rate of nitration of 1-methoxy-4-nitrobenzene is 50 times that of 1-methoxy3-nitrobenzene. By showing resonance forms for the respective nitration
intermediate(s), explain this rate difference. (10 pts)
OCH3
OCH3
HNO3
H2SO4
OCH3
OCH3
H
H
H
NO2
+
NO2
NO2
k1
NO 2
OCH3
NO2
1-methoxy-4-nitrobenzene
NO2
NO2
H
NO2
k1 /k2=50
4 resonance forms.
NO 2
OCH3
OCH3
HNO3
H2SO4
OCH3
H
OCH3
H
O2N
H
O2N
O2N
k2
O
NO2
NO2
O
N
N
very unstable O-
O-
OCH3
1-methoxy-3-nitrobenzene
H
4 resonance forms, one of which is very unstable
O 2N
NO 2
OCH3
OCH3
OCH3
OCH3
HNO3
H2SO4
k2
O
NO2
NO2
H
NO2
N
NO2
H
H
NO2
OCH3 very unstable
NO2
4 resonance forms, one of which is very unstable
1-methoxy-3-nitrobenzene has one very unstable resonance form for ortho or para substitution, and
thus the rate of nitration for this isomer is slower when compared to 1-methoxy-4-nitrobenzene,
which has no such unstable resonance form.
NO2
H
NO2
O-
Bonus (10 pts)
Draw a reaction energy profile diagram for problem 6. Be sure to include both
reactions on the same graph, and indicate the relative energy levels of the
intermediates. For simplicity, assume that the potential energy levels of both reactants
are the same, and that the potential energy levels of all the products are the same.
Ea2
O2N
H
OCH3
NO2
Ea2>Ea1, leading to the observed
rate difference
Ea1
OCH3
H
NO2
potential
energy
OCH3
OCH3
NO2
OCH3
OCH3
NO2
NO2
NO2
OCH3
O2N
NO2
NO2
NO2
NO2
reaction progress
For electrophilic aromatic substitution, the rate-determining step is the first step,
formation of the arenium ion intermediate; this means that the first step has the
highest activation barrier. Reactions that have lower activation barriers for the ratedetermining step proceed faster. Since the rate-determining step is endothermic, the
transition state will resemble the intermediate both in structure and in energy (the
Hammond postulate). Thus, more stable intermediates will have more stable
transition states leading to them, and to a first approximation we can say that the
reaction with the more stable intermediate will proceed faster. As can be seen from
the diagram above, the arenium ion intermediate formed from nitration of 1-methoxy4-nitrobenzene is more stable than that formed from 1-methoxy-3- nitrobenzene (as
discussed in the resonance form analysis for problem 6), and thus we expect the
activation barrier leading to the more stable intermediate to be lower. As a result, at a
fixed temperature, 1-methoxy-4-nitrobenzene undergoes nitration faster than 1methoxy-3-nitrobenzene, and the energy difference Ea2-Ea1 is approximately
proportional (neglecting differences in the Arrhenius factor A for reactions of
different isomers) to the natural logarithm of the ratio of the rate constants k1 and k2
(ln50).
Congratulations!
Score:
1.
2.
3.
4.
5.
6.
Bonus:
Total:
/20
/20
/15
/15
/20
/10
/10
/100