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CHEMISTRY 332
SUMMER 08
EXAM I
June 26-28, 2008
The following materials are permissible during the exam: molecular model kits, course
notes (printed, electronic, or hand written), textbook, calculator, Internet browser. You
are not permitted to receive assistance, in any way, from another person during the exam.
Receiving aid on an exam from any person, by any means of communication, is
considered academic misconduct and can result in a failing grade for that exam. Answers
must be submitted via ACE Organic during the scheduled examination period. The paper
copy of your exam will be collected, but not graded. Your goal is to answer every
question correctly, in as few attempts as possible.
Average Bond Energies (kcal mole-1)
a
H
C
N
O
F
Si
S
Cl
Br
I
104
99
93
111
135
76
83
103
87
71
H
83a
73b
39
86c
53e
47
116d
65
45
37
72
65
81
46
52
68
52
48
56
91
61
58
74
52
56
C
N
O
F
Si
S
Cl
Br
I
C=C 146, C≡C 200
108
135
53
60
b
C=N 147, C≡N 213
d
C=O 176 (aldehydes)
179 (ketones)
e
In CF4
46
c
In nitrites and nitrates
36
CHEMISTRY 332
Exam I, Summer 08
1) [4 pts.] Draw the Lewis structure of the diatomic molecule, nitrogen fluoride (NF). You are
expected to follow the standard practice showing an electron configuration that satisfies the octet
rule for all atoms and which contains no unpaired electrons (i.e., radicals). Be sure to indicate
formal charges (the overall charge on the molecule is neutral).
Figure 1 shows nitrogen fluoride molecular orbital images and its MO energy diagram.
In the images, fluorine is on the left side and nitrogen is on the right. In the energy
diagram, electrons have been not been included.
2) [4 pts.] Based on the MO images in Figure 1, place a check in each box that corresponds to an
image of an antibonding MO.
3) [4 pts.] Using the information provided in Figure 1, match the numbers in the energy level
diagram to the letters of the MO images.
4) [2 pts.] Which atomic orbital makes the greatest contribution to the molecular orbital
designated “c”? (select one box only)
5) [2 pts.] Which atomic orbital makes the greatest contribution to pi*? (select one box only)
6) [6 pts.] Check 4 boxes for this problem. The molecular orbital designated “b” in Figure 1 is
best described as a __________ (sigma-type or pi-type) interaction resulting from the ________
(constructive / destructive) combination of the ____________ (fluorine 2s, fluorine 2p) and
__________ (nitrogen 2s, nitrogen 2p) atomic orbitals?
1
CHEMISTRY 332
Exam I, Summer 08
7) [4 pts.] Which statement explains why level 4 in the energy diagram of Figure 1 is higher than
level 3?
8) [2 pts.] What is the bond order of nitrogen fluoride?
Figure 2 shows energy diagrams of atomic p orbitals and their resulting pi/pi* MOs for
the series of cations containing double or triple bonds in QUESTION 9.
1
2
3
4
5
9) [4 pts.] Match the numbers associated with the energy diagrams in Figure 2 to the
corresponding molecular structure of the cations.
10) [4 pts.] For which of the cations in QUESTION 9 is pi* not the LUMO (check all that
apply)?
11) [4 pts.] On the basis of the MO diagrams, rank the cations in QUESTION 9 from most to
least electrophilic (most = 1 and least = 5).
12) [4 pts.] For each of the cations in QUESTION 9, identify the most electrophilic atom in that
structure by placing the number “1” next to this atom. The MO diagrams in Figure 2 can help
you. To add the number "1", right-click (Safari and Netscape for Mac users: control- or optionclick) on the atom that needs to be labeled (or unlabeled), and choose Map → M1 (or Off).
13) [4 pts.] The structure shown in ACE was recently suggested as an intermediate in the
mechanism for an acyl transfer reaction (J. Am. Chem. Soc. 2006, 128, 4556). There are 3
nitrogen atoms in the structure of this molecule. Label the nitrogen atoms “1” through “3” to
indicate the most to least basic sites (assign “1” to the most basic nitrogen, “2” to the second
most basic nitrogen, and 3 to the least basic nitrogen). To add a number, right-click (Safari and
Netscape for Mac users: control- or option-click) on the atom that needs to be labeled (or
unlabeled), and choose Map → number (or Off).
2
CHEMISTRY 332
Exam I, Summer 08
Figure 3 shows a mechanism for the electrophile-induced cyclization of an alkynyl-2,3epoxy alcohol into an iodopyran-4-one. Use this figure to answer questions 14-18 (ref. J.
Org. Chem., 73, 4342, 2008).
Cl
I
I
CH3
QUESTION 14
H
H
CH3
O
O
O
CH3
O
CH3
Cl–
alkynyl-2,3-epoxy alcohol
QUESTION 15
I
H
CH3
I
CH3
H
O
O
O
O
CH3
QUESTION 16
CH3
QUESTION 17
H
H
H3O+
O
H
I
CH3
O
CH3
O
I
QUESTION 18
O
O
CH3
iodopyran-4-one
CH3
14) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in
the mechanism step labeled “QUESTION 14” of Figure 3. First, indicate the frontier orbitals
involved; your answer should be one of the nine possible combinations of filled→empty orbital
pairs (e.g., π → π*). Next, indicate whether the frontier orbitals are involved in a σ-type
(coaxial) or π-type (side-by-side) interaction.
15) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in
the mechanism step labeled “QUESTION 15” of Figure 3. First, indicate the frontier orbitals
involved; your answer should be one of the nine possible combinations of filled→empty orbital
pairs (e.g., π → π*). Next, indicate whether the frontier orbitals are involved in a σ-type
(coaxial) or π-type (side-by-side) interaction.
3
CHEMISTRY 332
Exam I, Summer 08
16) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in
the mechanism step labeled “QUESTION 16” of Figure 3. First, indicate the frontier orbitals
involved; your answer should be one of the nine possible combinations of filled→empty orbital
pairs (e.g., π → π*). Next, indicate whether the frontier orbitals are involved in a σ-type
(coaxial) or π-type (side-by-side) interaction.
17) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in
the resonance interaction labeled “QUESTION 17”. First, indicate the frontier orbitals involved;
your answer should be one of the nine possible combinations of filled→empty orbital pairs (e.g.,
π → π*). Next, indicate whether the frontier orbitals are involved in a σ-type (coaxial) or π-type
(side-by-side) interaction.
18) [4 pts.] Check two boxes. Base your answer to this question on the curved arrows drawn in
the mechanism step labeled “QUESTION 18” of Figure 3. First, indicate the frontier orbitals
involved; your answer should be one of the nine possible combinations of filled→empty orbital
pairs (e.g., π → π*). Next, indicate whether the frontier orbitals are involved in a σ-type
(coaxial) or π-type (side-by-side) interaction.
Figure 4 shows a reversible chemical reaction in which the hydrophobic effect drives
complex formation (Ref. Chem. Commun., 2008 DOI: 10.1039/b805446k). Use the information
in this figure to answer QUESTION 19.
complex
HO
Keq
estradiol
cavitand
complex
OH
estradiol
Structure of "cavitand" molecular capsules
side view
cavitand interior
19) [4 pts.] Check all of the boxes that are true based on your interpretation of Figure 4. The
reported Keq for estradiol at room temperature in aqueous solution is about 1 × 108 M-2.
4
CHEMISTRY 332
Exam I, Summer 08
20) [4 pts.] Use the table of bond energies to estimate the enthalpy change for the following
reaction (Ref. Org. Lett., 2008, 10, 2569).
O
O
O
O
O
O
O
O
+ H2O
O
21) [4 pts.] Three structures taken from the CSD are shown in ACE. For each structure, you are
to indicate the most likely geometry of the nitrogen or phosphorus atom. Specifically, label each
nitrogen or phosphorus with a number to indicate the expected geometry (1 = linear, 2 = trigonal
planar, 3 = pyramidal, or 4 = bent). To add a number in ACE, right-click (Safari and Netscape
for Mac users: control- or option-click) on the atom that needs to be labeled (or unlabeled), and
choose Map → number (or Off).
22) [2 pts.] A CSD search was performed similar to Quiz 5, only arsenic was used instead of
nitrogen or phosphorus. The search structure and histogram are provided. Given this data, what
geometry would you predict for the arsenic atom?
Histogram of <C-As-C> Valence Angles
25
15
10
5
10
0
10
2
10
4
10
6
10
8
11
0
11
2
11
4
11
6
11
8
12
0
12
2
96
98
0
92
94
Frequency
20
<C-As-C> Valence Angle (Degrees)
5
CHEMISTRY 332
Exam I, Summer 08
23) [2 pts.] A search of the CSD was performed similar to Quiz 5 only the carbons were
constrained to having a coordination number of 4. Which of the following histograms would you
expect as your result?
Histogram of <C-N-C> Valence Angles
Histogram of <C-N-C> Valence Angles
<C-N-C> Valence Angles (Degrees)
<C-N-C> Valence Angles (Degrees)
Histogram of <C-N-C> Valence Angles
Histogram of <C-N-C> Valence Angles
<C-N-C> Valence Angles (Degrees)
10
0
10
2
10
4
10
6
10
8
11
0
11
2
11
4
11
6
11
8
12
0
12
2
96
98
94
92
10
0
10
2
10
4
10
6
10
8
11
0
11
2
11
4
11
6
11
8
12
0
12
2
98
D
96
94
C
92
10
0
10
2
10
4
10
6
10
8
11
0
11
2
11
4
11
6
11
8
12
0
12
2
96
98
92
0
10
2
10
4
10
6
10
8
11
0
11
2
11
4
11
6
11
8
12
0
12
2
98
10
96
94
92
94
B
A
<C-N-C> Valence Angles (Degrees)
6
CHEMISTRY 332
Exam I, Summer 08
24) [4 pts.] When exposed to the reaction conditions indicated below, the trans isomer results in
both the fragmentation product and a mixture of E2 elimination products. The fragmentation
pathway requires that the sigma bond undergoing fragmentation overlap in a pi-type fashion with
sigma* of the C–Cl bond. One specific ring-flipped chair conformation of the trans-isomer is
ideally suited for fragmentation and the other ring-flipped chair conformation is ideally suited for
E2 elimination. Follow steps (i) – (iv) below to draw the ring-flipped chair conformation
that is best suited to undergo the fragmentation pathway.
i. To add a nitrogen or chlorine substituent to the ring, press the appropriate atom button.
ii. Click on an H atom in the structure to replace that H atom with the atom or group at the tip of
the cursor.
iii. Do not delete any of the atoms or bonds in the ring or directly attached to the ring with the
Erase button, and do not move atoms or groups directly attached to the ring by dragging
them.
iv. Don’t forget to draw the –CH3 groups on nitrogen (use the bond tool and click on the N atom
to add a methyl group).
CH3
H3C
CH3
H3C
N
fragmentation
product
Cl–
N
Cl
EtOH, Et3N, H2O
CH3
trans-isomer
H3C
CH3
N
H3C
N
+
mixture of E2 elimination products
7
CHEMISTRY 332
Exam I, Summer 08
25) [2 pts.] The mechanism for an amine (N) reacting with an acyl phosphate monoester (S) is
provided below. The rate-determining step is decomposition of the tetrahedral intermediate (I).
Which equation determines the overall rate of the reaction? (Ref. J. Org. Chem., 2008, 73, 4753).
O
O
P
O
OCH3
O
R
N
O
OCH3
k2
NHR
H2N
O
R
NH2
S
O
P
k1
k-1
+
O
O
O
+
I
P
O
P
HO
OCH3
26) [4 pts.] The rate coefficient that governs the rate expression in QUESTION 25 is directly
related to the pKa of the amine’s conjugate acid (the larger the pKa, the larger the rate
coefficient). Rank these amines from largest rate coefficient to smallest (1 = largest rate
coefficient, 5 = smallest rate coefficient).
27) [2 pts.] Assume that the [I] follows steady-state behavior; also assume that the values of k-1
and k2 have comparable magnitudes (unlike the case in your lecture notes). Given these
assumptions, derive an expression for the steady-state concentration of [I] in terms of [N] and
[S]. Select the equation that you derived.
28) [4 pts.] Based on the above information, check all the statements that are true.
8
ACE Organic homework
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Exam 1 - Summer 08
Maximum allowed tries per question: Unlimited
(1) QUESTION 1) [4
pts.] Follow the
instructions in the
exam.
(Question -1118)
(2) QUESTION 2) [4
pts.] Follow the
instructions in the
exam.
(Question -1119)
Option 1. a
Option 2. b
Option 3. c
Option 4. d
Option 5. e
Option 6. f
(3) QUESTION 3) [4
pts.] Follow the
instructions in the
exam.
(Question -1120)
Item 1. a
Item 2. b
Item 3. c
Item 4. d
Item 5. e
Item 6. f
(4) QUESTION 4) [2
pts.] Follow the
instructions in the
exam.
(Question -1121)
(5) QUESTION 5) [2
pts.] Follow the
instructions in the
exam.
(Question -1122)
(6) QUESTION 6) [6
pts.] Follow the
instructions in the
exam.
(Question -1123)
Option 1. A nitrogen 2p atomic orbital.
Option 2. A nitrogen 2s atomic orbital.
Option 3. A fluorine 2p atomic orbital.
Option 4. A fluorine 2s atomic orbital.
Option 1. A nitrogen 2p atomic orbital.
Option 2. A nitrogen 2s atomic orbital.
Option 3. A fluorine 2p atomic orbital.
Option 4. A fluorine 2s atomic orbital.
Option 1. sigma-type
Option 2. pi-type
Option 3. constructive
Option 4. destructive
Option 5. fluorine 2s
Option 6. fluorine 2p
Option 7. nitrogen 2s
Option 8. nitrogen 2p
(7) QUESTION 7) [4
pts.] Follow the
instructions in the
1 of 8
Option 1. The nitrogen 2s orbital makes a significant contribution to this
MO.
6/25/08 7:29 AM
ACE Organic homework
exam.
(Question -1124)
http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable...
Option 2. Coaxial interaction of a pair of p-orbitals provides less overlap
than side-by-side overlap.
Option 3. s/p mixing decreases the energy of the pi MO levels.
Option 4. All of the above.
(8) QUESTION 8) [2
pts.] Follow the
instructions in the
exam.
(Question -1125)
(9) QUESTION 9) [4
pts.] Follow the
instructions in the
exam.
Option 1. 1.0
Option 2. 1.5
Option 3. 2.0
Option 4. 2.5
Option 5. 3.0
Item 1.
(Question -1126)
Item 2.
Item 3.
Item 4.
Item 5.
2 of 8
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ACE Organic homework
(10) QUESTION 10) [4
pts.] Follow the
instructions in the
exam.
http://aceorganic.pearsoncmg.com/epoch-plugin/homework/printable...
Option 1.
(Question -1127)
Option 2.
Option 3.
Option 4.
Option 5.
(11) QUESTION 11) [4
pts.] Follow the
instructions in the
3 of 8
Item 1.
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ACE Organic homework
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exam.
(Question -1128)
Item 2.
Item 3.
Item 4.
Item 5.
(12) QUESTION 12) [4
pts.] Follow the
instructions in the
exam.
Fig. 1 of 1
(Question -1129)
(13) QUESTION 13) [4
pts.] Follow the
instructions in the
exam.
4 of 8
Fig. 1 of 1
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(Question -1141)
(14) QUESTION 14) [4
pts.] Follow the
instructions in the
exam.
(Question -1132)
Option 1. sigma → a
Option 2. sigma → pi*
Option 3. sigma → sigma*
Option 4. pi → a
Option 5. pi → pi*
Option 6. pi → sigma*
Option 7. n → a
Option 8. n → pi*
Option 9. n → sigma*
Option 10. coaxial (sigma-type) interaction
Option 11. side-by-side (pi-type) interaction
(15) QUESTION 15) [4
pts.] Follow the
instructions in the
exam.
(Question -1133)
Option 1. sigma → a
Option 2. sigma → pi*
Option 3. sigma → sigma*
Option 4. pi → a
Option 5. pi → pi*
Option 6. pi → sigma*
Option 7. n → a
Option 8. n → pi*
Option 9. n → sigma*
Option 10. coaxial (sigma-type) interaction
Option 11. side-by-side (pi-type) interaction
(16) QUESTION 16) [4
pts.] Follow the
instructions in the
exam.
(Question -1134)
Option 1. sigma → a
Option 2. sigma → pi*
Option 3. sigma → sigma*
Option 4. pi → a
Option 5. pi → pi*
Option 6. pi → sigma*
Option 7. n → a
Option 8. n → pi*
Option 9. n → sigma*
5 of 8
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Option 10. coaxial (sigma-type) interaction
Option 11. side-by-side (pi-type) interaction
(17) QUESTION 17) [4
pts.] Follow the
instructions in the
exam.
(Question -1135)
Option 1. sigma → a
Option 2. sigma → pi*
Option 3. sigma → sigma*
Option 4. pi → a
Option 5. pi → pi*
Option 6. pi → sigma*
Option 7. n → a
Option 8. n → pi*
Option 9. n → sigma*
Option 10. coaxial (sigma-type) interaction
Option 11. side-by-side (pi-type) interaction
(18) QUESTION 18) [4
pts.] Follow the
instructions in the
exam.
(Question -1136)
Option 1. sigma → a
Option 2. sigma → pi*
Option 3. sigma → sigma*
Option 4. pi → a
Option 5. pi → pi*
Option 6. pi → sigma*
Option 7. n → a
Option 8. n → pi*
Option 9. n → sigma*
Option 10. coaxial (sigma-type) interaction
Option 11. side-by-side (pi-type) interaction
(19) QUESTION 19) [4
pts.] Follow the
instructions in the
exam.
(Question -1137)
Option 1. As shown in Figure 4, the stoichiometry of the cavitand:estradiol
complex is 1:1.
Option 2. In the absence of solvent, ΔS for complex formation is positive.
Option 3. Ionization of the cavitand’s carboxylic acid groups will help
promote solubility of the cavitand in water.?
Option 4. In aqueous solution without added estradiol, the cavitand’s
interior is filled with ordered water molecules.?
Option 5. In aqueous solution, complex formation is entropically favored
because estradiol binding liberates water molecules from the cavitand’s
interior.?
Option 6. Higher temperature will bring about a decrease in Keq.
(20) QUESTION 20) [4
pts.] Follow the
instructions in the
exam.
(Question -1138)
6 of 8
Option 1. +258 kcal/mol
Option 2. +36 kcal/mol
Option 3. +4 kcal/mol
Option 4. -12 kcal/mol
Option 5. -18 kcal/mol
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Option 6. -162 kcal/mol
(21) QUESTION 21) [4
pts.] Follow the
instructions in the
exam.
Fig. 1 of 1
(Question -1142)
(22) QUESTION 22) [2
pts.] Follow the
instructions in the
exam.
(Question -1143)
(23) QUESTION 23) [2
pts.] Follow the
instructions in the
exam.
(Question -1144)
(24) QUESTION 24) [4
pts.] Follow the
instructions in the
exam.
Option 1. pyramidal
Option 2. trigonal planar
Option 3. linear
Option 4. bent
Option 1. A
Option 2. B
Option 3. C
Option 4. D
Fig. 1 of 1
(Question -1145)
(25) QUESTION 25) [2
pts.] Follow the
instructions in the
exam.
Option 1. rate = k1[N][S]
(Question -1146)
Option 4. rate = k1[N][S] - k2[I]
Option 5. rate = (k-1 - k2)[I]
(26) QUESTION 26) [4
pts.] Follow the
instructions in the
exam.
Item 1.
Option 2. rate = k-1[I]
Option 3. rate = k2[I]
(Question -1147)
Item 2.
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Item 3.
Item 4.
Item 5.
(27) QUESTION 27) [2
pts.] Follow the
instructions in the
exam.
Option 1. [I] = {k1/(k-1 + k2)}[N][S]
(Question -1149)
Option 4. [I] = {k1k 2/(k-1 + k2)}[N][S]
(28) QUESTION 28) [4
pts.] Follow the
instructions in the
exam.
Option 1. The most basic amino group will react fastest in a competitive
situation as long as the pH of the reaction solution is above the pK a of that
(Question -1148)
Option 2. [I] = {k1/(k-1 - k2)}[N][S]
Option 3. [I] = {k1k 2/(k-1 - k2)}[N][S]
of the amine.
Option 2. The most basic amino group will react slowest in a competitive
situation as long as the pH of the reaction solution is above the pK a of that
of the amine.
Option 3. The rate of the reaction will be independent of pH.
Option 4. In addition to the rate coefficient and concentrations of the
reactants, the relative amount of free vs protonated amine determines the
overall rate of amide formation.
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