CH341/CH540, Exam 1 Key, Fall 2009
______________________________________
Name
This exam consists of several parts of unequal difficulty. Look over the whole exam right now, and start
on something you know well. If you don’t know something, skip it, and come back to it later. Use your
time wisely.
Keep Your Eyes On Your Own Paper!
Grading Summary
Page 2. _______________ (16)
Page 6. _______________ (23)
Page 3. _______________ (14)
Page 7. _______________ (17)
Page 4. _______________ (12)
Bonus! _______________ (≤5)
Page 5. _______________ (18)
Total ________________ (100)
A Partial Periodic Table of the Elements
1A
IIA
IIIB
IVB
VB
VIB
VII
B
VIII
IB
IIB
IIIA
IVA
VA
VIA
VII
A
1
H
VIII
A
2
He
1.01
4.00
3
Li
4
Be
5
B
6
C
7
N
8
O
9
F
10
Ne
6.94
9.01
10.8
12.0
14.0
16.0
19.0
20.2
11
Na
12
Mg
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
23.0
24.3
27.0
28.1
31.0
32.1
35.5
40.0
19
K
20
Ca
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
39.1
40.1
69.7
72.6
74.9
79.0
79.9
83.8
37
38
49
50
51
52
53
I
54
21
22
23
24
25
26
27
28
29
30
39
40
41
42
43
44
45
46
47
48
127
2
1. Expand the following condensed structure. Show all atoms, all bonds, and all unshared pairs of
electrons. No atoms are charged.
NCCH2COCH3
N
C
H
O
H
C
C
C
H
CH3
H
2. Draw two structures that correspond to the formula C2H5NO2. Show all atoms, bonds, and unshared
pairs of electrons. No atoms may be charged. One structure must contain one ring, and the other
structure must not contain any rings.
There are many possibilities: only two are shown.
O
H
H
O
C
C
C
O
H
H
O
N
H
C
N
H
H
H
H
H
3. Draw one important resonance form of the following structure. Use curved arrows to show how
electrons would have to move to get from the original structure to the one you drew.
There are two possible good resonance forms of the original.
O
O
O
-
C
H
-
C
C
C
CH3
N
N
N
H
C
C
C
CH3
H
C
C
CH3
3
4a. Complete the following acid-base reaction. Draw curved arrows to show how bonds are broken and
formed.
H
H + H3C
S
Benzenethiol pka = 7
N
S
CH3
-
+
H3C
N
CH3
+
CH3
CH3
TMA
pka = 11
4b. The pKa of benzenethiol is 7. The pKa of the conjugate acid of TMA is 11. Which side of the above
reaction is favored at equilibrium? Explain your reasoning.
The favored side at equilibrium has the weaker acid. Since an acid with a pKa of 11 is weaker
than an acid with a pKa of 7, the products are favored.
5. Circle and label neatly each occurrence of the following functional groups in the structure below.
Alkene
Alkyne
Carboxylic acid
Ether
Ester
Alcohol
Aromatic
O
OH
C
CH
Ketone
O
CH2
C
CH
CH
HN
C
CH
Amide
CH
Aromatic
O
Ether
Aldehyde
CH2
CH
C
Ketone
Alcohol
CH
C
CH2
Amide
H2 C
C
Amine
CH
CH2
CH3
CH
Alkene
C
CH
Alkyne
4
6. Multiple Choice: Choose the best answer to each question.
1.
D
____
2.
A
____
3.
C
____
4.
How many pi-bonds are in the structure below?
C
N
A. 0
B. 1
C. 2
D. 3
When one compares the densities of hexane and water, one finds:
A. Hexane is less dense than water.
B. Hexane and water have the same density.
C. Hexane is more dense than water.
D. The relative densities of two immiscible liquids cannot be measured.
If a mixture contains 80% of one compound, and 20% of its enantiomer, then the optical
purity (enantiomeric excess) of the mixture is?
A. 100%
B. 80%
C. 60% D. 20%
Which line-angle formula corresponds to the condensed structure below?
HOCH2C(O)CH(CH3)2
A.
B.
C.
D.
B
H
O
HO
5.
HO
HO
O
O
O
Which of the following is a diastereomer of the structure below?
CH3
O
CH3
D
A.
H3C
B.
C.
D.
CH3
CH3
CH3
____
H3C
6.
Which of the following structures has the highest boiling point?
CH3
A.
B
____
O
H2C
CH2
CH2 CH2
H3C
B.
H2C
CH2
H2C
CH
CH3
C.
H2C
H2C
O
H
CH2
CH
D.
H 2C
H 2C
CH3
CH2
CH
NH2
5
7. What are the indicated bond angles and hybridization states in the following molecule? The angles in
the drawing may not be accurate, since the molecule is really 3-Dimensional.
E
H3C
B
H3C
C
C
C
CH3
C
CH2
C
C
H3C
A
D
O
F
Bond Angles
Hybridization States
A: 180°
D: sp
B: 109°
E: sp2
C: 120°
F: sp2
8. Draw the two chair forms of following structure. Label the groups in each form as either equatorial or
axial. Which form is more stable? Why
H
F
CH2CH2CH3
H
Equatorial
H
Axial
CH2CH2CH3
F Axial
Equatorial
H
CH3CH2CH2
F
Equatorial
H Axial
Equatorial
H Axial
Bigger groups are more stable in the equatorial position, since they bump into axial hydrogens
when they are axial. Since propyl is bigger than fluoro, the more stable from is the right one,
which has the propyl equatorial.
6
9. What is the IUPAC name of each of the following structures?
a.
H
F
CH2CH2CH3
H
cis-1-fluoro-2-propylcyclohexane
b.
CH3
H3C
CH2
CH
CH2
CH
CH2
CH3
Cl
5-chloro-2-methylheptane
c.
CH2CH2CH2CH3
H3C
CH3
CH
HC
H 3C
CH3
1-butyl-2,5-diisopropylcyclooctane
d. H3C
CH2
CH
CH2
CH2
CH3
CH
CH2
CH
H3C
H3C
2,3,7-trimethylnonane
CH3
Classify the indicated atom as either R or S.
4 H
HO
2
C
CH2 S
Br1
CH2
3
CH2
CH3
CH2
7
10. Star (*) all of the chirality centers in the following structure.
H3C
CH
CH2
H2C
CH2
CH3
CH
CH
CH2
H3C
CH
CH
CH2
CH3
Cl
11. Classify each pair of structures below as identical, constitutional isomers, enantiomers, or
diastereomers.
a.
H
Br
Br
H
C
H3C
C
CH2
H
and
CH3
Br
Br
H
C
H3C
C
CH2
CH3
Identical: they are just flipped over front to back.
b.
Br
H
C
H3C
Br
CH2
CH2
CH3
and
H3C
CH2
CH2
CH2
H
C
CH2
CH3
Enantiomers: they are mirror images that don’t superimpose.
c.
CH2
H2C
H
H3C
CH3
H3C
C
H
C
CH2
CH2
CH2
C
and
CH3
C
H
H2C
H
CH2
CH2
Constitutional isomers: one is a 1,4-dimethylcyclohexane, and the other is a 1,3dimethylcyclohexane.
8
Bonus:
The most stable chair conformation of the following molecule has the OH in an axial position. Explain
why.
O
O
O
H
O
H
O
O
When the OH is axial, it can hydrogen-bond with the Oxygens in the ring. Hydrogen-bonds
stabilize molecules by 20-30 kJ/mol. Also, with the two O’s in the ring, there are no axial H’s for
the OH to bump into!