Chemistry 241
Exam II Winter 2008
Chapters 2, 3, 12, & 13
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Chemistry 241
Exam II Winter 2008
Chapters 2, 3, 12, & 13
1. Name the molecules below using the IUPAC nomenclature system. (4 ea)
Br
Cl
8-bromo-8-chloro
-2-methyl-2-octene
3-isopropyl-6,6-dimethylcycloheptane
or 6,6-dimethyl-3-(1-methylethyl)
cycloheptane
(2E, 5Z)3-ethyl-8methyl-2,5-nonadiene
I
N
OH
cylcohex-2-en-1-ol
(E) 1-iodo-3-methyl-2-pentene
(E) 3-amino-N-ethylN-methyl-1,3-hexadiene
2. For each of the molecule pairs pictured below, indicate whether they are (3 ea)
a. constitutional isomers
b. stereoisomers
c. conformers
d. none of the above
a.
d.
Cl
Cl
c.
H
H
H
Br
CH2CH3
Br
CH3
a.
3. Rank each of these from #1 (most or highest) to #3 (least/lowest) in the indicated
category. (4 ea)
a. stability (hint: stability is the opposite of reactivity)
1
2
3
b. number of H-NMR signals
Br
2
1
3
c. number of 13C-NMR signals
Br
1
2
3
4. A molecule has a molecular formula of C12H18. It has only three H-NMR signals. (12)
a. at 7.1 ppm, 4 H’s, singlet
b. at 2.3 ppm, 2 H’s, septet
c. at 1.3 ppm, 12 H’s, doublet
What is the structure of this molecule? On your structure, indicate which H’s
correlate with which signal on the H-NMR.
4 on our IHD: benzene ring (confirmed by four identical H’s @ 7.1 ppm)
a. 4 identical CH’s on benzene ring next to 0 other kinds of H’s:
disubstitutued benzene ring and both substituents are identical
b. 1 CH2 or two CH’s next to benzene ring on one side and multiple (6) H’s on the other
side(s). given the symmetry of the molecule (identical substituents, above), it would be
difficult to have a single CH2
c. 12 identical H’s (4 CH3’s) each next to 1 H, so “b” has to be a CH
pieces:
CH3
CH3
CH
CH
CH3
CH3
put together:
c
a
a
b
c
c
b
a
a
c
5. There are two chair conformations for cis 1,2-dimethylcyclohexane and two
conformations for trans 1,2-dimethylcyclohexane. Draw all four. Label which are cis
and which are trans. Rate them in energy from highest to lowest. (15)
drawing: 2 ea
label cis/trans: 3
rate energy: 4
trans
#1
#4
cis
#2
#2
6. Draw a molecule with the formula C6H13BrO which is consistent with the information
below. (12)
IR - no absorptions in the range 3000 - 4000 cm-1
no absorptions in the range 1600 - 2800 cm-1
1-H NMR
a.
b.
c.
d.
e.
1.1 ppm, 3H, triplet
1.7 ppm, 3H, singlet
1.9 ppm, 2H, quartet
3.3 ppm, 3H, singlet
3.9 ppm, 2H, singlet
7. A molecule has a percent composition of 35.51% O, 9.15% H, and 54.53% C. This
small molecule only has three possible isomers. What is the formula, what are the three
structures, and how would you use IR to differentiate between them? (13)
formula: 4 pts
structures: 2
formula: C2H4O
1 double bond or ring.
can have C=C dble bond, C=O dble bond or 3 member ring
OH
strong, broad –OH at ~3300 cm-1
O
strong C=O at ~1700 cm-1, aldehyde sp2 CH’s @2700 & 2800 cm-1
O
none of the peaks listed on other molecules are here.
BTW, C-C, C-O, and sp3 CH’s are present on all molecules, and thus cannot be used to
distinguish between these three.