Chemistry 3720, Spring 2004
Exam 1 - Key
Name:
This exam is worth 100 points out of a total of 600 points for Chemistry 3720/3720L. You have 50 minutes to
complete the exam and you may use the spectroscopy data sheet as needed. Good Luck.
1. (10 pts) An unknown organic compound has the empirical formula C10H14. Given the following spectral
data, provide a structure for the unknown that agrees with the data. Match the 1H NMR data to your
answer.
1
H NMR (ppm)
13
C NMR (ppm)
1.24 (t, 6H, J = 7.0 Hz), 2.59 (q, 4H, J = 7.0 Hz), 7.05 (m, 4H)
14.6, 32.7, 125.0, 128.3, 128.6, 139.4 (4 x Ar-C is a clue to meta substitution)
Mass spectrum (m/z) 134.20 (M+) (matches formula, therefore C10H14 is actual formula)
Infra Red (cm-1)
720, 800 (clue to meta substitution)
6H triplet
1.24 ppm
4H quartet
2.59 ppm
H
H
H
H
H
ethyl groups
equivalent
4H multiplet
7.05 ppm
1
2. (10 pts) You have made a new organic compound that has the empirical formula C13H16O3. Knowing
that the mass spectrum shows M+ = 220.26, and being given the following spectral data, come up with
the structure of the unknown.
8
1
7
6
5
PPM
4
3
2
1
0
H NMR (ppm) 1.28 (s, 9H), 3.88 (s, 3H), 7.81 (d, 2H, J = 6.9 Hz), 8.02 (d, 2H, J = 6.9 Hz)
220
13
200
180
160
140
120
100
PPM
80
60
40
20
C NMR (ppm) 26.4, 44.4, 51.5, 128.7, 129.8, 134.6, 141.1, 166.0, 210.8
Answer:
O
H 3 CO
O
2
0
3. (15 pts) Give the major organic products from each of the following reaction sequences. Where there is
more than one step, a product from each is expected.
a.
O
O
1. LiAlH 4, THF
Al(OR)3 Li
1.
OH
2.
2. H 3O +
b.
O
O
OCH 3
1. LiAlH 4 , THF
1.
H
2. H 3O +
Al(OR)3 Li
OH
2.
H
H
H
c.
H
O
H
1. PhMgBr
1.
2. H 3 O +
H
OMgBr
H
2.
Ph
H
OH
H
Ph
d.
OH
OsO 4
OH
t-BuOOH, NaOH
e.
O
1. 2 CH 3Li, THF
1.
OLi
+
OCH 3 2. H 3O
3
2.
OH
4. (9 pts) Draw the structures of the expected products (in the boxes) in the following synthetic scheme.
O
OH
NaBH 4, CH 3OH
H
H 2SO 4, H 2 O
heat
CH 2I2, Zn, ether
clue: 1H signals at ~5.0 ppm
5. (10 pts) Draw the approximate 1H NMR spectrum (including the scale) for the molecule below. Include
approximate chemical shifts and show which signal corresponds to which hydrogen(s) in the molecule.
5.72
O
1.16
H
1.16
7.78
6.85
O
4.12
O
1.30
7
1.16
4.33
7.781.43
O
2.67
6
5
1.43
H
5.62
6.85
4
PPM
4
3
2
1
0
6. (10 pts) Which structure below do the following spectra belong to? Explain your answer by matching
the 1H NMR data to the structure you choose.
1
H NMR (ppm)
1.06 (t, 3H, J = 6.9 Hz), 1.22 (t, 3H, J = 7.1 Hz), 2.00 (q, 2H, J = 6.9 Hz), 2.09 (s,
3H), 3.12 (d, 2H, J = 7.3 Hz), 4.00 (q, 2H, J = 7.1 Hz), 4.59 (t, 1H, J = 7.3 Hz)
8.4, 15.6, 25.1, 31.0, 39.4, 62.6, 98.2, 154.4, 202.9
1630, 1720
13
C NMR (ppm)
IR (cm-1)
O
O
O
O
O
O
1.22 ppm
4.59 ppm
2.09 ppm
O
O
4.00 ppm
3.12 ppm
2.00 ppm
1.06 ppm
7. (10 pts) Give the final organic product from the following sequence and give a complete mechanism for
its formation.
O
O
1. 2 x CH 3CH 2MgBr
ether
OH
2. H 3 O +
BrMg
O
OMgBr
O
BrMg
5
OMgBr
8. (10 pts) Ibuprofen (structure given below) is the main ingredient in Motrin, Advil, and Nuprin.
Indicate which of the 1H NMR spectra shown below matches the structure of ibuprofen and match the
signals to the H’s in the structure.
HO
O
1
H NMR (ppm): 1.01 (d, 6H),
2.22 (m*, 1H), 2.35 (s, 3H),
2.51 (d, 2H), 3.49 (s, 2H), 6.90
(m, 3H), 11.0 (s, 1H)
m* = 9 lines in signal
10
8
6
PPM
4
2
0
1
H NMR (ppm): 1.01 (d, 6H),
1.34 (d, 3H), 2.21 (m*, 1H),
2.93 (quintet, 1H), 3.49 (s,
2H), 7.00 (m, 4H), 11.0 (s, 1H)
m* = 8 lines in signal
10
8
6
PPM
4
2
0
1.47
11.0
3.82
HO
1
7.07
7.07
O
7.07
2.22
7.07
H NMR (ppm): 1.01 (d, 6H),
1.47 (d, 3H), 2.22 (m*, 1H),
2.51 (d, 2H), 3.82 (q, 1H), 7.00
(m, 4H), 11.0 (s, 1H)
1.01
1.01
2.51
m* = 9 lines in signal
10
8
6
PPM
4
2
6
0
9. (8 pts) Suggest possibilities for the missing reagents or materials in the following syntheses.
a.
O
O
NaBH 4 , CH 3OH
or
LiAlH 4 , THF
then H 3 O +
OH
H
then H 3O +
MgBr
H
give reagents
give reagents
b.
O
OH
1. 2 x CH 3Li, THF
OCH 3
CH
CH 3 3
2. H 3 O +
O
1. CH 3Li, THF
CH 3
2. H 3 O +
starting material?
starting material?
10. (6 pts) The following compounds are isomers (formula = C11H14O2). In the space below give two
differences you expect to see in the 1H NMR, 13C NMR and IR spectra of each compound.
OH
O
O
O
A
B
Two 1H NMR differences:
A will have a 3H singlet at ~2.0 ppm for the CH3 group alpha to the ketone carbonyl, B will not.
B will have a 1H singlet at ~ 11.0 ppm for the carboxylic acid OH, A will not.
Two 13C NMR differences:
A will have 4 signals in the Ar-C region (110-150) since it is symmetrical, B will have 6.
A has a C=O signal at ~200 ppm for the ketone, B has a C=O signal at ~175 ppm for the acid.
Two IR differences:
A has a signal at ~1700 cm-1 for the C=O, B has a similar signal plus one at ~3200 for the OH.
A will have an absorbance at ~800 cm-1 (para-disubstituted), B will have 700 and 770 cm-1
absorbances (meta-disubstituted).
7