Name _____________________________________________
(please print)
Chemistry 331
Organic Chemistry I
Exam 4
The gas-phase IR spectrum of cocaine.
N
CH3
H
CO2CH3
O
O
H
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Organic Chemistry I Exam 4: IR and NMR Spectroscopy - Prof. William Setzer, Exams of Organic Chemistry
Information from exam 4 of organic chemistry i, focusing on the use of ir and nmr spectroscopy to identify and distinguish between various organic compounds. Topics include mass spectrometry, uv-vis spectra, resonance structures, ir spectroscopy, and 1h and 13c nmr spectroscopy.
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2011/2012
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Name _____________________________________________ (please print)
Chemistry 331
Organic Chemistry I
Exam 4
The gas-phase IR spectrum of cocaine.
N
CH 3
H
CO 2 CH 3
O
O
H
1. How would you distinguish between the following pair of compounds using mass spectrometry? (5 points):
N
2. Identify which of the following compounds is expected to have the largest λmax in its UV-vis spectrum. Briefly explain. (5 points):
5. How would you use 1 H NMR spectroscopy to distinguish between the following pair of compounds? (5 points):
O
O
6. Identify the number of signals expected in both the 1 H NMR and 13 C NMR spectra of each of the following compounds. (10 points): OCH (^3) CH 3
CH 3
Cl
CH 3
Br Br
O
O
1 H NMR
13 C NMR
1 H NMR
13 C NMR
7. Consider the following compound:
O
Predict the number of signals and the general location of the signals in a 13 C NMR spectrum of the compound. (5 points)
300 MHz 1 H NMR spectrum
(^13) C NMR spectrum
δ ppm Int. 145.89 378 135.11 378 129.05 975 126.31 965 33.78 443 24.12 1000 20.94 234
6H
3H
4H
1H
8. b) C 6 H 10 O cyclohexanone
IR spectrum
Mass spectrum
UV spectral data: λ max = 285 nm, log ε max = 1.
8. c) C 6 H 10 O 2 trans -2-hexenoic acid
IR spectrum
Mass spectrum
300 MHz 1 H NMR spectrum
(^13) C NMR spectrum
2H
bs m
1H
dt
1H
1H
m
2H
dt t
3H
172.50 (^) 152.
J = 15.6, 1.5 Hz
3. In their IR spectra, the νC=C for pent-3-en-2-one is 1600 cm-1^ while νC=C for pent- 4-en-2-one is 1650 cm-1^. Use resonance structures to explain why the C=C bond in the conjugated compound produces a signal at lower frequency. (5 points) O O O
O
Due to resonance, this bond has less double-bond character and resonates at lower frequency.
Resonance not possible for pent-4-en-2-one; this is just a regular double bond.
pent-3-en-2-one
4. Describe how IR spectroscopy can be used to monitor the progress of the following reaction. (5 points):
O O
H 2
Ni
The starting material is a conjugated enone and the νC=O will be at lower frequency (around 1680 cm-1^ ) than the product (a simple ketone with νC=O around 1720 cm -1^ ). So, look for the disappearance of the lower frequency carbonyl stretch with appearance of the higher frequency signal.
5. How would you use 1 H NMR spectroscopy to distinguish between the following pair of compounds? (5 points):
O
H H
H
H
O
H
H
H
H
H
H
H
H
Just count signals. Due to symmetry the first compound will have three signals, but the second compound will have six signals.
6. Identify the number of signals expected in both the 1 H NMR and 13 C NMR spectra of each of the following compounds. (10 points): OCH (^3) CH 3
CH 3
Cl
CH 3
Br Br
O
O
H
H
H
1 H NMR
13 C NMR
1 H NMR
13 C NMR
8. Determine the structure of each of the following compounds. Include a brief structural analysis and assignment of the key spectral features that led to your structure determination (20 points each).
a) C 10 H 14 p -cymene IR spectrum
Mass spectrum
νC-H ( sp^3 )
νC-H ( sp^2 )
aromatic overtones
M+
M – 15
M – 43
300 MHz 1 H NMR spectrum
(^13) C NMR spectrum
δ ppm Int. 145.89 378 135.11 378 129.05 975 126.31 965 33.78 443 24.12 1000 20.94 234
6H
3H
4H
1H
H 3 C CH
CH 3
CH 3
isopropyl pattern
methyl
aromatic
aromatic
90 MHz 1 H NMR spectrum
(^13) C NMR spectrum
δ ppm Int. 211.56 163 42.00 1000 27.11 946 25.07 516
4H
6H
O
H
H
H
H H
H
H
H
H
H
C C
C
C C
O C
H
H
H
H H
H
H
H
H
H
8. c) C 6 H 10 O 2 trans -2-hexenoic acid
IR spectrum
Mass spectrum