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Problem 1
Using the data given below, determine the structure of the compound that best fits all the
spectral data. Remember to record the answers to all questions.
0 12 34 5 PPMExchanges with D2OMultiplet (7 peaks)Doublet
70
60
50
40
PPM
30
20
10
0
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Additional Information:
Molecular Formula: C3H8O
Mass Spectrum Data:
Molecular Ion (M+)
Fragment
60 m/z
45 m/z (base peak)
UV-VIS Data:
No significant UV absorption > 220 nm.
Answer the following questions.
1. What is the degree of unsaturation for this compound?
a. 0
b. 1
c. 2
d. 3
2. What functional group is responsible for the broad, strong peak at 3354 cm-1?
a. carboxylic acid
b. amine
c. alcohol
d. phenol
3. What about the strong absorptions at 2971 cm-1?
a. Aliphatic C-H stretch (carbon atom is sp3 hybridized)
b. Allylic C-H stretch (carbon atom is sp2 hybridized)
c. Aldehyde hydrogen
d. Carboxylic acid hydrogen
4. Which proton readily exchanges with deuterium?
a. Aldehyde proton
b. Carboxylic acid proton
c. Aromatic ring proton
d. Hydroxyl proton
5. Which group in organic chemistry gives rise to the characteristic doublet-septet splitting
pattern in 1H NMR?
a. Ethyl group
b. Isopropyl group
c. Butyl group
d. Carbonyl group
6. How many nonequivalent carbon atoms are present in this compound?
a. 0
b. 1
c. 2
d. 3
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7. What would affect the chemical shift of the sp3 hybridized carbon atom that would cause
its peak to appear at ~ 64 ppm?
a. Hybridization
b. Resonance
c. Number of hydrogen atoms attached to the carbon
d. Electronegative group attached to the carbon
8. What fragmentation would give rise to the peak at m/z 45 in the Mass Spectrum?
a. Loss of hydroxyl group.
b. Loss of methyl group.
c. Loss of ethyl group.
d. Loss of isopropyl group.
9. What information is gained from the UV-VIS data?
a. Compound does not contain a conjugated system.
b. Compound does contain a conjugated system.
c. No information is gained from the UV-VIS data.
d. Need more information.
10. The identity of the unknown compound is ____________.
a. 1-Propanol
b. Cyclopropanol
c. 2-Propanol
d. Propanal
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Problem 2
Using the data given below, determine the structure of the compound that best fits all the
spectral data. Remember to record the answers to all questions.
(3H)
4
(3H)
3
180
160
140
2
PPM
120
100
PPM
1
80
60
40
0
20
0
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Additional Information:
Molecular Formula: C3H6O2
Mass Spectrum Data:
Molecular Ion (M+)
Fragment
74 m/z
43 m/z (base peak)
UV-VIS Data:
No significant UV absorption > 220 nm.
Answer the following questions.
11. What is the degree of unsaturation for this compound?
a. 0
b. 1
c. 2
d. 3
12. The peak at 1746 cm-1 on the IR spectrum tells us the compound contains a(n)
a. hydroxyl group
b. primary amine
c. carbonyl group
d. aromatic ring
13. The sharp intense peak at 1246 cm-1 is usually indicative of _____________ stretch.
a. carbon-carbon single bond
b. carbon-nitrogen single bond
c. carbon-hydrogen single bond
d. carbon-oxygen single bond
14. Based on the information from the IR spectrum, the major functional group in this
compound is a(n) _____________.
a. alcohol
b. carboxylic acid
c. ester
d. aldehyde
15. Based on the 1H NMR how many sets of non-equivalent protons are in this molecule?
a. 1
b. 2
c. 3
d. 4
16. The most deshielded peak in the 13C NMR is characteristic of the ______________
group.
a. vinyl carbon
b. sp-hybridized carbon atoms
c. carbonyl carbon atom
d. aromatic carbon atoms
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17. The base peak at m/z 43 in the mass spectrum is due to the loss of ___________ group.
a. OCH3
b. C=O group
c. CH3 group
d. O=CCH3
18. The peak at 3.4 ppm in the 1H NMR is more downfield than the peak at 2.0 ppm because
it is attached to _______________.
a. another carbon atom
b. an oxygen atom
c. more hydrogen atoms
d. more carbon atoms
19. The molecular ion peak at m/z 74 tells us the ______________ of the compound.
a. molecular formula
b. melting point
c. molecular weight
d. empirical formula
20. The name and structure of the compound that best fits this data is
O
a.
O
b.
H
H3CH2C
O
CH2CH3
H
propionic acid
O
ethyl formate
c.
O
O
d.
CH3
H3C
CH2OH
Hydroxyacetone
H3C
O
methyl acetate
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Problem 3
Using the data given below, determine the structure of the compound that best fits all the
spectral data. Remember to record the answers to all questions.
Triplet (3H)
Exchanges with D2O (2H)
Triplet
(2H)
(Multiplet (2H)
3
2
PPM
1
0
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45
40
35
30
25
PPM
20
15
10
5
0
Additional Information:
Molecular Formula: C3H9N
Mass Spectrum Data:
Molecular Ion (M+)
Fragment
59 m/z
30 m/z (base peak)
UV-VIS Data:
No significant UV absorption > 220 nm.
Answer the following questions.
21. What is the degree of unsaturation for this compound?
a. 0
b. 1
c. 2
d. 3
22. The pair of peaks at 3369 cm-1 and 3291 cm-1 in the IR spectrum is indicative of
____________________.
a. alcohol
b. secondary amine
c. amide
d. primary amine
23. Peaks that appear to the right of 3000 cm-1 in an IR spectrum are characteristic for which
stretching vibration?
a. sp2 carbon-hydrogen stretch
b. sp3 carbon-hydrogen stretch
c. sp carbon-hydrogen stretch
d. really are not all that important
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24. How many neighboring hydrogen atoms must the CH2 group (at 2.65 ppm in 1H NMR)
see in order to have its signal split into a triplet?
a. 0
b. 1
c. 2
d. 3
25. Which one of the following groups of functional groups does not have the ability to
undergo deuterium exchange in 1H NMR?
a. hydroxyl groups (OH)
b. amino groups (NH2 or NH)
c. thiol groups (SH2)
d. methylene groups (CH2)
26. What fragmentation must occur in this compound to produce a base peak at m/z 30 in the
mass spectrum?
a. loss of a methyl group
b. loss of an ethyl group
c. loss of a propyl group
d. loss of a butyl group
27. Why does this compound have an odd molecular weight according to the mass spectrum
data?
a. It has an odd number of hydrogen atoms.
b. It contains one nitrogen atom.
c. It has an odd number of carbon atoms.
d. It does not contain any oxygen atoms.
28. The regions of the IR spectrum between 1640 – 1560 cm-1 and 1350 – 1000 cm-1 are
characteristic for what type of absorptions?
a. C-H and N-H respectively
b. N-H and C-N respectively
c. C-N and N-H respectively
d. N-H and C-H respectively
29. What is the name of the compound that best fits this spectral data?
a. benzyl amine
b. benzyl alcohol
c. propyl amine
d. propyl alcohol
30. If this compound was dissolved in ether along with biphenyl (neutral organic
compound), the best way to separate the compounds is to _______________.
a. extract with a base
b. distill the mixture
c. extract with an acid
d. remove the neutral compound and ether using a roto-evaporator.
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Problem 4
Using the data given below, determine the structure of the compound that best fits all the
spectral data. Remember to record the answers to all questions.
A research student accidentally discarded her carboxylic acid product in the flammable waste
container. She was able to extract her compound from the flammable waste container using a
chemically active extraction procedure she learned about in organic chemistry laboratory.
31. How what this student able to recover her acidic product from the waste container?
a. Preformed a liquid-liquid extraction with an acid.
b. Preformed a liquid-liquid extraction with a base.
c. Preformed a liquid-liquid extraction with water.
d. Evaporated off all the flammable liquid and recrystallized the resulting solid.
Once the student was able to recover what she hopes is her acidic compound, she analyzed
the compound with GC/MS, UV-VIS, IR, 1H NMH and 13C NMR. She now needs help
analyzing her spectral data to determine if the compound she recovered was in fact her acidic
compound.
14
12
10
8
PPM
6
4
2
0
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0 20 40 60 80 100 120 140 160 180 PPM
Additional Information:
Molecular Formula:
Mass Spectrum Data:
Molecular Ion (M+)
Fragment
Fragment
C7H6O2
122 m/z
105 m/z (base peak)
77 m/z
UV-VIS Data:
max 241 nm
(log10  4.1)
Answer the following questions.
32. What is the HDI for this acidic compound?
a. 0
b. 1
c. 4
d. 5
33. This compound has an max at 241 nm. This implies that the compound
contains____________.
a. isolated double bonds
b. one double bond
c. conjugated system
d. no multiple bonds at all
34. The base peak in the mass spectrum for this compound indicated the compound lost a
fragment that has an m/z = 17. This fragmentation corresponds to the loss of a(n)
__________ group.
a. hydroxyl
b. carbonyl
c. methyl
d. ethyl
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35. The large fragmentation at m/z 77 is due to the ______________.
a. benzyl cation
b. acylium cation (R-CO+)
c. phenyl cation
d. cyclohexyl cation
36. What absorptions in the IR spectrum will help this student determine that she has
successfully extraction her acidic compound from the waste container?
a. The peak at 1696 cm-1
b. The broad peak in the range between 2500 – 3400 cm-1.
c. The peaks at 1251 and 1288 cm-1.
d. All of the above
37. The normal range for the carbonyl group of a carboxylic acid in the IR spectrum is 1725
– 1700 cm-1. Why is the carbonyl group of this compound appear at 1696 cm-1?
a. The carbonyl group is in conjugation with the aromatic ring.
b. The oxygen of the carbonyl must be protonated.
c. The carbonyl must be hydrogen bonding with water.
d. That is not the carbonyl peak at 1696 cm-1.
38. What part of this compound gives rise to the peak the furthest downfield (12.79 ppm) in
the 1H NMR spectrum?
a. Aromatic protons
b. Proton attached to the oxygen of the hydroxyl group
c. Carbonyl group
d. No protons absorb this far downfield, therefore it must be an impurity
39. The identify of the student’s acidic compound is ____________.
a. acetic acid
b. phenyl acetic acid
c. benzoic acid
d. diphenyl acetic acid
40. Does 13C NMR support your conclusion?
a. Yes, the peak at ~ 170 ppm is the carbon of the carbonyl and the peaks in the range of 128
– 134 ppm are the aromatic carbons.
b. No, there is not enough data on the 13C NMR to support my conclusion.
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Problem 5
Using the data given below, determine the structure of the compound that best fits all the
spectral data. Remember to record the answers to all questions.
10
8
6
4
PPM
2
0
[Type text]
200
180
160
140
120
100
PPM
Additional Information:
Molecular Formula:
C7H6O
Mass Spectrum Data:
Molecular Ion (M+)
Fragment
Fragment
106 m/z (base peak)
105 m/z
77 m/z
80
60
40
20
0
UV-VIS Data:
max 241 nm
(log10  4.1)
Answer the following questions.
41. What is the HDI for this compound?
a. 0
b. 1
c. 3
d. 5
42. Using your answer from question 41, what does an HDI of this magnitude indicate?
a. The compound contains 1 ring and 3 double bonds.
b. The compound contains 2 rings and three double bonds.
c. The compound contains 1 ring and 4 double bonds.
d. The compound contains 1 ring and two double bonds
43. Using your knowledge of IR and 1H NMR, what is the primary functional group of this
compound?
a. Ester
b. Aldehyde
c. Ketone
d. Amide
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44. What absorptions in the IR helped you to reach your conclusion in question 43?
a. Pair of peaks at 2738 and 2820 cm-1.
b. Peak at 1703 cm-1.
c. Both a and b.
d. Neither a nor b.
180
140 fingerprint
120
100 of the
80 IR spectrum,
60
40 do the
20 absorptions
0
In the
region
what
at 746 and 688 cm-1
PPM
tell you about the substitution pattern of this compound?
a. It is disubstituted aromatic ring with the substituents ortho to each other.
b. It is a monosubstituted aromatic ring.
c. It is disubstituted with the substituents meta to each other.
d. Peaks in the fingerprint region are unimportant and do not provide much detail concerning
this compound.
160
45.
46. Which proton is responsible for the downfield peak at 9.87 ppm?
a. Aldehyde proton
b. Carboxylic acid proton
c. Amide proton
d. Aromatic proton
47. The molecular ion peak for this compound happens to be the base peak at 106 m/z. The
first fragmentation results in the loss of a H and appears at m/z 105. What fragmentation
caused the peak at m/z 77?
a. benzyl cation
b. tropylium ion
c. hydronium ion
d. phenyl cation
48. Why does the 13C NMR show five different types of carbon atoms when the molecular
formula indicates the compound contains seven carbon atoms?
a. Only the carbon atoms that are attached to a hydrogen atom show a absorption.
b. There should only be three peaks in the 13C NMR, so the compound must contain
impurities.
c. The ortho and meta carbons of the ring are equivalent.
d. There are only five carbon atoms in this molecule. The molecular formula is a mistake.
49. Based on all the spectral information, the identity of this compound is __________.
a. Acetaldehyde
b. Benzaldehyde
c. Acetic acid
d. Ethyl benzoate
50. Does 1H NMR support your conclusion?
a. Yes, the group of peaks between 7.45 – 7.81 ppm are from the aromatic protons and the
peak at 9.87 ppm is from the aldehyde proton.
b. No, there is not enough data on the 1H NMR to support my conclusion.