Molecular Formula: C 7 H 6 O Mass Spectrum Data: UV-VIS Data: Molecular Ion (M +
ID: 1022194 • Letter: M
Question
Molecular Formula: C7H6O
Mass Spectrum Data: UV-VIS Data:
Molecular Ion (M+) 106 m/z (base peak) lmax 241 nm
Fragment 105 m/z (log10 e 4.1)
Fragment 77 m/z
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
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.
45. In the fingerprint region of the IR spectrum, what do the absorptions at 746 and 688 cm-1 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.
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.
Explanation / Answer
From the given data,
41. The HDI for this compound is,
d. 5
42. Using your answer from question 41, an HDI of this magnitude indicate,
c. The compound contains 1 ring and 4 double bonds.
43. Using your knowledge of IR and 1H NMR, the primary functional group of this compound,
b. Aldehyde
44. Absorptions in the IR helped you to reach your conclusion in question 43,
c. Both a and b.
45. In the fingerprint region of the IR spectrum, the absorptions at 746 and 688 cm-1 tells you about the substitution pattern of this compound,
b. It is a monosubstituted aromatic ring.
46. proton is responsible for the downfield peak at 9.87 ppm,
a. Aldehyde 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. The fragmentation caused the peak at m/z 77,
d. phenyl cation
48. The 13C NMR show five different types of carbon atoms when the molecular formula indicates the compound contains seven carbon atoms,
c. The ortho and meta carbons of the ring are equivalent.
49. Based on all the spectral information, the identity of this compound is __________.
b. Benzaldehyde
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.
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