CH14-15 sample problems Answers PDF

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1. For each of the following IR spectra, determine whether it is consistent with the structure of a ketone, an alcohol, a carboxylic acid, a primary amine, or a secondary amine. Answer)

2. Acetic acid has a mass spectrum showing a molecular ion peak at m/z 60. Butyric acid and pentanoic acid also show a peak at m/z 60. Show how this can occur. O

O OH

butyric acid

OH pentanoic acid

Answer) H

OH

O

+ OH

OH

m/z = 60

H

OH

O

+ OH

OH

m/z = 60

3. Explain how you would distinguish the following sets of compounds using 13C NMR spectroscopy. Answer)

13C

NMR: 4 peaks

13C

13C

NMR: 3 peaks

NMR: 5 peaks

4. Compound O (C6H8) reacts with two molar equivalents of hydrogen in the presence of Pd/C to produce P (C6H12). The broadband proton-decoupled 13C spectrum of O consists of two signals, one at d 26.0 and one at d124.5. In the DEPT-90, only one peak at d124.5 was observed, while DEPT-135 showed two peaks with one at d124.5 as a positive peak and the other at d 26.0 as a negative peak. Propose structures for O and P. Answer) δ26.0 δ124.5 2H2 Pd/C O

P

5. When dissolved in CDCl3, a compound (K) with the molecular formula C4H8O2 gives a 1H NMR spectrum that consists of a doublet at d 1.35 (3H), a singlet at d 2.15 (3H), a broad singlet at d 3.75 (1H), and a quartet at d 4.25 (1H). When dissolved in D2O, the compound gives a similar 1H NMR spectrum, with the exception that the signal at d 3.75 has disappeared. The IR spectrum of the compounds shows a strong absorption peak near 1720 cm-1. a) Propose a structure for compound K. b) Explain why the NMR signal at d 3.75 disappears when D2O is used as the solvent. Hint: Alcoholic H (in -OH) exchanges with D2O to form –OD which results in the disappearance of the NMR signal of alcoholic H. Answer) δ1.35

δ4.25

δ2.15

O

O

D2 O

CH3CHCCH3

CH3CHCCH3

OH

OD δ3.75

6. Treatment of compound A with an acid such as TsOH affords two products (M and N) with molecular formula C6H12. The 1H NMR spectra of M and N are given below. Propose structures for M and N and draw a mechanism to explain their formation. Note: the small peak from 7.26 ppm is from CHCl3 impurity in CDCl3. TsOH M and N OH TsOH = p-toluenesulfonic acid O S OH O

Answer)

OH

H

O H O S O

O S O O

H

OH2

H

H

N

M H O S O O...