Chapter 14: Q.21558-14-67P (page 567)
Question. Compound O has a molecular formula C10H12O and shows an IR absorption at 1687 . The 1H -NMR spectrum of O is given below. What is the structure of O?

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Chapter 14: Q.21558-14-67P (page 567)
Question. Compound O has a molecular formula C10H12O and shows an IR absorption at 1687 . The 1H -NMR spectrum of O is given below. What is the structure of O?

Answer

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For each compound, which of the protons on the highlighted carbons absorbs farther downfield?
a.

b.

c.

Compound A exhibits two signals in its 1H NMR spectrum at 2.64 and 3.69 ppm and the ratio of the absorbing signals is 2:3. Compound B exhibits two signals in its 1H NMR spectrum at 2.09 and 4.27 ppm and the ratio of the absorbing signals is 3:2. Which compound corresponds to dimethyl succinate and which compound corresponds to ethylene diacetate?

Question: The 1 H NMR spectrum of N,N-dimethylformamide shows three singlets at 2.9, 3.0, and 8.0 ppm. Explain why the two groups are not equivalent to each other, thus giving rise to two NMR signals.
Question: Identify the carbon atoms that give rise to the signals in the 13C NMR spectrum of each compound.
a. CH3CH2CH2CH2OH ; 13CNMR: 14, 19, 35, and 62 ppm
b. (CH3)2CHCHO ; 13C NMR: 16, 41, and 205 ppm
c. CH2=CHCHOHCH3 ; 13C NMR: 23, 69, 113, and 143 ppm
Question: Identify the structures of isomers A and B (molecular formula C9H10O ).
Compound A: IR peak at 1742 cm-1 ; 1 H NMR data (ppm) at 2.15 (singlet, 3 H), 3.70 (singlet, 2 H), and 7.20 cm-1(broad singlet, 5 H). Compound B: IR peak at 1688 ; 1 H NMR data (ppm) at 1.22 (triplet, 3 H), 2.98 (quartet, 2 H), and 7.28–7.95 (multiplet, 5 H).
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