# Compound A has molecular formula C7H14O and reacts with

Chapter 20, Problem 20.76

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QUESTION:

Compound A has molecular formula C7H14O and reacts with sodium borohydride in methanol to form an alcohol. The 1 H NMR spectrum of compound A exhibits only two signals: a doublet (I = 12) and a septet (I = 2). Treating compound A with 1,2-ethanedithiol (HSCH2CH2SH) under acidic conditions, followed by Raney nickel gives compound B. (a) How many signals will appear in the 1 H NMR spectrum of compound B? (b) How many signals will appear in the 13C NMR spectrum of compound B? (c) Describe how you could use IR spectroscopy to verify the conversion of compound A to compound B.

QUESTION:

Compound A has molecular formula C7H14O and reacts with sodium borohydride in methanol to form an alcohol. The 1 H NMR spectrum of compound A exhibits only two signals: a doublet (I = 12) and a septet (I = 2). Treating compound A with 1,2-ethanedithiol (HSCH2CH2SH) under acidic conditions, followed by Raney nickel gives compound B. (a) How many signals will appear in the 1 H NMR spectrum of compound B? (b) How many signals will appear in the 13C NMR spectrum of compound B? (c) Describe how you could use IR spectroscopy to verify the conversion of compound A to compound B.

Step 1 of 5

First, note that the molecular formula of compound A indicates a hydrogen deficiency index of one. So, one ring or one double bond should be present in the molecule. The fact that compound A reacts with sodium borohydride in methanol to form an alcohol suggests that it’s an aldehyde or ketone; this carbonyl group would explain the element of unsaturation.

The presence of both a doublet and a septet (integrations of 12 and 2, respectively) is very characteristic of isopropyl groups. The two isopropyl groups must be symmetrical, as their signals appear in the same region of the spectrum. Since each isopropyl group requires three carbons and seven hydrogens, we’re left with only one carbon and one hydrogen atom in the molecule. This indicates a ketone is present, and not an aldehyde. Hence, the structure of compound A is shown as follows: