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Spectroscopy in Structure Determination Questions and Answers

31.

The number of signals exhibited by fluoroacetone in 1H-NMR spectrum is :

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

Fluoroacetone (FCH2COCH3) shows two doublets.

It has two kinds of protons. The CH2F protons are split by the fluorine (19F, I = 1/2) on the same carbon, a large geminal coupling of about 47 Hz, giving a doublet. The CH3 protons are four bonds from the fluorine, through the carbonyl group, and show a small long-range H-F coupling of a few hertz, so they too appear as a doublet.

The two sets of protons are separated by the carbonyl carbon and do not couple with each other, so no triplet arises. Treating the methyl group as a singlet ignores the four-bond H-F coupling that the key counts.

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32.

The correct order for stretching frequency of carbonyl group of the following molecules is :

I. CH3-CO-O-CH=CH2
II. CH3-CO-CH2CH3
III. CH2=CH-CO-OCH3

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The order is I > III > II: vinyl acetate > methyl acrylate > butanone.

A ketone such as butanone (II) absorbs near 1715 cm-1. In an ester the alkoxy oxygen withdraws electrons inductively, which strengthens the C=O bond, so saturated esters absorb near 1740 cm-1.

Vinyl acetate (I) is an enol ester: the oxygen's lone pair is shared with the C=C bond, so it donates less to the carbonyl, and the C=O frequency rises to about 1770 cm-1. Methyl acrylate (III) is an alpha,beta-unsaturated ester: conjugation of C=C with C=O lowers its frequency to about 1725 cm-1, still above the ketone. So I > III > II, and the orders placing the ketone first or the acrylate above the enol ester are wrong.

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33.

The 19F-NMR spectrum of CF2Br - CBr2CN at - 98°C shows :

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The spectrum shows two doublets and a singlet.

At -98°C rotation about the C-C bond is slow on the NMR time scale, so each staggered rotamer gives its own spectrum. Looking along the C-C bond, the CF2Br carbon carries F, F and Br, and the CBr2CN carbon carries Br, Br and CN.

In the rotamer where the Br of CF2Br is anti to the CN group, a mirror plane makes the two fluorines equivalent, giving a singlet. In the other two rotamers that Br is anti to one of the bromines; these two are mirror images with identical spectra, and in them the two fluorines are non-equivalent and couple with each other, giving an AB pattern of two doublets. Two singlets or two doublets alone would leave out one kind of rotamer.

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