High-resolution ¹H NMR and the elucidation of structure: combining mass spectra and NMR to identify compounds

Overview

Combining mass spectra, including bromine's M and M+2 peaks, with ¹H NMR splitting to identify unknown compounds such as isomers of C₃H₆O₂ and C₄H₁₀O, and telling propanal from propanone.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   The mass spectrum of a compound shows two peaks of almost equal height at m/z 122 and 124. What does this tell you, and why?

   Model answer

   It tells you the molecule contains one bromine atom. Bromine has two isotopes, ⁷⁹Br and ⁸¹Br, of almost equal natural abundance, so molecules containing one bromine atom occur in two versions differing by two mass units and in almost equal numbers. The peak at 122 contains ⁷⁹Br and that at 124 contains ⁸¹Br. Had the element been chlorine the two peaks would have been in a ratio of about 3:1.

2. 2

   A compound of formula C₃H₆O₂ gives two singlets in its ¹H NMR spectrum, 3H at δ 2.0 and 3H at δ 3.7. Its infrared spectrum shows a strong absorption at 1740 cm⁻¹ and no broad absorption between 2500 and 3300 cm⁻¹. Deduce its structure, explaining your reasoning.

   Model answer

   The strong absorption at 1740 cm⁻¹ shows a C=O group, and the absence of a broad absorption around 2500 to 3300 cm⁻¹ rules out a carboxylic acid, so with two oxygen atoms the compound is an ester. Both signals are singlets, so neither methyl group has any hydrogen on an adjacent carbon; each must therefore be attached to an atom bearing no hydrogen. One methyl at δ 2.0 is attached to the carbonyl carbon; the other at δ 3.7 is shifted further downfield, so it is attached to oxygen. The compound is methyl ethanoate, CH₃COOCH₃.

3. 3

   A compound of formula C₃H₆O₂ shows peaks at δ 11.7 (1H, singlet), δ 2.4 (2H, quartet) and δ 1.1 (3H, triplet), and a very broad infrared absorption from 2500 to 3300 cm⁻¹ with a strong peak at 1710 cm⁻¹. Work out its structure.

   Model answer

   The very broad absorption from 2500 to 3300 cm⁻¹ together with the C=O peak at 1710 cm⁻¹ is characteristic of a carboxylic acid, and the 1H singlet at δ 11.7, far downfield, is the acidic −COOH hydrogen. The remaining signals are a classic ethyl group: a 3H triplet has two neighbours and a 2H quartet has three, so CH₃ and CH₂ are joined to one another. The CH₂ is at δ 2.4, next to the carbonyl. Putting an ethyl group on a carboxyl group gives CH₃CH₂COOH, propanoic acid, which fits C₃H₆O₂.

4. 4

   A compound of formula C₃H₆O₂ shows peaks at δ 8.1 (1H, singlet), δ 4.2 (2H, quartet) and δ 1.3 (3H, triplet), with a strong infrared absorption at 1725 cm⁻¹ and no broad absorption near 3000 cm⁻¹. Identify it.

   Model answer

   The 2H quartet and 3H triplet again show an ethyl group. The absence of the very broad acid absorption rules out a carboxylic acid, so with two oxygens and a C=O it is an ester. The CH₂ has moved downfield to δ 4.2, much further than the δ 2.4 of an acid, so it is attached to the single-bonded oxygen of the ester rather than to the carbonyl carbon; that makes the ethyl group the alcohol part. The remaining 1H singlet at δ 8.1 is then the hydrogen attached directly to the carbonyl carbon. The compound is ethyl methanoate, HCOOCH₂CH₃. Note that propanoic acid and methyl ethanoate share this molecular formula; they are told apart from ethyl methanoate by the presence of the very broad acid absorption in the first case, and by the CH₂ signal lying near δ 2.4 rather than δ 4.2 in the second.

5. 5

   Predict the high-resolution ¹H NMR spectrum of ethyl ethanoate, CH₃COOCH₂CH₃.

   Model answer

   There are three environments, integrating 3:2:3. The CH₃ attached to the carbonyl carbon gives a singlet at about δ 2.0, because the carbonyl carbon carries no hydrogen. The OCH₂ gives a quartet at about δ 4.1, split by the three hydrogens of the neighbouring methyl and moved downfield by the oxygen. The terminal CH₃ gives a triplet at about δ 1.3, split by the two hydrogens of the CH₂.

6. 6

   Propanal and propanone both have the molecular formula C₃H₆O. Explain how ¹H NMR would distinguish them.

   Model answer

   Propanone, CH₃COCH₃, is symmetrical, so all six hydrogens are equivalent and the spectrum is a single unsplit peak at about δ 2.1 integrating for 6H. Propanal, CH₃CH₂CHO, has three environments in the ratio 1:2:3. The −CHO hydrogen appears far downfield at about δ 9.7 as a triplet, split by the two hydrogens of the CH₂. The CH₃ at about δ 1.1 is a triplet, split by the CH₂. The CH₂ at about δ 2.4 is a multiplet with more than four lines, because it has two different sets of neighbours, the three hydrogens of the methyl and the one hydrogen of the CHO.

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