High-resolution ¹H NMR and the elucidation of structure: combining NMR with mass spectrometry

Overview

Identifying unknown compounds by combining molecular ions from mass spectrometry, infrared absorptions and ¹H NMR, aromatic proton shifts, coupling constants, and a systematic order for interpreting spectra.

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

   A different liquid, also C₄H₁₀O, gives only two peaks, a 4H quartet at δ 3.4 and a 6H triplet at δ 1.2, and nothing disappears on shaking with D₂O. Its infrared spectrum shows no absorption above 3000 cm⁻¹ other than C−H. Identify it.

   Model answer

   No exchangeable hydrogen and no broad O−H absorption mean the oxygen is not in an −OH group, so with the formula C₄H₁₀O it must be an ether. The 4H quartet and 6H triplet mean two identical ethyl groups, since each CH₂ is split by three hydrogens and each CH₃ by two. The CH₂ signals are downfield at δ 3.4, so both CH₂ groups are attached to the oxygen. The compound is ethoxyethane, CH₃CH₂OCH₂CH₃.

2. 2

   Why do hydrogen atoms attached to a benzene ring absorb at around δ 7?

   Model answer

   The delocalised π electrons of the ring circulate in the applied magnetic field and set up a small field of their own. Outside the ring, where the hydrogen atoms sit, this induced field reinforces the applied field, so those hydrogens feel a stronger field than they otherwise would. They are said to be deshielded, and they resonate well downfield, typically between δ 6.5 and δ 8.

3. 3

   An unknown compound has a molecular ion at m/z 120, with strong fragment peaks at m/z 105, 77 and 43. Its infrared spectrum has a strong absorption at 1685 cm⁻¹ and no broad O−H. Its ¹H NMR spectrum shows 5H between δ 7.4 and δ 8.0 and a 3H singlet at δ 2.6. Deduce the structure.

   Model answer

   The 5H signal near δ 7.5 and the fragment at m/z 77 both indicate a monosubstituted benzene ring, C₆H₅. The absorption at 1685 cm⁻¹ shows a C=O conjugated with the ring, and the absence of a broad O−H rules out an acid. The 3H singlet at δ 2.6 is a methyl group attached to something bearing no hydrogen, and its shift suits a methyl next to a carbonyl. The fragment at m/z 43 is CH₃CO⁺ and that at 105 is the molecular ion less 15, that is loss of CH₃ to give C₆H₅CO⁺. Assembling C₆H₅, CO and CH₃ gives C₈H₈O with Mr 120, matching the molecular ion. The compound is phenylethanone, C₆H₅COCH₃.

4. 4

   An unknown liquid has a molecular ion at m/z 60. Its infrared spectrum shows a very broad absorption from 2500 to 3300 cm⁻¹ and a strong peak at 1710 cm⁻¹. Its ¹H NMR spectrum has a 1H singlet at δ 11.4 and a 3H singlet at δ 2.1. Work out its structure.

   Model answer

   The very broad absorption together with the strong C=O peak is the signature of a carboxylic acid, and the 1H singlet far downfield at δ 11.4 is the −COOH hydrogen. The 3H singlet is a methyl group with no hydrogens on the next carbon, at a shift consistent with a methyl next to a carbonyl. A methyl group joined to a carboxyl group gives CH₃COOH, whose relative molecular mass is 12.0 + 3.0 + 12.0 + 32.0 + 1.0 = 60.0, matching the molecular ion. The compound is ethanoic acid.

5. 5

   An unknown compound has a molecular ion at m/z 88 with fragments at m/z 43 and 29. Its ¹H NMR spectrum shows a 2H quartet at δ 4.1, a 3H singlet at δ 2.0 and a 3H triplet at δ 1.25, and its infrared spectrum has a strong band at 1740 cm⁻¹ and no broad O−H. Identify it.

   Model answer

   A strong band at 1740 cm⁻¹ with no broad O−H indicates an ester. The 3H triplet and 2H quartet form an ethyl group, and since the CH₂ is downfield at δ 4.1 the ethyl is attached to the single-bonded oxygen. The 3H singlet at δ 2.0 is a methyl attached to the carbonyl carbon. That gives CH₃COOCH₂CH₃, of formula C₄H₈O₂ and relative molecular mass 88.0, in agreement with the molecular ion. The fragment at 43 is CH₃CO⁺ and that at 29 is C₂H₅⁺, both of which fit. The compound is ethyl ethanoate.

6. 6

   Explain how mass spectrometry, infrared spectroscopy and ¹H NMR each contribute to working out the structure of an unknown organic compound.

   Model answer

   Mass spectrometry gives the relative molecular mass from the molecular ion peak, and the fragment peaks suggest what pieces the molecule breaks into, while the pattern of peaks two mass units apart reveals chlorine or bromine. Infrared spectroscopy identifies which functional groups are present, since bonds such as O−H, N−H, C=O and C≡N absorb in characteristic wavenumber ranges. High-resolution ¹H NMR shows how many hydrogen environments there are, how many hydrogens are in each, what kind of group each is attached to, and, from the splitting, which groups are next to which. Together they establish the mass, the functional groups and the way the carbon skeleton is put together.

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