NMR spectroscopyEdexcel International A Level Chemistry: Revision notes

Nuclear magnetic resonance (NMR) spectroscopy detects atoms whose nuclei have spin, such as ¹³C and ¹H. In a strong magnetic field the nuclei absorb radio-frequency radiation, and the frequency depends on the chemical environment of the atom, meaning which atoms surround it. Atoms in the same environment give the same signal.

Positions are measured as the chemical shift, δ, in ppm, relative to tetramethylsilane (TMS), Si(CH₃)₄, at δ = 0. TMS is used because its 12 equivalent hydrogen atoms (and 4 equivalent carbon atoms) give one sharp peak at a lower δ than most other protons and carbons, it is inert and volatile so is easily removed. Samples are dissolved in solvents with no ¹H, such as CDCl₃ or CCl₄, so the solvent gives no ¹H signal.

Each different carbon environment gives one peak, so the number of peaks equals the number of different types of carbon. Equivalent carbon atoms, such as the carbons of a symmetrical group, give one peak: 2-methylpropan-2-ol, (CH₃)₃COH, has two peaks; butan-2-ol has four.

Typical chemical shifts (ppm), from the data booklet: C–C 5–40; C–Cl or C–Br 30–70; C–O 50–90; C=C and aromatic 100–150; C=O in esters and acids 160–185; C=O in aldehydes and ketones 190–220. To justify the number of peaks, state how many carbon environments there are and which carbons are equivalent.

Each different proton environment gives one peak. The chemical shift shows the type of proton, as electronegative atoms and groups pull electron density away and deshield the proton, raising δ. Typical values: R–CH₃ 0.7–1.2; CH₃–C=O 2.0–2.5; CH₂–Br 3.1–3.4; –O–CH₂– in esters 3.7–4.1; aldehyde –CHO 9.4–10.0; –COOH 10–12; aromatic H 6.4–8.5; O–H varies (about 0.5–5.5).

The peak area (integration) is proportional to the number of hydrogen atoms in that environment, so the ratio of areas equals the ratio of hydrogen atoms: ethanol, CH₃CH₂OH, gives areas 3 : 2 : 1.

Peaks split into multiplets because of non-equivalent protons on adjacent carbon atoms. A proton with n hydrogens on the adjacent carbon(s) is split into n + 1 peaks: singlet (n = 0), doublet (1), triplet (2), quartet (3), quintet (4), sextet (5), septet (6).

Example, 1-bromopropane, CH₃CH₂CH₂Br: CH₃ is a triplet (neighbour CH₂), the middle CH₂ is a sextet (3 + 2 = 5 neighbours), CH₂Br is a triplet. An ethyl group, –CH₂CH₃, shows a quartet and a triplet. Protons on the same carbon, or in the same environment, do not split each other. The O–H proton is usually a singlet.

Combine the evidence: the number of ¹³C peaks gives the number of carbon environments; the chemical shifts give the types of carbon and proton; the area ratio gives the numbers of hydrogens; the splitting shows neighbouring groups.

Example: C₄H₈O₂ with peaks δ 3.7 (singlet, 3H), δ 2.3 (quartet, 2H), δ 1.1 (triplet, 3H). The singlet at 3.7 is a CH₃ on oxygen (CH₃O–) with no neighbours. The quartet and triplet are CH₃CH₂–, with the CH₂ next to C=O (δ 2.3). The compound is methyl propanoate, CH₃CH₂COOCH₃. Always check the structure predicts the same number of peaks, areas and splittings as the data.