Topic 15 · page 25 of 28

Predicting and deducing from proton NMR

To predict a spectrum, find each hydrogen environment and give its shift, area and splitting. To deduce a structure, reverse the process and build pieces that fit every peak, then check them against the formula.

16 min read15.20(iv)15.20(iii)

Read first: Splitting and the (n + 1) rule

Predicting a spectrum 15.20(iv)

A prediction needs three facts for each environment: the chemical shift (from the data table), the relative area (the number of hydrogens) and the splitting (from the neighbouring hydrogens).

MethodPredicting a proton NMR spectrum

  1. Draw the displayed formula and mark the equivalent hydrogens (same environment).
  2. For each environment, write the number of H, which is the area.
  3. Use the shift table to give a δ range from what the hydrogens are next to.
  4. Count the hydrogens on the adjacent carbon(s) and apply the (n + 1) rule.

Worked exampleMethyl propanoate, CH₃CH₂COOCH₃

Predict the spectrum.

Hydrogens Number of H Next to δ / ppm Neighbouring H Pattern
CH₃ of the ethyl group 3 CH₂ 0.7–1.2, about 1.1 2 triplet
CH₂ next to C=O 2 C=O and CH₃ 2.0–2.7, about 2.3 3 quartet
CH₃ bonded to O 3 O only 3.3–4.3, about 3.7 0 singlet

The ratio of areas is 3 : 2 : 3.

Compare this with the isomer ethyl ethanoate, which has the same molecular formula C₄H₈O₂ and the same ratio 3 : 2 : 3, but a different pattern.

Two schematic proton NMR spectra of isomers with formula C4H8O2 on a reversed axis from 5 to 0 ppm. Spectrum X has a triplet at 1.1 with 3 H, a quartet at 2.3 with 2 H and a singlet at 3.7 with 3 H. Spectrum Y has a triplet at 1.2 with 3 H, a singlet at 2.0 with 3 H and a quartet at 4.1 with 2 H.
Same formula, same ratio of areas, different shifts and splitting. X is methyl propanoate and Y is ethyl ethanoate.
Methyl propanoate (X) Ethyl ethanoate (Y)
singlet 3 H at about 3.7 (OCH₃) 3 H at about 2.0 (CH₃CO)
quartet 2 H at about 2.3 (CH₂ next to C=O) 2 H at about 4.1 (OCH₂)
triplet 3 H at about 1.1 3 H at about 1.2

The quartet is the clue. In X the CH₂ is next to the C=O (δ about 2.3). In Y the CH₂ is next to the O (δ about 4.1).

Deducing a structure 15.20(iii)

Work through the spectrum one peak at a time, and write down what each peak tells you.

Worked exampleDeduce the structure

A compound has the molecular formula C₃H₆O₂. Its proton NMR spectrum has three peaks:

δ / ppm Area Splitting
about 1.1 3 triplet
about 2.3 2 quartet
about 11 1 singlet
  1. The area ratio 3 : 2 : 1 totals 6 hydrogens, matching C₃H₆O₂.
  2. The peak at δ about 11 is in the 10–12 range for a carboxylic acid, –COOH (1 H, singlet).
  3. The triplet (3 H) and quartet (2 H) together are an ethyl group, CH₃CH₂–. The CH₂ at δ about 2.3 is next to the C=O.
  4. Fit the pieces together: CH₃CH₂– plus –COOH.
  5. The compound is propanoic acid, CH₃CH₂COOH.

Always check the answer by predicting its spectrum again: propanoic acid gives a triplet (3 H), a quartet (2 H) and a singlet (1 H), which matches.

Check yourself

  1. Predict the shift, area and splitting of each peak in the proton NMR spectrum of butanone, CH₃COCH₂CH₃.
  2. Predict the spectrum of ethanol, CH₃CH₂OH, giving δ ranges, areas and splitting.
  3. A compound C₃H₆O₂ has two singlets, each of area 3, at δ = 2.0 and δ = 3.7. Deduce its structure.
  4. How does the proton NMR of ethyl ethanoate differ from that of methyl ethanoate?
  5. An aldehyde has a proton NMR spectrum with a doublet (3 H) at δ about 2.2 and a quartet (1 H) at δ about 9.8. Name it.

Answers cover these until you have tried the questions

  1. CH₃CO–: singlet, 3 H, δ about 2.1 (next to C=O, 2.0–2.7). –CH₂–: quartet, 2 H, δ about 2.4 (next to C=O, 2.0–2.7; 3 neighbouring H). –CH₃ of the ethyl group: triplet, 3 H, δ about 1.0 (0.7–1.2; 2 neighbouring H). Area ratio 3 : 2 : 3.
  2. CH₃: triplet (next to CH₂), 3 H, δ about 1.2. CH₂: quartet (next to CH₃), 2 H, δ about 3.7 (bonded to O, 3.3–4.3). O–H: singlet, 1 H, shift variable. Ratio 3 : 2 : 1.
  3. Two singlets mean no splitting: no hydrogens on adjacent carbons. δ = 2.0 is CH₃ next to C=O and δ = 3.7 is CH₃ bonded to O. The structure is methyl ethanoate, CH₃COOCH₃.
  4. Ethyl ethanoate shows three peaks: a singlet (3 H), a quartet (2 H) and a triplet (3 H), because of the ethyl group. Methyl ethanoate shows two singlets, each of area 3 (CH₃CO– and CH₃O–).
  5. A doublet for 3 H shows a CH₃ next to one hydrogen, and the quartet for 1 H at 9.8 is an aldehyde hydrogen next to a CH₃. The compound is ethanal, CH₃CHO.