Topic 20 · page 5 of 15

NMR splitting and solving structures

At high resolution each proton peak splits into n + 1 lines, where n is the number of hydrogens on the adjacent carbon atoms. Combine this with every other piece of evidence, and check your answer by predicting its spectrum.

15 min read20.1

Read first: NMR: environments, shifts and areas

The n + 1 rule 20.1

At high resolution a proton peak splits into a group of lines. The hydrogens on neighbouring carbon atoms change the magnetic field slightly, and each possible combination gives a line.

Hydrogens on adjacent carbons, n 0 1 2 3 6
Lines, n + 1 1 2 3 4 7
Name singlet doublet triplet quartet septet
Relative line heights 1 1 : 1 1 : 2 : 1 1 : 3 : 3 : 1 1 : 6 : 15 : 20 : 15 : 6 : 1

Two points decide which hydrogens count:

A schematic proton NMR spectrum of propan-2-ol with a reversed axis from 5 to 0 ppm and a TMS reference at zero. A doublet of area 6 hydrogens at 1.2, a septet of area 1 hydrogen at 4.0 with seven lines, and a singlet of area 1 hydrogen at 2.2 for the OH.
Propan-2-ol at high resolution. The two CH3 groups each have one neighbouring H, so a doublet. The CH has six neighbouring H, so a septet.

Worked exampleSplitting in propan-2-ol, (CH₃)₂CHOH

  1. Two CH₃ groups (6 H, δ about 1.2): the neighbouring carbon is the CH, with 1 H, so n = 1 and the peak is a doublet.
  2. CH (1 H, δ about 4.0): the neighbouring carbons are the two CH₃ groups with 6 H in total, so n = 6 and the peak is a septet. It is high because the carbon is bonded to O.
  3. OH (1 H, δ about 2.2): a singlet.

Some patterns appear so often that they are worth knowing on sight.

Pattern in the spectrum Group
a triplet (3 H) and a quartet (2 H) an ethyl group, –CH₂CH₃
a doublet (6 H) and a septet (1 H) an isopropyl group, –CH(CH₃)₂
a singlet (3 H) at about 2.0–2.1 CH₃ next to C=O with no H on the next atom
a singlet (3 H) at about 3.7 CH₃ bonded to O, as in a methyl ester

Putting the evidence together 20.1

Every question of this kind has the same shape. The data come in pieces and each piece answers one question.

MethodSolving a structure

  1. Formula. Use the combustion or percentage data and the molecular ion to get the molecular formula.
  2. Functional groups. Use the chemical tests and the infrared spectrum. Note what is absent.
  3. Carbon skeleton. Use the number of ¹³C peaks and any peak above 160.
  4. Proton detail. Use the number of peaks, the shifts, the area ratio and the splitting.
  5. Check. Draw your structure and predict its spectra. Every peak must be accounted for.

Worked exampleCompound Y, C₄H₈O₂

Illustrative data. Compound Y has MrM_\text{r} = 88 and the formula C₄H₈O₂ (page 1). It does not react with sodium carbonate solution. The other evidence is in the table. Identify Y.

Evidence Deduction
no reaction with Na₂CO₃(aq) not a carboxylic acid
IR: strong at 1740 cm⁻¹, no broad O–H C=O present, no O–H, so likely an ester
¹³C NMR: 4 peaks, one near 171 ppm four different carbons; 171 is the C=O of an ester
¹H NMR: δ 2.0, 3 H, singlet CH₃ next to C=O with no H on the next atom: CH₃CO–
¹H NMR: δ 4.1, 2 H, quartet CH₂ bonded to O, with 3 H next door
¹H NMR: δ 1.3, 3 H, triplet CH₃ with 2 H next door

The quartet and triplet are an ethyl group on oxygen, –OCH₂CH₃, and the singlet is CH₃CO–. Y is ethyl ethanoate, CH₃COOCH₂CH₃.

Check. Methyl propanoate, CH₃CH₂COOCH₃, has the same formula, but it would give a singlet of 3 H near 3.7 (OCH₃) and a quartet of 2 H near 2.3. Y has no singlet at 3.7, so it is not methyl propanoate.

A schematic proton NMR spectrum of ethyl ethanoate with a reversed axis from 5 to 0 ppm. A singlet of area 3 hydrogens at 2.0, a quartet of area 2 hydrogens at 4.1 and a triplet of area 3 hydrogens at 1.3.
Ethyl ethanoate: singlet 3 H at 2.0, quartet 2 H at 4.1, triplet 3 H at 1.3. Schematic, not measured.

Worked exampleCompound Q, C₄H₈O

Illustrative data. Compound Q has the formula C₄H₈O. Its infrared spectrum has a strong absorption at 1715 cm⁻¹ and nothing above 3000. Its ¹³C spectrum has 4 peaks, one at δ = 209. Its ¹H spectrum has peaks at δ 2.1 (3 H, singlet), δ 2.4 (2 H, quartet) and δ 1.0 (3 H, triplet). Identify Q.

  1. 1715 and no O–H: a C=O in an aldehyde or ketone.
  2. δ = 209 is in the 190–220 range for an aldehyde or ketone C=O. There is no peak at 9.4–10.0 in the proton spectrum, so there is no –CHO: Q is a ketone.
  3. The singlet at 2.1 (3 H) is CH₃CO–. The quartet (2 H) and triplet (3 H) are an ethyl group, next to the C=O.
  4. Q is butanone, CH₃COCH₂CH₃. It has four carbon environments, which matches the four ¹³C peaks.

Check yourself

  1. Predict the splitting of each proton peak in butanone, CH₃COCH₂CH₃, and give its area.
  2. Predict the splitting of each peak in the proton spectrum of ethanal, CH₃CHO.
  3. A compound C₃H₇Cl has a proton NMR spectrum with a doublet of area 6 and a septet of area 1. Identify it.
  4. A compound C₂H₄O₂ has two singlets in the ratio 3 : 1, at δ = 2.1 and δ = 11.5. Its infrared spectrum has a very broad absorption from 3300 to 2500 cm⁻¹ and one at 1710 cm⁻¹. Identify it.
  5. Compound R has MrM_\text{r} = 88 and the formula C₄H₈O₂. Its infrared spectrum has a strong absorption at 1735 cm⁻¹ and no broad O–H. Its ¹³C spectrum has 4 peaks. Its ¹H spectrum has δ 1.1 (3 H, triplet), δ 2.3 (2 H, quartet) and δ 3.7 (3 H, singlet). Identify R and explain why it is not ethyl ethanoate.
  6. State the number of lines in the CH peak of propan-2-ol and the ratio of their heights, and explain where the number comes from.

Answers cover these until you have tried the questions

  1. CH₃ next to C=O: singlet, 3 H, no hydrogens on the C=O carbon. CH₂: quartet, 2 H, the CH₃ next door has 3 H. CH₃ of the ethyl group: triplet, 3 H, the CH₂ next door has 2 H.
  2. CH₃: doublet, 3 H, because the CHO carbon has 1 H. CHO: quartet, 1 H, because the CH₃ has 3 H.
  3. A doublet of 6 H is two equivalent CH₃ groups next to a single H, and a septet of 1 H is that CH next to six H: an isopropyl group, –CH(CH₃)₂. The compound is 2-chloropropane, (CH₃)₂CHCl.
  4. The very broad O–H and the C=O show a carboxylic acid. The singlet at 11.5 (1 H) is the acid O–H and the singlet at 2.1 (3 H) is a CH₃ with no neighbouring H. The compound is ethanoic acid, CH₃COOH.
  5. C=O and no O–H: an ester. The singlet of 3 H at 3.7 is OCH₃. The quartet (2 H) and triplet (3 H) are an ethyl group next to the C=O, CH₃CH₂CO–. R is methyl propanoate, CH₃CH₂COOCH₃. Ethyl ethanoate would show a singlet at about 2.0 and a quartet at about 4.1 (an OCH₂ group), and R has neither.
  6. 7 lines, in the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1. The CH has n = 6 hydrogens on its two adjacent CH₃ groups, so it gives n + 1 = 7 lines.