Interpreting ¹H NMR Spectra

Introduction

Each feature of a ¹H NMR spectrum answers a different question. The number of signals tells you how many proton environments there are; the chemical shift tells you what each set of protons is near; the integration tells you how many protons are in each set; and the splitting tells you how many protons are on the neighbouring carbons. Interpreting a spectrum means combining all four clues, with the molecular formula, into one structure that explains every peak.

Core explanation

Step one: organise the data. Before drawing anything, build a signal table. For each peak record δ, the relative integration converted to numbers of hydrogens, the multiplicity (s, d, t, q, septet, m) and any coupling constant. Ignore peaks from the solvent and water.

Step two: use the molecular formula. Calculate the double bond equivalent (DBE):

DBE = (2C + 2 + N − H − X) ÷ 2

where C, H, N and X are the numbers of carbon, hydrogen, nitrogen and halogen atoms. A DBE of 1 suggests one C=O or C=C or a ring; a DBE of 4 or more with signals near δ 7 strongly suggests a benzene ring.

Step three: interpret each clue.

Clue What it tells you Example --- --- --- δ 0.9–1.5 alkyl CH next to other alkyl carbons CH₃ of an ethyl chain δ 2.0–2.7 CH next to C=O or an aromatic ring CH₃CO δ 3.3–4.3 CH on oxygen or halogen OCH₂, CH₂Cl δ 4.5–6.5 alkene CH CH=CH₂ δ 6.5–8.0 aromatic CH C₆H₅ δ 9–10 aldehyde CHO RCHO δ 10–13 (broad) carboxylic acid OH RCOOH integration number of H in each set 3H = CH₃ multiplicity number of H on neighbouring carbons quartet = next to CH₃

Step four: build fragments. Link coupled signals into groups: a 2H quartet and 3H triplet make CH₂CH₃; a 1H septet and 6H doublet make CH(CH₃)₂; singlets are isolated groups with no H on neighbouring carbons.

Step five: assemble and check. Subtract the atoms in the fragments from the molecular formula to find what is left — often C=O, O, N or Cl. Join the fragments so that the shifts make sense: a CH₂ at δ 4.1 must be attached to oxygen, a CH₃ singlet at δ 2.1 to a carbonyl. Finally, predict the complete spectrum of your structure and compare it with the data. If any signal is unexplained, revise.

Using other evidence. IR (for C=O or O–H), ¹³C NMR (for carbon count and carbonyl type) and a D₂O shake (for exchangeable protons) help confirm the answer. A structure is only secure when it explains all the data.

Formulae

DBE = (2C + 2 + N − H − X) ÷ 2, where oxygen atoms are ignored.

Step-by-step reasoning

A reliable routine for any ¹H spectrum:

1. Calculate the DBE from the molecular formula. 2. Tabulate δ, number of H and multiplicity for every signal. 3. Identify fragments from splitting patterns and integrations. 4. Use δ values to decide what each fragment is attached to. 5. Assemble the fragments with the leftover atoms and check the whole predicted spectrum.

Visual explanation

In the simulation, choose "Solve a spectrum" and load an unknown C₄H₈O₂. As each signal is clicked, it is added to a signal table and a matching fragment tile appears. Dragging the tiles together shows candidate structures, and the simulated spectrum of each candidate is overlaid on the real one until every peak matches.

Real-world analogy

Interpreting a spectrum is like solving a jigsaw with a picture on the box. The molecular formula is the box picture, each signal is a piece, splitting tells you which pieces interlock, and the chemical shift tells you whether a piece belongs at the edge or in the middle. The puzzle is finished only when no piece is left over.

Real-world example

Synthetic chemists run a ¹H NMR spectrum after almost every reaction. By checking that the expected new signals have appeared — for instance an OCH₂ quartet near δ 4.1 after making an ethyl ester — and that starting-material signals have gone, they confirm the product's identity within minutes.

Why?

Why is it essential to check the proposed structure against every signal? Several structures may fit two or three of the clues. For example, ethyl ethanoate and methyl propanoate both contain a CH₃, a CH₂ and a CH₃ environment with a quartet and triplet. Only the shifts — whether the CH₂ is on oxygen (δ 4.1) or next to C=O (δ 2.3) — decide between them.

Common misconception

"Splitting tells you how many hydrogens are in the group itself." A triplet does not mean a CH₃; it means the group has two neighbouring hydrogens. The integration, not the multiplicity, tells you how many hydrogens are in the group.

Worked example

Question: A compound C₄H₈O₂ gives δ 1.15 (t, 3H), 2.32 (q, 2H) and 3.67 (s, 3H). Identify it.

Reasoning: DBE = (8 + 2 − 8) ÷ 2 = 1, so one C=O. The triplet and quartet form an ethyl group; the CH₂ at δ 2.32 is next to C=O, not oxygen. The 3H singlet at δ 3.67 is an OCH₃. The remaining atoms, C, O and O, form an ester group joining them.

Answer: Methyl propanoate, CH₃CH₂COOCH₃.

Quick check

1. A 2H quartet at δ 4.1 and a 3H triplet at δ 1.3 appear together. What fragment do they show, and what is it attached to? Answer: An ethyl group, CH₃CH₂–, whose CH₂ is bonded directly to an oxygen atom.

Exam focus

Structure questions carry marks for each deduction, so write them down: DBE, each fragment with its evidence (δ, integration, multiplicity), and the final structure. A correct structure with no reasoning earns few marks; clear reasoning earns credit even if the final structure has a small error.

Advanced insight

When one-dimensional spectra are too crowded, chemists use two-dimensional techniques. COSY plots ¹H against ¹H and shows cross-peaks between coupled protons, mapping out whole chains of neighbours. HSQC correlates each proton with the carbon it is attached to. Together they allow complex molecules such as natural products and drugs to be assembled bond by bond.

Summary

To interpret a ¹H NMR spectrum, calculate the DBE, tabulate δ, integration and multiplicity for each signal, identify fragments from splitting and integration, place them using chemical shift, and assemble them with any leftover atoms. Always check that the proposed structure accounts for every signal, supported where possible by IR, ¹³C NMR or a D₂O shake.

Practice questions

1. A compound C₃H₆O shows a single ¹H NMR signal at δ 2.1. Identify it. Answer: Propanone, CH₃COCH₃: DBE = 1 for the C=O, and the six equivalent methyl protons next to C=O give one singlet near δ 2.1. 2. What does a 1H singlet at δ 9.7 suggest? Answer: An aldehyde CHO proton with no hydrogens on the adjacent carbon. 3. A compound C₈H₁₀ shows δ 7.2 (m, 5H), 2.6 (q, 2H) and 1.2 (t, 3H). Identify it. Answer: Ethylbenzene, C₆H₅CH₂CH₃: DBE = 4 and a 5H aromatic multiplet show a monosubstituted ring, and the quartet and triplet show an ethyl group attached to it. 4. Calculate the DBE of C₄H₉Cl and state what it tells you. Answer: DBE = (8 + 2 − 9 − 1) ÷ 2 = 0, so the molecule has no rings or π bonds. 5. Why must solvent peaks be identified before interpreting a spectrum? Answer: Residual solvent or water peaks do not belong to the compound, and assigning them would give a wrong structure or wrong integration ratios.