Organic

Reading an NMR spectrum: shift, integration, splitting

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A proton NMR spectrum is a compressed structural drawing of a molecule. Every hydrogen atom in the molecule contributes a peak — and the peak’s position, area, and multiplicity together tell you the atom’s electronic environment, how many equivalent atoms it has, and how many neighbors sit within three bonds. Learn to read three independent axes at once and NMR turns from an unfamiliar plot into the most information-dense analytical technique in the chemistry toolbox. This guide covers those three axes end-to-end, worked example first, then edge cases.

Assume familiarity with organic structure at the level of the SN1 vs SN2 guide.

The three axes: chemical shift, integration, splitting

Chemical shift (δ, ppm, x-axis) is the position of a peak on a scale referenced to tetramethylsilane (TMS = 0 ppm). Range: 0 to ~12 ppm for common organic molecules. Shift depends on the local electron density around the proton. Electron-rich neighborhoods (methyl group next to another alkyl) shield the proton, pushing its peak upfield (low δ, 0-2 ppm). Electron-poor neighborhoods (proton next to carbonyl, aromatic ring, or halogen) deshield the proton, pushing it downfield (high δ, 4-12 ppm). Shift categories are the top of the diagnostic pyramid — memorize the ranges and 80% of unknowns fall out on first inspection.

Integration (y-axis area) is the total signal area under a peak, proportional to the number of protons contributing to that peak. Modern spectrometers report an integration ratio directly under each peak. For a proton count of 12, a spectrum with integration ratios 3:2:6:1 tells you the four peaks come from 3, 2, 6, and 1 protons respectively — matching the count and pinpointing which peak is a methyl vs a methine.

Multiplicity (splitting, x-axis fine structure) counts adjacent hydrogens. The n+1 rule: a proton with n neighbors on adjacent carbons shows up as an (n+1)-line multiplet with predictable intensity ratios from Pascal’s triangle. A -CH₂- (2 neighbors) split by two adjacent -CH- (or -CH₂- with 2 protons each × 1… simpler if the neighbor is one H = doublet, two = triplet, three = quartet, six = septet.

The three axes work together: shift narrows down what functional environment the proton is in; integration says how many equivalent protons; splitting confirms who its neighbors are. A single peak, taken alone, is ambiguous. The combination is diagnostic.

Chemical shift ranges you must know

δ (ppm)EnvironmentExample
0.5-1.5Alkyl (CH₃, CH₂ near CH₃)ethyl, propyl
1.5-2.5Alkyl next to π-system or heteroatom-CH₂-CO-, -CH₂-Ar
2.0-3.0Alkyne (=C-H)HC≡CR
3.0-4.5-CH-O, -CH-N, -CH-X (halide)ethers, amines, alkyl halides
4.5-6.5Vinyl (=CH-)alkenes
6.5-8.5Aromaticbenzene ring 7.3
9-10Aldehyde -CHOacetaldehyde 9.8
10-13Carboxylic acid -COOH, very broadacetic acid 11.5

Values from Silverstein Ch. 3 Table 3.1. Real shifts drift ±0.3-0.5 ppm from these numbers depending on substituents; the rough band is what you memorize, and the functional groups cheatsheet has finer breakdowns.

Two subtle cases worth calling out:

  • Aromatic protons in electron-donating substituents (methoxy, amine) shift upfield to ~6.7 ppm because the ring is more electron-rich.
  • Aromatic protons ortho to electron-withdrawing substituents (nitro, carbonyl) shift downfield to ~8.2 ppm because the ring is depleted.

Substituent effects on the aromatic ring produce the ortho/meta/para pattern that lets you assign ring substitution positions from the spectrum alone — see Silverstein Ch. 3 for the shift-additivity tables.

Splitting rules and Pascal’s triangle

The n+1 rule assumes the neighbors are chemically equivalent and each contributes one three-bond coupling. Peak intensities follow Pascal’s triangle: doublet 1:1, triplet 1:2:1, quartet 1:3:3:1, septet 1:6:15:20:15:6:1. A memorable case: the ethyl group -O-CH₂-CH₃ shows a quartet (CH₂ split by 3 CH₃ H’s) and a triplet (CH₃ split by 2 CH₂ H’s) — the pattern is so distinctive that you can identify ethyl in a spectrum before reading any shift.

The rule breaks in three situations:

  1. Non-equivalent neighbors on both sides. A proton with 2 H’s on one side and 3 on the other yields a doublet-of-quartets, not a hexet.
  2. Complex second-order effects when Δν (the frequency difference between coupled peaks) is comparable to J (the coupling constant). At high-field spectrometers (400+ MHz) this is rare in undergraduate spectra.
  3. Exchange-broadened peaks for -OH, -NH, and -COOH protons. These are usually singlets and often show up as broad, poorly defined humps because the proton exchanges between molecules faster than it couples to neighbors.

Pavia Ch. 5 has a beautiful example of second-order coupling in the AB quartet pattern where two protons on adjacent carbons split each other but Δν is not much bigger than J.

Worked example — reading ethyl acetate

Worked example — 1H NMR of ethyl acetate (CH3COOCH2CH3)
Peak 1
4.12 ppm, integration 2H, quartet (J = 7 Hz)
Peak 2
2.05 ppm, integration 3H, singlet
Peak 3
1.26 ppm, integration 3H, triplet (J = 7 Hz)
Molecular formula
C4H8O2
  1. Formula
  2. Substitute
  3. Note

    Peak 3 (1.26 ppm, 3H triplet) is in the alkyl region (0.5-1.5 ppm), integrating for three protons, split into a triplet — a methyl group next to a CH2. Peak 1 (4.12 ppm, 2H quartet) is in the 'O-CH-' or 'N-CH-' region (3.0-4.5 ppm), integrating for two protons, split into a quartet by three neighbors. So peak 1 is an OCH2 next to the same methyl that made peak 3 — those two peaks form the classic ethyl-ether ethyl group. Peak 2 (2.05 ppm, 3H singlet) is in the 'alkyl next to pi-system' band (1.5-2.5 ppm), integrating for three protons with no neighbors visible — an isolated methyl group attached to a carbonyl. Piece it together: OCH2CH3 + isolated CH3-CO- fragment sums to C4H8O2 through an ester linkage. Structure: CH3COOCH2CH3, ethyl acetate. The 7 Hz coupling constant is textbook vicinal 3J(H,H) for a freely rotating ethyl group. Every ethyl ester on Earth gives this exact pattern; once you recognize it, ester identification becomes a matter of one glance.

  4. Result

Systematic identification workflow

Given a new spectrum, work in this order:

  1. Count peaks — number of distinct chemical environments in the molecule.
  2. Sum integrations — should match total H count from the molecular formula. If not, some peaks (usually OH/NH) may be missing or overlapped.
  3. Classify each peak by shift band — alkyl / heteroatom-adjacent / vinyl / aromatic / carbonyl.
  4. Count neighbors from multiplicity — use n+1 to deduce adjacent -CH- structures.
  5. Match to structural fragments — ethyl (quartet + triplet), isopropyl (septet + doublet), tert-butyl (singlet 9H), etc.
  6. Assemble fragments — each fragment is a substructure; the target molecule uses each once. Any leftover H count must come from OH/NH protons.
  7. Cross-check with degrees of unsaturation — a molecule with formula C₈H₁₀O has DBE = 4, which is consistent with a benzene ring plus an alcohol. The degrees of unsaturation calculator does this in one click.

When the spectrum lies

Three edge cases where the first-pass interpretation is wrong:

  • Broadened peaks with unusual shifts. -OH and -NH shifts move around unpredictably (0.5-6 ppm for alcohols, 0.5-4 ppm for amines) and often lack splitting. A shake with D₂O replaces exchangeable protons with deuterium and makes them disappear — the peak that vanishes on D₂O exchange is your exchangeable H.
  • Accidental degeneracy. Two chemically distinct protons can accidentally share the same shift, especially in aromatic systems. If a peak’s integration is unexpectedly high, degeneracy is a candidate.
  • Diastereotopic protons. A CH₂ group with an adjacent stereocenter has two magnetically inequivalent H atoms that appear as separate peaks with additional AB coupling between them. This is the reason a simple -CH₂- can look like a complex multiplet in a chiral molecule.

Claridge Ch. 3 has the treatment of second-order effects and accidental degeneracy.

Practice problems

Problem 1 Easy A peak appears at 7.3 ppm, integration 5H, appearing as a complex multiplet. What functional group?

Answer: Monosubstituted benzene ring — five aromatic protons at 7.3 ppm is the classic Ph-R (phenyl) signature. The multiplet is unresolved ortho/meta/para peaks that appear together in low-resolution spectra.

Problem 2 Intermediate A 1H NMR shows: 9.80 ppm (1H, singlet), 2.20 ppm (3H, singlet). Molecular formula CH2O (formaldehyde)... wait, 2 protons but formula shows 2. Actually, molecular formula C2H4O. What is the structure?

Answer: 9.80 ppm singlet 1H = aldehyde CHO. 2.20 ppm singlet 3H = methyl next to carbonyl (no neighbors, hence singlet). Formula C2H4O accounts for CH3-CHO. Structure: acetaldehyde. The aldehyde and its methyl are both isolated (no adjacent H's within 3 bonds), so both are singlets — quick identification once you see the 9.8 ppm signal.

Problem 3 Hard 1H NMR shows 4H (multiplet 7.2 ppm, 4 lines pattern), 2H (quartet, 2.6 ppm), 3H (triplet, 1.2 ppm). Formula C9H12. Structure?

Answer: 4H at 7.2 ppm = para-disubstituted benzene (the 4-line pattern is two overlapping doublets from the AB pattern of a symmetric para substitution). 2H quartet + 3H triplet at 2.6 / 1.2 ppm = ethyl group. The ethyl's CH2 at 2.6 ppm is benzyl-like (next to an aromatic ring). Formula C9H12 = C6H4 (para-substituted ring, 4H) + C2H5 (ethyl, 5H) + one more CH3? Actually 4 + 5 + 3 = 12 H total. So p-ethyltoluene: CH3-C6H4-CH2CH3. The second methyl (attached directly to the ring) would sit at 2.3 ppm as a 3H singlet — this may overlap with the CH2 quartet or appear as a fourth signal. Answer: 1-ethyl-4-methylbenzene (p-ethyltoluene).

Where to go next

  • 13C NMR complements 1H NMR by counting distinct carbon environments — Silverstein Ch. 4 for the shift ranges and DEPT/HSQC/HMBC 2D correlations.
  • Mass spectrometry cross-checks NMR by giving the molecular ion and fragmentation pattern; a combination of NMR + MS + IR is close to unambiguous for most small molecules.
  • The degrees of unsaturation calculator is a fast cross-check on any proposed structure — the DBE must match the formula.
  • The molar mass calculator helps confirm formula assignment when you have both NMR and mass-spec data.

NMR reading is a pattern-recognition skill built on top of three additive axes. Cover the shift table, drill the ethyl/isopropyl/tert-butyl/aromatic fingerprints, and 90% of undergraduate unknowns solve themselves.

Sources

  • Silverstein, Webster, Kiemle & Bryce, Spectrometric Identification of Organic Compounds, 8th ed. (2014), Ch. 3-4
  • Pavia, Lampman, Kriz & Vyvyan, Introduction to Spectroscopy, 5th ed. (2015), Ch. 3-5
  • Claridge, High-Resolution NMR Techniques in Organic Chemistry, 3rd ed. (2016), Ch. 1-3
  • Clayden, Greeves & Warren, Organic Chemistry, 2nd ed. (2012), Ch. 13 — 1H NMR
  • IUPAC Compendium of Chemical Terminology, 2nd ed. (1997) — 'chemical shift'