A Level Chemistry: NMR Spectroscopy — Revision Notes

Condensed for the final weeks. For the full explanation, use the NMR Spectroscopy study guide.

The principle

Nuclei with an odd mass number — ¹H and ¹³C — behave as tiny magnets. In a strong external field they align with or against it, and absorb radio-frequency radiation when flipping between the two states. The absorption frequency depends on the electronic environment of the nucleus.

TMS (tetramethylsilane) is the reference standard, defined as δ = 0. It is used because it is inert, volatile (easily removed), non-toxic, and gives a single sharp peak well away from most others — all twelve of its protons are equivalent.

Carbon-13 NMR

Number of peaks = number of chemically different carbon environments.

This is the simplest question type on the topic: count the distinct carbon environments, allowing for symmetry. Ethanol (CH₃CH₂OH) gives two peaks; benzene gives one, because all six carbons are equivalent.

There is no splitting in ¹³C NMR at this level, and peak heights are not quantitative.

Worked example: carbon-13 peak count

Predict the number of peaks for propan-1-ol, CH₃CH₂CH₂OH, and propan-2-ol, (CH₃)₂CHOH.

Propan-1-ol has three carbons, each in a genuinely different environment (terminal CH₃, central CH₂, CH₂–OH) — 3 peaks. Propan-2-ol also has three carbons, but its two CH₃ groups are equivalent by the molecule’s own symmetry, both attached identically to the central CHOH carbon — so only 2 distinct environments. Same molecular formula, different peak count: exactly why carbon-13 NMR distinguishes structural isomers other techniques might miss.

Proton NMR — three pieces of information

1. Chemical shift (δ) — identifies the environment:

The table below gives the thirteen rows of the Typical proton (¹H) NMR chemical shift values (δ) relative to TMS = 0 table printed in the Data section of the 9701 syllabus (2025–2027), which is provided in the examination — so it need not be memorised, only read correctly:

Environment of proton	Example	δ / ppm
alkane	–CH₃, –CH₂–, >CH–	0.9–1.7
alkyl next to C=O	CH₃–C=O, –CH₂–C=O, >CH–C=O	2.2–3.0
alkyl next to aromatic ring	CH₃–Ar, –CH₂–Ar, >CH–Ar	2.3–3.0
alkyl next to electronegative atom	CH₃–O, –CH₂–O, –CH₂–Cl	3.2–4.0
attached to alkene	=CHR	4.5–6.0
attached to aromatic ring	H–Ar	6.0–9.0
aldehyde	HCOR	9.3–10.5
alcohol	ROH	0.5–6.0
phenol	Ar–OH	4.5–7.0
carboxylic acid	RCOOH	9.0–13.0
alkyl amine	R–NH–	1.0–5.0
aryl amine	Ar–NH₂	3.0–6.0
amide	RCONHR	5.0–12.0

Note (as printed): δ values for O–H and N–H protons can vary depending on solvent and concentration. Ar is used to represent an aromatic ring.

O–H and N–H shifts are highly variable, depending on concentration, solvent and hydrogen bonding — this is exactly why they are excluded from the n+1 splitting rule (see below) and identified instead by the D₂O shake.

2. Integration — the relative area under each peak gives the ratio of hydrogens in each environment. Note it is a ratio, not an absolute count, so a 3:2:1 integration could be 3:2:1 or 6:4:2.

3. Splitting — the n+1 rule — a peak is split into (n+1) lines, where n is the number of hydrogens on adjacent carbons.

Neighbours	Pattern
0	Singlet
1	Doublet
2	Triplet
3	Quartet
4 or more (or a combination of non-equivalent neighbouring sets)	Multiplet

Splitting tells you about the neighbours, not the peak’s own environment. That is the conceptual point students most often invert. A CH₃ next to a CH₂ appears as a triplet — because the CH₂ has two hydrogens.

Ethanol is the classic worked example: CH₃ (triplet, 3H), CH₂ (quartet, 2H), OH (singlet, 1H).

Deducing structure from a spectrum combines all three pieces together: the number of peaks gives the number of distinct proton environments, chemical shift identifies what kind of environment each is, relative area gives how many protons are in each, and the splitting pattern reveals how many protons sit on each neighbouring carbon — usually enough to piece together the full structure, especially alongside the molecular formula from mass spectrometry.

OH and NH protons

These appear as broad singlets at variable shift, and they do not split or get split, because they exchange rapidly with the solvent.

The D₂O shake confirms them: adding D₂O replaces the OH/NH hydrogen with deuterium, which is not detected, so the peak disappears. That is the standard identification test.

Common mistakes to avoid

Including an OH or NH proton in a neighbouring group’s splitting count. Because these protons usually exchange rapidly, they don’t show the expected coupling to neighbouring C–H protons, and are excluded when applying the n+1 rule to an adjacent CH₂ or CH₃.

Assuming the D₂O shake changes the carbon skeleton. The exchange R–OH + D₂O ⇌ R–OD + HOD is a real, reversible reaction, but it swaps only the labile O–H/N–H hydrogen for deuterium; every C–H bond and the carbon connectivity are unchanged, which is why only the O–H/N–H peak disappears.

Solvents

Use deuterated solvents such as CDCl₃, because ordinary solvents would contribute their own large proton peaks and obscure the sample’s.

Exam traps

• Counting all carbons rather than distinct carbon environments.
• Applying the n+1 rule to the hydrogens on the same carbon.
• Treating integration as an absolute number of hydrogens.
• Expecting OH to split or be split.
• Forgetting why a deuterated solvent is required.
• Using splitting in ¹³C NMR.

Self-test

1. Why is TMS used as the reference standard?
2. How many peaks does ethanol give in ¹³C NMR, and why?
3. State the n+1 rule and say what n counts.
4. What does integration actually tell you?
5. What is the D₂O shake for, and what happens?
6. Predict the number of carbon-13 peaks for propan-2-ol, and explain why it differs from propan-1-ol.

Answers: 1. It is inert, volatile, non-toxic and gives a single sharp peak away from most others, because all twelve of its protons are equivalent. 2. Two — there are two chemically different carbon environments, CH₃ and CH₂. 3. A signal is split into n+1 lines, where n is the number of hydrogen atoms on the adjacent carbon atoms, not on the same carbon. 4. The ratio of the numbers of hydrogen atoms in each environment — not the absolute number. 5. It identifies OH and NH protons: adding D₂O exchanges the labile hydrogen for deuterium, which is not detected, so that peak disappears. 6. Two peaks: the two CH₃ groups are equivalent by the molecule’s symmetry, both attached identically to the central CHOH carbon, so they give a single combined peak — unlike propan-1-ol, where all three carbons occupy genuinely different environments and give three separate peaks.