HookThe scanner in every hospital dropped one word out of fear

When doctors first started putting patients inside the machine that images soft tissue, it was called nuclear magnetic resonance imaging — NMRI. Then the marketing departments got nervous. In the late 1970s the word nuclear still meant fallout and reactors to most people, and no one wanted to be told they were going into a nuclear scanner. So the industry quietly deleted the frightening word and rebranded the technique as MRI, magnetic resonance imaging. The physics did not change at all. The scanner and the spectrometer sitting in a chemistry lab are the same idea: put certain atomic nuclei in a very strong magnetic field, hit them with radio waves, and read what comes back.

That is because a nucleus with an odd arrangement of protons — a hydrogen-1 or a carbon-13 nucleus — behaves like a tiny bar magnet. In a strong external field it can line up with the field or against it, and radio-frequency radiation of exactly the right energy makes it flip between the two. The energy needed depends, very slightly, on the nucleus's chemical surroundings — on which other atoms are nearby and how they pull electrons around. NMR reads those tiny differences and turns them into a map of a molecule's skeleton. For A-level chemistry it is the most powerful structure-solving tool you meet: it tells you how many kinds of carbon or hydrogen a molecule has, how many of each, and what sits next door to what.

ModelHow NMR reads a molecule — and why TMS is the zero

Every peak in an NMR spectrum is measured against a single reference compound, tetramethylsilane — TMS, \(\text{Si(CH}_3)_4\) — whose peak is fixed at exactly zero on the scale. The position of any other peak, its chemical shift, is quoted in parts per million (ppm) and given the symbol \(\delta\), measured relative to that TMS zero.

TMS was not chosen at random; AQA expects you to justify it. All twelve of its hydrogen atoms are in one identical environment, and all four of its carbons are identical, so it produces a single sharp peak rather than a confusing cluster. That peak sits at one end of the scale, clear of almost every signal a real sample gives, so it never overlaps the peaks you are trying to read. TMS is chemically inert, so it does not react with the sample; it is non-toxic; and it is volatile (it boils at around 27 °C), so once the spectrum is run it evaporates away easily and the sample is recovered.

The reason chemical shift is useful is that a nucleus's environment changes the field it actually feels. Electrons around a nucleus shield it slightly from the external field; electron-withdrawing neighbours such as oxygen or a carbonyl group pull that shielding away — deshielding the nucleus — so it resonates at a higher chemical shift. You do not have to memorise the exact numbers: the exam gives you a data sheet of shift ranges, and your job is to match the peaks you see to the environments in the molecule.

ModelCarbon-13 NMR — counting the kinds of carbon

The simplest NMR spectrum to read is carbon-13 NMR, and its rule is beautifully direct: the number of peaks equals the number of different carbon environments in the molecule. Carbons that are equivalent by symmetry share one peak; carbons in genuinely different surroundings each give their own. There is no splitting to worry about in the carbon-13 spectra AQA uses, so every environment is simply one peak, and the chemical shift of each peak (read off the data sheet) tells you what kind of carbon it is — a carbonyl carbon sits far downfield near 170–200 ppm, a carbon attached to oxygen around 50–70 ppm, a plain alkyl carbon below about 50 ppm.

The skill the exam tests is spotting equivalent carbons. Symmetry collapses several carbons into one peak, so counting peaks is really counting distinct environments, not counting carbon atoms. This is why propanone, with its two identical methyl groups, gives only two peaks rather than three, and why benzene — six identical carbons — gives just one.

Worked example

Predict the number of peaks in the carbon-13 NMR spectrum of (a) propanone, \(\text{CH}_3\text{COCH}_3\), and (b) ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

(a) Propanone has three carbon atoms, but the two methyl groups are identical — the molecule is symmetrical, with a mirror plane through the C=O. So there are only two carbon environments: the two equivalent methyl carbons (one peak) and the carbonyl carbon (one peak). Answer: 2 peaks.

(b) Ethyl ethanoate has four carbons, and this time none of them is equivalent to another: the carbonyl carbon \(\text{C=O}\), the methyl attached to it (\(\text{CH}_3\text{CO}\)), the \(\text{OCH}_2\) carbon and the terminal \(\text{CH}_3\) of the ethyl group are all in different surroundings. Answer: 4 peaks. The lesson: never count carbon atoms — count environments, and always check for symmetry first.

MechanismProton NMR — shift, integration and the n+1 rule

Proton (hydrogen-1) NMR carries three separate pieces of information, and reading all three is what turns a spectrum into a structure. First, the number of groups of peaks tells you how many different hydrogen environments there are, exactly as with carbon. Second, the integration trace — the relative area under each set of peaks — gives the ratio of the numbers of hydrogens in each environment; a 3:2:1 integration means three environments containing hydrogens in that proportion.

Third, and uniquely, proton peaks are split. A hydrogen's peak is split by the hydrogens on the neighbouring carbon atoms, following the n+1 rule: if a hydrogen has \(n\) equivalent hydrogens on the adjacent carbon(s), its peak is split into \(n+1\) lines. No neighbours gives a single line (a singlet); one neighbour gives a doublet; two give a triplet; three give a quartet. This splitting is the tool that tells you what sits next to what — a triplet next to a quartet is the fingerprint of an \(-\text{CH}_2\text{CH}_3\) ethyl group.

So a full proton-NMR reading works down a checklist: how many environments (number of peak groups), how many hydrogens in each (integration ratio), what type each is (chemical shift from the data sheet), and what each is next to (splitting via n+1). Put those four together and the structure usually falls out.

Worked example

Interpret the proton NMR of ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\).

Environments: three — the \(\text{CH}_3\), the \(\text{CH}_2\) and the \(\text{OH}\) — so three groups of peaks.

Integration: the areas are in the ratio 3 : 2 : 1, matching the three hydrogens of \(\text{CH}_3\), the two of \(\text{CH}_2\) and the one of \(\text{OH}\).

Chemical shift: the \(\text{CH}_3\) sits near \(\delta\) 1.2 (a plain alkyl environment); the \(\text{CH}_2\) is pulled downfield to about \(\delta\) 3.7 because it is next to the electron-withdrawing oxygen; the \(\text{OH}\) appears somewhere in the broad 1–5 range.

Splitting (n+1): the \(\text{CH}_3\) has two hydrogens on the adjacent \(\text{CH}_2\), so \(n=2\) and it is a triplet (2+1). The \(\text{CH}_2\) has three hydrogens on the adjacent \(\text{CH}_3\), so \(n=3\) and it is a quartet (3+1). The \(\text{OH}\) proton normally appears as a singlet, because it exchanges too quickly to split with its neighbours — the very property the next block puts to work.

CaseUsing D2O to catch the OH and NH protons

The hydrogen atoms in \(-\text{OH}\) and \(-\text{NH}\) groups are awkward. Their chemical shift is variable — it drifts with concentration, temperature and solvent — and they often appear as a broad singlet rather than splitting neatly, because they swap places with other protons faster than the instrument can resolve. That same lability, though, gives chemists a clean trick to identify them.

Shake the sample with a few drops of deuterium oxide, \(\text{D}_2\text{O}\) (heavy water), and run the spectrum again. The labile \(-\text{OH}\) or \(-\text{NH}\) hydrogens exchange with deuterium: \(\text{R–OH} + \text{D}_2\text{O} \rightleftharpoons \text{R–OD} + \text{HOD}\). Deuterium does not show up in a proton NMR spectrum, so the peak due to that \(-\text{OH}\) or \(-\text{NH}\) hydrogen disappears (or shrinks sharply) in the second spectrum. Comparing the spectrum before and after the D2O shake tells you at once which peak was the \(-\text{OH}\) or \(-\text{NH}\) proton — it is the one that vanished.

Worked example

An unknown compound with molecular formula \(\text{C}_2\text{H}_6\text{O}\) gives a proton NMR spectrum with three peaks. When the sample is shaken with \(\text{D}_2\text{O}\), the peak at \(\delta\) 2.6 disappears while the other two remain. What does the D2O test tell you, and is the compound ethanol or methoxymethane?

The peak that vanishes on adding D2O must be a labile proton — an \(-\text{OH}\) or \(-\text{NH}\). The formula \(\text{C}_2\text{H}_6\text{O}\) has two possible structures: ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\), which contains an \(-\text{OH}\), and methoxymethane, \(\text{CH}_3\text{OCH}_3\), which does not. Only ethanol has an exchangeable proton, so the disappearing peak identifies the compound as ethanol. As a cross-check, ethanol has three hydrogen environments (giving three peaks, as observed), whereas methoxymethane's six hydrogens are all equivalent and would give a single peak. The D2O result and the peak count agree — a model structure-elucidation answer combines both.

VocabularyKey terms the mark scheme pays for

Chemical shift (δ)

The position of a peak in parts per million (ppm), measured relative to TMS at δ = 0. It indicates the environment of a nucleus, read against the data sheet in the exam.

TMS (tetramethylsilane)

The reference compound, Si(CH₃)₄, fixed at δ = 0. Chosen because it gives one sharp peak clear of other signals, is inert, non-toxic and volatile (easily removed).

Chemical (magnetic) environment

The set of surroundings of a nucleus. Nuclei in different environments resonate at different chemical shifts; nuclei made equivalent by symmetry share one peak.

Integration trace

In proton NMR, the relative area under each group of peaks, giving the ratio of the numbers of hydrogen atoms in each environment (e.g. 3 : 2 : 1).

Spin-spin splitting

The splitting of a proton's peak into several lines by the hydrogen atoms on neighbouring carbon atoms, which reveals what each group is bonded next to.

n+1 rule

A proton with n equivalent hydrogens on the adjacent carbon(s) is split into n+1 lines: 0 neighbours → singlet, 1 → doublet, 2 → triplet, 3 → quartet.

Equivalent nuclei

Atoms in the same environment, usually by symmetry. They give a single combined peak, so the number of peaks counts environments, not atoms.

D₂O exchange

Shaking a sample with deuterium oxide replaces labile –OH and –NH hydrogens with deuterium, which is invisible to ¹H NMR, so their peak disappears — identifying those protons.

TrapsMisconceptions that cost marks

“The number of peaks tells you the number of hydrogen (or carbon) atoms.”

Actually: It tells you the number of different environments. Symmetry makes several atoms equivalent, so benzene's six carbons give one ¹³C peak and propanone's two methyl groups give one. Count environments, not atoms, and check for symmetry first.

“Carbon-13 peaks are split by the n+1 rule like proton peaks.”

Actually: In the carbon-13 spectra used at A-level there is no splitting — each carbon environment gives a single peak. The n+1 splitting rule applies only to proton NMR, where a peak is split by hydrogens on the neighbouring carbons.

“The integration trace gives the exact number of hydrogens in each environment.”

Actually: It gives the ratio. A 3 : 2 : 1 trace could be 3, 2 and 1 hydrogens or 6, 4 and 2 — you scale the ratio to fit the molecular formula, which is why you use the formula alongside the spectrum.

ExamWhat examiners want

Read the data sheet, don't memorise shifts. The exam supplies the tables of chemical-shift ranges for both ¹H and ¹³C NMR, so quote a range from the sheet to justify each assignment rather than recalling a number — the mark is for matching the peak to an environment.

For carbon-13, state the rule explicitly (peaks = number of carbon environments) and check for symmetry before you count, because equivalent carbons collapse into one peak. For proton NMR, work through all three pieces of information every time: number of environments (peak groups), integration ratio (relative hydrogen numbers, scaled to the formula) and splitting by the n+1 rule (to identify neighbouring groups). A triplet-and-quartet pair should immediately suggest an ethyl group.

Know the TMS justification in full — one sharp peak, inert, non-toxic, volatile, and a signal clear of the others at δ = 0 — because it is a stock two- or three-mark question. Use D₂O to assign –OH and –NH protons: the peak that disappears after shaking with D₂O is the labile one. In full structure-elucidation questions, combine NMR with the molecular formula, infrared spectroscopy and mass spectrometry rather than relying on the NMR alone, and always name the environment you assign each peak to.