Related topics: Organic Synthesis
Why NMR? The structure-solving toolkit
No single instrument tells you a molecule’s structure. In practice chemists run a compound through a set of techniques and let each one contribute what it does best. You met the first two at AS in Organic analysis (3.3.6): mass spectrometry pins down the Mr (and, at high resolution, the molecular formula) while infrared spectroscopy identifies the functional groups from their absorptions.
That is usually where the trail goes cold: a molecular formula and a functional group still fit several isomers. Nuclear magnetic resonance (NMR) is the technique that finishes the job. It reports on the position of every 13C and 1H atom in the molecule: how many distinct environments there are, how the hydrogens are shared between them, and even what sits on the neighbouring carbon. Where infrared names a functional group and mass spectrometry weighs the molecule, NMR draws you the map.
The big idea
NMR looks at atoms with nuclear spin (13C and 1H) and reports on their environment. 13C NMR counts the carbon environments; 1H NMR is richer, telling you how many hydrogen environments there are, how many H are in each, and what sits on the neighbouring carbon. The exam skill is not theory: it is reading a spectrum backwards to a structure.
TMS, solvents & the δ scale
Every peak position is measured relative to a standard, tetramethylsilane (TMS, Si(CH3)4), which is assigned δ = 0. The horizontal axis is the chemical shift δ in parts per million (ppm); the further a signal is from electronegative atoms and double bonds, the closer it sits to TMS.
Why TMS is the standard
- Its 12 hydrogens are all equivalent, so it gives a single sharp signal.
- It is non-toxic and inert (won’t react with the sample).
- It is volatile (bp 26 °C), so it is easily removed afterwards.
- Its signal is upfield of almost all others (to the right, at δ 0), so it doesn’t overlap the sample’s peaks.
The sample is dissolved in a solvent with no 1H atoms. Otherwise the solvent’s own protons would appear. Common choices are CCl4 (no hydrogen at all) or deuterated solvents such as CDCl3, where the hydrogen is replaced by deuterium (2H), so there is no 1H present to give a signal in the 1H spectrum.
13C vs 1H at a glance
13C NMR is the simpler spectrum: one peak per carbon environment, positions from the Data Booklet. No splitting or integration to consider. 1H NMR adds two more layers of information: integration (how many H per environment) and splitting (what’s on the neighbouring carbon).
Exam questions
Q1
[1 mark]
Why does tetramethylsilane (TMS) give only one peak in a 1H NMR spectrum?
The four methyl groups are equivalent, so all twelve H atoms share one environment → one signal. (The low bp and inertness are reasons TMS is convenient, not why it gives one peak.)
Q2
[6 marks]
A compound is mixed with Si(CH3)4 and either CCl4 or CDCl3 before recording its 1H NMR spectrum. State why each substance is used and explain how its properties make it suitable.
Show answer
Si(CH3)4 (TMS) is the reference / standard, giving the peak at δ = 0. It has 12 equivalent H (one signal), is inert (won’t react with the sample), and is volatile / has a low bp (easily removed). Its signal is away from most others. up to 3 marks
CCl4 / CDCl3 are solvents. Both contain no 1H, so give no signals of their own (in CDCl3 the hydrogen is replaced by deuterium, 2H). CCl4 is non-polar (dissolves non-polar samples); CDCl3 is polar (dissolves polar samples). up to 3 marks
Source: AQA A-Level Chemistry past papers.
13C NMR: counting carbon environments
A 13C spectrum has one peak for each carbon environment. Carbons are in the same environment if they are equivalent by symmetry. So the first job is always to count how many different carbons the molecule has: that is the number of peaks.
The position of each peak (its chemical shift) tells you the type of carbon, read from Table C of the Data Booklet. As in Table B, each row describes one carbon, the red C below (bold in the booklet):
| Type of carbon | δ / ppm |
|---|---|
| C–C (alkyl) | 5–40 |
| R–C–Cl or R–C–Br | 10–70 |
| R–C–C=O (C next to a carbonyl) | 20–50 |
| R–C–N (amines) | 25–60 |
| –C–O (alcohols, ethers, esters) | 50–90 |
| C=C (alkene) | 90–150 |
| R–C≡N (nitrile) | 110–125 |
| Aromatic C | 110–160 |
| R–C=O (esters or acids) | 160–185 |
| R–C=O (aldehydes or ketones) | 190–220 |
Exam questions
Q3
[1 mark]
How many peaks are in the 13C NMR spectrum of propan-2-one (propanone), CH3COCH3?
The two CH3 groups are equivalent (one environment) and the C=O is a second, so 2 peaks, not 3. Symmetry reduces the count.
Source: AQA A-Level Chemistry past papers.
1H NMR: shift, integration & splitting
A 1H spectrum carries three pieces of information at once: where each signal sits (chemical shift), how big it is (integration), and how many lines it is split into (spin–spin splitting). Read them in that order for every signal: each layer answers a different question about the molecule.
1. Chemical shift: what kind of environment?
One signal per hydrogen environment, and the signal’s chemical shift δ tells you the type of environment, read from Table B of the Data Booklet. The pattern behind the table: the nearer the hydrogen is to an electronegative atom (O, Cl) or to a C=O, the more the electron density around it is pulled away and the larger its shift: the peak moves left, away from TMS.
Read each row carefully: it describes one hydrogen only, shown in red below (the Data Booklet prints it in bold). Everything else in the fragment is just scenery telling you where that H sits.
| Type of proton | δ / ppm |
|---|---|
| ROH | 0.5–5.0 |
| RCH3 | 0.7–1.2 |
| RNH2 | 1.0–4.5 |
| R2CH2 | 1.2–1.4 |
| R3CH | 1.4–1.6 |
| R–CO–C–H (H on a C next to a C=O) | 2.1–2.6 |
| R–O–C–H (H on a C bonded to O) | 3.1–3.9 |
| RCH2Cl or RCH2Br | 3.1–4.2 |
| R–CO–O–C–H (H on a C bonded to an ester O) | 3.7–4.1 |
| R2C=CH– (alkene) | 4.5–6.0 |
| R–CHO (aldehyde) | 9.0–10.0 |
| R–COOH (carboxylic acid) | 10.0–12.0 |
Interactive: find it on the δ ruler
Tap a hydrogen environment and watch where its signal lands.
2. Integration: how many H?
The area under each signal is proportional to the number of H atoms in that environment. On AQA papers the integration is usually given as a number printed above each peak (sometimes with a small integral step drawn over the peak instead). Convert the numbers to the simplest whole-number ratio: a 1.2 : 1.2 : 1.8 integration is a 2 : 2 : 3 ratio of hydrogens.
One trap: the ratio is not automatically the number of H. Check it against the molecular formula: if C4H8O shows a 3 : 2 : 3 ratio, the total (8) matches and the counts really are 3H, 2H, 3H; but a single-peak spectrum of C3H6O with “ratio 1” is six hydrogens, not one.
Interactive: integration to H count
3. Splitting: the n+1 rule
Each signal is split by the hydrogens on the neighbouring carbon (spin–spin coupling). The number of lines is one more than the number of adjacent, non-equivalent H:
| Signal | Lines | H on the adjacent carbon |
|---|---|---|
| singlet | 1 | 0 |
| doublet | 2 | 1 |
| triplet | 3 | 2 |
| quartet | 4 | 3 |
Interactive: the n+1 sandbox
Choose how many H sit on the carbon next door and watch the signal split. Red H give the signal; blue H are the neighbours doing the splitting.
Precision points: –OH and –NH do not split
The H of an –OH (alcohol or acid) or –NH usually does not couple: it appears as a singlet and does not split the signal on the next carbon. A lone 1H singlet is very often an –OH or –COOH.
Patterns worth memorising
- Ethyl group CH3CH2– → a triplet and a quartet, ratio 3 : 2.
- –CH2–CH2– → two triplets, 2 : 2.
- CH3–CH< → a doublet and a quartet, 3 : 1 (isopropyl gives a doublet, 6H).
- CH3– next to a carbon with no H → a singlet, 3H.
Exam questions
Q4
[3 marks]
For butanone, CH3COCH2CH3, deduce the splitting pattern of each of the three 1H signals.
Show answer
CH3CO (next to C=O, no H on the adjacent carbon) → singlet (3H). 1 mark
CH2 (adjacent CH3 has 3 H) → quartet (2H). 1 mark
CH3CH2 (adjacent CH2 has 2 H) → triplet (3H). 1 mark
Source: AQA A-Level Chemistry past papers.
Working back to a structure
Working back from a spectrum to a structure is the exam skill. You are given a spectrum (and usually the molecular formula) and must build the structure. Take one peak at a time and squeeze out all three pieces of information. The fragments then click together:
The method, per peak
- Integration → how many H are in this environment.
- Splitting → how many H on the neighbour (lines − 1, by the n+1 rule).
- Shift → the environment (Table B): is it near O, Cl, C=O, a ring?
Write each peak up as one line, then join the fragments. Here it is done in full for a compound C3H7ClO whose spectrum shows three peaks:
One worked spectrum, as exam-answer lines
- δ 3.35, 3H, singlet → CH3 next to O with no H on the next atom: fragment CH3–O–
- δ 3.65, 2H, triplet → O–CH2 next to a CH2: fragment –O–CH2–CH2–
- δ 3.95, 2H, triplet → Cl–CH2 next to a CH2: fragment –CH2–Cl
Join them: CH3–O–CH2–CH2–Cl.
Practise: the NMR analyser
This is how the method becomes automatic. Read a spectrum works exactly like the exam: take each peak in turn, fix its integration, splitting and Data-Booklet environment, collect the exam-answer lines, then sketch the structure on paper before revealing it. Predict a spectrum runs the drill in reverse: you fill in the analysis for a given molecule and the spectrum builds below as you go. The 13C drill does the lighter carbon version: count the environments, place the peaks.
Interactive: the NMR analyser
Data Booklet shift tables (open while you work)
Precision points: the marks that get dropped
- Counting environments wrong: equivalent atoms share one peak (symmetry lowers the count in both 13C and 1H).
- Reading splitting the wrong way: lines = neighbours + 1, so a triplet means 2 H next door, not 3.
- Splitting an –OH or –NH, or letting it split its neighbour: it usually doesn’t couple.
- Ignoring the molecular formula: the integration H-total must add up to it.
Examiner reports note that the strongest answers state the deduction, not just the label (“triplet, so 2 H on the adjacent carbon” and “δ 3.95, so next to an electronegative atom”) before drawing the structure. Show the reasoning for each peak.
🎯 Model answer: the C3H7ClO structure deduction
C3H7ClO gives three 1H peaks: δ 3.95 (2H, triplet), δ 3.65 (2H, triplet), δ 3.35 (3H, singlet). Deduce the structure, explaining each peak. Award yourself a mark for each:
- δ 3.95, triplet, 2H → Cl–CH2 next to a CH2.
- δ 3.65, triplet, 2H → O–CH2 next to a CH2.
- δ 3.35, singlet, 3H → CH3–O with no H on the adjacent atom.
- Structure: CH3OCH2CH2Cl.
3.3.15 NMR spectroscopy: Quick-reference summary
- Techniques combine: mass spectrometry fixes Mr and the molecular formula, infrared names the functional groups, and NMR maps the 13C and 1H environments. Together they confirm a structure.
- TMS (δ = 0): one signal, inert, volatile, upfield: the reference. Solvents CCl4 / CDCl3 have no 1H, so give no signal.
- 13C NMR: number of peaks = number of carbon environments; positions from Data Booklet Table C. Simpler: no splitting or integration.
- 1H NMR: number of signals = H environments; integration = ratio of H atoms; chemical shift from Table B.
- n+1 rule: a signal is split into (number of H on the adjacent carbon) + 1 lines: singlet, doublet, triplet, quartet. –OH and –NH usually don’t couple (appear as singlets).
- Solving a spectrum: for each peak read integration → H count, splitting → neighbours, shift → environment; assemble the fragments into the structure.