An introduction to proton NMR spectroscopy
How the hydrogen atoms in a molecule report their own surroundings, and how to turn those signals into a structure.
Proton NMR is the single most informative tool in the organic chemist’s toolkit. IR tells you which functional groups are present and mass spectrometry gives you the molecular formula; ¹H NMR tells you how the hydrogen atoms are arranged, and therefore how the carbon skeleton is put together. This page introduces where the signals come from and how to extract four kinds of information from every spectrum you read.
Where the signal comes from
A proton (a hydrogen nucleus) behaves like a tiny magnet because it has spin. Place the sample in a strong external magnetic field and each proton aligns either with the field, the lower-energy state, or against it, the higher-energy state. The energy gap between those two states is tiny, and it falls in the radiofrequency range.
If we irradiate the sample with radio waves and sweep the frequency range, a proton absorbs energy and flips to the higher state at exactly the frequency that matches its energy gap; the nucleus is said to be in resonance. The spectrometer records which frequencies were absorbed. The crucial point is that not every proton in a molecule has the same gap: the electrons around each proton shield it from the external field, so protons in different chemical environments come into resonance at slightly different frequencies. Those small differences are the whole spectrum.
Shielding and deshielding of protons
Electron density around a proton opposes the applied field, so a proton in an electron-rich environment feels a weaker field and absorbs at low frequency; it is shielded. Pull electrons away with an electronegative atom such as oxygen or a halogen and the proton is deshielded, absorbing further to the left of the spectrum. This is why chemical shift is a map of electron density.
The chemical shift scale
Absolute resonance frequencies depend on the strength of the magnet, which would make spectra from different instruments incomparable. To fix this, shifts are reported relative to a reference compound, tetramethylsilane (TMS), on a scale called delta (δ) in parts per million:
Because both numerator and denominator scale with the magnet, δ values are instrument-independent; a proton at 2.0 ppm on a 300 MHz machine is at 2.0 ppm on a 600 MHz machine. TMS is defined as 0 ppm, and almost all organic protons fall between 0 and 12 ppm. Note that the axis runs right to left: 0 ppm sits at the right edge, and moving left means more deshielded, or “downfield”.
The four pieces of information in a spectrum
Every ¹H NMR spectrum answers four separate questions. Learn to ask them in this order and interpretation becomes a routine.
Equivalent protons: counting the signals
Two protons are chemically equivalent if they sit in identical environments, and equivalent protons produce a single signal. In practice, look for symmetry: the three hydrogens of a freely rotating CH₃ are equivalent, and so are the two ortho hydrogens on either side of a symmetrically substituted benzene ring.
- Ethanol, CH₃CH₂OH, has three environments (CH₃, CH₂, OH) and therefore three signals.
- Propanone (acetone), CH₃COCH₃, has one environment; the two methyls are related by symmetry, so the whole spectrum is one singlet.
- 1,4-dimethylbenzene has two signals: the six methyl hydrogens, and the four equivalent hydrogens on the aromatic ring.
Characteristic chemical shifts
You do not need to memorise exact values; recognising the regions is enough to place a signal.
| Type of proton | δ (ppm) | Notes |
|---|---|---|
| Alkane C–H (R–CH₃, R₂CH₂) | 0.8 – 1.8 | The most shielded region; almost every spectrum has something here |
| H next to a carbonyl (CH₃–C=O) | 2.0 – 2.6 | Ketones, esters, aldehydes; often a clean singlet |
| Benzylic H (CH₂ next to a ring) | 2.3 – 2.9 | Deshielded by the ring’s circulating electrons |
| H on C–N (amine) | 2.2 – 2.9 | Nitrogen is less electronegative than oxygen, so less shifted |
| H on C–O (alcohol, ether, ester) | 3.3 – 4.5 | A reliable marker for a carbon bearing oxygen |
| Vinyl H (C=C–H) | 4.5 – 6.5 | Alkene protons; complex splitting is common |
| Aromatic H | 6.5 – 8.0 | Strongly deshielded by the ring current; a very distinctive region |
| Aldehyde H (R–CHO) | 9.5 – 10.5 | Almost diagnostic on its own; nothing else appears here |
| Carboxylic acid H (–COOH) | 10 – 12 | Far downfield, broad, exchangeable |
| O–H (alcohol), N–H (amine) | 0.5 – 5 (variable) | Position depends on solvent and concentration; usually broad and unsplit |
Why O–H and N–H misbehave
Protons on oxygen and nitrogen swap rapidly between molecules; this is exchange. The rapid swapping averages out their coupling, so such signals are typically broad singlets that neither split nor split their neighbours. Adding a drop of D₂O replaces them with deuterium and the signal disappears, which is a neat way to confirm an OH or NH.
Integration: counting the protons
The area under each signal, not its height, is proportional to the number of protons producing it. Spectrometers report this as a step curve or a printed number beside each peak. What matters is the ratio, not the absolute value: an integration of 2 : 3 : 3 in ethyl acetate accounts for all eight hydrogens of C₄H₈O₂.
Two habits will keep you out of trouble. First, reduce the reported areas to the smallest whole-number ratio, then scale that ratio up until the total matches the molecular formula; a 1 : 1 ratio could be 1H : 1H or 3H : 3H. Second, remember that integration counts hydrogens, so it puts a hard limit on your candidate structures.
Spin–spin splitting and the n + 1 rule
A proton feels the tiny magnetic fields of protons on neighbouring carbons. Those neighbours can each be aligned with or against the main field, which splits the signal into several closely spaced lines. The result is summarised by the n + 1 rule: a proton with n equivalent neighbouring protons appears as n + 1 lines.
| Neighbours (n) | Pattern | Line heights | Typical source |
|---|---|---|---|
| 0 | singlet (s) | 1 | CH₃ flanked by a carbonyl or quaternary carbon |
| 1 | doublet (d) | 1 : 1 | CH₃ next to a single CH |
| 2 | triplet (t) | 1 : 2 : 1 | CH₃ next to a CH₂ (the ethyl pattern) |
| 3 | quartet (q) | 1 : 3 : 3 : 1 | CH₂ next to a CH₃ (the ethyl pattern) |
| 4 | quintet | 1 : 4 : 6 : 4 : 1 | A CH₂ between two CH₂ groups |
| 6 | septet | 1 : 6 : 15 : 20 : 15 : 6 : 1 | The CH of an isopropyl group |
Three practical points. Splitting is mutual, so a triplet next to a quartet is the unmistakable signature of an ethyl group; wherever you see that pair, write down CH₂CH₃. Equivalent protons do not split each other, which is why the three hydrogens of a methyl group give one line rather than a multiplet among themselves. And the spacing between the lines of a multiplet, the coupling constant J measured in hertz, is identical for two protons that are coupled to each other; matching J values is how you prove which signals are neighbours.
How to interpret a spectrum, step by step
- 1
Get the formula and count the hydrogens. If a molecular formula is given, note the total number of H atoms; your integrations must add up to it.
- 2
Count the signals. That is the number of distinct hydrogen environments, and it immediately tells you how symmetric the molecule is or isn’t.
- 3
Convert integrations into proton counts. Reduce to whole numbers and scale to the formula, assigning 1H, 2H, 3H and so on to each signal.
- 4
Read each chemical shift. Use the regions in section 5 to name what each environment is attached to; look hard at anything above 6.5 ppm, which means aromatic or aldehyde or carboxylic acid.
- 5
Use splitting to connect the fragments. Apply n + 1 backwards: a quartet has three neighbours, so it sits beside a CH₃. Look for the triplet–quartet ethyl pair and the doublet–septet isopropyl pair.
- 6
Assemble and check. Join the pieces into a structure, confirm every hydrogen is accounted for, and test it against your IR and MS data.
Worked interpretation examples
Example 1. Reading Figure 1
Three signals mean three environments, and the integration 2 : 1 : 3 accounts for six hydrogens. The 3H triplet at 1.2 ppm and the 2H quartet at 3.7 ppm are a mutual pair, so an ethyl group is present: n + 1 says the triplet has two neighbours (a CH₂) and the quartet has three (a CH₃). The quartet’s shift near 3.7 ppm places that CH₂ on an oxygen. The remaining signal is a single hydrogen, a broad unsplit singlet at 2.6 ppm, which is the hallmark of an exchangeable O–H. Joining CH₃CH₂– to –OH gives ethanol, CH₃CH₂OH.
Example 2. C₃H₆O with a single signal
A compound of formula C₃H₆O gives one signal only: a 6H singlet at 2.1 ppm. One signal means one environment, so all six hydrogens are equivalent and the molecule must be symmetric. The singlet has no neighbours, and 2.1 ppm places it in the carbonyl-adjacent region, so the fragment is CH₃–C=O with an identical CH₃ on the other side: propanone (acetone), CH₃COCH₃. Its isomer propanal, CH₃CH₂CHO, would instead give three signals, including a 1H aldehyde signal near 9.7 ppm.
Example 3. Distinguishing two isomers of C₃H₈O
Propan-1-ol and propan-2-ol share a formula, and NMR separates them at a glance.
| Compound | Signals | The give-away |
|---|---|---|
| Propan-1-ol CH₃CH₂CH₂OH | Four: 3H t (0.9 ppm), 2H sextet (1.6 ppm), 2H t (3.6 ppm), 1H broad s | A 2H triplet near 3.6 ppm; the OCH₂ has two neighbours |
| Propan-2-ol (CH₃)₂CHOH | Three: 6H d (1.2 ppm), 1H septet (4.0 ppm), 1H broad s | A 6H doublet plus a 1H septet; the isopropyl signature |
Practice questions
Try each before opening the answer.
1How many signals would you expect in the ¹H NMR spectrum of 1,4-dimethylbenzene, and in what integration ratio?Show answer
Two signals in a 6 : 4 (3 : 2) ratio. The two methyl groups are equivalent by symmetry (6H singlet near 2.3 ppm), and the four ring hydrogens are all equivalent (4H singlet near 7.0 ppm). Neither signal is split, because equivalent protons do not split each other and the methyls have no H neighbours.
2Why does a chemical shift reported in ppm not change when a sample is moved from a 300 MHz to a 500 MHz spectrometer?Show answer
δ is a ratio. A stronger magnet increases the frequency difference from TMS (in Hz) and the spectrometer frequency (in MHz) by the same factor, so the two changes cancel and the ppm value stays the same.
3A spectrum contains a 6H doublet at 1.2 ppm and a 1H septet at 3.9 ppm. What structural fragment is present?Show answer
An isopropyl group, (CH₃)₂CH–. The 6H doublet is two equivalent methyls next to one H; the 1H septet is a CH with six neighbours. Its shift of 3.9 ppm suggests that CH is attached to an oxygen.
4An unknown of formula C₂H₄O₂ gives two singlets, 3H at 2.1 ppm and 1H at 11.4 ppm. Identify it and explain both shifts.Show answer
Ethanoic (acetic) acid, CH₃COOH. The 3H singlet at 2.1 ppm is a methyl next to a carbonyl, with no H neighbours to split it. The 1H singlet at 11.4 ppm is the carboxylic acid O–H, far downfield and unsplit because it exchanges.
5Two isomers of C₈H₁₀ are ethylbenzene and 1,4-dimethylbenzene. Which spectral feature separates them at once?Show answer
Splitting. Ethylbenzene shows the ethyl triplet–quartet pair (3H t near 1.2 ppm, 2H q near 2.6 ppm) plus a 5H aromatic signal. 1,4-dimethylbenzene shows only two singlets, 6H and 4H.
Key takeaways
- Protons in different chemical environments resonate at different frequencies because electrons shield them to different extents.
- Shifts are reported as δ in ppm relative to TMS, which makes them independent of the magnet; the axis runs right to left, and downfield means deshielded.
- Read four things in order: number of signals (environments), chemical shift (what each is), integration (how many H), splitting (who the neighbours are).
- The n + 1 rule converts multiplets into neighbour counts; a triplet with a quartet means an ethyl group, and a doublet with a septet means isopropyl.