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Lesson 8 · 35 min

NMR: listening to hydrogen nuclei

Put a sample in a strong magnet and its hydrogen nuclei absorb radio waves of precise frequencies, like tiny compass needles. The chemical shift tells a proton’s environment, the integral how many there are, and the splitting who its neighbours are — enough clues to recognise a molecule.

In this lesson

Explain why protons in a magnetic field absorb radio waves only at ν = γB₀/2π, and why the NMR signal is weak
Read the chemical shift as a measure of shielding: electronegative neighbours move a proton downfield (to higher δ)
Use integrals and the n + 1 rule to work out how many protons each group has and who its neighbours are
Explain why a higher-field instrument separates a crowded spectrum: shifts scale with the field, couplings do not

In 1951 Arnold, Dharmatti and Packard at Stanford put ethanol in a strong magnetic field and recorded three separate signals with areas in the ratio 3 : 2 : 1 — exactly the numbers of CH₃, CH₂ and OH hydrogens. The same kind of nucleus absorbs at slightly different frequencies depending on its chemical surroundings. Hospital MRI scanners rely on the same physics; in the chemistry laboratory nuclear magnetic resonance (NMR) spectroscopy is the most powerful tool for working out molecular structures: the chemical shift says what environment a proton is in, the integral how many there are, and the splitting who its neighbours are.

Predict first

In the ¹H NMR spectrum of ethanol, into how many lines is the CH₂ signal split?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

1 2 3 4

Nuclei in a magnet

A proton has spin and behaves like a tiny compass needle. In a magnetic field $B_0$ it can take only two orientations: along the field (slightly lower in energy) or against it (slightly higher). The energy difference corresponds to a radio-frequency photon whose frequency is proportional to the field, with the proton’s gyromagnetic ratio $\gamma$ as the constant: 42.58 MHz per tesla. A “400 MHz NMR” places the sample in a 9.4 T superconducting magnet — some two hundred thousand times the Earth’s field.

The energy difference is far smaller than thermal energy at room temperature, so the two orientations are almost equally populated: at 400 MHz and 25 °C only about 32 more protons per million sit in the lower state, and that small excess is the whole signal. That is why NMR is insensitive and needs milligrams of sample — and why ever stronger magnets keep being built.

\nu = \frac{\gamma}{2\pi}B_0,\qquad \delta = \frac{\nu - \nu_\mathrm{TMS}}{\nu_0}\times 10^{6}\ \text{ppm}

The resonance frequency is proportional to the field; the chemical shift δ is the frequency difference from the reference, tetramethylsilane (TMS), in parts per million of the instrument frequency.

$\nu$: resonance frequency
$\gamma/2\pi$: proton gyromagnetic ratio, 42.577 MHz T⁻¹
$B_0$: applied magnetic field
$\nu_0$: operating frequency of the instrument
$\delta$: chemical shift, ppm

Chemical shift: shielding by electrons

The electrons around a nucleus set up a small opposing field in the magnet, “shielding” the nucleus so that it feels slightly less than $B_0$ . Fewer electrons mean weaker shielding, a higher resonance frequency and a larger δ — conventionally, a shift “downfield”. The chemical shift is therefore almost a report on electron density:

a CH₃ bonded only to carbon sits near 1 ppm (ethanol’s CH₃, 1.24);
a CH₃ next to a carbonyl is near 2.1 (acetone 2.17, the CH₃CO of ethyl acetate 2.05);
a CH₂ or CH₃ bonded directly to oxygen is at 3.2–4.2 (ethanol 3.72, the OCH₂ of ethyl acetate 4.12);
two chlorines on the same carbon push CH₂Cl₂ to 5.30;
hydrogens on a benzene ring are near 7 (p-xylene 7.06), an effect of the ring current of the π electrons.

The protons of the reference, TMS, are strongly shielded by electron-rich silicon, so nearly every organic signal lies to their left.

Measured chemical shifts in CDCl₃
Molecule	Protons	δ / ppm	Splitting
$\ce{CH3CH2OH}$	CH₃ / CH₂ / OH	1.24 / 3.72 / 1.42	t / q / br s
$\ce{CH3COOCH2CH3}$	CH₃CO / OCH₂ / CH₃	2.05 / 4.12 / 1.26	s / q / t
$\ce{CH3COCH2CH3}$	CH₃CO / CH₂ / CH₃	2.14 / 2.46 / 1.06	s / q / t
$\ce{(CH3)2CHOH}$	CH / CH₃	4.03 / 1.21	sept / d
$\ce{CH3OC(CH3)3}$	OCH₃ / C(CH₃)₃	3.22 / 1.19	s / s
$\ce{CH2Cl2}$	CH₂	5.30	s
$\ce{C6H4(CH3)2}$	ArH / CH₃	7.06 / 2.31	s / s

Source: N. R. Babij et al., Org. Process Res. Dev. 2016, 20, 661–667, Table 1 (CDCl₃); J ≈ 7 Hz for the ethyl groups, 6.1 Hz for 2-propanol.

Integration and splitting

Integration: the area of a signal is proportional to the number of protons producing it. The step heights of the integral trace give the ratio of the groups — 3 : 2 : 1 for ethanol, 3 : 9 for methyl tert-butyl ether.

Spin–spin splitting: the field a proton feels also depends on the spins of neighbouring protons two or three bonds away. $n$ equivalent neighbours have $n + 1$ different total spins, so the signal splits into $n + 1$ lines with binomial (Pascal’s-triangle) intensities: a doublet 1 : 1, a triplet 1 : 2 : 1, a quartet 1 : 3 : 3 : 1 — and the CH of 2-propanol, split by six equivalent CH₃ protons, a 1 : 6 : 15 : 20 : 15 : 6 : 1 septet. The spacing of the lines is the coupling constant $J$ , measured in hertz and independent of the field.

Two points to remember: equivalent protons do not split one another (CH₂Cl₂ gives a single line), and an OH proton exchanging rapidly with water is usually neither split nor splitting.

Worked example

Reading the spectrum of ethyl acetate

Ethyl acetate, $\ce{CH3COOCH2CH3}$ , shows three signals: 4.12 (q, 2H), 2.05 (s, 3H) and 1.26 (t, 3H). Assign them to the protons of the molecule and explain why 4.12 is furthest downfield.

¹H NMR: chemical shift, integration and spin–spin splitting

NMR spectrometer

Here is the spectrum of one sample. Which one is it?

Magnet 9.39 T Integral trace TMS reference (0 ppm)

Your turn: the spectrometer holds an unknown sample. Use the positions, integrals and splittings to pick it from the list. Ethyl tert-butyl ether is a good test — switch to 60 MHz and look again at its 1.1–1.3 ppm region.

Concept check

What changes when the instrument is switched from 300 MHz to 600 MHz?

Both the shifts (ppm) and the couplings (Hz) double Neither shifts nor couplings change, but each multiplet occupies half the width in ppm, so overlapping signals separate The shifts double and the couplings stay the same The couplings halve, so the splitting disappears

Concept check

Which of these molecules shows a septet in its spectrum?

Ethanol 2-Propanol Ethyl acetate Methyl tert-butyl ether

Try the calculation

How strong a magnetic field does a 600 MHz NMR spectrometer need? (Proton γ/2π = 42.577 MHz T⁻¹)

Key takeaways

Protons in a field B₀ resonate at ν = (γ/2π)B₀ (42.58 MHz T⁻¹); the two spin states are almost equally populated, and the signal comes from an excess of about thirty per million.
The chemical shift δ reports shielding: protons attached near O, Cl or an aromatic ring move to higher δ (downfield).
Integrals give the ratio of protons in each group; n equivalent neighbours split a signal into n + 1 lines spaced by the field-independent coupling constant J.
The higher the field, the narrower the multiplets on the ppm axis, the more easily overlapping signals separate, and the closer the spectrum is to first order.

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Data and references

N. R. Babij, E. O. McCusker, G. T. Whiteker et al., “NMR Chemical Shifts of Trace Impurities: Industrially Preferred Solvents Used in Process and Green Chemistry”, Org. Process Res. Dev. 20, 661–667 (2016).
H. E. Gottlieb, V. Kotlyar, A. Nudelman, “NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities”, J. Org. Chem. 62, 7512–7515 (1997).
J. T. Arnold, S. S. Dharmatti, M. E. Packard, “Chemical effects on nuclear induction signals from organic compounds”, J. Chem. Phys. 19, 507 (1951).
E. Tiesinga, P. J. Mohr, D. B. Newell, B. N. Taylor, “CODATA recommended values of the fundamental physical constants: 2018”, Rev. Mod. Phys. 93, 025010 (2021) — proton gyromagnetic ratio.
D. A. Skoog, F. J. Holler, S. R. Crouch, Principles of Instrumental Analysis, 7th ed., Cengage, 2018, chapter 19.