NMR Spectroscopy

What is it?

Nuclear magnetic resonance (NMR) spectroscopy places a sample in a strong magnetic field and probes it with radio waves. Hydrogen nuclei (protons) in different chemical environments absorb at slightly different frequencies, so a ¹H NMR spectrum shows how the hydrogen atoms in a molecule are arranged.

A ¹H NMR spectrum gives four pieces of information:

Number of signals: how many different hydrogen environments there are.
Chemical shift, $\delta$ (in ppm): what kind of environment each one is.
Integration (relative area): how many hydrogens are in each environment.
Splitting (multiplicity): how many hydrogens are on the neighbouring carbons.

Hydrogens are in the same environment if they are interchangeable by symmetry: the six H atoms in propanone, $\ce{CH3COCH3}$ , are all equivalent, so propanone gives one signal.

Key idea

Each signal is a group of equivalent hydrogens. Its position tells you what is nearby (an oxygen, a C=O, a benzene ring), its area tells you how many H atoms it represents, and its splitting tells you how many H atoms sit on the next carbon along.

Why does it matter?

Structure determination. NMR is the most powerful tool chemists have for working out how atoms are connected in a new molecule.
Medicine. MRI (magnetic resonance imaging) is NMR of the hydrogen nuclei in water and fat in the body.
Industry and research. NMR checks the purity of medicines and follows reactions as they happen.

How does it work?

1. Chemical shift

Shifts are measured relative to a reference, TMS (tetramethylsilane, set at $\delta = 0$ ), and are given in parts per million so that they are the same on every spectrometer:

\small\begin{aligned} &\delta\ (\text{ppm}) \\[4pt] &= \frac{\text{shift from TMS (Hz)}}{\text{spectrometer frequency (MHz)}} \end{aligned}

Nearby electronegative atoms and C=O or aromatic rings pull electron density away from a hydrogen (“deshielding”), moving its signal to a higher δ:

Type of hydrogen	Typical δ (ppm)
alkyl, $\ce{R-CH3}$ , $\ce{R-CH2-R}$	0.9–1.7
next to C=O, $\ce{-CO-CH3}$	2.0–2.7
on carbon next to O, $\ce{-O-CH2-}$ , $\ce{-O-CH3}$	3.3–4.3
alkene, $\ce{C=C-H}$	4.5–6.5
aromatic, on a benzene ring	6.5–8.0
aldehyde, $\ce{-CHO}$	9–10
carboxylic acid, $\ce{-COOH}$	10–12
alcohol $\ce{-OH}$	1–5 (variable)

2. Integration

The area under each signal is proportional to the number of hydrogens it represents. Spectrometers print the areas as a ratio, such as 3 : 2 : 1 for ethanol.

3. Splitting: the n + 1 rule

A signal from hydrogens with n equivalent hydrogens on adjacent carbons is split into n + 1 peaks:

H on neighbouring carbons (n)	0	1	2	3
Peaks (n + 1)	1, singlet	2, doublet	3, triplet	4, quartet

So an ethyl group, $\ce{-CH2CH3}$ , always shows a quartet (the $\ce{CH2}$ , split by 3 H) and a triplet (the $\ce{CH3}$ , split by 2 H). An O–H hydrogen usually appears as a singlet and is not split by, nor does it split, its neighbours.

4. ¹³C NMR in brief

¹³C NMR shows one signal for each carbon environment (without splitting in the usual spectra). Typical ranges: alkyl carbons 0–50 ppm, carbons bonded to O 50–90 ppm, alkene and aromatic carbons 110–160 ppm, and C=O carbons 160–220 ppm.

Think of it like this

Reading an NMR spectrum is like working out where people live from a list of neighbours. Each signal is a household; its shift says which part of town it is in (near an oxygen, near a ring); its area says how many people live there; and its splitting says how many people live next door.

More precisely

The n + 1 rule works when the neighbouring hydrogens are all equivalent and couple equally; more complex molecules give multiplets. Adding a drop of $\ce{D2O}$ makes an O–H or N–H signal disappear, because deuterium replaces the hydrogen and does not show in ¹H NMR; this is a useful way to identify them. Spectrometers are described by their proton frequency (300 MHz, 400 MHz…); a higher frequency spreads the signals out more in Hz, but δ in ppm stays the same.

Visualise it

Schematic proton NMR spectrum of ethanol, CH3CH2OH, with chemical shift decreasing from left to right. A quartet near 3.7 ppm with relative area 2 (the CH2, next to oxygen and split by the CH3), a singlet near 2.6 ppm with area 1 (the OH), and a triplet near 1.2 ppm with area 3 (the CH3, split by the CH2). TMS is at 0 ppm. — ¹H NMR spectrum of ethanol (schematic): three environments in a 2 : 1 : 3 ratio.

Worked example

Worked example: Counting environments

Question: How many ¹H NMR signals do these give: (a) propanone, $\ce{CH3COCH3}$ (b) ethanol, $\ce{CH3CH2OH}$ (c) methyl ethanoate, $\ce{CH3COOCH3}$ ?

(a) Both $\ce{CH3}$ groups are equivalent: 1 signal (6H).
(b) $\ce{CH3}$ , $\ce{CH2}$ and $\ce{OH}$ : 3 signals (3 : 2 : 1).
(c) The $\ce{CH3}$ next to C=O and the $\ce{CH3}$ on O are different: 2 signals, at about 2.0 and 3.7 ppm, both singlets (no neighbouring H).

Worked example: Chemical shift in ppm

Question: On a 400 MHz spectrometer, a signal appears 1092 Hz from TMS. Calculate δ, and say what kind of hydrogen it might be.

\begin{aligned} &\delta = \frac{1092\ \text{Hz}}{400\ \text{MHz}} \\[4pt] &= 2.73\ \text{ppm} \end{aligned}

(1 Hz per MHz = 1 ppm.) At 2.73 ppm, it is probably a hydrogen on a carbon next to a C=O group.

Worked example: Predicting a spectrum: ethyl ethanoate

Question: Predict the ¹H NMR spectrum of ethyl ethanoate, $\ce{CH3COOCH2CH3}$ .

Group	δ (ppm)	Area	Neighbours (n)	Splitting
$\ce{CH3-CO}$	about 2.0	3	0	singlet
$\ce{O-CH2}$	about 4.1	2	3	quartet
$\ce{CH2-CH3}$	about 1.3	3	2	triplet

Worked example: Solving a structure

Question: A compound $\ce{C4H8O2}$ shows a strong IR absorption at 1740 cm⁻¹ (no O–H). Its ¹H NMR spectrum has a triplet at 1.2 ppm (3H), a quartet at 2.3 ppm (2H) and a singlet at 3.7 ppm (3H). Identify it.

IR: C=O with no O–H, and two O atoms: an ester.
Triplet (3H) + quartet (2H): an ethyl group. The quartet at 2.3 ppm means the $\ce{CH2}$ is next to C=O: $\ce{CH3CH2-CO-}$ .
Singlet at 3.7 ppm (3H): a $\ce{CH3}$ on oxygen with no neighbours: $\ce{-O-CH3}$ .
The compound is methyl propanoate, $\ce{CH3CH2COOCH3}$ (4 C, 8 H, 2 O ✓).

Common mistake

Common mistake: Counting the hydrogens on the same carbon

Splitting comes from hydrogens on the neighbouring carbons, not on the same carbon. The $\ce{CH3}$ in an ethyl group is a triplet because its neighbour, $\ce{CH2}$ , has 2 H.

Common mistake: Treating integration as an exact count

Integration gives a ratio. If the areas are 1 : 1.5, the actual numbers could be 2 : 3 or 4 : 6. Use the molecular formula to fix the totals.

Common mistake: Expecting O–H to be split

In most spectra the O–H hydrogen appears as a singlet and does not split neighbouring signals, because it exchanges rapidly between molecules.

Notation note

δ is read “delta”; NMR spectra are drawn with δ decreasing from left to right (like IR).
s = singlet, d = doublet, t = triplet, q = quartet, m = multiplet.

Remember this

Number of signals = number of H environments. Integration = relative number of H in each.
δ (ppm) = shift (Hz) ÷ spectrometer frequency (MHz); TMS = 0 ppm.
Higher δ near electronegative atoms, C=O and rings: about 1 (alkyl), 2–2.7 (next to C=O), 3.3–4.3 (on C–O), 6.5–8 (aromatic), 9–10 (CHO), 10–12 (COOH).
n + 1 rule: n neighbouring H give n + 1 peaks; ethyl = quartet + triplet.
Combine MS (mass), IR (functional groups) and NMR (C–H framework) to solve a structure.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/nmr-spectroscopy/