Exchangeable Protons and D₂O Shake

Introduction

Signals from hydrogen atoms bonded to oxygen or nitrogen are the troublemakers of ¹H NMR. Their positions wander from sample to sample, they are often broad, and they usually refuse to show splitting. The reason is that these protons are exchangeable : they hop rapidly between molecules. Chemists turn this behaviour into a diagnostic test — the D₂O shake — which makes O–H and N–H signals vanish and so identifies them with confidence.

Core explanation

Proton exchange. Hydrogen atoms in O–H and N–H groups are weakly acidic or basic and take part in hydrogen bonding. In solution they transfer continually between molecules, and to traces of water or acid present in the sample:

the hydrogen on one alcohol molecule may, a fraction of a second later, be sitting on a different alcohol molecule or on a water molecule. This transfer is often faster than the NMR measurement, which has important consequences.

Consequence 1: no splitting. Because an O–H proton moves from molecule to molecule, a neighbouring CH₂ group "sees" a constantly changing partner spin, and the coupling averages to zero. The O–H usually appears as a singlet , and it does not split its neighbours. In ethanol the CH₂ signal is therefore a simple quartet from the CH₃ group only.

Consequence 2: broad signals. Exchange at intermediate rates, combined with the quadrupolar ¹⁴N nucleus in amines, broadens these signals. N–H peaks in particular can be so broad that they are hard to spot above the baseline.

Consequence 3: variable chemical shift. The shift depends on hydrogen bonding, and hydrogen bonding depends on concentration, temperature and solvent.

Proton type Typical δ range Appearance --- --- --- Alcohol R–OH 0.5–5.0 broad singlet Phenol Ar–OH 4.5–7.0 broad singlet Carboxylic acid –COOH 10–13 very broad singlet Amine R–NH₂ 1.0–5.0 broad singlet Amide –CONH– 5.0–8.5 broad

The D₂O shake. Deuterium oxide, D₂O, is added to the NMR tube and the tube is shaken. Exchangeable protons are replaced by deuterium:

R–OH + D₂O ⇌ R–OD + HOD

Deuterium nuclei resonate at a completely different frequency from ¹H nuclei, so they give no signal in a ¹H spectrum. When the spectrum is run again, any peak that has disappeared (or shrunk sharply) belonged to an O–H or N–H proton. A new peak for HOD typically appears near δ 4.7–4.8, depending on the solvent. C–H protons are not acidic enough to exchange, so their signals are unchanged.

Step-by-step reasoning

To identify an exchangeable proton:

1. Record the ¹H spectrum in a deuterated solvent such as CDCl₃. 2. Note any broad singlet with an unexpected position or integration. 3. Add a small amount of D₂O, shake and re-record the spectrum. 4. Compare the two spectra: the peak that vanishes is from O–H or N–H. 5. Check that its integration matched the expected number of exchangeable protons.

Visual explanation

In the simulation, load the spectrum of propan-1-ol and press "Add D₂O". The broad singlet near δ 2.3 fades out, a new HOD peak appears near δ 4.8, and the remaining CH₃, CH₂ and CH₂O signals stay exactly where they were. The before-and-after overlay makes the vanished peak obvious.

Real-world analogy

Imagine a classroom register in which one pupil keeps swapping seats with visitors. If all visitors wear invisible-ink name badges, then after enough swaps that seat reads as empty. The O–H proton is the seat-swapper; deuterium is the invisible visitor, and the seat's signal disappears.

Real-world example

Natural-product chemists isolating new compounds from plants often find several broad signals and need to know which are hydroxyl groups. A quick D₂O shake identifies every O–H in a sugar or flavonoid, helping to count the hydroxyl groups before more advanced two-dimensional experiments are run.

Why?

Why does a C–H signal survive a D₂O shake while an O–H signal does not? Exchange requires a proton to be removed and replaced. O–H and N–H protons are attached to electronegative atoms, so their bonds are polar and the protons are labile. Ordinary C–H bonds are far less polar and do not exchange under these mild conditions.

Common misconception

"The O–H signal of an alcohol always appears in the same place." Its shift depends on concentration, temperature and solvent because hydrogen bonding changes with these conditions. A D₂O shake is more reliable than position for identifying it.

Worked example

Question: A compound C₂H₆O shows a 3H triplet at δ 1.2, a 2H quartet at δ 3.7 and a 1H broad singlet at δ 2.6. After a D₂O shake the singlet disappears. Identify the compound and explain the observations.

Reasoning: The triplet and quartet show an ethyl group on oxygen. The 1H singlet that disappears is exchangeable, so it is an O–H. It is a singlet, and it does not split the CH₂, because of rapid exchange.

Answer: Ethanol, CH₃CH₂OH.

Quick check

1. What happens to the carboxylic acid peak near δ 11 in ethanoic acid after a D₂O shake? Answer: It disappears, because the acidic O–H proton is replaced by deuterium, which gives no ¹H signal.

Exam focus

Explain the D₂O shake in two parts: deuterium replaces the O–H or N–H hydrogen, and deuterium does not produce a signal in ¹H NMR, so that peak disappears. Also be ready to explain why O–H signals appear as singlets.

Advanced insight

In very dry, pure samples, or in the solvent (CD₃)₂SO, exchange slows down and O–H protons can show coupling. For example, in dry dimethyl sulfoxide the OH of ethanol appears as a triplet and the CH₂ becomes a more complex multiplet. This allows primary, secondary and tertiary alcohols to be distinguished by the OH multiplicity (triplet, doublet and singlet respectively).

Summary

O–H and N–H protons exchange rapidly between molecules, so their signals are usually broad singlets with variable shifts, and they do not split neighbouring signals. Shaking the sample with D₂O replaces these protons with deuterium, which gives no ¹H signal, so the exchangeable peaks disappear and can be identified.

Practice questions

1. Why does the CH₂ signal of ethanol normally appear as a quartet rather than a more complex pattern? Answer: The O–H proton exchanges rapidly, so its coupling to the CH₂ averages to zero and only the three CH₃ protons split the CH₂ signal. 2. State what would be seen when D₂O is added to a sample of propylamine, CH₃CH₂CH₂NH₂. Answer: The broad NH₂ signal disappears and an HOD peak appears; the CH₃ and CH₂ signals are unchanged. 3. Why does deuterium not produce a peak in a ¹H NMR spectrum? Answer: Deuterium nuclei have a different magnetic moment and resonate at a very different frequency from ¹H nuclei, outside the range recorded. 4. Suggest why the shift of an alcohol O–H peak changes when the sample is diluted. Answer: Dilution reduces intermolecular hydrogen bonding, which increases shielding and moves the O–H signal to lower δ.