Why OH Protons Do Not Always Follow NMR Splitting Rules

Take ethanol, \(\ce{CH3CH2OH}\). The two hydrogen atoms in its \(\ce{CH2}\) group are next to the oxygen atom, so you might expect them to interact with the hydrogen in the \(\ce{OH}\) group. Surely the OH proton should produce a triplet, and the \(\ce{CH2}\) signal should receive extra splitting?

Usually, neither happens.

In a typical \(^{1}\mathrm{H}\) NMR spectrum, ethanol produces a methyl triplet, a methylene quartet, and a broad OH singlet. The reason is proton exchange: the hydrogen attached to oxygen can transfer between molecules. When this exchange happens rapidly, the magnetic interaction between the OH proton and neighbouring hydrogen atoms is averaged out.

The important point is that the \(n+1\) rule has assumptions. An OH proton does not always behave like a hydrogen attached to carbon, and its observed splitting depends on the conditions of the NMR experiment.

01The ethanol puzzle: why doesn’t OH produce a triplet?

Start with ethanol’s structure:

\[ \ce{CH3-CH2-OH} \]

There are three distinct proton environments:

• Three hydrogen atoms in the \(\ce{CH3}\) group.
• Two hydrogen atoms in the \(\ce{CH2}\) group.
• One hydrogen atom in the \(\ce{OH}\) group.

If you apply the introductory splitting rules mechanically, you might make these predictions:

• The \(\ce{CH3}\) signal is a triplet because it has two neighbouring \(\ce{CH2}\) hydrogen atoms.
• The \(\ce{CH2}\) signal is a quartet because it has three neighbouring \(\ce{CH3}\) hydrogen atoms.
• The OH signal is a triplet because it has two neighbouring \(\ce{CH2}\) hydrogen atoms.

The third prediction seems perfectly reasonable.

However, it ignores something important: the OH proton can exchange with other hydrogen-containing molecules.

What is proton exchange?

An alcohol contains an O-H bond. The hydrogen attached to oxygen is called an exchangeable proton because it can transfer to another molecule.

For example, an ethanol molecule can exchange its OH proton with:

• A water molecule.
• Another ethanol molecule.
• Other suitable proton-donating or proton-accepting species in the sample.

Trace amounts of water, acids, and bases can influence how quickly this exchange happens.

The important question is how quickly the proton exchanges compared with the timescale of the magnetic coupling interaction.

Think of a long-exposure photograph of a moving object. If the object moves quickly, its position becomes blurred into an average rather than appearing at one fixed location.

Something similar happens in NMR. When the OH proton exchanges rapidly, its magnetic interaction with a particular neighbouring \(\ce{CH2}\) group is averaged out.

As a result, the expected coupling is no longer resolved.

This gives us the usual observation:

• The OH signal appears as a broad singlet, or may be too weak to identify clearly.
• The neighbouring \(\ce{CH2}\) signal is not split by the OH proton.
• The \(\ce{CH2}\) signal remains a quartet due to the three \(\ce{CH3}\) hydrogen atoms.

The analogy describes the averaging, not the physical movement of a proton through the NMR instrument.

Does OH always appear as a singlet?

No. This is where the introductory model needs another layer of precision.

If proton exchange becomes sufficiently slow, the OH proton can remain associated with an individual ethanol molecule long enough for its coupling to the neighbouring \(\ce{CH2}\) hydrogen atoms to become observable.

Under suitable dry, acid-free conditions, the OH signal can therefore become a triplet.

The \(\ce{CH2}\) signal can also become more complicated because it couples to both the \(\ce{CH3}\) group and the OH proton.

We can compare the possibilities:

These are simplified predictions for ethanol. The slow-exchange patterns assume that the relevant coupling interactions are resolved and that the hydrogen atoms within each carbon-bound group are equivalent.

Notice what has changed: the splitting pattern.

The molecular structure has not changed. Ethanol still has three proton environments.

This is why you should not assume that every OH proton must produce a particular multiplicity.

02Four different features you must keep separate

Before going further, make sure you understand what each part of a proton NMR spectrum tells you.

These four features answer different questions.

1. Signal count: how many proton environments?

A signal represents a group of chemically equivalent hydrogen atoms.

Ethanol normally has three proton environments:

\[ \ce{CH3},\quad \ce{CH2},\quad \ce{OH} \]

Therefore, three signals are expected if the OH signal is observable and distinguishable.

However, an OH signal can be broad, weak, or difficult to detect. You might therefore see only two clearly identifiable signals in an ordinary spectrum.

Remember that a triplet is still one signal. It contains three lines, but those lines belong to the same proton environment.

If you need to revise how equivalent hydrogen atoms are grouped into environments, see How to Count Proton Environments from a Structural Formula.

2. Integration: how many hydrogen atoms contribute?

Integration measures the area underneath a signal.

For ethanol, the expected hydrogen ratio is:

\[ 3:2:1 \]

This represents:

• Three hydrogen atoms in \(\ce{CH3}\).
• Two hydrogen atoms in \(\ce{CH2}\).
• One hydrogen atom in \(\ce{OH}\).

However, the OH integration may not match the expected value.

Why?

Rapid exchange can broaden the OH signal. A broad signal can be difficult to integrate accurately because its edges blend into the baseline.

Water contamination and overlapping signals can make the problem worse.

For this reason, the integration of carbon-bound hydrogen atoms is often more reliable than the integration of an exchangeable OH proton.

Important: An OH signal integrating to 0.5 relative units does not mean the molecule contains half a hydrogen atom. It means the measured area may not accurately represent the expected proton count.

3. Chemical shift: where does the signal appear?

Chemical shift is represented by \(\delta\) and measured in parts per million (ppm).

It describes the electronic environment around a hydrogen nucleus.

In a conventional proton NMR spectrum:

• Higher ppm values appear further downfield, towards the left.
• Lower ppm values appear further upfield, towards the right.
• Tetramethylsilane (TMS) is commonly used as the reference at 0 ppm.

The OH chemical shift is particularly variable.

Hydrogen bonding, solvent, concentration, and temperature can all influence its position.

For example, an OH proton might appear around 2 ppm in one sample and at a noticeably different position in another.

Therefore, do not expect every alcohol to have its OH signal at one fixed chemical shift.

4. Splitting: how many neighbouring hydrogen atoms couple?

For simple introductory proton NMR, the \(n+1\) rule predicts:

\[ \text{Number of lines}=n+1 \]

Here, \(n\) is the number of equivalent neighbouring hydrogen atoms that couple to the proton environment being examined.

For example:

The rule works well for many ordinary carbon-bound hydrogen environments.

However, it assumes that the relevant coupling interactions are observable and that the splitting can be treated using the simple introductory model.

For OH protons, rapid exchange can prevent the coupling from being observed.

That is why you should not automatically include an OH proton when counting neighbouring hydrogen atoms.

05A contrasting case: OH does not always become a triplet

There is another important detail.

Even when proton exchange is slow, an OH proton does not automatically produce a triplet.

The number of neighbouring hydrogen atoms still matters.

Compare three alcohols:

Why are the results different?

In ethanol, the carbon bonded to oxygen has two hydrogen atoms. The OH proton can couple to both, producing a triplet.

In propan-2-ol, the carbon bonded to oxygen has only one hydrogen atom. Under slow-exchange conditions, the OH proton can therefore produce a doublet.

In 2-methylpropan-2-ol, the carbon bonded to oxygen has no hydrogen atoms.

There is no usual three-bond OH-to-carbon-bound-hydrogen coupling available through the adjacent carbon. Consequently, a singlet is expected in the simple introductory model, even if proton exchange is slow.

Any longer-range coupling is outside this simplified prediction.

This gives you an important decision rule:

A singlet OH signal does not automatically prove rapid proton exchange.

First check whether there are any suitable neighbouring hydrogen atoms available to produce coupling.

06How can an experiment confirm that a signal is from OH?

Suppose you are given an unknown proton NMR spectrum containing a broad signal.

How can you check whether it comes from an exchangeable OH proton?

One useful method is the deuterium oxide test, commonly called the D2O shake test.

The experimental setup

A solution of the compound is placed in a narrow NMR tube.

The tube is positioned inside the strong magnetic field of the NMR spectrometer. A radiofrequency coil surrounding the sample excites the hydrogen nuclei and detects the resulting magnetic signal.

First, the spectrum is recorded under the original conditions.

A small amount of deuterium oxide, \(\ce{D2O}\), is then added to the sample. The tube is mixed, returned to the instrument, and another proton NMR spectrum is recorded.

The key is to compare the same sample before and after adding \(\ce{D2O}\).

What should happen?

If the broad signal comes from an exchangeable OH proton, the hydrogen can be replaced by deuterium.

The simplified exchange reaction is:

\[ \ce{ROH + D2O <=> ROD + HOD} \]

Here:

• \(\ce{ROH}\) represents an alcohol.
• \(\ce{ROD}\) represents the alcohol after its OH hydrogen has been replaced by deuterium.
• \(\ce{HOD}\) is water containing one hydrogen and one deuterium atom.

A standard proton NMR experiment detects hydrogen-1 nuclei, not deuterium-2 nuclei.

Therefore, after exchange, the original OH signal disappears from the \(^{1}\mathrm{H}\) spectrum.

The carbon-bound hydrogen signals remain.

What does the experiment actually prove?

The disappearance of a signal supports the conclusion that it came from an exchangeable hydrogen.

However, it does not uniquely prove the presence of an alcohol.

Other groups, including \(\ce{-NH2}\) and \(\ce{-COOH}\), can also contain exchangeable hydrogen atoms.

There is another limitation.

If the OH signal was already too broad or weak to identify before adding \(\ce{D2O}\), its disappearance cannot be observed reliably.

The D2O test is therefore supporting evidence, not a complete method for identifying an unknown compound.

For a more detailed explanation of the exchange behaviour and D2O test, see Chemistry LibreTexts’ discussion of alcohol proton NMR.

07A practical decision process for OH splitting

When interpreting an OH signal in an HSC-style proton NMR question, use this sequence.

1. Identify the proton environments.

Count the distinct hydrogen environments from the structural formula. Include the OH proton if it is present, even if its signal is difficult to observe.

2. Examine the integration.

Use the carbon-bound signals to establish the more reliable hydrogen ratios.

Treat OH integration cautiously because exchange and broadening can affect its measured area.

3. Examine the chemical shift.

Remember that the OH position is variable. Do not rely on one fixed ppm value to identify it.

4. Examine the splitting.

For a typical introductory spectrum, assume rapid OH exchange unless the question gives evidence of slower exchange.

Do not automatically include the OH proton when applying the \(n+1\) rule to neighbouring carbon-bound hydrogen atoms.

5. Check for unusual conditions.

If the question describes a very dry sample, slow exchange, or a solvent that suppresses exchange, OH coupling may become observable.

In that case, count the equivalent hydrogen atoms on the carbon bonded to oxygen and consider whether reciprocal splitting of the carbon-bound signal is also expected.

The key is to use the experimental conditions provided rather than applying one splitting pattern to every alcohol.

08Questions and solutions

The following questions progress from ordinary OH behaviour to more demanding situations involving exchange, coupling, integration, and mixtures.

All numerical chemical shifts and coupling constants are illustrative unless otherwise stated.

Question 1

A typical \(^{1}\mathrm{H}\) NMR spectrum of propan-2-ol, \(\ce{(CH3)2CHOH}\), contains:

• A doublet integrating to 6H.
• A septet integrating to 1H.
• A broad OH signal.

Predict the usual splitting pattern of the OH signal.

Explain why the central \(\ce{CH}\) signal is normally a septet rather than a septet of doublets.

Solution 1

The OH signal is usually a broad singlet, and the central \(\ce{CH}\) signal remains a septet.

The two methyl groups in propan-2-ol are equivalent. Together, they contain six equivalent hydrogen atoms.

These six hydrogen atoms split the central \(\ce{CH}\) signal:

\[ 6+1=7 \]

Therefore, the central methine signal is a septet.

The central \(\ce{CH}\) proton also sits next to the oxygen atom through the C-O-H arrangement. Under sufficiently slow exchange, it could couple to the OH proton.

However, in a typical sample, rapid OH exchange prevents this coupling from being resolved.

Consequently, the central signal remains a septet rather than becoming a septet of doublets.

The methyl signal is a doublet because each methyl group is adjacent to one methine hydrogen.

Question 2

A very dry, acid-free sample of propan-2-ol is analysed under conditions where OH exchange is sufficiently slow for coupling to be observed.

The following coupling constants are measured:

• Coupling between the central \(\ce{CH}\) proton and the six methyl hydrogen atoms: \(7.0\ \mathrm{Hz}\).
• Coupling between the central \(\ce{CH}\) proton and the OH proton: \(4.5\ \mathrm{Hz}\).

The spectrometer operates at 500 MHz.

Determine:

a) The splitting pattern of the OH signal.

b) The splitting pattern and number of lines in the central \(\ce{CH}\) signal, assuming all coupling interactions are resolved.

c) The separation between adjacent OH doublet lines in ppm.

Solution 2

The OH signal is a doublet, the central methine signal is a septet of doublets containing 14 lines, and the OH line separation is \(0.0090\ \mathrm{ppm}\).

Part a: OH splitting

The OH proton couples to one neighbouring methine hydrogen.

Therefore:

\[ n+1=1+1=2 \]

The OH signal is a doublet.

Part b: Central methine splitting

The central \(\ce{CH}\) proton couples to six equivalent methyl hydrogen atoms.

This produces a septet:

\[ 6+1=7 \]

It also couples to one OH proton, which splits each septet line into two.

Therefore, the overall pattern is a septet of doublets.

The total number of lines is:

\[ 7\times2=14 \]

These 14 lines still belong to one proton signal representing the central \(\ce{CH}\) environment.

Part c: OH line separation

Use:

\[ \Delta\delta= \frac{J}{\nu_0} \]

Substitute the supplied values:

\[ \begin{aligned} \Delta\delta&=\frac{4.5}{500}\\ &=0.0090\ \mathrm{ppm} \end{aligned} \]

The adjacent OH doublet lines are separated by \(0.0090\ \mathrm{ppm}\).

This example demonstrates that slow exchange can introduce additional splitting without creating additional proton environments.

Question 3

A sample of 2-methylpropan-2-ol, \(\ce{(CH3)3COH}\), is prepared under very dry conditions.

Its OH signal appears as a singlet.

A student argues:

“The OH signal must still be undergoing rapid proton exchange. If exchange were slow, it would split into a multiplet.”

Evaluate this statement.

Solution 3

The student’s conclusion is not justified. The OH signal can remain a singlet even when proton exchange is slow.

The key is the structure of 2-methylpropan-2-ol:

\[ \ce{(CH3)3C-OH} \]

The carbon directly bonded to oxygen is attached to three methyl groups.

It has no hydrogen atoms of its own.

Therefore, there are no neighbouring carbon-bound hydrogen atoms available for the usual three-bond OH-to-C-H coupling.

The methyl hydrogen atoms are further away and do not normally produce resolved OH splitting in the introductory model.

Consequently, the OH signal can remain a singlet even when exchange is sufficiently slow.

The mistake is assuming that a singlet always indicates rapid exchange.

You must first check whether the molecular structure provides suitable neighbouring hydrogen atoms for coupling.

Question 4

An unknown alcohol produces the following illustrative \(^{1}\mathrm{H}\) NMR data.

After a small amount of \(\ce{D2O}\) is added, the broad signal disappears. The other three signals remain.

a) Suggest the most likely alcohol.

b) Explain why the broad signal integrates to only 0.3 rather than 1.0.

c) Explain what the D2O experiment establishes and identify one limitation of this conclusion.

Solution 4

The most likely alcohol is propan-1-ol, \(\ce{CH3CH2CH2OH}\). The broad signal is consistent with an exchangeable OH proton, and its low integration is not unexpected.

Part a: Identify the alcohol

Consider the structural formula:

\[ \ce{CH3-CH2-CH2-OH} \]

It contains three proton environments attached to carbon:

• A terminal \(\ce{CH3}\) group.
• A central \(\ce{CH2}\) group.
• An oxygen-bound \(\ce{CH2}\) group.

The terminal methyl group is split by the two central methylene hydrogen atoms, producing a triplet.

The central methylene group experiences coupling with hydrogen atoms on both neighbouring carbon groups, producing a more complicated multiplet.

The oxygen-bound methylene group is split by the two hydrogen atoms on the adjacent central carbon, producing a triplet.

These assignments match the supplied data.

The broad signal is consistent with the OH proton.

Part b: Explain the low integration

The OH proton is exchangeable.

Rapid exchange can broaden the signal, making its baseline difficult to identify and its area difficult to measure accurately.

Therefore, an integration of 0.3 does not contradict the expected presence of one OH hydrogen.

The carbon-bound hydrogen integrations provide a more reliable basis for interpreting the structure.

Part c: Interpret the D2O experiment

The disappearance of the broad signal after adding \(\ce{D2O}\) supports the assignment of an exchangeable proton.

The OH hydrogen has been replaced by deuterium, which is not detected in an ordinary \(^{1}\mathrm{H}\) NMR spectrum.

However, the experiment does not uniquely prove that the exchangeable proton belongs to an alcohol.

Other functional groups, such as amines and carboxylic acids, can also contain exchangeable hydrogen atoms.

The remaining carbon-bound signals provide the additional evidence supporting the proposed propan-1-ol structure.

Question 5

An unknown sample is found to contain two sets of carbon-bound proton environments.

The first set is consistent with ethanol:

\[ \ce{CH3CH2OH} \]

The second set is consistent with propan-1-ol:

\[ \ce{CH3CH2CH2OH} \]

Under very dry conditions and at a sufficiently low temperature, the sample produces two separate OH signals:

Both OH signals show coupling to their respective oxygen-bound \(\ce{CH2}\) groups.

A small amount of water is then added, and the sample is warmed slightly.

The new spectrum shows:

• One broad OH envelope instead of two separate OH signals.
• No resolved OH coupling to either \(\ce{CH2}\) group.
• The same carbon-bound signal positions and relative integrations as before.
• No new carbon-bound signals.

Assume that the added water increases the rate of intermolecular proton exchange sufficiently to average the two OH environments. There is no reagent or reaction condition intended to change either carbon skeleton.

A student concludes:

“The sample must have changed into one alcohol because the two OH signals have become one. The original two OH environments must have become chemically equivalent.”

Evaluate this conclusion.

Also explain whether a D2O shake would establish whether the sample contains one or two alcohols.

Solution 5

The student’s conclusion is incorrect. The sample still contains two different alcohols. The two OH signals have merged because rapid proton exchange averages their magnetic environments, not because the two molecular structures have become identical.

This problem requires you to distinguish between the behaviour of exchangeable protons and the behaviour of carbon-bound protons.

Step 1: Interpret the original dry spectrum

Under slow-exchange conditions, the OH proton in ethanol can couple to the two equivalent hydrogen atoms in its neighbouring \(\ce{CH2}\) group.

Therefore:

\[ 2+1=3 \]

Its OH signal is a triplet.

The same applies to propan-1-ol.

Its OH proton is also next to an oxygen-bound \(\ce{CH2}\) group containing two hydrogen atoms.

It therefore also produces a triplet.

The two different chemical environments produce two separate OH signals because their chemical shifts are distinguishable under the original conditions.

Step 2: Explain what happens after adding water

Adding water increases the rate of proton exchange.

The OH protons can now transfer between alcohol molecules and water molecules more rapidly.

Two effects follow.

First, the individual OH protons no longer remain associated with one particular alcohol molecule long enough for their original coupling interactions to be resolved.

The OH-to-\(\ce{CH2}\) coupling disappears.

Second, the two OH chemical environments are averaged by rapid exchange.

Instead of two separate resonances, the spectrum can display one broad, averaged OH envelope.

This is a change in the observed NMR behaviour, not evidence of a change in molecular structure.

Step 3: Check the carbon-bound signals

This is the crucial constraint in the question.

The carbon-bound signals retain their original positions and relative integrations.

That means the proton environments associated with the ethanol and propan-1-ol carbon skeletons remain distinguishable.

The two sets of signals have not disappeared or been replaced by a new set.

Therefore, the evidence still supports the presence of both alcohols.

A single OH envelope does not establish that there is only one alcohol species.

Step 4: Evaluate the D2O test

Adding \(\ce{D2O}\) would allow the exchangeable OH hydrogen atoms in both alcohols to be replaced by deuterium.

Both original OH contributions would consequently disappear from the ordinary proton NMR spectrum.

However, this would not establish whether the original sample contained one alcohol or two.

The carbon-bound signals would still be needed to distinguish the molecular structures.

The deeper lesson is that rapid exchange can change the number of distinguishable resonances without changing the number of compounds present.

A signal count is an observation made under particular experimental conditions. It is not, by itself, a guaranteed count of all the chemically distinct species in a sample.

09What to remember when interpreting an OH signal

The OH proton does not have one guaranteed splitting pattern.

Its observed behaviour depends on two things:

1. The number of suitable neighbouring hydrogen atoms available for magnetic coupling.
2. The rate of proton exchange under the experimental conditions.

For a typical introductory spectrum, expect a broad OH singlet because rapid exchange usually prevents observable coupling.

If exchange is sufficiently slow, the OH proton can split neighbouring carbon-bound hydrogen signals and can itself become a multiplet.

However, a singlet does not automatically prove rapid exchange. The molecular structure must also provide suitable neighbouring hydrogen atoms.

Once you have accounted for OH exchange, use the more reliable carbon-bound integrations and splitting patterns to identify structural fragments. In particular, recognising the characteristic 3H triplet and 2H quartet of an ethyl group is a useful next step when interpreting an unknown proton NMR spectrum.

For that skill, see How to Identify an Ethyl Group Using Proton NMR.