How to Use the n Plus One Rule Without Counting the Wrong Neighbours

A signal in a proton NMR spectrum has six lines. Does that mean the hydrogen group has five neighbouring hydrogens?

Possibly. But there’s a catch: those five hydrogens must be chemically equivalent neighbours for you to apply the simple \(n+1\) rule in one step.

This is where students often go wrong. They count every hydrogen on the nearby carbons, add them together, and apply \(n+1\). That works when all the neighbouring protons belong to one equivalent set. It can give the wrong answer when two different groups of protons are involved.

The reliable method is to choose one proton signal, identify the carbon atoms directly connected to the carbon bearing those protons, and count the neighbouring hydrogens in separate equivalent groups.

If there is one equivalent group containing \(n\) neighbouring hydrogens, the signal has \(n+1\) lines. If there are multiple non-equivalent groups, you need to consider their splitting separately.

Let’s build that method properly.

01First, separate the four types of information in a proton NMR spectrum

Before predicting splitting, make sure you know what the different parts of a spectrum actually tell you.

These features answer different questions.

For example, a quartet contains four individual lines, but it is still one signal. Its integration might be 2H, meaning that two equivalent hydrogen atoms contribute to that signal.

The integration measures the total area of the quartet, not the number of lines or the height of its tallest peak.

Chemical shift is not splitting

Chemical shift, represented by \(\delta\), describes where a signal appears on the horizontal axis.

It is measured in parts per million (ppm), relative to a reference substance, usually tetramethylsilane (TMS), assigned a value of 0 ppm.

On a conventional proton NMR spectrum:

• Higher positive chemical shifts appear further downfield, towards the left.
• Lower chemical shifts appear further upfield, towards the right.

A proton close to an electronegative oxygen atom is often less shielded and appears at a higher chemical shift than an ordinary alkyl proton.

However, chemical shift does not tell you how many neighbouring protons are present. That is what splitting helps you determine.

02What actually causes a signal to split?

An NMR spectrometer does not photograph hydrogen atoms or count the atoms surrounding a particular carbon. It measures the magnetic behaviour of nuclei in a strong external magnetic field.

It helps to follow the measurement in order.

1. A small sample of the compound is dissolved in a suitable solvent and placed in a narrow glass tube.
2. The tube is positioned inside a strong magnet.
3. A radiofrequency coil applies an electromagnetic pulse to the sample.
4. The hydrogen nuclei respond to the pulse and produce a changing magnetic signal as they return towards equilibrium.
5. A receiver detects this signal, and the instrument processes it to produce the NMR spectrum.

Why do neighbouring protons affect a signal?

Hydrogen-1 nuclei have nuclear spin, which gives them a magnetic property.

In the external magnetic field, neighbouring hydrogen nuclei can occupy different spin states. Their magnetic effects slightly change the effective magnetic field experienced by the proton group being observed.

Consequently, the observed group can experience several slightly different magnetic environments.

Instead of producing one resonance line, its signal splits into multiple lines.

This interaction is called spin-spin coupling.

Think of each neighbouring proton as providing a small magnetic nudge. Depending on its spin state, it can alter the local magnetic environment in one direction or the other.

The analogy has limits: nuclei are not literally tiny balls spinning like miniature compass needles. Their behaviour is quantum mechanical. However, the magnetic-nudge picture helps explain why neighbouring nuclei affect the observed signal.

The n+1 rule

For a simple first-order proton NMR spectrum, the number of lines in a signal is:

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

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

The common splitting patterns are:

The relative intensities follow Pascal’s triangle.

For a quartet, the four lines therefore have relative intensities of 1:3:3:1.

The central two lines are taller than the outer lines.

The important point is that \(n\) counts equivalent neighbouring protons, not the number of hydrogens in the molecule.

03How to count the correct neighbours

Use the following method whenever you are given a structural formula and asked to predict splitting.

Step 1: Choose the proton environment you are analysing

Do not try to predict the whole spectrum at once.

Choose one group, such as a \(\ce{CH3}\), \(\ce{CH2}\), or \(\ce{CH}\) group.

You are predicting the splitting of this particular group’s signal.

Step 2: Look at the carbon directly attached to your target carbon

This is the most important step.

For the introductory HSC model, count hydrogens attached to carbon atoms directly bonded to the carbon bearing your target protons.

These are usually the neighbouring protons involved in three-bond, or vicinal, coupling.

Do not count the hydrogens in your target group itself.

For example, consider:

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

If you are examining the \(\ce{CH3}\) group, look at the adjacent \(\ce{CH2}\) carbon.

It contains two hydrogens.

Therefore, the methyl signal is split into:

\[
n+1=2+1=3
\]

It is a triplet.

If you examine the \(\ce{CH2}\) group instead, its adjacent carbon is the \(\ce{CH3}\) carbon, containing three hydrogens.

Its signal is therefore a quartet.

Notice that the same bond connects the two groups, but each group counts the hydrogens on the other side.

Step 3: Check whether the neighbouring protons are equivalent

This is the step that prevents most incorrect predictions.

Suppose the target group is next to two different carbon groups:

• One neighbouring carbon contains three hydrogens.
• Another neighbouring carbon contains two hydrogens.

You might be tempted to add them:

\[
n=3+2=5
\]

Then predict a sextet.

But this is only justified if all five neighbouring protons are equivalent and produce the same coupling interaction.

If the two neighbouring groups have different chemical environments, treat them as separate sets.

The first group may split the signal into a quartet. The second may split each of those lines into a triplet.

The result is a more complicated multiplet, rather than a straightforward sextet.

Step 4: Apply the rule to each equivalent set

If you have one equivalent neighbouring group, apply \(n+1\) once.

If you have two distinct neighbouring groups, consider each coupling separately.

For two groups with different coupling constants, the number of ideal, fully resolved lines is the product of their individual multiplicities.

For example:

\[
4\times3=12\text{ lines}
\]

This would be a quartet of triplets, or a triplet of quartets, depending on the relative coupling constants and how the pattern is described.

You do not need to force every multiplet into a singlet, doublet, triplet, or quartet.

Step 5: Check the other spectral information

Finally, compare your prediction with:

• Integration: does the signal represent the expected number of hydrogen atoms?
• Chemical shift: is the position sensible for the group’s electronic environment?
• Other splitting patterns: do neighbouring groups show compatible coupling?

Two groups that split each other normally have the same coupling constant, \(J\).

This provides a useful cross-check when interpreting an unknown spectrum.

07Where the introductory n+1 rule can fail

The n+1 rule is extremely useful, but it is a simplified model.

It works best when:

• The observed protons form one chemically equivalent group.
• The neighbouring protons are equivalent within each coupling group.
• The spectrum follows first-order behaviour.
• The individual lines are sufficiently well resolved.

There are several situations where these assumptions need attention.

1. Different neighbouring groups

As we’ve seen, a proton group can interact with multiple sets of non-equivalent neighbouring protons.

Each set can produce its own splitting.

Do not combine their hydrogen counts into one value of \(n\) unless they are genuinely equivalent and have the same coupling interaction.

2. Exchangeable protons

Hydrogens in groups such as \(\ce{-OH}\) and \(\ce{-NH2}\) can exchange with other molecules.

This exchange can make their signals broad and can prevent their coupling with nearby protons from being observed.

For example, the hydroxyl hydrogen in an alcohol is often treated as a broad singlet in introductory questions.

You should not automatically count it as a neighbouring proton when predicting the splitting of an adjacent \(\ce{CH2}\) group.

However, under suitable conditions, slow exchange can allow OH coupling to be observed.

Always follow the conditions and experimental information provided.

3. Non-equivalent hydrogens within a CH2 group

A common introductory assumption is that the two hydrogens in a \(\ce{CH2}\) group are equivalent.

This is often appropriate for simple, flexible molecules.

However, a nearby stereogenic centre or restricted molecular movement can make the two hydrogens chemically different.

They may then produce separate signals and more complicated splitting.

This is an extension beyond the simplest n+1 model.

4. Strong coupling and complex spectra

The simple first-order rule also assumes that the chemical-shift difference between coupled groups is sufficiently large compared with their coupling constant.

When this condition is not satisfied, second-order effects can distort the expected line positions and relative intensities.

You may see patterns that do not match the neat Pascal’s triangle predictions.

Long-range coupling can also occur in some molecular structures, particularly those containing multiple bonds.

These effects explain why the n+1 rule is a prediction for suitable spectra, rather than a universal law for every proton in every molecule.

For a more advanced treatment, see OpenStax Organic Chemistry: Spin-Spin Splitting in ¹H NMR Spectra.

08Questions and solutions

Question 1

2,2-dimethylpropane has the structural formula:

\[
\ce{C(CH3)4}
\]

All four methyl groups are chemically equivalent.

Predict:

a) The number of proton environments.

b) The relative integration.

c) The splitting pattern of the proton signal.

Assume ordinary solution conditions and the introductory first-order model.

Solution 1

The molecule produces one proton environment, with an integration of 12H, and a singlet.

All four methyl groups have identical chemical surroundings, so all twelve hydrogen atoms contribute to the same signal.

Now consider splitting.

Each methyl carbon is directly attached to the central carbon.

That central carbon has no hydrogen atoms.

Therefore, the number of equivalent neighbouring hydrogens is:

\[
n=0
\]

Applying the rule:

\[
n+1=0+1=1
\]

The signal is a singlet.

The tempting mistake is to count the hydrogens in the other methyl groups as neighbours. However, equivalent protons do not split one another, and the central carbon carries no hydrogen atoms.

The final result is:

• Signal count: 1.
• Integration: 12H.
• Splitting: singlet.

Question 2

Consider 2-methylbutane:

\[
\ce{CH3-CH(CH3)-CH2-CH3}
\]

Focus on the central methine (\(\ce{CH}\)) proton.

A student counts eight neighbouring hydrogens:

• Six hydrogens from the two equivalent methyl groups.
• Two hydrogens from the adjacent methylene group.

The student predicts a nonet using:

\[
n+1=8+1=9
\]

Explain why this prediction is incorrect.

Assume that the two coupling constants are different and the resulting lines are fully resolved. Predict the correct multiplicity and total number of lines.

Solution 2

The central methine signal produces a septet of triplets, containing 21 lines under the stated assumptions.

The student has correctly counted eight neighbouring hydrogens but incorrectly assumed they all belong to one equivalent group.

Separate the two sets.

The first set: the two methyl groups

The two methyl groups attached to the central methine carbon are equivalent.

Together, they contain six equivalent hydrogen atoms.

These six hydrogens split the methine signal into a septet:

\[
n+1=6+1=7
\]

The second set: the methylene group

The adjacent \(\ce{CH2}\) group contains two hydrogens.

These are chemically different from the six methyl hydrogens.

They split each line of the septet into a triplet:

\[
n+1=2+1=3
\]

Combine the patterns

The total number of resolved lines is:

\[
7\times3=21
\]

Therefore, the methine signal is a septet of triplets, or a triplet of septets depending on the relative coupling constants.

A nonet would be appropriate only if all eight neighbouring hydrogens formed one equivalent coupling set with the same coupling interaction.

The important distinction is that counting eight hydrogens is not enough. You must establish how many equivalent neighbouring groups those hydrogens belong to.

Question 3

An unknown organic compound has the molecular formula:

\[
\ce{C5H10O2}
\]

Two possible structures are being considered:

Candidate A: Isopropyl ethanoate

\[
\ce{CH3COOCH(CH3)2}
\]

Candidate B: Ethyl propanoate

\[
\ce{CH3CH2COOCH2CH3}
\]

An illustrative proton NMR spectrum produces the following data:

Assume that all signals are fully resolved, the shifts are illustrative, and the spectrum follows the simple first-order model.

Identify the correct structure.

Assign each signal to its corresponding proton group and explain how the integration, chemical shift, and splitting support your answer.

Solution 3

The unknown compound is isopropyl ethanoate, Candidate A.

Its structure produces three proton environments with an integration ratio of \(6:1:3\), matching the supplied data.

Consider each signal separately.

Signal A: 1.20 ppm, doublet, 6H

The two equivalent methyl groups attached to the central methine carbon contribute six hydrogen atoms.

Each methyl group is next to one methine hydrogen.

Therefore, the six methyl hydrogens produce one doublet signal.

Their relatively low chemical shift is consistent with ordinary alkyl protons.

Signal B: 4.95 ppm, septet, 1H

The central methine group contains one hydrogen atom.

It is attached to an oxygen atom through the ester structure, making it relatively deshielded.

Its two neighbouring methyl groups contain six equivalent hydrogen atoms altogether.

Therefore:

\[
n+1=6+1=7
\]

The methine signal is a septet.

Signal C: 2.05 ppm, singlet, 3H

The \(\ce{CH3CO-}\) group contains three hydrogen atoms.

Its adjacent carbonyl carbon has no hydrogen atoms.

Therefore, there are no ordinary adjacent carbon-bound hydrogen neighbours to split this signal.

It appears as a singlet.

Its chemical shift is consistent with a methyl group adjacent to a carbonyl group.

Why does Candidate B not fit?

Ethyl propanoate contains two different ethyl groups:

\[
\ce{CH3CH2COOCH2CH3}
\]

The ethyl group beside the carbonyl and the ethyl group attached through oxygen have different chemical environments.

Its expected introductory splitting pattern would contain two triplets and two quartets, with relative integrations of:

\[
3:2:2:3
\]

That does not match the three signals in the supplied spectrum.

The evidence therefore supports isopropyl ethanoate.

Notice how the answer comes from combining several features rather than relying on the septet alone.

Question 4

Consider 2-methylpropan-1-ol:

\[
\ce{(CH3)2CHCH2OH}
\]

An illustrative proton NMR spectrum contains:

• One 6H doublet.
• One 2H doublet.
• One 1H complex multiplet.
• One broad OH signal.

Assume rapid hydroxyl proton exchange and that the two hydrogens in the \(\ce{CH2}\) group are equivalent.

A student argues that the \(\ce{CH2OH}\) signal should be a triplet because it has two neighbouring hydrogens: one on the adjacent carbon and one in the OH group.

a) Explain why the student’s prediction is incorrect under the stated conditions.

b) Predict the splitting of the central methine proton if both neighbouring coupling interactions are resolved and have different coupling constants.

c) Explain why the OH signal may not integrate accurately.

Solution 4

The \(\ce{CH2OH}\) signal is a doublet, not a triplet, under the stated rapid-exchange conditions. The central methine signal is a complex multiplet.

Part a: The \(\ce{CH2OH}\) group

The \(\ce{CH2}\) carbon is directly bonded to the central methine carbon.

That carbon contains one hydrogen.

The OH hydrogen is attached to oxygen, not to the adjacent carbon.

More importantly, the question specifies rapid OH exchange. This generally prevents its coupling with the neighbouring \(\ce{CH2}\) group from being resolved.

Therefore, the \(\ce{CH2}\) group has one observable equivalent neighbouring hydrogen:

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

Its signal is a doublet.

The student’s mistake is treating every hydrogen connected through the nearby atoms as an observable splitting neighbour.

Part b: The central methine group

The central \(\ce{CH}\) is next to two different proton groups:

• Two equivalent methyl groups containing six hydrogen atoms.
• One \(\ce{CH2OH}\) group containing two hydrogen atoms.

The six methyl hydrogens split the methine signal into a septet.

The two methylene hydrogens split it into a triplet.

Because the coupling constants are different and both interactions are resolved, the expected pattern is a septet of triplets, containing:

\[
7\times3=21\text{ lines}
\]

It is not simply a nonet.

Part c: The OH signal

The hydroxyl hydrogen can exchange with other hydrogen-containing molecules, including traces of water.

This exchange can broaden the signal and make its integration unreliable.

Consequently, the measured OH signal may not correspond neatly to one hydrogen in the integration data.

This is why exchangeable hydrogens require special care when interpreting proton NMR spectra.

Question 5

An unknown compound produces a signal with:

• Integration of 2H.
• Six equally spaced lines.
• Relative line intensities of \(1:5:10:10:5:1\).

A student concludes that the observed \(\ce{CH2}\) group must have five chemically equivalent neighbouring hydrogens.

Is this conclusion necessarily correct?

Consider the middle \(\ce{CH2}\) group in 1-propanol:

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

Assume:

• The two hydrogen atoms in the middle \(\ce{CH2}\) group are equivalent.
• The three hydrogens in the adjacent \(\ce{CH3}\) group are equivalent.
• The two hydrogens in the adjacent \(\ce{CH2OH}\) group are equivalent.
• The two neighbouring groups have exactly the same coupling constant with the middle \(\ce{CH2}\) group.
• The hydroxyl proton undergoes rapid exchange.

Explain whether this structure could produce the observed six-line pattern.

Your answer must address:

a) Why the five neighbouring hydrogens do not form one chemically equivalent group.

b) How the two separate splitting patterns combine.

c) Why observing a sextet does not, by itself, prove that five chemically equivalent neighbouring hydrogens are present.

Solution 5

No. The six-line pattern does not prove that five chemically equivalent neighbouring hydrogens are present.

The middle \(\ce{CH2}\) group in 1-propanol provides an alternative explanation, even though its five neighbouring hydrogens belong to two different chemical environments.

Part a: Identify the two neighbouring groups

The middle \(\ce{CH2}\) group is attached to:

• A \(\ce{CH3}\) carbon containing three equivalent hydrogens.
• A \(\ce{CH2OH}\) carbon containing two equivalent hydrogens.

The methyl and hydroxymethyl groups have different chemical environments.

Therefore, the five neighbouring hydrogens are not one chemically equivalent set.

Part b: Consider the two splitting interactions separately

The three methyl hydrogens split the middle \(\ce{CH2}\) signal into a quartet.

The relative intensities are:

\[
1:3:3:1
\]

The two hydrogens in the \(\ce{CH2OH}\) group split it into a triplet.

The relative intensities are:

\[
1:2:1
\]

Normally, if the two coupling constants differ, these interactions produce twelve separate lines.

However, the question specifies that the two coupling constants are exactly equal.

This changes what we observe.

The quartet and triplet have the same line spacing, so their individual lines overlap at the same frequency positions.

When the intensities combine, the resulting pattern is:

\[
\begin{aligned}
1:3:3:1\\
1:2:1
\end{aligned}
\]

Combining the overlapping intensities gives:

\[
1:5:10:10:5:1
\]

There are now six distinct line positions.

The resulting pattern is identical in line count and relative intensity to the simple sextet expected from five equivalent neighbouring protons.

Part c: Explain the limitation

The key distinction is between chemical equivalence and equal coupling constants.

The methyl hydrogens and hydroxymethyl hydrogens are chemically different, even though they have the same coupling constant with the middle \(\ce{CH2}\) group.

Equal coupling constants can make two separate splitting patterns overlap.

Therefore, a sextet does not uniquely establish that five equivalent neighbouring hydrogens are present.

To investigate the structure, you would also examine the neighbouring signals.

The terminal methyl group is coupled to the middle \(\ce{CH2}\) group, and the \(\ce{CH2OH}\) group is also coupled to it. Their different chemical shifts and integrations provide additional evidence that two distinct neighbouring groups are involved.

This is the limitation of relying on line count alone: different molecular arrangements can produce the same apparent multiplet.

09What to learn next

The n+1 rule becomes much more useful when you combine it with integration and chemical shift.

Once you can identify the correct neighbouring protons, you can use the splitting patterns to recognise common structural fragments and test possible molecular structures.

For the foundation behind identifying the different hydrogen environments, read How to Count Proton Environments from a Structural Formula.

Then practise combining integration and splitting to identify an ethyl group in How to Identify an Ethyl Group Using Proton NMR.

The next time you see a quartet, septet, or sextet, don’t start by counting lines. Start with the structure, choose the proton group, and identify exactly which neighbouring hydrogens can split it.