Why Different Structures Can Fit the Same Proton NMR Clue

Several different structures can fit one NMR clue because each clue tells you something different about a molecule, and none usually describes the entire structure. A signal at a particular chemical shift might suggest a certain functional group, but several structures can contain that group. Three signals tell you how many distinct hydrogen environments exist, not how those environments are connected. Even a triplet or a 3H integration can fit more than one molecule. The trick is to understand exactly what each piece of NMR evidence tells you, and what it cannot tell you on its own.

## Why one NMR clue is rarely enough

Imagine you are given a proton NMR spectrum with three signals. You might immediately suggest a molecule with three different hydrogen environments.

That's a reasonable starting point. But you haven't identified the molecule yet.

Consider two possible structures:

- Propan-1-ol: \(\mathrm{CH_3CH_2CH_2OH}\)
- Ethyl methyl ether: \(\mathrm{CH_3CH_2OCH_3}\)

Both contain three distinct proton environments under the usual introductory treatment. However, their hydrogen environments are not chemically identical, and their chemical shifts and splitting patterns provide further evidence that can help distinguish them.

[IMAGE: Side-by-side structural diagrams of propan-1-ol and ethyl methyl ether, showing their three distinct proton environments with matching colour highlights for each environment, without implying that corresponding colours represent equivalent chemical environments between the two molecules.]

This illustrates the central problem: a single observation can be consistent with multiple structures. You need to combine independent clues before deciding which structure is most plausible.

The four main types of information in proton NMR are:

| NMR feature | What it tells you | What it does not tell you by itself |
|---|---|---|
| Signal count | Number of distinct proton environments | How many hydrogens are in each environment |
| Integration | Relative number of hydrogens contributing to each signal | Which atoms those hydrogens are attached to |
| Chemical shift | Approximate electronic environment of the hydrogens | One unique functional group or structure |
| Splitting | Information about neighbouring, coupling hydrogens | The complete connectivity of the molecule |

These clues work together, but they answer different questions. Mixing them up is one of the easiest ways to eliminate a correct structure or accept an incorrect one.

## 1. Signal count: how many different environments?

A proton NMR signal represents hydrogens in the same chemical environment.

For example, consider ethane, \(\mathrm{CH_3CH_3}\). Its six hydrogen atoms are equivalent in the usual introductory model. They produce one proton NMR signal.

Now consider propane, \(\mathrm{CH_3CH_2CH_3}\). The two end methyl groups are equivalent, but the middle methylene group is different. Propane therefore produces two proton environments and two signals.

[IMAGE: Structural comparison of ethane and propane. Highlight all six equivalent hydrogens in ethane as one group; in propane, highlight the two equivalent terminal methyl groups together and the central CH2 hydrogens separately.]

The important word is *distinct*. Six hydrogens do not necessarily produce six signals. What matters is how many chemically different environments they occupy.

However, the introductory rule has a limitation. Equivalence depends on the molecule's symmetry and, more precisely, on whether replacing individual hydrogens would produce chemically distinguishable molecules. In more advanced chemistry, stereochemical relationships and molecular motion can complicate this analysis.

For standard introductory HSC questions, focus on identifying equivalent proton environments from the structural formula and the expected molecular symmetry.

**The key limitation:** signal count does not tell you the number of hydrogens in each environment. A molecule with three signals could have a ratio of 3:2:1, 3:3:3, or another ratio altogether.

That is why integration is a separate clue.

02Integration: how many hydrogens contribute to each signal?

Signal count tells you how many different environments exist. Integration tells you how many hydrogens belong to each environment.

Integration is based on the area underneath a signal. A signal representing three equivalent hydrogens has three times the integrated area of a signal representing one hydrogen, assuming the spectrum is properly acquired and processed.

For example, suppose an unknown compound produces three signals with integration values of:

\[ 1.5:1.0:1.5 \]

Divide every value by the smallest value:

\[ \frac{1.5}{0.5}: \frac{1.0}{0.5}: \frac{1.5}{0.5} =3:2:3 \]

The relative numbers of hydrogens are therefore 3:2:3.

Notice that the ratio doesn’t automatically tell you the absolute number of hydrogens.

A ratio of 3:2 could represent:

• One \(\ce{CH3}\) group and one \(\ce{CH2}\) group.
• Two equivalent \(\ce{CH3}\) groups and two equivalent \(\ce{CH2}\) groups, giving 6H:4H.
• Another combination with the same relative proportions.

You need the molecular formula, or other information about the molecule, to determine the actual numbers.

A common integration trap

Suppose an unknown compound has the formula \(\ce{C4H10O}\), meaning it contains ten hydrogen atoms.

Its proton NMR spectrum has just two signals, with an integration ratio of 3:2.

A student might immediately suggest one methyl group and one methylene group, accounting for only five hydrogens.

But the molecule contains ten.

The correct integration must be scaled:

\[ 3:2 \longrightarrow 6:4 \]

This could indicate two equivalent ethyl groups, each containing one \(\ce{CH3}\) and one \(\ce{CH2}\).

Diethyl ether, \(\ce{CH3CH2OCH2CH3}\), is a good example. Its two ethyl groups are equivalent, so the molecule produces only two proton environments:

• Six equivalent methyl hydrogens.
• Four equivalent methylene hydrogens.

This is why integration must always be checked against the molecular formula.

One limitation: exchangeable hydrogens, particularly those in alcohol \(\ce{-OH}\) groups, can produce broad signals and unreliable integration. Don’t force every OH signal to integrate perfectly to 1H.

03Chemical shift: where does the signal appear?

Chemical shift gives you information about the electronic environment surrounding a hydrogen nucleus.

It is measured in parts per million, written ppm, and represented by the symbol \(\delta\).

The basic idea is shielding.

Electrons surrounding a hydrogen nucleus partially shield it from the external magnetic field. Different chemical surroundings change the amount of shielding.

• More shielding generally means a lower chemical shift.
• Less shielding generally means a higher chemical shift.

Hydrogens near electronegative atoms such as oxygen are often less shielded than ordinary alkyl hydrogens. They therefore tend to appear further downfield, at higher ppm values.

Approximate proton chemical shift ranges

The following are useful introductory guides, not rigid boundaries.

Alcohol OH signals are particularly variable. Their positions and widths depend on factors such as solvent, concentration, and hydrogen bonding.

Why chemical shift still doesn’t identify one structure

Imagine you see a signal at approximately 1.2 ppm.

What does that tell you?

It suggests a relatively shielded hydrogen environment, such as an ordinary alkyl group. But countless organic molecules contain alkyl hydrogens in this region.

Similarly, a signal around 3.5-4.0 ppm suggests hydrogens in an environment influenced by an electronegative atom, often oxygen.

However, several different structures contain \(\ce{C-H}\) bonds near oxygen.

Chemical shift narrows the possibilities. It doesn’t give you a unique molecular structure.

A small measurement calculation: what does ppm mean?

Chemical shift is measured relative to a reference, conventionally tetramethylsilane (TMS), which is assigned 0 ppm.

For a proton NMR spectrum, the chemical shift can be calculated using:

\[ \delta = \frac{\nu_{\text{sample}}-\nu_{\text{reference}}} {\nu_0}\times 10^6 \]

Here:

• \(\nu_{\text{sample}}\) is the sample resonance frequency, in Hz.
• \(\nu_{\text{reference}}\) is the reference resonance frequency, in Hz.
• \(\nu_0\) is the spectrometer operating frequency for hydrogen, in Hz.
• \(\delta\) is the chemical shift in ppm.

Suppose a 400 MHz spectrometer detects a signal 480 Hz away from the TMS reference.

Since \(400\ \mathrm{MHz}=400\\,000\\,000\ \mathrm{Hz}\),

\[ \begin{aligned} \delta &= \frac{480}{400\\,000\\,000}\times 10^6\\\ &=1.20\ \mathrm{ppm} \end{aligned} \]

The signal therefore appears at 1.20 ppm.

This calculation explains the scale, not the structure. A value of 1.20 ppm is still compatible with many different alkyl environments.

04Splitting: how many neighbouring hydrogens?

Splitting describes how a signal is divided into multiple lines.

In the introductory model, neighbouring equivalent hydrogen nuclei can split a signal through spin-spin coupling.

The useful rule is called the \(n+1\) rule:

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

Here, \(n\) is the number of equivalent neighbouring hydrogens.

Let’s use an ethyl group:

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

It contains two different proton environments.

The methyl group, \(\ce{CH3}\):

It has two neighbouring hydrogens on the adjacent \(\ce{CH2}\) group.

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

Its signal is a triplet.

The methylene group, \(\ce{CH2}\):

It has three neighbouring hydrogens on the adjacent \(\ce{CH3}\) group.

\[ n+1=3+1=4 \]

Its signal is a quartet.

The result is:

Together, these signals provide strong evidence for an ethyl fragment.

The important distinction: lines are not signals

This is a particularly common mistake.

A triplet contains three lines, but it represents one proton environment.

A quartet contains four lines, but it also represents one proton environment.

Therefore, an ethyl group produces:

• Two proton environments.
• Two signals.
• Seven individual lines across the two multiplets, assuming all lines are resolved.

It does not produce seven different proton environments.

Splitting also has limits

The simple \(n+1\) rule works best when neighbouring hydrogens are equivalent and the signals behave as simple first-order multiplets.

There are two useful limitations to remember.

First, the OH proton in an alcohol often doesn’t show the expected splitting from neighbouring hydrogens. Rapid proton exchange can remove or blur the coupling.

For example, ethanol contains a \(\ce{CH2OH}\) group. You might predict that its OH proton should interact with the two neighbouring methylene hydrogens and form a triplet.

In practice, the OH signal is often a broad singlet because of proton exchange.

Second, if a hydrogen group has more than one chemically distinct set of neighbouring hydrogens, the splitting can become more complicated than a simple triplet or quartet.

For introductory HSC-style problems, use the \(n+1\) rule where its assumptions apply. Don’t force it onto every multiplet.

05Why combining the four clues works

Each piece of evidence narrows the possible structures in a different way.

Consider an unknown molecule with the following observations:

• Three proton NMR signals.
• Integration ratio of 3:2:3.
• One triplet, one quartet, and one singlet.
• A molecular formula of \(\ce{C4H8O2}\).

You might feel confident that you’ve identified the compound.

But consider two possible structures:

Candidate A: Ethyl ethanoate

\[ \ce{CH3COOCH2CH3} \]

Candidate B: Methyl propanoate

\[ \ce{CH3CH2COOCH3} \]

Both structures have:

• The same molecular formula.
• Three distinct proton environments.
• An integration ratio of 3:2:3.
• A triplet from a methyl group next to a methylene group.
• A quartet from a methylene group next to a methyl group.
• A singlet from the remaining methyl group.

So the signal count, integration, and splitting pattern all fit both structures.

What separates them?

Chemical shift.

Consider the following illustrative data. These are approximate teaching values, not measurements from an actual experiment.

In ethyl ethanoate, the quartet comes from \(\ce{-OCH2CH3}\). The methylene hydrogens are attached to a carbon bonded directly to oxygen, so they appear further downfield.

In methyl propanoate, the quartet comes from \(\ce{-CH2COOCH3}\). The methylene hydrogens are next to a carbonyl carbon rather than directly attached to a carbon bonded to oxygen.

The singlets also differ:

• Ethyl ethanoate has a \(\ce{CH3CO-}\) group.
• Methyl propanoate has an \(\ce{-OCH3}\) group.

Their different electronic environments produce different chemical shifts.

This is exactly why you cannot stop after recognising a triplet and quartet. The splitting identifies the neighbouring hydrogen arrangement, but the chemical shift helps establish what that arrangement is connected to.

05How the NMR experiment produces these observations

It helps to understand where the evidence actually comes from.

A proton NMR spectrometer doesn’t photograph a molecule or directly count its hydrogen atoms. It measures the response of hydrogen nuclei to a magnetic field and radiofrequency energy.

The experiment follows a sequence.

1. Prepare the sample.

The organic compound is dissolved in a suitable solvent and placed in a narrow glass NMR tube. A deuterated solvent is commonly used to reduce interference from ordinary hydrogen in the solvent.

2. Apply the magnetic field.

The tube is positioned inside the spectrometer’s strong magnetic field.

Hydrogen nuclei have nuclear spin, which gives them magnetic properties. The external field creates different allowed energy states.

3. Apply radiofrequency energy.

A coil applies a radiofrequency pulse to the sample.

Hydrogen nuclei respond to the energy at frequencies determined by their magnetic environments.

4. Detect the response.

The receiver coil detects an electrical signal produced by the changing transverse magnetisation of the nuclei.

This time-dependent signal is called the free induction decay.

5. Produce the spectrum.

A computer applies a mathematical operation called a Fourier transform to convert the time-domain signal into the familiar frequency-domain NMR spectrum.

The experiment therefore provides several types of evidence:

• Different resonance frequencies produce different chemical shifts.
• Signal areas provide integration information.
• Interactions between neighbouring nuclei produce splitting patterns.

The measured spectrum is an experimental result. Assigning its signals to particular parts of a molecule is the chemical interpretation.

There are also experimental limitations. Signals can overlap, weak signals may be difficult to detect, and temperature or solvent conditions can affect exchangeable hydrogens.

This is why an apparent signal count should not always be treated as an infallible count of every possible proton environment.

06A reliable decision process

When several structures appear to fit an NMR clue, use the following sequence rather than guessing from the most familiar peak.

A useful habit is to write down what each clue actually proves before proposing a structure.

For example:

For a more detailed explanation of identifying equivalent hydrogens from a structural formula, see How to Count Proton Environments from a Structural Formula.

If you’re comfortable with the basic clues and want to practise combining triplet-quartet patterns with integration, see How to Identify an Ethyl Group Using Proton NMR.

07Questions and solutions

The following questions move from straightforward signal interpretation to situations where several clues agree but still don’t produce a unique answer.

Treat all chemical shifts as illustrative. Unless stated otherwise, assume ordinary solution-state proton NMR, resolved signals, and the introductory splitting model.

Question 1

An unknown organic compound has the molecular formula \(\ce{C3H6O2}\).

Its proton NMR spectrum contains:

Two possible structures are:

• A: Ethyl methanoate, \(\ce{HCOOCH2CH3}\).
• B: Propanoic acid, \(\ce{CH3CH2COOH}\).

Identify the more likely structure and explain how the four pieces of NMR evidence support your answer.

Solution 1

Structure B, propanoic acid, is the correct answer. The chemical shifts and the broad OH signal distinguish it from ethyl methanoate.

Work through the evidence:

• The triplet integrating to 3H indicates a methyl group next to two equivalent hydrogens.
• The quartet integrating to 2H indicates a methylene group next to three equivalent hydrogens.
• Together, these signals support a \(\ce{CH3CH2}\) fragment.
• The broad signal around 11.1 ppm is consistent with a carboxylic acid OH proton.

The chemical shift of the methylene group is particularly useful.

In propanoic acid, the methylene group is adjacent to a carbonyl carbon:

\[ \ce{CH3CH2COOH} \]

A chemical shift around 2.35 ppm is consistent with this environment.

In ethyl methanoate, the methylene group is attached to oxygen:

\[ \ce{HCOOCH2CH3} \]

Its methylene signal would instead appear much further downfield, typically around 4.2 ppm. Ethyl methanoate would also have a formyl hydrogen signal rather than a carboxylic acid OH signal.

The tempting conclusion is that both candidates contain an ethyl fragment, so both should fit. The splitting evidence does support both, but the chemical shifts and the OH signal separate them.

Question 2

An unknown compound has the molecular formula \(\ce{C3H8O}\).

Its proton NMR spectrum contains three signals:

Two possible structures are:

• A: Propan-1-ol, \(\ce{CH3CH2CH2OH}\).
• B: Methoxyethane, \(\ce{CH3OCH2CH3}\).

Identify the correct structure. Explain why the two 3H signals cannot simply be treated as equivalent.

Solution 2

Structure B, methoxyethane, fits the spectrum. Its two methyl groups have different chemical environments and therefore produce separate signals.

Start with the triplet at 1.18 ppm.

It integrates to 3H and is consistent with a terminal \(\ce{CH3}\) group next to a \(\ce{CH2}\) group.

The quartet at 3.48 ppm integrates to 2H. It is consistent with a methylene group next to three equivalent methyl hydrogens.

The chemical shift suggests that this methylene group is attached to a carbon bonded to oxygen:

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

The remaining 3H singlet at 3.35 ppm is consistent with a methoxy group:

\[ \ce{CH3-O-} \]

There are no neighbouring hydrogens directly attached to the oxygen that would produce ordinary splitting of this methyl signal.

The complete structure is:

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

Now compare propan-1-ol.

Its three carbon-bound proton environments would be:

• Terminal \(\ce{CH3}\): 3H.
• Middle \(\ce{CH2}\): 2H.
• \(\ce{CH2OH}\): 2H.

It also has an exchangeable OH proton.

It cannot account for a separate 3H methoxy singlet.

The key distinction is that integration tells you the number of hydrogens, while chemical shift helps you work out what those hydrogens are attached to.

Question 3

An unknown compound has the molecular formula \(\ce{C4H8O}\).

Its proton NMR spectrum contains:

A student argues that the 3H singlet at 2.15 ppm proves the compound contains an ester-like \(\ce{CH3CO-}\) group, and therefore suggests ethyl ethanoate, \(\ce{CH3COOCH2CH3}\).

Evaluate this suggestion and identify the most likely structure using all the supplied evidence.

Solution 3

The unknown is butan-2-one, \(\ce{CH3COCH2CH3}\), not ethyl ethanoate. The 3H singlet is consistent with a methyl group next to a carbonyl, but that observation alone does not establish an ester.

Start with the molecular formula.

Ethyl ethanoate has the formula \(\ce{C4H8O2}\), containing two oxygen atoms.

The unknown has the formula \(\ce{C4H8O}\), containing only one oxygen atom.

Therefore, ethyl ethanoate is already inconsistent with the supplied formula.

Now interpret the NMR spectrum.

The triplet at 1.05 ppm integrates to 3H. This is consistent with a terminal methyl group next to a methylene group.

The quartet at 2.45 ppm integrates to 2H. It is consistent with a methylene group next to three equivalent hydrogens.

Its chemical shift suggests that the methylene group is adjacent to a carbonyl:

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

The remaining singlet at 2.15 ppm integrates to 3H.

This fits a methyl group attached directly to the carbonyl carbon:

\[ \ce{CH3-CO-} \]

The carbonyl carbon has no attached hydrogens, so the methyl group has no neighbouring hydrogens to split its signal in the introductory model.

Combining these fragments gives:

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

This is butan-2-one.

The misconception is treating a signal near 2.1 ppm as proof of an ester. Methyl groups next to carbonyl groups can occur in both ketones and esters.

The molecular formula and the chemical shift of the methylene group provide the additional evidence needed here.

Question 4

An unknown compound produces the following two proton NMR signals:

A student claims that these two signals prove the unknown compound is ethyl ethanoate.

Two possible structures are:

• Ethyl methanoate, \(\ce{HCOOCH2CH3}\).
• Ethyl ethanoate, \(\ce{CH3COOCH2CH3}\).

Explain what the two signals allow you to conclude, why they don’t establish the complete structure, and predict one additional signal that would help distinguish the candidates.

Solution 4

The two signals strongly support an ethoxy fragment, but they do not uniquely identify ethyl ethanoate. Both candidate molecules contain the same \(\ce{-OCH2CH3}\) group.

The triplet at 1.2 ppm integrates to 3H.

This is consistent with the terminal methyl group, which has two neighbouring methylene hydrogens.

The quartet at 4.1 ppm integrates to 2H.

Its splitting indicates three neighbouring methyl hydrogens. Its relatively high chemical shift suggests that the methylene group is attached to a carbon bonded to oxygen.

Together, these observations support:

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

Both structures contain this fragment.

The difference lies on the other side of the ester oxygen.

Ethyl methanoate:

\[ \ce{HCOOCH2CH3} \]

It contains a formyl hydrogen, which would typically produce a signal around 8.0-8.5 ppm.

Ethyl ethanoate:

\[ \ce{CH3COOCH2CH3} \]

It contains a methyl group next to the carbonyl, which would typically produce a 3H singlet around 2.0 ppm.

Therefore, the additional signal helps identify which carbonyl-containing group is present.

The tempting conclusion was that recognising an ethoxy group identifies the entire molecule. It doesn’t. The spectrum has identified a fragment, not the complete connectivity.

Question 5

An unknown compound has the molecular formula \(\ce{C4H8O2}\).

Its IR spectrum shows a strong carbonyl absorption, supporting the presence of a carbonyl-containing functional group.

The available proton NMR information is:

• Exactly three signals.
• Integration ratio of 3:2:3.
• One triplet integrating to 3H.
• One quartet integrating to 2H.
• One singlet integrating to 3H.

Unfortunately, the chemical shift axis has been removed, so the actual ppm positions are unavailable.

A student confidently identifies the compound as ethyl ethanoate, arguing that the triplet and quartet prove the presence of an ethyl group.

Consider these two possible structures:

A. Ethyl ethanoate

\[ \ce{CH3COOCH2CH3} \]

B. Methyl propanoate

\[ \ce{CH3CH2COOCH3} \]

Answer the following:

1. Can the available evidence distinguish the two structures?
2. Explain why both structures satisfy the signal count, integration, and splitting information.
3. The chemical shift data are recovered. The triplet is at 1.25 ppm, the quartet is at 4.12 ppm, and the singlet is at 2.05 ppm. Identify the compound.
4. Predict the approximate chemical shifts of the quartet and singlet if the unknown were methyl propanoate instead.

Solution 5

The original evidence cannot distinguish ethyl ethanoate from methyl propanoate. The restored chemical shifts identify ethyl ethanoate. The crucial difference is whether the methylene group is attached to oxygen or directly to the carbonyl carbon.

Part 1: Can the original evidence distinguish the structures?

No. Both structures satisfy every observation supplied.

They have the same molecular formula:

\[ \ce{C4H8O2} \]

They also both contain an ester functional group, consistent with the carbonyl absorption in the IR spectrum.

Neither the molecular formula nor the IR result distinguishes these two ester isomers.

Part 2: Compare their proton environments.

For ethyl ethanoate:

\[ \ce{CH3COOCH2CH3} \]

The three proton environments are:

• \(\ce{CH3CH2}\): 3H.
• \(\ce{OCH2}\): 2H.
• \(\ce{CH3CO}\): 3H.

For methyl propanoate:

\[ \ce{CH3CH2COOCH3} \]

The three proton environments are:

• Terminal \(\ce{CH3}\): 3H.
• \(\ce{CH2CO}\): 2H.
• \(\ce{OCH3}\): 3H.

Both structures therefore have three signals with an integration ratio of:

\[ 3:2:3 \]

Both also have the same basic splitting pattern.

In each structure, the \(\ce{CH3CH2}\) fragment produces:

• A 3H triplet.
• A 2H quartet.

The remaining methyl group produces a 3H singlet.

The student has correctly recognised an ethyl fragment but has assumed that the fragment must be attached to oxygen. That assumption is not supported by splitting alone.

Part 3: Use the recovered chemical shifts.

The recovered spectrum contains:

The quartet at 4.12 ppm is consistent with \(\ce{OCH2}\).

The methylene carbon is directly bonded to oxygen, which reduces shielding of its attached hydrogens.

The singlet at 2.05 ppm is consistent with \(\ce{CH3CO}\), a methyl group attached to a carbonyl carbon.

These assignments fit:

\[ \ce{CH3COOCH2CH3} \]

Therefore, the unknown is ethyl ethanoate.

Part 4: Predict the shifts for methyl propanoate.

In methyl propanoate, the methylene group is attached to the carbonyl carbon rather than directly to oxygen.

Its quartet would typically appear around:

\[ \delta \approx 2.3\ \mathrm{ppm} \]

The remaining methyl group is attached directly to oxygen:

\[ \ce{-OCH3} \]

Its singlet would typically appear around:

\[ \delta \approx 3.7\ \mathrm{ppm} \]

The terminal methyl triplet would remain around 1.1 ppm.

These approximate values give a different arrangement of chemical shifts, even though the signal count, integration, and splitting remain the same.

The deeper lesson is that recognising a familiar pattern is not the same as identifying a molecule. A triplet and quartet can establish a connected \(\ce{CH3CH2}\) fragment, but they cannot tell you everything attached to that fragment.

08What to remember

When several different structures fit one NMR clue, it usually means you’re asking that clue to provide more information than it actually contains.

Keep the four observations separate:

• Signal count tells you how many distinct proton environments are observed.
• Integration tells you the relative number of hydrogens in each environment.
• Chemical shift suggests the electronic environment surrounding those hydrogens.
• Splitting provides information about neighbouring hydrogens.

A convincing structure must explain all four, as well as the molecular formula and any additional spectroscopic evidence.

Once you’ve identified the proton environments and connected likely fragments, the next useful step is checking the carbon environments using carbon-13 NMR. This can help distinguish structures that have similar proton NMR patterns but different arrangements of carbon atoms.