How to Identify an Ethyl Group Using Proton NMR

An unknown compound produces a triplet integrating to 3H at \(\delta 1.20\ \mathrm{ppm}\) and a quartet integrating to 2H at \(\delta 4.10\ \mathrm{ppm}\). What can you deduce from those two signals?

The strongest conclusion is that the molecule contains an ethyl group, \(\ce{-CH2CH3}\). The triplet and quartet provide complementary evidence: the \(\ce{CH3}\) group has two neighbouring hydrogens, while the \(\ce{CH2}\) group has three. Their integration ratio, 3:2, confirms the number of hydrogens in each group.

This is one of the most useful patterns to recognise in proton nuclear magnetic resonance (\(^{1}\mathrm{H}\) NMR) spectroscopy. The trick is to combine four different observations without confusing them: the number of signals, integration, chemical shift, and splitting.

01What does an ethyl group look like in proton NMR?

An ethyl group consists of two connected carbon atoms:

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

The two carbon atoms have different numbers of attached hydrogens.

• The \(\ce{CH3}\) group contains three equivalent hydrogens.
• The \(\ce{CH2}\) group contains two hydrogens, treated as equivalent in the introductory HSC model.

Because these groups have different chemical environments, they normally produce two separate signals.

Now make a prediction.

If the three hydrogens in \(\ce{CH3}\) sit next to a carbon carrying two hydrogens, how many lines should their signal contain?

Three.

The two neighbouring hydrogens split the \(\ce{CH3}\) signal into three lines. This is a triplet.

The same reasoning works in reverse. The two hydrogens in \(\ce{CH2}\) sit next to three hydrogens in \(\ce{CH3}\), so their signal contains four lines. This is a quartet.

The resulting pattern is:

The important feature is that the splitting patterns agree with each other. The triplet suggests a neighbouring \(\ce{CH2}\), and the quartet suggests a neighbouring \(\ce{CH3}\).

Together, they strongly support the connected fragment \(\ce{-CH2CH3}\).

02How does the NMR experiment produce these signals?

Before interpreting a spectrum, it helps to understand what the instrument actually measures.

A sample of the organic compound is placed in a narrow glass tube. The tube sits inside a strong magnetic field in the NMR spectrometer.

A radiofrequency coil applies electromagnetic energy to the sample. Hydrogen nuclei respond to this energy at frequencies that depend on their local electronic environments. The instrument detects the response as the nuclei return towards equilibrium, and a computer processes the signal into a spectrum.

What should the experiment show?

If every hydrogen nucleus experienced exactly the same chemical environment, the spectrum would contain one signal. If the molecule contained several different hydrogen environments, separate signals would appear.

The experiment also reveals splitting caused by interactions between neighbouring hydrogen nuclei.

This is how the spectrum provides structural information. However, the instrument does not directly identify an ethyl group. That conclusion comes from interpreting the measured signals together.

03The four pieces of information you must keep separate

A proton NMR spectrum provides four different types of information. Each answers a different structural question.

These observations work together, but they are not interchangeable.

1. Number of signals: how many environments?

A signal represents a group of chemically equivalent hydrogen nuclei.

For example, an ethyl group normally produces two signals:

• One from the three equivalent hydrogens in \(\ce{CH3}\).
• One from the two equivalent hydrogens in \(\ce{CH2}\).

There are two signals, even though one is split into three lines and the other into four.

This is a common examination mistake. A triplet is one signal containing three lines, not three separate hydrogen environments.

2. Integration: how many hydrogens contribute?

Integration measures the relative area under a signal.

A larger integrated area means that more hydrogens contribute to that signal. It is the total area of the multiplet that matters, not the height of its tallest line.

For an ethyl group, the expected integration ratio is:

\[
3:2
\]

This means the triplet has an integrated area proportional to three hydrogens, while the quartet has an area proportional to two.

However, integration gives relative numbers, not automatically the absolute number of hydrogens.

Consider a spectrum with two signals whose integration values are 1.5 and 1.0 arbitrary units.

Divide both values by the smaller number:

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

This gives the familiar ethyl ratio.

But suppose the molecular formula tells you that the molecule contains ten hydrogens, and these are the only two signals. The actual integration must then be scaled:

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

The molecule could contain two equivalent ethyl groups rather than just one.

Keep this distinction in mind when interpreting integration.

3. Chemical shift: what is the hydrogen bonded near?

Chemical shift describes how the electronic environment affects the resonance of a hydrogen nucleus.

It is represented by \(\delta\) and measured in parts per million (ppm), relative to a reference compound, usually tetramethylsilane (TMS), assigned a chemical shift of 0 ppm.

Electronegative atoms, such as oxygen, withdraw electron density from nearby bonds. This reduces the shielding around nearby hydrogen nuclei.

These hydrogens are described as deshielded and generally appear at higher chemical shifts.

On a conventional proton NMR spectrum, higher chemical shifts appear further to the left.

For ethyl groups, the chemical shift of the \(\ce{CH2}\) signal is particularly useful because it helps identify what the ethyl group is attached to.

These are approximate guides, not fixed boundaries. The exact chemical shift depends on the surrounding molecular structure and experimental conditions.

For example:

• \(\ce{CH3CH2-O-}\) often has a \(\ce{CH2}\) signal around 3.5-4.2 ppm.
• \(\ce{CH3CH2-C(=O)-}\) often has a \(\ce{CH2}\) signal around 2.0-2.7 ppm.

Both structures contain an ethyl group. Their chemical shifts help distinguish how that group is connected.

For a broader explanation of how chemical environments relate to functional groups, see HSC Chemistry: How to Identify Organic Functional Groups.

4. Splitting: how many neighbouring hydrogens?

Splitting occurs when hydrogen nuclei interact with neighbouring, non-equivalent hydrogen nuclei.

For the simple splitting patterns used in introductory HSC problems, the number of lines follows the \(n+1\) rule:

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

Here, \(n\) is the number of equivalent neighbouring hydrogens on adjacent carbon atoms.

For an ethyl group:

• \(\ce{CH3}\) has two neighbouring hydrogens, producing a triplet.
• \(\ce{CH2}\) has three neighbouring hydrogens, producing a quartet.

The relative line intensities also follow predictable patterns.

A triplet has relative line intensities of \(1:2:1\), while a quartet has relative line intensities of \(1:3:3:1\).

One important detail: hydrogens do not split other hydrogens within their own equivalent group in this introductory model. They split signals from neighbouring groups.

Also, do not automatically count every hydrogen attached to nearby atoms. In the simple HSC model, the \(n+1\) rule is applied to equivalent neighbouring hydrogens on adjacent carbon atoms. Exchangeable \(\ce{OH}\) and \(\ce{NH}\) hydrogens are generally not counted.

04A reliable method for identifying an ethyl group

When you receive an unfamiliar proton NMR spectrum, work through the following sequence.

Step 1: Count the signals.

Identify the number of distinct hydrogen environments. Do not count individual lines within multiplets as separate signals.

Step 2: Read the integration values.

Look for a signal integrating to 3H and another integrating to 2H. If the values are given as arbitrary units, simplify them to a ratio.

Step 3: Check the splitting patterns.

Look for a triplet integrating to 3H and a quartet integrating to 2H.

The triplet indicates two neighbouring hydrogens, while the quartet indicates three. These are complementary clues.

Step 4: Match the chemical shifts.

Check whether the \(\ce{CH2}\) group is attached to oxygen, adjacent to a carbonyl group, or in an ordinary alkyl environment.

Step 5: Check the rest of the molecule.

Use the molecular formula and remaining NMR signals to test whether your proposed structure accounts for every hydrogen.

Do not assume that any triplet and quartet in the same spectrum must belong together. Their integration, chemical shifts, and neighbouring environments must make sense as a connected fragment.

05Worked examples

06When the simple ethyl pattern becomes more complicated

The triplet-and-quartet combination is a useful shortcut, but it is not an absolute requirement for every ethyl group.

There are three situations worth recognising.

1. The methylene group has more than one set of neighbours.

A \(\ce{CH2}\) group between a \(\ce{CH3}\) group and another \(\ce{CH2}\) group interacts with hydrogens on both sides.

If those neighbouring sets are not equivalent, the resulting splitting can be more complicated than a simple quartet.

2. Exchangeable hydrogens behave differently.

Hydrogens attached to oxygen or nitrogen can exchange with other molecules. Their signals are often broad and may not show reliable splitting.

For introductory HSC questions, treat an \(\ce{OH}\) signal as a separate environment and do not count it as a neighbouring hydrogen in the simple \(n+1\) calculation.

3. Equivalent ethyl groups can share signals.

If a molecule has two ethyl groups in identical chemical environments, their signals can combine.

Instead of a 3H triplet and a 2H quartet, the spectrum may show a 6H triplet and a 4H quartet.

The ratio is still 3:2, but the actual numbers of hydrogens are different.

These limitations are particularly important when interpreting unfamiliar molecules rather than simply recognising textbook examples.

Syllabus context: Proton and carbon-13 NMR are included in Module 8 of the Chemistry Stage 6 Syllabus (2017). The replacement Chemistry 11-12 Syllabus (2025) is scheduled to begin implementation in 2028, with its first HSC examination in 2029. Students should follow the syllabus applying to their cohort.

07Questions and solutions

The following questions progress from recognising a straightforward ethyl group to interpreting more complicated spectra. All chemical shifts and spectral data are illustrative.

Question 1

An unknown organic compound produces the following two proton NMR signals:

Assuming simple first-order splitting, identify the structural fragment responsible for these signals.

Explain which hydrogen group produces each signal and justify your answer using integration and splitting.

Solution 1

The two signals indicate an ethyl group, \(\ce{-CH2CH3}\).

The triplet at 1.08 ppm integrates to 3H, identifying a methyl group. Its three lines indicate that it has two neighbouring hydrogens.

The quartet at 3.55 ppm integrates to 2H, identifying a methylene group. Its four lines indicate that it has three neighbouring hydrogens.

These patterns are complementary:

• \(\ce{CH3}\) is split by the two hydrogens in \(\ce{CH2}\).
• \(\ce{CH2}\) is split by the three hydrogens in \(\ce{CH3}\).

The integration ratio is \(3:2\), which agrees with the hydrogen counts in an ethyl group.

The higher chemical shift of the methylene signal also suggests that this carbon is directly attached to an electronegative atom, potentially oxygen.

Question 2

An unknown organic compound has the molecular formula \(\ce{C6H12O2}\).

Its illustrative proton NMR spectrum contains five signals:

Determine the most likely molecular structure.

Identify the ethyl group and explain why the two triplets integrating to 3H should not automatically be paired with the same quartet.

Solution 2

The most likely structure is ethyl butanoate, \(\ce{CH3CH2CH2COOCH2CH3}\). The ethyl group is the \(\ce{-OCH2CH3}\) fragment.

Start by identifying the clearest complementary pair.

The quartet at 4.10 ppm integrates to 2H. This indicates a \(\ce{CH2}\) group with three neighbouring hydrogens. Its high chemical shift suggests that it is directly attached to oxygen.

The triplet at 1.25 ppm integrates to 3H. This indicates a \(\ce{CH3}\) group with two neighbouring hydrogens.

These two signals form the ethoxy fragment:

\[
\ce{-OCH2CH3}
\]

The other triplet at 0.95 ppm also integrates to 3H, but it belongs to a different part of the molecule.

The remaining signals can be assigned as follows:

• 0.95 ppm, triplet, 3H: terminal methyl group.
• 1.65 ppm, multiplet, 2H: middle methylene group.
• 2.30 ppm, triplet, 2H: methylene group adjacent to the carbonyl.

The methyl group at 0.95 ppm is connected to the middle methylene group, not directly to the oxygen-bound methylene at 4.10 ppm.

The middle methylene has neighbouring hydrogens on both sides. These are different sets of hydrogens, so its splitting is more complex than a simple quartet.

Finally, the total integration is:

\[
3+2+2+2+3=12\text{ H}
\]

This agrees with \(\ce{C6H12O2}\).

The structure \(\ce{CH3CH2CH2COOCH2CH3}\) accounts for all five signals and the molecular formula.

The important lesson is that a triplet and a quartet should be paired using their chemical environments and integration, not simply because both patterns appear in the same spectrum.

Question 3

A compound has the molecular formula \(\ce{C5H10O2}\) and produces four signals in its proton NMR spectrum.

Determine the most likely structure.

Your explanation must:

• Identify both ethyl-containing fragments.
• Match each quartet with its corresponding triplet.
• Explain why the two methylene groups have different chemical shifts.
• Account for all ten hydrogens.

Solution 3

The most likely structure is ethyl propanoate, \(\ce{CH3CH2COOCH2CH3}\).

The spectrum contains two separate ethyl groups, but they are in different chemical environments.

First, consider the quartet at 4.12 ppm.

It integrates to 2H, indicating a methylene group. Its high chemical shift suggests that it is directly attached to oxygen.

The triplet at 1.25 ppm integrates to 3H and indicates a methyl group next to this methylene.

Together, these signals identify:

\[
\ce{-OCH2CH3}
\]

Now consider the second quartet at 2.35 ppm.

It also integrates to 2H, but its lower chemical shift suggests that the methylene group is adjacent to a carbonyl carbon rather than directly attached to oxygen.

The triplet at 1.05 ppm integrates to 3H and identifies its neighbouring methyl group.

Together, these signals indicate:

\[
\ce{CH3CH2C(=O)-}
\]

Combining the two fragments gives:

\[
\ce{CH3CH2COOCH2CH3}
\]

The structure contains an ester functional group and two ethyl-containing fragments.

The integration also checks out:

\[
3+2+2+3=10\text{ H}
\]

This agrees with the molecular formula \(\ce{C5H10O2}\).

The key distinction is that the two methylene groups are attached to different parts of the ester structure. One is directly bonded to oxygen, while the other is adjacent to the carbonyl carbon. Their different chemical shifts allow you to distinguish them even though both produce quartets.

Question 4

An unknown compound has the molecular formula \(\ce{C3H6O2}\) and produces the following proton NMR signals:

Two possible structures are proposed:

Structure A: Ethyl methanoate

\[
\ce{HCOOCH2CH3}
\]

Structure B: Propanoic acid

\[
\ce{CH3CH2COOH}
\]

Determine which structure is consistent with the spectrum.

Explain why the triplet and quartet identify an ethyl group in both possible structures, but the chemical shifts allow you to distinguish between them.

Solution 4

Structure B, propanoic acid, is consistent with the spectrum.

Both proposed structures contain an ethyl group, so the integration and splitting patterns alone do not distinguish them.

The triplet at 1.10 ppm integrates to 3H. This indicates a methyl group with two neighbouring hydrogens.

The quartet at 2.35 ppm integrates to 2H. This indicates a methylene group with three neighbouring hydrogens.

Together, these signals identify:

\[
\ce{CH3CH2-}
\]

The important evidence is the chemical shift of the methylene group.

At 2.35 ppm, the methylene hydrogens are consistent with a carbon directly adjacent to a carbonyl carbon.

This supports the structure:

\[
\ce{CH3CH2COOH}
\]

The broad singlet at 11.3 ppm also supports a carboxylic acid group. Carboxylic acid hydrogens commonly appear at relatively high chemical shifts and produce broad signals.

Structure A would produce a different pattern of chemical shifts.

In ethyl methanoate, the \(\ce{CH2}\) group is directly attached to oxygen, so its signal would generally appear around 4 ppm rather than 2.35 ppm. Its formyl hydrogen would also produce a signal in a different chemical shift region, rather than the broad carboxylic acid signal.

The total integration confirms the molecular formula:

\[
3+2+1=6\text{ H}
\]

Therefore, propanoic acid is the better structural assignment.

This question demonstrates an important distinction: integration and splitting can identify the ethyl fragment, but chemical shift tells you more about its attachment to the rest of the molecule.

Question 5

A pure, acyclic organic compound has the molecular formula \(\ce{C5H12O}\).

Its proton NMR spectrum contains exactly four signals:

The broad singlet disappears after the sample is shaken with \(\ce{D2O}\).

Assume that there is no accidental signal overlap and that all twelve hydrogens are accounted for.

A student reduces the integration ratio \(6:4\) to \(3:2\) and concludes that the molecule contains only one ethyl group, with the remaining hydrogens belonging to other parts of the molecule.

Is this conclusion correct?

Determine the most likely molecular structure and explain:

• Why the 6H triplet and 4H multiplet indicate two equivalent ethyl groups.
• Why the 4H signal is not a simple quartet.
• What the 1H signal at 3.55 ppm and the disappearing broad signal reveal about the molecule.
• How the molecular formula and symmetry constrain the final structure.

Solution 5

The most likely structure is pentan-3-ol:

\[
\ce{CH3CH2CH(OH)CH2CH3}
\]

The student’s conclusion is incorrect because reducing an integration ratio does not tell you how many identical groups are present in the molecule.

This problem requires you to combine integration, symmetry, splitting, chemical shift, and molecular formula.

Step 1: Check the total number of hydrogens.

The integration values add to:

\[
6+4+1+1=12\text{ H}
\]

This matches the molecular formula \(\ce{C5H12O}\).

There are no additional hydrogens that need to be assigned to hidden or overlapping signals.

Step 2: Interpret the 6H triplet.

The triplet integrates to 6H.

A single methyl group contains only three hydrogens, so this signal represents two equivalent methyl groups.

Each methyl group has two neighbouring hydrogens, producing a triplet.

The two methyl groups must therefore be in identical chemical environments.

Step 3: Interpret the 4H multiplet.

The 4H signal represents two equivalent methylene groups.

Together, the two methyl groups and two methylene groups form two equivalent ethyl fragments:

\[
\ce{CH3CH2-}
\]

The important point is that the integration ratio \(6:4\) simplifies to \(3:2\), but the actual integration values show that the molecule contains two equivalent ethyl groups.

Symmetry explains why the two ethyl groups produce only two signals rather than four separate signals.

Step 4: Explain why the methylene signal is a multiplet rather than a quartet.

This is the most important complication in the question.

Each methylene group is connected to:

• A methyl group containing three hydrogens.
• A central methine group, \(\ce{CH(OH)}\), containing one hydrogen.

The methylene hydrogens therefore interact with two different sets of neighbouring hydrogens.

The three methyl hydrogens and the one methine hydrogen are not equivalent neighbours.

You cannot simply add them together and apply the \(n+1\) rule as though all four hydrogens were identical.

The resulting signal is more complicated than a simple quartet, so it is appropriately represented as a multiplet.

This is where the introductory splitting model becomes incomplete. The simple \(n+1\) rule works most directly when the neighbouring hydrogens belong to one equivalent set.

Step 5: Interpret the remaining two signals.

The 1H multiplet at 3.55 ppm is consistent with the central methine hydrogen attached to a carbon that also carries an \(\ce{OH}\) group.

The oxygen withdraws electron density, deshielding the hydrogen on that carbon and moving its signal to a higher chemical shift.

The broad 1H singlet disappears after the sample is shaken with \(\ce{D2O}\).

This identifies an exchangeable hydrogen, consistent with an \(\ce{OH}\) group.

The two remaining signals therefore support a secondary alcohol structure.

Step 6: Assemble the structure and check the constraints.

The molecule contains:

• Two equivalent terminal \(\ce{CH3}\) groups.
• Two equivalent \(\ce{CH2}\) groups.
• One central \(\ce{CH}\) group.
• One exchangeable \(\ce{OH}\) hydrogen.

The two ethyl groups must be attached to the same central carbon because they are chemically equivalent, and the central carbon is bonded to the oxygen-containing group.

The resulting structure is:

\[
\ce{CH3CH2CH(OH)CH2CH3}
\]

This is pentan-3-ol.

The molecular formula, integration values, symmetry, and chemical shifts all agree.

The tempting mistake was to treat the simplified \(3:2\) ratio as proof of one ethyl group. The full integration and signal count show that the molecule actually contains two equivalent ethyl groups. The complicated methylene splitting then provides further evidence about how those groups connect to the central carbon.

08What to do after identifying the ethyl group

Recognising \(\ce{-CH2CH3}\) is a starting point, not the final structural deduction.

Your next decision is to establish what the ethyl group is attached to. Use the methylene chemical shift to distinguish an oxygen-bound group from one adjacent to a carbonyl or an ordinary carbon chain.

Then account for every remaining signal and check that your proposed structure agrees with the molecular formula.

For the next stage of organic structure identification, revisit HSC Chemistry: How to Identify Organic Functional Groups. If you need to translate an identified carbon skeleton into its systematic name, see Naming Methyl and Ethyl Branched Chains for HSC Chemistry.