How to Distinguish Ethanol from Dimethyl Ether Using NMR

Both ethanol and dimethyl ether have the molecular formula \(\ce{C2H6O}\), but they have different structures. How can you tell which compound is in an unknown sample using NMR spectroscopy?

The answer comes from their proton environments. In a typical \(^1\mathrm{H}\) NMR spectrum, ethanol produces three signals, usually with integrations of 3:2:1, while dimethyl ether produces one signal integrating to 6H. Ethanol’s methyl group also produces a triplet, and its methylene group produces a quartet. Dimethyl ether produces a singlet.

The important distinction is that NMR detects chemically different environments, not simply the number of atoms in a molecule. Let’s work through how to use signal count, integration, chemical shift, and splitting to identify each compound, including what happens when the spectrum is less tidy than the textbook version.

01Start with the two structures

Before looking at a spectrum, predict which molecule should produce fewer signals. Consider the symmetry of each structure.

Ethanol has the structure:

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

Dimethyl ether, also called methoxymethane, has the structure:

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

Both contain two carbon atoms, six hydrogen atoms, and one oxygen atom. However, the atoms are connected differently.

Why ethanol produces three signals

Ethanol contains three different hydrogen environments:

• The three hydrogen atoms in \(\ce{CH3}\).
• The two hydrogen atoms in \(\ce{CH2}\).
• The one hydrogen atom in \(\ce{OH}\).

These groups experience different chemical environments.

The \(\ce{CH3}\) group is attached to a carbon atom, while the \(\ce{CH2}\) group is directly attached to an electronegative oxygen atom. The hydroxyl hydrogen is bonded directly to oxygen.

Consequently, the three groups generally produce three separate signals in a proton NMR spectrum.

Why dimethyl ether produces one signal

Dimethyl ether has a symmetrical structure:

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

The two methyl groups are chemically equivalent because they have identical surroundings. The six hydrogen atoms therefore experience the same chemical environment in an ordinary solution-phase proton NMR experiment.

All six hydrogen atoms contribute to one signal.

This gives us our first useful prediction:

Notice that dimethyl ether has two methyl groups but only one hydrogen environment. Counting the groups in a structural formula is not enough. You need to determine whether those groups are chemically equivalent.

02How the NMR experiment produces the spectrum

An NMR spectrum is an experimental measurement, not simply a diagram of a molecule.

To understand what the spectrum tells us, it helps to follow the sample through the instrument.

The measurement follows this sequence:

1. A small quantity of the compound is dissolved in a suitable solvent, usually a deuterated solvent to minimise interference from solvent hydrogen signals.
2. The solution is placed in a narrow glass tube, which is inserted into the probe inside a strong magnetic field.
3. A radiofrequency pulse is applied to the sample. Hydrogen nuclei respond to this energy according to their magnetic environments.
4. As the nuclei return towards equilibrium, the instrument detects a weak, decaying electrical signal called the free induction decay (FID).
5. A mathematical process called a Fourier transform converts this time-dependent signal into the familiar frequency-domain NMR spectrum.

The resulting spectrum allows us to distinguish hydrogen environments by their resonance positions and splitting patterns.

A useful description of the modern instrument is available in JEOL’s introduction to NMR.

What should the experiment show?

We can make two predictions before collecting any data.

If the sample contains ethanol, we expect three hydrogen environments, with the methyl and methylene groups producing recognisable signals.

If the sample contains dimethyl ether, we expect one signal representing all six equivalent hydrogen atoms.

The observed signal count and relative integration can then be compared with these predictions.

There is one complication. The hydroxyl hydrogen in ethanol can exchange with other hydrogen-containing molecules, particularly water. This can make its signal broad, weak, or difficult to detect.

Therefore, three signals are the usual introductory prediction for ethanol, but observing only two does not automatically rule it out.

03The four features you need to interpret

To distinguish the compounds reliably, examine four features of the spectrum.

1. Number of signals

The number of signals represents the number of chemically distinct hydrogen environments.

For ethanol:

• One signal comes from the three equivalent methyl hydrogen atoms.
• One signal comes from the two equivalent methylene hydrogen atoms.
• One signal may come from the hydroxyl hydrogen.

For dimethyl ether:

• All six hydrogen atoms are equivalent.
• Only one signal is produced.

Be careful with terminology. A signal can contain multiple individual lines.

For example, a triplet contains three lines, but it is still one signal representing one hydrogen environment.

This distinction prevents a common mistake: counting every line in a triplet or quartet as a separate hydrogen environment.

2. Integration: how many hydrogen atoms contribute?

Integration measures the area underneath an NMR signal.

The area is proportional to the relative number of hydrogen nuclei contributing to that signal.

For ethanol, the expected integration ratio is:

\[
3:2:1
\]

This corresponds to:

For dimethyl ether, all six hydrogen atoms contribute to one signal:

\[
6
\]

The integration therefore distinguishes the compounds even when their chemical shifts are relatively close.

Remember that integration gives relative areas, not necessarily absolute hydrogen counts. A spectrum might display areas of 30 and 20 units rather than 3 and 2. The important information is their ratio.

Also, the hydroxyl hydrogen may not integrate accurately because of proton exchange and signal broadening. The 3:2 ratio between ethanol’s carbon-bound hydrogen groups is often more reliable.

3. Chemical shift: where does each signal appear?

Chemical shift describes the position of a signal on the horizontal axis of an NMR spectrum.

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

Most introductory NMR spectra display higher chemical shifts on the left and lower chemical shifts on the right.

A higher chemical shift is described as downfield, while a lower chemical shift is described as upfield.

Why do different hydrogen environments appear at different positions?

Electrons surrounding a hydrogen nucleus partially shield it from the applied magnetic field. Electronegative atoms such as oxygen withdraw electron density from nearby bonds, reducing this shielding.

The hydrogen nuclei close to oxygen therefore generally appear at higher chemical shifts.

For our two compounds, the approximate positions are:

These values are illustrative ranges, not fixed constants. The hydroxyl signal is particularly sensitive to solvent, concentration, hydrogen bonding, and temperature.

There is an important trap here.

The oxygen-adjacent hydrogen signals of ethanol and dimethyl ether occur in similar regions. Ethanol’s \(\ce{CH2}\) signal is around 3.6 ppm, while dimethyl ether’s methyl signal is around 3.3 ppm.

Chemical shift alone is not enough to distinguish these compounds reliably.

You need to consider integration and splitting as well.

4. Splitting: how many neighbouring hydrogen atoms are present?

Splitting occurs because nearby hydrogen nuclei can interact magnetically with one another.

For introductory proton NMR, we commonly use the \(n+1\) rule:

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

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

The neighbouring hydrogens are normally on an adjacent carbon atom in the simplified model.

The two important splitting patterns in ethanol are the triplet and quartet.

Ethanol’s \(\ce{CH3}\) group:

The three methyl hydrogen atoms are next to a \(\ce{CH2}\) group containing two hydrogen atoms.

Therefore:

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

The methyl signal is a triplet.

Ethanol’s \(\ce{CH2}\) group:

The two methylene hydrogen atoms are next to a \(\ce{CH3}\) group containing three hydrogen atoms.

Therefore:

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

The methylene signal is a quartet.

The hydroxyl hydrogen is usually not counted in this simplified calculation because it exchanges rapidly with other molecules. Its signal is therefore commonly observed as a broad singlet, without noticeable coupling to the methylene group.

Dimethyl ether:

Each methyl group is attached to an oxygen atom, not to a carbon carrying neighbouring hydrogen atoms.

The equivalent methyl groups do not split one another in the introductory model.

Consequently, all six hydrogen atoms contribute to a singlet.

The four features can now be compared directly.

04The important exception: ethanol’s hydroxyl signal

The hydroxyl hydrogen is the part of this comparison most likely to cause confusion.

You might expect the \(\ce{OH}\) hydrogen to behave just like the hydrogen atoms attached to carbon. However, it can exchange with hydrogen atoms from other molecules.

For example, water in the sample can participate in proton exchange.

This exchange can broaden the hydroxyl signal or make it difficult to distinguish from the baseline.

It can also prevent the hydroxyl hydrogen from splitting the neighbouring \(\ce{CH2}\) signal.

This explains why the usual introductory ethanol spectrum contains:

• A methyl triplet.
• A methylene quartet.
• A broad hydroxyl singlet, which may be weak or absent.

How can you test for an OH group?

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

Deuterium oxide, \(\ce{D2O}\), contains deuterium instead of ordinary hydrogen.

When ethanol is mixed with \(\ce{D2O}\), the hydroxyl hydrogen can exchange with deuterium:

\[
\ce{CH3CH2OH + D2O <=> CH3CH2OD + HOD}
\]

The original hydroxyl hydrogen is replaced by deuterium.

Because a standard proton NMR experiment detects \(^1\mathrm{H}\), rather than deuterium, the original hydroxyl signal disappears.

The carbon-bound hydrogen signals remain.

This gives us an additional experimental prediction:

• Ethanol: an observable OH signal disappears after D2O exchange.
• Dimethyl ether: there is no OH hydrogen to exchange, so no OH signal disappears.

There is a limitation. If ethanol’s OH signal was already invisible before adding D2O, its absence afterwards does not prove anything. You need other evidence, such as the triplet and quartet, or a repeat measurement under conditions that make the OH signal easier to observe.

A further limitation of the \(n+1\) rule

The simple splitting model assumes that the hydroxyl proton exchanges rapidly enough that its coupling is not observed.

In very dry ethanol, proton exchange can be slower. The hydroxyl hydrogen may then couple to the neighbouring methylene hydrogen atoms.

Under suitable high-resolution conditions:

• The OH signal can appear as a triplet because it is coupled to two methylene hydrogen atoms.
• The CH2 quartet can be split further by the OH hydrogen, producing a quartet of doublets if the two coupling interactions are resolved.

This is why the \(n+1\) rule should not be treated as an instruction to count every hydrogen atom within a molecule.

It is a simplified model for predicting observable coupling under particular experimental conditions.

For the usual introductory comparison, use the triplet, quartet, and broad OH singlet model.

What if the question gives you a carbon-13 NMR spectrum?

The same compounds can also be distinguished using \({}^{13}\mathrm{C}\) NMR, but the interpretation changes.

You are now observing carbon environments rather than hydrogen environments.

Ethanol has two different carbon environments because its methyl and methylene carbons are not equivalent.

Dimethyl ether has only one because its two carbon atoms have identical chemical environments.

Do not use the proton integration ratio of 3:2:1 when interpreting a carbon-13 spectrum.

05A practical method for identifying the compound

When you are given an unknown spectrum and asked to distinguish ethanol from dimethyl ether, follow this sequence.

Step 1: Count the hydrogen environments.

Three signals usually suggest ethanol, while one suggests dimethyl ether.

Remember that ethanol’s OH signal may be missing.

Step 2: Check the integration.

Look for ethanol’s 3:2 ratio between its methyl and methylene signals, or a single 6H signal for dimethyl ether.

Step 3: Check the chemical shifts.

A signal around 1.2 ppm is characteristic of ethanol’s methyl group.

A signal around 3.3-3.7 ppm could represent either compound’s oxygen-adjacent hydrogens, so do not use this feature alone.

Step 4: Examine the splitting.

A triplet and quartet together are strong evidence for ethanol.

A single 6H singlet is consistent with dimethyl ether.

Step 5: Consider the OH signal.

If a broad OH signal is present, it supports ethanol. If it is missing, do not immediately reject ethanol.

This method uses several independent pieces of evidence rather than relying on one peak.

06Worked examples

07Questions and solutions

Question 1

A pure unknown compound has the molecular formula \(\ce{C2H6O}\). Its proton NMR spectrum contains one signal at \(\delta=3.32\) ppm.

The signal is a singlet and integrates to 6H.

Determine whether the compound is ethanol or dimethyl ether. Explain your reasoning.

Solution 1

The compound is dimethyl ether, \(\ce{CH3OCH3}\).

All six hydrogen atoms in dimethyl ether have equivalent chemical environments because the molecule is symmetrical.

They therefore produce one signal with an integration of 6H.

The singlet is also expected because the methyl groups do not have neighbouring nonequivalent hydrogen atoms that produce observable splitting in the introductory model.

Ethanol would normally produce separate methyl, methylene, and hydroxyl signals. It would not produce one singlet representing all six hydrogen atoms.

Question 2

A pure unknown compound has the molecular formula \(\ce{C2H6O}\).

Its proton NMR spectrum contains:

• A triplet at 1.20 ppm, integrating to 3H.
• A quartet at 3.60 ppm, integrating to 2H.

No third signal can be identified.

A student concludes that the compound cannot be ethanol because ethanol contains six hydrogen atoms, but the spectrum only accounts for five.

Is this conclusion correct? Explain what may have happened to the remaining hydrogen signal.

Solution 2

The compound is consistent with ethanol. The missing hydroxyl signal does not rule out its presence.

The triplet integrating to 3H identifies a methyl group next to two hydrogen atoms.

The quartet integrating to 2H identifies a methylene group next to three hydrogen atoms. Its chemical shift near 3.6 ppm indicates that it is attached to oxygen.

Together, these signals give the structure:

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

The remaining hydrogen belongs to the hydroxyl group.

It may not be visible because the OH proton exchanges rapidly with water or other hydrogen-containing molecules. This exchange can broaden the signal until it becomes difficult to distinguish.

The tempting mistake is to assume that every hydrogen atom must produce a clearly visible, measurable signal.

That is not always true, particularly for exchangeable hydroxyl hydrogens.

If an OH signal can be observed, adding \(\ce{D2O}\) provides a useful test because the signal should disappear following hydrogen-deuterium exchange. If the signal is already absent, the D2O test cannot establish its presence from that spectrum alone.

Question 3

An unknown pure compound with molecular formula \(\ce{C2H6O}\) produces two prominent proton NMR signals:

• A triplet at 1.18 ppm, integrating to 3H.
• A quartet at 3.34 ppm, integrating to 2H.

A student identifies the compound as dimethyl ether because the second signal occurs close to the expected chemical shift of its \(\ce{O-CH3}\) hydrogen atoms.

Explain why this identification is incorrect. Identify the compound and justify your answer using chemical shift, integration, and splitting.

Solution 3

The compound is ethanol, not dimethyl ether.

The chemical shift of 3.34 ppm is not enough to identify dimethyl ether because ethanol’s oxygen-adjacent methylene hydrogen atoms also appear in a similar region.

The integration and splitting provide the decisive evidence.

The 1.18 ppm signal is a triplet integrating to 3H. This identifies a methyl group next to two equivalent hydrogen atoms.

The 3.34 ppm signal is a quartet integrating to 2H. This identifies a methylene group next to three equivalent hydrogen atoms.

These two groups form the connected structure:

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

The higher chemical shift of the methylene group is consistent with its direct attachment to oxygen.

Dimethyl ether would instead produce one singlet integrating to 6H, rather than a separate 3H triplet and 2H quartet.

The key mistake is treating chemical shift as a unique identifier. Chemical shifts depend on the local electronic environment, and different compounds can have signals in similar regions. Integration and splitting must be considered alongside position.

Question 4

A sample of very dry ethanol is examined using a high-resolution proton NMR instrument.

Unlike the usual introductory spectrum, the OH hydrogen produces a triplet, while the methylene signal appears as a quartet that is further split into doublets.

The methyl signal remains a triplet.

Explain how this spectrum can still belong to ethanol. Why would the usual introductory \(n+1\) rule not predict these patterns?

Assume proton exchange is sufficiently slow and the separate coupling interactions can be resolved.

Solution 4

The spectrum can still belong to ethanol because slow OH proton exchange allows additional spin-spin coupling between the hydroxyl and methylene hydrogen atoms.

The usual introductory model assumes that the OH proton exchanges rapidly, preventing observable coupling with the neighbouring \(\ce{CH2}\) group.

Under the conditions in this question, that assumption no longer holds.

Consider each group separately.

The methyl group:

The methyl group is next to two methylene hydrogen atoms.

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

It remains a triplet.

The methylene group:

The methylene hydrogen atoms interact with two different sets of neighbouring hydrogen atoms:

• Three equivalent hydrogen atoms in the methyl group.
• One hydroxyl hydrogen.

The three methyl hydrogen atoms first split the methylene signal into a quartet.

The OH hydrogen then splits each of those four lines into a doublet.

The result is a quartet of doublets, containing eight lines when the two coupling interactions are resolved.

The hydroxyl group:

The OH hydrogen is next to two equivalent methylene hydrogen atoms.

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

It therefore produces a triplet.

This does not contradict the usual ethanol spectrum. It shows that the simple introductory splitting model depends on experimental conditions.

The important lesson is that the \(n+1\) rule cannot always treat every neighbouring hydrogen as part of one equivalent group. Different coupling interactions can produce separate layers of splitting.

Question 5

A sealed sample is known to contain only ethanol and dimethyl ether. It may contain either pure compound or a mixture of both.

Two proton NMR spectra are recorded under the same acquisition conditions.

Before adding \(\ce{D2O}\), the spectrum shows:

After adding \(\ce{D2O}\), the spectrum shows:

The broad shoulder disappears after adding \(\ce{D2O}\).

The broad shoulder is known to be an exchangeable OH signal, and its integration is not quantitatively reliable. The carbon-bound hydrogen signals have quantitative integrations with a common response factor.

Determine:

(a) Whether the sample contains pure ethanol, pure dimethyl ether, or a mixture.

(b) The mole ratio of ethanol to dimethyl ether.

(c) Why the original 37.0-unit region cannot be interpreted as one signal representing seven equivalent hydrogen atoms.

Explain how the two spectra provide enough evidence to identify both components, even though the OH signal overlaps with the dimethyl ether signal.

Solution 5

The sample is a 1:1 molar mixture of ethanol and dimethyl ether.

The key is to recognise that the original signal region near 3.32 ppm contains contributions from two different hydrogen environments. Adding \(\ce{D2O}\) removes the ethanol OH contribution but leaves the dimethyl ether signal intact.

Let’s establish the composition using the carbon-bound hydrogen signals.

Step 1: Identify the ethanol component.

The triplet at 1.20 ppm integrates to 18.0 units.

A methyl group in ethanol contains three hydrogen atoms, so its relative amount is proportional to:

\[
\frac{18.0}{3}=6
\]

The quartet at 3.60 ppm integrates to 12.0 units.

A methylene group in ethanol contains two hydrogen atoms:

\[
\frac{12.0}{2}=6
\]

Both signals give the same relative amount.

This is strong evidence that the triplet and quartet belong to the same compound: ethanol.

Step 2: Identify the dimethyl ether component.

After adding \(\ce{D2O}\), the narrow singlet at 3.32 ppm remains and integrates to 36.0 units.

Dimethyl ether contains six equivalent hydrogen atoms, so its relative amount is:

\[
\frac{36.0}{6}=6
\]

This is the same relative amount obtained from both ethanol signals.

The persistent singlet therefore identifies dimethyl ether.

Step 3: Calculate the mole ratio.

The relative amounts are:

\[
n(\text{ethanol})=6
\]

\[
n(\text{dimethyl ether})=6
\]

Therefore:

\[
\boxed{
\text{Ethanol}:\text{dimethyl ether}=1:1
}
\]

The sample contains equal mole amounts of the two compounds.

Step 4: Explain the disappearing shoulder.

Before adding \(\ce{D2O}\), the region near 3.32 ppm contains two overlapping contributions:

• A narrow singlet from the six equivalent hydrogen atoms in dimethyl ether.
• A broad signal from the exchangeable OH hydrogen in ethanol.

The combined region has an integration of 37.0 units.

After adding \(\ce{D2O}\), the OH hydrogen is replaced by deuterium. Its contribution disappears from the proton NMR spectrum.

The narrow dimethyl ether singlet remains with an integration of 36.0 units.

This explains the change in the spectrum.

Step 5: Identify the tempting incorrect interpretations.

There are two plausible mistakes.

The first is to identify the entire 37-unit region as one signal from seven equivalent hydrogen atoms.

That interpretation ignores the different line shapes and the response to deuterium exchange. The narrow singlet and broad shoulder represent different environments, even though they overlap in chemical shift.

The second is to identify the sample as pure ethanol because the triplet and quartet are clearly visible.

Those signals do identify ethanol, but they cannot explain the persistent 6H singlet after D2O addition.

Likewise, pure dimethyl ether cannot explain the 3H triplet and 2H quartet.

The full spectrum requires both compounds.

There is another important detail: we deliberately did not use the small OH integration to calculate the amount of ethanol. Exchangeable proton signals are not necessarily quantitative. The reliable carbon-bound hydrogen signals establish the mole ratio independently.

This question demonstrates why a good NMR interpretation must explain every significant feature of the spectrum, rather than selecting the first pattern that looks familiar.

08What to learn next

Once you can distinguish ethanol from dimethyl ether using proton NMR, the next useful step is combining NMR with infrared spectroscopy.

The two techniques provide different information:

• Proton NMR identifies hydrogen environments, relative numbers of hydrogen atoms, and neighbouring hydrogen atoms.
• Infrared spectroscopy helps identify bonds and functional groups through their characteristic absorptions.

For these two compounds, the presence of an O-H bond in ethanol provides an additional distinguishing feature. Ethanol produces a broad O-H absorption in its IR spectrum, while dimethyl ether does not contain an O-H bond.

Using both techniques gives you stronger evidence when identifying an unknown organic compound.

If you also need to revise how ethanol is produced, see Fermentation Production of Ethanol: HSC Chemistry Guide.

09Further reading

• Ausetute: Proton NMR Spectroscopy – typical ethanol chemical shifts and proton environments.
• Hebrew University: Proton NMR Reference – chemical shift variation, integration, and the effects of experimental conditions.
• JEOL: NMR Basics – sample preparation, radiofrequency excitation, signal detection, and spectrum processing.
• Proton Chemical Shifts in NMR: A Study of Ethers – experimental chemical shift data for ether compounds.