How to Distinguish Alcohol and Carboxylic-Acid O-H Bands

An unknown organic compound produces a broad O-H absorption in its infrared (IR) spectrum. Does it contain an alcohol or a carboxylic acid?

Both functional groups contain an O-H bond, and both can produce broad absorption bands. The difference is where the band appears, how far it spreads, and what other absorption features accompany it.

The key distinction is:

• Alcohol O-H: usually a broad absorption between \(3230\) and \(3550\ \mathrm{cm}^{-1}\).
• Carboxylic-acid O-H: a much broader absorption, commonly between \(2500\) and \(3000\ \mathrm{cm}^{-1}\), often extending towards \(3300\ \mathrm{cm}^{-1}\).
• Carboxylic-acid C=O: an additional strong absorption, usually within the \(1680\)-\(1750\ \mathrm{cm}^{-1}\) region.

The most reliable identification comes from combining the O-H band’s position and shape with evidence of a carbonyl group. Don’t classify a compound from the O-H band alone.

01Start with the structural difference

Before interpreting the spectrum, look at what makes the two functional groups different.

An alcohol contains a hydroxyl group attached to a carbon-containing group:

\[ \ce{R-OH} \]

A carboxylic acid contains a carboxyl group:

\[ \ce{R-C(=O)-OH} \]

Notice the important difference. A carboxylic acid contains both a carbonyl bond, \(\ce{C=O}\), and a hydroxyl bond, \(\ce{O-H}\), attached to the same carbon atom.

This gives us a useful prediction before looking at either spectrum.

An alcohol should produce an O-H absorption, but a simple alcohol should not produce a strong carbonyl absorption.

A carboxylic acid should produce both an O-H absorption and a strong C=O absorption.

There is one complication worth remembering: a molecule can contain an alcohol group and a separate carbonyl group. A hydroxyketone, for example, contains both. Therefore, finding C=O and O-H does not automatically establish that the compound is a carboxylic acid.

We need to examine the O-H band itself.

02What does an IR spectrum actually measure?

An IR spectrometer measures how a sample interacts with infrared radiation at different wavenumbers.

Chemical bonds are constantly vibrating. They can stretch, bend, and move in other ways. When infrared radiation has the appropriate frequency, it can transfer energy to a molecular vibration.

For a vibration to absorb infrared radiation, it must involve a change in the molecule’s dipole moment.

The instrument detects the radiation that passes through, or interacts with, the sample. It does not directly photograph individual bonds.

How the experiment works

Consider a simplified transmission experiment using a Fourier-transform infrared (FTIR) spectrometer.

The radiation follows this path:

1. An infrared source produces radiation.
2. An interferometer processes the radiation before it reaches the sample.
3. The radiation passes through a thin sample held in a suitable sample cell.
4. A detector measures the radiation that emerges.
5. A computer processes the measurements and produces the IR spectrum.

A background measurement helps account for absorption caused by the instrument, the atmosphere, and the solvent where applicable.

Suppose the unknown compound contains an O-H bond. We predict that some infrared radiation will be absorbed in the O-H stretching region.

The detector receives less radiation at those wavenumbers, and the computer produces an absorption feature in the spectrum.

However, the position and width of that feature depend on the molecular environment. This is particularly important when comparing alcohols and carboxylic acids.

Read the axes before interpreting the bands

The horizontal axis normally shows wavenumber, measured in reciprocal centimetres:

\[ \mathrm{cm}^{-1} \]

A conventional organic IR spectrum decreases from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(400\)-\(500\ \mathrm{cm}^{-1}\) on the right.

This means that:

• \(3500\ \mathrm{cm}^{-1}\) appears towards the left.
• \(3000\ \mathrm{cm}^{-1}\) appears further right.
• \(1700\ \mathrm{cm}^{-1}\) appears further right again.

Always check the actual numerical labels rather than assuming the axis direction.

The vertical axis also matters.

A downward trough on a percentage transmittance spectrum represents strong absorption, not weak absorption.

Don’t let the direction of the graph catch you out.

03Why are the two O-H bands different?

The main reason is hydrogen bonding.

Alcohols and carboxylic acids can both form hydrogen bonds between neighbouring molecules. However, carboxylic acids can form particularly strong, distinctive associations called hydrogen-bonded dimers.

Understanding this explains why their O-H absorptions have different shapes.

Alcohols: a broad O-H absorption

In an alcohol, the O-H bond is polar because oxygen attracts the shared electrons more strongly than hydrogen.

The hydrogen atom can therefore form an intermolecular hydrogen bond with an oxygen atom on another molecule.

These interactions affect the O-H bond’s vibration.

Think of the O-H bond as a spring connecting two atoms. A change in the interactions surrounding the bond changes its effective stiffness and therefore its vibrational frequency.

This analogy helps explain the shift in absorption, although real molecular vibrations involve more complex interactions than a mechanical spring.

In a liquid alcohol, many molecules experience slightly different hydrogen-bonding environments. This produces a range of O-H stretching frequencies rather than one perfectly sharp absorption.

The result is a broad band, usually around:

\[ 3230-3550\ \mathrm{cm}^{-1} \]

Alcohol O-H absorptions are therefore broad, but they generally remain in the higher-wavenumber region.

Carboxylic acids: a very broad O-H absorption

Carboxylic acids have an additional feature: they readily form hydrogen-bonded dimers.

A dimer consists of two carboxylic acid molecules held together by two hydrogen bonds.

Each molecule’s hydroxyl hydrogen interacts with the carbonyl oxygen of the other molecule. Together, these interactions produce a cyclic arrangement.

These interactions have a significant effect on the O-H stretching vibration.

The hydrogen bonding weakens the effective O-H bond, shifting its stretching absorption to a lower wavenumber.

Because the hydrogen-bonding environments vary, the absorption spreads across a much wider range.

This produces the characteristic very broad carboxylic-acid O-H absorption, often extending from approximately:

\[ 2500-3300\ \mathrm{cm}^{-1} \]

A common HSC-style reference table lists the carboxylic-acid O-H range as \(2500\)-\(3000\ \mathrm{cm}^{-1}\). The wider envelope extending towards \(3300\ \mathrm{cm}^{-1}\) is also commonly observed in full reference spectra.

The important feature is not an exact upper boundary. It is the unusually broad absorption extending into the lower-wavenumber region.

Why can the acid band overlap with C-H absorptions?

This is one of the most useful details when interpreting real spectra.

Alkyl C-H stretching absorptions commonly occur between approximately \(2850\) and \(3000\ \mathrm{cm}^{-1}\).

The carboxylic-acid O-H envelope overlaps this region.

As a result, the spectrum may contain several smaller, relatively narrow C-H absorptions sitting within the much broader O-H absorption.

You might see what looks like several small features emerging from a large, broad trough.

These smaller features do not necessarily represent additional O-H bonds. They can be C-H stretching absorptions superimposed on the acid O-H envelope.

This is a particularly useful clue when the spectrum looks less tidy than you expected.

04Compare the characteristic absorption ranges

The following table summarises the main features to recognise.

These are approximate ranges. Hydrogen bonding, molecular structure, concentration, solvent, and physical state can all affect the precise position and shape.

Use the reference data supplied in your examination question where applicable.

What should the spectra look like?

Consider two illustrative percentage-transmittance spectra.

The alcohol has a relatively rounded O-H absorption in the higher-wavenumber region, approximately \(3230\)-\(3550\ \mathrm{cm}^{-1}\).

The carboxylic acid has a much wider O-H envelope extending towards \(2500\ \mathrm{cm}^{-1}\). It also has a strong carbonyl absorption near \(1710\ \mathrm{cm}^{-1}\).

The key comparison is the combination of:

• The position of the O-H absorption.
• The total width of the O-H envelope.
• The presence or absence of a strong carbonyl absorption.

The C-O region can provide additional information, but it is not decisive. Both alcohols and carboxylic acids contain C-O bonds.

A useful calculation: comparing band widths

You can quantify the difference in the two illustrative spectra by calculating the approximate width of each O-H envelope.

The width is the difference between its higher and lower wavenumber boundaries:

\[ \Delta\tilde{\nu} = \tilde{\nu}_{\text{high}} - \tilde{\nu}_{\text{low}} \]

Suppose the acid envelope extends from \(2520\) to \(3220\ \mathrm{cm}^{-1}\).

\[ \begin{aligned} \Delta\tilde{\nu}_{\text{acid}} &=3220-2520\\ &=700\ \mathrm{cm}^{-1} \end{aligned} \]

Suppose the alcohol band extends from \(3290\) to \(3510\ \mathrm{cm}^{-1}\).

\[ \begin{aligned} \Delta\tilde{\nu}_{\text{alcohol}} &=3510-3290\\ &=220\ \mathrm{cm}^{-1} \end{aligned} \]

The acid envelope is approximately:

\[ \frac{700}{220}\approx 3.2 \]

times wider in this illustrative comparison.

This is not a universal numerical test. It simply demonstrates why the full width of the absorption is more informative than identifying one wavenumber.

05Why sample conditions can change the O-H band

There is an experimental complication worth understanding.

Suppose a chemist records the IR spectrum of a neat liquid alcohol. Its molecules form intermolecular hydrogen bonds, so the O-H absorption is broad.

Now the chemist dilutes the alcohol in a suitable non-polar solvent.

What should happen?

The alcohol molecules are further apart on average, reducing intermolecular hydrogen bonding. A sharper, higher-wavenumber O-H absorption can become more prominent.

Unassociated alcohol O-H groups commonly absorb around \(3600\)-\(3650\ \mathrm{cm}^{-1}\).

Carboxylic acids can also change their absorption patterns when diluted.

As their concentration decreases, hydrogen-bonded dimers become less favourable. The broad, low-wavenumber acid envelope can weaken, while a sharper, higher-wavenumber monomer O-H absorption emerges.

This means that a sharp feature at a high wavenumber does not automatically prove the presence of an alcohol.

Improving the experimental design

A single spectrum of a neat sample may not provide enough information about how hydrogen bonding affects the observed band.

A better investigation would:

1. Record a background spectrum or appropriate solvent blank.
2. Measure the sample at a controlled temperature.
3. Prepare a series of concentrations using a suitable dry solvent.
4. Keep the sample path length consistent.
5. Compare the O-H and C=O regions across the measurements.

For a carboxylic acid, dilution can shift the balance between associated dimers and individual acid molecules:

\[ 2\,\ce{RCOOH} \rightleftharpoons (\ce{RCOOH})_2 \]

At higher concentrations, dimer formation becomes more favourable. At lower concentrations, the equilibrium shifts towards individual molecules.

The O-H band changes because the molecular environment changes, not because the compound has gained or lost an O-H bond.

The carbonyl absorption can also shift as the acid dimer dissociates.

This is a useful experimental extension because it demonstrates why sample conditions matter. For routine HSC spectrum interpretation, however, focus first on the characteristic O-H ranges and supporting carbonyl evidence.

06A reliable method for distinguishing the two bands

When you are given an unfamiliar IR spectrum, follow this sequence.

Step 1: Check the axes.

Confirm that the horizontal axis shows wavenumber and determine whether the vertical axis is percentage transmittance or absorbance.

Step 2: Locate the entire O-H absorption.

Don’t just identify the lowest point of the trough. Look at where the absorption begins and ends.

Step 3: Examine its position and width.

• A broad absorption around \(3230\)-\(3550\ \mathrm{cm}^{-1}\) is consistent with an alcohol.
• A very broad absorption extending towards \(2500\)-\(3000\ \mathrm{cm}^{-1}\) strongly supports a carboxylic acid.

Step 4: Check for a carbonyl absorption.

Look for a strong absorption between \(1680\) and \(1750\ \mathrm{cm}^{-1}\).

Its presence supports a carboxylic acid when combined with the very broad acid O-H feature.

Remember that ketones, aldehydes, and esters also contain carbonyl groups.

Step 5: Check the complete pattern.

Make sure the proposed functional group explains the observed features together.

Do not treat an isolated C-O absorption or a single O-H wavenumber as definitive proof.

07Worked examples

08Questions and solutions

Question 1

An unknown pure organic compound produces the following IR absorption features:

• A broad absorption between \(3310\) and \(3490\ \mathrm{cm}^{-1}\).
• A strong absorption near \(1050\ \mathrm{cm}^{-1}\).
• No strong absorption between \(1680\) and \(1750\ \mathrm{cm}^{-1}\).

Is the compound more likely to be an alcohol or a carboxylic acid? Justify your answer using the absorption data.

Solution 1

The compound is more likely to be an alcohol.

The O-H absorption between \(3310\) and \(3490\ \mathrm{cm}^{-1}\) falls within the typical alcohol O-H range of \(3230\)-\(3550\ \mathrm{cm}^{-1}\).

The absorption near \(1050\ \mathrm{cm}^{-1}\) is consistent with C-O stretching, which provides additional support for an alcohol.

Most importantly, there is no strong carbonyl absorption between \(1680\) and \(1750\ \mathrm{cm}^{-1}\).

A carboxylic acid contains a C=O bond, so we would expect a strong absorption in this region.

Since the full spectrum is available and of sufficient quality, the absence of this feature makes a carboxylic acid less likely.

The combination of a higher-wavenumber O-H band, a C-O absorption, and no detectable carbonyl absorption supports an alcohol.

Question 2

An unknown organic compound produces the following features in a percentage-transmittance IR spectrum:

• A very broad absorption extending from \(2510\) to \(3210\ \mathrm{cm}^{-1}\).
• Three smaller, narrower absorptions at \(2960\), \(2930\), and \(2875\ \mathrm{cm}^{-1}\), superimposed on the broad absorption.
• A strong absorption at \(1714\ \mathrm{cm}^{-1}\).
• An additional absorption near \(1260\ \mathrm{cm}^{-1}\).

A student makes two claims:

1. The broad O-H absorption must belong to an alcohol because it extends above \(3000\ \mathrm{cm}^{-1}\).
2. The three smaller absorptions must represent three separate O-H bonds.

Evaluate both claims and identify the most likely functional group.

Solution 2

The compound is most likely a carboxylic acid, and both student claims are incorrect.

The first claim focuses on the upper boundary of the O-H absorption rather than its full width.

The absorption extends from \(2510\) to \(3210\ \mathrm{cm}^{-1}\), making it extremely broad. Its lower-wavenumber extension is characteristic of a carboxylic-acid O-H absorption.

The fact that the upper boundary exceeds \(3000\ \mathrm{cm}^{-1}\) does not make it an alcohol band.

The second claim incorrectly assigns every smaller absorption to an O-H bond.

The features at \(2960\), \(2930\), and \(2875\ \mathrm{cm}^{-1}\) are consistent with C-H stretching vibrations. These can appear as narrower features within the broad acid O-H envelope.

The strong absorption at \(1714\ \mathrm{cm}^{-1}\) provides further evidence of a carbonyl group.

Finally, the absorption near \(1260\ \mathrm{cm}^{-1}\) is consistent with C-O stretching, although this is not unique to carboxylic acids.

The combination of a very broad, low-wavenumber O-H envelope and a strong C=O absorption supports a carboxylic acid.

Question 3

Two possible structures are proposed for an unknown pure organic compound:

Structure A: Butanoic acid

\[ \ce{CH3CH2CH2COOH} \]

Structure B: 4-hydroxybutan-2-one

\[ \ce{CH3COCH2CH2OH} \]

The IR spectrum contains:

• A broad O-H absorption between \(3280\) and \(3520\ \mathrm{cm}^{-1}\).
• A strong C=O absorption at \(1718\ \mathrm{cm}^{-1}\).
• A C-O absorption near \(1115\ \mathrm{cm}^{-1}\).
• Several narrower C-H absorptions between \(2850\) and \(3000\ \mathrm{cm}^{-1}\), without a continuous, very broad O-H envelope extending down towards \(2500\ \mathrm{cm}^{-1}\).

Determine which structure is more consistent with the spectrum.

Explain why the presence of both O-H and C=O absorptions is not sufficient, by itself, to identify a carboxylic acid.

Solution 3

Structure B, 4-hydroxybutan-2-one, is more consistent with the observed spectrum.

The O-H absorption between \(3280\) and \(3520\ \mathrm{cm}^{-1}\) is characteristic of an alcohol O-H stretching vibration.

The absence of a very broad absorption extending down towards \(2500\ \mathrm{cm}^{-1}\) makes the carboxylic-acid O-H assignment less likely.

However, there is also a strong carbonyl absorption at \(1718\ \mathrm{cm}^{-1}\).

This is where the structural comparison becomes important.

Structure A contains a carboxylic acid group. It should therefore produce a very broad acid O-H absorption as well as a carbonyl absorption.

Structure B contains a separate alcohol group and a ketone group. Its alcohol produces the higher-wavenumber O-H absorption, while its ketone produces the strong C=O absorption.

Therefore, the two observed bands do not necessarily belong to the same functional group.

The C-O absorption near \(1115\ \mathrm{cm}^{-1}\) is consistent with Structure B, but it is not decisive because both structures contain C-O bonds.

The combination of an alcohol-like O-H band and a carbonyl absorption supports 4-hydroxybutan-2-one.

This example demonstrates why the exact position and shape of the O-H absorption matter even when a carbonyl group is present.

Question 4

A laboratory report contains an incomplete IR spectrum.

Only the region between \(3600\) and \(3100\ \mathrm{cm}^{-1}\) is displayed.

Within this region, a broad absorption appears near \(3380\ \mathrm{cm}^{-1}\). The absorption is still extending downwards when the plotted spectrum ends at \(3100\ \mathrm{cm}^{-1}\).

A student concludes that the compound is an alcohol because no carboxylic-acid O-H absorption is visible.

Can this conclusion be justified?

Explain which additional spectral regions should be measured before making a reliable identification.

Solution 4

No. The available spectrum does not contain enough information to distinguish an alcohol from a carboxylic acid reliably.

The main problem is that the spectrum has been cropped.

A characteristic carboxylic-acid O-H absorption can extend from approximately \(2500\) to \(3300\ \mathrm{cm}^{-1}\).

The displayed spectrum stops at \(3100\ \mathrm{cm}^{-1}\), so it does not show the full region needed to examine the lower-wavenumber extension of the absorption.

The visible band near \(3380\ \mathrm{cm}^{-1}\) is consistent with an alcohol O-H absorption, but a cropped spectrum cannot establish that the characteristic acid absorption is absent.

There is also no information about the carbonyl region.

Before identifying the compound, the laboratory should obtain a more complete spectrum covering approximately \(4000\) to \(500\ \mathrm{cm}^{-1}\).

Two regions are particularly important:

• \(2500\)-\(3300\ \mathrm{cm}^{-1}\): examine the full width and lower-wavenumber extension of the O-H absorption.
• \(1680\)-\(1750\ \mathrm{cm}^{-1}\): check for a strong carbonyl absorption.

A reliable identification requires the complete pattern of evidence.

The absence of an absorption from a region that was never measured cannot be used to rule out a functional group.

Question 5

A chemist is investigating an unknown compound that is known to be a single pure substance rather than a mixture.

Two IR spectra are recorded at the same temperature using the same sample path length and a dry, non-polar solvent.

The following observations are obtained:

Two possible structures are being considered:

• Model A: A monofunctional carboxylic acid containing one O-H bond.
• Model B: A hydroxyketone containing one alcohol O-H bond and one ketone C=O bond. Assume the ketone carbonyl is not involved in hydrogen bonding.

A student confidently argues that the appearance of the narrower, higher-wavenumber O-H absorption proves that the molecule must contain a second type of hydroxyl group, meaning Model B is correct.

Is this conclusion justified?

Determine which model better explains all the observations. Your answer must explain the relationship between concentration, hydrogen bonding, O-H absorption, and the change in carbonyl wavenumber.

Solution 5

Model A, the monofunctional carboxylic acid, better explains the complete set of observations. The narrower O-H absorption does not require the presence of a second hydroxyl group.

The key is that the same O-H bond can produce different absorption features depending on its molecular environment.

Step 1: Interpret the concentrated sample.

At higher concentration, carboxylic acid molecules can associate to form hydrogen-bonded dimers.

The equilibrium is:

\[ 2\,\ce{RCOOH} \rightleftharpoons (\ce{RCOOH})_2 \]

In the dimer, the O-H bonds participate in strong intermolecular hydrogen bonding with neighbouring carbonyl oxygen atoms.

This weakens the effective O-H bonds and shifts their stretching absorption to lower wavenumbers.

The variety of hydrogen-bonding environments produces a very broad absorption extending from \(2520\) to \(3260\ \mathrm{cm}^{-1}\).

This is characteristic of associated carboxylic acids.

Step 2: Explain what happens when the sample is diluted.

Dilution shifts the association equilibrium towards individual carboxylic acid molecules.

The proportion of hydrogen-bonded dimers decreases, so the very broad, low-wavenumber O-H envelope becomes weaker.

The individual acid molecules now produce a narrower, higher-wavenumber O-H absorption.

This explains why an absorption near \(3535\ \mathrm{cm}^{-1}\) becomes prominent.

It does not mean that a new hydroxyl group has appeared.

The same O-H bond is absorbing at a different wavenumber because its hydrogen-bonding environment has changed.

Step 3: Examine the carbonyl shift.

The carbonyl absorption also changes:

\[ \begin{aligned} \Delta\tilde{\nu}_{\ce{C=O}} &=1735-1710\\ &=25\ \mathrm{cm}^{-1} \end{aligned} \]

The carbonyl absorption has shifted to a higher wavenumber by \(25\ \mathrm{cm}^{-1}\).

This is consistent with the change from associated acid dimers to individual acid molecules.

The carbonyl group’s molecular environment changes as the hydrogen-bonded dimer dissociates, affecting its stretching vibration.

This is particularly important because it provides independent evidence that the O-H and C=O changes are related to the same acid-association process.

Step 4: Test the competing model.

Model B contains an alcohol O-H group and a separate ketone C=O group.

Dilution could reduce intermolecular hydrogen bonding between alcohol molecules, causing the alcohol O-H absorption to become sharper and shift higher.

So the change in O-H absorption alone does not eliminate Model B.

However, under the stated assumption that the ketone carbonyl is not involved in hydrogen bonding, its absorption should not undergo the same concentration-dependent shift associated with acid dimer dissociation.

Model B therefore does not explain the coordinated change in both the O-H and C=O absorptions as well as Model A.

Step 5: Reach the conclusion.

Model A is the better explanation.

The concentrated sample contains hydrogen-bonded carboxylic acid dimers, producing a very broad, low-wavenumber O-H absorption and a carbonyl absorption near \(1710\ \mathrm{cm}^{-1}\).

Dilution favours individual acid molecules, producing a narrower, higher-wavenumber O-H absorption and shifting the carbonyl absorption towards \(1735\ \mathrm{cm}^{-1}\).

The important conceptual point is that one functional group can produce different absorption features when its molecular environment changes.

A new absorption band does not necessarily mean a new bond or functional group has appeared.

09Where to go next

Once you can distinguish alcohol and carboxylic-acid O-H absorptions, the next step is combining this evidence with carbonyl and fingerprint-region absorptions to identify the most likely organic structure.

For more practice choosing between competing organic structures using multiple IR features, continue with How to Choose Between Organic Structures Using Two IR Features.

If you want to strengthen your understanding of why carboxylic acids form hydrogen-bonded dimers and how this affects their properties, see HSC Chemistry: Carboxylic Acids, Amines and Amides Explained.