How to Identify a Carbonyl Group in an IR Spectrum

An unknown liquid gives a strong absorption band at \(1722\ \mathrm{cm}^{-1}\). Does it contain a carbonyl group?

Probably. A carbonyl group, \(\ce{C=O}\), produces a characteristic, usually strong infrared absorption between \(1680\) and \(1750\ \mathrm{cm}^{-1}\). However, before making that conclusion, you need to check the spectrum’s axes, the direction of the absorption band, and any other features that support or challenge your interpretation.

The reliable method is to find a strong absorption in the carbonyl region, confirm that it represents absorption rather than transmission, and compare it with other characteristic bands.

One important distinction: identifying a carbonyl group does not automatically tell you whether the compound is an aldehyde, ketone, carboxylic acid, or ester.

01What does a carbonyl group look like in an IR spectrum?

A carbonyl group consists of a carbon atom double bonded to an oxygen atom:

\[ \ce{C=O} \]

When analysing an IR spectrum, look for a strong absorption between:

\[ \boxed{1680-1750\ \mathrm{cm}^{-1}} \]

This is the characteristic carbonyl absorption range given on the HSC Chemistry data sheet.

Why does a carbonyl bond absorb infrared radiation?

Atoms in a molecule are constantly vibrating. Their bonds can stretch, bend, and move in other ways.

Imagine two atoms connected by a spring. Pull the atoms apart, and the spring stretches. Release them, and they move towards each other again.

A chemical bond behaves somewhat like this spring. The analogy is useful for picturing molecular vibrations, although real bonds follow quantum mechanical rules and their vibrations can interact with other bonds in the molecule.

A carbonyl bond is particularly useful in IR spectroscopy because it is strongly polar. When the \(\ce{C=O}\) bond stretches, the molecule experiences a substantial change in its dipole moment.

This allows the bond to absorb infrared radiation effectively.

The instrument detects this absorption as a strong band in the carbonyl region.

What actually happens inside the instrument?

In a basic transmission IR experiment, the process works as follows:

1. An infrared source produces radiation across a range of wavenumbers.
2. The instrument directs this radiation towards a sample.
3. The radiation passes through the sample, where certain frequencies are absorbed by molecular vibrations.
4. A detector measures how much radiation passes through the sample.
5. The instrument processes the measurements to produce an IR spectrum.

Modern FTIR instruments use additional optical components to collect and process the signal, but the essential measurement principle is the same.

What should the instrument detect if the sample contains a carbonyl group?

It should detect a noticeable reduction in transmitted infrared radiation around \(1680-1750\ \mathrm{cm}^{-1}\).

That is the experimental evidence supporting the presence of a carbonyl group.

However, a raw detector signal can also contain contributions from the instrument, the sample cell, the solvent, and the surrounding atmosphere. A background measurement helps account for these contributions.

This is why a reliable carbonyl assignment should be based on the sample’s absorption after appropriate background correction, rather than simply identifying an interesting-looking feature in a raw signal.

02Read the IR spectrum correctly before identifying carbonyl

You can recognise the correct absorption range and still get the question wrong if you misread the axes.

There are two axes to check.

The horizontal axis: wavenumber

The horizontal axis of a typical IR spectrum represents wavenumber, measured in reciprocal centimetres:

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

Wavenumber is the reciprocal of wavelength. A higher wavenumber corresponds to a shorter wavelength.

A typical HSC IR spectrum has a horizontal axis that decreases from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(400\ \mathrm{cm}^{-1}\) on the right.

This means:

• \(3500\ \mathrm{cm}^{-1}\) appears towards the left.
• \(1700\ \mathrm{cm}^{-1}\) appears closer to the middle.
• \(1100\ \mathrm{cm}^{-1}\) appears further towards the right.

This direction is important. Do not assume that the numbers increase as you move right.

The carbonyl region, \(1680-1750\ \mathrm{cm}^{-1}\), sits between the higher-wavenumber O-H region and the lower-wavenumber fingerprint region.

The vertical axis: transmittance or absorbance?

This is one of the most common sources of mistakes in IR questions.

An IR spectrum can display either percentage transmittance or absorbance. The same absorption looks different depending on which measurement is plotted.

If the spectrum shows percentage transmittance, a strong carbonyl absorption appears as a deep downward trough.

If the spectrum shows absorbance, the same carbonyl absorption appears as a strong upward peak.

Notice that the absorption itself has not changed. Only the way it is represented has changed.

A downward trough on a transmittance spectrum is not a weak absorption. It indicates that less radiation has passed through the sample.

Always check the vertical axis before interpreting the height or direction of a band.

A small calculation: converting wavenumber into wavelength

You might see a wavenumber such as \(1715\ \mathrm{cm}^{-1}\) and wonder what wavelength of infrared radiation it represents.

The relationship is:

\[ \tilde{\nu}=\frac{1}{\lambda} \]

where:

• \(\tilde{\nu}\) is wavenumber, in \(\mathrm{cm}^{-1}\).
• \(\lambda\) is wavelength, in centimetres.

For a carbonyl absorption at \(1715\ \mathrm{cm}^{-1}\):

\[ \begin{aligned} \lambda &= \frac{1}{1715}\\ &=5.83\times10^{-4}\ \mathrm{cm}\\ &=5.83\times10^{-6}\ \mathrm{m}\\ &=5.83\ \mu\mathrm{m} \end{aligned} \]

This means the carbonyl bond absorbs infrared radiation with a wavelength of approximately \(5.83\ \mu\mathrm{m}\).

You do not normally need to convert wavenumbers into wavelengths to identify functional groups in HSC questions. The calculation simply shows what the wavenumber represents physically.

03Know the carbonyl absorption range and its competitors

The HSC Chemistry data sheet gives the characteristic absorption ranges for several important bonds.

The most useful ones for identifying carbonyl groups are:

These ranges match the 2026 NESA Chemistry HSC data sheet.

For students sitting the 2026 or 2027 HSC, the relevant course is the Chemistry Stage 6 Syllabus (2017). NESA schedules the 2025 syllabus for implementation from 2028, with its first HSC examination in 2029. See the NESA syllabus information.

Why is the C=O absorption different from C=C?

Both bonds are double bonds, but they do not absorb at exactly the same wavenumber.

The HSC data sheet gives:

• C=O: \(1680-1750\ \mathrm{cm}^{-1}\)
• C=C: \(1620-1680\ \mathrm{cm}^{-1}\)

This makes the distinction useful when interpreting a spectrum containing several bands.

For example, a strong absorption at \(1720\ \mathrm{cm}^{-1}\) is consistent with a carbonyl group. A medium absorption at \(1650\ \mathrm{cm}^{-1}\) is more consistent with a carbon-carbon double bond.

The ranges are approximate, so do not treat their boundaries as perfectly sharp dividing lines. Molecular structure and experimental conditions can affect the precise position of an absorption.

Why should you also examine O-H and C-O?

A carbonyl absorption tells you that a \(\ce{C=O}\) bond is likely present. Other absorptions help narrow down which functional group might be responsible.

For example:

• A strong C=O absorption together with a very broad O-H absorption at \(2500-3000\ \mathrm{cm}^{-1}\) supports a carboxylic acid.
• A strong C=O absorption together with strong C-O absorptions at \(1000-1300\ \mathrm{cm}^{-1}\), but without the characteristic acid O-H band, supports an ester.
• A strong C=O absorption without an O-H absorption could be consistent with a ketone or aldehyde.

One important detail: an alcohol O-H band is usually broad in the \(3230-3550\ \mathrm{cm}^{-1}\) region. A carboxylic acid O-H band is very broad and appears at lower wavenumbers, approximately \(2500-3000\ \mathrm{cm}^{-1}\).

These features can overlap with other absorptions, particularly C-H stretching. Interpret the complete pattern rather than treating every broad feature as a separate functional group.

Also, a C-O absorption alone does not establish that a carbonyl group is present. You need evidence in the carbonyl region.

04A reliable method for identifying carbonyl

When given an unfamiliar IR spectrum, use this sequence.

Step 1: Check the horizontal axis.

Confirm that it represents wavenumber in \(\mathrm{cm}^{-1}\). Check whether the values decrease from left to right.

Step 2: Check the vertical axis.

Determine whether the spectrum shows percentage transmittance or absorbance.

Remember:

• %T: absorption points down.
• Absorbance: absorption points up.

Step 3: Find the carbonyl region.

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

Step 4: Examine the strength and shape of the band.

A strong, relatively distinct absorption in this region supports the presence of a carbonyl bond.

Do not reject a carbonyl assignment simply because the band is not exactly at \(1715\ \mathrm{cm}^{-1}\).

Step 5: Check the other absorptions.

Look for O-H, C-O, and C=C bands that support or challenge your interpretation.

Step 6: State your conclusion carefully.

A strong absorption in the carbonyl region is evidence for a C=O bond. It does not, by itself, identify the complete molecule or distinguish every carbonyl-containing functional group.

05Worked examples

06Questions and solutions

Question 1

An unknown organic compound produces a percentage transmittance IR spectrum with a strong downward trough at \(1732\ \mathrm{cm}^{-1}\).

Identify the bond most likely responsible for this absorption. Explain your reasoning using the HSC Chemistry data sheet.

Solution 1

The compound most likely contains a carbonyl bond, \(\ce{C=O}\).

The absorption occurs at \(1732\ \mathrm{cm}^{-1}\), which lies within the characteristic carbonyl range of \(1680-1750\ \mathrm{cm}^{-1}\).

The vertical axis shows percentage transmittance, so the downward trough indicates that infrared radiation has been absorbed by the sample.

The strong absorption is consistent with the polar C=O bond undergoing stretching vibrations.

However, the absorption does not identify the specific carbonyl-containing functional group.

Question 2

Two IR spectra have been obtained from the same organic sample.

• Spectrum A displays percentage transmittance and has a downward trough at \(1718\ \mathrm{cm}^{-1}\).
• Spectrum B displays absorbance and has an upward peak at \(1718\ \mathrm{cm}^{-1}\).

A student claims that only Spectrum A shows a carbonyl group because carbonyl absorptions must appear as downward troughs.

Evaluate this claim. Explain why the two spectra can represent the same molecular absorption.

Solution 2

Both spectra show evidence of a carbonyl group. The student’s claim is incorrect because the direction of an absorption feature depends on the vertical axis.

In Spectrum A, the downward trough indicates that less infrared radiation is transmitted through the sample at \(1718\ \mathrm{cm}^{-1}\).

In Spectrum B, the upward peak indicates increased absorbance at exactly the same wavenumber.

Both measurements describe the same event: the sample absorbs infrared radiation associated with the C=O stretching vibration.

Furthermore, \(1718\ \mathrm{cm}^{-1}\) falls within the characteristic carbonyl range of \(1680-1750\ \mathrm{cm}^{-1}\).

The important distinction is between the two ways of displaying absorption, not between the chemical properties of the samples.

The absolute heights of peaks on absorbance and percentage transmittance graphs also cannot be compared directly as though they used the same vertical scale.

Question 3

An unknown organic sample produces the following features in its percentage transmittance IR spectrum:

(a) Identify the absorption most likely associated with the carbonyl group.

(b) Identify the likely bonds responsible for the other two absorptions.

(c) A student concludes that the unknown must be an ester. Evaluate this conclusion.

Solution 3

(a) The strong absorption at \(1740\ \mathrm{cm}^{-1}\) is most likely due to a carbonyl group.

It lies within the characteristic C=O range of \(1680-1750\ \mathrm{cm}^{-1}\). The strong downward trough on the percentage transmittance spectrum indicates substantial infrared absorption.

(b) The other two absorptions are consistent with:

• \(1652\ \mathrm{cm}^{-1}\): C=C stretching, within the \(1620-1680\ \mathrm{cm}^{-1}\) range.
• \(1170\ \mathrm{cm}^{-1}\): C-O stretching, within the \(1000-1300\ \mathrm{cm}^{-1}\) range.

(c) An ester is a plausible interpretation, but the evidence does not prove that the compound must be an ester.

An ester contains both a carbonyl bond and a C-O bond, so the strong absorptions at \(1740\) and \(1170\ \mathrm{cm}^{-1}\) support this possibility.

The absence of the very broad carboxylic acid O-H absorption between \(2500\) and \(3000\ \mathrm{cm}^{-1}\) makes a carboxylic acid less likely.

However, the presence of C=O and C-O absorptions does not prove that these bonds are connected in an ester functional group. Other combinations of functional groups could produce similar evidence.

If the only two candidate structures were an ester and a carboxylic acid, the ester would be the better-supported choice.

This is an important distinction: IR spectroscopy identifies characteristic bonds and functional groups, but a single spectrum does not always establish a unique molecular structure.

Question 4

A student analyses an unknown organic liquid dissolved in solvent S.

The student records two percentage transmittance spectra:

• A blank spectrum containing only solvent S.
• A spectrum containing the unknown compound dissolved in solvent S.

Both raw spectra show a downward trough at \(1710\ \mathrm{cm}^{-1}\) with almost identical depth.

After background correction, the unknown sample has no distinct absorption between \(1680\) and \(1750\ \mathrm{cm}^{-1}\). However, it shows a new strong absorption at \(1120\ \mathrm{cm}^{-1}\).

Assume the sample concentration and measurement quality are adequate.

(a) Can the original trough at \(1710\ \mathrm{cm}^{-1}\) be used as evidence for a carbonyl group in the unknown?

(b) What does the new absorption at \(1120\ \mathrm{cm}^{-1}\) suggest?

(c) Explain one improvement that could be made to the experimental method.

Solution 4

(a) No. The original \(1710\ \mathrm{cm}^{-1}\) trough cannot reliably be attributed to the unknown compound.

The blank containing only solvent S already produces an absorption at the same wavenumber with almost identical depth.

This means the feature is associated with the background measurement, rather than providing evidence that the unknown compound itself contains a carbonyl group.

After background correction, there is no distinct absorption in the carbonyl region.

Given the stated assumption of adequate measurement quality, the experiment provides no convincing evidence for a C=O bond.

(b) The absorption at \(1120\ \mathrm{cm}^{-1}\) is consistent with a C-O stretching vibration.

It lies within the characteristic C-O range of \(1000-1300\ \mathrm{cm}^{-1}\).

However, this absorption alone does not establish the presence of a carbonyl group.

(c) The student could repeat the experiment using a solvent that does not absorb significantly in the carbonyl region.

A correctly matched background measurement should then allow the instrument to distinguish absorptions caused by the unknown compound from those caused by the solvent and experimental setup.

The key experimental principle is that an absorption assigned to the unknown must be attributable to the sample, not simply present in the measurement system.

Question 5

An unknown liquid produces an absorbance spectrum with the following features:

• A strong absorption at \(1715\ \mathrm{cm}^{-1}\).
• A broad absorption between \(3230\) and \(3550\ \mathrm{cm}^{-1}\).
• A strong absorption at \(1080\ \mathrm{cm}^{-1}\).
• No very broad absorption between \(2500\) and \(3000\ \mathrm{cm}^{-1}\).

Two possible explanations have been proposed.

Model A: The sample is a single compound, 4-hydroxybutan-2-one:

\[ \ce{CH3COCH2CH2OH} \]

Model B: The sample is a mixture of propanone and ethanol:

\[ \ce{CH3COCH3} \]

and

\[ \ce{CH3CH2OH} \]

A student argues that Model A must be correct because the IR spectrum shows both a carbonyl absorption and an alcohol O-H absorption.

Evaluate the student’s argument.

In your answer:

• Identify the bonds responsible for the three main absorptions.
• Explain why both models can produce the observed pattern.
• State what additional information would be needed to distinguish the models.
• Explain how your conclusion would change if the sample were independently confirmed to be a single pure compound.

Solution 5

The student’s conclusion is not justified. The IR spectrum establishes that the sample contains evidence of both carbonyl and alcohol-type O-H bonds, but it does not prove that these bonds belong to the same molecule.

First, identify the three main absorptions.

The strong absorption at \(1715\ \mathrm{cm}^{-1}\) is consistent with a carbonyl bond.

It lies within the characteristic C=O range of \(1680-1750\ \mathrm{cm}^{-1}\).

The broad absorption between \(3230\) and \(3550\ \mathrm{cm}^{-1}\) is consistent with an alcohol O-H bond.

The strong absorption at \(1080\ \mathrm{cm}^{-1}\) is consistent with a C-O stretching vibration.

The absence of the very broad carboxylic acid O-H feature between \(2500\) and \(3000\ \mathrm{cm}^{-1}\) makes a carboxylic acid less likely.

Now compare the two models.

Model A contains both a ketone and an alcohol within the same molecule.

Its ketone produces the C=O absorption, while its alcohol produces the broad O-H absorption and contributes to the C-O absorption.

Model B contains two separate compounds.

Propanone produces the C=O absorption, while ethanol produces the alcohol O-H and C-O absorptions.

Both models therefore explain the observed IR features.

The tempting conclusion is that seeing both functional groups in one spectrum proves they belong to one molecule. That is the exact mistake.

An IR spectrum records absorptions from everything present in the sample. It does not automatically reveal which individual bonds belong to which molecule.

What additional evidence would distinguish the models?

Mass spectrometry could help establish the molecular masses of the compounds present.

Model A has a molecular formula of \(\ce{C4H8O2}\), whereas Model B contains propanone, \(\ce{C3H6O}\), and ethanol, \(\ce{C2H6O}\).

The different molecular masses provide a way to test the competing explanations.

NMR spectroscopy could provide further information about the different carbon and hydrogen environments and help establish how the atoms are connected.

Chromatography could also help determine whether the original sample contains multiple components.

Finally, consider what happens if the sample is confirmed to be pure.

If independent testing establishes that the sample contains only one pure compound, Model B can be eliminated.

The carbonyl and alcohol O-H groups must then belong to the same molecule.

However, even this does not prove that the compound is specifically 4-hydroxybutan-2-one. Other molecular structures could contain both functional groups and produce similar IR features.

The IR spectrum establishes the presence of characteristic bonds. Additional evidence is needed to determine their exact arrangement.

This is the limitation of IR spectroscopy: it is excellent for identifying functional groups, but it does not necessarily provide a complete molecular structure.

06What should you learn next?

Once you have identified a carbonyl group, the next task is determining which carbonyl-containing functional group is present.

Start by comparing the carbonyl absorption with other diagnostic features:

• A very broad acid O-H absorption supports a carboxylic acid.
• A strong C-O absorption without the characteristic acid O-H band supports an ester.
• The absence of an O-H absorption can help narrow down the possibilities towards an aldehyde or ketone, depending on the other evidence.

For a focused comparison, continue with How to Distinguish Esters and Carboxylic Acids Using IR.

If you are given several candidate structures and need to combine different spectral features, read Mass Spectrometry, IR and NMR: How to Identify a Structure.

The important decision is to establish the carbonyl first, then use the remaining spectral evidence to work out which structure best explains the complete pattern.