Why IR Spectroscopy Identifies Functional Groups, Not Molecules

An unknown liquid produces a strong absorption band at \(1718\ \mathrm{cm}^{-1}\) in its infrared (IR) spectrum. You might identify it as propanone, but there’s a problem: aldehydes, ketones, carboxylic acids, and esters can all produce strong absorptions in a similar region because they all contain a carbonyl bond, \(\ce{C=O}\).

The spectrum gives you useful evidence, but it doesn’t tell you exactly which molecule is in the sample.

This is why IR spectroscopy is particularly good at suggesting functional groups without necessarily identifying an entire organic molecule. Different molecules can contain the same bonds, so they can produce similar characteristic absorption bands. Identifying the complete structure requires more information about how the atoms are connected.

The important skill is knowing exactly what a spectrum allows you to conclude, and where the evidence stops.

01Why different molecules produce similar IR absorptions

Imagine comparing the structural formulae of four organic compounds:

• Propanone, a ketone.
• Ethanal, an aldehyde.
• Ethanoic acid, a carboxylic acid.
• Methyl ethanoate, an ester.

Their structures are different, but all four contain a carbonyl bond, \(\ce{C=O}\).

Before looking at their spectra, make a prediction.

If all four molecules contain a carbonyl bond, should their IR spectra contain a similar absorption?

Yes. All four should produce a characteristic absorption in approximately the \(1680\)-\(1750\ \mathrm{cm}^{-1}\) region.

The precise position may differ because the surrounding atoms and molecular environment influence the vibration. However, the shared carbonyl bond gives us a useful starting point.

This is called a characteristic group frequency: a particular type of bond tends to absorb infrared radiation within a recognisable range of wavenumbers.

The crucial limitation is that a characteristic group frequency identifies a type of bond, not necessarily the complete molecule containing it.

02What is actually happening during IR spectroscopy?

To understand why this works, we need to look at what the instrument measures.

Atoms in molecules are constantly moving. Chemical bonds can stretch, bend, and vibrate in different ways.

A useful starting analogy is to imagine two balls connected by a spring.

Pull the balls apart slightly and release them. The spring pulls them towards each other, and the system oscillates.

A chemical bond behaves somewhat like this. Different bonds have different strengths and involve atoms with different masses, so they vibrate at different frequencies.

This analogy is useful for predicting why different bonds absorb at different frequencies. However, real molecular vibrations are more complicated: atoms can move together, and several bonds can contribute to the same vibrational mode.

How infrared radiation interacts with a molecule

When infrared radiation passes through a sample, certain frequencies can be absorbed by molecular vibrations.

For absorption to occur, the radiation must have an appropriate frequency to excite a vibrational mode. The vibration must also involve a change in the molecule’s dipole moment.

A dipole moment describes the separation of positive and negative charge within a molecule.

When a bond vibrates and changes this charge distribution, it can interact with infrared radiation.

For example, the carbonyl bond is strongly polar. Its stretching vibration produces a substantial change in dipole moment, which helps explain why carbonyl absorptions are usually strong.

Different bonds therefore produce characteristic absorption patterns.

However, the instrument does not directly photograph the bonds or identify their names. It measures changes in the infrared radiation passing through, or interacting with, the sample.

03How the experiment produces an IR spectrum

Consider how a chemist would investigate an unknown liquid.

The question is straightforward:

Does the unknown compound absorb infrared radiation at frequencies associated with particular molecular vibrations?

A simplified transmission experiment uses the following arrangement.

1. An infrared source produces radiation across a range of frequencies.
2. An optical system directs the radiation towards the sample.
3. The radiation passes through a thin sample held between suitable infrared-transparent windows.
4. A detector measures the radiation that emerges from the sample.
5. A computer processes the measurements and produces an IR spectrum.

Modern Fourier-transform infrared (FTIR) instruments use an interferometer to collect information across many wavenumbers. A mathematical operation called a Fourier transform converts the recorded interference signal into the familiar spectrum.

What should happen if the sample contains a carbonyl group?

Suppose the unknown contains a \(\ce{C=O}\) bond.

We would predict that it absorbs infrared radiation in the characteristic carbonyl region, approximately \(1680\)-\(1750\ \mathrm{cm}^{-1}\).

When this radiation is absorbed, less reaches the detector at those wavenumbers.

The resulting spectrum therefore contains an absorption feature in that region.

But there is an experimental problem.

The instrument can also detect effects from the surrounding atmosphere and its own optical components. Water vapour and carbon dioxide in the air, for example, can absorb infrared radiation.

A sample measurement alone cannot automatically distinguish these contributions from absorptions caused by the unknown compound.

A background scan helps address this problem. The instrument first measures the signal without the sample, then measures the signal with the sample present. Comparing the two measurements helps account for background absorption and the instrument response.

This makes the resulting spectrum more representative of the sample itself.

The important distinction is between the experimental observation and its chemical interpretation:

The instrument gives us evidence about molecular vibrations. We use chemistry to interpret that evidence.

04Reading the axes and band shapes correctly

Before identifying any functional group, check both axes.

This sounds basic, but misreading an IR spectrum can completely reverse your interpretation.

The horizontal axis: wavenumber

The horizontal axis usually represents wavenumber, measured in reciprocal centimetres:

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

Wavenumber is the reciprocal of wavelength:

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

A higher wavenumber corresponds to a shorter wavelength and a higher frequency.

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.

For example:

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

Do not assume that the numbers increase as you move right.

The vertical axis: transmittance or absorbance?

There are two common ways to display an IR spectrum.

On a percentage-transmittance spectrum, a deep downward trough represents strong absorption.

On an absorbance spectrum, the same absorption appears as a tall upward peak.

The chemistry has not changed. Only the way the measurement is displayed has changed.

Why band shape matters

Identifying a band involves more than finding its position.

Three features are particularly useful:

• Position: The wavenumber at which the absorption occurs.
• Width and shape: Whether the absorption is sharp, broad, or spread across a large region.
• Intensity: How strongly the sample absorbs at that frequency.

Consider an O-H bond.

An alcohol O-H absorption is generally broad, partly because hydrogen bonding creates a range of bonding environments.

Carboxylic acids produce an especially broad O-H absorption, typically extending across approximately \(2500\)-\(3000\ \mathrm{cm}^{-1}\) on the HSC data sheet.

This broad shape is useful evidence. It helps distinguish different types of O-H environments.

However, broadness alone does not identify a complete molecule.

Also, do not use the depth of a trough to count how many bonds are present. Absorption intensity depends on factors including the strength of the vibrational absorption, sample concentration, and sample thickness.

A deeper trough does not automatically mean that the molecule contains more of that type of bond.

05Characteristic absorption ranges: what can you recognise?

The following approximate ranges are useful when interpreting common organic functional groups.

These are approximate ranges, not fixed boundaries. Molecular structure, hydrogen bonding, and measurement conditions can affect the precise position and shape of a band.

Notice that several different functional groups can share absorption regions.

For example, a C-O absorption can occur in alcohols, ethers, esters, and carboxylic acids.

Finding a C-O absorption therefore does not automatically identify an ester or an alcohol.

You need to consider what other bonds are present.

06Why the fingerprint region adds more information

So far, we have focused on characteristic bond frequencies.

But an IR spectrum contains considerably more information than these familiar group frequencies.

The region below approximately \(1500\ \mathrm{cm}^{-1}\) is commonly called the fingerprint region.

Why?

Because this region contains many complex molecular vibrations, including bending and stretching movements involving several atoms.

These vibrations depend on the overall molecular structure, not just one isolated functional group.

Two compounds with the same functional group can therefore have similar absorptions in the higher-wavenumber region but different patterns in the fingerprint region.

Consider two alcohols:

• Butan-1-ol: \(\ce{CH3CH2CH2CH2OH}\)
• Butan-2-ol: \(\ce{CH3CH(OH)CH2CH3}\)

Both contain an alcohol hydroxyl group.

Both should produce a broad O-H absorption and a C-O absorption.

However, the hydroxyl group is attached to a different carbon in each molecule. Their molecular structures are different, so their complete vibrational patterns are also different.

This gives us an important distinction:

Characteristic group frequencies help identify functional groups. The fingerprint region can help distinguish complete molecular structures.

The fingerprint region can be highly useful when comparing an unknown compound against a reference spectrum.

However, it is not a magical identification code. Its interpretation can be complicated by overlapping absorptions, mixtures, and differences in experimental conditions.

A full-spectrum comparison is much more informative than simply identifying one or two characteristic bands.

07A particularly important limitation: IR does not always tell you which bonds belong to the same molecule

Consider an unknown sample that produces:

• A strong carbonyl absorption.
• A broad alcohol-like O-H absorption.
• An absorption in the C-O region.

You might conclude that the sample contains a hydroxyketone: a molecule containing both a ketone and an alcohol group.

That is one possible explanation.

But consider another possibility.

The sample could contain a mixture of a ketone and an alcohol. The ketone produces the carbonyl absorption, while the alcohol produces the O-H absorption.

The spectrum contains evidence of both types of bonds, but it does not automatically establish that they belong to the same molecule.

This is particularly important when interpreting spectra from samples that have not been confirmed as pure.

To establish the identity of the individual compounds, a chemist may need to separate the mixture and analyse its components.

Other analytical techniques also provide complementary information:

No single technique needs to answer every question.

The strength comes from combining independent evidence.

If you need to refresh how functional groups are recognised from structural formulae, see HSC Chemistry: How to Identify Organic Functional Groups.

08A practical method for interpreting IR evidence

When you are given an unfamiliar spectrum in an HSC question, work through the following sequence.

Step 1: Check the axes.

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

Step 2: Identify the strongest diagnostic features.

Look for characteristic absorptions such as C=O or O-H.

Step 3: Examine the band shapes.

Consider whether an O-H absorption is broad or very broad, and whether other features support your interpretation.

Step 4: Look for supporting evidence.

A C-O absorption, for example, may support an oxygen-containing functional group, but it is not unique to one functional group.

Step 5: Compare the evidence with possible structures.

Ask whether each proposed structure explains the observed absorptions.

Also consider whether a proposed structure should produce a strong characteristic absorption that is absent from a complete, good-quality spectrum.

Step 6: State the strength of your conclusion.

There is a difference between saying:

• The spectrum suggests a carbonyl group.
• The spectrum supports a ketone.
• The spectrum identifies propanone.

These statements make progressively stronger claims.

Your conclusion must not go further than the available evidence.

09Questions and solutions

Question 1

An unknown organic compound produces a strong absorption at \(1730\ \mathrm{cm}^{-1}\) in its percentage-transmittance IR spectrum.

(a) Identify the bond most strongly suggested by this absorption.

(b) Explain why the absorption does not establish that the compound is an ester.

Solution 1

(a) The absorption suggests a carbonyl bond, \(\ce{C=O}\).

The characteristic carbonyl absorption occurs approximately between \(1680\) and \(1750\ \mathrm{cm}^{-1}\).

The observed absorption at \(1730\ \mathrm{cm}^{-1}\) falls within this range. Because the spectrum displays percentage transmittance, the strong downward trough represents strong absorption.

(b) The absorption does not establish that the compound is an ester because several functional groups contain carbonyl bonds.

Aldehydes, ketones, carboxylic acids, and esters can all produce characteristic absorptions in the carbonyl region.

The absorption therefore supports the presence of a carbonyl group, but further evidence is needed to distinguish between these functional groups.

For example, an ester would be expected to show supporting C-O absorptions, whereas a carboxylic acid would also be expected to show a very broad O-H absorption.

The important point is that identifying a shared bond does not identify the complete functional group.

Question 2

An illustrative percentage-transmittance spectrum contains two absorption bands:

A student claims that the compound must contain more C-O bonds than C=O bonds because band B is deeper than band A.

Evaluate this conclusion.

Solution 2

The student’s conclusion is incorrect. Band B represents stronger absorption, but its greater depth does not establish that the compound contains more C-O bonds.

On a percentage-transmittance spectrum, lower transmittance indicates stronger absorption.

Band B reaches 15% transmittance, compared with 35% for band A. Therefore, more infrared radiation is absorbed at the wavenumber associated with band B.

However, absorption intensity depends on several factors, including:

• How strongly the particular vibrational mode interacts with infrared radiation.
• The concentration of the sample.
• The thickness of the sample through which radiation passes.
• The extent to which different vibrational absorptions overlap.

Different bonds do not necessarily produce absorptions of equal intensity.

The band at \(1720\ \mathrm{cm}^{-1}\) supports the presence of a carbonyl group, while the band at \(1120\ \mathrm{cm}^{-1}\) supports a C-O-containing structure.

The relative depths cannot be used to count the number of C=O and C-O bonds.

A strong absorption is evidence about the strength of the measured interaction, not a direct count of the bonds responsible.

Question 3

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

Two possible structures are:

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

An illustrative, high-quality percentage-transmittance spectrum shows:

• An absorption near \(1120\ \mathrm{cm}^{-1}\).
• No broad absorption in the alcohol O-H region.
• No strong absorption in the carbonyl region.

Which candidate is more strongly supported? Explain why the C-O absorption alone is insufficient to make this decision.

Solution 3

Methoxyethane is more strongly supported because the spectrum contains a C-O absorption but does not show the characteristic broad alcohol O-H absorption.

Both proposed structures contain a C-O bond.

Propan-1-ol contains a C-O bond as part of its alcohol group, while methoxyethane contains a C-O bond within an ether functional group.

Therefore, the absorption at \(1120\ \mathrm{cm}^{-1}\) does not distinguish the two candidates.

The more useful evidence is the absence of a broad O-H absorption.

Propan-1-ol contains an O-H bond, so a sufficiently concentrated, pure sample should produce a characteristic broad absorption in approximately the \(3230\)-\(3550\ \mathrm{cm}^{-1}\) region.

Methoxyethane does not contain an O-H bond, so this absorption is not expected.

The absence of the O-H feature therefore favours methoxyethane, assuming the spectrum is complete and of sufficient quality to detect the expected absorption.

This question demonstrates why identifying a bond that two candidates share is less useful than finding a feature that distinguishes them.

Question 4

An unknown liquid produces the following IR absorptions:

• A strong absorption near \(1715\ \mathrm{cm}^{-1}\).
• A broad absorption near \(3400\ \mathrm{cm}^{-1}\).
• An absorption in the \(1000\)-\(1300\ \mathrm{cm}^{-1}\) region.

A student identifies the compound as 4-hydroxybutan-2-one, a molecule containing both a ketone and an alcohol group.

Explain why the available IR evidence does not establish this complete molecular structure.

Your answer should distinguish between a pure compound containing both functional groups and a mixture of two different compounds.

Solution 4

The spectrum supports the presence of carbonyl and hydroxyl groups in the sample, but it does not prove that both groups belong to the same molecule.

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

The broad absorption near \(3400\ \mathrm{cm}^{-1}\) supports an alcohol O-H bond.

The C-O absorption provides further supporting evidence for an oxygen-containing structure.

One possible explanation is that the unknown is a hydroxyketone, such as 4-hydroxybutan-2-one.

However, another possibility is that the sample contains two separate compounds:

• A ketone, which produces the carbonyl absorption.
• An alcohol, which produces the broad O-H absorption.

Both possibilities can produce the observed types of absorption.

IR spectroscopy measures the vibrational absorptions of the molecules present in the sample. It does not automatically associate every observed absorption with one particular molecule.

To investigate further, the chemist could separate the sample into individual components using a suitable separation technique, then analyse the isolated compounds.

Alternatively, NMR spectroscopy could provide information about the different carbon and hydrogen environments, helping establish whether the proposed functional groups belong to the same molecular structure.

The student’s proposed structure is possible, but the IR evidence alone does not establish it.

Question 5

An unknown pure compound has the molecular formula \(\ce{C4H10O}\). The only two candidate structures are butan-1-ol and butan-2-ol.

A chemist records an IR spectrum of the unknown dissolved at low concentration in a non-polar solvent.

The following illustrative observations are obtained:

The two reference spectra were recorded using neat liquid samples, rather than dilute solutions.

A student argues:

“The unknown must be butan-1-ol because its C-O absorption is closer to the butan-1-ol reference. The O-H absorption is at a different wavenumber, but that doesn’t matter because the C-O absorption is the important one.”

Evaluate the student’s reasoning.

Explain why the different measurement conditions matter, whether the available evidence establishes the identity of the isomer, and how the chemist could make a more reliable comparison.

Solution 5

The student has not established that the unknown is butan-1-ol. The different sample conditions can change the observed absorption features, and one C-O band is insufficient to identify the isomer.

There are two important problems with the student’s reasoning.

First, the O-H absorption is affected by hydrogen bonding.

In a neat liquid sample, alcohol molecules can form intermolecular hydrogen bonds with neighbouring molecules.

These interactions affect the O-H bond’s vibrational frequency and produce a broader absorption, often at a lower wavenumber than an isolated O-H bond.

In a dilute solution in a non-polar solvent, alcohol molecules are more separated.

There is less intermolecular hydrogen bonding between alcohol molecules, so the O-H absorption generally becomes narrower and shifts towards a higher wavenumber.

This explains why the unknown can show an O-H absorption near \(3520\ \mathrm{cm}^{-1}\), while the neat reference samples have broader absorptions at lower wavenumbers.

The difference does not establish that the unknown is a different compound.

It may simply reflect the different physical conditions under which the spectra were recorded.

Second, the C-O absorption does not uniquely identify the isomer.

The unknown produces a C-O absorption at \(1085\ \mathrm{cm}^{-1}\).

This is closer to the butan-1-ol reference at \(1060\ \mathrm{cm}^{-1}\) than the butan-2-ol reference at \(1120\ \mathrm{cm}^{-1}\).

However, this numerical closeness is not sufficient to establish the identity of the compound.

The position of an absorption can be influenced by the molecular environment and experimental conditions.

More importantly, the C-O absorption is a characteristic feature of both alcohols. It does not independently establish where the O-H group is attached to the carbon chain.

How should the chemist improve the comparison?

The chemist should obtain reference spectra for both butan-1-ol and butan-2-ol under the same conditions as the unknown.

For example:

1. Prepare dilute solutions of both reference alcohols using the same non-polar solvent.
2. Use comparable concentrations and measurement conditions.
3. Record an appropriate solvent background.
4. Compare the complete spectra, paying particular attention to the fingerprint region.

The full fingerprint patterns may provide evidence that distinguishes the two isomers.

If the IR comparison remains inconclusive, proton or carbon-13 NMR spectroscopy can provide further information about the molecular connectivity.

Final conclusion: The unknown is consistent with an alcohol, and the molecular formula and restricted candidate list are consistent with either butanol isomer. However, the available comparison does not establish whether it is butan-1-ol or butan-2-ol. The difference in O-H absorption is reasonably explained by hydrogen bonding and sample conditions, while the C-O absorption alone cannot resolve the structural difference.

10Where to go next

The main skill in IR interpretation is distinguishing evidence of a bond from evidence of an entire molecular structure.

A carbonyl absorption suggests a carbonyl group. A broad O-H absorption supports a hydroxyl group. A combination of features can narrow down the possible functional groups.

But when two candidate molecules contain the same functional groups, you need additional structural information.

The next useful step is learning how to combine multiple IR features when comparing possible organic structures. See How to Choose Between Organic Structures Using Two IR Features.

Remember the distinction when writing HSC responses: IR spectroscopy provides evidence about molecular bonds and functional groups. Identifying the complete molecule requires evidence about the arrangement of those groups and the connectivity of the atoms.