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How does infrared spectroscopy identify the functional groups present in a molecule?

Interpret infrared spectra to identify functional groups from characteristic absorption bands

A focused answer to the WACE Year 12 Chemistry dot point on infrared spectroscopy, how molecular bonds absorb infrared radiation at characteristic wavenumbers, how to identify functional groups such as O-H, C=O and N-H from absorption bands, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Infrared (IR) spectroscopy identifies the functional groups in a molecule. Different bonds absorb infrared radiation at characteristic energies, so the pattern of absorptions acts as a fingerprint of the bonds present.

How it works

When a molecule absorbs infrared radiation, its bonds vibrate (stretch and bend) more vigorously. A bond absorbs only at the frequency that matches its natural vibration, which depends on the masses of the atoms and the stiffness of the bond. The spectrum plots absorption (or transmittance) against wavenumber (in cm $^{-1}$ ), and each absorption is recorded as a dip or peak.

Diagnostic absorption bands

The bands you must recognise (values are approximate ranges from the data table) include:

O-H (alcohol): broad, around 3200 to 3550 cm $^{-1}$ .
O-H (carboxylic acid): very broad, around 2500 to 3300 cm $^{-1}$ .
N-H (amine, amide): around 3300 to 3500 cm $^{-1}$ .
C=O (carbonyl, in aldehydes, ketones, acids, esters): strong and sharp, around 1670 to 1750 cm $^{-1}$ .
C-H: around 2850 to 3100 cm $^{-1}$ .

The fingerprint region

The region below about 1500 cm $^{-1}$ is the fingerprint region, a complex pattern unique to each compound. It is too complex to assign band by band, but it can confirm identity by matching against a reference spectrum of a known compound.

Reading a spectrum in practice

A reliable routine for an examination spectrum is to scan from high wavenumber down. First look above $3000\ \text{cm}^{-1}$ : a broad band there signals O-H (alcohol or, if very broad and extending lower, a carboxylic acid) or N-H (amine or amide). Next look around $1700\ \text{cm}^{-1}$ : a strong sharp band there signals a C=O carbonyl, present in aldehydes, ketones, carboxylic acids, esters and amides. Then combine the evidence: a C=O on its own (no O-H) points to an aldehyde or ketone; a C=O together with a very broad O-H points to a carboxylic acid; a C=O together with N-H points to an amide. The C-H bands near $2850$ to $3100\ \text{cm}^{-1}$ are present in almost all organic molecules and are not diagnostic on their own. Always quote the wavenumber, the band shape and intensity, and the bond it indicates, because markers reward the full reasoning rather than just naming a group.

Why this matters

Infrared spectroscopy complements mass spectrometry and NMR: mass spectrometry gives the molar mass, IR identifies the functional groups present, and NMR reveals the carbon-hydrogen framework. Used together they let chemists determine a complete structure, which is the core skill of the instrumental analysis dot points.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksTwo isomers of

\text{C}_3\text{H}_6\text{O}

, propanal (an aldehyde) and propan-1-ol (an alcohol), are to be distinguished by infrared spectroscopy. (a) State the key absorption band expected for each compound and the bond responsible. (b) Explain how the two spectra would differ. (c) Explain why infrared alone cannot reliably distinguish propanal from propanone.

Show worked answer →

A 6 mark question rewards the diagnostic bands, the spectral difference, and the limitation.

(a) Propanal: a strong sharp band near $1715\ \text{cm}^{-1}$ from the C=O carbonyl. Propan-1-ol: a broad band around $3200$ to $3550\ \text{cm}^{-1}$ from the O-H of the alcohol.

(b) The alcohol spectrum shows the broad O-H band above $3000\ \text{cm}^{-1}$ but no strong carbonyl peak near $1700\ \text{cm}^{-1}$ . The aldehyde spectrum shows the strong sharp C=O band near $1715\ \text{cm}^{-1}$ but no broad O-H band. The presence or absence of these two bands distinguishes them.

(c) Propanal and propanone are both carbonyl compounds, so both show a strong C=O band near $1700$ to $1715\ \text{cm}^{-1}$ in the same region. Infrared shows the functional-group type (a carbonyl) but cannot easily tell an aldehyde from a ketone, so another technique (such as NMR or a chemical test) is needed.

Markers reward the C=O band for the aldehyde and broad O-H for the alcohol, the presence/absence difference, and the shared-carbonyl limitation.

WACE 20204 marksExplain, in terms of bond vibration, why different bonds absorb infrared radiation at different wavenumbers, and why the fingerprint region is useful even though it is too complex to assign band by band.

Show worked answer →

A 4 mark answer needs the vibration reasoning and the fingerprint-region role.

A bond absorbs infrared radiation only at the frequency that matches its natural vibration (stretching or bending). This frequency depends on the masses of the bonded atoms and the stiffness of the bond, so different bonds (for example C=O versus O-H) vibrate at different frequencies and therefore absorb at different wavenumbers. Stronger or stiffer bonds and lighter atoms vibrate at higher wavenumbers.

The fingerprint region (below about $1500\ \text{cm}^{-1}$ ) contains many overlapping bands from the whole molecular skeleton. It is too complex to assign individually, but its overall pattern is unique to a particular compound, so matching it against a reference spectrum of a known substance confirms identity.

Markers reward the dependence of vibration frequency on atom mass and bond stiffness, and the use of the fingerprint region for identity-matching.