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Infrared spectroscopy: WACE Year 12 Chemistry

Syllabus dot point

“Interpret infrared spectra to identify functional groups from characteristic absorption bands”

8 min read

Quick answer

Infrared (IR) spectroscopy identifies functional groups. Bonds absorb infrared radiation at characteristic frequencies (wavenumbers) that make them vibrate. By matching the absorption bands in a spectrum against a data table, you identify which bonds and therefore which functional groups are present. A broad O-H band, a sharp strong C=O band, and an N-H band are the most diagnostic.

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^{-1}), and each absorption is recorded as a dip or peak.

Wavenumber

The position of an infrared absorption, measured in reciprocal centimetres (cm−1^{-1}). It is proportional to the energy of the radiation absorbed and is characteristic of a particular type of bond.

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^{-1}.
  • O-H (carboxylic acid): very broad, around 2500 to 3300 cm−1^{-1}.
  • N-H (amine, amide): around 3300 to 3500 cm−1^{-1}.
  • C=O (carbonyl, in aldehydes, ketones, acids, esters): strong and sharp, around 1670 to 1750 cm−1^{-1}.
  • C-H: around 2850 to 3100 cm−1^{-1}.
Use the data table, look for the key bands

You are given a table of characteristic absorptions, so you do not memorise exact numbers. Look for the most diagnostic bands: a strong sharp peak near 1700 cm−1^{-1} signals a C=O; a broad band above 3000 cm−1^{-1} signals O-H or N-H. The combination identifies the functional group.

The fingerprint region

The region below about 1500 cm−1^{-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.

Identifying a functional group from IR

Question

An organic compound shows a strong sharp absorption near 1715 cm−1^{-1} and a very broad absorption between 2500 and 3300 cm−1^{-1}. What functional group is present, and what family is the compound likely to belong to?

Solution

The strong sharp band near 1715 cm−1^{-1} is characteristic of a carbonyl group, C=O. The very broad absorption spanning roughly 2500 to 3300 cm−1^{-1} is characteristic of the O-H of a carboxylic acid (much broader and lower than an alcohol O-H).

Having both a C=O and a carboxylic-acid O-H together indicates a carboxylic acid (-COOH), so the compound is most likely a carboxylic acid.

Reading a spectrum in practice

A reliable routine for an examination spectrum is to scan from high wavenumber down. First look above 3000 cm−13000\ \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 cm−11700\ \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 28502850 to 3100 cm−13100\ \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.

Confusing the alcohol and acid O-H, or ignoring band shape

A common error is treating every O-H band the same; an alcohol O-H is broad around 3300 cm−1^{-1}, while a carboxylic acid O-H is very broad and extends much lower (2500 to 3300 cm−1^{-1}). Another is reporting a band position without using its shape and intensity; a strong sharp peak near 1700 cm−1^{-1} is C=O, and this combination of position and shape is what identifies the group.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks
Two isomers of C3H6O\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 cm−11715\ \text{cm}^{-1} from the C=O carbonyl. Propan-1-ol: a broad band around 32003200 to 3550 cm−13550\ \text{cm}^{-1} from the O-H of the alcohol.

(b) The alcohol spectrum shows the broad O-H band above 3000 cm−13000\ \text{cm}^{-1} but no strong carbonyl peak near 1700 cm−11700\ \text{cm}^{-1}. The aldehyde spectrum shows the strong sharp C=O band near 1715 cm−11715\ \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 17001700 to 1715 cm−11715\ \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 2020-style4 marks
Explain, 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 cm−11500\ \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.

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

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