IR, mass spec and NMR analysis (SACE Stage 2 Chemistry)

Syllabus dot point

“Interpret infrared spectra, mass spectra and proton NMR spectra to determine the structure of organic molecules.”

10 min readUpdated 2 Jun 2026

Quick answer

Three instrumental methods together pin down an organic structure. Infrared (IR) shows which functional groups are present from characteristic absorption bands (e.g. $\text{C}=\text{O}$ near $1715\ \text{cm}^{-1}$ , broad $\text{O-H}$ near $3300\ \text{cm}^{-1}$ ). Mass spectrometry (MS) gives the molar mass from the molecular ion and structural clues from fragment $m/z$ values. Proton NMR reveals the number of hydrogen environments (peaks), how many hydrogens are in each (integration), and which groups are adjacent (splitting).

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What this dot point is asking
Lead worked calculation
Infrared spectroscopy
Mass spectrometry
Proton NMR
Combining the evidence
Why it matters
Exam-style questions

What this dot point is asking

SACE expects you to read each spectrum and combine the evidence to deduce a structure, justifying each step from the data.

Lead worked calculation

Deducing a structure from three spectra

A compound has molecular formula $\text{C}_3\text{H}_8\text{O}$ . Its IR shows a broad band near $3300\ \text{cm}^{-1}$ but no $\text{C}=\text{O}$ . Its mass spectrum has a molecular ion at $m/z = 60$ and a fragment at $m/z = 45$ . Its proton NMR shows four peaks. Deduce the structure.

Use the IR

The broad band near $3300\ \text{cm}^{-1}$ indicates an $\text{O-H}$ group; the absence of a $\text{C}=\text{O}$ band means it is an alcohol, not a carboxylic acid.

Use the mass spectrum

$M(\text{C}_3\text{H}_8\text{O}) = 60$ , matching the molecular ion. Loss of $15$ (a $\text{CH}_3$ group) to give $m/z = 45$ shows at least one methyl group attached near the rest of the molecule.

Use the NMR

Four peaks means four different hydrogen environments. Propan-1-ol ( $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ ) has four environments ( $\text{CH}_3$ , two different $\text{CH}_2$ , $\text{O-H}$ ); propan-2-ol has only three. So the structure is propan-1-ol.

Conclude

Only by combining all three spectra (alcohol from IR, mass from MS, four environments from NMR) is the isomer uniquely identified.

Infrared spectroscopy

IR absorption and functional groups

IR radiation makes bonds vibrate; each bond type absorbs at a characteristic wavenumber, so the spectrum is a fingerprint of the functional groups present. Key SACE bands: broad $\text{O-H}$ (alcohol) around $3200$ to $3550\ \text{cm}^{-1}$ ; very broad $\text{O-H}$ (carboxylic acid) around $2500$ to $3300\ \text{cm}^{-1}$ ; $\text{C}=\text{O}$ (carbonyl) near $1700\ \text{cm}^{-1}$ ; $\text{C-H}$ around $2850$ to $3000\ \text{cm}^{-1}$ . The presence or absence of a band identifies a group.

Mass spectrometry

The sample is ionised and the ions sorted by mass-to-charge ratio ( $m/z$ ).

The molecular ion ( $\text{M}^+$ ), the highest significant $m/z$ , equals the molar mass of the compound, confirming the molecular formula.
Fragment ions form when the molecular ion breaks apart; the difference between $m/z$ values reveals the lost group (e.g. loss of $15$ is $\text{CH}_3$ , loss of $17$ is $\text{OH}$ , loss of $29$ is $\text{CHO}$ or $\text{C}_2\text{H}_5$ ).

Proton NMR

The three pieces of NMR information

Number of peaks = number of different hydrogen (chemical) environments. Integration (peak area ratio) = relative number of hydrogens in each environment. Splitting (the n+1 rule) = number of hydrogens on adjacent carbons; a hydrogen with $n$ neighbours appears as $n + 1$ peaks (a triplet means two neighbouring hydrogens). Together these locate every hydrogen and reveal which groups are next to each other.

For ethanol, $\text{CH}_3\text{CH}_2\text{OH}$ : three environments give three peaks in a $3:2:1$ area ratio; the $\text{CH}_3$ is split into a triplet by the two $\text{CH}_2$ hydrogens, and the $\text{CH}_2$ into a quartet by the three $\text{CH}_3$ hydrogens.

Combining the evidence

No single technique is conclusive; structure determination combines them. IR names the functional groups, MS fixes the molar mass and flags lost fragments, and NMR maps the hydrogen environments and connectivity. Working through them in that order, group, then mass, then arrangement, usually pins down the structure and distinguishes between isomers.

Reading NMR peak number as hydrogen count

A common SACE error is treating the number of NMR peaks as the number of hydrogens; it is the number of different environments. The integration (area ratio) gives the relative hydrogen counts. Another error is mis-assigning IR: a broad band near $3300\ \text{cm}^{-1}$ is $\text{O-H}$ , while $\text{C}=\text{O}$ sits near $1700\ \text{cm}^{-1}$ ; do not confuse them. With mass spectra, take the molecular ion as the highest significant $m/z$ , not the tallest peak.

Why it matters

IR, MS and NMR are the standard tools for identifying unknown organic compounds in research, forensics, pharmaceuticals and quality control. Learning to combine their evidence is the practical skill that ties together functional groups, isomerism and the reactions studied throughout this topic.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2022-style5 marks

An organic compound has molecular formula

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

. Its IR spectrum shows a strong absorption near

1715\ \text{cm}^{-1}

but no broad band near

3300\ \text{cm}^{-1}

. Its proton NMR shows a single peak. (a) Identify the functional group from the IR data. (b) Deduce the structure consistent with all the data. (c) Explain how the single NMR peak supports your structure.

Show worked answer →

(a) A strong absorption near $1715\ \text{cm}^{-1}$ indicates a $\text{C}=\text{O}$ (carbonyl) group; the absence of a broad $3300\ \text{cm}^{-1}$ band rules out an $\text{O-H}$ , so it is not a carboxylic acid or alcohol. (2 marks)

(b) With $\text{C}_3\text{H}_6\text{O}$ , a carbonyl and no $\text{O-H}$ , the compound is propanone ( $\text{CH}_3\text{COCH}_3$ ), a ketone. (2 marks)

(c) Propanone has two $\text{CH}_3$ groups in identical (equivalent) environments, so all six hydrogens are equivalent and give a single NMR peak, consistent with the spectrum. (1 mark)

SACE 2020-style5 marks

A compound

\text{C}_2\text{H}_6\text{O}

gives a mass spectrum with a molecular ion at

m/z = 46

and major fragments at

m/z = 31

and

m/z = 15

. (a) Confirm the molecular ion is consistent with the formula. (b) Identify the species lost to give the

m/z = 31

fragment. (c) The proton NMR shows three peaks in a

3:2:1

ratio; deduce the structure and assign the peaks.

Show worked answer →

(a) $M(\text{C}_2\text{H}_6\text{O}) = 2(12) + 6(1) + 16 = 46$ , matching the molecular ion at $m/z = 46$ . (1 mark)

(b) Loss from $46$ to $31$ is a loss of $15$ , corresponding to a $\text{CH}_3$ group (mass $15$ ); the $m/z = 15$ fragment is the $\text{CH}_3^+$ ion. (1 mark)

(c) Three peaks in $3:2:1$ means three hydrogen environments with $3$ , $2$ and $1$ hydrogens: this is ethanol, $\text{CH}_3\text{CH}_2\text{OH}$ . The peaks are $\text{CH}_3$ ( $3\text{H}$ ), $\text{CH}_2$ ( $2\text{H}$ ) and $\text{O-H}$ ( $1\text{H}$ ). (3 marks)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board