SACE Board SACE 2022-style practice: An organic compound has molecular formula C_3H_6O. Its IR spectrum shows a strong absorption near 1715\ cm^-1 but no broad band near 3300\ 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.

(a) A strong absorption near 1715\ cm^-1 indicates a C=O (carbonyl) group; the absence of a broad 3300\ cm^-1 band rules out an O-H, so it is not a carboxylic acid or alcohol. (2 marks) (b) With C_3H_6O, a carbonyl and no O-H, the compound is **propanone** (CH_3COCH_3), a ketone. (2 marks) (c) Propanone has two 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)

§-Syllabus dot point

SAChemistrySyllabus dot point

How do IR, mass spectrometry and NMR help identify an organic compound?

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

Using IR absorption bands, mass-spectrum fragments and molecular ion, and proton NMR (number of environments, integration, splitting) to deduce an organic structure, with a fully worked SACE-style spectra-combination example.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

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

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

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

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.

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 practice questions

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

SACE 20225 marksAn 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 20205 marksA 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)