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How do infrared and mass spectra help identify organic molecules?

Infrared spectroscopy and the absorptions of bonds and functional groups, the fingerprint region, mass spectrometry of organic molecules including the molecular ion and fragmentation, and the use of both to identify and determine the structure of compounds.

A CCEA A-Level Chemistry answer on infrared and mass spectrometry, covering how infrared absorptions identify bonds and functional groups, the fingerprint region, the molecular ion and fragmentation in mass spectra, and how both techniques are combined to determine the structure of organic compounds.

Generated by Claude Opus 4.88 min answer

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  1. What this dot point is asking
  2. Infrared spectroscopy
  3. Mass spectrometry of organic molecules
  4. Combining the techniques
  5. Examples in context
  6. Try this

What this dot point is asking

CCEA wants you to explain how infrared spectroscopy identifies bonds and functional groups, recognise the fingerprint region, interpret the mass spectra of organic molecules including the molecular ion and fragmentation, and use both techniques together to determine the structure of a compound.

Infrared spectroscopy

The fingerprint region below about 1500 cm11500\ \text{cm}^{-1} contains a complex pattern unique to each molecule, so a spectrum can be matched against a database to confirm identity. The reason bonds absorb at characteristic frequencies is that each bond behaves like a spring connecting two masses: stronger bonds and lighter atoms vibrate at higher frequencies, so they absorb at higher wavenumbers. This is why a C-H bond absorbs higher than a C-Cl bond. In practice, candidates use a data sheet of characteristic absorptions: the presence or absence of the broad O-H, the sharp C=O, and the C-H stretches is usually enough to identify the functional groups present, while the fingerprint region confirms the exact molecule.

Mass spectrometry of organic molecules

When the molecular ion breaks apart, it gives fragment ions. The mass differences between peaks reveal the groups lost, for example a loss of 1515 indicates a CH3\text{CH}_3 group, a loss of 1717 indicates OH\text{OH}, a loss of 2929 indicates CHO\text{CHO} or C2H5\text{C}_2\text{H}_5, and a loss of 4545 indicates COOH\text{COOH}. Equally useful is the mass of the fragment itself: a peak at m/z=15m/z = 15 is CH3+\text{CH}_3^+, at 2929 is CHO+\text{CHO}^+ or C2H5+\text{C}_2\text{H}_5^+, at 4343 is CH3CO+\text{CH}_3\text{CO}^+ (the acylium ion from a methyl ketone), and at 7777 is C6H5+\text{C}_6\text{H}_5^+ (a phenyl group). Small peaks one or two mass units above the molecular ion (the M+1 and M+2 peaks) come from the heavier isotopes such as carbon-13 and chlorine-37, and the size of an M+2 peak can indicate the presence of chlorine or bromine.

Combining the techniques

Examples in context

Example 1. Monitoring vehicle emissions and breath alcohol. Modern roadside breath testers and laboratory emission analysers use infrared absorption: ethanol and carbon dioxide each absorb infrared at their own characteristic wavenumbers, so the instrument measures how much radiation of those frequencies is absorbed and converts it directly to a concentration. This is the everyday application of the idea that each bond absorbs at a fixed frequency, the same principle a CCEA candidate uses to read a laboratory spectrum.

Example 2. Confirming the structure of a synthesised product. When a chemist makes a new compound, mass spectrometry gives the relative molecular mass from the molecular ion peak and clues to the structure from the fragments, while infrared confirms which functional groups are present. Together they distinguish isomers that share a formula: ethanol and methoxymethane both have M+=46\text{M}^+ = 46, but only ethanol shows the broad O-H absorption and the m/z=31m/z = 31 fragment. This combined-technique approach is exactly what CCEA structure-determination questions assess.

Try this

Q1. State what the molecular ion peak tells you about a compound. [1 mark]

  • Cue. Its mass-to-charge ratio equals the relative molecular mass.

Q2. State the approximate wavenumber of the C=O absorption in infrared spectroscopy. [1 mark]

  • Cue. About 1700 cm11700\ \text{cm}^{-1}.

Exam-style practice questions

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

CCEA 20216 marksA compound X has the molecular formula C3H6O. Its mass spectrum shows a molecular ion peak at m/z = 58 and a strong fragment at m/z = 43. Its infrared spectrum shows a strong absorption at about 1715 cm-1 but no broad absorption between 2500 and 3550 cm-1. Deduce the identity of X, explaining your reasoning.
Show worked answer →

Markers want the molecular ion confirmed, the infrared interpreted, the fragment explained, and a final structure with reasoning.

The molecular ion at m/z=58m/z = 58 matches the relative molecular mass of C3H6O\text{C}_3\text{H}_6\text{O} (3×12+6×1+16=583 \times 12 + 6 \times 1 + 16 = 58), confirming the formula.

The strong infrared absorption at about 1715 cm11715\ \text{cm}^{-1} is a C=O\text{C}{=}\text{O} bond, so X is a carbonyl compound (an aldehyde or a ketone). The absence of any broad band between 25002500 and 3550 cm13550\ \text{cm}^{-1} rules out an O-H group, so X is not an alcohol or a carboxylic acid.

The fragment at m/z=43m/z = 43 is a loss of 1515 from the molecular ion, which is the loss of a CH3\text{CH}_3 group, leaving CH3CO+\text{CH}_3\text{CO}^+ (the acylium ion, mass 4343). This loss of a methyl group from each side fits propanone, CH3COCH3\text{CH}_3\text{COCH}_3.

So X is propanone (acetone). Markers reward confirming the molecular mass, identifying the C=O\text{C}{=}\text{O}, ruling out O-H, explaining the loss of 15 as CH3\text{CH}_3, and naming propanone.

CCEA 20193 marksExplain how infrared spectroscopy could be used to confirm that a sample of ethanol had been completely oxidised to ethanoic acid.
Show worked answer →

A short interpretation question linking functional groups to absorptions.

In ethanol, the infrared spectrum shows a broad O-H absorption between about 32003200 and 3550 cm13550\ \text{cm}^{-1} (the alcohol O-H) but no strong C=O\text{C}{=}\text{O} absorption.

If oxidation to ethanoic acid is complete, the spectrum should show two new or changed features: a strong C=O\text{C}{=}\text{O} absorption at about 1700 cm11700\ \text{cm}^{-1}, and a very broad O-H absorption stretching from about 25002500 to 3300 cm13300\ \text{cm}^{-1} that is characteristic of a carboxylic acid (much broader and lower than the alcohol O-H).

Seeing both the carbonyl peak and the broad carboxylic-acid O-H, and the loss of the sharper alcohol O-H, confirms the oxidation is complete.

Markers reward identifying the carboxylic acid C=O\text{C}{=}\text{O} peak, the broad carboxylic acid O-H, and the contrast with the alcohol O-H of the starting material.

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