Home · A-Level Chemistry 9701 · Analytical techniques · Infrared spectroscopy · Practice
AS · Practice questions · Infrared spectroscopy

Read the dips.

Six original Cambridge-style questions on infrared spectroscopy.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Alcohol O–H, acid O–H, C=O, C–O.

01
Explain
[3 marks]

Explain why carbon dioxide absorbs infrared radiation and how this contributes to the greenhouse effect.

  • The C=O bonds in CO2 vibrate; bending and asymmetric stretching change the dipole of the molecule. ✓
  • So CO2 absorbs infrared radiation at the frequencies of these vibrations, including radiation emitted by the warm Earth's surface. ✓
  • The energy is re-emitted in all directions, some back towards the Earth, which warms the lower atmosphere. ✓
02
Analysis
[3 marks]

A compound C3H6O shows a strong absorption at 1715 cm−1 and no absorption between 3200 and 3600 cm−1. It does not reduce Fehling's solution. Identify it.

  • 1715 cm−1 is within 1670–1740: a C=O bond is present. ✓
  • No broad band at 3200–3600: not an alcohol (no O–H). ✓
  • C3H6O with C=O that is not an aldehyde: propanone, CH3COCH3. ✓
03
Recall
[2 marks]

State the wavenumber range and appearance of the O–H absorption in (a) an alcohol and (b) a carboxylic acid.

  • (a) 3200–3600 cm−1, broad. ✓
  • (b) 2500–3000 cm−1, very broad, often overlapping the C–H absorptions. ✓
04
Analysis
[3 marks]

Describe how the IR spectra of ethanol, ethanoic acid and ethyl ethanoate differ in the regions 2500–3600 cm−1 and 1670–1740 cm−1.

  • Ethanol: broad O–H at 3200–3600, no C=O. ✓
  • Ethanoic acid: very broad O–H at 2500–3000 and a strong C=O near 1710. ✓
  • Ethyl ethanoate: no O–H, strong C=O near 1740 (all three also show C–H near 2900). ✓
05
Explain
[2 marks]

Two isomers, butan-1-ol and butan-2-ol, give almost identical spectra above 1500 cm−1. Explain how infrared spectroscopy can still distinguish them.

  • Both contain the same bonds (O–H, C–H, C–O), so the functional-group absorptions match. ✓
  • Their fingerprint regions (below 1500 cm−1) differ; comparing each with database spectra identifies which isomer is present. ✓
06
Analysis
[3 marks]

A compound shows a sharp absorption at 2230 cm−1, C–H near 2900 cm−1 and no absorption at 1670–1740 or 3200–3600 cm−1. Its Mr is 55. Suggest its identity.

  • 2230 cm−1 lies in the 2200–2250 range: a C≡N bond, so it is a nitrile. ✓
  • No C=O and no O–H. ✓
  • C2H5CN: Mr = 3(12.0) + 5(1.0) + 14.0 = 55.0, so propanenitrile. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.