Infrared spectroscopyEdexcel A-Level Chemistry: Revision notes
Section 1
How infrared spectroscopy works
Covalent bonds vibrate by stretching and bending. A bond absorbs infrared radiation of the frequency that matches its vibration, and this depends on the atoms joined and the bond strength. The absorption is recorded as a wavenumber in cm⁻¹.
A spectrum plots transmittance against wavenumber, so each absorption appears as a downward peak. Different functional groups absorb at different, characteristic wavenumbers, so infrared spectra are used to identify functional groups.
Section 2
Absorptions you need to use
Use the data booklet ranges (typical values in cm⁻¹):
- C–H stretching (alkanes, alkenes, aldehydes): 2850–3100
- C=C stretching (alkenes): 1620–1669
- C=O stretching (aldehydes, ketones, acids): 1680–1750
- O–H stretching in alcohols: 3200–3600
- O–H in carboxylic acids: 2500–3300, very broad
- N–H stretching in amines: 3300–3500
You are given wavenumber data in an exam, so the skill is matching bonds to absorptions, not memorising numbers.
A carboxylic acid is the only common group with both a C=O absorption and a very broad O–H absorption.
Section 3
Alcohols, carbonyls and carboxylic acids
- An alcohol shows O–H at 3200–3600 and no C=O.
- An aldehyde or ketone shows C=O at 1680–1750 and no O–H absorption. Infrared cannot tell an aldehyde from a ketone, because both have C=O.
- A carboxylic acid shows C=O at 1680–1750 and a very broad O–H at 2500–3300, overlapping the C–H region.
Oxidation of ethanol can be followed this way: ethanol (O–H at 3200–3600) → ethanal (C=O only) → ethanoic acid (C=O plus broad O–H).
Do not choose an alcohol because you see C=O. The O–H of an alcohol is a separate absorption above 3200 cm⁻¹.
Section 4
Alkenes, alkanes and amines
- An alkene has C=C at 1620–1669 as well as C–H absorptions.
- An alkane shows only C–H stretching (2850–3100) in the group range, with no C=C, C=O, O–H or N–H.
- An amine shows N–H stretching at 3300–3500.
N–H (3300–3500) and alcohol O–H (3200–3600) overlap, so a peak near 3350 cm⁻¹ alone does not decide between an amine and an alcohol. Use the other absorptions, or the formula.
Section 5
Interpreting a spectrum
- Check the region above 3000 cm⁻¹ for O–H or N–H.
- Check 1680–1750 for C=O.
- Check 1620–1669 for C=C.
- Check 2500–3300 for a very broad O–H (acid).
- Combine with the molecular formula and the reaction, and state which groups are present and which are absent.
Example: C₃H₆O with a strong absorption at 1715 and none at 3200–3600 has C=O and no O–H, so it is an aldehyde or ketone. The one oxygen atom rules out an acid.
State both what is present and what is absent. Marks are often given for the absence of O–H or C=O.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Infrared spectroscopy
- An organic compound P has the molecular formula C₃H₆O. A student records its infrared spectrum and finds a strong absorption at 1715 cm⁻¹ and no absorption between 3200 and 3600 cm⁻¹.Explain why the infrared spectrum cannot distinguish between propanal and propanone, and describe a chemical test that does.2 marks
- A chemist is given two colourless liquids, ethanol and ethanoic acid, whose labels have fallen off. She records an infrared spectrum of each.Describe how the two spectra differ, so that the chemist can identify each liquid.2 marks
- A student heats propan-1-ol with acidified potassium dichromate(VI) and distils off the product as soon as it forms. The infrared spectrum of the distillate shows absorptions at 2980 cm⁻¹ and 1725 cm⁻¹ only, apart from the fingerprint region.Identify the bonds responsible for the two absorptions, and state what the absence of any absorption at 3200–3600 cm⁻¹ shows about the distillate.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).