Infrared spectroscopyAQA A-Level Chemistry: Revision notes
Section 1
Bonds absorb infrared radiation
The bonds in a molecule vibrate (stretch and bend). A bond absorbs infrared radiation at a characteristic frequency, which is quoted as a wavenumber in cm⁻¹. Different bonds absorb at different wavenumbers.
An infrared spectrum plots the transmittance against the wavenumber, so each absorption appears as a downward trough. A strong, narrow trough shows a bond such as C=O; broad troughs show O–H bonds.
Section 2
Identifying bonds and functional groups
Use the data sheet to match each absorption to a bond. Typical ranges:
- O–H (alcohols) 3230–3550 cm⁻¹, broad
- O–H (carboxylic acids) 2500–3000 cm⁻¹, very broad
- N–H 3300–3500 cm⁻¹
- C–H 2850–3100 cm⁻¹
- C≡N 2220–2260 cm⁻¹
- C=O 1680–1750 cm⁻¹, strong
- C=C 1620–1680 cm⁻¹
- C–O 1000–1300 cm⁻¹
A strong C=O absorption with no O–H absorption suggests an aldehyde or ketone. A C=O absorption with a very broad O–H absorption suggests a carboxylic acid.
Always quote the wavenumber and the bond, e.g. 'the absorption at 1715 cm⁻¹ is the C=O bond'.
Section 3
Telling alcohols from carboxylic acids
Both contain O–H, but the absorption differs:
- Alcohol O–H: broad, at 3230–3550 cm⁻¹
- Carboxylic acid O–H: very broad, at 2500–3000 cm⁻¹, overlapping the C–H absorptions
A carboxylic acid also shows a strong C=O absorption, but an alcohol does not.
Do not say the O–H absorption of any compound is at 3300. A carboxylic acid O–H is much lower, at 2500–3000 cm⁻¹.
Section 4
Fingerprinting
The region below about 1500 cm⁻¹ contains many overlapping absorptions that are unique to each compound. This fingerprint region is compared with the spectra of known compounds in a database. An exact match identifies the molecule, so fingerprinting can identify a compound even when the functional groups alone cannot.
Section 5
Detecting impurities
An impure sample gives extra absorptions that are not in the spectrum of the pure compound. For example, an O–H absorption near 3300 cm⁻¹ in a sample that contains no O–H bond suggests water or alcohol solvent. Comparison with a reference spectrum can show the impurity and help decide whether to purify the sample again.
Section 6
Infrared radiation and global warming
The Earth's surface absorbs radiation from the Sun and re-emits some as infrared radiation. Greenhouse gases, including carbon dioxide, methane and water vapour, have bonds (C=O, C–H, O–H) that absorb infrared radiation. The molecules re-emit the energy in all directions, and some goes back to the surface, so the atmosphere warms.
Burning fossil fuels and agriculture increase the amounts of carbon dioxide and methane, so more radiation is absorbed and the global warming effect increases.
Must Know
- Bonds absorb infrared radiation at characteristic wavenumbers
- Use the data sheet to match absorptions: C=O 1680–1750, O–H alcohols 3230–3550, O–H acids 2500–3000, C–H 2850–3100
- Fingerprint region (below about 1500 cm⁻¹) identifies a compound by comparison with a database
- Extra absorptions show impurities
- CO₂, CH₄ and H₂O absorb infrared radiation and cause global warming
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Infrared spectroscopy
- A student records the infrared spectrum of an organic liquid with the molecular formula C₃H₆O. The spectrum has a strong absorption at 1715 cm⁻¹ and no broad absorption anywhere between 2500 and 3600 cm⁻¹. Data: C=O 1680–1750 cm⁻¹; O–H (alcohols) 3230–3550 cm⁻¹; O–H (carboxylic acids) 2500–3000 cm⁻¹; C=C 1620–1680 cm⁻¹.Explain how the spectrum shows that the compound is neither an alcohol nor a carboxylic acid.2 marks
- A pharmaceutical chemist checks the purity of a solid drug product that was made in a reaction using ethanol as the solvent. The drug molecule contains no O–H bond. The chemist compares the infrared spectrum of each batch with the spectrum of a pure reference sample. One batch has an extra broad absorption at 3350 cm⁻¹. Data: C–H 2850–3100 cm⁻¹; O–H (alcohols) 3230–3550 cm⁻¹; O–H (carboxylic acids) 2500–3000 cm⁻¹; C=O 1680–1750 cm⁻¹.Explain how the spectrum shows that this batch is impure and what the impurity is likely to be.2 marks
- Two isomers, P and Q, both have the molecular formula C₃H₆O₂. The infrared spectrum of P has a very broad absorption from 2500 to 3000 cm⁻¹ and a strong absorption at 1710 cm⁻¹. The spectrum of Q has a strong absorption at 1740 cm⁻¹ and a strong absorption at 1200 cm⁻¹, but no broad absorption above 2500 cm⁻¹. Data: C=O 1680–1750 cm⁻¹; C–O 1000–1300 cm⁻¹; O–H (carboxylic acids) 2500–3000 cm⁻¹; O–H (alcohols) 3230–3550 cm⁻¹.Deduce the functional group present in P and in Q, referring to the bonds responsible for the absorptions.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).