How does infrared spectroscopy identify the functional groups present in an organic molecule?
Explain the origin of infrared absorption from bond vibrations, use characteristic absorption ranges from the data booklet to identify functional groups, and distinguish compounds such as alcohols, carbonyls and carboxylic acids from their spectra
A focused answer to the H2 Chemistry learning outcome on infrared spectroscopy. The origin of IR absorption in bond vibrations, using the data-booklet absorption ranges to identify functional groups, and distinguishing alcohols, carbonyls and carboxylic acids from their characteristic absorptions.
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What this dot point is asking
SEAB wants you to explain why molecules absorb infrared radiation (bond vibrations), use the characteristic absorption ranges in the data booklet to identify functional groups, and distinguish compounds such as alcohols, carbonyls and carboxylic acids from their spectra. Reading functional groups off an IR spectrum is a core structure-determination skill.
The answer
Why molecules absorb infrared
Covalent bonds vibrate (stretch and bend) at frequencies in the infrared region. A bond absorbs infrared radiation when the radiation's frequency matches the bond's natural vibration frequency, provided the vibration changes the molecule's dipole moment. Different bonds absorb at different, characteristic wavenumbers (measured in cm), so the spectrum is a map of the bonds present.
Using the data booklet
The SEAB Data Booklet lists characteristic absorption ranges. The most useful for identification:
| Bond | Wavenumber range / cm | Group |
|---|---|---|
| O-H (alcohol) | 3200 to 3550 (broad) | alcohols |
| O-H (carboxylic acid) | 2500 to 3300 (very broad) | acids |
| N-H | 3300 to 3500 | amines, amides |
| C-H | around 2850 to 3100 | hydrocarbons |
| C=O | 1670 to 1740 (strong, sharp) | aldehydes, ketones, acids, esters |
| C=C | around 1620 to 1680 | alkenes |
| C-O | 1000 to 1300 | alcohols, esters |
You should never memorise these values; learn how to use the booklet quickly.
Distinguishing the key groups
- Alcohol: broad O-H around 3200 to 3550, no C=O.
- Aldehyde or ketone: strong C=O around 1715, no broad O-H.
- Carboxylic acid: strong C=O around 1700 to 1725 and a very broad O-H around 2500 to 3300 (the two together are diagnostic).
- Ester: strong C=O and a strong C-O, but no broad O-H.
The fingerprint region
The region below about 1500 cm is the fingerprint region, a complex pattern unique to each molecule. It is not used to assign individual bonds but can confirm a compound's identity by matching against a known reference spectrum.
Examples in context
Example 1. Monitoring a reaction. Chemists follow the oxidation of an alcohol to a carboxylic acid by watching the IR spectrum: the broad alcohol O-H gives way to a strong C=O and the very broad acid O-H. SEAB uses this to test whether candidates can link the appearance and disappearance of absorptions to a chemical change.
Example 2. Breath alcohol screening. Some breathalysers use infrared absorption by the C-H and C-O bonds of ethanol to measure alcohol concentration in breath. This everyday application illustrates that the strength of absorption can be used quantitatively, not just to identify a group.
Try this
Q1. State the approximate wavenumber and the group responsible for the absorption that confirms a ketone. [2 marks]
- Cue. About cm; the C=O (carbonyl) stretch.
Q2. Explain how the IR spectra of ethanol and ethanoic acid differ. [2 marks]
- Cue. Ethanol has a broad O-H (3200 to 3550) and no C=O; ethanoic acid has a strong C=O (around 1715) and a very broad acid O-H (2500 to 3300).
Q3. Explain why the fingerprint region is useful even though individual bonds in it are hard to assign. [2 marks]
- Cue. It is a unique pattern for each molecule, so matching it against a reference spectrum confirms the compound's identity.
Exam-style practice questions
Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Specimen (9729)4 marksTwo compounds A and B both have the molecular formula C3H6O. The infrared spectrum of A shows a strong absorption at about 1715 cm-1 but no broad absorption near 3300 cm-1. The spectrum of B shows a broad absorption near 3350 cm-1 and a peak near 1650 cm-1. Suggest structures for A and B, explaining your reasoning.Show worked answer →
Use the data-booklet absorption ranges to assign each peak.
Compound A: the strong absorption at about 1715 cm-1 is a C=O stretch (carbonyl). No broad O-H absorption near 3300 means no alcohol or acid. So A is a carbonyl compound: propanal (CH3CH2CHO) or propanone (CH3COCH3).
Compound B: the broad absorption near 3350 cm-1 is an O-H stretch (alcohol), and the peak near 1650 cm-1 suggests a C=C. So B is an unsaturated alcohol: prop-2-en-1-ol (CH2=CHCH2OH).
Markers reward assigning the C=O for A with the absence of O-H, the O-H and C=C for B, and consistent structures.
2023 (style)3 marksExplain how infrared spectroscopy could be used to confirm that the oxidation of ethanol to ethanoic acid had occurred, referring to specific absorptions.Show worked answer →
Compare the infrared spectra of the starting ethanol and the product ethanoic acid.
Ethanol shows a broad O-H absorption around 3200 to 3550 cm-1 (alcohol O-H) and no C=O absorption.
Ethanoic acid shows a strong C=O absorption around 1700 to 1725 cm-1 (from the carboxylic acid carbonyl) as well as a very broad O-H absorption around 2500 to 3300 cm-1 (carboxylic acid O-H).
The appearance of the strong C=O peak in the product confirms a carbonyl-containing group has formed, consistent with oxidation to a carboxylic acid.
Markers reward the alcohol O-H in ethanol, the appearance of the C=O (and broad acid O-H) in the product, and the conclusion that oxidation occurred.
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