OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 20

Identifying Functional Groups Using Infrared Spectroscopy

What this question tests

This question assesses your ability to interpret an infrared (IR) spectrum by identifying characteristic absorption peaks and matching them to specific functional groups present in organic molecules. You must use your knowledge of wavenumber data from the Data Sheet to systematically rule out incorrect structural isomers.

Question 20 (Multiple Choice)

Identifying the Compound from its IR Spectrum

✅ Correct Answer

B (4-hydroxybutan-2-one / HO-CH₂-CH₂-CO-CH₃)

Marks: 1 / 1 (AO2.5)

💡 Key Knowledge

Broad peak at ~3200–3600 cm⁻¹: Indicates an O-H stretch (alcohol).
Sharp, strong peak at ~1680–1750 cm⁻¹: Indicates a C=O stretch (carbonyl / ketone / aldehyde / carboxylic acid).
Absence of very broad 2500–3300 cm⁻¹ peak: Rules out carboxylic acids.
Absence of C=C stretch (~1620–1680 cm⁻¹): Rules out alkenes.

🧠 Exam Technique

Elimination is key in spectroscopic multiple-choice questions. Scan the spectrum for the two most prominent features first: the very broad trough around 3300 cm⁻¹ (O-H) and the sharp trough around 1715 cm⁻¹ (C=O). Match these features against the options to quickly cross out distractors.

❌ Common Errors

Confusing the broad alcohol O-H stretch with the extremely broad carboxylic acid O-H stretch that overlaps with C-H regions.
Mistaking a carbonyl (C=O) peak for a C=C alkene stretch.
Failing to check all functional groups present in complex isomers (e.g. missing that compound A has a carboxylic acid group while B has both an alcohol and a ketone).

Examiner Analysis & Step-by-Step Breakdown

Step-by-Step Spectrum Analysis

Look for the O-H (alcohol) stretch: There is a very broad and strong absorption band centered around 3300 cm⁻¹ . This confirms the presence of an alcohol (OH) group, immediately eliminating compound D (which has an alkene and ketone, but no alcohol).
Look for the C=O (carbonyl) stretch: There is a very sharp, deep absorption peak around 1715 cm⁻¹ . This confirms a carbonyl group (C=O). Compound C has an alcohol and an alkene, but lacks a carbonyl group, so C is eliminated.
Distinguish between A and B (Carboxylic Acid vs Alcohol + Ketone):
- Compound A is a carboxylic acid. Carboxylic acids display a very characteristic, exceptionally broad O-H stretch that typically blankets the region from 2500 to 3300 cm⁻¹ , merging with the C-H stretches.
- Looking closely at the spectrum, the absorption around 3000 cm⁻¹ has sharp individual spikes characteristic of C-H stretches standing separate from the O-H trough, and there is no massive broad spread down to 2500 cm⁻¹. Furthermore, the carbonyl peak matches a standard ketone/aldehyde rather than a hydrogen-bonded carboxylic acid dimer.
- Therefore, the spectrum belongs to B, which contains both an alcohol (-OH) and a ketone (C=O).

Examiner Commentary: Top-level responses quickly identified both the alcohol and carbonyl peaks from the data sheet values. Common mistakes included selecting compound A because students saw both "O-H" and "C=O" listed in a carboxylic acid without carefully evaluating the exact breadth and shape of the O-H absorption region.