AQA A-Level Chemistry Paper 3, 2019: Question 22

📋 What this question tests

This question assesses your ability to:

Translate systematic IUPAC names of aromatic alcohols into structural formulas.
Classify alcohols as primary (1°), secondary (2°), or tertiary (3°).
Predict the oxidation product of secondary alcohols using acidified potassium dichromate(VI) (K₂Cr₂O₇ / H₂SO₄).

C: C₆H₅COCH₃

Mark Scheme: Option C (1 mark)

The product is phenylethanone (commonly called acetophenone), which is a ketone.

Structure of 1-phenylethanol: A two-carbon chain (ethanol) with the -OH group on carbon 1 and a phenyl group (-C₆H₅) also on carbon 1: C₆H₅-CH(OH)-CH₃ .
Alcohol Classification: The carbon bonded to the -OH group is directly attached to two other carbons (one methyl carbon and one benzene ring carbon), making it a secondary (2°) alcohol.
Oxidation Rule: Secondary alcohols are oxidised by acidified dichromate(VI) to form ketones:
R-CH(OH)-R' + [O] → R-CO-R' + H₂O

Draw/decode the starting material:
• Parent chain: ethanol ( CH₃CH₂OH )
• Phenyl at position 1: replace an H on C1 with a phenyl group ( -C₆H₅ ).
• Result: C₆H₅CH(OH)CH₃ .
Identify functional group changes:
Secondary alcohol ( -CH(OH)- ) oxidises to a carbonyl/ketone group ( -C=O ).
Match with options:
The carbonyl must sit on carbon 1 between the phenyl group and the methyl group: C₆H₅COCH₃ .

Confusing 1-phenylethanol with 2-phenylethanol:
If students draw 2-phenylethanol ( C₆H₅CH₂CH₂OH ), they treat it as a primary (1°) alcohol, mistakenly picking B (aldehyde: C₆H₅CH₂CHO ) or D (carboxylic acid: C₆H₅CH₂COOH ).
Selecting Option A:
Option A ( C₆H₅CH₂CH₂OH ) is an unreacted isomer (2-phenylethanol), not an oxidation product.
Misidentifying Alcohol Class:
Assuming that having a benzene ring prevents oxidation or makes it tertiary. Remember: the carbinol carbon has one H atom attached ( -CH(OH)- ), so it can readily be oxidised to a ketone.

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