AQA A-Level Chemistry AS Paper 2, June 2025: Question 11

📌 What this question tests

This question assesses your ability to:

Identify carbonyl functional groups (specifically aldehydes vs ketones) from condensed structural formulae.
Apply principles of primary, secondary, and tertiary alcohol oxidation using acidified potassium dichromate(VI).
Convert between condensed structural formulae and systematic IUPAC nomenclature for branched organic molecules.

Question 11 • Multiple Choice

D — 3-methylbutan-2-ol

1 Mark awarded (AO3) for correctly identifying option D.

Primary (1°) alcohols oxidise to aldehydes, then carboxylic acids.
Secondary (2°) alcohols oxidise directly to ketones.
Tertiary (3°) alcohols do not oxidise readily because there is no hydrogen atom on the carbon carrying the –OH group.
The product (CH₃)₂CHCOCH₃ contains a –CO– group flanked by two carbon chains; hence, it is a ketone and must come from a secondary alcohol.

Analyse the product formula:
(CH₃)₂CHCOCH₃ has the structure:
CH₃–CH(CH₃)–C(=O)–CH₃
Total carbon count = 5 carbons. The carbonyl group ( C=O ) is at carbon-2.
Reverse the oxidation (reduction):
Converting the ketone carbonyl group ( –C(=O)– ) back to the secondary alcohol group ( –CH(OH)– ) gives:
(CH₃)₂CH–CH(OH)–CH₃
Name the alcohol systematically:
• Longest continuous carbon chain = 4 carbons ( butan- )
• Number chain from right to give –OH the lowest locant → C2 ( butan-2-ol )
• Methyl branch is at C3 → 3-methylbutan-2-ol.

You can solve this rapidly using simple deduction:

Check total carbon count: (CH₃)₂CHCOCH₃ has (2 × 1) + 1 + 1 + 1 = 5 carbons.
Option A has 4 carbons (methylpropanol = 4 C) → Eliminate A .
Check alcohol classification: The product is a ketone, which requires a secondary alcohol.
- Option B is a 1° alcohol (–1-ol) → yields aldehyde/acid → Eliminate B .
- Option C is a 3° alcohol (branch and –OH at same position: 2-methyl...-2-ol) → cannot oxidise → Eliminate C .
- Option D is a 2° alcohol (–OH on C2, branch on C3) → oxidises cleanly to a ketone → Confirmed D.

Confusing numbering priorities: Naming the alcohol from the wrong end as 2-methylbutan-3-ol. Remember that functional groups (like the –OH group) have priority over alkyl branches and must receive the lowest possible carbon number.
Overlooking alcohol classification: Attempting to oxidise 2-methylbutan-2-ol. Because the carbon bearing the –OH has three alkyl groups attached, it has no C–H bond available to break during oxidation.
Misreading the carbonyl type: Confusing the ketone group –CO– with an aldehyde ( –CHO ) or ester ( –COO– ).

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