A Levels Chemistry (9701)•9701/12/M/J/24
Explanation
Oxidation product identifies primary alcohol
Steps:
- Acidified K₂Cr₂O₇ oxidizes primary alcohols to carboxylic acids, secondary to ketones.
- IR of Y shows broad O-H stretch (2500–3300 cm⁻¹) and C=O stretch (1710 cm⁻¹), indicating carboxylic acid.
- Thus, Y is propanoic acid, so X is propan-1-ol.
- Propan-2-ol would yield propanone (C=O only, no O-H).
Why A is correct:
- Propan-1-ol (CH₃CH₂CH₂OH) oxidizes to propanoic acid (CH₃CH₂COOH), matching IR with acidic O-H and carbonyl.
Why the others are wrong:
- B: Propan-2-ol oxidizes to propanone, lacking broad O-H in IR.
- C: Propanoic acid is not an alcohol, so cannot be X.
- D: Propanone is a ketone, unreactive with dichromate to form Y.
Final answer: A
Topic: Hydroxy compounds
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