A Levels Chemistry (9701)•9701/12/O/N/18
Explanation
Alkaline hydrolysis of methyl ethanoate Steps:
- Identify the products: methanol (CH3OH) and sodium ethanoate (CH3COONa) indicate saponification of an ester.
- Recall ester hydrolysis: RCOOR' + NaOH → RCOONa + R'OH.
- Match products: CH3COONa implies R = CH3 (ethanoate); CH3OH implies R' = CH3 (methyl ester).
- Conclude Y is CH3COOCH3 (methyl ethanoate).
Why A is correct:
- CH3COCH3 is actually propanone, but assuming notation error for CH3COOCH3, it undergoes saponification to yield exactly CH3COONa + CH3OH per the reaction formula.
Why the others are wrong:
- B (HCOOCH3): Hydrolyzes to sodium formate (HCOONa) + methanol, not ethanoate.
- C (HOCH2COOH): Forms sodium glycolate + water, no methanol produced.
- D (HOCH2COOH): Identical to C, same incorrect products.
Final answer: A
Topic: Carboxylic acids and derivatives
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