A Levels Chemistry (9701)•9701/12/O/N/22
Explanation
Alkaline hydrolysis of methyl propanoate Steps:
- Recognize the ester as CH3CH2COOCH3 (methyl propanoate), where the acyl group is from propanoic acid.
- In alkaline hydrolysis with NaOH, the ester undergoes saponification: RCOOR' + NaOH → RCOONa + R'OH.
- The carboxylate salt formed is from the acid part: CH3CH2COONa (sodium propanoate).
- The alcohol from the alkoxy part is CH3OH (methanol), confirming the salt product.
Why D is correct:
- Sodium propanoate (CH3CH2COONa) is the carboxylate salt produced from the propanoyl group in the ester, per the saponification reaction mechanism.
Why the others are wrong:
- A: Ethanol (C2H5OH) would form from an ethyl ester, not methyl.
- B: Propan-1-ol (C3H7OH) requires a propyl ester group, absent here.
- C: Sodium methanoate (HCOONa) derives from methanoic acid, not propanoic.
Final answer: D
Topic: Carboxylic acids and derivatives
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