O Levels Chemistry (5070)•5070/12/M/J/18
Explanation
Compound J is a carboxylic acid identified by its reactions
Steps:
- Reaction with sodium carbonate producing CO2 shows J is an acid that liberates CO2 from carbonates, characteristic of carboxylic acids.
- Reaction with ethanol yielding a sweet-smelling liquid indicates esterification, where J (as acid) reacts with alcohol to form an ester.
- Reaction with sodium hydroxide producing a salt confirms neutralization, typical for acids forming carboxylate salts.
- Matching properties to ethanoic acid, a common carboxylic acid fitting all reactions.
Why A is correct:
- Ethanoic acid (CH3COOH) reacts with Na2CO3 to give CO2 (2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2), with C2H5OH to form ethyl ethanoate (sweet ester), and with NaOH to form sodium ethanoate salt.
Why the others are wrong:
- B. Ethanol (alcohol) does not react with Na2CO3 to produce CO2.
- C. Ethyl ethanoate (ester) is neutral and does not liberate CO2 from carbonates.
- D. Ethyl methanoate (ester) lacks acidic properties to react with Na2CO3 for CO2.
Final answer: A
Topic: Carboxylic acids
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