Compound J is a carboxylic acid

Steps:

• Reaction with Na₂CO₃ producing CO₂ shows acidic properties, as acids liberate CO₂ from carbonates.
• Reaction with ethanol yielding sweet-smelling liquid indicates esterification, forming an ester.
• Reaction with NaOH producing a salt confirms neutralization to form a carboxylate.
• These match carboxylic acid behavior, specifically ethanoic acid.

Why A is correct:

• Ethanoic acid (CH₃COOH) reacts with Na₂CO₃ to give CO₂ (2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ + H₂O), esterifies with ethanol to ethyl ethanoate (fruity smell), and neutralizes with NaOH to sodium ethanoate.

Why the others are wrong:

• B. Ethanol: Alcohol; does not liberate CO₂ from Na₂CO₃ or form ester with itself.
• C. Ethyl ethanoate: Ester; hydrolyzes with NaOH but does not react with Na₂CO₃ for CO₂ or ethanol for sweet liquid.
• D. Ethyl methanoate: Ester; similar to C, lacks acidic reaction with Na₂CO₃.

Final answer: A