Science — CBSE Class 10 board question

(i)
Compound A = Ethanoic acid (CH₃COOH)
Nature: It is a weak organic acid.
Functional group: Carboxyl group (–COOH)

Reaction with a base (e.g., NaOH):

$$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$

(Sodium ethanoate + Water)

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(ii)
(1) Compound B = Ethanol (C₂H₅OH)
Compound C = Ethyl ethanoate / Ethyl acetate (CH₃COOC₂H₅) — a sweet-smelling ester.

(2) The acid (concentrated H₂SO₄) acts as a catalyst; it speeds up the esterification reaction without being consumed.

(3) Chemical equation:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\Delta}{\overset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

This reaction is called esterification.

Source: Chapter 4, Carbon and its Compounds

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• Molecular formula C₂H₄O₂ with acid properties → ethanoic acid (acetic acid); students must recognise the –COOH functional group.
• C₂H₆O is ethanol; together with ethanoic acid it undergoes esterification to give ethyl ethanoate (sweet smell is the key clue).
• Examiners expect: correct identification, nature, functional group name, role of acid as catalyst, and balanced equations with correct conditions (conc. H₂SO₄, heat, reversible arrow).
• The reversible arrow (⇌) is important — esterification is a reversible reaction.