Mohan heated ethanol with a compound 'X' in the presence of a few drops of conc. H₂SO₄ and observed a sweet smelling compound 'Y' is formed. When 'Y' is treated with sodium hydroxide it gives back ethanol and a compound 'Z'.
(i) Identify 'X', 'Y' and 'Z'.
(ii) Write the role of conc. H₂SO₄ in the reaction.
(iii) Write the chemical equations involved and name the reactions.
Generated by claude-sonnet-4-6 · 2026-06-15 06:46 · grounding rag
Model Answer
(i) Identification:
- X = Ethanoic acid (CH₃COOH)
- Y = Ethyl ethanoate / Ethyl acetate (CH₃COOC₂H₅) — sweet smelling ester
- Z = Sodium ethanoate (CH₃COONa)
(ii) Role of conc. H₂SO₄:
It acts as a catalyst and also as a dehydrating agent, removing water to drive the esterification reaction forward.
(iii) Chemical Equations:
Reaction 1 — Esterification:
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
Reaction 2 — Saponification (hydrolysis of ester):
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}$$
The first reaction is called Esterification; the second is called Saponification.
Source: Chapter 4, Section 4.4.2 (Ethanoic Acid — Reactions)
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Explanation
- Examiners expect all three identifications (X, Y, Z) stated clearly with formulae — 1 mark.
- The role of H₂SO₄ as catalyst/dehydrating agent is a standard 1-mark point.
- Both balanced equations must be written with correct formulae; half-marks are often given per equation.
- Naming both reactions (esterification and saponification) is essential — commonly missed.
- "Sweet smelling" is the clue for an ester; NaOH breaking an ester is always saponification.