Science (086) — AI-generated practice question

(i) The reaction is esterification. The product formed is ethyl ethanoate (an ester) and water.
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(ii) Here, conc. H₂SO₄ acts as an acid catalyst. When ethanol alone is heated with excess conc. H₂SO₄ at 443 K, H₂SO₄ acts as a dehydrating agent, removing water to form ethene.

(iii) The ester (ethyl ethanoate) is identified by its sweet smell. When treated with NaOH solution, it undergoes saponification, breaking back into ethanol and sodium ethanoate (sodium salt of ethanoic acid):
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}$$

Source: Chapter 4, Section 4.4.1 & 4.4.2

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• Examiners expect the equation for esterification and saponification — even in short answers, these earn marks.
• The key contrast for part (ii) is: catalyst vs. dehydrating agent — this distinction is frequently tested.
• For identification, "sweet smell" is the textbook answer; always link it to the ester's property.
• "Saponification" as the name of the NaOH reaction is a scoring keyword — include it.