Generated by claude-sonnet-4-6 · 2026-06-15 06:47 · grounding rag
Model Answer
(i)
- X = Ethanol (C₂H₅OH) — a good solvent
- Y = Sodium ethoxide (C₂H₅ONa)
- Z = Hydrogen gas (H₂) — used in hydrogenation of vegetable oil to make vegetable ghee
Reaction of X with sodium:
$$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$$
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(ii)
(I) Combustion of ethanol:
$$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{heat and light}$$
(II) Dehydration at 443 K with excess conc. H₂SO₄:
$$C_2H_5OH \xrightarrow{443\text{K, conc. H}_2SO_4} CH_2=CH_2\uparrow + H_2O$$
(Ethanol → Ethene + Water)
(III) Reaction with acidified potassium dichromate (oxidation):
$$C_2H_5OH \xrightarrow{K_2Cr_2O_7/H^+} CH_3COOH$$
(Ethanol is oxidised to ethanoic acid; orange dichromate turns green.)
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Explanation
- Part (i): The key clue is "Z converts vegetable oil to ghee" → Z = H₂ (hydrogenation). Since H₂ is produced with sodium, X must be an alcohol. As a "good solvent," X is ethanol. Y is sodium ethoxide (the sodium salt of ethanol).
- Part (ii): Examiners expect balanced equations. For combustion: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. For dehydration: excess H₂SO₄ at 443 K gives ethene (elimination). For oxidation with acidified K₂Cr₂O₇: ethanol → ethanoic acid (colour change from orange to green is a bonus point). Write product names alongside formulae for clarity.