Science — CBSE Class 10 board question

(i) Ethanol is a commercially important carbon compound with the functional group –OH.
Molecular formula: C₂H₅OH

(ii) Chemical equations:

(1) Reaction with Sodium metal:
$$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$$
Product: Sodium ethoxide + Hydrogen gas

(2) Reaction with Excess conc. H₂SO₄ (at 170°C):
$$C_2H_5OH \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 170°C} CH_2=CH_2 + H_2O$$
Product: Ethene (dehydration)

(3) Reaction with Ethanoic acid (esterification):
$$C_2H_5OH + CH_3COOH \xrightarrow{\text{H}^+} CH_3COOC_2H_5 + H_2O$$
Product: Ethyl ethanoate (an ester) — sweet-smelling compound

(4) Reaction with Acidified Potassium Dichromate (oxidation):
$$C_2H_5OH \xrightarrow{\text{acidified K}_2\text{Cr}_2\text{O}_7} CH_3COOH$$
Product: Ethanoic acid
(Oxidising agent turns from orange to green — this is an oxidation reaction.)

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• The examiner expects ethanol as the answer to (i) since it is the most commercially important alcohol (–OH group) covered in the chapter.
• For (ii), all four reactions must be written with balanced equations and product names — each reaction carries roughly 1 mark.
• Key points: Na reaction releases H₂; excess H₂SO₄ at 170°C causes dehydration (gives alkene); esterification gives sweet-smelling ester; acidified K₂Cr₂O₄ is an oxidising agent that converts alcohol → carboxylic acid (colour change orange → green is a bonus point).
• Do not confuse excess H₂SO₄ at 170°C (dehydration → ethene) with H₂SO₄ at 140°C (gives diethyl ether).