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Carbon and its Compounds — Science STD 10 — Question

CBSE BoardEnglish MediumSTD 10ScienceCarbon and its Compounds5 Marks
Question
A compound C (molecular formula, C2H4O2) reacts with Na-metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, C3H6O2). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A.

Identify C, R, A, S and write down the reactions involved.

✓

Answer

Compound C with molecular formula C2H4O2 contains two oxygen atoms so it can be either ester or carboxylic acid.

Since it reacts with sodium metal to form compound R and evolves a gas which burns with pop sound, therefore it should be a carboxylic acid which forms sodium alkanoate and hydrogen gas with sodium metal.

$2\text{CH}_3\text{COOH}+2\text{Na}\xrightarrow{\ \ \ \ \ }2\text{CH}_3\text{COONa}+\text{H}_2\uparrow$

The gas which burns with pop sound is hydrogen gas.

Reaction of ethanoic acid with alcohol in the presence of an acid (Conc. H2SO4) forms sweet smelling ester. So compound S that is formed due to reaction of ethanoic acid and methanol (A) is methyl ethanoate with molecular formula C3H6O2 and structural formula CH3COOCH3.

$\text{CH}_3\text{COOH}+\text{CH}_3\text{OH}\xrightarrow{\ \ \text{Conc. H}_2\text{So}_4\ \ }\text{CH}_3\text{COOCH}_3+\text{H}_2\text{O}$

Hence compound C = Ethanoic acid (CH3COOH), R = Sodium ethanoate (CH3COONa), A = Methanol (CH3OH) and S = Methyl merhanoate (CH3COOCH3)

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