A compound 'C' (molecular formula, $C_{2} H_{4} O_{2}$ ) 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, $C_{3} H_{6} O_{2}$ ). 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.

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Solution

Compound 'C' is ethanoic acid. It reacts with sodium to form sodium ethanoate. Hence. compound 'R' is sodium ethanoate or sodium acetate. We know that hydrogen gas burns with a pop sound. This reaction can be given by the following equation:

$2 C H_{3} C O O H + 2 N a \to 2 C H_{3} C O O N a + H_{2}$
(C) (R)

When ethanoic acid reacts with methanol in the presence of an acid, we get (methyl ethanoate) ester which is a sweet-smelling substance. Hence, compound S is methyl ethanoate and A is methanol. This reaction can be given as follows:

$C H_{3} C O O H + C H_{3} O H A c i d - - \to C H_{3} C O O - C H_{3} + H_{2} O$
(C) (A) (S)

When sodium hydroxide is added to ethanoic acid, it gives sodium ethanoate and water.

$C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$
(C) (R)
When methyl ethanoate is treated with NaOH solution, it gives back methanol and sodium ethanoate.

$C H_{3} C O O - C H_{3} + N a O H \to C H_{3} O H + C H_{3} C O O N a$
(S) (A) (R)

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