A compound C (molecular formula, `C_(2)H_(4)O_(2)`) reacts with Na metal to form a compounds R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in the presence of an acid form 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.

Explanation

From the available information, it is evident that
(i) Compound C with moleculr formula `C_(2)H_(4)O_(2)` is ethanoic acid `(CH_(3)COOH)`
(ii) The compound R is sodium ethanoate and has formula `CH_(3)COONa`
(iii) Since the compound S has only three carbon atoms `(C_(3)H_(6)O_(2))` and has been formed by the action of an alcohol on compound `C(C_(2)H_(4)O_(2))`, this means that the alcohol A has only one carbon atom. it is methanol `(CH_(3)OH)`.
(iv) The compound S with a sweet smell is methyl ethanoate with formula `CH_(3)COOCH_(3)`.
The chemical reactions involved are as follows:
(a) `underset((C))(2CH_(3)COOH)+2Natounderset((R))(2CH_(3)COONa)+H_(2)`
(b) `underset((C))(CH_(3)COOH)+underset((A))(CH_(3)OH)overset("conc. "H_(2)SO_(4))tounderset((S))(CH_(3)COOCH_(3))+H_(2)O`
(c) `underset((C))(CH_(3)COOH)+NaOHtounderset((R))(CH_(3)COONa)+H_(2)O`
(d) `underset((S))(CH_(3)COOCH_(3))+NaOHtounderset((R))(CH_(3)COONa)+underset((A))(CH_(3)OH)`