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

Compound C with molecular formula `C_(2)H_(4)O_(2)` contains two oxygen atoms so it can be either ester or carboxylic acid.
As it reacts with sodium metal to form compound R and evolves a gas which burns with pop sound, hence it should be a carboxylic acid which forms sodium alkanoate and hydrogen gas with sodium metal.
`2CH_(3)COOH+2Nato2CH_(3)COONa+H_(2)uarr`
The gas which burns with op sound is hydrogen gas .
Reaction of ethanoic acid with alcohol in the presence of an acid (Conc. `H_(2)SO_(4))` forms sweet smelling ester. Therefore ,compound S that is formed due to reaction of ethanoic acid and methanol (A) is methyl ethanoate with molecular formula `C_(3)H_(6)O_(2)` and structural formula `CH_(3)COOCH_(3)`.
Hence , C- Ethanoic acid
R- Sodium salt of ethanoic acid (sodium acetate ) and gas evolved is hydrogen
A- Methanol
S- Ester (Methly acetete)
Reaction involved are :
(i) `2CH_(3)underset((C))(COOH)+2Nato2CH_(3)underset((R))(COONa)+H_(2)`
(ii) `CH_(3)COOCH_(3)+NaOHtounderset((R))(CH_(3)COONa)+underset((A))(CH_(3)OH)`
(iii) `CH_(3)underset((C))(COOCH)+NaOHtounderset((R))(CH_(3)COONa)+Na+H_(2)O`
(iv) `CH_(3)COOCH_(3)+NaOHtounderset((R))(CH_(3))COONa+underset((A))(CH_(3)OH)`