Q. 524.1( 18 Votes )

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 :

C is ethanoic acid or acetic acid (CH3COOH).


R is sodium acetate or sodium ethanoate (CH3COONa) and the gas evolved is hydrogen.


A can be methanol (CH3OH) or ethanol (C2H5OH).


S is ester or methyl acetate (CH3COOCH3), an ester.


The reactions involved in this process are as follows:


1. 2CH3COOH + 2Na 2CH3COONa + H2


2. CH3COOH + CH3OH + Conc. H2SO4 CH3COOCH3 + H2O


3. CH3COOH + NaOH CH3COONa + H2O


4. CH3COOH + NaOH CH3COONa+ CH3OH


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