question_answer 80)

    A compound C (molecular formula, 021-1402) reacts with Na metal to forma compound 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 forms a sweet smelling compound S (molecular formula, ). On addition of 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:

                  (i) Since, compound contains two oxygen atoms, therefore most probably it may be a carboxylic acid, i.e, ethanoic acid . (ii) Since, an acid, i.e. ,ethanoic acid reacts with a base, i.e., Na metal to evolve a gas which burns with a pop sound along with the formation of compound (R), therefore, fl must be a salt, i'.e., sodium ethanoate and the gas which burns with a pop sound must be  gas. gas burns with 'pop' sound. Compound R is sodium ethanoate. This is also supported by the observation that when ethanoic acid reacts with NaOH, it gives R, is sodium ethanoate and water. Since compound C on treatment with an alcohol A in presence of acid forms a sweet smelling compound C(M.F.), therefore, S is methyl ethanoate (ester).Since, ester S has three carbon atoms and the acid C has two carbon atoms, therefore alcohol A must contain one C atom, i.e., A is methanol. Thus,