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A compound C (molecular formula, C_2H_4O_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_3H_6O_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. 

Answers (1)

Since Compound C has two oxygen atoms it can be an ester or carboxylic acid. 

Given that the gas evolved (on the reaction with sodium) burns with a pop sound in an indication that evolved gas is hydrogen.

Compound R is going to be a salt with sodium being the positive ion and the rest of the compound C which lost hydrogen must be negatively charged. This is only possible with carboxylic acid and not ester.

So C is carboxylic acid and based on number of carbons it is Ethanoic acid.

Compound C is CH_3COOH.

Compound R is CH_3COONa..

Compound A is CH_3OH.

Compound S is CH_3COOCH_3

Reactions involved:

2CH_3 COOH + 2Na \rightarrow 2CH_{3}COONa+H_{2}

       (C)                             (R)

CH_3COOH + CH_{3}OH \rightarrow CH_3COOCH_{3}+H_2O?
       (C)                     (A)                                        (R)

CH_3COOOH +NaOH\rightarrow CH_{3}COONa?+H_{2}O

       (C)                                   (R)

CH_3COOCH_{3}+NaOH\rightarrow CH_{3}COONa +CH_{3}OH?

   (R)                                (A)


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