Question

Base hydrolysis of an ester with NaOH gives a carboxylic acid whose sodium salt on Kolbe’s electrolysis yields ethane. The ester is
A.Ethyl methanoate
B.Methyl ethanoate
C.Phenyl benzoate
D.Ethyl propanoate

Answer 1

Hint: Hydrolysis is the process of simple reaction of the addition of water to a substance. Base hydrolysis of an ester with a strong base yields a carboxylate salt and alcohol. Kolbe’s electrolysis gives the decarboxylated product which is an alkane along with the production of carbon dioxide by a free radical mechanism.

Complete step by step answer:
We know that Kolbe’s electrolysis of a molecule gives the dimer products by combining free radicals after decarboxylation to form alkane. So we have been given ethane through Kolbe’s electrolysis which means that we have the alkyl group as methyl group in a carboxylic acid.
Further, as only methyl groups are present in carboxylic acid so the carboxylic acid here would be ethanoic acid. Now the carboxylic acid is ethanoic acid then the ester will be formed by removing the hydrogen in ethanoic acid, that is $C{H_3}COOC{H_3}$ according to the given conditions. The ester name is methyl ethanoate.
The reaction involved in the following processes are as follows
$C{H_3}COOC{H_3} + NaOH \to C{H_3}COONa + C{H_3}OH$. Now this sodium salt of ethanoic acid undergoes Kolbe’s electrolysis to form methyl radical which dimerises to form alkane. The reaction involved is given below
$C{H_3}CO{O^ - } \to C{H_3}CO{O^ \bullet }$
$C{H_3}CO{O^ \bullet } \to C{H_3}^ \bullet + C{O_2}$
$2C{H_3}^ \bullet \to C{H_3} - C{H_3}$
Therefore from the above reaction, we infer that the required ester is methyl ethanoate.
Hence the correct answer is option B.

Note:
Esters can be cleaved back into a carboxylic acid and alcohol by reaction with water and base. This reaction is called a saponification reaction which is a characteristic reaction of esters. Also as basic conditions are there, a carboxylate ion is made rather than a carboxylic acid.