Base hydrolysis of an ester with NaOH gives a carboxylic acid whose sodium salt on Kolbe’s electrolysis yields ethane. The ester isA.Ethyl methanoateB.Methyl ethanoateC.Phenyl benzoateD.Ethyl propanoate

Let's analyze the given reaction sequence step by step.

Base hydrolysis of an ester with NaOH yields a carboxylic acid and an alcohol. This reaction is known as saponification.

Kolbe's electrolysis of the sodium salt of a carboxylic acid leads to the decarboxylation of the acid, producing an alkane with two fewer carbon atoms than the original acid.

Now, let's look at the options:

A. Ethyl methanoate (CH3COOCH2CH3)
The base hydrolysis of ethyl methanoate would yield methanoic acid (formic acid, HCOOH), and Kolbe's electrolysis of its sodium salt would lead to the formation of methane (CH4), not ethane (C2H6).

B. Methyl ethanoate (CH3COOCH3)
The base hydrolysis of methyl ethanoate would yield ethanoic acid (acetic acid, CH3COOH), and Kolbe's electrolysis of its sodium salt would lead to the formation of ethane (C2H6). This aligns with the given reaction sequence.

C. Phenyl benzoate (C6H5COOC6H5)
Base hydrolysis of phenyl benzoate would yield benzoic acid (C6H5COOH), and Kolbe's electrolysis of its sodium salt would lead to the formation of benzene (C6H6), not ethane (C2H6).

D. Ethyl propanoate (CH3CH2COOCH2CH3)
Base hydrolysis of ethyl propanoate would yield propanoic acid (propionic acid, CH3CH2COOH), and Kolbe's electrolysis of its sodium salt would lead to the formation of propane (C3H8), not ethane (C2H6).

So, the correct answer is B. Methyl ethanoate.