A carboxylic acid molecular formula $C_{2} H_{4} O_{2}$ reacts with an alcohol in the presence of an acid catalyst to form a compound ‘X’. The alcohol on oxidation with alkaline $KMn O_{4}$ followed by acidification gives the same carboxylic acid $C_{2} H_{4} O_{2}$ .
Write the name of the compound X involved?

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Ethanoic acid

Ethanol

Ethyl ethanoate

Ethanone

answer is C.

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Detailed Solution

The name of the compound X is ethyl ethanoate.

Esterification is a chemical reaction that produces at least one ester product by reacting an organic acid with an alcohol to create an ester and water. The esterification of an alcohol and a carboxylic acid produces ester.
$C H_{3} COOH + C H_{3} C H_{2} OH + H_{2} S O_{4} \to \underset{X}{C H_{3} C O O C_{2} H_{5}} + H_{2} O$
In the presence of alkaline $KMn O_{4}$ the reaction taking place is $C H_{3} C H_{2} OH \overset{[O]}{\to} C H_{3} COOH$
Ethanoic acid is the carboxylic acid, ethanol is the alcohol and X is ethyl ethanoate.