An ester is boiled with $\mathrm{KOH}$. The product is cooled and acidified with concentrated HCl. A white…

An ester is boiled with $\mathrm{KOH}$. The product is cooled and acidified with concentrated HCl. A white crystalline acid separates. The ester is
  1. methyl acetate
  2. ethyl acetate
  3. ethyl formate
  4. ethyl benzoate

Solution

Ethyl benzoate on hydrolysis gives benzoic acid which is a solid, other esters give $\mathrm{CH}_{3} \mathrm{COOH}$ and HCOOH, both of which are liquids.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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