Question

An ester is boiled with KOH. The product is cooled and acidified with conc. HCL. A white crystalline acid separates. The ester is:

Options

• methyl acetate

• ethyl acetate

• ethyl formate

• ethyl benzoate

Answer 1

ethyl benzoate

Explanation:

\[\begin{array}{cc}
\phantom{}\ce{\underset{(Ethyl benzoate)}{C6H5COOC2H5} ->[KOH][Boiling] \underset{}{C6H5COOK + C2H5OH}->[H+/H2O] \underset{(Benzoic acid)}{C6H5COOH}}\\
\end{array}\]