CBSE Class 10 Science 2019 · Paper 31/3/1 · Question 17

(a) Esterification: a carboxylic acid reacts with an alcohol, when warmed with a few drops of concentrated sulphuric acid (an acid catalyst), to form a sweet-smelling ester and water: ${\text{CH}}_3\text{COOH}+{\text{C}}_2{\text{H}}_5\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$ (ethanoic acid + ethanol give ethyl ethanoate + water). Saponification: an ester reacts with an alkali such as sodium hydroxide to give back the alcohol and the sodium salt of the carboxylic acid: ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}\to {\text{C}}_2{\text{H}}_5\text{OH}+{\text{CH}}_3\text{COONa}$ (ethyl ethanoate + sodium hydroxide give ethanol + sodium ethanoate). So esterification forms an ester from an acid and an alcohol in an acidic medium, while saponification breaks an ester with a base and is used to make soap. (b) Activity: take $1$ mL of ethanol and $1$ mL of glacial ethanoic acid in a test tube and add a few drops of concentrated sulphuric acid; warm the test tube in a water bath for at least $5$ minutes; pour the mixture into a beaker containing $20$ to $50$ mL of water and smell it. A sweet, fruity smell shows that an ester (ethyl ethanoate) has been formed.