CBSE Class 10 Science 2017 · Paper 31/1 · Question 8

Esterification: a carboxylic acid reacts with an alcohol, in the presence of an acid catalyst (a few drops of concentrated sulphuric acid) on warming, to form an ester and water. $${\text{CH}}_3\text{COOH}+{\text{C}}_2{\text{H}}_5\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4,\text{warm}}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$$ Saponification: an ester reacts with an alkali (sodium hydroxide) and is converted back to the alcohol and the sodium salt of the carboxylic acid. $${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶{\text{CH}}_3\text{COONa}+{\text{C}}_2{\text{H}}_5\text{OH}$$ Differences: reactants — acid + alcohol in esterification, ester + alkali in saponification; condition — an acid catalyst in esterification, an alkali in saponification; products — ester + water in esterification, alcohol + sodium salt of the acid in saponification; esterification forms an ester, saponification breaks one down. Uses: (i) esters are sweet-smelling and are used in making perfumes and as flavouring agents; (ii) saponification is used in the preparation of soap.