CBSE Class 10 Science 2019 · Paper 31/4/1 · Question 16(a)

(a) Esterification: a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (a few drops of concentrated sulphuric acid) to form a sweet-smelling ester and water: ${\text{CH}}_3\text{COOH}+{\text{C}}_2{\text{H}}_5\text{OH}\stackrel{{\text{conc. H}}_2{\text{SO}}_4}{\to }{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+{\text{H}}_2\text{O}$ Saponification: an ester reacts with an alkali such as sodium hydroxide to give back the alcohol and the sodium salt of the carboxylic acid; with fats and oils this reaction makes soap: ${\text{CH}}_3{\text{COOC}}_2{\text{H}}_5+\text{NaOH}⟶{\text{C}}_2{\text{H}}_5\text{OH}+{\text{CH}}_3\text{COONa}$ Thus esterification forms an ester from an acid and an alcohol with an acid catalyst, while saponification breaks an ester with a base into an alcohol and a salt. (b) Take 1 mL ethanol, 1 mL glacial ethanoic acid and a few drops of concentrated sulphuric acid in a test tube. Warm it in a water bath for about five minutes (see the labelled diagram). 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.