Question

Distinguish between esterification and saponification reactions of organic compounds.

Answer 1

In the esterification reaction, an acid reacts with alcohol in the pressure of conc. $H_{2}S{O}_4$ to form an ester with a pleasant or fruity smell. For example,

$\underset{Ethanoicacid}{C{H}_{3}COOH}$ + $\underset{Ethanol}{C{H}_{3}C{H}_{2}OH}$ $\stackrel{conc.H_{2}S{O}_4}{\to }$ $\underset{Ethylethanoate\left(ester\right)}{C{H}_{3}COO{C}_2{H}_5}$ + $H_{2}O$

Saponification is quite different from esterification because in this case an ester reacts with an alkali
(NaOH or KOH) to form a salt of acid and alcohol. For example,

$\underset{Ethylethanoate}{C{H}_{3}COO{C}_2{H}_5}$ + $NaOH\to$ $\underset{Sodiumethanoate}{C{H}_{3}COONa}$ + $\underset{Ethanol}{C_2{H}_{5}OH}$