Question

When ethanoic acid is treated with ethanol in presence of $H_{2}S{O}_4$, then the products formed are:

• A. ${\mathrm{HCOOC}}_2{H}_5\mathrm{and}{H}_{2}O$
• B. $C_2{H}_5{\mathrm{COOCH}}_3\mathrm{and}{H}_{2}O$
• C. ${\mathrm{CH}}_3{\mathrm{COOC}}_2{H}_5\mathrm{and}{H}_{2}O$
• D. $C_2{H}_5{\mathrm{COOC}}_2{H}_5\mathrm{and}{H}_{2}O$

Answer 1

The correct option is C ${\mathrm{CH}}_3{\mathrm{COOC}}_2{H}_5\mathrm{and}{H}_{2}O$

• Carboxylic acid reacts with alcohol to form an ester.
• Ethanoic acid will react with ethanol in presence of sulphuric acid to form ethyl ethanoate and water.${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{CH}}_3\mathrm{COOH}\to {\mathrm{CH}}_3{\mathrm{COOC}}_2{H}_5\mathrm{and}{H}_{2}O$