Question

A compound C (molecular formula, C₂H₄O₂) reacts with Na – metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, C₃H₆O₂). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A. Identify C, R, A, S and write down the reactions involved.

Answer 1

• Compound C is ethanoic acid. When it reacts with sodium metal, it forms sodium ethanoate and hydrogen gas is evolved which burns with a pop sound. Hence, compound R is sodium ethanoate.
  \[\ce{\underset{ethanoic acid}{2CH3COOH} + 2Na -> \underset{Sodium ethanoate}{2CH3COONa} + H2}\]
• Ethanoic acid on treatment with methanol in the presence of an acid forms an ester which is a sweet-smelling compound.
  \[\ce{CH3COOH + CH3OH \overset{Acid}{-> }CH3COOCH3 + H2O}\]
  Hence, alcohol A is methanol and compound S is methyl ethanoate.
• On addition of NaOH to ethanoic acid, it gives sodium ethanoate and water.
  \[\ce{NaOH + CH3COOH -> CH3COONa + H2O}\]
• On addition of NaOH to ethanoic acid, it gives sodium ethanoate and water.
  \[\ce{CH3COOCH3 + NaOH -> CH3OH + CH3COONa}\]