Question

An organiccompound 'c' (molecular formula c2h4o2) reacts with sodium metal to form a compound 'r' and evolves a gas which burns with a pop sound. Compound 'c' on treatment with alcohol 'a' in the presence of an acid forms a sweet-smelling compound 's' (molecular formula c3h6o2). Addition of naoh to 'c'also gives 'r' and water. 's' on treatment with naoh solution gives back 'r' and 'a'. Identify 'c', 'r', 'a'and's',and write the reactions involved.

Answer 1

C: Ethanoic acid.

R: Sodium ethanoate.

A: Ethanol.

S: Ester or Ethyl ethanoate.

Answer 2

Answer:

2 CH₃COOH + 2 Na ⇒  2 CH₃COONa + H₂ ↑

      ( A )                                   ( B )

CH₃COOH + C₂H₅OH ⇒  CH₃COOC₂H₅ + H₂O

                        ( C )                 ( D )

CH₃COOC₂H₅ + NaOH  ⇒ C₂H₅OH + CH₃COONa

Therefore we identified

A - CH₃COOH

B - CH₃COONa

C - C₂H₅OH

D - CH₃COOC₂H₅