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:

  • Compound C is CH3COOH
  • Compound R is CH3COONa
  • Compound A is C2H5OH
  • Compound S is CH3COOC2H5

(1) Compound C is Ethanoic acid (CH3COOH). It reacts with sodium metal to form a compound called R where R is Sodium Ethanoate (CH3COONa).

2CH3COOH + 2Na → 2CH3COONa + H2

(2) Compound S is Ester or Ethyl ethanoate(CH3COOC2H5) and compound A is Ethanol (C2H5OH).

CH3COOH + C2H5OH → CH3COOC2H5 + H20

(3) Compound R is again Sodium Ethanoate

CH3COOC2H5 + NaOH → CH3COONa + C2H5OH

So compound C is Ethanoic acid.

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