Question

A compound 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 an alcohol A in presence of an acid forms a sweet smelling compounds S (moleculat formula C3H6O2) On addition of NaOH to C it slwo gives R and water S on treatment with NaOh solution gives back R and A Identify C , R, A, S and write the reactions involved

Answer 1

1)Compound C is Ethanoic acid[Acetic acid]
It reacts with sodium metal to form a compound called R . R is Sodium Ethanoate.
2CH3COOH  +2 Na--------------> 2CH3COONa+H2

                                               conc, H2SO4
2) CH3COOH  +C2H5OH       ------ -------------> CH3COOC2H5 + H20
so compound S is Ester or Ethyl ethanoate and compound A is Ethanol
3)CH3COOC2H5+NaoH--------> CH3COONa    + C2H5OH.
so compound R is again Sodium Ethanoate.
So compound C is Ethanoic acid
A is Ethanol
S is Ester
R is sodium 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