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 R and A?

Answer 1

Compound R is identified as sodium acetate (CH₃COONa), and compound A is ethanol (C₂H₅OH). These identifications are based on the chemical reactions provided involving compound C, which is acetic acid, that produces R and evolves hydrogen gas.

Step-by-step explanation:

Based on the given information, compound C (C₂H₄O₂) reacts with Na metal to evolve a gas which burns with a 'pop' sound, indicating the gas is hydrogen (H₂).

This suggests compound C could be acetic acid (CH₃COOH) because it reacts with sodium (Na) to produce sodium acetate (CH₃COONa), which is compound R, and hydrogen gas (H2).

When acetic acid reacts with an alcohol (A) in the presence of an acid, it undergoes an esterification reaction to form a sweet-smelling ester (S), which has the formula C₃H₆O₂.

Given that S upon hydrolysis with NaOH yields the same sodium acetate (R) and alcohol (A), it is clear that alcohol A is ethanol (C₂H₅OH), and the sweet-smelling ester (S) is ethyl acetate (CH₃COOC₂H₅).

The reactions can be summarized as follows:

• CH₃COOH + Na → CH₃COONa + H2 (gas)
• CH₃COOH + C₂H₅OH →CH₃COOC₂H5 (ethyl acetate) + H₂O
• CH₃COOC₂H₅ + NaOH → CH₃COOHNa + C₂H₅OH

Therefore, R is sodium acetate (CH₃COONa) and A is ethanol (C₂H₅OH).