Question

An organic compound ‘A’ with molecular formula C₄H₈O₂ undergoes acid hydrolysis to form two compounds ‘B’ and ‘C’. Oxidation of ‘C’ with acidified potassium permanganate also produces ‘B’.

Sodium salt of ‘B’ on heating with soda lime gives methane.

(1) Identify ‘A’, ‘B’ and ‘C’.

(2) Out of ‘B’ and ‘C’, which will have higher boiling point?

Answer 1

Compound A should be

\(CH_3 - \underset{(A)}{\overset{O}{\overset{||}C}}- O + C_2H_5\)

A undergoes acid hydrolysis to form acetic acid and ethanol as follows

\(CH_3 - \overset{O} {\overset{||}C} - O \ C_2 H_5 \xrightarrow{H_3O^+} \underset{(B)}{CH_3COOH}+\underset{(C)}{ C_2H_5OH}\)

Ethanol on oxidation with acidified potassium permanganate gives acetic acid

\(C_2H_5OH \xrightarrow{KMnO_4/H^+} CH_3COOH\) 

Sodium salt of B i.e. CH₃COONa gives methane on heating with soda lime as follows

\(CH_3COONa \underset{\Delta} {\xrightarrow {Soda lime}}CH_4\) 

Boiling point of acetic acid (B) is \(118^\circ C\) whereas that of ethanol (C) is \(78^\circ C\). This is because acetic acid has intermolecular hydrogen bonding.