Question

An organic compound 'A' with the 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? Give reason.

Answer 1

1. A = C₄H₈O₂ [CH₃COOC₂H₅] ester Reactions involved are
   \[\ce{\underset{A}{CH3-COOC2H5} + H2O ->[dil H2SO4] \underset{B}{CH3COOH} + \underset{C}{CH3CH2OH}}\]
   Then
   \[\ce{\underset{C}{CH3CH2OH} ->[KMnO4][{[O]}] \underset{B}{CH3COOH}}\]
   So A = CH₃COOC₂H₅, B = CH₃COOH, C = CH₃CH₂OH
2. B has a higher boiling point than C. Because of their propensity to generate intermolecular H-bonds, carboxylic acids have higher boiling temperatures than alcohols.