Q. The product formed in the reaction n-hexanamide $+ Br_{2} + KOH,\$is
A) Hexanamine
B) Propanamine
C) Butanamine
D) pentanamine
Explanation
(d)
When amide is heated with a mixture of $Br_{2}$ in the presence of NaOH or KOH amine is formed which has one carbon atom less than original amide. This is called Hofmann’s degradation reaction.
Hexanamide$+ Br_{2} + 4KOH \rightarrow Pentanamine + K_{2}CO_{3} + 2KBr + 2H_{2}O$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.