The reaction of RCONH_2 with bromine and KOH | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given reaction is the Hoffmann-Bromamide degradation reaction. In this reaction,the amide $(RCONH_2)$ reacts with $Br_2$ and $KOH$ to form an

Vedclass · Accepted Answer

$RN=C=O$

Vedclass · Answer

(C) The given reaction is the Hoffmann-Bromamide degradation reaction. In this reaction,the amide $(RCONH_2)$ reacts with $Br_2$ and $KOH$ to form an $N$-bromamide intermediate $(RCONHBr)$,which then undergoes rearrangement to form an isocyanate intermediate $(RN=C=O)$. Finally,the isocyanate is hydrolyzed to form the primary amine $(RNH_2)$. Both $RCONHBr$ and $RN=C=O$ are intermediates,but $RN=C=O$ (alkyl isocyanate) is the key intermediate formed during the rearrangement step.

The reaction of $RCONH_2$ with bromine and $KOH$ gives $RNH_2$ as the end product. Which one of the following is the intermediate product formed in this reaction?

Similar Questions

Which among the following is the strongest Bronsted base?

In the above reactions,product $A$ and product $B$ respectively are:

$o$-Phenylenediamine $\stackrel{HNO_2}{\longrightarrow}$ '$X$'

The reaction of benzylamine with nitrous acid $(HNO_2)$ produces product $A$. Identify product $A$.

Which of the following is the correct decreasing order of basic strength for the given amines and ammonia?

Vedclass Test Series

Exam Paper Generator

Online Exam Module