When R-CO-NH_2 is treated with Br_2 and KOH,i | vedclass

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

Question

(A) The reaction described is the $Hofmann$ $Bromamide$ $Degradation$ reaction.In this reaction,the amide $(R-CO-NH_2)$ reacts with $Br_2$ in the pres

Vedclass · Accepted Answer

$R-CO-NHBr$

Vedclass · Answer

(A) The reaction described is the $Hofmann$ $Bromamide$ $Degradation$ reaction.In this reaction,the amide $(R-CO-NH_2)$ reacts with $Br_2$ in the presence of a strong base $(KOH)$ to form an isocyanate intermediate $(R-N=C=O)$.However,the initial step involves the formation of $N-bromamide$ $(R-CO-NHBr)$ by the replacement of a hydrogen atom on the nitrogen with a bromine atom.Thus,$R-CO-NHBr$ is the key intermediate formed during the process.

When $R-CO-NH_2$ is treated with $Br_2$ and $KOH$,it gives $R-NH_2$ as the major product. Which intermediate is involved in this reaction?

Similar Questions

In the given reaction sequence,what is $Z$?
Aniline $\xrightarrow{CH_3COCl} X$ $\xrightarrow{Br_2/AlBr_3} Y$ $\xrightarrow{H_3O^+} Z$

Give the major product of the following reaction:

$(CH_3)_2CHCONH_2$ $\xrightarrow{Br_2/NaOH} A$ $\xrightarrow{COCl_2} B$. In this reaction sequence,$B$ is:

When aniline is treated with sodium nitrite and hydrochloric acid at $0\,^{\circ}C$,it gives:

The strongest conjugate acid will result from:

Vedclass Test Series

Exam Paper Generator

Online Exam Module