RCONH_2 is converted into RNH_2 by means of H | vedclass

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(B) $1.$ The conversion of amide $(RCONH_2)$ to $N$-bromoamide $(RCONHBr)$ is achieved using $Br_2$ in the presence of a base like $KOH$. Thus,$(D)$ i

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$D, D, B$

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(B) $1.$ The conversion of amide $(RCONH_2)$ to $N$-bromoamide $(RCONHBr)$ is achieved using $Br_2$ in the presence of a base like $KOH$. Thus,$(D)$ is the correct reagent.$2.$ The rate-determining step in the Hofmann bromamide degradation is the formation of the nitrene intermediate or the rearrangement step,which corresponds to the formation of the isocyanate $(R-N=C=O)$ from the $N$-bromoamide anion $(iii)$. In the given mechanism,this is the formation of $(iv)$.$3.$ The Hofmann bromamide degradation is an intramolecular reaction. When a mixture of $m$-deuterobenzamide $(i)$ and $^{15}N$-benzamide $(ii)$ is treated,the $D$ atom remains on the ring and the $^{15}N$ atom remains in the amine group. Thus,$m$-deuteroaniline and $^{15}N$-aniline are formed. This corresponds to option $(B)$.

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