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(D) The Hoffmann rearrangement involves the following steps:$1$. Formation of $N$-bromamide by the reaction of amide with $Br_2$ and $NaOH$.$2$. Forma

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All of the above are involved in the reaction.

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(D) The Hoffmann rearrangement involves the following steps:$1$. Formation of $N$-bromamide by the reaction of amide with $Br_2$ and $NaOH$.$2$. Formation of an acyl nitrene or direct rearrangement of the $N$-bromoamide anion to an isocyanate.$3$. The isocyanate is then hydrolyzed to the amine.Looking at the provided mechanism in the solution image,the species involved are:$(a)$ $N$-bromocyclohexanecarboxamide$(b)$ The conjugate base of the $N$-bromoamide$(c)$ Cyclohexyl isocyanateSince all these species are intermediates or products in the mechanism of the Hoffmann rearrangement,option $D$ is the correct answer.

List-$I$	List-$II$
$A$. Amide	$I$. Carbylamine reaction
$B$. Nitrile	$II$. Hinsberg's reagent
$C$. $C_6H_5SO_2Cl$	$III$. Hofmann's bromamide reaction
$D$. $1^{\circ}$-Amine	$IV$. $LiAlH_4$

Column $I$	Column $II$
$A$. $CH_3CH_2CH_2CN$	$p$. Reduction with $Pd-C/H_2$
$B$. $CH_3CH_2OCOCH_3$	$q$. Reduction with $SnCl_2/HCl$
$C$. $CH_3-CH=CH-CH_2OH$	$r$. Development of foul smell on treatment with chloroform and alcoholic $KOH$
$D$. $CH_3CH_2CH_2CH_2NH_2$	$s$. Reduction with diisobutylaluminium hydride $(DIBAL-H)$
	$t$. Alkaline hydrolysis

Which of the following species would not be involved in the Hoffmann rearrangement shown below?

Similar Questions

Match the following:
List-$I$ List-$II$
$A$. Amide $I$. Carbylamine reaction
$B$. Nitrile $II$. Hinsberg's reagent
$C$. $C_6H_5SO_2Cl$ $III$. Hofmann's bromamide reaction
$D$. $1^{\circ}$-Amine $IV$. $LiAlH_4$

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