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(A) The reaction sequence is as follows:$1$. $A$ is Nitrobenzene $(C_6H_5NO_2)$.$2$. Reduction of $A$ gives $B$,which is Aniline $(C_6H_5NH_2)$.$3$. A

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Nitrobenzene

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(A) The reaction sequence is as follows:$1$. $A$ is Nitrobenzene $(C_6H_5NO_2)$.$2$. Reduction of $A$ gives $B$,which is Aniline $(C_6H_5NH_2)$.$3$. Aniline reacts with trichloromethane $(CHCl_3)$ and caustic potash $(KOH)$ to form Phenyl isocyanide $(C_6H_5NC)$ as compound $C$ (Carbylamine reaction).$4$. Catalytic reduction of $C$ $(C_6H_5NC)$ gives $N$-methylbenzenamine $(C_6H_5NHCH_3)$.Thus,compound $A$ is Nitrobenzene.

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	$IV$. $LiAlH_4$

An organic compound $A$ on reduction gives compound $B$,which on reaction with trichloromethane and caustic potash forms $C$. The compound $C$ on catalytic reduction gives $N$-methylbenzenamine. The compound $A$ is,

Similar Questions

The incorrect statement regarding the reaction shown below is:

What is the major product of the following reaction? $C_2H_5-CONH_2$ $\xrightarrow{Br_2 / KOH} A$ $\xrightarrow{CHCl_3 / KOH} \text{Major Product}$

The major product of the following reaction is:

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 $IV$. $LiAlH_4$

In the given reaction,the end product $(C)$ is:

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