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(C) Step $1$: $Nitrobenzene$ reacts with $Br_2/FeBr_3$ to give $m-bromonitrobenzene$.Step $2$: Reduction with $Sn/HCl$ converts the $-NO_2$ group to $

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$5$

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(C) Step $1$: $Nitrobenzene$ reacts with $Br_2/FeBr_3$ to give $m-bromonitrobenzene$.Step $2$: Reduction with $Sn/HCl$ converts the $-NO_2$ group to $-NH_2$, yielding $m-bromoaniline$.Step $3$: $pH$ neutralisation is performed.Step $4$: $m-bromoaniline$ reacts with $Br_2/H_2O$ to give $2,4,6-tribromo-3-bromoaniline$ (which is $2,3,4,6-tetrabromoaniline$).Step $5$: Diazotization with $NaNO_2/HBr$ at $0-5^{\circ}C$ converts $-NH_2$ to $-N_2^+Br^-$.Step $6$: Sandmeyer reaction with $CuBr/NaBr$ replaces the diazonium group with a $Br$ atom.The final product is $1,2,3,4,5-pentabromobenzene$ (or $1,2,3,4,6-pentabromobenzene$ depending on numbering, but it contains $5$ bromine atoms).Thus, the number of $Br$ atoms in the final product $(P)$ is $5$.

List-$I$ (Amine)	List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline	$I$. $9.30$
$B$. Aniline	$II$. $8.92$
$C$. $N$-Ethylethanamine	$III$. $9.38$
$D$. $N$-Methylaniline	$IV$. $3.00$

Consider the above sequence of reactions. The number of bromine atom$(s)$ in the final product $(P)$ will be:

Similar Questions

Match the following:
List-$I$ (Amine) List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline $I$. $9.30$
$B$. Aniline $II$. $8.92$
$C$. $N$-Ethylethanamine $III$. $9.38$
$D$. $N$-Methylaniline $IV$. $3.00$

The correct answer is

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Product $A$ is

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