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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-$I$ (Compounds)	List-$II$ (Reactions)
$A. \ CH_3(CH_2)_3NH_2$	$(i)$ Alkaline hydrolysis
$B. \ CH_3C\equiv CH$	$(ii)$ With $KOH$ (alc.) and $CHCl_3$ produces bad smell
$C. \ CH_3CH_2COOCH_3$	$(iii)$ Gives white ppt. with ammoniacal $AgNO_3$
$D. \ CH_3CH(OH)CH_3$	$(iv)$ With Lucas reagent cloudiness appears after $5 \ minutes$

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	$IV$. $LiAlH_4$

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

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

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 $IV$. $LiAlH_4$

Product $(E)$ is

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