C_{6}H_{5}NO_{2} xrightarrow{Sn+HCl} A xright | vedclass

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(A) $1$. The reduction of nitrobenzene $(C_{6}H_{5}NO_{2})$ with $Sn+HCl$ yields aniline $(C_{6}H_{5}NH_{2})$,which is compound $A$.$2$. Aniline then

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(A) $1$. The reduction of nitrobenzene $(C_{6}H_{5}NO_{2})$ with $Sn+HCl$ yields aniline $(C_{6}H_{5}NH_{2})$,which is compound $A$.$2$. Aniline then undergoes an electrophilic aromatic substitution reaction (coupling reaction) with benzenediazonium chloride $(C_{6}H_{5}N_{2}^{\oplus}Cl^{\ominus})$ in a weakly acidic medium.$3$. The diazonium ion acts as an electrophile and attacks the para-position of the aniline ring to form $p$-aminoazobenzene,which is a yellow-colored azo dye $(P)$.

$C_{6}H_{5}NO_{2}$ $\xrightarrow{Sn+HCl} A$ $\xrightarrow{C_{6}H_{5}N_{2}^{\oplus}Cl^{\ominus}} P$
(Yellow coloured compound)
Consider the above reaction,the Product "$P$" is:

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$C_{6}H_{5}NO_{2}$ $\xrightarrow{Sn+HCl} A$ $\xrightarrow{C_{6}H_{5}N_{2}^{\oplus}Cl^{\ominus}} P$(Yellow coloured compound)Consider the above reaction,the Product "$P$" is: