Aniline reacts with mixed acid (conc. HNO_3 a | vedclass

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(C) $1$. The reaction of aniline with mixed acid at $288 \ K$ produces $o$-nitroaniline $(P)$,$p$-nitroaniline $(Q)$,and $m$-nitroaniline $(R)$. Thus,

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$1,3,5-$tribromobenzene

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(C) $1$. The reaction of aniline with mixed acid at $288 \ K$ produces $o$-nitroaniline $(P)$,$p$-nitroaniline $(Q)$,and $m$-nitroaniline $(R)$. Thus,$R$ is $m$-nitroaniline.$2$. The sequence for $R$ ($m$-nitroaniline) is:a) Acetylation: $m$-nitroaniline reacts with $Ac_2O$/pyridine to form $N$-($3$-nitrophenyl)acetamide.b) Bromination: Bromination occurs at the $p$-position relative to the $-NHCOCH_3$ group (which is the $4$-position relative to the acetamido group,i.e.,the $6$-position relative to the nitro group).c) Hydrolysis: $H_3O^+$ removes the acetyl group to yield $4$-bromo-$3$-nitroaniline.d) Diazotization and Deamination: $NaNO_2/HCl$ followed by $EtOH/\Delta$ replaces the $-NH_2$ group with $H$,yielding $1$-bromo-$2$-nitrobenzene.e) Reduction: $Sn/HCl$ reduces the $-NO_2$ group to $-NH_2$,yielding $2$-bromoaniline.f) Bromination: Excess $Br_2/H_2O$ leads to tribromination at the $o, o, p$ positions relative to the $-NH_2$ group,resulting in $2,4,6$-tribromoaniline.g) Diazotization and Deamination: $NaNO_2/HCl$ followed by $H_3PO_2$ removes the $-NH_2$ group,yielding $1,3,5$-tribromobenzene.

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