Scheme 1 and 2 describe the conversion of P t | vedclass

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(C) Step $1$: Conversion of $P$ (aniline) to $Q$ ($2$,$4$,$6$-tribromobenzoyl chloride).Aniline reacts with excess $Br_2/H_2O$ to form $2,4,6$-tribrom

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$4$

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(C) Step $1$: Conversion of $P$ (aniline) to $Q$ ($2$,$4$,$6$-tribromobenzoyl chloride).Aniline reacts with excess $Br_2/H_2O$ to form $2,4,6$-tribromoaniline. Diazotization followed by reaction with $CuCN/KCN$ gives $2,4,6$-tribromobenzonitrile. Hydrolysis gives $2,4,6$-tribromobenzoic acid,which reacts with $SOCl_2$ to form $Q$ ($2,4,6$-tribromobenzoyl chloride).Step $2$: Conversion of $R$ (benzene) to $S$ ($4$-bromophenol).Benzene reacts with oleum to form benzenesulfonic acid,which on fusion with $NaOH$ followed by acidification gives phenol. Bromination of phenol with $Br_2/CS_2$ at $273 \ K$ gives $4$-bromophenol $(S)$ as the major product.Step $3$: Synthesis of $T$ from $Q$ and $S$.$S$ ($4$-bromophenol) reacts with $NaOH$ to form sodium $4$-bromophenoxide,which acts as a nucleophile and attacks the carbonyl carbon of $Q$ ($2,4,6$-tribromobenzoyl chloride) to form an ester $T$ ($4$-bromophenyl $2,4,6$-tribromobenzoate).Structure of $T$ contains $3$ $Br$ atoms from the benzoyl part and $1$ $Br$ atom from the phenoxy part.Total number of $Br$ atoms in $T = 3 + 1 = 4$.

Scheme $1$ and $2$ describe the conversion of $P$ to $Q$ and $R$ to $S$,respectively. Scheme $3$ describes the synthesis of $T$ from $Q$ and $S$. The total number of $Br$ atoms in a molecule of $T$ is. . . . . . . .

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