The major products Q and R from the following | vedclass

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(B) $1$. The reaction of benzoic acid with $Br_2 / FeBr_3$ is an electrophilic aromatic substitution. The $-COOH$ group is a meta-directing group,so t

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(B) $1$. The reaction of benzoic acid with $Br_2 / FeBr_3$ is an electrophilic aromatic substitution. The $-COOH$ group is a meta-directing group,so the product $P$ is $m$-bromobenzoic acid.$2$. In the next step,$m$-bromobenzoic acid reacts with $B_2H_6$. Diborane $(B_2H_6)$ is a selective reducing agent that reduces the carboxylic acid group $(-COOH)$ to a primary alcohol $(-CH_2OH)$ without affecting the bromine atom on the ring. Thus,$Q$ is $m$-bromobenzyl alcohol.$3$. In the third step,$m$-bromobenzoic acid reacts with $MeOH$ in the presence of $HCl$ (gas). This is a Fischer esterification reaction,where the carboxylic acid reacts with an alcohol to form an ester. Thus,$R$ is methyl $m$-bromobenzoate.

The major products $Q$ and $R$ from the following reactions,respectively are:

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