Compound A is:

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Solution

The correct option is B
$ \underset{(1 \text { eqv.) }}{\stackrel{ Br _{2}}{ FeBr _{3}}} $ (A)
$C C l_{3}$ being electron withdrawing group is a meta directing group for electrophilic substitution reaction.

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Compound A is:

The correct option is B \( \underset{(1 \text { eqv.) }}{\stackrel{ Br _{2}}{ FeBr _{3}}} \) (A) CCl3 being electron withdrawing group is a meta directing group for electrophilic substitution reaction.

The correct option is B
\( \underset{(1 \text { eqv.) }}{\stackrel{ Br _{2}}{ FeBr _{3}}} \) (A)
$C C l_{3}$ being electron withdrawing group is a meta directing group for electrophilic substitution reaction.