The major product formed in the following reaction is
$C_6H_5CH=CH_2$ $\xrightarrow[(ii) H_3O^+]{(i) KMnO_4, KOH, \Delta}$ $\xrightarrow{(iii) Br_2/FeBr_3} \text{Product}$

  • A
    $p$-bromophenylacetic acid
  • B
    $o$-bromobenzoic acid
  • C
    $m$-bromoacetophenone
  • D
    $m$-bromobenzoic acid

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