$CH_3-CH_2-CH=CH_2 \xrightarrow[CCl_4]{Br_2} \text{Product}$

  • A
    $CH_3-CH_2-CH(Br)-CH_2Br$
  • B
    $CH_3-CH_2-CH_2-CH_2Br$
  • C
    $CH_3-CH(Br)-CH_2-CH_3$
  • D
    $CH_3-CH_2-C(Br)_2-CH_3$

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