$CH_3-CH_2-CH_2-CH_3 \xrightarrow[HBr]{AlCl_3}$ Product in the above reaction is:

  • A
    $CH_3-CH(Br)-CH_2-CH_3$
  • B
    $CH_3-CH(CH_3)-CH_3$
  • C
    $Br-CH_2-CH_2-CH_2-CH_3$
  • D
    All of these

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