$^{14}CH_2=CH-CH_3 \xrightarrow[\text{or high temp}]{\text{low conc of } Br_2} (?)$
Product of the above reaction is

  • A
    $^{14}CH_2=CH-CH_2-Br$
  • B
    $CH_2=CH-^{14}CH_2-Br$
  • C
    $Br-^{14}CH_2-CH(Br)-CH_3$
  • D
    both $(A)$ and $(B)$

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