MCQ
$C{H_3} - CH = C{H_2}\xrightarrow[{(low\,\,conc.)}]{{B{r_2}/hv}}(A)$ ; Product $(A)$ of the reaction is
  • A
    $\begin{array}{*{20}{c}}
    {C{H_3} - CH - C{H_2} - Br}\\
    {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
    {Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
    \end{array}$
  • $H_2C = CH - CH_2 -Br$
  • C
    $\begin{array}{*{20}{c}}
    {C{H_3} - C = C{H_2}}\\
    |\\
    {\,\,\,Br}
    \end{array}$
  • D
    $Br - CH_2 - CH_2 - CH_2 -Br$

Answer

Correct option: B.
$H_2C = CH - CH_2 -Br$
b
$(b)\, Br_2/hv$ (low cone.) (or) $N.B.S,$ Allylic free radical substitution reaction.

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