MCQ
$\begin{array}{*{20}{c}}
  {{H_3}C - CH - CH = C{H_2}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,} 
\end{array}$  $+HBr \rightarrow A$

$A$ (predominantly) is

  • A
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,} \\ 
      {\,\,\,Br\,\,\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$
  • B
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - CH - C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,} \\ 
      {\,C{H_3}\,\,\,\,\,\,\,Br\,\,} 
    \end{array}$
  • C
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH - C{H_2} - C{H_2}Br} \\ 
      {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} 
    \end{array}$
  • $\begin{array}{*{20}{c}}
      {Br\,\,\,\,\,} \\ 
      {|\,\,\,\,\,\,\,\,} \\ 
      {C{H_3} - C - C{H_2}C{H_3}} \\ 
      {|\,\,\,\,\,\,\,\,\,} \\ 
      {C{H_3}\,\,\,} 
    \end{array}$

Answer

Correct option: D.
$\begin{array}{*{20}{c}}
  {Br\,\,\,\,\,} \\ 
  {|\,\,\,\,\,\,\,\,} \\ 
  {C{H_3} - C - C{H_2}C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,} \\ 
  {C{H_3}\,\,\,} 
\end{array}$
d

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