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STD 11 - 9. Hydricarbon — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 9. Hydricarbon4 Marks
MCQ
$\begin{array}{*{20}{c}}
  {A\xleftarrow{{HBr}}Ph - CH = C - C{H_3}\xrightarrow[{Peroxide}]{{HBr}}B,} \\ 
  {\,\,|} \\ 
  {\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$

$A$ and $B$ is

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

Answer

Correct option: B.
$\begin{array}{*{20}{c}}
  {Ph - CH - CH - C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\ 
  {\,\,\,Br\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,} 
\end{array}$  and  $\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
  {\,\,\,Br} \\ 
  | 
\end{array}\,\,\,\,} \\ 
  {Ph - C{H_2} - C - C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$
b

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