MCQ
$Ph - CH _{2}- CH = CH - CH _{3}$ $\xrightarrow[{{\text{(ii) }}Alc.\,KOH}]{{{\text{(i) B}}{{\text{r}}_2}}}$
  • A
    $Ph - CH = CH - CH = CH _{2}$
  • B
    $\begin{array}{*{20}{c}}
      {Ph - C{H_2} - CH - CH - C{H_3}} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,} \\ 
      {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,\,OH\,\,\,} 
    \end{array}$
  • $Ph - CH _{2}- C \equiv C - CH _{3}$
  • D
    $Ph - C \equiv C - CH _{2}- CH _{3}$

Answer

Correct option: C.
$Ph - CH _{2}- C \equiv C - CH _{3}$
c
The final product formed is shown below:

$Ph - CH _{2}- CH = CH - CH _{3}$ $\xrightarrow{{(i)\,B{r_2}}}$$\begin{array}{*{20}{c}}
  {\,\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} 
\end{array}} \\ 
  {Ph - C{H_2} - CH - CH - C{H_3}} \\ 
  {|\,\,\,\,\,\,} \\ 
  {Br\,\,\,} 
\end{array}$ $\xrightarrow[{ - 2HBr}]{{{\text{Alc}}{\text{. KOH}}}}$ $Ph - CH _{2}- C \equiv C - CH _{3}$

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