MCQ
$C{H_3} - C{H_2} - C{H_2} - C \equiv CH\xrightarrow[{Excess}]{{HCl}}$ ?
  • A
    $CH_3-CH_2-CH_2-CH_2-CHCl_2$
  • $CH_3-CH_2-CH_2-CCl_2-CH_3$
  • C
    $\begin{array}{*{20}{c}}
    {C{H_3} - CH - C{H_2} - C{H_2} - C{H_2} - Cl}\\
    {|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
    {Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
    \end{array}$
  • D
    $\begin{array}{*{20}{c}}
    {Cl}\\
    {|\,\,}\\
    {C{H_3} - C{H_2} - C - C{H_2} - C{H_3}}\\
    {|\,\,}\\
    {\,\,Cl\,\,\,}
    \end{array}$

Answer

Correct option: B.
$CH_3-CH_2-CH_2-CCl_2-CH_3$
b

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