Giải bài 4 trang 138 SGK Hoá học 11

\(\begin{align} & a)\,C{{H}_{3}}-C{{H}_{3}}\xrightarrow{{{500}^{o}C}}C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}} \\ & C{{H}_{2}}=C{{H}_{2}}+C{{l}_{2}}\to \underset{\left| {} \right.}{\mathop{C{{H}_{2}}}}\,-\underset{\left| {} \right.}{\mathop{C{{H}_{2}}}}\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,Cl \\ & b)\,\underset{\left| {} \right.}{\mathop{C{{H}_{2}}}}\,-\underset{\left| {} \right.}{\mathop{C{{H}_{2}}}}+KOH\,\xrightarrow{ancol\,}CH_2=CHCl+KCl+H_2O \\ & \,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align} \)

\(CH_2=CHCl+HCl→CH_3−CHCl_2\)

\(CH_2Cl−CH_2Cl+KOH\to CH_2=CHCl+KCl+H_2O\)