Giải bài 2 trang 145 SGK Hoá học 11

$\begin{align} & a)C{{H}_{3}}-C\equiv CH+{{H}_{2}}\xrightarrow[Pd/PbC{{O}_{3}}]{{{t}^{o}}}C{{H}_{3}}-CH=C{{H}_{2}} \\ & b)C{{H}_{3}}-C\equiv CH+2B{{r}_{2}}\to C{{H}_{3}}-CB{{r}_{2}}-CHB{{r}_{2}} \\ & c)\,C{{H}_{3}}-C\equiv CH+AgN{{O}_{3}}+N{{H}_{3}}\to C{{H}_{3}}-C\equiv CAg\downarrow +N{{H}_{4}}N{{O}_{3}} \\ & d)\,1.C{{H}_{3}}-C\equiv CH+HCl\xrightarrow{HgC{{l}_{2}}}C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{C}}\,=C{{H}_{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ & 2.\,C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{C}}\,=C{{H}_{2}}+HC{{l}_{(dư)}}\to C{{H}_{3}}-\underset{\left| {} \right.}{\overset{\left| {} \right.}{\mathop{C}}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ \end{align}$