Đáp án:
Giải thích các bước giải:
a)
$\begin{gathered}
C{H_3}COONa + NaOH\xrightarrow[{{t^o}}]{{CaO}}C{H_4} + N{a_2}C{O_3} \hfill \\
2C{H_4}\xrightarrow[{lamlanhnhanh}]{{{{1500}^o}C}}{C_2}{H_2} + 3{H_2} \hfill \\
{C_2}{H_2} + HCl\xrightarrow[{{t^o}}]{{HgC{l_2}}}C{H_2} = CHCl \hfill \\
nC{H_2} = CHCl\xrightarrow{{{t^o},p,xt}}{\left( { - C{H_2} - CHCl - } \right)_n} \hfill \\
\end{gathered} $
b)
$\begin{gathered}
2C{H_4}\xrightarrow[{lamlanhnhanh}]{{{{1500}^o}C}}{C_2}{H_2} + 2{H_2} \hfill \\
2{C_2}{H_2}\xrightarrow[{{{100}^o}C}]{{CuCl,N{H_4}Cl}}C{H_2} = CH - C \equiv CH \hfill \\
C{H_2} = CH - C \equiv CH + {H_2}\xrightarrow[{{t^o}}]{{Pd/PbC{O_3}}}C{H_2} = CH - CH = C{H_2} \hfill \\
nC{H_2} = CH - CH = C{H_2}\xrightarrow{{{t^o}.p.xt}}{\left( { - C{H_2} - CH = CH - C{H_2} - } \right)_n} \hfill \\
\end{gathered} $
c)
$\begin{gathered}
3{C_2}{H_2}\xrightarrow[{{{600}^o}C}]{C}{C_6}{H_6} \hfill \\
{C_6}{H_6} + C{l_2}\xrightarrow[{{t^o}}]{{Fe}}{C_6}{H_5}Cl + HCl \hfill \\
{C_6}{H_5}Cl + 2NaOH\xrightarrow{{{t^o}cao,pcao}}{C_6}{H_5}ONa + NaCl + {H_2}O \hfill \\
{C_6}{H_5}ONa + HCl \to {C_6}{H_5}OH + NaCl \hfill \\
\end{gathered} $
