Đáp án đúng:
Giải chi tiết:
Phương trình hóa học:
\(\begin{gathered} 2C{H_4}\xrightarrow[{lam\,lanh\,nhanh}]{{{t^o}}}{C_2}{H_2} + 3{H_2} \hfill \\ {C_2}{H_2} + {H_2}\xrightarrow[{}]{{Pd/PbC{O_3}}}{C_2}{H_4} \hfill \\ {C_2}{H_4} + {H_2}O\xrightarrow[{}]{{{H^ + }}}{C_2}{H_5}OH \hfill \\ {C_2}{H_5}OH + {O_2}\xrightarrow{{men\,giam}}C{H_3}COOH + {H_2}O \hfill \\ C{H_3}COOH + {C_2}{H_2}\xrightarrow{{HgS{O_4},{H_2}S{O_4},{{80}^o}C}}C{H_3}COOCH = C{H_2} \hfill \\ nC{H_3}COOCH = C{H_2}\xrightarrow{{{t^o},xt,p}}{( - C{H_2} - CH - )_n} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,| \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}COO \hfill \\ 2{C_2}{H_5}OH\xrightarrow{{A{l_2}{O_3},{{450}^o}C}}C{H_2} = CH - CH = C{H_2} + {H_2} + 2{H_2}O \hfill \\ nC{H_2} = CH - CH = C{H_2}\xrightarrow{{{t^o},xt,p}}{( - C{H_2} - CH = CH - C{H_2} - )_n} \hfill \\ \end{gathered} \)
