Cho sơ đồ sau: \(etilen\xrightarrow{{ + {H_2}O/xt}}\,X\,\xrightarrow{{xt,{t^0}}}Y\xrightarrow{{xt\,Na\,,{t^0}}}\,po\lim e\,M.\) Vậy M là:

A. polietilen.

B. polibutađien.

C. poli ( vinyl clorua).

D. poliisopren.

Hướng dẫn

Chọn đáp án là: B

Lời giải chi tiết:

\(\begin{gathered}

etilen(C{H_2} = C{H_2})\xrightarrow{{ + {H_2}O/xt}}\,C{H_3}C{H_2}OH(X)\xrightarrow{{xt,{t^0}}}C{H_2} = CH – CH = C{H_2}(Y) \hfill \\

\xrightarrow{{xt\,Na\,,{t^0}}}\,( – C{H_2} – CH = CH – C{H_2} – )po\lim e\,M. \hfill \\

C{H_2} = C{H_2} + {H_2}O\xrightarrow{{{H^ + },{t^0}}}C{H_3}C{H_2}OH \hfill \\

2C{H_3}C{H_2}OH\xrightarrow{{A{l_2}{O_3},C{r_2}{O_3},{{450}^0}C}}C{H_2} = CH – CH = C{H_2} + 2{H_2}O + {H_2} \hfill \\

nC{H_2} = CH – CH = C{H_2}\xrightarrow[{Na}]{{xt,{t^0},p}}\underbrace {{{( – C{H_2} – CH = CH – C{H_2} – )}_n}}_{poli\,butadien} \hfill \\

\end{gathered} \)

Đáp án B