Hoàn thành chuỗi phản ứng sau: a) CH4 –(1)—> C2H2 —(2)–> CH2=CHCL ——(3)–> PVC b) C2H2 —-(1)–> CH2=CH2 —–(2)–> PE

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1. a,

   (1) $2CH_4\xrightarrow{{1500^oC}} C_2H_2+3H_2$

   (2) $C_2H_2+HCl\xrightarrow{{HgCl_2, t^o}} CH_2=CHCl$

   (3) $nCH_2=CHCl\xrightarrow{{t^o, p, xt}} (\kern-6pt-CH_2-CHCl-\kern-6pt)_n$

   b,

   (1) $C_2H_2+H_2\xrightarrow{{Pd/PbCO_3, t^o}} C_2H_4$

   (2) $nCH_2=CH_2\xrightarrow{{t^o, p, xt}} (\kern-6pt-CH_2-CH_2-\kern-6pt)_n$

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2. Các phản ứng xảy ra:

   a)

   \(2C{H_4}\xrightarrow{{{{1500}^o}C;lln }}{C_2}{H_2} + 3{H_2}\)

   \({C_2}{H_2} + HCl\xrightarrow{{}}C{H_2} = CHCl\)

   \(nC{H_2} = CHCl\xrightarrow{{{t^o},xt}}{( – C{H_2} – CHCl – )_n}\)

   b)

   \(CH \equiv CH + {H_2}\xrightarrow{{Pd/PbC{O_3},{t^o}}}C{H_2} = C{H_2}\)

   \(nC{H_2} = C{H_2}\xrightarrow{{{t^o},xt}}{( – C{H_2} – C{H_2} – )_n}\)

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