Hoàn thành chuỗi phản ứng sau:
a) CH4 –(1)—> C2H2 —(2)–> CH2=CHCL ——(3)–> PVC
b) C2H2 —-(1)–> CH2=CH2 —–(2)–> PE
Hoàn thành chuỗi phản ứng sau: a) CH4 –(1)—> C2H2 —(2)–> CH2=CHCL ——(3)–> PVC b) C2H2 —-(1)–> CH2=CH2 —–(2)–> PE
By Alexandra
By Alexandra
Hoàn thành chuỗi phản ứng sau:
a) CH4 –(1)—> C2H2 —(2)–> CH2=CHCL ——(3)–> PVC
b) C2H2 —-(1)–> CH2=CH2 —–(2)–> PE
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) $nCH_2=CH_2\xrightarrow{{t^o, p, xt}} (\kern-6pt-CH_2-CH_2-\kern-6pt)_n$
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)
\(CH \equiv CH + {H_2}\xrightarrow{{Pd/PbC{O_3},{t^o}}}C{H_2} = C{H_2}\)
\(nC{H_2} = C{H_2}\xrightarrow{{{t^o},xt}}{( – C{H_2} – C{H_2} – )_n}\)