Câu 6, 7 như hình.
Câu 8:
(1) $CaC_2+2H_2O\to Ca(OH)_2+C_2H_2$
(2) $2C_2H_2\xrightarrow{{CuCl, NH_4Cl, t^o}} CH\equiv C-CH=CH_2$
(3) $CH\equiv C-CH=CH_2+H_2\xrightarrow{{Pd/PbCO_3, t^o}} CH_2=CH-CH=CH_2$
(4) $nCH_2=CH-CH=CH_2\xrightarrow{{t^o, p, xt}} -\kern-6pt(CH_2-CH=CH-CH_2 )\kern-6pt-_n$