Từ đất đèn điều chế P.E , etyl axetat , phenol , dimetyl ete , cao su Buna 07/10/2021 Bởi Amara Từ đất đèn điều chế P.E , etyl axetat , phenol , dimetyl ete , cao su Buna
\(\begin{array}{l} Ca{C_2} + 2{H_2}O \to {C_2}{H_2} + Ca{(OH)_2}\\ {C_2}{H_2} + {H_2} \xrightarrow{xt,t^0} {C_2}{H_4}\\ nC{H_2} = C{H_2} \xrightarrow{xt,t^0} {( – C{H_2} – C{H_2} – )_n}\\ {C_2}{H_4} + {H_2}O \xrightarrow{xt,t^0} {C_2}{H_5}OH\\ C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5}\\ 3{C_2}{H_2} \xrightarrow{xt,t^0} {C_6}{H_6}\\ {C_6}{H_6} + B{r_2} \to {C_6}{H_5}Br + HBr\\ {C_6}{H_5}Br + NaOH \to {C_6}{H_5}ONa + HBr\\ {C_6}{H_5}ONa + C{O_2} + {H_2}O \to {C_6}{H_5}OH + NaHC{O_3}\\ {C_2}{H_2} \xrightarrow{xt,t^0} C{H_2} = CH – C \equiv CH\\ C{H_2} = CH – C \equiv CH + {H_2} \xrightarrow{xt,t^0} C{H_2} = CH – CH = C{H_2}\\ nC{H_2} = CH – CH = C{H_2} \xrightarrow{xt,t^0} {( – C{H_2} – CH = CH – C{H_2} – )_n} \end{array}\) Bình luận
Điều chế PE, etyl axetat ($CH_3COOC_2H_5$), phenol ($C_6H_5OH$), cao su Buna như hình.
\(\begin{array}{l} Ca{C_2} + 2{H_2}O \to {C_2}{H_2} + Ca{(OH)_2}\\ {C_2}{H_2} + {H_2} \xrightarrow{xt,t^0} {C_2}{H_4}\\ nC{H_2} = C{H_2} \xrightarrow{xt,t^0} {( – C{H_2} – C{H_2} – )_n}\\ {C_2}{H_4} + {H_2}O \xrightarrow{xt,t^0} {C_2}{H_5}OH\\ C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5}\\ 3{C_2}{H_2} \xrightarrow{xt,t^0} {C_6}{H_6}\\ {C_6}{H_6} + B{r_2} \to {C_6}{H_5}Br + HBr\\ {C_6}{H_5}Br + NaOH \to {C_6}{H_5}ONa + HBr\\ {C_6}{H_5}ONa + C{O_2} + {H_2}O \to {C_6}{H_5}OH + NaHC{O_3}\\ {C_2}{H_2} \xrightarrow{xt,t^0} C{H_2} = CH – C \equiv CH\\ C{H_2} = CH – C \equiv CH + {H_2} \xrightarrow{xt,t^0} C{H_2} = CH – CH = C{H_2}\\ nC{H_2} = CH – CH = C{H_2} \xrightarrow{xt,t^0} {( – C{H_2} – CH = CH – C{H_2} – )_n} \end{array}\)