Viết các phương trình hoá học thực hiện dãy chuyển hoá sau (ghi rõ điều kiện phản ứng)
\({C_2}{H_4}\buildrel {(1)} \over
\longrightarrow {C_2}{H_5}OH\buildrel {(2)} \over
\longrightarrow {C_2}{H_5}Na\buildrel {(3)} \over
\longrightarrow {C_2}{H_5}OH\buildrel {(4)} \over
\longrightarrow C{H_3}COOH \\ \buildrel {(5)} \over
\longrightarrow {(C{H_3}COO)_2}Ba\buildrel {(6)} \over
\longrightarrow C{H_3}COONa\buildrel {(7)} \over
\longrightarrow C{H_3}COOH\buildrel {(8)} \over
\longrightarrow C{H_3}COO – C{H_2}C{H_3}.\)
Các phương trình hóa học:
\(\begin{array}{l}(1)\,{C_2}{H_4} + {H_2}O \to {C_2}{H_5}OH(axit)\\(2)\,2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}ONa + {H_2} \uparrow \\(3)\,{C_2}{H_5}ONa + {H_2}O \to {C_2}{H_5}OH + NaOH\\(4)\,{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O(men\,giấm)\\(5)\,Ba{(OH)_2} + 2C{H_3}COOH \to {(C{H_3}COO)_2}Ba + 2{H_2}O\\(6)\,{(C{H_3}COO)_2}Ba + N{a_2}S{O_4} \to BaS{O_4} \downarrow + 2C{H_3}COONa\\(7)\,C{H_3}COONa + HCl \to C{H_3}COOH + NaCl\\(8)\,C{H_3}COOH + HO – C{H_2} – C{H_3} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COO – C{H_2} – C{H_3} + {H_2}O({H_2}S{O_4}đặc,{t^0})\end{array}\)