Từ C2h4, cac2, c2h2 điều chế ch3cl kiểu j ạ

Đáp án:

$C_2H_4+\dfrac{1}{2}O_2\xrightarrow{{CuCl_2, PdCl_2, t^o}} CH_3CHO$

$CH_3CHO+\dfrac{1}{2}O_2\xrightarrow{{Mn^{2+}, t^o}} CH_3COOH$

$CH_3COOH+NaOH\to CH_3COONa+H_2O$

$CH_3COONa+NaOH\xrightarrow{{CaO, t^o}} CH_4+Na_2CO_3$

$CH_4+Cl_2\xrightarrow{{askt, 1:1}} CH_3Cl+HCl$