Theo chiều từ trái sang, từ trên xuống nhé
\(C_2H_2+H_2\underrightarrow{t^o,Pd/PbCO_3}C_2H_4\)
\(C_2H_4+H_2O\underrightarrow{H^+,t^o}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)
\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(2CH_3COOH+MgO\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)
\(C_2H_2+H_2\xrightarrow[t^o]{Pd}C_2H_4\\ C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\\ C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_4đặc}CH_3COOC_2H_5+H_2O\\ CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ 2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
