$HCOOH+C_2H_5OH\xrightarrow{t^o,H_2SO_4đ}HCOOC_2H_5+H_2O$ $HCOOC_2H_5+NaOH\xrightarrow{t^o}HCOONa+C_2H_5OH$ $C_2H_5OH+CuO\xrightarrow{t^o}CH_3CHO+Cu+H_2O$ $CH_3CHO+Br_2+H_2O\xrightarrow{}CH_3COOH+2HBr$ Sửa $HCOOCH_3$ thành $CH_3COONa$ vì ko thể từ $CH_3COOH$ ra $HCOOCH_3$ $CH_3COOH+NaOH\xrightarrow{}CH_3COONa+H_2O$ $CH_3COONa+NaOH\xrightarrow{CaO, t^o}CH_4+Na_2CO_3$
