$HCOOH+C_2H_5OH\xrightarrow{t^o,H_2SO_4đ}HCOOC_2H_5+H_2O$ $HCOOC_2H_5+NaOH\xrightarrow{t^o}HCOONa+C_2H_5OH$ $C_2H_5OH+CuO\xrightarrow{t^o}CH_3CHO+Cu+H_2O$ $CH_3CHO+Br_2+H_2O\xrightarrow{}CH_3COOH+2HBr$ Sửa $HCOOCH_3$ thành $CH_3COONa$ vì ko thể từ $CH_3COOH$ ra $HCOOCH_3$ $CH_3COOH+NaOH\xrightarrow{}CH_3COONa+H_2O$ $CH_3COONa+NaOH\xrightarrow{CaO, t^o}CH_4+Na_2CO_3$
$CH_4+Cl_2\xrightarrow{áskt,1:1}CH_3Cl+HCl$
$CH_3Cl+NaOH\xrightarrow{}CH_3OH+NaCl$
$CH_3OH+CuO\xrightarrow{t^o}HCHO+Cu+H_2O$
$HCHO+\dfrac{1}{2}O_2\xrightarrow{t^o,xt}HCOOH$
$HCOOH+C_2H_5OH\xrightarrow{t^o,H_2SO_4đ}HCOOC_2H_5+H_2O$
$HCOOC_2H_5+NaOH\xrightarrow{t^o}HCOONa+C_2H_5OH$
$C_2H_5OH+CuO\xrightarrow{t^o}CH_3CHO+Cu+H_2O$
$CH_3CHO+Br_2+H_2O\xrightarrow{}CH_3COOH+2HBr$
Sửa $HCOOCH_3$ thành $CH_3COONa$ vì ko thể từ $CH_3COOH$ ra $HCOOCH_3$
$CH_3COOH+NaOH\xrightarrow{}CH_3COONa+H_2O$
$CH_3COONa+NaOH\xrightarrow{CaO, t^o}CH_4+Na_2CO_3$
$CH_4+Cl_2\xrightarrow{{as}} CH_3Cl+HCl$
$CH_3Cl+NaOH\xrightarrow{{t^o}} CH_3OH+NaCl$
$CH_3OH+CuO\xrightarrow{{t^o}} HCHO+Cu+H_2O$
$HCHO+\dfrac{1}{2}O_2\xrightarrow{{Mn^{2+}, t^o}} HCOOH$
$HCOOH+C_2H_5OH\buildrel{{H_2SO_4, t^o}}\over\rightleftharpoons HCOOC_2H_5+H_2O$
$HCOOC_2H_5+KOH\xrightarrow{{t^o}} HCOOK+C_2H_5OH$
$C_2H_5OH+CuO\xrightarrow{{t^o}} CH_3CHO+Cu+H_2O$
$CH_3CHO+Br_2+H_2O\to CH_3COOH+2HBr$
Hai PT cuối không xảy ra.