hoàn thành phương trình sau

PTHH:

$\rm e, HCOOCH=CH_2+H_2O \buildrel{{H^{+},t^o}}\over\rightleftharpoons HCOOH+CH_3CHO\\f, CH_3COOH+C_2H_5OH \buildrel{{H_2SO_4(đ),t^o}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O \\ g, CH_3COOCH=CH_2+Br_2\to CH_3COOCHBr-CH_2Br \\ h, CH_3COOCH=CH_2 \xrightarrow{xt,p,t^o}

(\kern-6pt- \, CH(OOCCH3)-CH_2 -\kern-6pt)_n\\i, CH_3COOC_6H_5+2NaOH\xrightarrow{t^o} CH_3COONa+C_6H_5ONa+H_2O \\ k, (C_{15}H_{31}COO)_3C_3H_5+3H_2O \buildrel{{H_2SO_4(đ),t^o}}\over\rightleftharpoons 3C_{15}H_{31}COOH+C_3H_5(OH)_3 \\ l, (C_{17}H_{35}COO)_3C_3H_5+3KOH\xrightarrow{t^o} 3C_{17}H_{35}COOK+C_3H_5(OH)_3$ 

Đáp án:

$\rm e.\ HCOOCH=CH_2+H_2O\overset{H^+,\ t^°}\leftrightarrows HCOOH+CH_3CHO\\f.\ CH_3COOH+C_2H_5OH\overset{H_2SO_4,\ t^°}\leftrightarrows CH_3COOC_2H_5+H_2O\\g.\ CH_3COOCH=CH_2+Br_2\to CH_3COOCHBr=CH_2Br\\h.\ nCH_3COOCH=CH_2\xrightarrow{t^°,\ xt,\ p}(\kern-6pt- CH(OCOCH_3)-CH_2 -\kern-6pt)_n\\i.\ CH_3COOC_6H_5+2NaOH\xrightarrow{t^°}CH_3COONa+C_6H_5ONa+H_2O\\k.\ (C_{15}H_{31}COO)_3C_3H_5+3H_2O\overset{H^+,\ t^°}\leftrightarrows 3C_{15}H_{31}COOH+C_3H_5(OH)_3\\l.\ (C_{15}H_{31}COO)_3C_3H_5+3NaOH\xrightarrow{t^°}3C_{15}H_{31}COONa+C_3H_5(OH)_3$