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Đáp án:

Giải thích các bước giải:

Bài 2 :

$CH_3COOCH_3 + NaOH → CH_3COONa + CH_3OH$

$CH_3COOCH_3 + H_2O \buildrel{{H^+}}\over\rightleftharpoons CH_3COOH + CH_3OH$

$HCOOC_2H_5 + NaOH → HCOONa + C_2H_5OH$

$HCOOC_2H_5 + H_2O \buildrel{{H^+}}\over\rightleftharpoons HCOOH + C_2H_5OH$

$C_2H_5COOCH_3 + NaOH → C_2H_5COOCH_3 + CH_3OH$

$C_2H_5COOCH_3 + H_2O \buildrel{{H^+}}\over\rightleftharpoons C_2H_5COOH + CH_3OH$

$(CH_3COO)_2C_2H_4 + 2NaOH → 2CH_3COONa + C_2H_4(OH)_2$

$(CH_3COO)_2C_2H_4+2H_2O\buildrel{{H^+}}\over\rightleftharpoons2CH_3COOH+ C_2H_4(OH)_2$

$(COOC_2H_5)_2 + 2NaOH → (COONa)_2 + 2C_2H_5OH$

$(COOC_2H_5)_2+2H_2O\buildrel{{H^+}}\over\rightleftharpoons(COOH)_2 + 2C_2H_5OH$

Bài 3 :$C_5H_{10}O_2$ không phải là este no,đơn chức

Bài 3' :

$a/$

$CH_3COOCH=CH_2 + Br_2 → CH_3COOCHBr-CH_2Br$

$b/$

$CH_3COOCH=CH_2 + NaOH → CH_3COONa + CH_3CHO$

$c/$

$CH_3COOCH=CH_2 + H_2 \xrightarrow{t^o,xt} CH_3COOCH_2CH_3

$d/$

$C_6H_5COOH + NaOH → C_6H_5COONa + H_2O$
$e/$

$CH_3COOCH_2-CH_2-OOCC_2H_5 + 2NaOH → CH_3COONa + C_2H_5COONa + C_2H_4(OH)_2$

$g/$

$(C_{17}H_{35}COO)_3C_3H_5 + 3NaOH → 3C_{17}H_{35}COONa + C_3H_5(OH)_3$