Em tham khảo nhé!
Bài 8-a/
$\begin{array}{l}
HCOO{C_2}{H_5}({A_1}) + NaOH \to HCOONa({A_2}) + {C_2}{H_5}OH\\
HCOONa⇄HCOOH(A3)\\
HCOOH + 2AgN{O_3} + 4N{H_3} + {H_2}O \to {(N{H_4})_2}C{O_3}({A_4}) + 2Ag + 2N{H_4}N{O_3}\\
{(N{H_4})_2}C{O_3} + 2NaOH \to N{a_2}C{O_3} + N{H_3} \uparrow + {H_2}O\\
{(N{H_4})_2}C{O_3} + 2HCl \to N{H_4}Cl + {H_2}O + C{O_2} \uparrow
\end{array}$
Bài 8-b/
$\begin{array}{l}
+ )C{H_2} = C(C{H_3})COO{C_2}{H_5} + NaOH \to \underbrace {C{H_2} = C(C{H_3})COONa}_D + \underbrace {{C_2}{H_5}OH}_B\\
{C_2}{H_5}OH + [O] \to C{H_3}COOH(C)\\
C{H_3}COOH + {C_2}{H_5}OH ⇄ C{H_3}COO{C_2}{H_5}\\
+ )C{H_2} = C(C{H_3})COONa \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} \underbrace {C{H_2} = C(C{H_3})COOH}_E\\
C{H_2} = C(C{H_3})COOH + C{H_3}OH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} \underbrace {C{H_2} = C(C{H_3})COOC{H_3}}_F\\
C{H_2} = C(C{H_3})COOC{H_3}{( - CH2 - C(C{H_3})(COOC{H_3}) - )_n}
\end{array}$
Câu 8-c/
$\begin{array}{l}
C{H_4} \to \underbrace {{C_2}{H_2}}_{{A_1}} \to \underbrace {{C_2}{H_4}}_{{A_2}} \to \underbrace {{C_2}{H_5}OH}_{{A_3}} \to C{H_3}CHO \to C{H_3}COOH\\
{C_2}{H_2}({A_1}) + {H_2}O \to C{H_3}CHO\\
{C_2}{H_5}OH({A_3}) + [O] \to C{H_3}COOH({A_4})\\
C{H_3}COOH({A_4}) + N{H_3} \to C{H_3}COON{H_4}({C_2}{H_7}{O_2}N)\\
C{H_3}COOH({A_4}) + {C_2}{H_5}OH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_3}COO{C_2}{H_5}({C_4}{H_8}{O_2})\\
C{H_3}COOH({A_4}) + CH \equiv CH \to C{H_3}COO{C_2}{H_3}({C_4}{H_6}{O_2})
\end{array}$
