Hoàn thành dãy chuyển chuyển hóa sau bằng các phương trình hóa học:
Natri axetat—>CH4—>Axetilen—>Vinyl axetilen—>Butan—>Propen—>Propan-2-ol
Axetilen—>Vinyl clorua—>anđehitaxetic—>Axit axetic—>etyl axetat
Hoàn thành dãy chuyển chuyển hóa sau bằng các phương trình hóa học:
Natri axetat—>CH4—>Axetilen—>Vinyl axetilen—>Butan—>Propen—>Propan-2-ol
Axetilen—>Vinyl clorua—>anđehitaxetic—>Axit axetic—>etyl axetat
Đáp án:
Giải thích các bước giải:
$a/$
$CH_3COONa + NaOH \xrightarrow{t^o,CaO} Na_2CO_3 + CH_4$
$2CH_4 \xrightarrow{t^o,lln} C_2H_2 + 3H_2$
$C_2H_2 \xrightarrow{t^o,xt,p} CH≡C-CH=CH_2$
$C≡C-CH=CH_2 + 3H_2 \xrightarrow{t^o,Ni} CH_3-CH_2-CH_2-CH_3$
$C_4H_{10} \xrightarrow{t^o,Ni} C_3H_6 + CH_4$
$CH_2=CH-CH_3 + H_2O \xrightarrow{t^o.H^+} CH_3-CH(OH)-CH_3$
$b/$
$CH≡CH + HCl \xrightarrow{t^o,HgCl_2} CH_2=CH-Cl$
$CH_2=CH-Cl + NaOH → NaCl + CH_3-CHO$
$CH_3-CHO + \dfrac{1}{2}O_2 \xrightarrow{t^o} CH_3COOH$
$CH_3COOH + C_2H_5OH ⇄ CH_3COOC_2H_5 + H_2O$
Giải thích các bước giải:
a,
\(\begin{array}{l}
C{H_3}COONa + NaOH \to N{a_2}C{O_3} + C{H_4}\\
2C{H_4} \to {C_2}{H_2} + 2{H_2}\\
2{C_2}{H_2} \to CH \equiv C – CH = C{H_2}\\
CH \equiv C – CH = C{H_2} + 3{H_2} \to C{H_3}C{H_2}C{H_2}C{H_3}\\
{C_4}{H_{10}} \to C{H_4} + {C_3}{H_6}\\
{C_3}{H_6} + {H_2}O \to {C_3}{H_7}OH
\end{array}\)
b,
\(\begin{array}{l}
{C_2}{H_2} + HCl \to C{H_2}CHCl\\
C{H_3}CHO + {O_2} \to C{H_3}COOH\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O
\end{array}\)