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Câu hỏi: Viết các phương trình hóa học để hoàn thành các dãy biến hóa sau:
a. ${C_2}{H_4} \Rightarrow C{H_3}CHO\xrightarrow{{B{{\text{r}}_2},{H_2}O}}A\xrightarrow{{ + B}}{C_4}{H_8}{O_2}\xrightarrow{{LiAl{H_4},t^\circ }}B$
b. $C{H_3}C{H_2}COOH \Rightarrow C{H_3}CHBrCOOH \Rightarrow C{H_2} = CHCOOK$
$ \Rightarrow C{H_2} = CHCOOH \Rightarrow C{H_2} = CHCOOC{H_3} \Rightarrow po\lim e$
c. $C{H_2} = C{H_2} \Rightarrow C{H_3}CHO \Rightarrow C{H_3}COOH \Rightarrow C{H_3}COOCH = C{H_2}$
$ \Rightarrow po\lim e 1 \Rightarrow po\lim e 2$
a. ${C_2}{H_4} \Rightarrow C{H_3}CHO\xrightarrow{{B{{\text{r}}_2},{H_2}O}}A\xrightarrow{{ + B}}{C_4}{H_8}{O_2}\xrightarrow{{LiAl{H_4},t^\circ }}B$
b. $C{H_3}C{H_2}COOH \Rightarrow C{H_3}CHBrCOOH \Rightarrow C{H_2} = CHCOOK$
$ \Rightarrow C{H_2} = CHCOOH \Rightarrow C{H_2} = CHCOOC{H_3} \Rightarrow po\lim e$
c. $C{H_2} = C{H_2} \Rightarrow C{H_3}CHO \Rightarrow C{H_3}COOH \Rightarrow C{H_3}COOCH = C{H_2}$
$ \Rightarrow po\lim e 1 \Rightarrow po\lim e 2$
Phương pháp giải
Dựa vào tính chất hóa học, phương trình điều chế của các hợp chất hữu cơ.
Lời giải chi tiết
B)
$C{H_3}C{H_2}COOH + B{r_2}\buildrel {{t^0}} \over
\longrightarrow C{H_3}CHBrCOOH + HBr$
$C{H_3}CHBrCOOH + 2KOH\buildrel {{C_2}{H_5}OH,{\rm{ }}{t^0}} \over
\longrightarrow C{H_2} = CHCOOK + KBr + 2{H_2}O$
CH2=CHCOOK+HCl(l)→ CH2=CHCOOH+KCl
$C{H_2} = CHCOOH + C{H_3}OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_2} = CHCOOC{H_3} + {H_2}O$
$nC{H_2} = CHCOOC{H_3}\buildrel {{t^0},xt,p} \over
\longrightarrow - {\left(C{H_2} - CH - COOC{H_3}\right)_n} - $
c) $2C{H_2} = C{H_2} + {\rm{ }}{O_2}\buildrel {{t^0},xt} \over
\longrightarrow {\rm{ }}2C{H_3} - CHO$
$C{H_3}COOH + C{H_2} = CH - OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_3}COOCH = C{H_2} + {H_2}O$
$nC{H_3}COOCH = C{H_2}\buildrel {{t^0},xt,p} \over
\longrightarrow - \left( {C{H_2} - CH - OCOC{H_3}} \right){ - _n}$.
Dựa vào tính chất hóa học, phương trình điều chế của các hợp chất hữu cơ.
Lời giải chi tiết
B)
$C{H_3}C{H_2}COOH + B{r_2}\buildrel {{t^0}} \over
\longrightarrow C{H_3}CHBrCOOH + HBr$
$C{H_3}CHBrCOOH + 2KOH\buildrel {{C_2}{H_5}OH,{\rm{ }}{t^0}} \over
\longrightarrow C{H_2} = CHCOOK + KBr + 2{H_2}O$
CH2=CHCOOK+HCl(l)→ CH2=CHCOOH+KCl
$C{H_2} = CHCOOH + C{H_3}OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_2} = CHCOOC{H_3} + {H_2}O$
$nC{H_2} = CHCOOC{H_3}\buildrel {{t^0},xt,p} \over
\longrightarrow - {\left(C{H_2} - CH - COOC{H_3}\right)_n} - $
c) $2C{H_2} = C{H_2} + {\rm{ }}{O_2}\buildrel {{t^0},xt} \over
\longrightarrow {\rm{ }}2C{H_3} - CHO$
$C{H_3}COOH + C{H_2} = CH - OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_3}COOCH = C{H_2} + {H_2}O$
$nC{H_3}COOCH = C{H_2}\buildrel {{t^0},xt,p} \over
\longrightarrow - \left( {C{H_2} - CH - OCOC{H_3}} \right){ - _n}$.