Bài 4.6 trang 9 SBT Hóa học 12

Đề bài

Viết các phương trình hóa học để hoàn thành các dãy biến hóa sau:

a. \({C_2}{H_4} \to C{H_3}CHO\xrightarrow{{B{{\text{r}}_2},{H_2}O}}A\xrightarrow{{ + B}}{C_4}{H_8}{O_2}\xrightarrow{{LiAl{H_4},t^\circ }}B\)

b. \(C{H_3}C{H_2}COOH \to C{H_3}CHBrCOOH \to C{H_2} = CHCOOK\)

\( \to C{H_2} = CHCOOH \to C{H_2} = CHCOOC{H_3} \to po\lim e\)

c. \(C{H_2} = C{H_2} \to C{H_3}CHO \to C{H_3}COOH \to C{H_3}COOCH = C{H_2}\)

\( \to po\lim e\,\,1 \to po\lim e\,\,2\)  

Lời giải chi tiết

a)

\(2C{H_2} = C{H_2} + {O_2}\xrightarrow{{xt,{t^^\circ }}}2C{H_3}CHO\)

\(C{H_3}CHO + B{{\text{r}}_2} + {H_2}O \to C{H_3}COOH + 2HB{\text{r}}\)

\(C{H_3}COOH + {C_2}{H_5}OH\overset {H{ & _2}S{O_4},{t^^\circ }} \leftrightarrows C{H_3}COO{C_2}{H_5} + {H_2}O\)

\(C{H_3}COO{C_2}{H_5}\xrightarrow{{LiAl{H_4},{t^^\circ }}}2C{H_3}C{H_2}OH\)

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

\(C{H_2} = CHCOOH + C{H_3}OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_2} = CHCOOC{H_3} + {H_2}O\)

\(nC{H_2} = CHCOOC{H_3}\buildrel {{t^0},xt,p} \over
\longrightarrow - {(C{H_2} - CH - COOC{H_3})_n} - \)

c) \(2C{H_2} = C{H_2} + {\rm{ }}{O_2}\buildrel {{t^0},xt} \over
\longrightarrow {\rm{ }}2C{H_3} - CHO\)

\(C{H_3}COOH + C{H_2} = CH - OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_3}COOCH = C{H_2} + {H_2}O\)

\(nC{H_3}COOCH = C{H_2}\buildrel {{t^0},xt,p} \over
\longrightarrow - \left( {C{H_2} - CH - OCOC{H_3}} \right){ - _n}\).

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