Viết các phương trình hoá học để hoàn thành các dãy biến hoá sau :
a) \({\rm{ }}{C_2}{H_4} \to C{H_3}CHO\buildrel {B{r_2},{H_2}O} \over
\longrightarrow A\buildrel { + B} \over
\longrightarrow {C_4}{H_8}{O_2}\buildrel {LiAl{H_4},{\rm{ }}{t^0}} \over
\longrightarrow B\)
b) CH3CH2COOH→CH3CHBrCOOH→CH2=CHCOOK→ CH2=CHCOOH →CH2=CHCOOCH3 →polime
c)CH2=CH2→CH3CHO→CH3COOH→CH3COOCH=CH2→polime 1→ polime 2
a)
\(2C{H_2} = C{H_2} + {\rm{ }}{O_2}\buildrel {{t^0},xt} \over
\longrightarrow {\rm{ }}2C{H_3} - CHO\)
CH3-CHO + Br2 +H2O → CH3-COOH + 2HBr
CH3-COOH + C2H5OH ↔ CH3COOC2H5 + H2O
\(C{H_3}COO{C_2}{H_5}\buildrel {LiAl{H_4},{\rm{ }}{t^0}} \over
\longrightarrow 2C{H_3}C{H_2}OH\)
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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) \(2C{H_2} = C{H_2} + {\rm{ }}{O_2}\buildrel {{t^0},xt} \over
\longrightarrow {\rm{ }}2C{H_3} - CHO\)
\(C{H_3}COOH + C{H_2} = CH - OH\buildrel {{t^0},{\rm{ }}{H_2}S{O_4}\left( \right)} \over
\longrightarrow C{H_3}COOCH = C{H_2} + {H_2}O\)
\(nC{H_3}COOCH = C{H_2}\buildrel {{t^0},xt,p} \over
\longrightarrow - \left( {C{H_2} - CH - OCOC{H_3}} \right){ - _n}\).