Bài 10: Hoàn thành sơ đồ chuyển hóa sau (mỗi mũi tên là một PTHH), ghi rõ điều kiện:
a. Metan —–> Axetilen —–>Etilen—–> Etanol—–> Etilen —–>polietilen(P.E)
b. Canxi cacbua ——> Axetilen—–> vinylaxetilen ——>butađien—–> caosu buna(polibutađien)
c. Axetilen——> bạc axetilua——> axetilen——> anđehit axetic ——> Etanol
Câu 11: Viết đồng phân ankin có CTPT sau và gọi tên
a. C3H4 b. C4H6
c. C5H8 d. C6H10
mng giúp mk với, mk cảm ơn!
Bài 10:
a.
$2C{H_4}\xrightarrow[{làm\,\,lạnh\,\,nhanh}]{{{{1500}^o}C}}{C_2}{H_2} + 3{H_2}$
${C_2}{H_2} + {H_2}\xrightarrow{{Pd/PbC{O_3}}}{C_2}{H_4}$
${C_2}{H_4} + {H_2}O \to {C_2}{H_5}OH$
${C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},\,\,{{170}^o}C}}{C_2}{H_4} + {H_2}O$
$nC{H_2} = C{H_2}\xrightarrow{{{t^o},\,\,xt}}{( – C{H_2} – C{H_2} – )_n}$
b.
$Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}$
$2CH \equiv CH \to CH \equiv C – CH = C{H_2}$
$CH \equiv C – CH = C{H_2} + {H_2}\xrightarrow{{Pd/PbC{O_3}}}C{H_2} = CH – CH = C{H_2}$
$nC{H_2} = CH – CH = C{H_2}\xrightarrow{{{t^o},xt,p}}{( – C{H_2} – CH = CH – C{H_2} – )_n}$
c.
$CH \equiv CH + 2AgN{O_3} + 2N{H_3} \to CAg \equiv CAg + 2N{H_4}N{O_3}$
$CAg \equiv CAg + 2HCl \to CH \equiv CH + 2AgCl$
$CH \equiv CH + {H_2}O \to C{H_3}CHO$
$C{H_3}CHO + {H_2}\xrightarrow{{Ni,\,\,{t^o}}}C{H_3}C{H_2}OH$
Câu 11:
* ${C_3}{H_4}$:
$CH \equiv C – C{H_3}$: propin
* ${C_4}{H_6}$:
$CH \equiv C – C{H_2} – C{H_3}$: but-1-in
$C{H_3} – C \equiv C – C{H_3}$: but-2-in
* ${C_5}{H_8}:$
$CH \equiv C – C{H_2} – C{H_2} – C{H_3}$: pent-1-in
$C{H_3} – C \equiv C – C{H_2} – C{H_3}$: pent-2 -in
$CH \equiv C – CH(C{H_3}) – C{H_3}$: 3 – metylbut-1-in
* $C_6H_{10}$:
$CH \equiv C – C{H_2} – C{H_2} – C{H_2} – C{H_3}$: hex-1-in
$C{H_3} – C \equiv C – C{H_2} – C{H_2} – C{H_3}$: hex-2-in
$C{H_3} – C{H_2} – C \equiv C – C{H_2} – C{H_3}$: hex-3-in
$CH \equiv C – C{H_2} – CH(C{H_3}) – C{H_3}$: 4-metyl-pent-1-in
$CH \equiv C – CH(C{H_3}) – C{H_2} – C{H_3}$: 3-metyl-pent-1-in
$C{H_3} – C \equiv C – CH(C{H_3}) – C{H_3}$: 4-metyl-pent-2-in
$CH \equiv C – C{(C{H_3})_2} – C{H_3}$: 3,3-đi metylbut-1-in
Đáp án: