Đáp án:
\({C_2}{H_2};{\text{ }}{{\text{C}}_3}{H_4}\)
Giải thích các bước giải:
a) Gọi công thức chung của 2 ankin là \({C_n}{H_{2n - 2}}\)
Phản ứng xảy ra:
\({C_n}{H_{2n - 2}} + (1,5n - 0,5){O_2}\xrightarrow{{}}nC{O_2} + (n - 1){H_2}O\)
\(C{O_2} + Ca{(OH)_2}\xrightarrow{{}}CaC{O_3} + {H_2}O\)
\(\to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,1n = \frac{{25}}{{100}} = 0,25{\text{ mol}} \to {\text{n = 2}}{\text{,5}}\)
Nhận thấy 2<2,5<3 và 2 ankin liên tiếp nên 2 ankin là \({C_2}{H_2};{\text{ }}{{\text{C}}_3}{H_4}\)
CTCT: \(CH \equiv CH;{\text{ C}}{{\text{H}}_3} - C \equiv CH\)
Tên gọi: A: etin B : propin
b) \(3{C_2}{H_2}\xrightarrow{{{t^o},xt}}{C_6}{H_6}\)
\({C_2}{H_2} + {H_2}\xrightarrow{{Pd,{t^o}}}{C_2}{H_4}\)
\({C_2}{H_4} + {H_2}\xrightarrow{{}}{C_2}{H_6}\)
\({C_2}{H_2} + 2AgN{O_3} + 2N{H_3}\xrightarrow{{}}A{g_2}{C_2} + 2N{H_4}N{O_3}\)
\({C_2}{H_2} + HCl\xrightarrow{{}}{C_2}{H_3}Cl;{\text{ n}}{{\text{C}}_2}{H_3}Cl\xrightarrow{{}}{({C_2}{H_3}Cl)_n}\)
\(2{C_2}{H_2}\xrightarrow{{}}{C_4}{H_4};{\text{ }}{{\text{C}}_4}{H_4} + {H_2}\xrightarrow{{Pd}}{C_4}{H_6};{\text{ n}}{{\text{C}}_4}{H_6}\xrightarrow{{{t^o},xt}}{( - C{H_2} - CH = CH - C{H_2} - )_n}\)
c) \(C{H_3} - C \equiv CH + {H_2}\xrightarrow{{Pd,{t^o}}}C{H_3} - CH = C{H_2}\)
\(C{H_3} - C \equiv CH + 2B{r_2}\xrightarrow{{}}C{H_3} - CB{r_2} - CHB{r_2}\)
