Giải bài 6 trang 182 - SGK môn Hóa học lớp 11 nâng cao

$\begin{align} & C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\xrightarrow{xt,{{t}^{0}}}C{{H}_{3}}-CH=CH-C{{H}_{3}}(A)+{{H}_{2}} \\ & C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\xrightarrow{xt,{{t}^{0}}}C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{3}}(B)+{{H}_{2}} \\ & C{{H}_{3}}-CH=CH-C{{H}_{3}}(A)\xrightarrow{-{{H}_{2}}}C{{H}_{2}}=C=CH-C{{H}_{3}}(C)+{{H}_{2}} \\ & C{{H}_{3}}-CH=CH-C{{H}_{3}}(A)\xrightarrow{-{{H}_{2}}}C{{H}_{3}}-C\equiv C-C{{H}_{3}}(D)+{{H}_{2}} \\ & C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{3}}(B)\xrightarrow{-{{H}_{2}}}CH\equiv C-C{{H}_{2}}-C{{H}_{3}}(E) \\ & C{{H}_{2}}=C=CH-C{{H}_{3}}(C)+B{{r}_{2}}\to C{{H}_{2}}Br-CBr=CH-C{{H}_{3}} \\ & C{{H}_{2}}=C=CH-C{{H}_{3}}(C)+B{{r}_{2}}\to C{{H}_{2}}Br-CBr=CH-C{{H}_{3}} \\ & C{{H}_{3}}-C\equiv C-C{{H}_{3}}(D)+B{{r}_{2}}\to C{{H}_{3}}-CBr=CBr-C{{H}_{3}} \\ & CH\equiv C-C{{H}_{2}}-C{{H}_{3}}(E)+AgN{{O}_{3}}+N{{H}_{3}}\to CAg\equiv C-C{{H}_{2}}-C{{H}_{3}}+N{{H}_{4}}N{{O}_{3}} \\ \end{align}$

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Ankin trong phân tử còn H ở nguyên tử C nối ba có thể tham gia phản  ứng thế ion kim loại.