Cho sơ đồ sau :                                   ...

B

- Hướng dẫn giải

- Các phản ứng xảy ra:

\(\begin{array}{l}C{H_2} = C(C{H_3}) - C{\rm{OO}}{C_2}{H_5}\;(M) + NaOH \to C{H_2} = C(C{H_3}) - C{\rm{OO}}Na\;(Y) + {C_2}{H_5}OH\;(X)\\{C_2}{H_5}OH\;(X) \to C{H_2} = C{H_2}\;({X_1}) + {H_2}O\\nC{H_2} = C{H_2}\;({X_1}) \to C{H_2} - C{H_2}{_n}\\2C{H_2} = C(C{H_3}) - C{\rm{OO}}Na\;(Y) + {H_2}S{O_{4\,}} \to 2C{H_2} = C(C{H_3}) - C{\rm{OOH}}\;({Y_1}) + N{a_2}S{O_4}\\C{H_2} = C(C{H_3}) - C{\rm{OOH}}\;({Y_1}) + C{H_3}OH \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} C{H_2} = C(C{H_3}) - C{\rm{OOC}}{{\rm{H}}_3}\;({Y_2})\\nC{H_2} = C(C{H_3}) - C{\rm{OOC}}{{\rm{H}}_3}\;({Y_2}) \to C{H_2} = (C{H_3})C(C{\rm{OOC}}{{\rm{H}}_3}){_n}\end{array}\)