Giải bài 6 trang 213 SGK Hóa học 11
Viết phương trình hóa học của các phản ứng hoàn thành dãy chuyển hóa sau:
\(\begin{align} & \underset{{}}{\mathop{C{{H}_{2}}=CHC{{H}_{3}}}}\,\xrightarrow{\left( 1 \right)}C{{H}_{3}}\underset{\left| {} \right.}{\mathop{C}}\,HC{{H}_{3}}\xrightarrow{\left( 2 \right)}C{{H}_{3}}\underset{\left\| {} \right.}{\mathop{C}}\,C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\ & \underset{{}}{\mathop{C{{H}_{2}}=CHC{{H}_{3}}}}\,\xrightarrow{\left( 3 \right)}C{{H}_{2}}=CHC{{H}_{2}}Cl\xrightarrow{\left( 4 \right)}C{{H}_{2}}CHC{{H}_{2}}OH\xrightarrow{\left( 5 \right)}C{{H}_{2}}=CHCH=O \\ \end{align} \)
\(\begin{align} & 1.C{{H}_{2}}=CHC{{H}_{3}}+{{H}_{2}}O\xrightarrow{{{H}^{+}}}C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\ & 2.C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{3}}+CuO\to C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{C}}\,-C{{H}_{3}}+Cu+{{H}_{2}}O \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O \\ & 3.C{{H}_{2}}=CH-C{{H}_{3}}+C{{l}_{2}}\xrightarrow[{{500}^{o}}C]{{{t}^{o}}}C{{H}_{2}}=CH-C{{H}_{2}}Cl+HCl \\ & 4.C{{H}_{2}}=CH-C{{H}_{2}}Cl+NaOH\to C{{H}_{2}}=CH-C{{H}_{2}}OH+NaCl \\ & 5.C{{H}_{2}}=CH-C{{H}_{2}}OH+CuO\xrightarrow{{{t}^{o}}}C{{H}_{2}}=CH-CHO+Cu+{{H}_{2}}O \\ \end{align} \)