Giải bài 1 trang 195 SGK Hóa học 11

$\begin{align} & {{C}_{4}}{{H}_{9}}Cl \\ & CTCT:\,C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}Cl\left( butyl\,clorua \right) \\ \end{align}$

Đồng phân:

$\begin{align} & C{{H}_{3}}-C{{H}_{2}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{3}}\left( \sec -butyl\,clorua \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ & C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{2}}-Cl\left( \,isobutyl\,clorua \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\ & C{{H}_{3}}-\underset{\left| {} \right.}{\overset{\left| {} \right.}{\mathop{C}}}\,-C{{H}_{3}}\left( \,tert-butyl\,clorua \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,CH_3 \\ \end{align}$

$\begin{align} & {{C}_{4}}{{H}_{10}}O \\ & C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-OH\left( ancol\,butylic \right) \\ & C{{H}_{3}}-C{{H}_{2}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{3}}(ancol\,\sec -butylic) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\ & C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{2}}-OH(ancol\,\text{is}obutylic) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\ & C{{H}_{3}}-\underset{\left| {} \right.}{\overset{\left| {} \right.}{\mathop{C}}}\,-C{{H}_{3}}(ancol\,t\text{er}t-butylic) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ \end{align}$

$\begin{align} & {{C}_{4}}{{H}_{8}}O \\ & C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{2}}-OH \\ & C{{H}_{3}}-CH=CH-C{{H}_{2}}-OH \\ & C{{H}_{2}}=\underset{\left| {} \right.}{\mathop{C}}\,-C{{H}_{2}}-OH \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}} \\ & C{{H}_{2}}=CH-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\ \end{align}$