Giải bài 6 trang 187 SGK Hóa học 11

$\begin{align} & a)\underset{a\,mol}{\mathop{{{C}_{x}}{{H}_{y}}O}}\,+\left( x+\frac{y}{4}-\frac{1}{2} \right){{O}_{2}}\to xC{{O}_{2}}+\frac{y}{2}{{H}_{2}}O \\ & {{n}_{C{{O}_{2}}}}=ax=\frac{1,32}{44}=0,03\,mol \\ & {{n}_{{{H}_{2}}O}}=\,\frac{ay}{2}=\frac{0,72}{18}=0,04mol \\ & \Rightarrow \frac{x}{y}=\frac{3}{8} \\ \end{align}$

b) Công thức phân tử tổng quát: $(C_3H_8)_nO$

Biện luận: 

$\begin{align} & y\le 2x+2 \\ & 8n\le 6n+2 \\ & n=1 \\ & CTPT:{{C}_{3}}{{H}_{8}}O \\ & CTCT:C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}-\underset{\left| {} \right.}{\mathop{CH}}\,-C{{H}_{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH \\ \end{align}$

c) Để tạo anđehit, ancol A phải bậc I: $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH:propan-1-ol$