Giải bài 6 trang 233 - SGK môn Hóa học lớp 11 nâng cao

$\begin{align} & {{m}_{\,\text{dd}NaOH}}=V.D=14,4.1,11\approx 16\,gam \\ & {{m}_{NaOH}}=\dfrac{16,10\%}{100\%}=1,6\,gam \\ & {{n}_{NaOH}}=0,04\,mol \\ & {{m}_{B{{r}_{2}}}}=\dfrac{300.3,2\%}{100\%}=9,6\,gam \\ & {{n}_{B{{r}_{2}}}}=0,06\,mol \\ & {{C}_{6}}{{H}_{5}}OH+3B{{r}_{2}}\to {{C}_{6}}{{H}_{2}}OH{{(Br)}_{3}}\downarrow +3HBr \\ \end{align}$

${{C}_{6}}{{H}_{5}}-CH=C{{H}_{2}}+B{{r}_{2}}\to {{C}_{6}}H{_{5}}-CHBr-C{{H}_{2}}Br$

$HBr+NaOH\to NaBr+{{H}_{2}}O$

${{n}_{{{C}_{6}}{{H}_{5}}OH}}=\dfrac{1}{3}{{n}_{HBr}}={{n}_{NaOH}}=\dfrac{0,04}{3}\,mol \\ $

${{n}_{{{C}_{6}}{{H}_{5}}-CH=C{{H}_{2}}}}={{n}_{B{{r}_{2}}}}-3{{n}_{{{C}_{6}}{{H}_{5}}OH}}=0,06-\dfrac{0,04}{3}.3=0,02\,mol \\ $

$\%{{m}_{{{C}_{6}}{{H}_{5}}OH }}=\dfrac{\dfrac{0,04}{3}.94}{\dfrac{0,04}{3}.94+0,02.106}.100\%\approx 37,15\% \\ $

$\%{{m}_{{{C}_{6}}{{H}_{5}}-CH=C{{H}_{2}}}}=62,85\% \\ $

Ghi nhớ:

Phenol phản ứng tạo kết tủa với dung dịch brom theo tỉ lệ mol 1 phenol: 3 brom