MCQ
Cyclohexanol and phenol can be distinguish by
- A$Br_2/H_2O$
- BNeutral $FeCl_3$
- C$Ph\mathop {{N_2}}\limits^ \oplus C{l^\Theta }$
- ✓All of these
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$\begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{{C_6}{H_5} - CH - C{H_2} - C - C{H_3}}
\end{array}\mathop {\xrightarrow{{(i)\,NaOBr}}}\limits_{(ii)\,{H_2}O/{H^ + }\,(iii)\,\Delta } $ product
product will be :
$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - Br}\\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{{C_2}{H_5}\,\,\,\,\,\,\,\,\,}
\end{array}$
$CH_3COOH + PCl_5 \longrightarrow (A)$ $\xrightarrow[(2){{H}_{2}}O]{(1)\,C{{H}_{3}}MgBr}\left( B \right)$ Product $B$ would be
