Alcohol, Phenol and Ether MCQ Questions & Answers in Organic Chemistry | Chemistry

\[\underbrace{CO+{{H}_{2}}}_{\text{water gas}}+{{H}_{2}}\xrightarrow[{{300}^{\circ }}C]{C{{r}_{2}}{{O}_{3}}-ZnO}\underset{\text{Methanol}}{\mathop{C{{H}_{3}}OH}}\,\]

Phenol has activating ( electron releasing ) $$–OH$$  group and bromine water supplies $$B{r^ + }\,ion$$   easily, hence under such conditions reaction does not stop at monobromo or dibromo stage but a fully brominated ( 2, 4, 6, - tribromophenol ) compound is the final product.

\[\begin{align} & \underset{{{1}^{\circ }}\text{-alcohol}}{\mathop{RC{{H}_{2}}OH}}\,\xrightarrow[\Delta ]{Cu}\underset{\text{ Aldehyde}}{\mathop{RCHO}}\, \\ & \underset{{{2}^{\circ }}\text{-alcohol}}{\mathop{R\overset{\begin{smallmatrix} OH \\ |\,\,\,\, \end{smallmatrix}}{\mathop{-CH-}}\,}}\,R\xrightarrow[\Delta ]{Cu}\underset{\text{Ketone}}{\mathop{R-\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{C}}\,-R}}\, \\ & \underset{{{3}^{\circ }}\text{-alcohol}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,OH}}\,\xrightarrow[\Delta ]{Cu}\underset{\text{alkene}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C=}}\,C{{H}_{2}}}}\, \\ \end{align}\]

$$\left( {\text{i}} \right)\,C{H_3}C{H_2}C{H_2}C{H_2}C{H_2}OH$$

$$\left( {{\text{iv}}} \right)\,{\left( {C{H_3}} \right)_3}CC{H_2}OH$$

The order of stability of benzyl carbocations is :

The electron withdrawing effect of $$ - N{O_2}$$  group is greater than that of $$- Cl$$  group.
Thus, the correct increasing order of reactivity with $$HBr/HCl$$   is : (ii) < (iii) < (i).

$$Cr{O_3}/$$  pyridine is mild oxidising agent.
$${1^ \circ }$$ Alcohol → Aldehyde
$${2^ \circ }$$ Alcohol → Ketone ;