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

$$Mn{O_2}$$  being a mild oxidising agent stops the oxidation of $$ - C{H_2}OH$$   group at aldehyde stage.

\[\underset{\text{2-Methylpropanal}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\, \\ C{{H}_{3}}\, \end{smallmatrix}}{\mathop{-CH-}}\,\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}\,=}}\,\]    \[\underset{\text{Isopropyl magnesium iodide}}{\mathop{O+{{\left( C{{H}_{3}} \right)}_{2}}CHMgI\to }}\,\]    \[{{\left( C{{H}_{3}} \right)}_{2}}\,\underset{\begin{smallmatrix} |\,\,\,\,\,\, \\ \,\,\,\,\,\,\,\,\,CH{{\left( C{{H}_{3}} \right)}_{2}} \end{smallmatrix}}{\overset{\begin{smallmatrix} H\,\,\,\, \\ |\,\,\,\,\, \end{smallmatrix}}{\mathop{CH-C-OMgI}}}\,\]     \[\xrightarrow{{{H}_{2}}O}\underset{\text{2, 4-Dimethylpentan-3-o1}}{\mathop{{{\left( C{{H}_{3}} \right)}_{2}}\,\underset{\begin{smallmatrix} | \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,CH{{\left( C{{H}_{3}} \right)}_{2}} \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{CH-C-OH}}}\,}}\,\]

Primary alcohols readily form ether when heated with conc. $${H_2}S{O_4}.$$

Tertiary alcohols react fastest with conc. $$HCl$$  and anhydrous $$ZnC{l_2}$$  (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation.

\[\underset{\left( A \right)}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,\xrightarrow[{{H}_{2}}S{{O}_{4}}]{\Delta }\]      \[\underset{\left( B \right)}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}+{{H}_{2}}O}}\,\]
\[\underset{\left( B \right)}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\,+HCl\to \]      \[\underset{\begin{smallmatrix} \,| \\ \,Cl \\ \left( C \right) \end{smallmatrix}}{\mathop{C{{H}_{3}}CH-}}\,C{{H}_{3}}\]
\[C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\,\, \\ Cl\,\,\,\,\,\, \\ \left( C \right) \end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{3}}+KO{{H}_{\left( aq. \right)}}\]      \[\to C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\, \\ OH\,\, \\ \left( D \right) \end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{3}}\]
\[A\text{ and }D\text{ are position isomers}\text{.}\]

\[{{C}_{3}}{{H}_{8}}{{O}_{3}}:\underset{\begin{smallmatrix} |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ C{{H}_{2}}OH\,\,\,\,\, \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{2}}OH\,\, \\ |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{smallmatrix}}{\mathop{CHOH}}}\,\]

Let the formula be $${C_x}{H_y}{O_z}$$
Molecular weight $$ = 12x + y + 16z;$$
$$\% $$  of oxygen $$ = 100\, - $$  $$\left[ {\,\,\% } \right.$$  of carbon $$ + \,\,\% $$  of hydrogen $$\left. {} \right]$$
$$\eqalign{ & = 100 - \left[ {60 + 13.3} \right] \cr & = 26.7\% \cr} $$
$$\% $$  of carbon $$ = \frac{{12x}}{{12x + y + 16z}} \times 100 = 60\% $$
$$\% $$  of hydrogen $$ = \frac{y}{{12x + y + 16z}} \times 100 = 13.3\% $$
$$\% $$  of oxygen $$ = \frac{{16z}}{{12x + y + 16z}} \times 100 = 26.7\% $$
On solving the above equations, we get the ratio of $$x:y:z = 3:8:1$$     (approximate)
∴ Formula of compound is $${C_3}{H_8}O.$$