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

The $$C – O$$  bond length in alcohols is $$142\,pm$$  and in Phenol it is $$136\,pm.$$  The $$C – O$$  bond length in phenol is shorter than methanol it is due to the conjugation of unshared pair of electrons on oxygen with the ring, which imparts double bond character to the $$C – O$$  bond.

An organic compound form yellow precipitate of iodoform with $${I_2}$$ in presence of alkali, if it has $$C{H_3}CO - $$  group directly or it has
$$\left( {\text{A}} \right)C{H_3}CH\left( {OH} \right)C{H_3} + {I_3}$$      \[\xrightarrow{NaOH}C{{H}_{3}}COC{{H}_{3}}\]     $$ + 2HI + 3NaI + C{H_3}CO{O^ - }{N^ + }a$$        $$ + 3{H_2}O$$
$$C{H_3}COC{H_3} + 3{I_2} + 4NaOH$$     $$ \to \mathop {CH{I_3} \downarrow }\limits_{{\text{Yellow ppt}}{\text{.}}} + 3NaI + C{H_3}CO{O^ - }{N^ + }a$$         $$ + 3{H_2}O$$
$$\left( {\text{B}} \right)C{H_2} - C{H_2}CH\left( {OH} \right)C{H_3} + {I_2} \to $$      
$${\text{It gives iodoform test}}$$

$$\mathop {CH{I_3} \downarrow }\limits_{{\text{Yellow ppt}}} 3NaI + C{H_3}C{H_2}COONa + 3{H_2}O$$
$$\left( {\text{C}} \right)C{H_3}OH + {I_2} \to HCHO + 2HI$$
It does not have methyl ketonic group, so it does not give yellow $$ppt.$$  with $${I_2}$$  in presence of alkali.

Due to the presence of $$ - COC{H_3}$$   group, it gives Haloform test.

First write down the structure of the required compound, observe the nature of functional group,

and review the memory for preparing the required functional group. Here the required compound is a $${3^ \circ }$$  alcohol having three ethyl groups, so only combination given in option $$c$$  will introduce three ethyl groups.

$${C_6}{H_5}Br + NaO - C{H_2} - CH = $$       $$C{H_2} \to {C_6}{H_5} - O - $$     $$\mathop {C{H_2} - CH = C{H_2}}\limits_{{\text{Allyl phenyl ether}}} + NaBr$$

NOTE :
Fehling solution is a weak oxidising agent which can oxidise aldehyde but not ketone.
Primary alcohols undergoes oxidation with alkaline $$KMn{O_4},$$  acidic dichromate and conc. $${H_2}S{O_4}$$  to give acids, whereas with $$Cu$$  they give aldehydes.
\[\begin{align} & \underset{\text{Propalol-1}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,\xrightarrow[heat]{Cu}\,\underset{\underset{\text{(responds Fehling solution)}}{\mathop{\text{Propanal}}}\,}{\mathop{C{{H}_{3}}C{{H}_{2}}CHO}}\, \\ & \underset{\text{Propalol-2}}{\mathop{C{{H}_{3}}CHOHC{{H}_{3}}}}\,\to \underset{\underset{\left( \text{negative to Fehling solution} \right)}{\mathop{\text{Propanon}}}\,}{\mathop{C{{H}_{3}}COC{{H}_{3}}}}\, \\ \end{align}\]

The equimolar mixture of concentrated hydrochloric acid and anhydrous $$ZnC{l_2}$$  is called Lucas reagent. Lucas reagent is used to distinguish between $${1^ \circ },{2^ \circ }$$ and $${3^ \circ }$$ alcohols.

Tertiary halide can involve elimination of $$HX$$  to give alkene in presence of a base. Hence, a tertiary alkoxide and primary alkyl halide is taken for the reaction.
$$C{H_3}Br + \mathop {NaOC{{\left( {C{H_3}} \right)}_3}}\limits_{{\text{sod}}{\text{. }}t{\text{ - butoxide}}} \to $$      $$\mathop {C{H_3}O - C{{\left( {C{H_3}} \right)}_3}}\limits_{{\text{ methyl }}t{\text{ - butyl ether }}} + NaBr$$