Alkyl and Aryl Halide MCQ Questions & Answers in Organic Chemistry | Chemistry

\[\begin{align} & C{{H}_{3}}-C{{H}_{2}}-CHC{{l}_{2}}\xrightarrow[\Delta ]{NaN{{H}_{2}}\left( Sodamide \right)} \\ & C{{H}_{3}}-CH=CHCl\xrightarrow[\Delta ]{NaN{{H}_{2}}}\underset{\text{Final product}}{\mathop{C{{H}_{3}}-C\equiv CH}}\, \\ \end{align}\]

$${S_N}2$$  reaction is favoured by small groups on the carbon atom attached to halogen.
So, the order of reactivity is
$$C{H_3}Cl > {\left( {C{H_3}} \right)_2}CHCl > {\left( {C{H_3}} \right)_3}CCl > {\left( {{C_2}{H_5}} \right)_2}CHCl$$
NOTE : $${S_N}2$$  reaction is shown to maximum extent by primary halides. The only primary halides given is $$C{H_3}Cl$$   so the correct answer is (D).

The industrial preparation of chloroform involves the following steps :
$$\left( {\text{i}} \right)CaOC{l_2} + {H_2}O \to Ca{\left( {OH} \right)_2} + C{l_2}$$
$$\left( {{\text{ii}}} \right)C{H_3}COC{H_3} + 3C{l_2} \to $$      $$CC{l_3}COC{H_3} + 3HCl$$

Grignard reagents react with compounds containing active hydrogen to form hydrocarbons corresponding to alkyl ( or aryl ) part of the Grignard reagent.
$${C_6}{H_5}MgBr + M{e_3}COH \to {C_6}{H_6} + M{e_3}COMgBr$$

This is Reimer-Tiemann reaction. So finally $$ - CHO$$  group is introduced

Aprotic solvents like $$DMF$$  increases the reactivity of nucleophile and favours $${S_N}2$$  reaction. The relative reactivity of alkyl halides towards $${S_N}2$$  reactions is as follows
$$C{H_3} - X > {\text{Primary}} > {\text{Secondary}} > {\text{Tertiary}}$$
However, if the primary alkyl halide or the nucleophile/base is sterically hindered the nucleophile will have difficulty to getting the back side of the $$\alpha $$ - carbon as a result of this, the elimination product will be predominant. Here, $$C{H_3}C{H_2}Br$$   is the least hindered, hence it has the highest relative rate towards $${S_N}2$$  reaction.

Chloroform on reaction with nitric acid gives chloropicrin (nitro chloroform). Its reaction is shown below
\[\underset{\text{Chloroform}}{\mathop{CHC{{l}_{3}}}}\,+HN{{O}_{3}}\to \underset{\begin{smallmatrix} \text{Nitre chloroform } \\ \text{(chloropicrin)} \end{smallmatrix}}{\mathop{C\left( N{{O}_{2}} \right)C{{l}_{3}}}}\,+{{H}_{2}}O\]

$$C{H_3}{O^ - }$$  is a strong base and strong nucleophile, so favourable condition is $${S_N}2/E2.$$
The given alkyl halide is $${2^ \circ }$$ and $$\beta $$  carbons are $${4^ \circ }$$ and $${2^ \circ },$$ so sufficiently hindered, thus $$E2$$  dominates over $${S_N}2.$$
Also, polarity of $$C{H_3}OH$$  (solvent) is not as high as $${H_2}O$$  so $$E1$$ is also dominated by $$E2.$$

Kharasch effect or peroxide effect is only observed in case of addition of $$HBr$$  to unsymmetrical alkenes, so the addition of $$HCl$$  with propene takes place as usual by Markownikoff's rule

\[BrC{{H}_{2}}-C{{H}_{2}}Br\xrightarrow{Alc.\,\,KOH}\]       \[C{{H}_{2}}=CHBr\xrightarrow{NaN{{H}_{2}}}CH\equiv CH\]
NOTE: Elimination of \[HBr\]  from \[C{{H}_{2}}=CHBr\]    requires a stronger base because here, $$Br$$  acquires partial double bond character due to resonance.