Electrophilic Aromatic Substitution (Haloalkanes and Haloarenes) MCQ Questions & Answers in Organic Chemistry | Chemistry

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In (i) is major product while is minor product according to Saytzeff's rule.

Small difference in boiling points of $${C_2}{F_6}$$  and $${C_2}{H_6}$$  is due to the fact that (i) the $$F$$  atom is slightly larger than $$H,$$ and (ii) $$F$$  has low polarizability

\[\underset{\text{2-Bromo -3, 3-dimethylbutane}}{\mathop{{{H}_{3}}C\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,\underset{\begin{smallmatrix} |\,\,\,\,\,\,\,\,\,\, \\ Br\,\,\,\,\,\,\,\,\, \end{smallmatrix}}{\mathop{CH-}}\,C{{H}_{3}}}}\,\xrightarrow{aq.\,KOH}\]       \[\underset{\text{2-3-Dimethylbutan-2-o1}}{\mathop{{{H}_{3}}C-\overset{\begin{smallmatrix} C{{H}_{3}}\,\,\, \\ |\,\,\,\,\,\,\,\,\,\, \end{smallmatrix}}{\mathop{CH\,\,-}}\,\underset{\begin{smallmatrix} |\,\,\,\,\,\,\, \\ OH\,\,\,\, \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,C{{H}_{3}} \\ |\,\,\,\,\, \end{smallmatrix}}{\mathop{C\,\,-}}}\,C{{H}_{3}}}}\,\]

$$Cl$$  in 2, 4, 6-trinitrochlorobenzene is activated by three $$N{O_2}$$  groups at $$o,$$ and $$p$$ - positions and hence undergoes hydrolysis most readily.

\[\begin{align} & {{C}_{2}}{{H}_{5}}I\xrightarrow{alc.\,KOH}C{{H}_{2}}=C{{H}_{2}}\xrightarrow{B{{r}_{2}}} \\ & BrC{{H}_{2}}-C{{H}_{2}}Br\xrightarrow{KCN}CNC{{H}_{2}}.C{{H}_{2}}CN \\ \end{align}\]

Unstable $${C^ + }$$ has general tendency to rearrange to more stable $${C^ + }.$$

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In the first case the reaction gives side chain substitution product which is easier in $$A.$$  In the second case the reaction will proceed by benzyne mechanism for which ortho position $$w. r.$$  to $$Cl$$  must have $$H$$ - atoms.

$$\eqalign{ & {\text{Rate of}} \cr & {S_N}2 \propto \frac{1}{{{\text{Steric crowding near to reaction centre}}}}. \cr} $$