question_answer

\[{{C}_{6}}{{H}_{5}}CHClC{{H}_{3}}\] is hydrolysed more easily with KOH than \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl\] . Explain.

Answer:

                Both \[{{C}_{6}}{{H}_{5}}CHClC{{H}_{3}}\]and \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl\]are benzylic halides and hence undergo hydrolysis in aqueous KOH by \[{{S}_{N}}1\] mechanism. But in \[{{S}_{N}}1\]reactions, \[{{C}_{6}}{{H}_{5}}CHClC{{H}_{3}}\xrightarrow{\text{Ionization}}\,\,\underset{\begin{smallmatrix}  {{\text{2}}^{\text{o}}}\,\text{Carbocation}\,\text{(I)} \\  \,\,\,\,\,\,\,\text{(more}\,\,\text{stable)} \end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}+C{{l}^{-}}}}\,\] \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl\xrightarrow{\text{Ionization}}\,\,\underset{\begin{smallmatrix}  {{1}^{\text{o}}}\,\text{Carbocation}\,\text{(II)} \\  \,\,\,\,\,\,\,\text{(less}\,\,\text{stable)} \end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}-\overset{+}{\mathop{C}}\,{{H}_{2}}+C{{l}^{-}}}}\,\] reactivity depends upon the stability of the intermediate carbocations. Now \[{{C}_{6}}{{H}_{5}}-CHClC{{H}_{3}}\] on ionization gives a \[2{}^\circ \] carbocation, \[{{C}_{6}}{{H}_{5}}\overset{+}{\mathop{C}}\,HC{{H}_{3}}(I)\] which is stabilized both by + R-effect of the \[{{C}_{6}}{{H}_{5}}\] group and +1-effect of the \[C{{H}_{3}}\] group but \[{{C}_{6}}{{H}_{5}}C{{H}_{2}}Cl\] on ionization gives a \[1{}^\circ \] carbocation \[{{C}_{6}}{{H}_{5}}\overset{+}{\mathop{C}}\,{{H}_{2}}(II)\] which is only stabilized by +R-effect of the \[{{C}_{6}}{{H}_{5}}\overset{+}{\mathop{C}}\,{{H}_{2}}\]. Since carbocation \[(I)\] is more stable than carbocation\[(II)\], therefore, \[{{C}_{6}}{{H}_{5}}CHClC{{H}_{3}}\] undergoes hydrolysis more easily than \[{{C}_{6}}{{H}_{5}}\,C{{H}_{2}}Cl\].