question_answer

Arrange the following halides in the decreasing order of \[{{S}_{N}}1\] reactivity. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl\,(I),\,C{{H}_{2}}=CHCHClC{{H}_{3}}(II)\]\[and\,\,C{{H}_{3}}C{{H}_{2}}CHClC{{H}_{3}}(III)\]

Answer:

                (i) In \[{{S}_{N}}1\] reactions, carbocations are the intermediates. Obviously more stable the carbocation, more reactive is the alkyl halide. Since alkyl halide \[(II)\] gives an allylic carbocation which is stabilized by resonance, therefore, alkyl halide \[(II)\] is the most reactive. (ii) Out of alkyl halides \[(I)\] and \[(III)\], \[(III)\]gives a more stable \[2{}^\circ \] carbocation while \[(I)\] gives a less stable \[1{}^\circ \]carbocation, therefore, alkyl halide \[(III)\] is more reactive than alkyl halide\[(I)\] . \[\underset{\text{III}}{\mathop{C{{H}_{3}}C{{H}_{2}}-CHCl-C{{H}_{3}}\xrightarrow{\text{Ionization}}}}\,\]\[\underset{2{}^\circ \,Butyl\,carbocation\,\,(more\,stable)}{\mathop{C{{H}_{3}}C{{H}_{2}}-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}+C{{l}^{-}}}}\,\] \[\underset{I}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl\xrightarrow{\text{Ionization}}}}\,\]\[\underset{{{\text{1}}^{\text{o}}}\,\text{Propyl}\,\,\text{carbocation}\,\,\text{(less}\,\text{stable)}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-\overset{+}{\mathop{C}}\,{{H}_{2}}+C{{l}^{-}}}}\,\] Thus, from the above discussion, it follows that the overall reactivity in \[{{S}_{N}}1\]reaction follows the sequence : \[II>III>I.\]