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Predict the order of reactivity of the following compounds in S_(N^(1)) and S_(N^(2)) reactions :(i) The four isomeric bromobutanes(ii) C_(6)H_(5)CH_(2)Br, C_(6)H_(5)CH(C_(6)H_(5))Br, C_(6)H_(5)CH(CH_(3))Br, C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br |
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Answer» Solution :(i)`CH_(3)CH_(2)CH_(2)CH_(2)BR lt (CH_(3))_(2)CHCH_(2)Br lt CH_(3)CH_(2)CH(Br)CH_(3)lt (CH_(3))_(3)CBr(S_(N^(1)))` `CH_(3)CH_(2)CH_(2)CH_(2)Br gt (CH_(3))_(2)CHCH_(2)Br gt CH_(3)CH_(2)CH(Br)CH_(3)gt (CH_(3))_(3)CBr(S_(N^(2))` Of the two primary bromides, the CARBOCATION intermediate derived from `(CH_(3))_(2) CHCH_(2)Br` is more stable than derived from `CH_(3)CH_(2)CH_(2)CH_(2)Br` because of greater electron donating inductive effect of `(CH_(3))_(2)CH-` group. Therefore, `(CH_(3))_(2)CHCH_(2)Br` is more reactive than `CH_(3)CH_(2)CH_(2)CH_(2)Br` in `S_(N^(1))` reactions. `CH_(3)CH_(2)CH(Br)CH_(3)` is a secondary bromide and `(CH_(3))_(3)C-Br` is a tertiary bromide. Hence the above order is followed in `S_(N^(1))`. The reactivity in `S_(N^(2))` reactions follows the reverse order as the steric hinderance AROUND the ELECTROPHILIC carbon increases in that order. (ii)`C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br gt C_(6)H_(5)CH(C_(6)H_(5))Br gt C_(6)H_(5)CH(CH_(3))Br gt C_(6)H_(5)CH_(2)Br(S_(N^(1))` `C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br lt C_(6)H_(5)CH(C_(6)H_(5))Br lt C_(6)H_(5)CH(CH_(3))Br lt C_(6)H_(5)CH_(2)Br (S_(N^(2)))` Of the two secondary bromides, the carbocation intermediate obtained from `C_(6)H_(5)CH(C_(6)H_(5))Br` is more stable than obtained from `C_(6)H_(5)CH(CH_(3))Br` because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in `S_(N^(1))` reactions. A phenyl group is BULKIER than a methyl group. Therefore, `C_(6)H_(5)CH(C_(6)H_(5))Br` is less reactive than `C_(6)H_(5)CH(CH_(3))Br` in `S_(N^(2))` reactions. |
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