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Which compound in each of the following pairs will react faster by S_(N)1 reactions in aqueous KOH. Give reasons. (a) CH_(3)CH_(2)-Br or CH_(3)CH_(2)-Cl (b) CH_(3)CH_(2)CH_(2)CH_(2)-X or (CH_(3))_(3)C-X |
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Answer» Solution :(a) `S_(N)1` reactions occur by carbocation intermediates. Since, `CH_(3)CH_(2)Br` or `CH_(3)CH_(2)Cl` give the same carbocation, `CH_(3)CH_(2)^(+)`, therefore, stability of the carbocation does not affect the rate of the REACTION. Further, since, `Br^(-)` ion is a BETTER leaving group than `Cl^(-)` ion, therefore, `CH_(3)CH_(2)-Br` will react faster than `CH_(3)CH_(2)-Cl` with aq. KOH by `S_(N)1` mechanism. (b). `S_(N)1` reactions occur by carbocation intermediates  Since, `3^(@)` carbocation in more stable than `1^(@)` carbocation, therefore, `(CH_(3))_(3)C-X` will react faster than `CH_(3)CH_(2)CH_(2)CH_(2)-X` with aq. KOH by `S_(N)1` mechanism. |
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