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Arrange the following compounds in the decreasing order of reactivity towards S_(N)2 displacement reaction and give reasons in support of your answer: (a) C_(2)H_(5)Br , C_(2)H_(5)I, C_(2)H_(5)Cl (b) (CH_(3))_(3)CBr , CH_(3)CH_(2)CHBrCH_(3), CH_(3)CH_(2)CH_(2)CH_(2)Br . |
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Answer» Solution :(a) `C-I ` I bond is weaker than C - Br bond which is weaker than C-Cl bond. The second step in `S_(N)2`REACTION is the BREAKING of carbon-halogen bond. Weaker the bond, easier is it to break it and greater will be the reactivity of the `S_(N)2`reaction. Hence, the decreasing order of reactivity towards `S_(N)2`displacement reaction will be `C_(2)H_(5)I gt C_(2)H_(5)Br gt C_(2)H_(5)Cl ` (b) `underset("Tertiary halide") ((CH_(3))_(3)CBr)"" underset("SECONDARY halide") (CH_(3)CH_(2)CHBrCH_(3)) "" underset("PRIMARY halide")(CH_(3)CH_(2)CH_(2)Br) ` Primary halide Primary halides offer the least steric HINDERANCE to the attachment of the nucleophile to the carbon atom, followed by secondary halide followed by tertiary halide. Hence, the decreasing order of reactivity towards `S_(N)2`reaction will be: `CH_(3)CH_(2)CH_(2)CH_(2)Br gt CH_(3)CH_(2)CHBrCH_(3) gt (CH_(3))_(3)CBr` |
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