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 .

Class 12 Chemistry in Class 12 3 years ago

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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`

Posted on 04 Dec 2021, this text provides information on Class 12 related to Chemistry in Class 12. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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