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Which alkyl halide from the following pairs would you expect to react more rapidly by S_(N^(2)) mechanism? . |
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Answer» Solution :If the leaving group is the same in different isomers of a particular molecular formula, the reactivity of the isomers towards `S_(N^(2))` mechanism decrases with the increase in steric hindrance. In the light of above, the reactivity order in different cases is: (i). `CH_(3)CH_(2)CH_(2)CH_(2)Br` is a primary alkyl HALIDE `(1^(@))`. it is more reactive than the other isomer which is a secondary `(2^(@))` alkyl halide becausae less steric hindrance is caused by primary alkyl group as COMPARED to secondary alkyl group. (ii). `CH_(3)CH_(2)underset(Br)underset(|)(C)HCH_(3)` is a secondary alkyl halide `(2^(@))`. it is more reactive than the other isomer which is a tertiary alkyl halide `(3^(@))` the explanation is the same. (III). Here both the isomers are primary alkyl halides `(1^(@))`. However, the isomer with `CH_(3)` group at `C_(2)` atom exerts more steric hindrance to the attacking nucleophile at `C_(1)` atom as compared TOT he other isomer in which a `CH_(3)` group is attached to `C_(3)` atom. it is , therefore, less reactive.  .
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