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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) In `S_(N)1` reactions, the order of reactions, the order of reactivity depends upon the stability of the intermediate CARBOCATIONS, therefore, `(CH_(3))_(3)CB r` which gives a `3^(@)` carbocations, i.e., `(CH_(3))C^(+)` is the most reactive. `CH_(3)CH_(2)CH(Cl)CH_(3)` gives a `2^(@)` carbocation, i.e., `CH_(3)CH_(2)overset(+)(C)HCH_(3)` and hence is less reactive that `(CH_(3))_(3)CB r`. Out of the `CH_(3)CH_(2)CH_(2)CH_(2)^(+)` due to greater +I-effect of `(CH_(3))_(2)CHCH_(2)`-group w.r.t. `CH_(3)CH_(2)CH_(2)CH_(2)`-group and increasing reactivity of the four ISOMERIC bromobutanes towards `S_(N)1` reaction follows the order: `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) CB r` The reactivity in `S_(N)2` reactions, however, follows the reverse order, i.e., `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)CB r`. Since the steric hindrance around the electrophilic carbon (i.e., `ALPHA`-carbon) increase in the order (ii) Since the reactivity in `S_(N)1` reactions increases as the stability of the intermediate carbocations increases, therefore, `(C_(6)H_(5))C(CH_(3))(C_(6)H_(5))Br` which gives `3^(@)` carbocation, i.e., `(C_(6)H_(5))_(2)overset(+)(C)(CH_(3))` is the most reactive. Of the two `2^(@)` bromides, the carbocation intermediate derived from `C_(6)H_(5)overset(+)(C)HCH_(3)` obtained from `C_(6)H_(5)CH(CH_(3))Br` because it is stabilized by two phenyl groups due to resonance. the fourth alkyl BROMIDE, i.e., `C_(6)H_(5)CH_(2)Br` which gives the `1^(@)` carbocation, i.e., `C_(6)H_(5)CH_(2)^(+)` is, however, the least reactive. THUS, the overall reactivity of these alkyl bromides towards `S_(N)1` reactions follows the order: `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` In `S_(N)2` reactions, it is the steric hindrance which determines the reactivity. since a phenyl group is much bulkier than a methyl group, therefore, `C_(6)H_(5)C(CH_(3))(C_(6)H_(5))` Br is the least reactive followed by `C_(6)H_(5)CH(C_(6)H_(5))Br` and `C_(6)H_(5)CH(CH_(3))Br` while `C_(6)H_(5)CH_(2)Br` is the most reactive. thus, the overall reactivity of these alkyl bromides towards `S_(N)2` reactions follows the order: `C_(6)H_(5)CH_(2)Br gt C_(6)H_(5)CH(CH_(3))Br gt C_(6)H_(5)CH(C_(6)H_(5))Br gt C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br`. |
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