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Which alkyl halide from the following pairs would you expect to react more rapidly by S_(N)2 mechanism? Explain your answer. (i) CH_(3)CH_(2)CH_(2)CH_(2)Br or CH_(3)CH_(2)-underset(Br)underset(|)(C)H-CH_(3) (ii) CH_(3)CH_(2)-underset(Br)underset(|)(C)H-CH_(3) or H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-Br (iii) CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)CH_(2)Br or CH_(3)CH_(2)-underset(CH_(3))underset(|)(C)H-CH_(2)Br |
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Answer» Solution :(i) `CH_(3)CH_(2)CH_(2)CH_(2)Br` is a `1^(@)` ALKYL halide while `CH_(3)CH_(2)-CHBr-CH_(3)` is a `2^(@)` alkyl halide. Since there will be some steric hindrance `2^(@)` alkyl halides than in `1^(@)` alkyl halides, therefore, `CH_(3)CH_(2)CH_(2)CH_(2)Br` will react faster than `CH_(3)CH_(2)-CHBr-CH_(3)` in `S_(N)2` reaction. (ii) `CH_(3)CH_(2)-CHBr-CH_(3)` is a `2^(@)` alkyl halie while `(CH_(3))_(3)C-Br` is `3^(@)` alkyl halide. since due to lesser steric hindrance, `2^(@)` alkyl halides react faster than `3^(@)` alkyl halides, therefore, `CH_(3)CH_(2)-CHBr-CH_(3)` will react faster than `(CH_(3))_(3)CBR` in `S_(N)2` reaction. (iii) `overset(4)(C)H_(3)-underset(underset(I)(C)H_(3))underset(|)overset(3)(C)H-overset(2)(C)H_(2)overset(1)(C)H_(2)Br""overset(4)(C)H_(3)overset(3)(C)H_(2)-underset(underset(II)(C)H_(3))underset(|)overset(2)(C)H-overset(1)(C)H_(2)Br` Both I and II are `2^(@)` alkyl halides. but in alkyl halide (II), the `CH_(3)` group at `C_(2)` is closer to the Br ATOM while in alkyl hlaide (I), the `CH_(3)` group at `C_(3)` is little away from the Br atom. as a result, alkyl halide (I) will suffer GREATER steric hindrance than alkyl halide (I). therefore, `(CH_(3))_(2)CHCH_(2)Br` will react faster than `CH_(3)CH_(2)CH(CH_(3))CH_(2)Br` in `S_(N)2` reaction. |
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