InterviewSolution
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InterviewSolution
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Which of the compound in each of the following pairs will react faster in `SN^2` reaction with `overset (Ө) O H` (A) `MeBr (I) and MeI (II)` (B) `Me_3 C - Cl (III) and MeCl (IV)` .A. `overset ((A)) ((I)) overset ((B)) ((III)) overset ((C))((V))`B. `overset ((A)) ((I)) overset ((B)) ((IV)) overset ((C))((V))`C. `overset ((A)) ((II)) overset ((B)) ((III)) overset ((C))((VI))`D. `overset ((A)) ((II)) overset ((B)) ((IV)) overset ((C))((VI))` |
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Answer» Correct Answer - D `(A) (II), I^(Ө)` is a better leaving group than `Br^(Ө)`. `(B)(V),I^@ RX` undergoes `SN^2` reaction than `3^@ RX` `(A) (VI)`, Vinyl halide `(V)` does not undergo `SN^1` or `SN^2` reaction , `(VI)` is `1^@ RX`, therefore undergoes `SN^2`. |
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