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Analyze the generalized rate data: RX + M^(ɵ) rarr Product |{:("Experiment",[RX] "Substrate",[M^(ɵ)] "Attaking species","Rate"),(I,0.10 M,0.10 M,1.2 xx 10^(-4)),(II,0.20 M,0.10 M,2.4 xx 10^(-4)),(III,0.10 M,0.20 M,2.4 xx 10^(-4)),(IV,0.20 M,0.20 M,4.8 xx 10^(-4)):}| For the reaction under conisderation, 3^(@) alkyl has been found to be the most favourable alkyl group. Which of the following attacking species (M^(ɵ)) will give the best yield in the reaction ? |
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Answer» `(CH_(3))_(2)CH-O^(ɵ)` The reaction under conisderation, `3^(@)RX` is favourable. So reaction Would be `E_(2)` not `S_(N)2`. `3^(@)RX gt 2^(@)RX gt 1^(@)RX RARR Favours E_(2)` `1^(@)RX gt 2^(@)RX gt 3^(@)RX rArr Favours S_(N)2`. Hence `E_(2)` is favoured by strong bulkyl bronsted base. Acidic order: `H_(2)O gt C_(2)H_(5)OH gt (CH_(3))_(2)CH-OH gt (CH_(3))_(3) C-OH`. Baisc order: `overset(ɵ)(OH) lt C_(2)H_(5)O^(ɵ) lt (CH_(3))_(2) CH-O^(ɵ) lt (CH_(3))_(3)C-O^(ɵ)` Therefore, stronger base `(CH_(3))_(3)C-O^(ɵ)` will GIVE the best yield of the reaction. Hence the answer is (b). Note: If the rreaction proceeds via `S_(N)2` mechanism, then `(M^(ɵ))` acts as nucleophilie. Since all of the species have same nuclepohile centre, so stronger the base stronger will be the nucleophile. `:.` Baisc order and nucleophile order should be: `(CH_(3))_(3)C-O^(ɵ) gt (CH_(3))_(2)CH - O^(ɵ) gt C_(2)H_(5)O^(ɵ) gt overset(ɵ)(OH)` So `(CH_(3))_(3)C-O^(ɵ)` should act as a stronger nucleophile. But the actual order of nucleophile is different. `:.`Nucleophile order: `C_(2)H_(5)O^(ɵ) gt (CH_(3))_(2)CH-O^(ɵ)gt(CH_(3))_(3)C-O^(ɵ) gt overset(ɵ)(OH)` This isbecause of its bulkiness, `(CH_(3))_(3)C-O^(ɵ)` is a poorer nucleophile because of steric hindrance, baiscity, and nucleophilicity may diverge. Thus, if `RX` would have been `1^(@)RX`, then `S_(N)2` reaction would have taken placed and attacking species `(C_(2)H_(5)O(ɵ))` would gove the best yield of the reaction. |
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