mechreview_key

mechreview_key - OWLS: Mechanism Review Solutions 1. When...

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OWLS: Mechanism Review Solutions 1. When considering S N 2 versus S N 1, we consider S N 2 first because it involves the lower energy (faster) rate-determining step. If the S N 2 requirements are reasonably met, we predict the reaction proceeds via the S N 2 pathway. If the S N 2 requirements are not met then we consider the S N 1 mechanism. (a) S N 2 requirements: Moderate or better leaving group: Iodide ion is a superior leaving group due to its large atomic radius. Moderate or better nucleophile: Cyanide is a good nucleophile due to the small atomic radius of carbon and the formal negative charge. Methanol is a protic solvent and thus will decrease nucleophilicity, but probably not so much as to prevent the S N 2 mechanism. Not tertiary: The carbon bearing the leaving group is primary and thus not too sterically hindered for the nucleophile to approach. Solvent: An S N 2 reeaction between a negatively charged nucleophile and a neutral electrophile is fastest in a nonpolar, aprotic solvent. Methanol is moderately polar ( ε = 33) and protic. This disfavors, but does not necessarily prevent, this S N 2 reaction. Conclusion: The formation of product A is predicted to occur by the S N 2 mechanism. I C N C N (b) That the carbon bearing the leaving group does not end up bonded to the nucleophile suggests this cannot be an S N 2 reaction. Therefore we consider the S N 1 mechanism. S N 1 requirements: Moderate or better leaving group: Iodide ion is a superior leaving group due to its large atomic radius. Stable carbocation: The carbocation formed by ionization of the carbon-
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mechreview_key - OWLS: Mechanism Review Solutions 1. When...

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