I ask this because SN2 requires a strong nucleophile, and I have read $\ce{OH-}$ isn't a strong nucleophile in a polar protic solvent like water. So, how can $\ce{OH-}$ replace, say $\ce{I-}$ from $\ce{CH3I}$ (methyl iodide)?
Answer
When you are considering a particular SN2 reaction, you have to consider the nucleophilicity as well as the nature of the leaving group of both the groups.
Here as you mentioned, $\ce{OH-}$ and $\ce{I-}$ (and other halide ions, except $\ce{F-}$, which is a very bad nucleophile) have comparable nucleophilicity in protic solvents. But, as for their leaving nature, $\ce{I-}$ (as well as other halides) is far more better leaving group than $\ce{OH-}$, because we know weak bases are better leaving groups. That is why the reaction is feasible.
No comments:
Post a Comment