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)?

  • $\begingroup$ Even though its weaker in protic solvents, I won't say its not a strong nucleophile. $\endgroup$
    – Sawarnik
    May 30, 2017 at 4:34
  • $\begingroup$ With methyl iodide there is no opportunity for E2 elimination. Unhindered primary halides (Cl,Br,I) favor SN2 over E2. $\endgroup$
    – user55119
    Mar 30, 2018 at 21:23

1 Answer 1


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.


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