# How can aqueous KOH give alcohol by SN2 of alkyl halide?

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)?

• Even though its weaker in protic solvents, I won't say its not a strong nucleophile. – Sawarnik May 30 '17 at 4:34
• With methyl iodide there is no opportunity for E2 elimination. Unhindered primary halides (Cl,Br,I) favor SN2 over E2. – user55119 Mar 30 '18 at 21:23

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.