# Why does hydroxide attack the carbon atom of methyl iodide (in an Sn2 reaction)?

As the electronegativity difference between carbon and iodine is negligible, why does $\ce{OH-}$ attack the carbon atom (in an $\mathrm{S_N2}$ mechanism) even though carbon doesn't seem to be electrophilic? (Since the $\ce{C-I}$ bond is not polar enough to give a sufficiently positive partial charge on carbon)

$$\ce{OH- + CH3-I -> HO-CH3 + I-}$$
While the attack onto the only very slightly polarised carbon in itself may not be particularly beneficial, iodide is a very good leaving group as it is able to distribute the resulting negative charge well. (Large size, many electrons, small charge per volume ratio.) Likewise, if we consider thermodynamics, the $\ce{C-I}$ bond is not a particularly strong bond while a $\ce{C-O}$ bond is in the region of $\ce{C-C}$ or $\ce{C-H}$ bonds — so overall we are gaining a much stronger bond for a rather weak one.
Consider what had happened had oxygen attacked the iodide side of iodomethane. While oxidising iodide is something rather common in chemistry, the leaving group would have to be $\ce{CH3-}$ — a terribly non-stabilised leaving group, the conjugate base of an extremely weak acid; something that simply does not leave. Thus, there is no way hydroxide would attack the iodide side in an $\mathrm{S_N2}$ reaction — even aside from the reasons above left for the reader to figure out.