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)


1 Answer 1


Of course, just looking at methyl iodide and the partial charges, one would assume that there is no real reason for any nucleophile to attack the carbon atom for the reasons you stated. I’ll just point out and leave as an exercise to the reader that there is equally hardly any reason for the nucleophile to attack iodide, though.

However, the reaction also has a product side. Remember that what is happening is the following:

$$\ce{OH- + CH3-I -> HO-CH3 + I-}$$

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.

  • $\begingroup$ I would rate this answer even more highly if we were to connect the stability of the leaving group to the energy of the transition state. $\endgroup$
    – Zhe
    Nov 17, 2016 at 1:19

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