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What is more nucleophilic, $\ce{OH-}$ or $\ce{CH3O-}$? When I look it from the viewpoint of steric hindrance, the answer seems to be $\ce{OH-}$, but when I see it from the viewpoint of extra electron density provided by hyperconjugation of $\ce{CH3}$, the answer seems to be $\ce{CH3O-}$.

bon
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user247855
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3 Answers3

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Alkoxides tend to be stronger bases than the hydroxide ion. Adjacent atoms in an alkane chain will donate electron density to oxygen, allowing it to be more polarizable. An increase in polarizability will allow oxygen to more easily donate electrons.

The oxygen in a hydroxide ion does not experience the inductive effect. Therefore its valence electrons do not "reach" out as far.

Methoxide is a relatively unhindered base, as is hydroxide. But as explained above, methoxide experiences induction whereas hydroxide does not. So one would expect methoxide to be a better nucleophile than hydroxide because it is a stronger base than hydroxide and still remains unhindered.

T. Kent
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When negative charge is present on same type of atom in different nucleophiles, the order of nucleophilicity is determined by taking the strength of their conjugate acids. If the acid is strong its nucleophilicity is less. $\ce{CH3OH}$ is a slightly more acidic ($\pu{pKa = 15.5}$) than $\ce{H2O}$ ($\pu{pka = 15.74}$). Therefore hydroxide ion is slightly more nucleophilicity than methoxide ion

airhuff
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My answer would be same as the above mentioned. I would just add that of the reduction in nucleophilic character of methoxide due to steric factors is 10 then an increase due to inductive factor is 40 which makes a net +30 better than hydroxide. Note- above values have no physical significance they are just made up to explain the comparative difference.

  • 10 what? As you said, these numbers are meaningless and come from nowhere - so they are not useful as an explanation. – bon Mar 28 '16 at 12:30