Lets say there is $\ce{CH3-OH + NaH}$. The reaction leads to $\ce{CH3-O- Na+ + H2}$. However $\ce{H-}$ from the metal hydroxide is also a strong nucleophile, so why can't the reaction lead to $\ce{CH4 + OH-}$? I realize that $\ce{OH-}$ is a bad leaving group, but then again, an alkoxide isn't a good leaving group either. An alkoxide's $\mathrm{p}K_\mathrm{a}$ is even higher than $\ce{OH-}$, so its a worse leaving group.

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    – Jan
    Nov 13 '15 at 19:14
  • $\begingroup$ Also, hydride is a pretty bad nucleophile. Much better base than nucleophile. $\endgroup$
    – Jan
    Nov 13 '15 at 19:15
  • $\begingroup$ @Jan This answers my question. And thanks for the edit! $\endgroup$
    – Abid Rizvi
    Nov 13 '15 at 19:23

In this simple case it is pretty obvious that a hydride ion is a much better base than it is a nucleophile. But that doesn’t answer everything; we could consider cases with much better nucleophiles ($\ce{Br-}$).

Another reason is that hydroxide is an even lousier leaving group than hydride is a nucleophile. It is literally impossible to do $\mathrm{S_N}$ displacements of uncharged oxygen atoms unless you are opening an epoxide (in which case you lose ring strain making the overall process favourable. It does not matter whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$; $\ce{RO-}$ will never leave. To remove an oxygen atom, you always need to use strong acids that can protonate it, giving the far better leaving group $\ce{ROH+-R}$ which gives the free alcohol/water when displaced. Thus, the following reaction works well due to the very strong hydroiodic acid (I tried it):

$$\ce{R-CH2-OH + HI -> R-CH2-OH2+ + I- -> R-CH2-I + H2O}$$


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