When $\ce{CH3CH2OH}$ reacts with $\ce{NaH}$, why are the products $\ce{CH3CH2ONa + H2}$? Why not $\ce{CH3CH3 + NaOH}$? Can you please explain in detail?
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1$\begingroup$ Think about the mechanism. $\endgroup$ – bon May 16 '16 at 7:40
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$\begingroup$ chemistry.stackexchange.com/questions/30982/… $\endgroup$ – Mithoron May 16 '16 at 11:00
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$\begingroup$ @bon Shouldn't Na combine better with OH-? (I sound dumb I guess, but I need a bit of a detailed explanation :)) $\endgroup$ – Ram Bharadwaj May 17 '16 at 15:54
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There are a number of reasons why the reaction you suggest in much less favourable than the deprotonation. Firstly, hydride is a terrible nucleophile because its 1s orbital is too small to effectively overlap with anything except hydrogen.
Secondly, hydroxide is a poor leaving group for an $S_N2$ reaction and so you need a very good nucleophile in order to displace hydroxide from an alcohol group.
Thirdly, the reaction of sodium hydride as a base often produces a quantitative deprotonation because the product, $\ce{H2}$ is lost as a gas, which drives the reaction forward.
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Consider firstly the ease of reaction: the hydrogen in $\ce{NaH}$ is basic $\ce{H-}$ and the hydrogen in ethanol is acidic $\ce{H+}$. The easiest reaction then, is a simple acid-base neutralization.
The reaction products you propose requires that a $\ce{C-O}$ bond is broken, or a situation wherein which the hydride attacks a carbon center which then ejects a hydroxide ion $\ce{OH-}$. In reality, however, you would have to eject $\ce{O^{2-}}$ which is much harder to remove.
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$\begingroup$ @RamBharadwaj if you draw out the mechanism (assuming $S_N2$) it will look like $O^-$ is the leaving group, so after leaving it will be $O^{2-}$ because you would start out with ethoxide rather than ethanol. $\endgroup$ – IT Tsoi May 18 '16 at 1:02
