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A TA talked about this today in the context of ketones.

If you stick an acetone and $\ce{NaBH4}$ together in a protic solvent, you get 2-propanol.

If you stick acetone and $\ce{NaH}$ together, you get an enolate. What's the difference? They're both hydrides.

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First of all, borohydride as a group is not same as hydride. So the mechanisms are very different.

Because a simple $\ce{H-}$ anion is more basic than an enolate $\ce{C=C-O^{-}}$(product of $\alpha$ hydrogen deprotonation of a ketone), it just deprotonate before reduction will happen. Deprotonated ketones (enolates) will no longer be a substrate for reduction.

Borohydride $\ce{BH4-}$ is not as basic as enolate, so deprotonation hardly happen. Thus a reduction will be the predominant reaction.

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To complement the answer provided by @Ian Fang and edited by @santiago, sodium hydride's action as a Brønsted base differs from other bases.

"Normal" acid base reactions are reactions of an equilibrium, like the partial deprotonation of malonic esters by 5 m% sodium carbonate in some Knoevenagel reactions:

$\ce{R-H} + \mbox{base} \rightleftharpoons{} \ce{R-} + \mbox{protonated base}$

Deploying sodium hydride as base however often intends a quantitative deprotonation. The equilibrium is driven to the right hand side, in favour of the protonated base ($\ce{H2}$) immediately leaving the reaction mixture -- so the back reaction can't take place.

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  • $\begingroup$ The solvents are different for the two reagents. NaBH4 uses protic solvents while NaH requires aprotic solvents. This raises a question. Does NaH reduce benzophenone to the alkoxide or is a radical anion formed which couples along with the liberation of hydrogen? Or do they not react at all? $\endgroup$ – user55119 Mar 4 '18 at 3:37

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