Lithium aluminium hydride is an important reducing agent in organic chemistry.

Since I cannot find the half-equations telling how this reducing agent performs its action, I am wondering, is lithium a dummy role in redox reactions involving the reducing agent? Or does it contribute to the stability of the compound?

So, are compounds like sodium aluminium hydride available to perform the same thing?

  • $\begingroup$ There is also sodium analogue, but cation makes some difference. It coordinates to hydrides in the compounds and also to lone pairs in org. compounds. $\endgroup$
    – Mithoron
    Sep 20 at 14:42
  • $\begingroup$ Sodium Aluminium hydride has different solubility characteristics to LiAlH4 which somewhat reduces its utility but it does similar reactions en.wikipedia.org/wiki/Sodium_aluminium_hydride $\endgroup$
    – Waylander
    Sep 20 at 17:02

1 Answer 1


Contrasting to reductions with elemental sodium (or other metals), the reduction by $\ce{LiAlH4}$ is not an electrochemical one by mere electron transfer only. Instead, the complex anion ($\ce{[AlH4]^-}$) acts e.g., on the $\ce{C=O}$ double bond of a carbonyl group as a $\sigma$-bond nucleophile to yield an alcoholate. The subsequent (light acidic) aqueous workup protonates to eventually yield an alcohol:

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Mechanistically similar is the reduction with $\ce{NaBH4}$ (and related compounds, e.g. $\ce{Na[BH3CN]}$, sodium cyanoborohydride) where again the reduction is not by sodium, but by the hydride.


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