Wikipedia lists this reaction for preparation of lithium aluminium hydride:
$\ce{4 LiH + AlCl_3 -> Li[AlH_4] + 3 LiCl}$

But is it possible to prepare it by reacting aluminium hydride with lithium chloride?
$\ce{Al_2H_6 + LiCl ->Li[AlH_4]}$
(I have not written all the products which may form)

If no, why?

  • $\begingroup$ First, at least try to enumerate the formal reaction, before asking if it is possible to do so. $\endgroup$
    – Poutnik
    Jan 22 at 9:59
  • $\begingroup$ Could you expand this comment into an answer if you are trying to hint that it won't be possible? I couldn't guess what products may form. $\endgroup$
    – Dodo
    Jan 22 at 17:53

1 Answer 1


First off, aluminum hydride is not a molecular compound. It has a macromolecular structure with empirical formula $\ce{AlH3}$. Since the compound is not molecular, the empirical formula is the one we use.

You should, indeed, include all the products in the proposed reaction in order to understand it properly. There must be a chloride product, and we can reasonably assume it's lithium chloroaluminate. Upon adding this product and balancing the equation the following is obtained:

$\ce{4AlH3 + 4LiCl -> 3LiAlH4 + LiAlCl4}$

The problem here is you are requiring the aluminum hydride to "disproportionate" its hydride ligands, with one mole surrendering all the hydride ligands to the other three moles. In other words, you require the same material to act as both a hydride ion donor and a hydride ion acceptor. This generally does not happen.

Instead, we need to be adding hydride ions from an external source that provides them more readily than aluminum hydride (whose metal atom is accepting the hydride ions). Lithium hydride is predominantly ionic (thus a good hydride ion source) and relatively easy to dissolve into polar organic solvents (which solvate lithium ions well). Hence the reaction between lithium hydride and aluminum chloride us the one to use.


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