Discussion
Reaction of LiAlH4 with one equivalent of Me3N·HCl gives a tetrahedral compound, which reacts with another equivalent of Me3N·HCl to give compound N. The compound N and its geometry, respectively, are:
Answer: C
N is AlH3(NMe3)2, trigonal bipyramidal.
Me3N·HCl delivers HCl: the first equivalent converts LiAlH4 into LiCl, H2 and AlH3, which is captured by trimethylamine as the tetrahedral adduct AlH3·NMe3. The second equivalent supplies another amine ligand, and aluminium expands to five-coordination: AlH3(NMe3)2.
With five electron pairs and no lone pair, the shape is trigonal bipyramidal, the two bulky amines axial and the three hydrides equatorial. VSEPR gives no pentagonal arrangement for five bonding pairs, and LiAlH4NMe3 or Li2AlH4Cl would not form once HCl has destroyed the AlH4− ion.
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