Product of reaction between aluminium chloride and trimethylamine [closed]








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Can anyone tell the product of the reaction between $\ce{AlCl3}$ (aluminium chloride) and $\ce{N(CH3)3}$ (trimethylamine)?




Here are a few tips to get you started:


1. Lewis acids are electron pair acceptors; Lewis bases are electron pair donors.
2. This looks like a Lewis acid-base reaction. You mention that this *is* an acid-base reaction but there are no easily accessible protons in either reactant. Aluminum trichloride has **no** protons; trimethylamine has does have protons but these are $\ce{sp^3~C-H}$ type protons - which are extremely weak. Also, there is no path to stability; deprotonation of one of the methyl groups leaves a *localized* negative charge on a carbon.


There are other heuristics one can use. For example, note that aluminum trichloride has an electron-deficient aluminum atom (the Al doesn't have an octet). Acceptance of an electron pair *will* provide it with an octet. This heuristic works fairly well; consider boron trichloride for example. Boron is similarly electron-deficient. And boron trichloride is also a Lewis acid. 


Also note that the amine has a lone pair. Keeping in mind that Lewis acid-base reactions are about the movement of electrons (rather than protons) then we might reasonably suspect this to be a Lewis acid-base reaction. Lone pairs generally indicate Lewis basicity, as these are generally available for donation to a Lewis acid. Delocalization of lone pairs can weaken the lone pair's Lewis basicity. Thus, the presence of a lone pair only serves as an indicator of Lewis basicity rather than the *strength* of Lewis basicity. And of course you can also donate bonding electrons rather than lone pairs (halogenation of ethene would be an example; the $\ce{C=C}$ double bond is broken and the electrons previously involved in that double bond are put into the antibonding orbital of some diatomic halogen)


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