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The problem is such: enter image description here

I know that the t-Bu locks the ring in equatorial position as I have drawn out: enter image description here Could someone please point me in the right direction as how I can get rid of a Me in the NMe3+ group?

Is it possible that tBuO- is acting as a nucleophile in this case?

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  • $\begingroup$ t-butoxide can certainly act as a nucleophile, but it would much prefer to do an elimination. Why do you think the elimination in this case to form a cyclohexene is not favoured? What do you need for an elimination reaction (specifically an E2 reaction) to occur? $\endgroup$ – orthocresol♦ Dec 30 '15 at 14:25
  • $\begingroup$ @orthocresol we need the NMe3+ to be axial and anti to the H removed which it is not $\endgroup$ – justbehappy Dec 30 '15 at 14:31
  • $\begingroup$ Due to steric interactions, the bulky base – triethylamine cannot readily abstract the proton. In these situations, a less sterically hindered proton is preferentially abstracted instead. As a result, the Hofmann product is typically favored when using bulky bases. $\endgroup$ – Sujith Sizon Dec 30 '15 at 15:10
  • $\begingroup$ @SujithSizon there is no lone pair on N in the quaternary ammonium salt, there is no elimination going on in this reaction. $\endgroup$ – orthocresol♦ Dec 30 '15 at 15:39
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As you point out, the tert-butyl group wants to remain in an equatorial position, else both the tert-butyl group and the trimethylammonium group end up being axial which is disfavoured for two bulky groups.

The obvious role of tert-butoxide:

The tert-butoxide would obviously rather act as a base than a nucleophile (in-fact under ordinary circumstances we would say that tert-butoxide is a non-nucleophilic base in the same way we refer to LDA).

In order for the tert-butoxide to act as a base, and do the elimination, the conformation of the ring would first have to put both groups axial, which we have already said isn't favourable.

There is an additional reason why this is unable to happen however, since the protons that you need to abstract to get elimination are 1,3-diaxial to the tert-butyl group. This would be highly sterically disfavoured and so is unlikely to happen.

The less obvious role of tert-butoxide:

Since the tert-butoxide can't act as a base, there is the possibility of it attacking one of the methyl groups on the trimethylammonium group to give the product as drawn in an SN2 type reaction. Whilst this seems a little odd, it does have the advantage of neutralising the charges on both the tert-butoxide and the trimethylammonium group to give two neutral molecules.

enter image description here

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