In my Chemistry class, we were given this question to solve, and I'm not sure I fully understood the solution.

Reaction of N-ethyl-N,N,1-trimethylcyclohexan-1-aminium with hydroxide ion

I understand that since the substrate is bulky, Saytzeff elimination can't happen, so it has to be Hofmann. What I don't understand is how it happens here. Where does the base attack the substrate? Essentially, I'd like to know the mechanism of the reaction.

Also, our teacher made us write that had all three substituents on the nitrogen atom been methyls, Saytzeff elimination would have been followed as there wouldn't have been any other choice. However, even if all three substituents were methyls, wouldn't the substrate still be bulky? Why would it follow Saytzeff elimination?


Wow, this is an amazing question. The expected reactivity is strongly dependent on the exact structure.

For starters, trimethylammonium is about the size of a tert-butyl group. So, expect an A value around 4.9, which is very large. So, put that group equatorial on the ring.

Mechanisms of two possible eliminations from the preferred conformation

Once you do this, you see that there are only 2 hydrogen atoms that are anti-periplanar for an E2 elimination. One is on the axial methyl group (red). The other is on the N-ethyl group (blue). Only these two can easily be deprotonated for elimination. Of these two hydrogen atoms, the one on the axial methyl is very sterically crowded. Specifically, any base has to get around the ring structure and the encroaching flag pole hydrogen atoms. On the other hand, the hydrogen atom on the ethyl group sees no such steric encumberance.

Of the three elimination products, I expect to see:

  1. Almost exclusively ethylene
  2. Maybe a small amount of the methylidenecyclohexane
  3. No methylcyclohexene

EDIT: @Martin-マーチン was kind enough to compute the structure below. Notice the structural features we've already discussed: the ring conformation and the position of the anti-periplanar hydrogen atoms. Computed structure

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.