# How does Hofmann elimination work in this reaction?

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

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.

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.

Here in this compound, you have a positive formal charge on $\ce{N}$ atom. When you add $\ce{OH^-}$ to the compound, it just gets attached near the positive charge on $\ce{N}$, so an ionic interaction just exists between them, but the $\ce{OH^-}$ still remains labile.

So, after this addition of $\ce{OH^-}$, if you heat the compound, then β-elimination (syn) occurs. That hydrogen is abstracted by $\ce{OH^-}$ (which acts as a base in this case), which is β to that nitrogen atom. So, in this compound there are three choices (without looking into the stereochemistry of the compound) for β-$\ce{H^+}$ abstraction:

1. Abstraction from the ethyl group which is attached to the nitrogen produces ethylene, and
2. abstraction from the cyclohexane ring produces 1-methylcyclohexene, and
3. abstraction from the methyl group attached to cyclohexane produces methyledene cyclohexane.

But note that the methylidene cyclohexane formation is not totally a syn-elimination. So, that elimination product is not going to be a major product.

Thus, two products are possible, ethylene and 1-methylcyclohexene, out of which ethylene is a very less substituted alkene, and is hence favoured in Hoffmann Elimination. But the other one, i.e. 1-methyl cyclohexene is more substituted product and minor in the elimination. Ethylene is the major product and 1-methylcyclohexene is the minor product.

A possible mechanism for the products formed by this Hoffmann elimination (syn) is as follows:

• @Soumik There's a third product - methylidene cyclohexane. Why isn't this compound the major product? – FreakyLearner Apr 5 '18 at 9:29
• Hofmann and Saytzeff are just rules, not Laws, for the E2 elimination of TRIMETHYL ammonium salts. Clearly, the rate of the ethylene-forming is faster than the othet two options. – user55119 Apr 5 '18 at 11:59
• @Soumik The methyl attached to cyclohexane has 3 hydrogen atoms. I think any one of the H atom can always be syn to the N atom. Then why is it "not totally a syn elimination"? – FreakyLearner Apr 5 '18 at 14:10
• In that case also, product will be more substituted and minor. – Soumik Das Apr 5 '18 at 14:18
• There's no good reason why the bulky N group would adopt an axial conformation – it's at least as big as a t-Bu group, locking the conformation. – NotEvans. Aug 4 '18 at 9:57