# E2 Elimination; Anti-periplanar beta hydrogen

I'm confused by this picture I found online on a very popular organic chemistry website:

I agree that the product is a product.

However, can't there be another product? Specifically, can't the methyl group by the bromine have an anti and beta hydrogen relative to the bromine?

So shouldn't there be two products?

Specifically, this one?

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For some odd reason, introduction of $sp^2$ center into six- and five- rings is strongly unfavored. I never cared much to find the exact reason, but guess that single $sp^2$ center induces more strain than two of them. ] –  permeakra Jul 30 '14 at 21:09
Would that have to do with angle strain? I mean in any case the author of the graphic says that a double bond in a 6-membered ring is a "major" product; I don't see how what I drew can't also be a "major" or at least "minor" product; my product only introduces one sp2 carbon into a ring. –  Dissenter Jul 30 '14 at 21:17

## 3 Answers

A reasonable chemist would be on the lookout for two products, especially given that E2 elimination from 1-bromo-1-methylcyclohexane produces some methylenecyclohexane

In the transition state for the E2 elimination the pi bond is already partially developed. This means that the relative thermodynamic stabilities of the products will play a role in determining the product distribution. Since more highly substituted olefins are thermodynamically more stable than less substituted olefins, one would predict that the product with the double bond internal to the ring would predominate over the analogue with the external double bond.

Reaction temperature, base size, steric constraints around the various hydrogens that might be involved and the numbers of the various hydrogens all factor into the final product ratio, but these effects are usually secondary unless you change circumstances so much that the reaction mechanism changes and shifts away from E2.

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So the picture I posted is incorrect? –  Dissenter Jul 30 '14 at 23:03
Your picture is correct, I would have expected some of that product based on the methylenecyclohexane example I provided in my post, albeit I would have expected it to be the minor product. Despite our predictions, according to the link you provided, the isomer with the exo-methylene group apparently doesn't form. That's why we make predictions and then run experiments. –  ron Jul 30 '14 at 23:17
Oops I meant the first one - the one from the site. I think we can both agree with the site that the non-exocyclic double bond containing molecule is probably not the "only product!" At least from our predictions. Perhaps the author of the site actually ran the experiment and found that it was the "only product!" –  Dissenter Jul 30 '14 at 23:20
Yes, that's what I was referring to - shouldn't have called it "your" picture. I assume it is correct - that is what the link clearly says. That's why I said, given my example I was surprised by their result, but I assume it is correct. –  ron Jul 30 '14 at 23:25
So would I. I'd add that I expected it to be the minor isomer. –  ron Jul 30 '14 at 23:35

If the elimination occurs via an ${\mathrm{E_2}}$ mechanism, the other endocyclic double bond (i.e., the one between the two methyl bearing carbons) cannot be formed, as the $\ce{C_2}$ $\beta$-hydrogen will not be anti-periplanar to the $\ce{Br-}$ leaving group. That said, the other possible elimination product you've identified, with the exocyclic double bond, is entirely possible. However, this would be the Hofmann product, as it contains the less-substituted double bond. The major product of an elimination reaction is typically the Zaitsev product, which contains the most-substituted double bond. The Zaitsev product is the thermodynamic one (as the more-substituted alkene is typically more stable), while the Hofmann product is the kinetic one (as it results from a base abstracting a more sterically accessible or otherwise more acidic hydrogen). The Hofmann product can be obtained if a sterically bulky base is used, as this raises the energy of the transition state due to steric hindrance between the approaching base and the substrate molecule, and puts the reaction under kinetic control (assuming reaction temperature is controlled such that the $E_a$ for the thermodynamic pathway is prohibitively high). $\ce{NaOEt}$ is not bulky, so we expect the Zaitsev product (as given in the reaction scheme you cite).

If an $\mathrm{E_1}$ mechanism occurs, a tertiary carbocation intermediate is generated, and the double-bond is likely to form between the two methyl bearing carbons, as this would give the maximally stable tetrasubstituted alkene, and the requirement for anti-periplanarity is lifted. That said, $\mathrm{E_1}$ reactions are not favorable given a high concentration of strong base, so I would expect $\ce{E_2}$ to predominate, assuming at least a stoichiometric amount of $\ce{NaOEt}$ is present.

(Note that, with cyclohexanes, the issue is slightly more complex, as the anti-periplanarity requirement of the $\mathrm{E_2}$ mechanism requires the leaving group to be axial, which in turn necessitates a ring flip that has an associated energy barrier. Hence, in theory, the reaction with a cyclic molecule may be more likely to be kinetically controlled under a wider range of conditions.)

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The ring flip is very important, very much agreed! Pre equilibria can mess up a lot of reaction mechanisms... –  Martin - マーチン Jul 31 '14 at 3:56

Yes that should be correct. Your product is less favoured but still likely. In fact I guess, ( I am not near a database to confirm that but you could try it) in reality you get the E1 product too. So a "nice" mixture of at least 3 products

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What's the 3rd product? –  Dissenter Jul 30 '14 at 23:03
Oh nvm you mean the SN1 product right? –  Dissenter Jul 30 '14 at 23:38
@Dissenter The third product via E1 forms the double bond between C1 and C2 as greg also pointed this out. –  Martin - マーチン Jul 31 '14 at 3:57