# Competition between substitution and elimination (uni- or bimolecular)

Whats is/are the main product/s of the following reaction:

I think the main product would be the one obtained by $S_N2$, and then the one pair of enantiomers obtained by $E2$. Is $E1$ with subsequent rearrangement possible?

• I believe that E1 mechanism is not possible, as there is a good base present and reaction is carried out in diluted base on room temperature. – Uros Nov 26 '15 at 13:54
• Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. Marko improved the typesetting of your formulae with MathJax, for more information on how to do so yourself, check out the help center, this meta post or this one. – Jan Nov 26 '15 at 14:37
• "E2 gives enantiomers" - are you sure? Draw that molecule in the chair conformation. – orthocresol Nov 26 '15 at 14:43
• Jan, thank for the welcome. @orthocresol If you take the beta-proton from the left side, or the beta proton from the right side, you get two products, which are enantiomers. – Uros Nov 26 '15 at 14:50
• Oh, ok, my bad :D – orthocresol Nov 26 '15 at 15:29

For an $\mathrm{E1}$ mechanism, we would need to generate a carbocation. Our halide is attached to a secondary carbon so the generated carbocation would be secondary. A quick check of neighbouring stabilising groups is negative. And as a professor of mine once put it (possible slightly oversimplified but with some general truth in it):
Thus, $\mathrm{E1}$ is not possible since the carbocation will not form in any measurable amounts. For the same reason, $\mathrm{S_N1}$ is not an option. For everything else, let’s consider the molecule’s 3D structure:
The tert-butyl group acts as a conformation ‘anchor’ meaning that the other chair form is not viable. Luckily, bromine is pointing upwards, which gives us access both to $\mathrm{S_N2}$ and $\mathrm{E2}$ mechanisms without having to invert or use a boat conformation.
Our base/nucleophile is methanolate which is both a good enough base and a good enough nucleophile, able to access both pathways. The centre for nucleophilic attack is also well accessable. If I had to choose, I would indeed go with your choice of saying the main products will be $\mathrm{S_N2}$ adduct and two enantiomers of elimination product will also be formed. (They stem from either abstracting the proton I drew or the enantiotopic proton on the back half of the six-membered ring.)