# Determine mechanism: E1 vs. E2

I am asked to indicate what is the product of this reaction, and the mechanism by which it is formed:

3-Bromo-3-ethylpentane + sodium hydroxide in methanol, 50 ºC

I identify it as an elimination reaction. I would say that it is a bimolecular elimination, E2, because a tertiary alkyl halide and a strong base (NaOH) are used. However, my teacher told me that it was unimolecular, E1, and I don't understand why.

• If there is strong base present, then E2. If only methanol and no base, then its E1. Dec 7, 2020 at 23:27
• I would argue that you can posit a guess, but without experience, you cannot be certain unless you have experimental evidence. Methoxide/hydroxide is strong-ish, but it would be a better base if it weren't in a protic solvent. Heating is frequently indicative of E1. I would lean that direction, given that the electrophile is somewhat bulky, but again, you can't be 100% sure a priori.
– Zhe
Dec 8, 2020 at 0:19
• @user55119 but it is that in the reaction are the two conditions Dec 8, 2020 at 8:21

3-Bromo-3-ethylpentane is so bulky that an E2 elimination seems to be unlikely as the $$\ce{OH}^{-}$$, despite being a strong base, has a very low chance of directly attacking the beta $$\ce{H}$$. Moreover, proceeding by the E1 mechanism, we have a tertiary carbocation which is quite stable compared to primary and secondary carbocations.
The only catch I find in this question is that, it’s given that you are using methanol as your solvent and that’s a protic solvent . It will be the case that the base $$\ce{OH^{-}}$$ would be solvated by methanol and so it cannot efficiently grab a proton at the literal time of the leaving group from the $$\beta$$ carbon (we also know $$\ce{Br^{-}}$$ is a very good leaving group, so it leaves out fast). So the reaction favours E1 mechanism.