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.
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.
However, as @Zhe has pointed out, looking at the experimental data gives the best answer. A general rule of thumb (although exceptions may exist) is that tertiary alkyl halides tend to follow E1 mechanism rather than E2 even under strong bases.