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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.

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  • $\begingroup$ If there is strong base present, then E2. If only methanol and no base, then its E1. $\endgroup$
    – user55119
    Commented Dec 7, 2020 at 23:27
  • $\begingroup$ 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. $\endgroup$
    – Zhe
    Commented Dec 8, 2020 at 0:19
  • $\begingroup$ @user55119 but it is that in the reaction are the two conditions $\endgroup$
    – Carlos
    Commented Dec 8, 2020 at 8:21

2 Answers 2

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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.

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  • $\begingroup$ Well, I thought that the unimolecular elimination reaction, E1, takes place when a tertiary alkyl halide or a secondary alkyl halide reacts in the presence of a weak base (iodide anion, water...) $\endgroup$
    – Carlos
    Commented Dec 8, 2020 at 8:20
  • $\begingroup$ The unimolecular elimination reaction, E1, takes place when a tertiary alkyl halide is treated with a base (both weak or strong) or when a secondary alkyl halide reacts in the presence of a weak base. $\endgroup$
    – Adiboy
    Commented Dec 8, 2020 at 12:13
  • $\begingroup$ @Adiboy it’s not true that if we have strong basic conditions then E1 would be favoured for tertiary alkyl halides. E2 dominates in that case as in the transition state, you form a more substituted alkene for a E2 and also E2 requires strong base. The only thing that would be required to see is if the solvent is protic or aprotic. $\endgroup$
    – Floatoss
    Commented Jan 6, 2021 at 8:36
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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.

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