Why doesn't this reaction use an E2 mechanism?

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SUMMARY

The reaction of tert-butyl bromide with MeOK in methanol does not proceed via an E2 mechanism due to steric hindrance and the nature of the reactants. The tertiary carbon in tert-butyl bromide creates significant steric hindrance, making SN2 unlikely. Although MeOK is a strong nucleophile and base, the reaction favors E1 or SN1 pathways over E2 due to the stability of the carbocation formed. Consequently, E2 is not the predominant mechanism in this scenario.

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  • Understanding of E2, E1, and SN1 reaction mechanisms in organic chemistry.
  • Knowledge of steric hindrance and its effects on reaction pathways.
  • Familiarity with the properties of tert-butyl bromide and MeOK.
  • Basic concepts of nucleophiles and bases in organic reactions.
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  • Study the differences between E1 and E2 mechanisms in organic chemistry.
  • Research the effects of steric hindrance on nucleophilic substitution reactions.
  • Learn about the stability of carbocations and their role in reaction mechanisms.
  • Explore the properties and reactivity of alkyl halides in elimination reactions.
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gauss44
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tert-butyl bromide + MeOK in MeOH ----->

I would think that MeOK would dissociate in solution making it MeO- which would be strong enough to yank a proton off tert-butyl bromide and initiate an E2 reaction. Where am I going wrong?

PS - I am a tutor who hasn't done organic chemistry for a while, so please leave this here AND do NOT send me to the ¨Homework¨ section. This is NOT homework.
 
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The central carbon atom in tBut-Br is tertiary, which gives rise to high steric hinderance, which means that we can exclude Sn2, leaving E1, E2 or Sn1. Continuing, if we consider the other reactant Me-OK, it is negatively charged and thus a strong nucleophile/base. However, since it is strong Sn1 or E1 are unlikely to happen, leaving us only E2 to happen. Don't forget though that Sn1, E1 and Sn2 will happen, but it is to a such small extend that it can be approximately unconsidered.
 

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