[OrganicH2O]

I am thinking about a chemical reaction of 2-Bromo-2-methylbutane with tert-butyl alcohol and heat, assuming elimination products.  I am thinking this reaction will be E1 because it  is a tertiary halide and tert-butanol is a very weak base, and fairly polar and also protic. If it is a pure E1 reaction, would we see the same favoring of the Hoffman product that is seen with an E2 reaction with tert-butoxide?

I have heard that E1 reactions give Zaitsev products regardless of steric hindrance on the base, perhaps because the transition state to form the alkene is very late due to Hammond's postulate. But I don't know if this is always true.  And perhaps in this case the mechanism is a concerted hybrid of E2 and E1, with an asynchronous transition state, because tert-butanol is not polar enough to allow for carbocation formation. If I just knew the experimental outcome I could backwards justify it.

[hollytara]

So your plan is to make a solution of the bromo compound in t-butanol and heat it?  So in the E1, it will make the carbocation - for that substrate there will be quite a bit less steric hindrance for each of the protons that might be lost, compared to the hindrance in the E2 transition state. 

But why not just do the E2? Add some t-butoxide to the t-butanol.  This will also reduce side reactions that are SN1!

[OrganicH2O]

I strongly agree with your statement, but this is specifically related to an exam question I saw recently. The professor said that the correct answer was Hoffman product, but I am suspicious and wanted to get some feedback from knowledgeable individuals.

[OrganicH2O]

I made a goof in my original explanation. The transition state would be very early due to Hammond's postulate, for deprotonation of the carbocation to form the alkene. Going from carbocation --> alkene is obviously very exothermic. This is arguing for steric hindrance being less important, although this would also make the relative stability of the two alkenes less important as well.