[blueblueblue]

I am studying a solvolysis reaction of 2-bromo-2-methylhexane by potassium tert-butoxide in tert-butyl alcohol.

I need to come up with all of the products that form from this reaction.

1) E2 reaction because tert-butoxide is a strong base. 
However, I am not sure where to put a double bond because tert-butoxide is a bulky group.  And the tertiary alkyl-halide is bulky also.  I am thinking Zaitsev's rule does not apply to this double bond formation.  Would I put a double bond between C1 & C2?  Or, C2 & C3?

I am pretty sure that I am correct, but I cannot find any information on this reaction.  Would any of you be able to provide insight?

Thanks in advance for your help.

[AC Prabakar]

Yes your correct!
Zaitsev's rule does not apply to this double bond formation and the other one(Hoffman product-with less substituted alkene) will be the major product.

The below attachment may be helpful to u...

[orgopete]

This is how I think of them. Elimination reactions can be thought of as push-pull reactions. If the leaving group leaves before the proton is lost, the carbocation (and leaving group) pull electrons toward it. These reactions give the more substituted alkenes because the carbon or hydride best able to donate electrons is the more substituted.

In a pull reaction, the leaving group does not make the same contribution. In order for the reaction to occur, it is acidity of the hydrogen that dominates the products. So poor leaving groups and poor conditions for carbocation formation favor Hoffman products.

This reaction is difficult to predict and probably difficult to get a large degree of selectivity because bromides are such good leaving groups. If you had the chloride, then I'd think the Hoffman product would be in a larger ratio. In this case, I think the Hoffman product will be major, but with an appreciable amount of the Zaitsev product.

[blueblueblue]

Thank you very much.  I learned about the Hoffman reaction just now, and I am going to make a new post, so if you have time please check it out =D

Thanks again!