# Will t-butoxide still form less substituted alkene when there is a possiblity for more stabilisation?

Consider this reaction:

Will a usual Hoffman elimination take place or will a carbocation form by leaving of $\ce{Br}$ and rearrange to give a conjugated alkene like this one:

I have heard that reactions with t-Butoxide are E2 and hence carbocation are not formed. So wouldnt the major product still be the least substituted alkene?

• t-Butoxide is likely too strong of a base/nucleophile to be starting an E1 reaction; I don't think a carbocation will form here. In this case, since t-Butoxide is bulky, the Hoffman product will be favored over the Zaitsev for E2 elimination. – John Smith Mar 12 '16 at 17:46