Consider this reaction:

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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:

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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?

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    $\begingroup$ 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. $\endgroup$ – John Smith Mar 12 '16 at 17:46

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