Organic Chemistry: E2 reaction mechanism?

potassium tert-butoxide is a strong and exceptionally hindered base and therefore is only capable of abstracting primary (terminal) hydrogens in non-cyclic alkyl halides. The E2 mechanism is concerted, so there are no rearrangements.

hmm, my prof said it had nothin to do with steric hindrance. and 1-butene is also the major product for the reaction with KOH, which is not sterically hindered. Thanks though, I appreciate it.

I think anoop is right. Maybe your prof meant that sterics are irrelevant in terms of the stability of the final product (which they are in this case)—but I think the steric hindrance of the potassium tert-butoxide is a factor. I’m too rusty to draw out the reaction mechanism, but try it and see what happens—if you push the electrons around correctly your problem should solve itself.

“The Zaitsev Rule is a good predictor for simple elimination reactions of alkyl chlorides, bromides and iodides as long as relatively small strong bases are used. Thus hydroxide, methoxide and ethoxide bases give comparable results. Bulky bases such as tert-butoxide tend to give higher yields of the less substituted double bond isomers, a characteristic that has been attributed to steric hindrance.”
http://www.cem.msu.edu/~reusch/VirtualText/alhalrx3.htm

Alright, thanks guys. I think my prof may have just misspoke. In the review packet it has trans-2-butene listed as the major product of KOH plus 2-bromobutane, which is what I expected, but in lab he said otherwise. Also in lab, when we did the reaction, 1-butene was the major product. I don’t know, I’m still confused, because everyone got 1-butene as major product, but theoretically it shouldn’t be?

I think the bulkiness of tert-butoxide is what’s causing the difference between the rxn in your original question and the one with KOH. Just not sure exactly how.