Question

(4) Draw the three elimination products formed by reaction of 2-bromo-4-methyl pentane with potassium

tert-butoxide in 2-methyl-2-propanol. Circle the major product. Does this reaction follow Zaitsev’s or

Hofmann’s rule?

Answer 1

Under normal conditions, we’re taught that eliminations always favor the more substituted alkene (Zaitsev’s rule) because they are more thermodynamically stable. However, when a bulky base is used (t-butoxide or LDA) we observe that the major products instead result from removal of a proton at the least hindered carbon (Hofmann Product).

The key lesson here is that steric effects are destabilizing, and the transition state leading to the Zaitsev product is necessarily going to be the most sterically hindered. If a bulky base is used, the destabilizing steric effects of the Zaitsev transition state start to outweigh the stabilizing effect of forming the more substituted double bond, and the Hofmann transition state is favored.

below are the three products: MAJOR is Hoffmann's product