[nj_bartel]

I came across a practice passage recently that stated the poorer the halide leaving group (F -> Cl -> Br -> I) the higher the percentage of Hoffmann elimination.  Intuition tells me it would be the other way around.  What's going on here?

[ufalynn88]

well, by definition the Zaitsev product is the most favorable, and the hoffman often the least....thus, the better the halide as a LG the higher percentage of zaitsev product..and the worse the halide, the higher percentage of hoffman (:

[Dan]

I came across a practice passage recently that stated the poorer the halide leaving group (F -> Cl -> Br -> I) the higher the percentage of Hoffmann elimination.  Intuition tells me it would be the other way around.  What's going on here?

A poor leaving group will push the reaction towards the E1cb mechanism - so you will have a transition state somewhere between an E2 and E1cb - with the weakening of the C-H bond before weakening of the C-F bond. This means you get build up of partial negative charge in the transition state. For a simple haloalkane the best carbon to accommodate this partial negative charge is the least substituted (primary carbanions are more stable than secondary etc.) resulting in Hofmann elimination proceeding via a lower energy transition state => major product - it's a kinetic effect.

[nj_bartel]

Awesome, got it! thanks