Question

Br
Alc. KOH
Product(s)
H
Сн,
The above reaction follows bimolecular paths. Find the major product of the following:
(A) Saytzeff alkene
(B) Hoffmann alkene
(C) Alcohol (secondary)
(D) Alcohol (tertiary)​

Answer 1

Answer:

The Hofmann Elimination is an elimination reaction of alkylammonium salts that forms C-C double bonds [pi bonds]. [note] It proceeds through a concerted E2 mechanism. In contrast with most elimination reactions that yield alkenes, which follow the Zaitsev (Saytzeff) rule, the Hofmann elimination tends to provide the less substituted alkene. In this post we go through the difference between Hofmann elimination and Zaitsev elimination and explain the key features in the Hofmann degradation mechanism that result in its preference for the “less substituted” alkene.