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In the following reaction, Hofmann elimination is preferred.
This reaction takes place by E1 mechanism. I thought the Zaitsev product would be major due to 1 extra hyperconjugating structure. In the book, it is given that it is due to sterically compressed alkene, so Hoffmann product is major. I am unable to understand why the sterical hinderance would lead to the instability in alkene as the three methyl groups is one bond away from alkene.
The Hoffmann product is formed as a major product generally in four cases. They are:
Case 1:
It happens in the example given in the question the hindrance of base in abstraction of proton favors Hoffman product
Case 2:
It happens with poor leaving groups such as flourine (based on the carbanion character on the carbon from which proton is abstracted in the trasition state)
Case 3:
It happens in the Hoffmann's exhaustive methylation reaction where the bulky leaving group is a 3° amine group
Case 4:
It happens in the treatment of a 3° amine with Caro's acid followed by heating and in heat treatment of xanthate esters
Lastly I want to add that these are general results the yields may vary based on reactant.
Look at the carbon which has more α-hydrogens and thus will be stabilized more by resonance when it is attacked by the $\ce{OH-}$ ions. So the carbon atom on the right hand side that has two methyl groups has more α-hydrogens with it and is stabilized by resonance and $\ce{OH-}$ is attacked there. Hence the major product has carbon with the double bond.
