Why is hofmann product prefered here? 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:


When the ${\gamma}$ carbon is 4°
When the leaving group is poor leaving group  
When the leaving group is bulky leaving group
When the elimination is internal elimination (${E_i}$ mechanism)    


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.