Doubt in Alkyl Halide

Please explain the elimination reaction product formation i.e. Product A. The correct answer is option (c).

Thank you.

First Compound A will be Hoffmann product of alkene and second bromine elimination will take place chlorine atom will not be replaced

Sorry Bromine will not be eliminated NmE3 will be eliminated i am uploading all A and B product

Why hoffman ? Why not satyazeff?

Because of d orbital resonance of N we treat Nme3 as a bulky group which would yield Hoffmann product as major product

Is it because Chlorine being more electronegative it will stabilize conjugate base further and make the hydrogen attached to it more acidic and that's why it is abstracted and the double bond is formed ? Is this reason behind the formation of the double bond with carbon containing the chlorine group ?


But bhai NMe3 already nikal chuka h ,ab uska kya lena dena hoffman se ?

Yes correct in this hoffman and saytzeff products are same due to even carbon atoms that too 4 but when Nme3 is there Hoffmann product is formed in major