What will be the product after first step (question from elimination)


The solution which I have

As tbu-OK is a bulky base that's why hoffmann product is formed i.e. less no. of alpha-H.

In which step are you facing difficulty ? Addition of HBr or elimination product ?

What is the exact definition of hoffmann product, least stable product or product having least number of alpha H?
Elimination product @Shwetanshu_2018

Carbanion will be formed on benzene conjugated carbon in this case due to resonance stability

Can you tell why hoffmann product (less stable one) don't be formed in the reaction?

I think the double bond should have been formed on the other side rather than on the side as shown in the solution because this base is bulky so elimination should take place from lesser bulky side

the product is decided due to double bond not by carbocation
stability of double bond decided according to number of +h group b/w them hence in most of cases hofman product are formed but when bullky bases come in contact the product changes because of high repulsion b/w bond of element..
hence satzeff product are mainly stable..