The correct option is
D Hoffman product means the alkene with less no of carbon attached to it. Here, the elimination takes place via.
E2 mechanism in which anti elimination takes place.So, in order of explaining each and every option:
1. In option 1, both H and F are on same side. So, H abstraction will take place from
−CH3.
2.In option 2, Sodium tert butoxide(non-nucleophilic base) , being a bulkier base couldn't be able to abstract proton from inner C. So, proton will be astracted from the outer
−CH3.
3. In option 3 and 4, The H attached at benzyl position will be abstracted because the negative charge formed will undergo delocalisation with the aromatic ring (Ph) attached to that carbon.Hence, making the negative charge more stable.
As a result, saytzaff product is formed in option 3.
Hoffman (less carbon substitued product ) product will be formd in option 4.