JEE Main 2019 — Hydrocarbons Question with Solution

Alcoholic KOH performs $$\alpha$$ - $$\beta$$ elimination with halides.

The carbon which is attached to Br atom is called $$\alpha$$ carbon.

Here, two $$\alpha$$ carbon presents. Let's call them $$\alpha$$₁ and $$\alpha$$₂.

In case of $$\alpha$$₁ carbon, H elimination can take place either from $$\beta$$₁ or $$\beta$$₂ carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $$\beta$$₁ carbon then a $$\pi$$ bond is created which will perticipate in resonance with benzene ring and product will be more stable. But if H is removed from $$\beta$$₂ carbon then the created $$\pi$$ bond will not perticipate in resonance with benzene ring so product will be less stable.

In case of $$\alpha$$₂ carbon, H elimination can take place either from $$\beta$$₂ or $$\beta$$₃ carbon. But as we know always more stable product is formed from a reaction. Here if H is removed from $$\beta$$₂ carbon then a $$\pi$$ bond is created which will perticipate in resonance with the $$\pi$$ bond associated with the $$\alpha$$₁ carbon so product will be more stable. But if H is removed from $$\beta$$₃ carbon then the created $$\pi$$ bond will not perticipate in any resonance so product will be less stable.