# Elimination reaction on 2-fluorobutane using alc.KOH

Question: Find the major product formed when 2-fluorobutane reacts with alcoholic $\ce{KOH}$

My answer: Elimination of $\ce{HF}$ takes place and a more substituted alkene should have formed. (Zaitsev product). I think the mechanism is E2 since strong base is used.

but answer given is a less substituted 1-butene. Is the answer wrong or am I making any mistake?

E1CB elimination takes place because of a poor leaving group. This is because of the high $\ce{C - F}$ bond strength. Stronger bonds are more difficult to break, making them less reactive.