# 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?

## 1 Answer

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.

If the reaction was E2, then zaitsev product would have been formed. But since the mechanism is E1CB, the reaction proceeds via carbanion.

From wikipedia: The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favor an E1cB mechanism.

I hope you know that for carbanions, the stability order is primary > secondary > Tertiary.

Now that you know the stability order of carbanions, the preferred position for carbanion is a primary carbon. So base abstracts a proton from a primary carbon.