# Leaving group ability vs Stability of Transition state for determining rate of E2 elimination reaction [closed]

Source: Teacher's Assignment. (not a homework doubt, but a big conceptual doubt about the question)

In this question we have to compare rate of E2 reactions and the answer is C2H4I2 > C2H4Br2 > C2H4Cl2 > C2H4F2

I get it that we checked leaving group ability for getting the answer, but shouldn't we also check the stability of transition state. In C2H4I2, due to high repulsion between Iodine and Iodine, it will nearly always stay in anti conformer. So getting to the transition state involves (i)getting to gauche conformer(conformer for which E2 will happen) and also (ii)partially breaking bonds. During the Transition State, Iodine and Bromine will face repulsion making the Transition state less stable, while Chlorine will face less repulsion and it also a decent leaving group at the same time. If we are only looking at "leaving group ability" we are only considering the point(ii).

Here's a free energy curve to illustrate:

The blue part is high and red part is low for C2H4I2. As we go from I to F, the blue part decreases,(even negative for fluorine) and the red part keeps increase(due to higher bond energy)

C2H4I2 will have the least probability to be in the gauche form while C2H4F2 will have the highest (more favoured by $$\sigma$$- $$\sigma$$* hyperconjugation, already bringing double bong character). But it is a poor leaving group. So, how do we decide the stability of transition state? The easibiliy to break bonds, or minimising repulsion during Transition State?

Are there experimental values so that we can verify the solution.

Help would be very much appreciated.

• BTW you should be more careful with formatting and punctuation. Commented May 16 at 15:36
• OK, still, this q. is poorly written. "stability of transition state"? There's no such thing, there is activation energy. Commented May 16 at 15:59

The transition state you are comparing isn't the actual transition state. They are the original molecules themselves. The 'stability of transition state' you are referring to is used to decide the configuration of the product (E-Z, R-S, etc.) and not compare the rates of two different molecules. You yourself said that the compounds containing iodine and bromine are less stable due to steric hindrance thus their tendency to undergo $$E_2$$ elimination is more i.e. have a lower activation energy.