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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:

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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.

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

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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.

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  • $\begingroup$ I am having trouble understanding your statement. We do look at the conformer for determining (E/Z). Check this image for reference. i.imgur.com/KnyTm2S.png . I did say that Iodine and Bromine compounds are less stable, but I meant that in transition state. Anti conformer of I and Br is much more stable than gauche form. For E2, there has to be a Hydrogen in the plane and also opposite to iodine (gauche form). But to get to that form, we have to give extra energy (for going from anti conformer to gauche), thus increasing the activation energy, as shown in the graph in the question. $\endgroup$
    – EagerToLearn
    Commented May 17 at 20:55
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    $\begingroup$ The groups can and always rotate about the single bond freely despite the energy differences. All the conformers will be present in a given sample and they would continuously interconvert. $\endgroup$
    – JackSparrow
    Commented May 18 at 2:03
  • $\begingroup$ Yes, bond rotation is not restricted. But we also take the percentage occurence of eclipsed forms nearly 0 percent due to the repulsion. So suppose the Iodine compound stays in gauche form 5% and in anti form 95% (not actual values) then the activation energy per mole, would have 95% contribution from the anti conformer, which requires a higher activation energy. I think the formula of effective activation energy would be: (n1E1+n2E2)/(n1+n2) where n1 is percent of anti form and E1 is the activation energy of anti form. n2 and E2 correspond to gauche form. as n1 is high, Anti form dominates $\endgroup$
    – EagerToLearn
    Commented May 18 at 6:06
  • $\begingroup$ The difference in activation energies here will be very less. (you might know that we have to provide heat in most cases for elimination to take place). $\endgroup$
    – JackSparrow
    Commented May 18 at 7:15
  • $\begingroup$ The mechanism followed will depend on the base taken. If a strong base is taken then the hydrogen will have to be on the anti side. I agree that the anti conformers are going to be the major substrate but they are in an equilibrium with the other forms. As the no. of molecules in Gauche form decrease due to consumption, the equilibrium will shift forwards. $\endgroup$
    – JackSparrow
    Commented May 18 at 7:15

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