Selectivity and kinetics of reactions are somehow linked to the activation energies of reactions. However, there's something not totally clear in my mind about the selectivity. I will get a general example as many systems can fit to this.

Let's consider a reaction between A and B. The reaction between A and B can proceed through different pathways (1 & 2). A selectivity of 100% is observed for pathway 1 and it is demonstrated that the activation energy for this pathway is slightly lower than pathway 2.

Let's consider than a system of C and D. The reaction between C and D can also proceed through 2 pathways, let's name them 3 & 4. In this scenario, a selectivity of 75% for example is observed for pathway 3. The activation energy for pathway 3 is lower than pathway 4 in a more marked way than when we considered the reaction of A and B.

My question is then. Why are sometimes observed a total selectivity when the difference in activation is low, while sometimes the products result from different selectivities when the difference in activation energies between the pathways is rather high.


1 Answer 1


It's not just the activation energy that determines the selectivity of a reaction. There is a thermodynamic factor too. This is the classic kinetic-vs-thermodynamic control that you will find in most chemistry textbooks at a degree level.

  • Reactions under thermodynamic control have outcomes that depend on the position of the equilibrium and therefore the relative stability of the possible products
  • Reactions under kinetic control have outcomes that depend on the rate at which the reaction processds, and therefore on the relative energies of the transition states leading to the alternative products.

[Organic Chemistry, 2nd ed., J. Clayden, N. Greeves, S. Warren, OUP] page 264

You are considering only activation energies ($\Delta G^\ddagger$) i.e. the difference in energy between the transition states and the reactants.

activation energy and reaction energy

If your reaction has two pathways, then the pathway with the lower activation energy is faster. This means that the product of that pathway is formed faster. This is kinetic control.

However, the reaction can be reversible. The activation energy for the reverse reaction is $\Delta G^\ddagger + \Delta_r G$. Therefore, if the reaction energy (i.e. $\Delta_r G$) is not too large, then there would be backward reaction.

In this condition, the products and reactants are in equilibrium, so the side with the lowest Gibbs free energy is favoured. If the product from one pathway is more stable (i.e. more negative $\Delta_r G$) than the other product, then the stable product will the be the one with higher yield when the equilibrium is reached. This is thermodynamic control.

In the above diagram for example, pathway 1 is faster as TS1 is lower in energy. So product 1 is the kinetic product. Pathway 2 is slower, but the product is lower in energy, so product 2 is the thermodynamic product.

Whether a reaction is under thermodynamic or kinetic control would depend on the reaction conditions, the specific identity of the reactants, and how much time is allowed for the reaction.

  • $\begingroup$ This is a good answer, but it might help even more to have on the diagram a second pathway with lower activation energy but higher energy final products to show more clearly how one product can be favored thermodynamically and another kinetically. $\endgroup$
    – Andrew
    May 15, 2021 at 23:29

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