I am currently building a reaction mechanism through Gaussian calculations. I successfully found each TS and intermediates through the different jobs (opt, freq, irc) for my 3-step mechanism from reactants to my products.

However, I was questioning myself about one step in particular which could involve one extra molecule. Here's a scheme of my system:

$$\ce{A + B -> C <=> D -> E}$$

Where A and B are in same equivalence experimentally.

All steps are characterized by stable intermediates and reasonable energy barriers confirming experimental data, except one step. The step from C to D is a simple tautomerism from enol to ketone. However, this proton exchange needs to be driven by a proton acceptor like B to have a energy barrier that is closer to the reality. Then, for my system, an extra molecule B could be implied in the mechanism only from this C to D step.

Practically, you perform optimization of your TS and complex of products and reactants involving all molecules for the reaction step. However, I was wondering if I can draw a reaction pathway from A and B to E without the extra molecule B and involving it only for the step where it is required (as a sort of catalyst). Let's make it more clear:

  1. I take the step A and B to C as already calculated
  2. I perform C to D with an extra molecule B acting as proton acceptor for that step
  3. I take the step from D to E

In this way, the extra B is used only in the second step of the mechanism and I can get from there the relative energies. However, in that case, how can I draw the mechanism plot with all relative energies ? C or C + B have totally different energies. Am I forced to re optimize all my system with an extra B molecule ?

  • $\begingroup$ Are these A, B, C, D and E substances corresponding to real molecules ? Or are they here simply for the fun of writing kinetics formula ? Is it possible to imagine that for example$\ce{A = C2HR, B = H2O, C = RCHCHOH, D = RCH2-CHO}$ and $\ce{E}$ is something else ? So, why thinking that a second B is necessary ? $\endgroup$
    – Maurice
    Feb 20, 2021 at 11:22
  • $\begingroup$ It is as example instead of providing a real system. The other B would act as a sort of catalyst for only one step of the mechanism and would not be involved in the other steps (except in the first one but as reactant). $\endgroup$
    – thchem
    Feb 22, 2021 at 10:14


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.