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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 ?

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  • $\begingroup$ 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
    Commented 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
    Commented Feb 22, 2021 at 10:14

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