[habetsth]

Hi everyone !

As DFT calculation is an excellent tool for confirming and understanding experimental results, I got deep into this field as organic chemist. I successfully isolated interesting transition states (TS) for my system and performed frequencies, IRC, and optimization calculations.

However, I have a practical question for the interpretation of the relative energies obtained from the calculations. My system can be understood as a simple two-step reaction:

[tex]\ce{A + B -> C -> D + E}[/tex]

I did all calculations for the first step toward the formation of C and I was able to isolate several geometries for the TS for the second step. From all possible geometries tested, two have shown to be more favorable as the difference of energy between the reactants complex (optimization of IRC in reverse direction) and the TS were the most little (around 200 kJ/mol). The reaction pathway with relative energies is shown on the attached figure.



I have the following questions:

1. When having the choice between these two pathways, which could be the best choice as the difference of energy between reactant complex and TS for this step are the same? Can the orange pathway be preferred for the reaction for its more little TS absolute energy?

2. I want to compare this reaction to the same reaction with a different substrate. With a first substrate, I get a TS at an energy of the same order than the blue pathway, an alternative geometry like the orange giving the same energy. The difference of energy between complex and TS is also of 200 kJ/mol.
When I use my new substrate, I expect to get an activation barrier that is lower, which could fit with the orange pathway. May this new orange geometry with a lower TS energy explain why the reaction is faster with this new substrate (saw experimentally)?

Thanks !

[Orcio_87]

1. When having the choice between these two pathways, which could be the best choice as the difference of energy between reactant complex and TS for this step are the same? Can the orange pathway be preferred for the reaction for its more little TS absolute energy?

That it is preferred does not mean that it is the sole one, as the difference is rather small (about 19 kJ / mol).

When I use my new substrate, I expect to get an activation barrier that is lower, which could fit with the orange pathway. May this new orange geometry with a lower TS energy explain why the reaction is faster with this new substrate (saw experimentally)?

True, as the reaction constant is described as:

k = A exp (-E_a / RT)

that is - increases when E_a falls down.