As a simple example, say I wish to calculate the product of a reaction between methyl iodide and triethylamine, then compare it with the product of benzyl iodide and triethylamine. My goal is to compare the products' relative stabilities (in a larger system this could be useful to compare reactivity options on a common substrate). I couldn't just calculate triethylmethylammonium's enthalpy and compare it with benzyltriethylammonium, right? I would expect the system with more electrons/bonds to always have a lower enthalpy value.

Alternatively, I could calculate all the reactants as well as products, including the iodide as a non-coordinating anion, then subtract the reactants from the products. This difference I could then use to compare the relative stability of the two products, am I correct in this? My worry with this approach is whether DFT can model the non-coordinating anion accurately, as well whether its result would mean anything without solvent effects. It feels like the rabbit hole can always go deeper, and I guess I am curious at what point I obtain qualitatively useful results.

  • $\begingroup$ Could you make the question more precise? The actual title question is unclear. Anything can be compared in principle. Perhaps you wanted to ask whether a difference of some energies has some particular physical significance? Then in the body you mix two other questions, the first one is not really DFT-specific, but general, and the other doesn't have much to do with the title question, but is about DFT accuracy. $\endgroup$ – jhrmnn Jun 26 '17 at 6:20
  • $\begingroup$ Hi, sorry for the confusion. In DFT the energy of the system can be calculated, for example the enthalpy or free energy from a frequency calculation. I understand these to provide insight into the thermal stability of two compounds, for example in a complex to compare different coordination stereoisomers (eg. different ligands). My question is a) whether such values can be used if the molecules under comparison aren't isoelectronic and b) how I could otherwise go about comparing two molecules' relative stabilities. $\endgroup$ – Blaise Jun 26 '17 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.