2
$\begingroup$

I've been performing Gaussian 16 DFT optimization and single-point energy calculations on different conformers of a rather large organic molecule. My first series of calculations was without solvent model, the second one was with dichloromethane solvent effects included via SMD model. When I analyzed the results, I saw that the absolute energy of the conformers became lower in solvent (as expected due to solvent interactions) but also the energy differences between the conformers became smaller. The order of conformers according to energy however did not change a lot. This would mean that conformers having a higher energy in gas phase than other conformers would experience more stabilization by the solvent. Is this possible? How could this be caused?
If more information is needed about my problem, feel free to ask.

Thanks in advance

$\endgroup$
5
$\begingroup$

One way this can happen is if the different conformers have different dipole moments. Those with largest dipoles might have higher energy due to greater charge separation. However those conformations with larger dipoles will also interact most strongly with the surrounding solvent medium (dielectric), and thereby the solvent will reduce their energy by a greater amount.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you for your answer, I will check the dipole moments and see if this is indeed the case. $\endgroup$ – Bub May 18 at 12:56
  • 1
    $\begingroup$ @Bub note that my answer might be an oversimplification given that you speak of a "rather large organic molecule" in which case other moments and local charges might matter. Your focus is on how a simple continuum dielectric model of the solvent can have this effect, I provided one simple explanation. $\endgroup$ – Buck Thorn May 18 at 13:56
  • $\begingroup$ For my data, there is no visible relation between the dipole moment and the gas phase energy so probably other properties of the conformation indeed matter $\endgroup$ – Bub May 18 at 14:03
  • 1
    $\begingroup$ In any case you can be certain that the effect of the solvent is purely electrostatic (and not H bonds say), so focus your attention on the charge distribution in the conformers. If charges become more exposed to the solvent (rather than being buried) that is liable to lower the energy as well. $\endgroup$ – Buck Thorn May 18 at 14:20

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.