Looking at nelfinavir, I get variable results from different compchem packages on whether the phenol hydroxyl oxygen should be rotatable (SP3) or non-rotatable (SP2). Which one is it?


Feasibly, there are resonance forms where double bonds hop over to the oxygen from the benzene. The crystal structures are all pretty consistent, showing a hydrogen bond from the hydroxyl that keeps the hydrogen planar with the benzene ring (consistent with SP2).

RDKit claims this oxygen is SP2:

import rdkit
from rdkit import Chem
mol = Chem.MolFromSmiles('Cc1c(cccc1O)C(=O)N[C@@H](CSc2ccccc2)[C@@H](C[N@@]3C[C@H]4CCCC[C@H]4C[C@H]3C(=O)NC(C)(C)C)O')
atom = mol.GetAtomWithIdx(7)
print(atom.GetAtomicNum(), atom.GetHybridization())

out: 2022.09.1 \n 8 SP2

However, some packages claim SP3. Here's one example (not picking on PIKAChU here):

from pikachu.general import read_smiles
s = read_smiles('Cc1c(cccc1O)C(=O)N[C@@H](CSc2ccccc2)[C@@H](C[N@@]3C[C@H]4CCCC[C@H]4C[C@H]3C(=O)NC(C)(C)C)O')
pika_at = s.atoms[7]

out: sp3

Similarly, some paid packages optimize the coordinates of the bound molecule in the crystal structures, and treat the hydroxyl as rotatable. As a result of the optimization, the H-bond is no longer in-plane with the benzene, which would indicate SP3.


  • $\begingroup$ I'm not an expert on these two libraries. The only question that jumps into my mind is: are you sure you're referring to the same atom among the two libraries? Do the two libraries start counting the atoms' indexes by 0? $\endgroup$ Feb 7, 2023 at 8:37

1 Answer 1


The experimental findings (and from higher theory levels) are:

The phenolic OH group is in plane with the aromatic ring (best lone pair overlap), with the barrier for rotation around 3.5 kcal/mol. The situation is similar in 2,6-dimethylphenol. In your case, the system is less symmetric so that you can have a preference for one position of the O-H group.

Regarding the empirical assignment, SP2 is definitely wrong. The O-H is in plane of aromatic ring for different reasons, than SP2 hybridization.

  • 1
    $\begingroup$ this is great, thank you! Could you fill in a few gaps please? What expmt findings are you referring to (to support the 3.5 kcal/mol barrier)? How does 'best lone pair overlap' explain this barrier - is it avoiding overlap or maximising overlap, and between what atoms? $\endgroup$
    – lewiso1
    Feb 7, 2023 at 18:22
  • $\begingroup$ to play devil's advocate here, WolframAlpha says 'sp2' (as does my old first year textbook): wolframalpha.com/… $\endgroup$
    – lewiso1
    Feb 7, 2023 at 23:06
  • $\begingroup$ -OH has a withdrawing group in its meta position so the pull is not as strong compared to ortho and para. A complete charge-transferred sp2 is a correct but less abundant resonance form. Look at the carbon and proton NMRs of nelfinavir in the SI here: doi.org/10.1021/acs.joc.8b00039 $\endgroup$
    – Ardalan
    Feb 8, 2023 at 15:04
  • $\begingroup$ @leviso1 The barrier for C-O rotation can be determined by different methods, one quite accurate and esoteric is rotational spectroscopy, see doi.org/10.1016/0022-2860(69)85029-5 The lone pairs of oxygen should overlap with the pi-system of the aromatic ring. This is best achieved when the oxygen is sp3, then both pairs are ~19 degrees from perpendicular to the benzene plane. Would you use sp2 O, one lone pair is perpendicular to benzene plane, but the second is in plane, therefore no resonance. $\endgroup$
    – ssavec
    Mar 10, 2023 at 11:57

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