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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?

pic

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
print(rdkit.__version__)
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]
print(pika_at.hybridisation)

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.

Thanks

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  • $\begingroup$ 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

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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.

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    $\begingroup$ 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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