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StructuralGenomicsConsortium / CNP4-Nsp13-C-terminus-B

An SGC Open Chemical Networks Project Devoted to a site on the SARS-CoV-2 protein nsp13
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Choosing 10 compounds to buy #6

Closed edwintse closed 1 year ago

edwintse commented 2 years ago

PastedGraphic-1

Compound code SMILES
Z1980543937 O=C(Nc1ccc(Cl)cc1)N1CCC(c2nc(=O)[nH][nH]2)CC1
Z29215694 O=C(Cc1ccc(Cl)cc1)Nc1ccc2c(c1)NC(=O)CO2
Z16219632 O=C1CSc2ccc(C(=O)OCC(=O)c3ccc(F)cc3)cc2N1
Z1732170310 O=C(Nc1ccc(F)cn1)NC1CCN(CC2CCOC2)CC1
Z1866425571 O=C(NC1CCN(CC2CCOC2)CC1)c1ncc(F)cc1F
Z1023120704 CN(CC(=O)N1CCCC(c2ccn[nH]2)C1)c1ccc(Cl)cn1
Z999226296 COCCN1CCC(NC(=O)Cc2c[nH]c3cc(F)ccc23)CC1
Z1732184996 O=C(Nc1ccc(F)cn1)N1CCN(c2cccc(O)c2)CC1
Z1681587653 O=C(CCC(=O)c1ccc(Cl)cc1)NCc1csc(=O)[nH]1
Z433872406 O=C(Nc1cccc2cn[nH]c12)C1CCCN(C(=O)Cc2ccc(F)cc2)C1
Z1625847743 CC(=O)Nc1c[nH]nc1C1CCCN(C(=O)Cc2c[nH]c3cc(F)ccc23)C1
Z2053840098 O=C(Cc1c[nH]c2cc(F)ccc12)N1CCCC(O)(c2c[nH]nn2)C1
Z1282042436 O=C(NCC(O)c1cccs1)c1c[nH]cc1-c1ccc(F)cc1
Z738966512 COc1ccc(NC(=O)c2cc[nH]n2)cc1NC(=O)c1ccc(Cl)cc1F
2332187921 CC(CC(=O)N1CCCC(c2nc(=O)[nH][nH]2)C1)c1ccc(F)cc1
Z1743139940 O=C(NCc1ccc(Cl)cc1F)N1CCCC(c2ccn[nH]2)C1
Z989732710 CC(C(=O)Nc1ccc(Cl)cn1)N1CCC(c2ccn[nH]2)CC1
Z1282042436 O=C(NCC(O)c1cccs1)c1c[nH]cc1-c1ccc(F)cc1
mattodd commented 2 years ago

OK, so we need to get down to 10? Numbering from left to right, it looks like we have some similar pairs, from which we can select one without losing too much. e.g. 13 and 16. 2 and 6 Others? @TomkUCL @edwintse @H-agha @jemimahaque @kipUNC

edwintse commented 2 years ago

Also: 5 and 9 1 and 8

H-agha commented 2 years ago

I think these pairs are somehow closely related as well 1 and 3 2 and 11 4 and 12

NB: five different compounds 7 is unique in 3D 10 has two substituted aromatic rings at both ends (one substituted with para fluorine and the other has meta hydroxyl). 14 (most compounds that has two aromatics at both ends one of them is phenyl and the other is five membered ring but compound 14 has phenyl and bulkier fused ring "1H-indazole") 17 has a linear linker between two aromatics (no heterocycle - flexible structure) 15 has ortho and para substituted phenyl ring and has extra aromatic ring in the middle (total three aromatic rings)

Based on that, I would strongly recommend the selection of these five unique compounds (7, 10, 14, 15, 17) From all the pairs mentioned before, I would select ( 1, 2, 4, 5, 13). I excluded closely similar structures as follow (3 & 8, 6 &11, 12, 9, 16 ).

Any thoughts!

image
TomkUCL commented 2 years ago

6 and 11, 11 has two halogens on its terminal aryl group and is slightly lengthier. 2 and 11 1 and 8 5 and 9 7, 10, 14, 15, 17 look unique to me as well.

H-agha commented 2 years ago

Good point Tom!

do you mean 6 & 16 have two halogens? 15 as well has two halogens.

In this case, I think it is better to select 6 instead of 2, to see how much space do we have in the pocket.

mattodd commented 1 year ago

Closing. Page linked from Story So Far page. Compounds were bought and evaluated.