C19H15FN4O — CID 71537849
3-fluoro-4-(2,3,5,12-tetrazatetracyclo[8.6.1.02,7.013,17]heptadeca-1(16),4,6,8,10,13(17)-hexaen-14-yl)phenol (PubChem CID 71537849) has the molecular formula C19H15FN4O and a molecular weight of 334.35 g/mol. Its IUPAC name is 3-fluoro-4-(2,3,5,12-tetrazatetracyclo[8.6.1.02,7.013,17]heptadeca-1(16),4,6,8,10,13(17)-hexaen-14-yl)phenol.
| Compound Name | 3-fluoro-4-(2,3,5,12-tetrazatetracyclo[8.6.1.02,7.013,17]heptadeca-1(16),4,6,8,10,13(17)-hexaen-14-yl)phenol |
|---|---|
| PubChem CID | 71537849 |
| Molecular Formula | C19H15FN4O |
| Molecular Weight | 334.35 g/mol |
| Exact Mass | 334.12 |
| IUPAC Name | 3-fluoro-4-(2,3,5,12-tetrazatetracyclo[8.6.1.02,7.013,17]heptadeca-1(16),4,6,8,10,13(17)-hexaen-14-yl)phenol |
| SMILES | Oc1ccc(C2CC=C3c4c(c[nH]c42)C=CC2=CN=CNN23)c(F)c1 |
| InChI | InChI=1S/C19H15FN4O/c20-16-7-13(25)3-4-14(16)15-5-6-17-18-11(8-22-19(15)18)1-2-12-9-21-10-23-24(12)17/h1-4,6-10,15,22,25H,5H2,(H,21,23) |
| InChIKey | MAOVTXUUMMMUGV-UHFFFAOYSA-N |
| XLogP | 3.45 |
| TPSA | 63.65 Ų |
| H-Bond Donors | 3 |
| H-Bond Acceptors | 4 |
| Rotatable Bonds | 1 |
| Heavy Atoms | 25 |
| Complexity | — |
Passes Rule of Five
| Rule | Value |
|---|---|
| MW ≤ 500 | 334.35 |
| LogP ≤ 5 | 3.45 |
| H-Bond Donors ≤ 5 | 3 |
| H-Bond Acceptors ≤ 10 | 4 |