C19H15FN2O — CID 71570799
6-(4-fluorophenyl)-4-phenyl-1,2,4,5-tetrahydroindazol-3-one (PubChem CID 71570799) has the molecular formula C19H15FN2O and a molecular weight of 306.34 g/mol. Its IUPAC name is 6-(4-fluorophenyl)-4-phenyl-1,2,4,5-tetrahydroindazol-3-one.
| Compound Name | 6-(4-fluorophenyl)-4-phenyl-1,2,4,5-tetrahydroindazol-3-one |
|---|---|
| PubChem CID | 71570799 |
| Molecular Formula | C19H15FN2O |
| Molecular Weight | 306.34 g/mol |
| Exact Mass | 306.12 |
| IUPAC Name | 6-(4-fluorophenyl)-4-phenyl-1,2,4,5-tetrahydroindazol-3-one |
| SMILES | O=c1[nH][nH]c2c1C(c1ccccc1)CC(c1ccc(F)cc1)=C2 |
| InChI | InChI=1S/C19H15FN2O/c20-15-8-6-12(7-9-15)14-10-16(13-4-2-1-3-5-13)18-17(11-14)21-22-19(18)23/h1-9,11,16H,10H2,(H2,21,22,23) |
| InChIKey | CWTFHGKITHGPEK-UHFFFAOYSA-N |
| XLogP | 3.92 |
| TPSA | 48.65 Ų |
| H-Bond Donors | 2 |
| H-Bond Acceptors | 1 |
| Rotatable Bonds | 2 |
| Heavy Atoms | 23 |
| Complexity | — |
Passes Rule of Five
| Rule | Value |
|---|---|
| MW ≤ 500 | 306.34 |
| LogP ≤ 5 | 3.92 |
| H-Bond Donors ≤ 5 | 2 |
| H-Bond Acceptors ≤ 10 | 1 |