C12H15FN2O4 — CID 118128954
(1R,2R,3R,4R)-1-(6-fluoro-1H-benzimidazol-2-yl)pentane-1,2,3,4-tetrol (PubChem CID 118128954) has the molecular formula C12H15FN2O4 and a molecular weight of 270.26 g/mol. Its IUPAC name is (1R,2R,3R,4R)-1-(6-fluoro-1H-benzimidazol-2-yl)pentane-1,2,3,4-tetrol.
| Compound Name | (1R,2R,3R,4R)-1-(6-fluoro-1H-benzimidazol-2-yl)pentane-1,2,3,4-tetrol |
|---|---|
| PubChem CID | 118128954 |
| Molecular Formula | C12H15FN2O4 |
| Molecular Weight | 270.26 g/mol |
| Exact Mass | 270.10 |
| IUPAC Name | (1R,2R,3R,4R)-1-(6-fluoro-1H-benzimidazol-2-yl)pentane-1,2,3,4-tetrol |
| SMILES | C[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)c1nc2ccc(F)cc2[nH]1 |
| InChI | InChI=1S/C12H15FN2O4/c1-5(16)9(17)10(18)11(19)12-14-7-3-2-6(13)4-8(7)15-12/h2-5,9-11,16-19H,1H3,(H,14,15)/t5-,9-,10+,11+/m1/s1 |
| InChIKey | IAKCUAYVXBYQJX-SJXNIKTHSA-N |
| XLogP | -0.16 |
| TPSA | 109.60 Ų |
| H-Bond Donors | 5 |
| H-Bond Acceptors | 5 |
| Rotatable Bonds | 4 |
| Heavy Atoms | 19 |
| Complexity | — |
Passes Rule of Five
| Rule | Value |
|---|---|
| MW ≤ 500 | 270.26 |
| LogP ≤ 5 | -0.16 |
| H-Bond Donors ≤ 5 | 5 |
| H-Bond Acceptors ≤ 10 | 5 |