CAS 823817-55-6 (3S)-3-Amino-1-(3-(trifluoromethyl)-5,6-dihydro(1,2,4)triazolo(4,3-a)pyrazin-7(8H)-yl)-4-(2,4,5-trifluorophenyl)butan-1-one (HCID 11432118)

基本情報

CAS: 823817-55-6
標準SMILES: N[C@H](CC(=O)N1CCn2c(nnc2C(F)(F)F)C1)Cc1cc(F)c(F)cc1F
InChIKey: MFFMDFFZMYYVKS-VIFPVBQESA-N
分子式: C16H15F6N5O
平均分子量: 407.31 g/mol

名称と識別子

優先名称: (3S)-3-Amino-1-(3-(trifluoromethyl)-5,6-dihydro(1,2,4)triazolo(4,3-a)pyrazin-7(8H)-yl)-4-(2,4,5-trifluorophenyl)butan-1-one
IUPAC: (3S)-3-amino-1-[3-(trifluoromethyl)-6,8-dihydro-5H-[1,2,4]triazolo[4,3-a]pyrazin-7-yl]-4-(2,4,5-trifluorophenyl)butan-1-one
英語名: Sitagliptin Impurity 1
分子式: C16H15F6N5O
分子量: 407.31

名称と別名

823817-55-6(S)-SITAGLIPTINWQC5W43W24Sitagliptin phosphate impurity A [EP](3S)-3-Amino-1-(3-(trifluoromethyl)-5,6-dihydro(1,2,4)triazolo(4,3-a)pyrazin-7(8H)-yl)-4-(2,4,5-trifluorophenyl)butan-1-one(3S)-3-Amino-1-(5,6-dihydro-3-(trifluoromethyl)-1,2,4-triazolo(4,3-a)pyrazin-7(8H)-yl)-4-(2,4,5-trifluorophenyl)-1-butanone1,2,4-Triazolo(4,3-a)pyrazine, 7-((3S)-3-amino-1-oxo-4-(2,4,5-trifluorophenyl)butyl)-5,6,7,8-tetrahydro-3-(trifluoromethyl)-1-Butanone, 3-amino-1-(5,6-dihydro-3-(trifluoromethyl)-1,2,4-triazolo(4,3-a)pyrazin-7(8H)-yl)-4-(2,4,5-trifluorophenyl)-, (3S)-SITAGLIPTIN PHOSPHATE IMPURITY A [EP IMPURITY]SITAGLIPTIN PHOSPHATE IMPURITY A (EP IMPURITY)1-Butanone, 3-amino-1-[5,6-dihydro-3-(trifluoromethyl)-1,2,4-triazolo[4,3-a]pyrazin-7(8H)-yl]-4-(2,4,5-trifluorophenyl)-, (3S)-RefChem:69332

名称をさらに表示

Sitagliptin S-EnantiomerSitagliptin-13(S)?-?SitagliptinSitagliptin EP Impurity A CRSSitagliptin impurity SKY-8-ISO

物性

構造計算による物性

XLogP: 0.7
極性表面積: 77 Å²
水素結合供与体数: 1
水素結合受容体数: 10
回転可能結合数: 4
重原子数: 28
形式電荷: 0
複雑度: 566

物理化学的性質

沸点: 529.9 · 529.9±60.0 °C(Predicted)
比重(密度): 1.61 · 1.61±0.1 g/cm3(Predicted)
酸解離定数(Pka): 7.2 · 7.20±0.10(Predicted)
InChI: InChI=1S/C16H15F6N5O/c17-10-6-12(19)11(18)4-8(10)3-9(23)5-14(28)26-1-2-27-13(7-26)24-25-15(27)16(20,21)22/h4,6,9H,1-3,5,7,23H2/t9-/m0/s1
InChIKey: MFFMDFFZMYYVKS-VIFPVBQESA-N
SMILES: C(N1CCN2C(C(F)(F)F)=NN=C2C1)(=O)C[C@@H](N)CC1=CC(F)=C(F)C=C1F