313369-25-4
中文名稱
313369-25-4
英文名稱
(6bR,10aS)-Ethyl 3-methyl-2-oxo-2,3,6b,7,10,10a-hexahydro-1H-pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8(9H)-carboxylate
CAS
313369-25-4
分子式
C17H21N3O3
分子量
315.37
MOL 文件
313369-25-4.mol
更新日期
2026/04/02 11:23:18
313369-25-4 結(jié)構(gòu)式
基本信息
中文別名
(6BR,10AS)-3-甲基-2-氧代-2,3,6B,7,10,10A-六氫-1H-吡啶[3',4':4,5]吡咯[1,2,3-DE]喹喔啉-8(9H)-羧酸鹽(6BR,10AS)-乙基3-甲基-2-氧代-2,3,6B,7,10,10A-六氫-1H-吡啶[3',4':4,5]吡咯[1,2,3-DE]喹啉-8(9H)-羧酸鹽
(6BR,10AS)-2,3,6B,9,10,10A-六氫-3-甲基-2-氧代-1H-吡啶基[3',4':4,5]吡咯并[1,2,3-DE]喹喔啉-8(7H)-羧酸乙酯
乙基(6BR,10AS)-3-甲基-2-氧代-2,3,6B,9,10,10A-六氫-1H-吡啶基[3',4':4,5]吡咯并[1,2,3-DE]喹喔啉-8(7H)-羧酸鹽
(6BR,10AS)-3-甲基-2-氧代-2,3,6B,7,10,10A-六氫-1H-吡啶并[3',4':4,5]吡咯并[1,2,3-DE]喹喔啉-8(9H)-羧酸叔丁酯
(6BR,10AS)-乙基3-甲基-2-氧代-2,3,6B,7,10,10A-六氫-1H-吡啶基[3',4':4,5]吡咯并[1,2,3 -DE]喹喔啉-8(9H)-羧酸鹽
英文別名
LUMAT-006(6BR,10AS)-ETHYL 3-METHYL-2-OXO-2,3,6B,7,10,10A-HEXAHYDRO-1H-PYRIDO[3',4':4,5]PYRROLO[1,2,3-DE]QUINO
ethyl 3-methyl-2-oxo-2,3,6b,7,10,10a-hexahydro-1H-pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8(9H)-carboxylate
Ethyl (6bR,10aS)-3-methyl-2-oxo-2,3,6b,9,10,10a-hexahydro-1H-pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8(7H)-carboxylate
Ethyl (6bR,10aS)-2,3,6b,9,10,10a-hexahydro-3-methyl-2-oxo-1H-pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8(7H)-carboxylate
(6bR,10aS)-Ethyl 3-methyl-2-oxo-2,3,6b,7,10,10a-hexahydro-1H-pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8(9H)-carboxylate
(6bR,10aS)-3-methyl-2-oxo-2,3,6b,9,10,10a-hexahydro-1H,7H-pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8-carboxylic acid ethyl ester
1H-Pyrido[3',4':4,5]pyrrolo[1,2,3-de]quinoxaline-8(7H)-carboxylic acid, 2,3,6b,9,10,10a-hexahydro-3-methyl-2-oxo-, ethyl ester, (6bR,10aS)-
所屬類別
有機(jī)原料：雜環(huán)化合物物理化學(xué)性質(zhì)
沸點(diǎn)525.6±50.0 °C(Predicted)
密度1.34±0.1 g/cm3(Predicted)
儲(chǔ)存條件Sealed in dry,Room Temperature
酸度系數(shù)(pKa)2.95±0.20(Predicted)
InChIInChI=1S/C17H21N3O3/c1-3-23-17(22)19-8-7-13-12(9-19)11-5-4-6-14-16(11)20(13)10-15(21)18(14)2/h4-6,12-13H,3,7-10H2,1-2H3/t12-,13-/m0/s1
InChIKeyFAGHUWIRBRMPCV-STQMWFEESA-N
SMILESN1(C)C2=C3C([C@]4([H])CN(C(OCC)=O)CC[C@]4([H])N3CC1=O)=CC=C2
