반응식
[CH3:1][C@:2]12[C@@H:11]3[CH2:12][CH2:13][C@@:14]4([O:19][C@@H]5O[C@H](CO)[C@@H](O)[C@H](O)[C@H]5O)[C:16]([CH2:18][C@@:10]3([CH2:15]4)[CH2:9][CH2:8][C@@H:7]1[C@@:6]([C:32]([OH:34])=[O:33])([CH3:31])[CH2:5][CH2:4][CH2:3]2)=[CH2:17].[Na].P([O-])([O-])([O-])=O.SC[C@H]([C@@H](CS)O)O.O[C@H]1O[C@H](CO)[C@@H](O)[C@H](O)[C@H]1O>OP([O-])(O)=O.[K+].CC#N.C(Cl)(Cl)Cl.CO.O>[CH3:1][C@:2]12[C@@H:11]3[CH2:12][CH2:13][C@@:14]4([OH:19])[C:16]([CH2:18][C@@:10]3([CH2:15]4)[CH2:9][CH2:8][C@@H:7]1[C@@:6]([C:32]([OH:34])=[O:33])([CH3:31])[CH2:5][CH2:4][CH2:3]2)=[CH2:17]참가 물질
실험 절차
If Steviolmonoside, 3-sulfopropyl ester, sodium salt prepared in Part D. would be incubated anaerobically for three days at 37° C. with 5 wt% fresh rat cecal contents, at concentrations of 0.25, 0.5, and 1.0 mg/ml in sterile Krebs-Ringer 0.25 M phosphate buffer (pH 7.4) containing 0.25 mg/ml dithiothreitol and 0.25 mg/ml α-D-glucose, TLC [silica gel F-254; CHCl3 :CH3OH:H2O (15:10:2)] and HPLC [30 cm C-18 on μ-Bondapak; 15 min linear gradient of 10-40% CH3CN in 0.005 M KH2PO4 (pH 3.45); 200 nm] analysis would show all the initial ester (Rf=0.37; tR =14.6 min) to have been consumed within 24 h to yield apparently only the sulfopropyl ester of steviol, (Rf =0.63; tR =16.3 min). No steviol (Rf =0.95; tR =31.3 min) would be detected. Thereafter the bacterial cells would be sedimented by centrifugation. The sediment would be extracted (THF), as would the supernatant, after lyophilization. HPLC analysis for steviol of the two THF extracts should show none to be detectable, with a detection limit of 0.5 μg.
후처리
- customtR =14.6 min) to have been consumed within 24 h
출처 및 근거
- ORD 기록
- ord-ea07c47774fa4f2396ff80434709698b
- 출처 데이터셋
- uspto-grants-1983_09
- 특허 번호
- US04402990
연결된 노트
불러오는 중…