Triphenylphosphine → HCID 59093854, 화합물

HRID 1167682노트

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HRID 1167682 반응식
[O:1]=[C:2]1[NH:6][C:5](=[O:7])[C:4]2([CH2:12][CH2:11][N:10]([C:13]([O:15][C:16]([CH3:19])([CH3:18])[CH3:17])=[O:14])[CH2:9][CH2:8]2)[NH:3]1.[CH3:20][O:21][C:22](=[O:46])[C@H:23]([CH2:44]O)[NH:24][C:25]([C:38]1[CH:43]=[CH:42][CH:41]=[CH:40][CH:39]=1)([C:32]1[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=1)[C:26]1[CH:31]=[CH:30][CH:29]=[CH:28][CH:27]=1.C1(P(C2C=CC=CC=2)C2C=CC=CC=2)C=CC=CC=1.N(C(OCC)=O)=NC(OCC)=O>C1COCC1>[CH3:20][O:21][C:22]([C@@H:23]([NH:24][C:25]([C:38]1[CH:43]=[CH:42][CH:41]=[CH:40][CH:39]=1)([C:26]1[CH:27]=[CH:28][CH:29]=[CH:30][CH:31]=1)[C:32]1[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=1)[CH2:44][N:6]1[C:5](=[O:7])[C:4]2([CH2:8][CH2:9][N:10]([C:13]([O:15][C:16]([CH3:19])([CH3:18])[CH3:17])=[O:14])[CH2:11][CH2:12]2)[NH:3][C:2]1=[O:1])=[O:46]

참가 물질

6

실험 절차

A solution of 3.6 g (13.37 mmol) of (34.2) in 15 ml of absolute THF is added to a solution of 3.72 g (10.3 mmol) of (34.3) and 3.5 g (13.34 mmol) of triphenylphosphine in 20 ml of absolute THF. 2.11 ml (13.37 mmol) of diethyl azodicarboxylate (DEAD) are added to this solution and the reaction mixture is stirred at room temperature until the reaction has come to an end. The solvent is removed in vacuo and the residue is chromatographed through silica gel using heptane/ethyl acetate. 6 g (95%) of (34.4) are obtained after concentrating the product fractions.

후처리

  1. customThe solvent is removed in vacuo
  2. customthe residue is chromatographed through silica gel

출처 및 근거

ORD 기록
ord-5ea45d0e79e44886a907e357ec190980
출처 데이터셋
uspto-grants-2003_09
특허 번호
US06620820B2

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