Water → Boc-Phe(4-Cl)-N-Me-Val-N-Me-Tyr(3-tBu)-NH2

HRID 1988234노트

반응식

HRID 1988234 반응식
[NH:1]([C:14]([O:16][C:17]([CH3:20])([CH3:19])[CH3:18])=[O:15])[C@H:2]([C:11]([OH:13])=O)[CH2:3][C:4]1[CH:9]=[CH:8][C:7]([Cl:10])=[CH:6][CH:5]=1.[NH:21]([CH3:46])[C@H:22]([C:26]([N:28]([CH3:45])[C@H:29]([C:42]([NH2:44])=[O:43])[CH2:30][C:31]1[CH:36]=[CH:35][C:34]([OH:37])=[C:33]([C:38]([CH3:41])([CH3:40])[CH3:39])[CH:32]=1)=[O:27])[CH:23]([CH3:25])[CH3:24].O>C1COCC1>[NH:1]([C:14]([O:16][C:17]([CH3:20])([CH3:19])[CH3:18])=[O:15])[C@H:2]([C:11]([N:21]([CH3:46])[C@H:22]([C:26]([N:28]([CH3:45])[C@H:29]([C:42]([NH2:44])=[O:43])[CH2:30][C:31]1[CH:36]=[CH:35][C:34]([OH:37])=[C:33]([C:38]([CH3:41])([CH3:39])[CH3:40])[CH:32]=1)=[O:27])[CH:23]([CH3:25])[CH3:24])=[O:13])[CH2:3][C:4]1[CH:5]=[CH:6][C:7]([Cl:10])=[CH:8][CH:9]=1

참가 물질

5

실험 절차

To a solution of Boc-Phe(4-Cl)-OH (354 mg, 1.18 mmol), N-Me-Val-N-Me-Tyr(3-tBu)-NH2 (0.33 g, 0.908 mmol) and CMPI (301 mg, 1.18 mmol) in THF (8 ml), TEA (0.38 ml, 2.72 mmol) was added under cooling with ice and stirred at room temperature overnight. The reaction mixture was mixed with water and extracted with ethyl acetate. The organic layer was washed with saturated brine, dried over anhydrous magnesium sulfate and evaporated to remove the solvent under reduced pressure; the thus obtained residue was subjected to silica gel column chromatography (developing solvent: chloroform:methanol:aqueous ammonia=40:1:0.05) to give Boc-Phe(4-Cl)-N-Me-Val-N-Me-Tyr(3-tBu)-NH2 (0.45 g, 77%).

후처리

  1. temperatureunder cooling with ice
  2. extractionextracted with ethyl acetate
  3. washThe organic layer was washed with saturated brine
  4. dry with materialdried over anhydrous magnesium sulfate
  5. customevaporated
  6. customto remove the solvent under reduced pressure

출처 및 근거

ORD 기록
ord-a54ac501124f43a9811ee1750d928657
출처 데이터셋
uspto-grants-2009_06
특허 번호
US07553969B1

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