4-Fluorophenol → HCID 58757382, compound

HRID 916699Notes

Reaction equation

HRID 916699 reaction equation
[CH2:1]([N:3]1[C:7]([C:8]2[CH:9]=[C:10]([C:14]([NH:17][S:18]([CH2:21][C:22]([F:25])([F:24])[F:23])(=[O:20])=[O:19])([CH3:16])[CH3:15])[CH:11]=[CH:12][CH:13]=2)=[CH:6][C:5]([CH:26]([OH:28])[CH3:27])=[N:4]1)[CH3:2].[F:29][C:30]1[CH:35]=[CH:34][C:33](O)=[CH:32][CH:31]=1.C1C=CC(P(C2C=CC=CC=2)C2C=CC=CC=2)=CC=1.CC(OC(/N=N/C(OC(C)C)=O)=O)C>>[F:29][C:30]1[CH:35]=[CH:34][C:33]([O:28][CH:26]([C:5]2[CH:6]=[C:7]([C:8]3[CH:9]=[C:10]([C:14]([NH:17][S:18]([CH2:21][C:22]([F:23])([F:24])[F:25])(=[O:20])=[O:19])([CH3:16])[CH3:15])[CH:11]=[CH:12][CH:13]=3)[N:3]([CH2:1][CH3:2])[N:4]=2)[CH3:27])=[CH:32][CH:31]=1

Participants

5

Reactant 4

HCID 97324-FluorophenolHCID 11776TriphenylphosphineHCID 5363146Diisopropyl azodicarboxylateHCID 58757367HCID 58757367, compound

Product 1

HCID 58757382HCID 58757382, compound

Procedure

prepared following the procedure of Example 1 step 6 using the alcohol from Step 2 (121 mg, 0.29 mmol), 4-fluorophenol (49 mg, 0.44 mmol), PPh3 (113 mg, 0.44 mmol), and DIAD (85 mg, 0.44 mmol). The crude mixture was purified by column chromatography on silica eluting with 70% Et2O/iso-hexane and then by normal phase preparative HPLC to yield the ether (2.0 mg, 1%). 1H NMR (CDCl3, 500 MHz) δ 7.60-7.45 (3H, m), 7.36-7.30 (1H, m), 7.00-6.90 (4H, m), 6.28 (1H, s), 5.40 (1H, q, J=6.6 Hz), 4.90 (1H, s), 3.53-3.42 (4H, m), 1.82 (6H, s), 1.69 (3H, d, J=6.5 Hz), 1.40 (3H, t, J=7.2 Hz). MS (ES+) C24H27F4N3O3S requires: 513, found: 514 (M+H+, 100%).

Workup

  1. customprepared
  2. customThe crude mixture was purified by column chromatography on silica eluting with 70% Et2O/iso-hexane
  3. customby normal phase preparative HPLC to yield the ether (2.0 mg, 1%)

Sources & evidence

ORD record
ord-b895afeef5cb418eb27dfc31672aa167
Source dataset
uspto-grants-2007_09
Patent number
US07271191B2

Reaction notes

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