[Cl:1][C:2]1[CH:3]=[CH:4][C:5]([OH:8])=[N:6][CH:7]=1.[H-].[Na+].Br[CH2:12][C:13]1[CH:22]=[C:21]2[C:16]([CH:17]=[C:18]([C:27]([O:29][CH2:30][CH3:31])=[O:28])[CH:19]([C:23]([F:26])([F:25])[F:24])[O:20]2)=[CH:15][C:14]=1[Cl:32]>>[Cl:32][C:14]1[CH:15]=[C:16]2[C:21](=[CH:22][C:13]=1[CH2:12][N:6]1[CH:7]=[C:2]([Cl:1])[CH:3]=[CH:4][C:5]1=[O:8])[O:20][CH:19]([C:23]([F:26])([F:25])[F:24])[C:18]([C:27]([O:29][CH2:30][CH3:31])=[O:28])=[CH:17]2The formation of the ester was performed according to the procedure outlined in Example 157 Step 4 starting with 97 mg of 5-chloro-2-pyridinol, 20.7 mg of sodium hydride (60% dispersion in mineral oil) and 300 mg of the bromide from Example 157 Step 3. The compound was purified on the FlashMaster® chromatography system eluting with 25% EA/hexane then 50% EA/hexane to give 128 mg (38%) of the desired compound.
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