[F:1][C:2]1[CH:7]=[CH:6][C:5]([N:8]2[C:16]3[C:11](=[CH:12][C:13]([CH:17]([C:22]4[CH:27]=[CH:26][CH:25]=[CH:24][CH:23]=4)[CH2:18][C:19](O)=[O:20])=[CH:14][CH:15]=3)[CH:10]=[N:9]2)=[CH:4][CH:3]=1.[NH2:28][C:29]1[S:30][CH:31]=[N:32][N:33]=1>>[F:1][C:2]1[CH:3]=[CH:4][C:5]([N:8]2[C:16]3[C:11](=[CH:12][C:13]([CH:17]([C:22]4[CH:23]=[CH:24][CH:25]=[CH:26][CH:27]=4)[CH2:18][C:19]([NH:28][C:29]4[S:30][CH:31]=[N:32][N:33]=4)=[O:20])=[CH:14][CH:15]=3)[CH:10]=[N:9]2)=[CH:6][CH:7]=1Example 136 was prepared from 3-(1-(4-fluorophenyl)-1H-indazol-5-yl)-3-phenylpropanoic acid (504 mg, 1.4 mmol) and 2-amino-1,3,4-thiadiazole using General Coupling Method A to give 310 mg (50% yield). MS found: (M+H)+=444. Resolution of this compound into its enantiomers could be accomplished using chiral HPLC as described above.
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