Sodium Iodide → HCID 57586567, compound
Reaction equation
[F:1][C:2]1[CH:3]=[C:4]([C:8]2[CH:13]=[C:12]([O:14]C)[CH:11]=[C:10]([I:16])[CH:9]=2)[CH:5]=[CH:6][CH:7]=1.[I-].[Na+].C[Si](Cl)(C)C>>[F:1][C:2]1[CH:3]=[C:4]([C:8]2[CH:9]=[C:10]([I:16])[CH:11]=[C:12]([OH:14])[CH:13]=2)[CH:5]=[CH:6][CH:7]=1Participants
4
Procedure
A similar procedure as described in Example 12, step 6 was used, starting from 3′-fluoro-3-iodo-5-methoxy-biphenyl (1.1 g, 3.35 mmol), sodium iodide (5.024 g, 33.52 mmol), and trimethylsilyl chloride (2.12 mL, 16.76 mmol) to give 3′-fluoro-5-iodo-biphenyl-3-ol (1.03 g, 98%) as an amorphous light brown oil: ES(+)-HRMS m/e calculated for C12H8FIO (M+H)+ 314.9677, found 314.9676.
Sources & evidence
- ORD record
- ord-1f0c2b8c5f9a4a62836fe75c15899e16
- Source dataset
- uspto-grants-2012_01
- Patent number
- US08093385B2
Reaction notes
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