Lithium borohydride → (1-(4-fluorophenyl)-6-isopropyl-1H-indazol-5-yl)methanol

HRID 563373Notes

Reaction equation

HRID 563373 reaction equation
[F:1][C:2]1[CH:7]=[CH:6][C:5]([N:8]2[C:16]3[C:11](=[CH:12][C:13]([C:20](OC)=[O:21])=[C:14]([CH:17]([CH3:19])[CH3:18])[CH:15]=3)[CH:10]=[N:9]2)=[CH:4][CH:3]=1.[Li+].[BH4-]>C1COCC1>[F:1][C:2]1[CH:3]=[CH:4][C:5]([N:8]2[C:16]3[C:11](=[CH:12][C:13]([CH2:20][OH:21])=[C:14]([CH:17]([CH3:19])[CH3:18])[CH:15]=3)[CH:10]=[N:9]2)=[CH:6][CH:7]=1

Participants

4

Reactant 2

HCID 4148881Lithium borohydrideHCID 67005486Methyl 1-(4-fluorophenyl)-6-isopropyl-1H-indazole-5-carboxylate

Product 1

HCID 67005739(1-(4-fluorophenyl)-6-isopropyl-1H-indazol-5-yl)methanolYield 93%

Reagents, catalysts & solvents 1

HCID 8028TetrahydrofuranSolvent2×

Procedure

Methyl 1-(4-fluorophenyl)-6-isopropyl-1H-indazole-5-carboxylate (175 mg, 0.56 mmol) was dissolved in 5 mL THF and 1M LiBH4 in THF (3 mmol) was added. Heated at reflux. After 12 h, the reaction was quenched with MeOH, poured into brine and extracted with EtOAc×3. The organic layers were dried over MgSO4, filtered, and concentrated. The residue was purified on SiO2 by MPLC using 1:3 EtOAc/hexane. Obtained 148 mg (93% yield) of (1-(4-fluorophenyl)-6-isopropyl-1H-indazol-5-yl)methanol. MS found: (M+H)+=285.

Workup

  1. temperatureHeated
  2. temperatureat reflux
  3. customthe reaction was quenched with MeOH
  4. additionpoured into brine
  5. extractionextracted with EtOAc×3
  6. dry with materialThe organic layers were dried over MgSO4
  7. filtrationfiltered
  8. concentrationconcentrated
  9. customThe residue was purified on SiO2 by MPLC

Sources & evidence

ORD record
ord-34c7fef8de8e4afc844832f1e5f82e73
Source dataset
uspto-grants-2011_11
Patent number
US08067447B2

Reaction notes

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