Bromine → HCID 11220991, compound

HRID 393545Notes

Reaction equation

HRID 393545 reaction equation
[Br:1]Br.[O:3]1[C:11]2[C:6](=[N:7][CH:8]=[CH:9][CH:10]=2)[N:5]=[C:4]1[N:12]1[CH:18]2[CH2:19][CH2:20][N:15]([CH2:16][CH2:17]2)[CH2:14][CH2:13]1.C([O-])(=O)C.[Na+]>O.C(O)(=O)C>[Br:1][C:9]1[CH:10]=[C:11]2[O:3][C:4]([N:12]3[CH:18]4[CH2:17][CH2:16][N:15]([CH2:20][CH2:19]4)[CH2:14][CH2:13]3)=[N:5][C:6]2=[N:7][CH:8]=1

Participants

6

Reactant 3

HCID 24408BromineHCID 517045Sodium AcetateHCID 11311114HCID 11311114, compound

Product 1

HCID 11220991HCID 11220991, compound

Reagents, catalysts & solvents 2

HCID 176Acetic AcidSolventHCID 962WaterSolvent

Procedure

Bromine (0.12 mL, 2.29 mmol) was added to a solution of 4-oxazolo[4,5-b]pyridin-2-yl-1,4-diaza-bicyclo[3.2.2]nonane (560 mg, 2.29 mmol, prepared in Example 27) and sodium acetate (2.26 g, 27.5 mmol) in water (12 mL) and acetic acid (12 mL). The resulting mixture was heated to reflux for 2 h. The mixture was cooled and extracted with ethyl acetate (3×). The combined organic layers were washed with water (2×) and brine (1×) and dried over sodium sulfate, filtered and concentrated. The crude residue was purified by chromatography (Biotage, 25M) using a gradient elution from 4% MeOH/CHCl3 containing 0.1% NH4OH to 8% MeOH/CHCl3 containing 0.1% NH4OH giving 578 mg (78%) of the title compound as an oil: 1H NMR (CDCl3, 400 MHz) δ 8.23 (d, 1H, J=1.7 Hz), 7.50 (d, 1H, J=1.7 Hz), 4.51 (br s, 1H), 3.92 (br s, 2H), 3.16-3.04 (m, 4H), 3.02-2.94 (m, 2H), 2.17-2.01 (m, 2H), 1.83-1.74 (m, 2H); MS (Cl) m/z 325.0/323.0 (M+1).

Workup

  1. temperatureThe resulting mixture was heated
  2. temperatureto reflux for 2 h
  3. temperatureThe mixture was cooled
  4. extractionextracted with ethyl acetate (3×)
  5. washThe combined organic layers were washed with water (2×) and brine (1×)
  6. dry with materialdried over sodium sulfate
  7. filtrationfiltered
  8. concentrationconcentrated
  9. customThe crude residue was purified by chromatography (Biotage, 25M)
  10. washa gradient elution from 4% MeOH/CHCl3 containing 0.1% NH4OH to 8% MeOH/CHCl3 containing 0.1% NH4OH giving 578 mg (78%) of the title compound as an oil

Sources & evidence

ORD record
ord-a65a6450e58d4c498d865b95e604af81
Source dataset
uspto-grants-2004_10
Patent number
US06809094B2

Reaction notes

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