Fluorene → HCID 601495, compound

HRID 275503Notes

Reaction equation

HRID 275503 reaction equation
[CH:1]1[C:13]2[CH2:12][C:11]3[C:6](=[CH:7][CH:8]=[CH:9][CH:10]=3)[C:5]=2[CH:4]=[CH:3][CH:2]=1.[CH2:14]([Li])[CH2:15][CH2:16][CH3:17].Br[CH:20](Br)[CH3:21]>O1CCCC1>[CH:1]1[C:13]2[CH:12]([CH2:17][CH2:16][CH:15]3[C:14]4[CH:21]=[CH:20][CH:8]=[CH:7][C:6]=4[C:5]4[C:4]3=[CH:3][CH:2]=[CH:1][CH:13]=4)[C:11]3[C:6](=[CH:7][CH:8]=[CH:9][CH:10]=3)[C:5]=2[CH:4]=[CH:3][CH:2]=1

Participants

5

Reactant 3

HCID 6853FluoreneHCID 112011,1-DibromoethaneHCID 178418339Butyllithium

Product 1

HCID 601495HCID 601495, compoundYield 70%

Reagents, catalysts & solvents 1

HCID 8028TetrahydrofuranSolvent2×

Procedure

About 8.3 g (0.05 m) of fluorene was dissolved in 150 ml of tetrahydrofuran. Then 31.8 ml (0.05 m) of butyl lithium (1.6 molar in hexane) was added dropwise to this solution. After one hour, 2.3 ml (0.25 m) of dibromoethane in 25 ml of tetrahydrofuran was added. The solution was stirred for 3 hours. The yellow solution was washed with 50 ml of an aqueous NH4Cl solution (5 g NH4Cl/50 ml H2O), then washed with 50 ml of water and then the organic phase was dried over Na2SO4. Then the solvent was removed in vacuo. The light yellow residue was washed twice with 25 ml of pentane. The resulting product was white. The yield was 12.5 g, i.e. a yield of about 70%, based on the moles of fluorene reacted. The product was confirmed through 1H NMR, 13C NMR, mass spectroscopy, and gas chromatography.

Workup

  1. washThe yellow solution was washed with 50 ml of an aqueous NH4Cl solution (5 g NH4Cl/50 ml H2O)
  2. washwashed with 50 ml of water
  3. dry with materialthe organic phase was dried over Na2SO4
  4. customThen the solvent was removed in vacuo
  5. washThe light yellow residue was washed twice with 25 ml of pentane
  6. customreacted

Sources & evidence

ORD record
ord-5fbdde780d8246f08d93d5e899ea3cdf
Source dataset
uspto-grants-1995_02
Patent number
US05393911

Reaction notes

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