2,2,2-Trifluoroethanol → Aluminum trifluoroethoxide

HRID 569601Notes

Reaction equation

HRID 569601 reaction equation
[F:1][C:2]([F:6])([F:5])[CH2:3][O-:4].[F:7][C:8]([F:12])([F:11])[CH2:9][O-:10].C([Al+2:15])C.[F:16][C:17]([F:21])([F:20])[CH2:18][OH:19].[Al]>C(Cl)Cl>[F:1][C:2]([F:6])([F:5])[CH2:3][O-:4].[F:7][C:8]([F:12])([F:11])[CH2:9][O-:10].[F:16][C:17]([F:21])([F:20])[CH2:18][O-:19].[Al+3:15]

Participants

5

Reactant 3

HCID 64092,2,2-TrifluoroethanolHCID 5359268AluminumHCID 21290501Ethylaluminum Di-(2,2,2-trifluoroethoxide)

Product 1

HCID 19419412Aluminum trifluoroethoxide

Reagents, catalysts & solvents 1

HCID 6344DichloromethaneSolvent2×

Conditions

Temperature
40 °C

Procedure

In a flask, under a nitrogen blanket, were placed 7.43 g (0.29 mole) of the product of Example 2 and 50 ml methylene chloride. Then, 3.07 g (0.031 mole) of 2,2,2-trifluoroethanol was added in a way so as to maintain the temperature at about 40° C. After the addition was complete, the flask was placed in an oil bath at 60° C. and most of the methylene chloride was evaporated. The product was allowed to stand at the same temperature for one hour and the remainder of the solvent stripped off under vacuum. There was obtained 9.09 g (95% of theoretical yield) of a white crystalline solid. Analysis for aluminum showed 8.36% Al. The theoretical value was 8.33%.

Workup

  1. customIn a flask, under a nitrogen blanket, were placed
  2. additionAfter the addition
  3. customwas evaporated
  4. customThere was obtained
  5. custom9.09 g (95% of theoretical yield) of a white crystalline solid

Sources & evidence

ORD record
ord-f1113ec866cb4cae87f808184b2052d2
Source dataset
uspto-grants-1988_03
Patent number
US04732993

Reaction notes

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