4-Methoxyphenylacetic Acid → 4-Methoxyphenethyl alcohol

HRID 1106097Notes

Reaction equation

HRID 1106097 reaction equation
[CH3:1][O:2][C:3]1[CH:8]=[CH:7][C:6]([CH2:9][C:10](O)=[O:11])=[CH:5][CH:4]=1.[H-].[Al+3].[Li+].[H-].[H-].[H-]>CCOCC>[CH3:1][O:2][C:3]1[CH:8]=[CH:7][C:6]([CH2:9][CH2:10][OH:11])=[CH:5][CH:4]=1

Participants

4

Reactant 2

HCID 76904-Methoxyphenylacetic AcidHCID 21226445Aluminium lithium hydride

Product 1

HCID 697054-Methoxyphenethyl alcoholYield 85%

Reagents, catalysts & solvents 1

HCID 3283Diethyl EtherSolvent2×

Procedure

A solution of 100 g (0.60 mol) of p-methoxyphenylacetic acid in ether is introduced into a 4 l reactor, under nitrogen, which contains 28 g (0.74 mol) of lithium aluminum hydride and 600 ml of anhydrous ether. The mixture is heated under reflux for 3 hours and then cooled in a bath of ice and salt and the excess hydride is removed by cautiously adding 25 ml of water, followed by 25 ml of 15% strength NaOH and, finally, 20 ml of water. The white precipitate is removed by filtration, and after evaporation of the ether, 80 g of an oil are obtained. This oil, when distilled under 0.07 mm Hg at a temperature of 115° C., gives 77 g of 2-(p-methoxyphenyl)-ethan-1-ol. Yield: 85%.

Workup

  1. temperatureThe mixture is heated
  2. temperatureunder reflux for 3 hours
  3. temperaturecooled in a bath of ice and salt
  4. customthe excess hydride is removed
  5. additionby cautiously adding 25 ml of water
  6. customThe white precipitate is removed by filtration
  7. customafter evaporation of the ether, 80 g of an oil
  8. customare obtained
  9. distillationThis oil, when distilled under 0.07 mm Hg at a temperature of 115° C.

Sources & evidence

ORD record
ord-055fa6731bad4e53bd99dd1c523c6b80
Source dataset
uspto-grants-1982_01
Patent number
US04311708

Reaction notes

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