4-Methoxyphenylacetic Acid → 4-Methoxyphenethyl alcohol

HRID 1414144Notes

Reaction equation

HRID 1414144 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-fc280f9001c14f208079d5d378c202f5
Source dataset
uspto-grants-1981_02
Patent number
US04252984

Reaction notes

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