Hydrochloric Acid → 4-(1-Ethoxyethyl)benzoic acid

HRID 2170467Notes

Reaction equation

HRID 2170467 reaction equation
[CH3:1][O:2][CH:3]([C:5]1[CH:14]=[CH:13][C:8]([C:9]([O:11]C)=[O:10])=[CH:7][CH:6]=1)[CH3:4].[OH-].[Li+].Cl.[CH3:18]O>O>[CH2:1]([O:2][CH:3]([C:5]1[CH:14]=[CH:13][C:8]([C:9]([OH:11])=[O:10])=[CH:7][CH:6]=1)[CH3:4])[CH3:18]

Participants

6

Reactant 4

HCID 313Hydrochloric AcidHCID 887MethanolHCID 3939Lithium HydroxideHCID 14597192Methyl 4-(1-methoxyethyl)benzoate

Product 1

HCID 189137424-(1-Ethoxyethyl)benzoic acidYield 97%

Reagents, catalysts & solvents 1

HCID 962WaterSolvent

Procedure

To the solution of methyl 4-(1-methoxyethyl)benzoate (181 mg, 0.95 mmol) in methanol and water (50 mL/10 mL) was added lithium hydroxide (50 mg, 2.1 mmol), the solution was stirred at room temperature for 12 hours. Then the solution was acidified by hydrogen chloride (1 N/mol) pH to 6, extracted with dichloromethane (50 mL*3), evaporated the solvent to give 4-(1-ethoxyethyl)benzoic acid (166 mg, 97%).

Workup

  1. extractionto 6, extracted with dichloromethane (50 mL*3)
  2. customevaporated the solvent

Sources & evidence

ORD record
ord-a47b5653508e463892dc129ae473cb7c
Source dataset
uspto-grants-2015_06
Patent number
US09051269B2

Reaction notes

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