Methanol → Methyl 3-bromo-4-fluorobenzoate

HRID 851638

Reaction equation

HRID 851638 reaction equation
[F:1][C:2]1[CH:10]=[CH:9][C:5]([C:6](F)=[O:7])=[CH:4][C:3]=1[Br:11].[CH3:12][OH:13]>N1C=CC=CC=1>[CH3:12][O:13][C:6](=[O:7])[C:5]1[CH:9]=[CH:10][C:2]([F:1])=[C:3]([Br:11])[CH:4]=1

Participants

4

Reactant 2

HCID 887MethanolHCID 129227073-Bromo-4-fluorobenzoyl fluoride

Product 1

HCID 7018022Methyl 3-bromo-4-fluorobenzoateYield 93.1%

Reagents, catalysts & solvents 1

HCID 1049PyridineCatalyst

Procedure

221 g (1 mole) of 4-fluoro-3-bromobenzoyl fluoride were added dropwise to a mixture of 96 g (3 moles) of methanol and 1 g of pyridine. The reaction mixture was then boiled for a further 2 hours under reflux and was subsequently worked up by distillation. 217 g (93% of theory) of 4-fluoro-3-bromobenzoic acid methyl ester were thus obtained in the form of a colorless oil which solidified when cold and had a boiling point of 124° C./22 mm Hg (melting point 29° to 30° C.).

Workup

  1. temperatureunder reflux
  2. distillationwas subsequently worked up by distillation

Sources & evidence

ORD record
ord-115a8ce7d24449d0a15d336ad898641d
Source dataset
uspto-grants-1983_05
Patent number
US04386035

Linked notes

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