Methylparaben → HCID 57743438, compound

HRID 1095121Notes

Reaction equation

HRID 1095121 reaction equation
F[C:2]1[CH:9]=[CH:8][C:7]([I:10])=[CH:6][C:3]=1[CH:4]=[O:5].[OH:11][C:12]1[CH:21]=[CH:20][C:15]([C:16]([O:18][CH3:19])=[O:17])=[CH:14][CH:13]=1.C([O-])([O-])=O.[K+].[K+]>CN(C)C(=O)C>[CH3:19][O:18][C:16](=[O:17])[C:15]1[CH:20]=[CH:21][C:12]([O:11][C:2]2[CH:9]=[CH:8][C:7]([I:10])=[CH:6][C:3]=2[CH:4]=[O:5])=[CH:13][CH:14]=1

Participants

5

Reactant 3

HCID 7456MethylparabenHCID 11430Potassium CarbonateHCID 162170582-Fluoro-5-iodobenzaldehyde

Product 1

HCID 57743438HCID 57743438, compoundYield 69.1%

Reagents, catalysts & solvents 1

HCID 31374DimethylacetamideSolvent

Procedure

In a manner similar to the method described in Example 50a, 2-fluoro-5-iodobenzaldehyde (3.6 g, 14.4 mmol) (Aldrich) was reacted with methyl 4-hydroxybenzoate (2.62 g, 17.3 mmol) and K2CO3 in N,N-dimethylacetamide to give 4-(2-formyl-4-iodo-phenoxy)-benzoic acid methyl ester as a yellow oil (Yield 3.8 g, 69%).

Sources & evidence

ORD record
ord-a582058af6634970aaecc9f633338009
Source dataset
uspto-grants-2012_03
Patent number
US08134001B2

Reaction notes

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