DL-2-(4-chlorophenyl)propanoic acid → 2-(4-Chlorophenyl)-1-propanol
Reaction equation
[Cl:1][C:2]1[CH:7]=[CH:6][C:5]([CH:8]([CH3:12])[C:9](O)=[O:10])=[CH:4][CH:3]=1.[H-].[Al+3].[Li+].[H-].[H-].[H-]>>[Cl:1][C:2]1[CH:3]=[CH:4][C:5]([CH:8]([CH3:12])[CH2:9][OH:10])=[CH:6][CH:7]=1Participants
Procedure
The synthesis of intermediates leading to different values of m and X is within the capability of one skilled in the art. Example 3 details the method of making the compound wherein X is --CH(CH3)-- and m is 2. This synthesis starts with the reduction of 2-(4-chlorophenyl)propionic acid with lithium aluminum hydride, producing 2-(4-chlorophenyl)propanol. Oxidation of this alcohol with pyridinium chlorochromate in methylene chloride produces 2-(4-chlorophenyl)propionaldehyde. This aldehyde is then reacted with (carbethoxymethylene)triphenylphosphorane, yielding ethyl 4-(4-chlorophenyl)-2-pentenoate. Hydrogenation of this ester in the presence of platinum oxide produces ethyl 4-(4-chlorophenyl)pentanoate which is then hydrolyzed with sodium hydroxide to the corresponding pentanoic acid (II). The remaining steps of this synthesis are shown in Schema 1.
Workup
- customThe synthesis of intermediates
Sources & evidence
- ORD record
- ord-abff4d5dbdf543f1aa97eaa952bd0040
- Source dataset
- uspto-grants-1996_03
- Patent number
- US05502054
Reaction notes
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