Grignard Synthesis of Chiral Amphetamines-Rhodium

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Rhodium
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05-31-04 12:15
No 510550

Grignard Synthesis of Chiral Amphetamines
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Asymmetric ?-Substituted Phenethylamines. V.
Synthesis of Chiral 1-Alkyl-2-phenylethylamines via Grignard Reaction of 4-Phenyl-1,3-oxazolidines
Hiroshi Takahashi,Yasuhiro Chida, Kimio Higashiyama, and Hiraku Onishi
Chem. Pharm. Bull. 33(11), 4662-4670 (1985) (https://www.rhodium.ws/pdf/----.grignard.oxazolidine.pdf)

Abstract
Chiral N-methyl-4-phenyl-1,3-oxazolidines (2a-e) having a methyl, ethyl, benzyl, isopropyl, and cyclohexyl group at the 2-position of the 1,3-oxazolidine ring were synthesized. Reactions of 2a-e with Grignard reagents gave (1R,1'R)- and (1S,1'R)-1-alkyl- and 1-cycloalkyl-N-2'-hydroxy-1'-phenylethyl-2-phenylethylamines (3a, 3b, 3d, 3e). The absolute configurations of (1R,1'R)-3a and -3e were determined. (R)-1-Methyl- and (R)-1-cyclohexyl-2-phenylethylamines (4a, 4e) were obtained in high yield by hydrogenolysis of (1R,1'R)-3a and -3e.

All in all, this reaction may look very complicated, but it really isn't anything else than the procedure in Post 202649 (Rhodium: "---- via grignard rxn (Gazz Chim Italiana)", Novel Discourse) with the difference being that the aldehyde has been made into an oxazolidine by reaction with a cheap chiral aminoalcohol. The good thing about using this oxazolidine is that 1) The yields are much higher and 2) the result isn't racemic ----, but a single enantiomer. In the article they make l-methedrine, but by choosing the other aminoalcohol enantiomer, the reaction will instead produce d-methedrine.

How do you think this reaction could be streamlined? There is really just a single obstacle in this procedure, and that is the removal of the protection group in the final step, for which they use catalytic hydrogenation. What other aminoalcohols could be used, which would be easier to remove?

Could something else be used (which would be a breeze to remove afterwards) if one would be satisfied with racemic product?

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