where please?
The "SUMMARY OF THE INVENTION" includes the following:
According to one embodiment of the invention, the amine (VI) is selected from the group consisting of ephedrine (R4 = OH, R5 = phenyl, R2 = methyl, R6 = R3 methyl), isoetharine (R4 = OH, R5 = 3,4-dihydroxyphenyl, R2 = ethyl, R6 = R3 = isopropyl), ritodrine (R4 = OH, R5 = 4- hydroxyphenyl, R2 = methyl, R6 = R3 = 2- (4- hydroxyphenyl) ethyl), methamphetamine (R4 = H, R5 = phenyl, R2 = methyl, R6 = R3 = methyl), fenfluramine (R4 = H, R5 = 3-trifluoromethylphenyl, R2 = methyl, R6=R3= ethyl) and propylhexedrine (R4 = H, R5 = cyclohexyl, R2 = methyl, R6 R3 = methyl). These compounds are preferably formed using hydrogen and a catalyst as the reductant, although they can be formed using a hydride reducing agent.
According to an alternative embodiment of the invention, the amine (VI) is selected from the group consisting of amphetamine, methoxamine, phenylpropanolamine, hydroxyamphetamine, ethylnorepinepherine and metaraminol. These compounds may be formed using a hydride reducing agent as the reductant.
I'm a Newbee, so please forgive and correct my errors. The imine that is formed from the L-PAC is 2-(methylamino)-1-phenyl-1-propanol, which needs to be reduced twice to get to methamphetamine. However, it looks to me like there will be ephedrine in situ because it is the result of a single reduction. How is this taking two steps forward and one step back?
Another question, why should this be done under strong acidic conditions?