@G.Patton I believe one of the drawn intermediates on this synthesis (and from the original rhodium article) is incorrect.
(3), the representation of the amphetamine-formaldehyde imine intermediate possesses the same structure as the phenylacetone-methylamine imine that can be seen elsewhere
this does not seem correct to me for two reasons: first, drawing the standard mechanism for imine formation on the substrate produces an imine wherein the double bond is between N and the would-be methyl group on the methamphetamine, and NOT between N and the alpha-carbon as in the listed structure. and think about it: the double bond is generated when the lone pair of the amine kicks off the water formed on the formaldehyde oxygen as a leaving group, which must be done on the side of the formaldehyde (the would-be methyl group)
second, the synthesis is reported as stereospecific based on the starting substrate. that is, the reduction of d-amph will yield d-meth and the same for the l-isomer. structure as is shown (3) possesses no chiral centers, and its reduction is equivalent to the standard reductive amination of phenylacetone which yields a racemic product. the proposed structure I've given for the intermediate possesses a chiral center, which would be the same as the starting material's and therefore match the observed stereospecificity of the reaction.
if i am correct, this would clear up the issues regarding the stereospecificity of the reaction earlier in the thread. if i have made a mistake somewhere, i apologize, but i was curious as to why the reduction of the non-chiral (3) leads to a non-racemic mixture of products when a non-racemic amphetamine starting material is used.
(3), the representation of the amphetamine-formaldehyde imine intermediate possesses the same structure as the phenylacetone-methylamine imine that can be seen elsewhere
this does not seem correct to me for two reasons: first, drawing the standard mechanism for imine formation on the substrate produces an imine wherein the double bond is between N and the would-be methyl group on the methamphetamine, and NOT between N and the alpha-carbon as in the listed structure. and think about it: the double bond is generated when the lone pair of the amine kicks off the water formed on the formaldehyde oxygen as a leaving group, which must be done on the side of the formaldehyde (the would-be methyl group)
second, the synthesis is reported as stereospecific based on the starting substrate. that is, the reduction of d-amph will yield d-meth and the same for the l-isomer. structure as is shown (3) possesses no chiral centers, and its reduction is equivalent to the standard reductive amination of phenylacetone which yields a racemic product. the proposed structure I've given for the intermediate possesses a chiral center, which would be the same as the starting material's and therefore match the observed stereospecificity of the reaction.
if i am correct, this would clear up the issues regarding the stereospecificity of the reaction earlier in the thread. if i have made a mistake somewhere, i apologize, but i was curious as to why the reduction of the non-chiral (3) leads to a non-racemic mixture of products when a non-racemic amphetamine starting material is used.
- By G.Patton
Hi. Firstly, a-carbon has more electron density than methyl carbon. Typical mechanism lead to such intermediate (hidden under Breaking Bad watermark, sry):
Secondly, do you have any reliable data about stereo-specific result of this synthesis?
Secondly, do you have any reliable data about stereo-specific result of this synthesis?
↑View previous replies…
- By NexusPrime
rhodium is a mistake, the double bond is not there, see the example with benzaldehyde below
