Isocyanate addition -> MDMA?

[Rhodium]

Check out the sub-header ORGN 615. "New Synthesis of Oxazolines" at

http://www.albmolecular.com/features/tekreps/vol04/no27/

A LAH reduction of the formed iodo-isocyanate would definitely give the methylamino derivative. If we used the reaction on safrole, LAH reduction would give MDMA in one step after the addition. Perhaps even NaBH4 in diglyme would do the job, it is said to reduce alkyl iodides under those conditions. What do you think?

[PrimoPyro]

Ha!  

Look closely at those oxazoline rings! The nitrogen is on the alpha carbon, and the oxygen on the beta. When I presented one of my pretty pictures on the couch, I had mistakenly made this "error" and was called on it  

How ironic. Does it even make a difference (psychedelically) which way those two atoms are ordered off the chain?

Also, can we assume that the propenyl benzene would produce the 4-Me derivative oxazoline ring?

                                                 PrimoPyro
Vivent Longtemps la Ruche! STRIKE For President!

[Rhodium]

No, I was thinking that I-NCO [from AgNCO + I2 -> INCO + AgI(s)] would be added across the double bond of safrole, making the iodine sticking off the terminal carbon, and the -NCO  sticking out from the 2-position. LAH reduction would dehalogenate the alkyl iodide to the 2-isocyanatosafrole, and at the same time reducing the isocyanate group to H2O and -NHCH3.

My idea was not to make the oxazolidine, just perform the iodoisocyanation and then reduce.

Also, in regards to your question about the N/O-orientation, G. I. Poos says in US Pat 3278382 that 2-amino-4-phenyl-oxazolidines does not have the same central activity as 2-amino-5-phenyl-oxazolidines (aminorex), so I guess it is very important that the molecule stays a phenylethylamine (with the other N/O orientation, it would be a benzylamine).