Hey everyone. I was bored one night, so I began to read my handy O Chem textbook (I'm still studying, still a student). I then realized something.
Pyrocatechol (catechol) is a very available compound. Rhodium describes a synthesis of safrole via an allyl ether. The allyl ether was then heated to above 200 Celsius to mediate a Claisen rearrengement to form 4-allyl catechol. This is used to synthesize safrole. However, it might not be necessary to go to safrole from pyrocatechol. What if pyrocatechol was propionated using propionic anhydride to form the propionate ester. Treat this with AlCl3 to undergo Fries rearrangement to the ketone. Unfortunately, either the 3 or 4 position would receive the ketone. I suppose one could avoid this by acquiring the corresponding aniline derivative (catechol with the amine group on the 3 position). Synthesize the aryl diazonium salt to remove this group (not before the Fries rearrangement). However, my book (which isn't 100% correct, but I am terrible at searching for references outside Rhodium or the book) states that the Fries rearrangement achieves a 64% yield for the para isomer (not of the catechol, something else). From your ketone you could easily add the methylene group to form the methylenedioxy.
Am I simply mentally masturbating? Am I way off? I will draw up a synthesis scheme on Chemsketch once I get home... Otherwise. Feedback? References? Already been done?
Pyrocatechol (catechol) is a very available compound. Rhodium describes a synthesis of safrole via an allyl ether. The allyl ether was then heated to above 200 Celsius to mediate a Claisen rearrengement to form 4-allyl catechol. This is used to synthesize safrole. However, it might not be necessary to go to safrole from pyrocatechol. What if pyrocatechol was propionated using propionic anhydride to form the propionate ester. Treat this with AlCl3 to undergo Fries rearrangement to the ketone. Unfortunately, either the 3 or 4 position would receive the ketone. I suppose one could avoid this by acquiring the corresponding aniline derivative (catechol with the amine group on the 3 position). Synthesize the aryl diazonium salt to remove this group (not before the Fries rearrangement). However, my book (which isn't 100% correct, but I am terrible at searching for references outside Rhodium or the book) states that the Fries rearrangement achieves a 64% yield for the para isomer (not of the catechol, something else). From your ketone you could easily add the methylene group to form the methylenedioxy.
Am I simply mentally masturbating? Am I way off? I will draw up a synthesis scheme on Chemsketch once I get home... Otherwise. Feedback? References? Already been done?