idontlie
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Proposed 5,6-(Methylenedioxy)-l-indanone from Caffeic Acid
Caffeic Acid is a pretty cool molecule all by its lonesome. Both a catechol and a carboxylic acid. It's structurally similar to cinnamic acid. But it
could be so much more interesting.
So what needs to be done to our molecule? Methylatenate and indane ring closure.
Methylenation occurs under basic conditions (NaOH, DCM, EtOH)... Will the Sodium Caffeic Salt ruin our chances of proceding or will the ions react
correctly in ring closure?
Ring closure is done by chlorinating the carboxylic acid with thionyl chloride then closed with Tin (IV) Chloride (SnCl4). Perhaps there are better
ways to perform this closure but I have not seen them. This will replace the OH's with Cl (chlorinate our catechol =( ) off of the phenyl ring. I
doubt there is any simple method for changing our two chlorines back to alcohols or to form the methylenedioxy group. [1]
From what I understand the thionyl chloride is not particular about which OH it wants to replace; If it was preferential to carboxylic acids we could
simply use the thionyl chloride as our limiting reactant added slowly. Unfortunately, this doesn't seem to be the case and doing the slow addition
would probably give yields around 5% of the wanted product.
This forces us to start methylenation. Methylenation will need to be done under inert atmosphere in basic conditions with halomethane, may be mixed.
Most likely Ar Gas, with EtOH-NaOH and DCM and may benefit from a PTC such as Aqualat 336.
My question is how could the methylenated salt be used to form the chlorinated acid intermediate and will the salt affect the reaction adversely? Or
could the salt be returned to its acidic state and continued with?
References:
[1] Nichols, David E. J. Med. Chem. 1990,33, 703-710
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Edit: Before you are able to chlorinate for ring close you must hydrogenate the double bond. This can be done using 10% Pd-C, raney nickel, ect at 60
psig H2 for 5h. [1, see prep for 3-[2,3-(Methylenedioxy)phenyl]propanoic Acid]. This may be the ideal first step of this reaction chain. followed by
methylenation, acidifing the salt, then carry onto the indanone formation.
[Edited on 21-1-2010 by idontlie]
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Bolt
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Quote: Originally posted by idontlie  |
Ring closure is done by chlorinating the carboxylic acid with thionyl chloride then closed with Tin (IV) Chloride (SnCl4). Perhaps there are better
ways to perform this closure but I have not seen them. This will replace the OH's with Cl (chlorinate our catechol =( ) off of the phenyl ring. I
doubt there is any simple method for changing our two chlorines back to alcohols or to form the methylenedioxy group. [1]
From what I understand the thionyl chloride is not particular about which OH it wants to replace; If it was preferential to carboxylic acids we could
simply use the thionyl chloride as our limiting reactant added slowly. Unfortunately, this doesn't seem to be the case and doing the slow addition
would probably give yields around 5% of the wanted product.
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Thionyl chloride will not chlorinate aryl alcohols, which is what you seem to be saying.
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idontlie
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Ah ha, Thank you!
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Sandmeyer
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Hi idontlie
If you want to make 5,6-Methylenedioxy-1-indanone from caffeic acid you might want to start by first reducing the double bond.
Then esterifying the carboxylic acid (goold ol' Fisher esterification with cat. H2SO4), methylenation for the bridge, basic hydrolysis of the ester.
Now you have 3,4-methylenedioxydihydrocinnamic acid - it can be chlorinated with thionyl chloride and converted to indanone (Friedel-Crafts). Keep in
mind that cinnamic acids can be made one-pot from aldehydes using a reaction called Perkin condensation, but you will need acetic anhydride (and
piperonal in this case).
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FreeMirage
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Sorry to dig up such an old thread but I just had to add that in the process of a Friedel-Crafts the methylenedioxy moiety will get cleaved.
The methylenation reaction, however, won't effect the aldehyde so save yourself some time and instead of dealing with an esterification and hydrolysis
just do the acid chlorification first and follow up with the Friedel-Craft and a methylenation.
[Edited on 3-10-2013 by FreeMirage]
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