Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Lonestar on March 09, 2008, 12:45:24 AM
-
hi guys, just a quick question, why does a sonogashira reaction need to be done under inert gas atmosphere of nitrogen or argon?
it has something to do with the oxidation state of the active catalyst with i'm thinking is the oxidative insertion of the organohalide to the palladium complex.... is it cause Oxygen will interfer with this process and instead of the oranohalide complexing with the palladium, the oxygen will instead?
hence the use of unreactive gasses nitrogen or argon?
-
ok i had a look at wiki on this subject and it mentions "palladium unstable in air, it promotes the formation of homocoupled acetylenes"
but not too sure what that means or how it does it...
-
Oxygen can quite easily oxidize Pd(0) to Pd(II), which would kill the reaction since you need Pd(0) to initiate the mechanism. The bis(alkyne) products come from the Glaser coupling reaction, which doesn't involve Pd at all, but does involve Cu and molecular oxygen.
-
I have seen dimerization of alkynes in a Cu free sonogashira coupling a number of times. Depending on Pd source (Pd0 or Pd(II)) you can also see reagent dimerization in the Stille reaction.
-
The text book suggested that the Sonogashira reaction carried out under inert atmosphere though, you do not need.
Of course, you can get the product in high yield when you do under N2 or Ar. Pd(0) is active species and Pd(II) is inactive one. The O2 can make Pd(0) to be Pd(II) though, it takes much more time. You have to worry about the oxidation of ligand when you use phosphine as ligand. Alkyphosphine is more easily converted to phosphine oxide than arylhphsophine. Therefore, you have to do experimental under N2 or Ar when you used alkylphosphine.
-
these guys are right. oxygen will oxidize your pd(O) to pd(II) and therfore it would be unfavorable to undergo oxidative insertion into the aryl-halide or other bond in the starting material. doing this would form pd(IV) which is a highly unfavorable state for palladium to be in, altough it does happen in some reactions.