Dan_Ny Posted March 14, 2011 Share Posted March 14, 2011 (edited) Beare, N. A.; Hartwig, J. F., J. Org. Chem. 2002, 67, 541. Well, inspired by the methodology from the paper above, I tried to make a palladium-catalyzed intramolecular cyclization on a substrate like this: When I use KOtBu as the base, nearly half of the product is the hydrodehalogenated one (the one below). Cyclization occurs quantitative, though and without side-products, when I use NaOtBu. I totally do not understand that. The solvent is toluene. Na3PO4 does not work at all and K3PO4 works as good as KOtBu. Can anybody explain these strange results? Thanks, Dan Edited March 14, 2011 by Dan_Ny Link to comment Share on other sites More sharing options...
Horza2002 Posted March 14, 2011 Share Posted March 14, 2011 While I am not sure, I cant think if several potential issues with a reaction like this. The first one will be the length of the linking carbon chain...if it is too short, then the reaction will be very slow because the resulting ring will be very strained. From reading the paper, they observer that the different reactions need different bases. One potential reason is the solubility of the resulting anion. I know that various sodium salts are pretty insoluble in THF as a solvent...and I would imagine they will be even more insoluble in toluene. I don't have much experience with potassium salts but I would expect that they are similar. Additionally, I would guess that there is a difference between the sodium and potassium enolate reaction rate with the palladium complex. In the case of the potassium case, the rate of palladium insertion is much faster than the rate of deprotonation. As a result, there is not enough enolate anion present to react and so the palladium complex is reduced to give the reduced product. In thase case of the sodium, the rate of deprotonation is faster and so there is enough of the enolate anion to react with the palladium complex. Link to comment Share on other sites More sharing options...
mississippichem Posted March 14, 2011 Share Posted March 14, 2011 Additionally, I would guess that there is a difference between the sodium and potassium enolate reaction rate with the palladium complex. In the case of the potassium case, the rate of palladium insertion is much faster than the rate of deprotonation. As a result, there is not enough enolate anion present to react and so the palladium complex is reduced to give the reduced product. In thase case of the sodium, the rate of deprotonation is faster and so there is enough of the enolate anion to react with the palladium complex. I'm going with this explanation. Solubility of enolates in organic solvents is usually a borderline case anyway. It would be interesting to know what the active form of the catalyst was though which might have some implications as to how it could interact with counter-ions. Link to comment Share on other sites More sharing options...
Dan_Ny Posted March 15, 2011 Author Share Posted March 15, 2011 (edited) Thank you for the answers, guys. Yeah, that sounds reasonable. Do you think that sodium would coordinate more to the enolate than potassium and thus being more soluble in toluene? I should try to find out if the reaction works faster with lithium, then... Edited March 15, 2011 by Dan_Ny Link to comment Share on other sites More sharing options...
Horza2002 Posted March 15, 2011 Share Posted March 15, 2011 The sodium will be coordinated more than the potassium and the lithium will be even more aswell. However, I seriously doubt any of them will dissolve in toluene as they are all ionic. Enolate salts are sparingly soluble in THF but is soluble in DMF. Another version you could try is to use a strong base (e.g. LDA or NaHMDS) to completey deprotonate the enolate; this would mean that the rate of reaction will be dependant on the rate of palladium insertion into the C-Br bond (which is very fast). You would need to use a very bulky base, both the ones I suggested should be good, to reduce the amount of addition to the nitrile. I know they said in the paper that strongbases didn't work, but they point out that eeach reaction is different. Link to comment Share on other sites More sharing options...
Dan_Ny Posted March 15, 2011 Author Share Posted March 15, 2011 (edited) The sodium will be coordinated more than the potassium and the lithium will be even more aswell. However, I seriously doubt any of them will dissolve in toluene as they are all ionic. Enolate salts are sparingly soluble in THF but is soluble in DMF. Even though that is logical, the reaction proceeds perfectly if toluene is used. Edited March 15, 2011 by Dan_Ny Link to comment Share on other sites More sharing options...
Horza2002 Posted March 15, 2011 Share Posted March 15, 2011 Yes it will work, a small amount of the enolate will be in solution to allow the reaction to occur. Link to comment Share on other sites More sharing options...
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