(Hive Bee)
06-18-02 11:32
No 322761
(Rated as: excellent)
For those of you who don't know, the wacker is composed of two simultaneous reactions. The first rxn is
R-CH=CH + PdCl2 + ROH __> R-C(OR)2-CH2 + Pd0 + 2HCl
The second rxn regenerates Pd+2 and varies depending on the system you use. For the benzo wacker
Pd0 + 2HCl + benzo __> PdCl2 + hydro
For the nitrite wacker
Pd0 + 2HCl + RONO __> PdCl2 + 2ROH + 2NO
And for the O2 wacker
Pd0 + 2HCl + 1/2O2 __> PdCl2 + H2O
In all of the above cases, cupric ion can be used as a catalyst for the reoxidation (thus the name cocatalyst) by the following mechanism.
Pd0 + 2CuCl2 __> PdCl2 + 2CuCl
The cupric ion is then regenerated by the reoxidant in the same manner that the Pd would have been regenerated in the absence of Cu.
In some cases, such as the benzo wacker, the reoxidation is so fast compared to the reaction rate of the alkene, that it is by no means rate determining. In the case of the O2 wacker and the nitrite wacker to a lesser extent, the reoxidation does influence the overall rate. The addition of copper, however, increases this rate so much that the reoxidation no longer limits the overall rate.
Now comes the tricky part. As you can see by the above equations, HCl is needed for reoxidation but is given off by, and greatly inhibits, the alkene oxidation. This inhibition arises because the alkene must displace the chloride ion from the Pd rxn center in order to form the rate limiting intermediate. Protons inhibit the reaction because deprotonation must occur to form this same complex. This is a very complex rxn, and I wrote a very detailed post on it earlier.
The point is, however, that you must balance the two rxns with the HCl concentration. If it is too high, the overall rxn will procede too slowly. If it is too low, the Pd will not be reoxidized fast enough and will polymerize and not participate in the rxn, even if more HCl is added.
As far as the nitrite wacker goes, after about 3 hrs of rxn a brownish-green precipitated falls out and the rxn stops. My current theory is that the precipitate is some sort of Cu+ salt (which are mostly insoluble). This happens because for some reason the HCl is depleted during the rxn, preventing the reoxidation of the copper. This in turn stops the reoxidation of the Pd, and the rxn grinds to a halt.
So where does the HCl go? My theory is that it escapes with the NO gas as NOCl formed by the following rxn.
RONO + HCl __> ROH + NOCl
Whether this rxn is favorable or not, the removal of product as gas will drive it to the right. Most likely, though, it has a very slow rate since H+ is probably needed to catalyze it. One thing that I have noticed is that doubling the volume and catalyst with the same amount of other reagents almost doubles the yield. Doubling the catalyst without doubling the volume does not affect yields on the same magnitude. I believe that this effect is caused because a dilution of the rxn contents slows the production of NOCl. These are my reasons
1) Double Volume and Catalyst - rxn rate doubles b/c of doubled catalyst. If NOCl production is first order wrt [RONO] and [HCl], it decreases b/c [RONO] decreases and [HCl] does not change.
2) Double volume only. Rxn rate does not change b/c catalyst amount does not change. Since [RONO] and [HCl] both decrease, NOCl production rate decreases and the rxn lasts longer.
3) Double catalyst only. Rxn rate doubles. [HCl] also doubles and [RONO] stays the same so NOCl production also doubles. Therefore, the increase and decrease offset eachother, the the overall rate is largely unaffected.
Currently, I can produce 50 g ketone in a 2L flask completely filled up regardless of how much safrole I add (assuming over 50 g, probably 60 g). This volumetrically sucks! No better than a fucking nitro amination.
Here's my solution. All of the HCl in the rxn comes from Cl- ion added as PdCl2 and CuCl2. Therefore, by adding more catalyst, you are also affecting another rxn variable, [HCl]. I probably have the optimum [HCl] concentration at the start b/c the rxn hums along nicely for the first few hours. In order to offset the [HCl] depletion, I should simply add more HCl after about two hours. I should also add some HCl inaddition to the intitial catalyst, since the optimum [HCl] is 0.1 M according to literature (This value is for ethylene, however, so may not be correct since ethylene reacts must faster). Currently, I only start with 0.08 M of HCl. I'm about to try it out and will post soon.
One more thing. My proposed rxn for the formation of NOCl is only an assumption. It could be catalyzed by metal ions such as the following.
NO + CuCl2 __> NOCl + CuCl
CuCl + RONO + HCl __> NO + ROH + CuCl2
Well, all this is hypothetical so far, so don't make any life or death decisions based on it. Now if only I could get some MeAm, my capacity would skyrocket!
(Stoni's sexual toy)
06-18-02 13:17
No 322790
I've seen a ref. where non-complexing strong acids (like HClO4) were added to the Wacker, and the reaction went MUCH faster.
Excessive halogenides are bad, because they will form a rather stable complex with the Pd, and this will slow down the reaction.
I'm not fat just horizontally disproportionate.
(Pioneer Researcher)
06-18-02 15:06
No 322823
Yes a very interesting reference that seems unexplored:
JOC 1990, 55, 2924-2927. They worked with Pd(AcO)2 and thenitrate, sulfate and chloride, PdCl2 didn't work.
I was thinking around to use Pd(AcO)2 in the nitrite, so acetic acid could be added to help the process.
(Stoni's sexual toy)
06-18-02 15:21
No 322830
I've also seen examples where using different Palladium salts resulted in different reaction products, e.g. phenylacetaldehyde or acetophenone from styrene. Something to consider when changing the catalyst!
I'm not fat just horizontally disproportionate.
(Pioneer Researcher)
06-18-02 18:44
No 322890
Yes, acording to that reference and the one cited by Osmium, the alcoholic wacker, the nitrate salt seems to be more selective than the acetate or the chloride in teh formation of methyl ketones.
(Hive Bee)
06-19-02 08:07
No 323072
Too much H+ is also bad. [H+] is 1st order inhibition. [Cl-] is 2nd order inhibition. Then again, you need both for reoxidation, so you must find the right balance.
Sunlight: Are the examples you are refering to done in acetic acid solvent? PdCl2 should work in alcoholic solvent if a small amount of acetic acid is added. The reaction in acetic acid proceeds by a completely different mechanism, giving vinyl esters. In alcoholic solvent, most of the acetic acid will be in its ester form anyway and will not affect [H+]. In fact, if the O2 wacker is run in ethanol, acetic acid is a byproduct.
The reason that different Pd salts affect the reaction differently arises from the fact that the cations are very important in the reaction mechanism. So is the solvent type. Therefore, every variation in these parameters results in a different mechanism.
Does anyone know if Pd(OH)2 can be isolated? If so, you could add Pd(OH)2 and Cu(OH)2 plus the acid corresponding to the cation that you want. This would give you more precise control over the rxn conditions.
(Pioneer Researcher)
06-19-02 10:13
No 323102
The idea of acetic acid was mine, using alcoholic solvente with a bit of water and some acetic in excess, and Pd(AcO)2, I've not tested it. The other rxns are in laoholic media, or the improved benzo in 7:1 acetonitrile water, and a 0.2 M concentration of HClO4, or nitric, sulfuric but not HCl.
If I remember correctly, in a patent posted by PolyethyleneSam, I read that PdCl2 was dissolved in a HCl solution, then NaOH solution added, the predipitated hydroxyde was washed several times with water to remove the Cl ions, then HNO3 added (and may be even acetic), to form a solution, bsified one more time and wsahed to remove the last traces of Cl, and filtered hydroxydes and then made the salt in that easy way again. Evaporation will give the nitrate , and if the acetate is not formed, addition of acetic acid to the nitrate solution and evaporation adding more acetic yield the acetate (I made this part time ago).
I was surprised becuase when I made the acetate, I precipitated Pd metal with formaldehyde, this way is easier, more direct.
(Hive Bee)
06-22-02 18:25
No 324236
I have the impression that after the first reaction, the final product is still not the ketone but the glycol or something like that. How does it becomes ketone, whithout having to reflux ? Sorry if I misunderstood something, who knows more please post.