calcium carboxylic acid salt coupling reaction

[ning]

I would like to know more about the nature of calcium or barium or (iron?)-based coupling reactions. I know of two that are commonly employed here on the HIVE:

1. calcium (acetate,phenylacetate)--> P2P
2. calcium adipate --> cyclopentanone

I guess in both cases the other product is calcium carbonate.

Ca (OCOR1)(OCOR2) --> R1COCR2 + CaCO3

What is the limitation of this reaction?
What is its typical yield?

The idea I had was for yet another way of making ketamine. This one would work like so:

Molecule:

coupling reaction ("c1ccccc1C(=O)[O-].C2CCCC2C(=O)[O-].[Ca2+]>>c1ccccc1C(=O)C2CCCC2")



Both of the precursors are conceivably produceable from OTC materials (cyclohexanone --> cyclopentyl carboxylate, toluene --> benzoic acid), and no really nasty chemicals are involved. However, the reaction's success obviously depends on the effectiveness of the cross coupling reaction.

The two reactions involved are:
2A + 2B --> 2AB
2A + 2B --> AA + BB

How likely are each to occur, and what factors affect success?

Well, bees, what say you?

[Rhodium]

The two reactions involved are:
2A + 2B --> 2AB
2A + 2B --> AA + BB

How likely are each to occur?

It's directly proportional to the concentration of each reactant. If you use A/B in a 1:1 ratio you will probably get AA/BB/AB in a 1:1:1 ratio. With an A/B ratio of 3:1 you will probably get AA/AB/BB in ratios similar to 2.50:1.25:0.25.

What is its typical yield?

Here is an article describing the synthesis of P2P using this method, and the yields in it are twice those I've seen for any other unsymmetrical ketone:

https://www.thevespiary.org/rhodium/Rhodium/chemistry/p2p.paa-ca.html

Symmetrical ketones are all right to make using this method (if the reactants/products can take the harsh rxn conditions), but it's very hard to optimize fror unsymmetrical ketones.

What is the limitation of this reaction?

I believe that is the projected product distribution and overall yield.

[ning]

I'm not so sure about your math, Rhodium. Wouldn't it be the case like this?

Example using PhAcOH and AcOH:

1 mole Ca(OAcPh)2
1 mole Ca(OAc)2

divides into four:
Ca(X)(Y)

PhAcO- , PhAcO-
PhAcO- , AcO-
AcO-   , PhAcO-
AcO-   , AcO-

in amounts of 1/4 mole each.

However, the middle two are the same thing! So really, one would arrive at a ratio of 2:1:1, with the mixed product being 2, or ideally 50% yield. Of course, other mechanisms would reduce it from this ideal.

Correct me if I'm missing something here, of course

[Osmium]

It also depends on the reaction rates, you assumed that each reaction was taking place at the same speed.