[Nescafe]

Hello,

So I read Cu(I) is the catalytically active form whereas Cu(II) is not. Cu(I) has 10 electrons in the d orbital whereas Cu(II) has 9 electrons in the d orbital. I dont understand why this means that Cu(I) is catalytically active while Cu(II) is not.

Can someone please explain this to me,

Thanks,

Sina. 

[Archer]

Can you elaborate on "catalytically active" please.

What particular reactions are you referring to?

[Nescafe]

Referring to copper catalyzed click chemistry where I have read that Cu(II) gets reduced to the catalytically active form Cu(I) by sodium ascorbate.

Thanks,

Nescafe.

[Archer]

I am not an inorganic chemist and my molecular orbital theory is a bit rusty so I could be completely wrong here but these are my thoughts.

I believe it has something to do with the occupation of the 3d-orbitals. In Cu(I) there are 10 electrons which makes them more accessible to the alkyne orbitals. in Cu(II) there are only 9 electrons in the 3d orbital which probably changes the orbitals in such a way that they do not interact with the alkyne orbitals so the reaction cannot occur.

I am very happy to be corrected on this by my colleagues with a better understanding of orbital theory and/or d-block chemistry.

[kriggy]

Ok I dont know much about click chemistry but I was thinking about this question since yesterday.

 according to http://www.organic-chemistry.org/namedreactions/click-chemistry.shtm there is Cu(III) formed. They dont discuss how but I would expect some sort of oxidative addition to the Cu atom. That might be the reason, Cu(I)  Cu(III) is possible, Cu(II) Cu(IV) i think not.
thoughts?

And the, what is the shape of the catalyst species? Square planar? Octahedral? It could have efects on how likely will the compound move to transition state (Like Cu(II) octahedral (or rather square bypiramidal)  Cu(II) pentagonal bipyramidal seems rather difficult due to ligand field stabilization, same for sq. bypiramidal to sq. pyramidal)

[Dan]

They dont discuss how but I would expect some sort of oxidative addition to the Cu atom. That might be the reason, Cu(I)  Cu(III) is possible, Cu(II) Cu(IV) i think not.
thoughts?

I think this is the reason. Cu(I)/Cu(III) cycles (involving oxidative addition of Cu(I)) are widely accepted mechanisms for Cu catalysis in many reactions (including Michael addition with Gilman reagents, and incidentally gives a much more satisfying explanation of the chemoselectivity of this reaction than the hard/soft argument I was taught as an undergraduate).

The failure of Cu(II) to catalyse these reactions is likely to be the higher energy barrier associated with access to Cu(IV).

Re: whether Cu(IV) can exist: Cs₂CuF₆ does exist (and there may be others).

[Nescafe]

Oxidative addition to Cu(III) makes sense.

I was under the impression that the Cu in the click chemistry remained at an oxidation state of +1 for majority of the first few steps. Looking at the mechanism cycle starting with CuLn at the start, the copper O.S. is at +1. This gives us 18 electrons with L = H2O and n = 4. After the first step in the cycle once the H is displaced and the Cu-sigma bond with the alkyne is formed we are still at the Cu O.S. of +1 are we not? What about when the Nitrogen of the azide comes in an the Cu is actually negatively charge. I guess I am not really understanding the O.S. of Copper in this cycle and any help would be appreciated

[Nescafe]

Any ideas Dan? Or anyone else on my last question

[kriggy]

Just to be clear oxidative addition to Cu(III) would mean oxidation of Cu to Cu(V). I think you mean oxidative addition to Cu(I). (and the more I think about this, the less it looks like oxidative addition because that would mean increase of coordination number by 2, but this doesnt happen in this case)

Anyway I was thinking about exactly the same thing as you are asking.
I briefly read the article and there is no mention about Cu in negative oxidation state.
I think after the coordinatnion of alkene there is still Cu(I). Then when the azide coordinates I think its still Cu(I). And total oxidation number of the complex is 0 because azide have both positive and negative charges. Then copper donates two electrons to form the dobule bond and gets oxidized to Cu(III) to form the 6 membered ring. In the next step is reduced back to Cu(I) and product is then released.

Note that oxidation number of metal is just a formal number and has nothing to do with the ligands. I mean if you have M²⁺ and you add 6 negatively charged ligands then you will have complex with total oxidation number of -4 but the metal will be in 2+ state. (unless it gets oxidized/reduced by the conditions). Some ligands can stabilize the low oxidation states of metals but I dont think that azide ligand has the properties for it.

[Nescafe]

Thanks Kriggy.

Sorry that is what I meant to say when I said oxidation to Cu(III) (meaning from Cu (I)) but I understand the confusion.

I agree with you that Cu stays at an oxidation state of I until it forms that ring where it becomes Cu(III). So it goes oxidation addition only at this step right? The first steps are not really oxidative addition as the oxidation remains at (I). I mean Cu displaces the H in the first step, it does not really under go OA.

[kriggy]

I dont think its oxidative addition, oxidative addition increases coordination number by 2 like
[PtCl₄]^2-+ Cl₂ → [PtCl₆]²⁻
and that doesnt seem to be the case here  because the coordintion number of Cu is 4 in all steps
F. Himo, T. Lovell, R. Hilgraf, V. V. Rostovtsev, L. Noodleman, K. B. Sharpless, V. V. Fokin, J. Am. Chem. Soc., 2005, 127, 210-216.

[Nescafe]

Hmmm I see what you mean, but Copper eventually does go from CuLn to R-Cu(Ln)-R though, but not in a single step but sequentially. I guess you are right, so it just oxidizes to Cu(I) to Cu(III) according to the mechanism?

[kriggy]

Its R-(CuL_n-2)-R or at least its shown in scheme here
http://www.organic-chemistry.org/abstracts/lit2/773.shtm
I think yes, it just gets  oxidized from I to III