[BCP_Che_Engineer]

Hi everyone,

So I'm a chemical engineering grad student and oddly enough I have been tasked with doing chemistry (strange I know). I'm currently doing atom-transfer radical polymerization (ATRP) using a metal (CuCl  or CuBr) and a ligand (dNbpy or PDMETA). My monomer is a methacrylate group with a tert-butyl dimethyl silyl tail on it. Currently, I'm having an issue with my polymerization and I think it's caused by catalyst (CuCl) poisoning, but I was hoping someone here might be able to shed some light on the situation.


Current State:
I have CuCl, lightish green so it's oxidized a little bit, but I assume it's no big deal. I place in roughly a 2:1 molar ratio of ligand (dNbpy) and metal (CuCl). I then proceed to put in my monomer and initiator (the initiator is a 1:1 molar ratio with the metal compound). The dNbpy is solid and so is the CuCl, the monomer and initiator being liquids. As soon as the liquids meet with the solids, I get a reddish color that will (over the course of about ten to twenty minutes) go from red, to brown, to green. I know that Cu(II)Cl2 is a green color, so I assume that once it becomes green, the majority of the metal is in its oxidized state. For ATRP, the key (as I understand it) is for propagation to be slow and initiation to be fast. Control is possible through the activation of a radical by the initiator (or halogen capped group) and the Cu(I)Cl reacting to form a radical and Cu(II)Cl2. This is a reversible reaction so assuming that activation and deactivation occur at roughly the same speeds, I'd expect the reaction to be red with all Cu(I)Cl present, brown with a mixture of Cu(I)Cl and Cu(II)Cl2 present, and green when pure Cu(II)Cl2 is present. So I figured my solution, during polymerization, should be brown...however...the solution is green! This makes worried, especially since the polymerization is going 7 times slower than it normally should.

Today I attempted to "very quickly" add the monomer and initiator and then freeze-pump-thaw (I always do this before polymerization) three times before the solution turned green. After three freeze/pump/thaw cycles, I saw no bubbles in the solution and assumed all the oxygen was gone. The solution was still red, and I then placed it in a heated oil bath (75oC) for the polymerization. 15-20 minutes later I checked on the polymerization (it's a bulk polymerization by the way) and the solution was green. I run all my polymerizations under vacuum, rather than over nitrogen or argon.

Can anyone tell me if the color of ATRP-type polymerizations with CuCl is suppose to red, brown, or green during the polymerization? Catalyst poisoning is the only criminal I can think of for my failed reactions.

[phth]

Not sure if you did, but the CuCl will complex with itself (μ-Cl bridges) in the solid and the liquid.  Cl is a good ligand for this type of conformation.  Dissolve the ligand and CuCl first.  Go by monitoring the reaction with UV-VIS abs spec.  That will tell you what your photocatalyst is i.e. Cu(II) vs Cu(I) and how much, kinetics of the rxn, etc. You can also measure the active photocatalyst by cyclic voltammetry; this will tell you how good your catalyst is, and what modification of the ligands does to the catalyst.

[BCP_Che_Engineer]

Thank you for the reply. I do recall reading that CuCl can oxidize in light. Our lab though is fitted with UV filters I believe (it's yellow lighting) since we do photoresist chemistry in the lab area. I guess the UV-Vis would tell me for certain if Cu(I)Cl or Cu(II)Cl2 is present in my reaction but I guess what I'm more uncertain about is whether Cu(II)Cl2 is suppose to be more present or not. Does the reduction of Cu(II)Cl2 proceed at a faster rate than the oxidation of Cu(I)Cl? This redox reactions controls the chain growth in ATRP, so I suspect that a well controlled reaction has the reduction of Cu(II)Cl2 (capping of the propigating radical) happening at a faster rate than the oxidation of Cu(I)Cl (formation of a radical for propagation).

I've used CuBr before and it reacts much more readily than CuCl, so I was warned that CuCl would be a slower reaction. I was also told that CuCl should give me tight control over the molecular weight distribution than CuBr. This makes me think that the reduction of the metal should occur (capping of radical) more often, meaning that my solution should have a majority of Cu(I)Cl and have a brownish color to it during the reaction, however I still see my reaction as green. 

[phth]

You can look at the rates with the UV-VIS to answer that question.  Calculation seems like it would take a long time.  You could try varying the amount of catalyst.  if 10% get poisioned of 10% eq then you get 9% effective; if 10% gets poisioned of 20% eq then you get 18% active etc.  You could also try to intentionally poision the catalyst. You can use an indicator like benzophenone to test your oxygen removal.