Photo Engineer said:
He {me} is on the right track and most everyone is missing three essential factoids about this developer. Maybe more.
OK - I pulled out the big organic books and my brand new, personnal inscribed copy of "Modern Photographic Processing" by Grant Haist and here's what I come up with. (By the way, I thought we were trying to determine the source of the purple or blue color when making it up wrong from the Tylenol. I hope I'm on track now!)
Sorry if this gets a bit chemically for some people's tastes - it is sunny here and I should be outside, if not taking photos, at least mowing the lawn one last time for the year or bicycling...
Here's my take - starting with Gerald Koch's formulation and procedure:
Gerald Koch said:
Solution 1
Distilled water ......................... 625 ml
Paraminophenol HCl ...................... 100.0 g
Potassium metabisulfite ................. 300.0 g
Add a pinch of the bisulfite to the water and then add the p-aminophenol hydrochloride and when dissolved add the rest of the metabisulfite.
Solution 2
Distilled water (10°C) .................. 300 ml
Sodium hydroxide ........................ 200 g
Distilled water to make (10°C) .......... 400 ml
Method of Preparation
Concentrated paraminophenol formulas are prepared differently from other
developers. Follow the directions below carefully.
Add 280 ml of Solution 2 to Solution 1 with stirring. A precipitate of paramino-
phenol will form. Slowly add more Solution 2 with constant stirring until there
is a sudden darkening in color. At this point, begin adding Solution 2 drop by
drop until all but a small amount of the precipitate dissolves.
IT IS IMPORTANT THAT NOT ALL OF THE PRECIPITATE DISSOLVES.
Looking at that formula - one can see there is 0.687 mol of p-aminophenol and 0.97 mol of sulfite, a slight excess. I'm sure Spock would immediately see some thing interesting, it is a 2:3 ratio.
Adding a pinch of the potassium bisulfite to the water in solution one will remove most of the dissolved oxygen by reacting the sulfite that forms from the solution of the bisulfite, forming sulfate which is non-reactive.
Dissolving the p-aminophenol will free the HCl from it by ionization in solution. This leaves the pap in solution. It will react with some oxygen from the water/air interface (unless we protect the solution with a nitrogen bath). This oxidized form of pap is called quinoneimine - similar to quinone that forms when hydroquinone reacts with oxygen, but with the amine group on one side of the molecule, hence the "-imine" in the name. Note that a "pinch" of potassium bisulfite will not be enough to protect that much pap when it is going into solution.
The formation of the quinonimine also forms hydrogen peroxide (H2O2) as a byproduct (the oxygen pulls the hydrogen off the -OH (phenol) end of the pap). The H2O2 will react with some of the sulfite in solution too (like any oxygen) to make more sulfate ion and consume any excess sulfite.
So now once we have dissolved all the pap into the water, we should have water, pap, quinonimine with a pretty much equal amount of hydrogen peroxide, and some free chloride, sulfate, and potassium ions floating around. The lack of excess oxygen scavangers in solution will allow the quinonimine to react with the strong oxidizer hydrogen peroxide and this will form some hydroxyquinonimine. (Structure wise, this is a benzene ring with a "=O" (oxygen double bonded) at the 1-position, a -OH at the 2-position, and a -NH2 at the 4-position.) I suspect that the hydroxyquinonimine will polymerize with another hydroxyquinonimine to make a dimer - the simplist form of a polymer is when only two of the polymeric structures combine (hence one of PE's hints earlier), and I suspect it could even go on to polymerize into a multilinked compound.
Looking at hydroquinone, we can see that it can do this as well, going through the oxidation steps of hydroquinone <=> quinone <=> hydroxyquinone <=> polyhydroxyquinone. This polyhydroxyquinone is a member of the class of compounds called tannins and lignans, which are polyphenolic compounds. Some, like tannic acid can have a molecular wieght of around 2000 AMU. (Another class of compounds that I have experience analyzing as an inorganic chemist.)
So back to the p-aminophenol - when we get to the hydroxyquinonimine, that compound is an aldehyde. As I mentioned in the posts above, if these aldehydes make a dimer of a certain structure, then we can get the purple colored byproducts. Another example of the polymerization of aldehyde compounds can be seen with the simplest aldehyde - formaldehyde, H2C=O. It can polymerize into a compound called paraformaldehyde. It simply connects from one molecule of the formaldehyde to the next by breaking the aldehyde C=0 bond and then making series of [-O-C(H2)-O-C(H2)-O-C(H2)-O-] links. This linking is reversable with the paraformaldehyde and it can regenerate formaldhyde by simply reacting with water and hydrolyzing.
So now once we have dissolved all the pap into the water, we should have water, pap, quinonimine with some hydrogen peroxide, some hydroxyquinoneimine, and maybe some polyhydroxyquinonimine of various polymer lengths, and of course our free chloride, sulfate, and potassium ions floating around.
All this stuff must be going on before the potassium bisulfite is dissolved (in any significant amounts). Otherwise, we would not get the formation of the hydroxyquinonimine and any dimers or polymers of the hydroxyquinonimine. If we had lots of free sulfite in solution, then we would have the hydrogen peroxide that we released (when the quinonimine formed) turned into sulfate. The reaction from sulfite to sulfate pretty much goes in one direction, so any sulfate formed will stay as sulfate and not be converted back into sulfite at any point.
When the full amount of potassium bisulfite is dissolved into Solution 1, then we will get some of the quinonimine reacting with the free sulfite ions, SO3=, to form p-aminophenol monosulfonate. If we have a large excess of sulfate, then we can get a pap disulfonate to form. Again, we can see the same reactions happen with hydroquinone. (Hydroquinone + O2 <=> quinone, quinone + sulfite <=> quinone monosulfonate, quinone monosulfonate + sulfite <=> quinone disulfonate.) We don't have a large amount of sulfite ions to work with, so I suspect that we don't get a lot of the disulfonate forming in this formulation.
Notice that nowhere does the sodium hydroxide in the Rodinal come into play here, which is why the approach of dissolving the NaOH into solution before the pap is wrong. Also, to promote any polymerization of the pap, we need to dissolve it before the sulfite ions are released into the solution.
So now once we have dissolved all the pap and bisulfite into the water, we should have water, pap, quinonimine, some hydroxyquinoneimine, and maybe some polyhydroxyquinonimine of various polymer lengths, some p-aminophenol monosulfonate, some free sulfite, and of course our chloride, sulfate, and potassium ions floating around.
There is a race between the oxidation of the pap to form the quinonimine and related compounds and the formation of the monosulfonate at this point in the solution... Since the formation of the hydrogen peroxide would be pretty quickly consumed by the large excess of the sulfite in the solution at this point, this would drive the reaction into the formation of the p-aminophenol monosulfonate direction, especially since the sulfate formed from the reaction with the peroxide is non-reversable.
Also, since there is a slight excess of sulfite, that may drive the formation of the monsulfonate only. But remember, it is not a huge excess - only 50% more sulfite than pap.
Now, as Gerald pointed out, we add the hydroxide solution carefully so we do not overtitrate all the precipitated pap compounds we worked so hard to form. The hydroxide would pull the hydrogen off the phenol group (-OH) on the p-aminophenol monosulfonate as well as any unreacted pap since the -OH group is an organic acid. The sodium salt salt that forms would be very water soluble. This would allow us to make a very saturated solution that will have the maximum developer concentration possible.
Side note: the NaOH is a 50% w/w solution in Gerald's Solution #2. This is a commonly available, strong (and I think the strongest) solution of Sodium Hydroxide that is routinely commercially available. Maybe this was a benefit for Agfa in that they could buy it by the rail car and then pump it into the plant as needed...
PE - Did I happen to hit on the three essential factoids in your estimation?
1) Formation of beneficial polymerization of the pap oxidation byproducts (and not non-beneficial ones that form colored compounds which are unusable as developers.)
2) Maximum developer concentration possible for pap-based developers.
3) An absolute minimum of free sulfite in solution.
OK - now I got to go and mow the lawn...
Kirk -
www.keyesphoto.com
