the one (*still*) test that was positive so far involved adding a small amount of some kind of salt with an iodine ion, and there was significant generation of HI vapour which tested positive to starch.
I think the key to the reaction mechanism is that formic acid must donate two hydrides to the iodine at once. When formic acid is ionised in water, it only has one.
so what's the answer? I bet everyone is already formulating the solution....
Ammonium formate.
ammonium formate is a salt that can have the water removed, while retaining the ability to donate 2 hydrogens.
any other salt you use will leave you with a metal ion which has no hydrogen to donate.
so the reaction runs as several steps, like this:
NH4.HCOOH + I2 --> CO2 + HI.NH4 + HI
thus producing 1 mole of HI. It's probably important to know how much water this reaction can tolerate, and i would think that would relate directly to the water of crystallisation. I was looking at all the hydrates that i'm aware of, sodium sulphate decahydrate, magnesium sulphate heptahydrate and some ammonium hydrates... and it's quite clear to me that hydration is directly related to hydrogen bonds with the exposed oxygens, and the other place is occluded between the ions (i noticed that if there wasn't an oxy-acid present, that there generally was no more water of crystallisation than the number of ionic bonds). So i would venture to guess that ammonium contributes one, and the HI contributes whatever is left.
the amount of moles of water to HI present in fuming 57% HI would be related to the weight ratio directly, so what is that?
HI mol wt= 128
inverse of 57% is 43% thus 128x.43=55.04
so that works out to 55/18 - about 3, in other words, a tri-hydrate. I might be wrong about the rule of thumb regarding hydration that i have guessed on (based on the principle of water of hydration and occluded H+/OH- water between each bond) then ammonium hydriodide is a tetrahydrate.
so, with the water on the product side first and then balanced onto the left... The amount of water of crystallisation is important here because, as i said, we do not want there to be any available to break the source of two hydrogens, the ammonium formate. I think, if it is unable to dissolve because more water acquisitive salts have it all, that this will drive the reaction forwards.
NH4.HCOOH + I2 + (H2O)7 --> CO2 + HI.NH4.(H2O)4 + HI.(H2O)3
On the left hand side there, where it says 7 water molecules, we should count an extra one for the hydration water in the ephedrine.HCl. I think that the ammonium formate probably occludes one too, so we can count that one out of the additional water to add to allow the reaction to occur.
so, molar ratios and weights of everyting from the left hand side should be established so as to form a sound theory for a reaction, also i am clearly displaying all of the ions as ions. I am working on the assumption that ammonium formate forms a monohydrate.
NH4+.HCOO-.H2O = 18 + 45 + 18 = 81g/mol
I2 = 254g/mol
ephedrine H+.Cl-.H2O = 201 + 18 = 219g/mol
(H2O)5 = 5x18 = 90g/mol
The theory i have is that if the water of hydration levels are used, then the end product of the reaction will be maximum concentration HI, and the inactive salt will be in a semi-solid state, totally saturated with water, but no more, which means that the ammonium does not participate in the solution beyond it's "dry" crystal hydration level, and because the HI is likewise only hydrated to it's "wettest" possible level, there's no more water left to dilute the HI below working concentration. The water that is there, is specifically to keep the ammonia ionised as, if it were to boil off readily by being driven out of solution it would not provide its unique combo with formic acid of being able to hydrogenate other compounds when heated, because both have the ability to donate one hydrogen each to a suitable reactant.
you know something funny, this is like the combining of leuckart and HI reduction. the ammonium formate reduces the I2, binding one HI and leaving one free. The free iodine liberated then...
ooh this means a correction. the formula has to have twice as much ammonium formate as ephedrine because the ammonium formate/I2 reaction only produces one mole of HI, and two are needed to react the ephedrine. Thus the HI generation reaction has to be performed twice for each. The formic acid is consumed in the reaction, but the iodine comes back one molecule for every two moles of ammonium formate reacting with it. This means also that we must use one mole less additional water, as one mole is occluded in the ammonium formate.
(NH4+.HCOO-.H2O)2 = (18 + 45 + 18)x2 = 162g
I2 = 254g
ephedrine H+.Cl-.H2O = 201 + 18 = 219g
(H2O)4 = 3x18 = 72g
I think there is only one obstacle to this reaction working, and that is that quite clearly the ammonia is going to build up, and very likely dissociate and vent, making stinky catpiss smells. So i would think the condenser should have a hose attached to a container of acidic water, preferably a strong acid to ensure it ionises completely before escaping. this would probably also demand a suckback trap. i guess you would say 'push pull', but for once with a real purpose, to catch lost ammonia and probably hydriodic acid too. I would suspect that there might be some risk of the formation of ammonia HI crystals in the condenser.
Well, if i've figured this right, then the thread title should be: "substituting NH4-.HCOO+ for P(red)"
something else about this reaction that i am seeing is that there is no need for heat. the water catalyses it when it is present in the correct proportion, because ammonium formate is readily reacted with iodine with the right amount of water, as the ammonium first donates it's H, as it is a base which is already known to directly ionise iodine into hydriodic acid, forming ammonium iodide (which i've been referring to as ammonium hydriodide, which is probably incorrect). Then the ammonium is hungry for another hydrogen, so it steals it from the formic acid, which then disappears as carbon dioxide.
ok, i'm getting a little hazy here, here is the deal:
when the NH4 ionises the I2, it first of all changes the iodine into iodide ions (this is because it becomes NH3HI initially). this makes the formic acid vulnerable to ionic iodine, of which one has already been liberated by the ammonia, which attracts the hydrogen more strongly than the formic acid, causing it to decarboxylate, producing a HI molecule in the process, and venting carbon dioxide.
Thus the mechanism of the reaction is like this (I'm leaving the water out because the ratios calculated before still apply):
NH4-.HCOO+ + I2 --> NH4-.I+ + I+ + HCOO+ --> NH4-.I+ + HI+ + CO2
the ammonium splits the I2 by ionising one of the I atoms, thus ionising the other, which then seeks out a proton, which is most readily available from the formic acid, because it can then form carbon dioxide if it gives away it's hydrogen, and the carbon also helps by being able to take up the spare electron of the formic acid that cleavage caused by the iodine ion.
