posted 09-20-1999 03:49 PM         

??

Anyway, my understanding of the bubbling of the H2S is to generate SO3 radical.

H2S has its share of hazards in use and smell.

COULD Na thiosulfate, which generates

.....S2O3->SO3 + S.......... be produced IN the mixture through gradual acidification, or, perhaps through electrolytic means?

If so, and if pressurization would be an asset to the effectiveness of the process, this could be the cat's meow, or the molasses' revenge:

easy for me to get excited: in my mind it works perfectly, being unconstrained by nuisance considerations of chemical minutia.

Might need a particular anode: might need a 1/2 cell: might need good agitation.

Might be ridiculous.

I oughta change my name to "slap me if I need it" for these far out what if posts.

posted 09-20-1999 04:02 PM         

"OK, I'm a dumb shit"
"Slap me if I need it"
"Not too smart but I try"
"??but I try??
"Comeon: have some pity!!"

posted 09-21-1999 12:11 PM         

------------------
Regards,Hematite.

posted 09-21-1999 08:01 PM         

Hell, if I thought about it I woulda picked "Petunia" or something sweet and nice like I am, really!
===================

Now if I could JUST get Wiz or Rev. to comment on the thiosulphate idea.

I'm willing to experiment if the idea is not TOO ludicrous: but do not want to waste my time if its obviously a non-runner.

hey THHHHHHpitterdude(dodge the spray): howabout using your not inconsiderable influence with the Reverend to inveigle him into critiqueing this possibility??

TTTHHHHHHanx!

posted 09-23-1999 02:37 AM         

The quantitative conversion of the thiosulphate, S2O3(-2) ion to tetrathionate, S4O6(-2) ion is unique with iodine, I2; OTHER oxidizing agents Cl2 and Br2, tend to carry the oxidation further, to sulphate ions or a mixture of sulphate ions and tetrathionate ions.

The thiosulphate, S2O3(-2) ion tend to decompose to give sulphur and hydrogen sulphite ion: S2O3(-2) + H(+) <==> HSO3(-) + S
The rate of the decompostion reaction increases markedly with increasing acidity. When thiosulphate, S2O3(-2) is added to a strong acidic solution, the decompostion is immediate and the solution turns cloudy due to sulphur precipitating out.

I believe addition of thiosulphate, S2O3(-2) to iodoephedrine will result in dimerization of the iodoephedrine, due to radical formation and coupling. The thiosulphate, S2O3(-2) will release 2 electrons, BUT NO protons H(+) to react with the I(-) ion formed and the C6H5-CH(..)-CH(NHCH3)-CH3 radical anion formed.

Hydrogen sulphide, H2S and Hydrazine, N2H4 will release electrons and protons for the reduction of iodoephedrine.

C6H5-CH(I)-CH(NHCH3)-CH3 + H2S ==>> C6H5-CH2-CH(NHCH3)-CH3 + S + HI

posted 09-23-1999 03:02 PM         

Then as Wizard-x notes in detail the thiosulfate degenerates.

Once again then I return to the question of why use Sodium thiosulfate to clear the iodide out of the finished reaction mixture.

Because as soon as the alkalizing begins with Sodium hydroxide the same thing happens the Na+ ionizes then hooks up with I- to form
NaI.

posted 10-10-1999 04:47 AM         

The way I read the hydrazine reaction would mean that I would add one half of a mole of N2H4 for every mole of I2 to produce the initial HI, and I would add one-half of a mole of N2H4 for every mole of p/efed to remove the I and replace it with an H, and regenerate the HI.

This process looks like it would generate H2O as well.

Have you a suggestion for actually running this reaction?

Combine ingredients in the proper ratios with a minimum of water and reflux for a long time? A short time?

This sounds too interesting to ignore.

posted 10-14-1999 07:30 AM         

The generation on N2 gas would be an indicator of the reaction's progress.

posted 11-13-1999 05:11 PM         

Dont everybody understand the removal point being considered by making the freebase ya are removing the Haloid from the mixture and it is gona co [do not miss this is not a typo idiot]Carbon monoxide is formed from the classical, or shall we also consider the solvent burning or well we could complicate this further by discussing the phosphor,,,NO THANKS solvate with a strong base such as lye in the water layer the lesser base the amine will swim into the solvent where there is alot of hydrogen trapped in the matrix, or lattice therin, the free base of it should be void of at least one thing, the haloid interaction, wheater or not this problem of the ripping of the oxygen of the molecule can be preformed another way is a different and excitting idea, but it seems there are simpler ways, thus if ussing Iodine ya can form a salt simmiliar to HCL and thus to it can also form HI and what ya are hopping to do is twist off one of the oxygen atoms containing the oh group, the halid is capable of this under increased tempature and friction is ussed in these classical therums to rip the thing away, Now how this happens is a matter of theory, the best explainations are formed from metal addation in my opinion such as that sulfer is combined to the oxygen atom in a cascade that eventually rips way too the activly transiting site, such that other metals are often employed such as lithium or aluminium or sodium, any of these metals considering whatever reaction might be employed, the real question I find paculiar about HI is this that the final critique ends up being as from florine working your way through the respective haloids bromine clorine and finally the iodine has the most metaloid like properties and therefore is a catalys that preforms, it is the not so preffered catalyst by any stretch of the immagination it just works very well under a great varity of conditions and this is good for begeners. Now someone out there does know an easy and gentle way to be shure and the final work up of the product so happens to conclude with a shared acid bond formation paticuliar to the subject at hand being discussed for educational purposes, thus the sulferuos conponent is a shared acid that the isomeric amines share between them, this results in dont trust the litature it aint always right nor will it ever be fixed, hum war inside your head, nothing to be scared of except your imagination in any case.....hummm so the molecule is at an odd charge is it??? the sharring of from the two results in an odd charge dont it????

IT AN ODD CHARGE AINT IT???????