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	All 11 posts	Subject: The reactions in a RP phudo Iodine mix

	12345x (Stranger) 11-01-02 16:18 No 375378		The reactions in a RP phudo Iodine mix	Bookmark Reply
			A few things are going on in the pot and its not HI + phudo--> meth first thing that reacts is red phos and Iodine ie: P4 + I2 ---> PI3 balanced reaction:    (P4) + 3(I2)--->4(PI3) and if Phos is 37gm/m and Iodine is 127 gm/m then  35gms P uses 3*127gms I =apx 10 gms Iodine for every gm Phos for the initial start reaction 1 part P4 10 parts I2  PI3 reacts with eather phudo  to form iodioephedine    or with water to form HI equations: A. PI3 + 3(phudo)---> 3(iodioephedrine)    A. says phudo in a 3:1 ratio with PI3 forms iodioephed    but more importanly Iodine in PI3 reacts in a 1:1 ratio    with phudo to form iodioephed I to phudo 1:1 to form (I-phudo) B. 2(PI3) + 3(H2O) --> 6(HI)   1 molucule of water "has two atoms hydrogen " with 2 atoms iodine form thru PI3   into 2 molucles HI that contane 2 atoms Iodine   so water weighs 18 relitive to iodine weighs 127 then 18 wt parts water to 2*127 wt parts iodine or apx 14 wt parts  I    to    1 wt part   water C. HI + Iodioephed --> meth + I2    1 part HI and 1 part Iodioephed form 1 part meth +                                          2 parts Iodine note in this reaction 1:1 ratio the iodoephed but iodine is also generated in a two to 1 ratio this tells us that we could use less then the amount need  in the begining ie: say we used half of the iodine needed at the start then half of the phudo would be converted to meth and that would relise all the iodine that would start reactioning as in the beginining with the other half witch would finish all the rest of the phudo intro meth then we would need the amount of 1 part of phos to to parts of this iodine to form HI witch would then be aclear solution then done. this means you can use less then the amounts of iodine needed to react all the phudo in one pass and then it just has to loop that many more times to cook all the phudo. i figured that if you used enough I to do it in one pass it would take around 4mins  sum up these reactions 1 part Phos 10 parts iodine ---> PI3 1 part water 14 parts Iodine thru PI3 --> HI 127 parts iodine thru PI3 168 ? parts phudo ---------> 205 parts iodioephed   127 parts HI   168 parts phudo to iodioephed ---> 152 meth +254 Iodine these are all wt parts ie:127gms HI and 168 gm phudo forms 152 gms meth and 254 gms iodine in the end because all the reactions must use PI3 then the initial ratio I to P is the ratio your Phos and Iodine get consumed and because the excess is always Phos if not you would have a black pot. you will need at least 1gm of phos for every 10 gms of Iodine. any more phos then this will be extra for the next batch. ill post some ratios in the next post.
	12345x (Stranger) 11-02-02 12:04 No 375691		PI3 reacts with .....	Bookmark Reply
			ok we put red p and iodine together they unite forming PI3 PI3 is a  brown waxy solid that fumes in air and melts around the same temp as candle wax. PI3 swaps 1 at a time its 3 iodine atoms for all things hydoxy (OH) ie: with water H-O-H  + I-P-I --> + HO-P-I + H-I once                  |            |                       I            I --------------------------------------------------------- twice   H-O-H   +  HO-P-I -->  HO-P-OH  + H-I                       |           |                       I           I    --------------------------------------------------------- three   H-O-H   +  HO-P-OH-->  HO-P-OH  + H-I times                 |           |                       I           OH    ---------------------------------------------------------- the result is phosphorous acid and 3 hydrogen iodide   molucules or hydodic acid when desolved in water. PI3 also reacts with the hydoxy group on phudo:    //-P-I  + HO-C-\\ --> //-P-OH   + I-C-\\ shown is just the parts of the PI3 and phudo molucules that get changed. PI3 can do this 3 times with 3 phudos just like it does with water above. the name of the phudo with iodine attached (HO) group removed is iodioephedrine the other halogens also form haloephedrines when using the correct halogenating reagent: ie bromoephedrine,and chloroephedrine question ... what are the conditions for the PI3 /water reaction what are the conditions for the PI3/phudo reaction PI3 /water : pI3 will react with water very quickly "as fast as it it formed" at all temps and pressures when the water is a gas and will also react very quickly with liquid water when the water is in a state such as the water isnt saturated with ions. let me explain : if you desolve acid as a gas into water the water desolves the acid gas by breaking the acid into a pair of ions  1 positve ion and a negitive ion it takes a certon number of water molucules to keep those two ions shelded away from each other if they get back together they form a gas and cant stay as a liquid if you add more acid gas molucles to the water more water molucules get tied up sheilding them if you keep doing this a point is reached where there isnt any more free water moluclues avalable to sheild any more ions.at this point we say the solution is saturated this is why pool acid can only be so stong ie: 35% in water ditto for HI 57% in water .why is hi 57% and poolacid 35%  because HCL and HI are differnt sizes and there for it takes a differnt number of water molucules to sheild them if you try to add more gas to 35% pool acid or 57% HI solution the gas  wount stay in the water because there isnt any water molucules that arnt already shilding acid ions ,that would be avalable to break the acid gas into ions and sheild them from each other. so think of water as being in 2 differnt forms 1. water molucules that arnt sheilding any ions 2. water molucules that are shielding ions.      now... PI3 will react with liquid water when the water isnt tied up shielding ions. that means if the water has acid in it but its not fuming which means theres free water molucles avalable to desolve more acid the pI3 will react very quickly with these "free to ionize water molucules" if the water is fuming "an indacation there isnt any free to ionize water molucules" the PI3 wont react but will desolve in the water unreacted now if the water is saturated with ions that were fromed from HI gas and the fact that PI3 if allowed to react with water will form HI gas.you have whats called a fuming acid solution it is specal in the sense that if you have PI3 desolved but not reacted due to the solution is saturated with HI and you use some of the HI up. the result would be dilution that would free up some water molucules then your PI3 would react with these free water molucules forming more HI and would continue untill the water is saturated again. this is a self stablelizing acid strenth solution. it replaces any used up acid automaticly. fuming nitric acid and fuming sulfuric acid do the same thing they have things inthem "nitric oxides for nitric sulfer oxides for sulfuric" that will make the acid if the solution acid level falls below saturation. now for PI3 and phudo what are the conditions..... if there is PI3 avalble and phudo they will react and form iodioephedrine iodioephedrine looks alot like PI3 except it doesnt fume in contact with moist air like pI# does its soluble in acid satuated solutions that contain iodide ions ie HI in water and some what in salts of iodide in water it peaty much is soluble in what iodine is soluble in. the only importaint reaction that iodioephedrine is involved in here except the reaction with hI forming meth.is the reaction where iodioephedrine reacts back into phudoephedrine thia occors when iodioephedrine is in contact with hot stong lye water solutions. that brown goo that is mixed with your phos that doesnt desolve but is stuck to sides of the flask when you stop a reaction to early is iodioephedrine ----------------------------------------------------------- the last reaction iodioephed + HI --> meth + iodine will be covered in the next post. im tired and got no shit good nite
	zibarium (Naked) 11-03-02 03:15 No 375878		excellent	Bookmark Reply
			all ears, mate
	cthulhujr (Hive Bee) 11-03-02 03:39 No 375892		a question from the special ed. section	Bookmark Reply
			you will need at least 1gm of phos for every 10 gms of Iodine. any more phos then this will be extra for the next batch. Is there any kind of a "surface area", for red phos to be accounted for here?  does additional phos need to be added so the reaction could consume the needed amount. Swim always assumed that was the reason for adding so much in a rp/i rxn, or does the lack of "organics" gumming up the rp, in this situation make it simply unnecessary? 1 part Phos 10 parts iodine ---> PI3 If PI3 were created without pfudo and water etc. Iä-R'lyeh! Cthulhu fhtagn! Iä Iä!
	RepVip (Hive Bee) 11-03-02 08:55 No 375961		cthulhujr: Is this kinda what you are asking ...	Bookmark Reply
			cthulhujr: Is this kinda what you are asking about surface area? Rhodium: "If it is true that an excess of pure RP gives a ..." (Stimulants) 12345x:  Come on bud!  It's been almost 20 hours since the last post! Where's the follow up?  I'm eatin this info up with fork and spoon!
	12345x (Stranger) 11-03-02 08:59 No 375962		iodioephedrine to meth	Bookmark Reply
			in the last post we got up iodioephedrine now the reaction of iodioephedrine with hi to meth I-phudo =iodioephedrine I-phudo + HI (gas) ---> meth +I2 note the hi is in the gas phase. to get HI gas out of the pot where its desolved in as two ions we will boil full strengh hydodic acid that will form bubbles of 57% hi in 43% steam. the HI will react where the molucules are contacting the surface of the liquid. untill the bubble pops and the HI is carried up into the air space above the reaction vessal. where it is condensed back or caught in a water trap there is a side reaction with I-phudo and PI3 that consumes an amount that equials 3 phudos for every 2 meth molucules made. this side reaction occures when PI3 is inexcess in a saturated with HI pot. if one keeps the PI3 to a minunium. and the pot boiling so lots of HI gas is generated. then when PI3 is formed it reacts with phudo forming I-phudo and then because the amounts of PI3 are small ... the chances of the I-phudo reacting with a HI gas molucle are much greater then the chanch of I-phudo reacting with anouther PI3 . when it reacts with HI your yealds are going twords !00% when it reacts with anouther PI3 your yealds are going twords 33% so the way to have a pot with optemum ratios is to make 57% hi in water. then add enough p so it is clear . then add phudo and boil hard. this will take 72 hrs but your yealds are the best most would rater give up some yealds for faster times then from this point just add a small amount of iodine. if you have it at 57% already the addion of iodine will blacken the liquid and it will not clear. how much iodine to add ..... if you want it done in less then a day 1% of the amount you used to make 57% will shorten it to less then a day anouth way of doing it is to after making the 57% hi filter out all the rp and add the phudo then boil untill its dark "apx 5 mins" then add the rp . this last way will finish in around 4 hrs with yealds still as high as 65 -70% if you get in ahurry you can throw so more iodine in which will shorten the time but drop yealds if you put enough iodine to finish in a hour you will get around 1/2 yeald back .more iodine and your cook time drops to under an hour and yealds drop to under 50% at the extreem you will cook in 5- 10 mins ,get the letter envalopes cristals in the neck and get back a thurd. 57% hi is made by taking 43ml waterand adding all the phos your going to use . then warm up. then add 57gms iodine and bring to boil. now you can add your phudo "upto 1:1 phudo to iodine" and boil. then if you want it faster then 72 hrs just spike it with a little iodine. when the liquid clears up its done. if you dilute it with water its stops cooking untill the acid strengh is back up to 57% because of evaporation due to boiling. or addtion of iodine to bring stengh back up to 57%       0               57              PI3 and no water om 0% |<------------->||&lt;----------------------------&gt;       |  no reactions || cooks faster and faster/yealds         |               / \                        drop        |              |100%|       |              |72hr|                      |zone|       0%             | 57%|  57% with more and more PI3
	ballzofsteel (Miss High & Mighty) 11-03-02 09:07 No 375963		Why do these posts not have a rating as yet?	Bookmark Reply
			Why do these posts not have a rating as yet?
	cthulhujr (Hive Bee) 11-03-02 11:07 No 375981		another question from the short bus	Bookmark Reply
			so what swi12345x is saying basically, is no more or less h20 should ever be needed than that required to maintain 57%HI? And by getting the rxn going before adding pfed, minimizes the pfeds multiple exposure to PI3, reducing the production of azirdines, benzaldehydes, p2p and all that stuff? Swim has certainly heard other Bee's mention this, and now it makes perfect sense to swim. This sounds like a better dream for swim. Although swim is thinking, with less than 1ml of h20 per gram of pfed, why is this reaction taking 72 hours. Swim has always speculated long reflux times are a result of excess h20, used to keep PI3 broken up into HI and friend, to keep these 'side' reactions to a minimum, then slowly dehydrate the mix to get to '57%ishness'. If this reaction is jam-packed with nice hot HI, from the start is the iodioe simply forming so slowly as to slow reaction times. Hence adding iodine to speed it up. Funny how swim is forced to ask obvious questions, but for some reason they remain cloudy. Or maybe swim simply needs to react longer to start with. iodeeodeeodio-eee! Sounds kinda like yodeling. Yep RepVip, something like yonder post is just what swim was thinking of. Thanx for posting all this stuff swi12345x, reel cool. Iä-R'lyeh! Cthulhu fhtagn! Iä Iä!
	b159510 (Professional Student) 11-03-02 11:15 No 375985		Good effort	Bookmark Reply
			First, I sometimes found the post difficult to read, so I can't comment on many points that were made. Second, much of this has been covered before, so you should be able to find the info either here at the hive or at Rhodiums. Third, the reaction described here between PI 3 and ephedrine is surely not the mechanism by which iodoephedrine is formed in 57% hydriodic acid. I also question whether it is the mechanism by which it occurs in the 'dry' rP/I2 reaction. Anyway, it seems like genuine effort was made here, so I would congratulate 12345x for that. There is also information here that is accurate and informative. I would find it difficult to rate this post if I were among those who did such things (thankfully I'm not). You don't fool me, Oilman
	ballzofsteel (Miss High & Mighty) 11-03-02 11:38 No 375987		HI------> meth.Why not?	Bookmark Reply
			So in a RP/HI rxn,in which HI is produced seperately,(for instance Wizard Xs synth),and no free iodine is present, the PI3 must be formed from the Iodine molecule from the HI right? So what happens to the hydrogen?Do the just lay a waiting until a position is vacant? Oh they must hook up with the OH that was displaced by the I from the PI3 ,to form H20 which mixes with your PI3 to create HI again(Looks LIke A recycling agent to me) If heat is applied in order to dissasociate the hydrogen and Iodine(my understanding as to why heat is applied),why would the Iodine feel the need to go to the P in order to just transfer to the Ephedrine molecule? I understand why,but isnt it for no better option? Take P out of the equation and it has but one option,the I from the HI. So the PI3 gives up its Iodine to ephedrine to form i-phudo,displacing the -OH. This leaves P and some OH`s which attract to form H3PO4. So the HI reduces the I-phudo to meth ,liberated I2 + water. This I2 goes back to the P to Form PI3 which reacts with water to form HI(doesnt it?} So why wouldnt the I2 from the HI once dissasociated, go straight onto the OH position as the I from the PI3 does? Surely the PI3 would have to give up its I before it could latch onto the free OH. Wouldnt this mean that an external influence e.g heat, would need to be present in order for the iodine to be freed up just as is the case with HI? Without P present why wouldnt the free OH go to the Free H From Your dissasociated HI to form H20,and the free I latch onto the vacant OH position forming I_phudo, and then inturn get reduced by the excess HI libberating I2 as you have explained? Sure,you may not end up with a crystal clear sollution as free Iodine would be present,but without any PI3 what concern would there be of the Meth reverting back to I-Phudo? Does not HI disasociate at higher temps?I think it does. Hope You are now all as confused as I2. Ater much contemplation,I must concurr with the above conclusion from the super student Please Keep it up 12345X,I feel your posts are just what this forum needed,and are like a fresh gust of wind to unlearned ones like myself,wishing to understand why.
	b159510 (Professional Student) 11-03-02 12:01 No 375989		no OH	Bookmark Reply
			There would not be any 'free' OH generated in this rxn. Meth reverting back to I-Phudo? This would not happen in any case in this rxn. You don't fool me, Oilman

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