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	All 36 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 Rhodium's. 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
	ballzofsteel (Miss High & Mighty) 11-03-02 12:19 No 375991		Sorry, I meant not reducing to meth.	Bookmark Reply
			Ahh,so the Oh group is not displaced but rather added to? Is that the case?Then does the whole thing come off and be replaced by H? Sorry guys,Ill stop posting bullshit ,which I do not understand.I spose I could go read a book or somethin` Im just a lazy fucker. Sorry, I meant not reducing to meth,due to excess PI3 as I think 12345x was stating.
	cthulhujr (Hive Bee) 11-03-02 12:34 No 375992		starte3d my beedy brain ticking!	Bookmark Reply
			When ephedrine is heated with hydriodic acid, with red phosphorus (Caution!, Ref. 3) or without, initially the hydroxyl is replaced with iodine (to give iodoephedrine). This can be accomplished by allowing iodine to react with phosphorus to form phosphorus triiodide which decomposes in the presence of water to phosphorous acid and hydrogen iodide. In this way, by adding phosphorus to the reaction mixture, hydrogen iodide is recycled and the reducing efficiency of hydriodic acid is enhanced [5]. after reading yon posts and thinking about this nutsyness, these are from Rhodium's Site. The halo compound may undergo an internal substitution reaction, whereby nitrogen replaces iodine to give an aziridine, which can decompose to give the compounds A (N-methylbenzylamine), B (benzaldehyde), C (propiophenone) and D (phenyl-2-propanone). This too, and how is it this occurs anyways, where's the nitrogen from, air? impurity? swim still thinks he members reading a PI3 related to meth impurity thing someplace, and has had the idea in his head, previously. Iä-R'lyeh! Cthulhu fhtagn! Iä Iä!
	b159510 (Professional Student) 11-03-02 14:27 No 376007		It says internal	Bookmark Reply
			The halo compound may undergo an internal substitution reaction So clearly the nitrogen containing entity is not from the air or an impurity, but rather the methylamino portion of the ephedrine molecule itself. Again, I'm sure this has already been discussed to some degree here at the hive, UTFSE for more info. You don't fool me, Oilman
	RepVip (Hive Bee) 11-03-02 14:32 No 376008		Everybody's got questions...	Bookmark Reply
			For example: what mechanism does pseudo follow to from iodo-ephed, either with HI or PI3? That's what I want to know.  The pseudo carbon is secondary and attached to a benzene ring... making it somewhat hindered.. and it's either gotta be SN2 or SN1.  If it's SN1, ionization to the carbocation could be slow, which could explain why the reaction takes 3 days..... but I thought it was SN2... and I don't like carbocations And I would have to agree that iodo-ephed IS made by the HI(aq) acid.  Protonated hydroxyl groups make for good leaving groups.... so I hear.  What mechanism it follows though..... does anybody know? 12345x sure sounds like he knows his shit.  I wish someone would/could disprove him with some actual reasons why, instead of just saying "nope, not like that."
	b159510 (Professional Student) 11-03-02 14:48 No 376015		RepVip:	Bookmark Reply
			You say you think the iodoephedrine is formed from hydriodic acid. 12345x seems to indicate s/he thinks it is formed from direct rxn with PI3. If you would UTFSE, you would find b159510: "Re: RP/I reaction" (Chemistry Discourse) among others regarding the reduction of iodoephedrine to meth. Everybody's got questions... and TFSE is the first place to look for answers, because you are probably not the first bee to ask the question. [EDIT] Sorry, thought you were also asking about the reduction of iodoephedrine. For the formation of iodoephedrine, try the thread starting with Rhodium: "Rxn parameters of HI/P reduction affecting purity" (Chemistry Discourse), though I'm sure there are others. and it's either gotta be SN2 or SN1 Have you not yet gotten to the chapter on elimination rxn's ? just kidding If it's SN1, ionization to the carbocation could be slow, which could explain why the reaction takes 3 days Or there could be an assload of water around, which would slow subsequent rxn's in a number of ways. 'Assload' of course is a technical word used to describe a large quantity of something. Anyway I don't want to get off topic here in 12345x's thread, so uhh, later.
	Jetson (Title Addict/Eraser) 11-04-02 19:41 No 376392		shweew...	Bookmark Reply
			correct me if i'm wrong, but swij's take on things is basically this: the rxn between the rp/i gives off HI acid and PI3 and the PI3 is what gives you the iodoephedrine.  so one would then think that the best route would be to firstly react the rp/i and h2o to form the HI and PI3 and let them react until no more PI3 is evolved then add the psuedo and heat.  possible or would the presence of excess rp in the rxn constantly recycle PI3?  if that were the case then a prefilter of the acid to rid it of the excess rp should do the trick eh?  well that's just swij's take on it but he knows one thing for sure that by making the acid first is most beneficial in any case. ^the devil is so lonely
	RepVip (Hive Bee) 11-04-02 23:03 No 376447		Jetson	Bookmark Reply
			In reply to: so then also what if the "white fog" in the rxn vessel during a rp/i/e rxn isn't HI afterall and it's actually PI3?  hmm... I'm pretty sure PI3 is a solid and not a gas.. and the white smoke is HI gas.  There seems to be some disagreement on whether the PI3 reduces pfed into iodo-pfed.. I believe it does.. oh wait I think I get what you were saying! You are saying that no PI3 means no iodo-meth, and that 57% HI will reduce pseudo to meth right?  I'm pretty sure it doesn't happen that way.  From what I understand, with some major help from b159510 and the two threads he posted above, the reaction goes like this: 1)the PI3, HI, or maybe even I2 reacts with pseudo, forming iodo-ephed (or iodo-meth, same thing) 2)HI then reduces the iodo-pfed into meth. So the iodo-meth is actually an intermediate in the reaction.  The PI3 is used, for the most part, to generate the HI which does the reducing.  b159510 and the threads he posted suggest the mechanism follows a radical-style reaction.  It may follow some sort of chain-reaction with radicals.. I think that's what radical reactions generally due.. with three steps: 1)initiation - forming the radical which starts the rxn. 2)propagation - radical reacts with reagents, producing another radical, which keeps the rxn going 3)termination - radical reacted with something that didn't produce a propagation radical. I'm not quite sure about filtering out the extra RP used to make the 57% acid.  I think it's a great idea.  12345x seems to know what he's talking about, and I believe he posted signs to look for and also what to do.  But I also believe that you might need some to produce more HI, to keep the concentration up as the reaction proceeds.... Does any of that make sense
	MnkyBoy78 (Stranger) 11-05-02 03:17 No 376544		57% HI Concentration needed	Bookmark Reply
			Yes, the rp is used to recycle the PI3 in the rxn.  If the rp was to be filtered off, a continous fresh supply of 57% HI would need to be added to the rxn to allow for the complete conversion to take place. As far as making the HI first, then add the E...Over at Rhodiums, http://www.rhodium.ws/chemistry/meth.hi-p.html, under Procedure 2, the HI is made first.  Recalling all the write-ups from the ancient ones, they all state that first HI is made, then E is added to the mix. Its all there...Its all been done before...Ironic, isn't it (or just pulling our heads out of our ass's): It took how long before we have come to the same conclusion on how to carry out this rxn. Why 78...I just dont know? But the voices said it sounded good
	zibarium (Naked) 11-05-02 03:35 No 376550		hypothetical confusion seeks feedback:	Bookmark Reply
			is zib lost, to suspect, that, IF it was possible, this rxn would work better if a tiny, yet continuous flow of HI and 'phudo' were introduced to each other, and sent on their way, out of the 'flask', while the next tiny spray of reactants  entered the flask? is their some innate advantage if ever having a large mass of reactants together at the same time? (other than the possible impossibility of arranging a system such as what i'm suggesting?) seems like, as redx takes place, the reduced matter only serves to slow the bubbles of HI, which must travel thru masses of now irrelevant molecules to do their thing. hopefully, zib's totally out to lunch, and won't even need to ponder 'fuel injection' systems.
	ballzofsteel (Miss High & Mighty) 11-05-02 08:15 No 376621		Missing the point.	Bookmark Reply
			So,Mnkyboy,you confirm that ephedrine will reduce with HI and without the need for PI3? That is the whole motivation of his postings,not to explain ratios or to show what order to mix things,or the ideal rxn environment,(this is all readily available here or at rhods`),But to disprove the method of a HI no P reduction,and to illustrate the way in which phudo becomes i-phudo through PI3 as aposed to HI. Thats what I think anyway. Lets keep it on track,or else start a new thread. Call it : Ideal rxn perammeters considering what 12345xhas to offer. seems like, as redx takes place, the reduced matter only serves to slow the bubbles of HI, which must travel thru masses of now irrelevant molecules to do their thing. Have we come to the conclusion that the whole reduction takes place on the surface of these HI gas bubbles? Or does it happen in situ as previously thought?
	ballzofsteel (Miss High & Mighty) 11-05-02 08:28 No 376627		um	Bookmark Reply
			b12345, I have searched for many an hour to find a definative or conclusive answer to the questions asked here. The threads you refer to do give me niether. As far as explaining "whats going on in the pot",12345x has given the most convincing account to date.The crux of this argument is whether or not iodo-ephedrine is produced via PI3 or HI,and if either,or,can produce the desired results,and not that HI reacts with Iodo-ephedrine. You said it your self in that first thread.It aint real definative.
	b159510 (Professional Student) 11-05-02 10:05 No 376642		As far as explaining "whats going on in the ...	Bookmark Reply
			As far as explaining "whats going on in the pot",12345x has given the most convincing account to date First, I respect what 12345x has to offer. It seems like s/he has at least some knowledge of chemistry. If his or her argument is that PI3 is the primary way in which iodoephedrine is formed in a 57% hydriodic acid solution, I disagree. Second, ballzofsteel, it is obvious to me that you do not have the required knowledge of chemistry to make any conclusions here. Please do not take this as an insult. Clearly you have a lot of company in this regard. As I have already spent much time considering what does in fact take place in this rxn, and have to my own satisfaction determined the most feasible mechanistic pathways, I bid this thread and this forum a fond farewell. You don't fool me, Oilman
	ol8dbee (Stranger) 11-05-02 14:23 No 376688		12345x's posts...car 54 where are you?	Bookmark Reply
			stomp, stomp, stomp,SLAM, aaarrrghhhaaa!! The sounds you have just heard are some poor ol8d running madly from her cabin in the woods, white hair flying, a look of confusion on her weathered face. Ok, now that I have waxxed poetic...The ol8d has been reading TFSE until she is so very confused. First let me state the obvious, she is chemically challenged in the worst way. Nontheless, she is attempting to learn new tricks.  From what little she has learned, 12345x's posts seem to make sense. NOW can she please get one of her personal heros like VE, Geeze or Scottydog to break all this down into a "Chemistry for Dummies" version/proceedure?  It's not that she doesn't appreciate the brainoids discussion, she's aware that they are the backbone of this forum, but due to the lack of her own education in chem., she just can't grok it all.  Poor thing was going to proceed with a standard (if there is such a thing) EIP rxn and a nice long reflux, but in light of 12345x's post, she's going to hold off a bit longer.  But, baby, that clean 10gs of E is just dying to bee morphed! SLAM, stomp stomp stomp...ol8d has just returned from her walk in the woods with that look of grim determination on her face again..Will look forward to some KISS(keep it simple stupid) for the chemically challenged.
	12345x (Stranger) 11-05-02 14:49 No 376692		the reason is	Bookmark Reply
			the reason it takes so long is the only source of iodine is from the iodine generated when the iodioephedrine turns to meth and relises 2 atoms of iodine. that and the little that decomposes when hi is boiled but this amount is all you want ... why because to keep the undisiarble side reaction from occuring. you want very little pi3 in the pot as if iodioephedrine "made from pi3 and phudo" contacts a pi3 molucule before it cintacts a HI gas molucule . the insueing reaction will pull your yealds down twords 33% to keep any iodioephed from contacting PI3 we keep the iodine concintraion to minium and have lots of HI boiling. yes with this small amount of PI3 being generated or remaning around in the pot with 57% HI the PI3, is whats slowing every thing down. its like it waits for PI3 then PI3 gets formed as soon as I is formed then PI3 + phudo --> iodioephed fast then iodioephed + HI gas --> meth +iodine fast then iodine +phos-->PI3     fast 1/2 pi3 +water ---> HI  "replacing the HI used to make the meth above" then wait for 1/2 pi3 to find ......... phudo --->iodioephed or  PI3 and iodioephed --->meth 33% yealds but because iodioephed +HI -->meth  "fast" that 2nd  PI3 aint going to find any iodioephed befor it finds phudo. which is good because if it finds iodioephed first then the yealds go down that makes a point.... if hi + phudo ---> meth how come a pot full of HI boiling doesnt make any meth for 3 days? and adding just 1% iodine will convert your phudo to meth in less than 1 day. because HI doesnt react with phudo or it would. whats the differnce in a 57% hi sol. and the same thing with 10% by wt PI3 aadded? its not a stronger HI solution cuz HI willnot stay in water at consintrations greater then 57% it cant.t here isnt any water left to keep it desolved as ions. you think the PI3 enriched sol is stonger but its still just 57% HI it looks stronger because the PI3 is reacting with fresh mosture in the air forming HI which cant go in the water because its saturated 57% so it fumes the room out. now back to the question if HI changed phudo to meth why does reaction rates increase by 100 when something is added that doesnt change the HI consintraion not even 1% and why if your HI is just 10% weaker say 50% vs 57% it wont make any meth at all cuz at 50% all the PI3 is used making HI there isnt any PI3 around long enough to react with phudo and no reaction with phudo no reaction to meth  then bring it up to 57% and its in 3days "why 57% because that is the strenght that allows PI3 to stay in the liquid without reacting instantly with water" then add 1% I and P and its in 5hrs "even thou doing this doesnt change HI concintration in water" add 10% more I and P and its in 20 mins  "ditto here also HI consintration still 57% " why the big change in reactions when the HI never varyed more then 7% ? first realization hey i know how to make it react faster! everytime ! then comes fact #2 when there are large amounts of PI3 in the pot the yealds drop my guess .... when pot is PI3 rich ..... PI3 will find iodioephed and lower your yealds. to prevenmt this make 57% excactly "57 gms I2 +43 ml water" heat with rp so its clear then add phudo while irs clear. use a reflux colluim so you dont lose water. dont spike any iodine nor let it sit with the rp sitting on the bottom as the liquid will darken and that will speed up the time to cook and lower yealds also if you get tired of waiting 72 hrs throw some iodine in it. and it will finish as soon as you want it to. but know you going to lose some for hurring. at least you can dictate the situation now. and spread the truth "HI doesnt make meth from phudo" Iand P make PI3 i and P make PI3 then with water to HI "when the pot is less then then 57%" PI3 and phudo make Iodioephedrine iodioephedrine and HI as a gas "by boiling" make meth and more iodine return to the top with your iodine and repeat untill there isnt any phudo left. no phudo no iodioephedrine no iodioephedrine red color goes away! it clears up. your done not sure dont know how to pulland wash up a taist in 5 mins? before killing the reaction with the water dilution? just stab it with some Iodine it gets dark if its done it will clear back up in 2 mins if it stays dark its not done keep boiling untill it does clear up. what could be simpler?
	12345x (Stranger) 11-05-02 16:08 No 376716		by what reaction does PI3 react with phudo	Bookmark Reply
			by what .... the pi3 bumps the back side of the carbon to which the hydoxy group is attached and yes there is steiro invirsion."shit so i cant spell" i mean the iodine is not where the (OH) group was it is where the hydrogen was and the hydrogen is now where the (OH) was. (those that dont understain words like SN1 SN2 or what a p orbital is forget this post it will just confuse you.and doesnt provide any thing usefull. this post is for those who think they know chemistry because they went to collage and for those who think HI can react with phudo ... then why does HI 57% take 3 days. and if i add I or make PI3 then the reaction drops down to a few hours and note when starting with 57% it is saturated. add some pi3 and the liquid is still 57% 57% is all the HI that the water can ionize it can hold no more. it looks stronger because your nosy mind lifts the cork then fresh mosture reacts with that PI3 forming HI but that HI gas formed  doesnt make the liqud any stronger "it was already satuated at 57%" so it fumes your nose,or colors the walls ,or desolves in your water trap or kitty litter. is the pot not still the same HI concintration as it was before you added the iodine spike so if hi is doing the reaction how come there is a huge differnce in the reaction rate 72 hrs to a few mins when something is added that doesnt change the HI concintration. well..... and if the HI concitration is cut to say 40% it wont make any meth ever."and thats a long time!" if it was the HI then how come reducing the ratio of HI to water by 20% stops the reaction cold? well..... smarty pants you explain why the reaction rates change wildly by 100 fold with addtion of things "PI3 added to 57% HI sol" that dont affect the stated reactent "ie HI in solution" nor its concuntration could it be that PI3 is involved in a primary way to the reaction? well MR wave mechanic smarty pants! better see that sn2 substatuion with steiro inversion better see iodio halogen redution with boiling hydrogen iodide better see that HI used lowers acid stringh below 57% then the two iodine atoms formed "i-phudo  + HI->I2 +meth react with rP --> PI3 one I from the pI3 forms HI "the 57% was diluted when it was used. now the HI is back up to 57% so the other I from the PI3 react wath a phudo to form iodioephedrine "that was used in the reaction above" now both Iodine atoms from the formation of one meth . was used 1. to replace the HI used to form the two iodines in the first place 2. to replace the iodioephed that was also used to form the two iodine atoms plus one meth molucule was formed one phudo molucule was used 2/3 of a phosporous atom was consumed. so thers your balenced equation. need i say any more.   can you write a balelenced equation with known reactions... using just HI and phudo HA HA HA BOW DOWN or continue to rant about things that JUST ARE NOT THE TRUTH! and show WHAT SCHOOL HAS TAUGHT YOU HAS YOUR EDUCATION HELPED GIVE YOU THE ABILTY TO SEE THE FUCKING TRUTH. WHEN IT GOES AGENST WHAT YOUVE BEEN HEARING ABOUT P ,I AND E "ON THE GOD DAMMED EVEL WEB!" BUT WHEN LOOKED AT , DOESNT REALY MAKE ANY SENSE WHEN STUDYED IN A LITTLE DEPTH
	b159510 (Professional Student) 11-05-02 17:28 No 376726		nice	Bookmark Reply
			I like the insults too. Based on your last post, I would give the entire thread a misinforming rating, and this post a flaming. You're too funny. Much of what you write is hard for me make sense of. Maybe that's why no other bees who know about this subject are commenting. Whatever. You did make it clear though that you think ephedrine will not react to form iodoephedrine using just hydriodic acid. That, without any doubt, is wrong. I could have told you as much long before I came to the hive or earned a degree in chemistry. And this isn't so much about what you called 'wave mechanics', or what a chemist would call molecular orbital theory (both phosphorus triiodide and hydriodic acid are capable of reacting with ephedrine), but simply about chemical kinetics (the thermodynamic formulation of activated-complex theory as applied to solution and gas-phase kinetics). Have a nice day. You don't fool me, Oilman
	zibarium (Naked) 11-05-02 18:17 No 376736		where does this leave zib's hypo-thetical ?	Bookmark Reply
			injection system? 72 hour reflux? (or back to mystical incantations?)
	b159510 (Professional Student) 11-05-02 18:28 No 376741		K.I.S.S.	Bookmark Reply
			I would suggest the long reflux, but would love to see your design, zib. You don't fool me, Oilman
	cthulhujr (Hive Bee) 11-05-02 20:37 No 376784		this has been a good thread for swim	Bookmark Reply
			swim has tweaked on this thread since posted, and has been trying to grasp the subject in reading all available material. In that sense it has been greatly informative, to commoner such as swimself, even if it was silly or misinforming to others. It seems the general concensus is HI alone will react pfed and form Iodipfed. Swim has also discovered the need to draw a thaumaturgic triangle around the reaction vessle may be completely unfounded.  The chem hack has only one way to be sure. Try it and see. It's only that simple. Iä-R'lyeh! Cthulhu fhtagn! Iä Iä!
	12345x (Newbee) 11-06-02 05:56 No 376921		ya and what reaction is that	Bookmark Reply
			HI reacts with ephedrine? and what reaction are you talking about? i know if you boil HI with ephedrine for 24 hrs and of course have rp in it only to keep the hi from decomposing.we dont want any Iodine in there to react with P4 .that would contaminate the test but under these conditions .... you end up with a pot of unreacted ephedrine. now you mite call that reacting well i dont. or mabie you mean HI will decompose upon boiling. the IOdine formed will react with the rp PI3 is formed. and PI3 reats with phudo to form iodioephedrine. so one could say the HI reacted? is that what you mean when you say HI will react with phudo?   im sorry for insulting you . but do you realy stand by the staitment HI is reacting with phudo and that PI3 doesnt /isnt reacting with phudo forming iodioephedrine? well do you?
	12345x (Newbee) 11-06-02 07:00 No 376935		let me clairify	Bookmark Reply
			when i said HI wont react to iodoephed i meant that HI doesnt enter into a reaction with phudo the result being iodioephed now what the one how begs to differ said that i said HI wont form iodioephed from phudo. well in that wording yes HI will form iodioephed i explain how..... if you boil HI some of it will decompose into free iodine than that iodine reacts with rp to form PI3 then the PI3 reacts with phudo to form iodioephedrine so to the one who says it does i agree. but the point of my post was to explain what is realy going on in the pot so as people could using these facts have a better hold on knowing what to do to get the reaction to react and know why they got the yealds they got. they say this info has already bean posted. ya it has but not in a way that the person reading it will get the concept. thats called running people arouind in circles but covering there butts if someone finds out the way it realy works and thinks gee how come those experts didnt tell me about PI3? then comes the alabi we did read the past posts. thats how you keep the dumb dumb and defend your self when someone finds out the truth. you can go look at the dea sites that explain about this reaction. they do the same thing. its called explaining the unimportant things as very important. and also "so no one says there lieing" include a three line sentence at the end of the abstact "a compound belived to be PI3 mite also react with phudo" but when you read it after you know how it works then it is apparent the persons who wrote the abstract arnt realy tring to inform whoever how and why RP and I form meth from phudo. and in realty they have husseled me also as they have goated me into defending my veiws. my goal was to inform people how phudo gets turned into meth. and probely because i know that noname gurus are keeping everyone in the dark as they give there half truths and strait bullshit.as expert advice .... and with that in mind ...... i say to do this...... make 57% HI in water "add rp to 43ml of hot water " "add 57gms iodine" "bring to boil" "add dry phudo up to a ratio of 1:1 phudo to Iodine" "boil with reflux or vented to water trap" expect to boil 3 days if you need it to finish sooner "with a drop in yealds" add some iodine the amounts are as such add 2% by wt of iodine used to make acid to shorten to less then 1 day. add 10% iodine to shorten time to less then 5 hrs add 20% iodine to shortan to less then 1 hr add 50% iodine to shortan to less then 30mins of course you have enough RP in the pot "some is always there" upto 5-10% extra iodine reflux will sufice higher then 10% and a water trap will be nessary to keep the oders down to a mininium" you boil it hard regardless what temp that happens to be "throw out the thermometer" you dont stop after a set time you stop when it clears up after it was dark. unless you are going for the 72 hr way as this never gets dark. in this case and realy should be in all cases you test a little before you end the reaction. if after you test and its done you add 2 times the volume with water "only do this when your sure its done" and bring back to boil for 5 mins then let cool now to test some before ending the main reaction you test by pouring just a pinch of the cool but not diluted pot contents into a glass jar. you add 3 times the volume water to the glass jar. you warm so the cloudyness created by adding the water clears up. you decant the liquid into glass jar #2 leaving the rp injar #1 you then add around 1/8 cup np to your jar#2 you then add lye with stirring "no metels please" untill the white clouds come. you then add 1/4 tsp more lye "when doing this with the main pot add a lot more lye you cant add to much . anyway you then pull your stirring wand out and touch your finger you then rub your finger to make sure they are slippery and then rince your finger and note the slippery fealing dosnt rince off."this tells you you have enough lye in the test jar."just wipe your finger on a dry rag to remove the slippey fealing after rincing in water" any way you transfer the #2 jar contents to a sept funnel. or squezze bottle.shake and let layers seporate. run the bottom layer back into the jar to the top layer add 2 times its volume pure water shake and run bottom layer including the muck at the line between the two layers ...into the drain. add 1/8 cup water to sept funnel contents then add 3 drops pool acid. shake then after setting run bottom layer into vision pan and dry. then test in your body. if you smelled no acid when drying repeat the water poll acid run into the pan and dry etc. repeat this untill you smell acid as its drying. after dry taist this pull carfully as this one is where any uncooked phudo is going to be. if you dont detect any phudo in the last acid smelling as it drys up pull. then your pot is done.dilute with water as stated above.and you can wash the main pot the same way you wased this tester. belive me the more water you use inthe pool acid shake and run off to dry the cleaner looking your shit will be. also use less drops of acid so you will have many small pulls instead of one big one with acid smell as it drys because the last one with the acid smell is the one with any impuraties. all preavise pulls are excactly the correct ph and are pure the bull shit follows the acid smell pull. save your unused phos and add it first to the next job so it "the used" gets used up before adding new so any left afterwords will be new only used once if you want you can grind moist rp to decrease its particals size "the smaller the particals the more surface area"ie less phos acts like more phos because its been ground into dust. have fun
	b159510 (Professional Student) 11-06-02 08:29 No 376969		HI reacts with ephedrine?	Bookmark Reply
			HI reacts with ephedrine? and what reaction are you talking about? Are you saying you don't know the different mechanisms by which a halogen (Cl, Br, I) can replace a hydroxyl group? If you don't know, then why do you say it doesn't happen? What analytical methods have you used to determine that every o-chem text and p-chem text is wrong - IR, MS/GC, ^1H NMR? One thing is for sure - hydriodic acid can react with ephedrine to form iodoephedrine. So if all your conclusions are based on your belief that this can't happen , then how can we take you seriously? So, to answer your question, the hydroxyl group would be protonated in the warm strongly acidic solution. This protonated hydroxyl group can then undergo substitution and/or elimination to be replaced with iodine. And I don't see how iodoephedrine reacting with PI 3 can form some of the reported side rxn products either.
	ballzofsteel (Miss High & Mighty) 11-06-02 10:06 No 377000		Final word.	Bookmark Reply
			b1591510, It is a pleasure to have you grace us with your presence in this;the lowest of forums. I have`nt the first clue when it comes to chemistry,this I have not proclaimed.I will be schooled in the future.With this should come some understanding.Until then all I can rely on,is the little I grasp from the books at home,and what I learn here. I have asked many a question regarding"whats realy going on" and HI reduction-no phosphorus ,over the past couple of months,and until 12345x`s posts have been satisfied that the answers from respected bees, have given me somewhat of an understanding. Would you rather me just say "yeah,well hes been around a while,he must be correct"and blindly believing it? To see these posts go unrated,and unresponded to by the ones I respect,makes me kinda query the validity of the previous school of thought. Ive seen mbrp threads shot down in less than 30 seconds by the top guns,yet,god forbid,something that resembles chemistry get mentioned in this forum,and everyone runs for cover. You should be heartend by the fact that something other than the usual is being discussed. If it is that far below you,why even click on this forum? My statements are simply their to evoke response.I think I have done alright sofar.Please disregard anything chemistry related that comes from my mouth,its realy just myself thinking out aloud and trying to grasp,by rehashing in my own words,what has been stated before. I have personal reasons to seek this information,even if I dont understand it all.A lot of time and effort,not to mention bucks,has been poured into my latest project,and without this kind of discussion,and without them Id be no closer to realising my dream. Again,I thank you for your input,and urge you to stick aroud to continue this debate.You seem to be the only one sofar. Anyhow,I think the ball is now rolling.I shall make no more comments here.Sorry if I pissed anyone off.Not my intention at all. yada yada yada ballz editAlso,after going over these threads a couple more times,I feel(in my mind)that what I have said is not to dissimular to what you have said,even if I did have a few free molcules floating around with nowhere tot go,te-he.I just used my dummie speak,while you knew the correct words to use. Basically,you have confirmed my thoughts/suspicions on the topic.This kida makes me all warm and fuzzy inside.
	zibarium (Naked) 11-06-02 10:23 No 377004		no bee really knows what an electron is	Bookmark Reply
			ballsofsteel; your response is understandable...and quite humble, really. when bees go blank, they're thinking...that's all we can assume. the tougher the question, the more thought delay. over time, zib has realized that when his questions aren't answered...or even flamed...that its a good sign that the question was good. these disputes between the knowledgeable, regardless of who is victorious in the long run, are the juice. (btw, zib  don't know shit)
	b159510 (Professional Student) 11-06-02 11:22 No 377010		ballzofsteel	Bookmark Reply
			Please don't leave the thread on my account. I don't want to discourage you or any other bee. I was just trying to point out that it is not prudent to make a conclusion if you don't have the knowledge of chemistry. Something might look or sound like a reasonable explanation, but in fact it may not be reasonable at all. We are all here to learn and share ideas. Ideas must be investigated as to their validity. Sometimes it can get a little confrontational. Part of that has to do with the medium we are using to communicate. Anyway, we surely can all get along here. I respect what you and all bees have to offer. Don't give up. Keep working and studying and please continue to share ideas. Moreover, by all means, don't take what I have to say as the final word on anything! If I say the sky is blue, you need to ask around, and better yet, look and see for yourself! You don't fool me, Oilman

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