Super High Potentcy Push/Pull dope

[Worlock]

That's too funny, arguing the same point
just misunderstanding each other

My fault,  I catch so much backwash from this kinda thing
that I assumed you were flaming. If I would read what you wrote, carefully this could be avoided, One blast leads to a rebuttle flame, 
Time for my  reality check..    lucky we did not have guns.

har har har

[MnkyBoy78]

SWIM just reported that a crystal garden has formed on the flask walls.  They are of shapes not seen before.  The base (part attached to the flask walls) are rectanglular.  The portion extending inside the flask are triangle like.  (Shaped like rooftops).  What would be your opinion as far as how much longer.  The temp has been ~91.3 F for 24 hours, no additional h2o has been added.

[ballzofsteel]

The only correlation between this and 12345x theories are the time periods involved.
He mentions longer cooking times,e.g 7days ect.
I dont think much else is relevant,except perhaps the order in which he introduces his reagents,and maybe his ratios.
Forget the rest.

[Worlock]

His theory that to compensate for the shit they put in pills these days ,
you need to a ratio  1(E) : 1.75(I2) because the E is not E,
Is pretty wild, but I believe him.

Then he pulls these two equations out of Organic Chemistry books and neither one of them makes sense, they look good , but when you attempt to take them apart, it can't be


123455z  makes the HI from RP and I2 and water  then   adds the
E to that which makes sense.
But when 12345x tried to support this fact , he hit a brick wall.
there are major inconsistencies in the Organic chem literature regarding the reaction. 12345x found these two references, the man is diabolically funny with his hidden humor. These equations are like stage props , they look good from a certain angle but they are fake.

1. P seens to be oxidized? (loss of electrond
while I2 is reduced  and Ephed is reduced. that can't be right
6E-OH + 2P + 3I2 --> 6E-I + 2P(OH)3

2.in this one I is oxidized and E is reduced, let's just ignore P
and forget about water too
2(E-OH) + 2(HI) --> 2(M-H) + I2

but 2 should be
2(E-OH) + 2(HI) --> 2(M-H) + I2 + 2(OH)


I messed around with these equations for a couple of hours, and could not get them to break down into a system of equatioms that make sense.
The numbers add up almost but the individual steps are a joke . maybe its just me ,

 12345x handled it this nutty way...

if HI reduced phudo it would go like this :
     2(R-OH) + 2(HI) --> 2(R-H) + I2
now you find this reaction in a book ...
 and the book states is a valid reaction 
with not just primary
 but also secondary or tetrary alcohols
"phudo has a secondary type alcohol group "

anyway find this reaction

and ill ........

ill.........

ill say

well hot damm
your the phudoephedrine man! 

ha ha he he  "well hot damn"

[Organikum]

no additional water. Count the phosphoric as inert addon and the 15% water (if 85% H3PO4 is used) counts to the water you would add anyways. Also 92% H3PO4 works.

And here I fucked something up with the equatations. I will join Worlock chained to the heating balancing equatations.

till later
ORG

[Worlock]

Mnkyboy 78

From what I read and have been told,
once SWIY are at this point
crystals and a color change.
add a little water ,
turn the heat up  + 5F
run it 6 more hours and SWIY will be there,
To be extra sure toss in a little bit more I2
It is not quite cooked fuilly yet
 SWIY must have had some good reagents

[Organikum]

To get the reaction to completion a raise in th amount of reducing agent present should be the way to go. And this is what Mnkyboy gets by adding H3PO3. I2 should be be enough there from all the pseudo(ephedrine) already reduced. It should be still in the flask if not emigrated to flee the WOD.
H3PO3 is used up. Time for a stop and refilling the tank. You might check the tire pressure now too.  

ORG

[Worlock]

Now that is something that many
newebees think of using H3PO4.
And for them it never works, because the environment
they put it in needs the high energy  oxidation number that is supplied by R4, This Phos is an amazing substance. A living Cell uses P-P bonds as the energy source that powers a cell, .
When it comes down to it every human has one function to make ATR the cells form of currency( cash.) Stop making ATP and you are  dead.

And the gov't wants to control it, then there will be a black market in phos
In response to TV propaganda, the Cartels came into existence the day the gov't put restrictions on a substance, cause effect, the gov;t maskes criminals,

You think they would learn after prohibition created the mafia
Well gov't did learn and they learned how to line there own pockets with drug money, that what this is all about, some greesy fingered elected, appointed, officual on the take,

God bless America
Land that I love
Stand beside her,
but don't bend over 

Unless your into ButterFingers

[Organikum]

yes thats it, and it mediates the temperature peaks and holds HI solvated for not to vanish as white smoke.

Worlock what are you referring to in your last post?

I am confused by your advise to Mnkyboy on adding I2 to his reaction for bringing it to completeness. I would think on adding H3PO3?
What is the point I overlook?

The equatations are almost complete.
Soon.

ORG

[ballzofsteel]

Wouldnt the heat generated by the reaction between the relatively dry reactants dissapate more readily through a solvent and therefore perhaps not allow enough energy to kick the reduction off?
Im talking in relation to the fact that very little exteranal heat/energy is applied to this cold
push/pull and that it relies on the heat generated within.

Like,the reagents wouldnt react so readily if their molecules are buffered from each other in an inert solvent.If they were in solvent and heat was applied, well thats more along reflux lines.
A variation on an old theme.
We cant forget the ephedrine and what it needs to perform.
Lights?

This is not the same as a long HI reflux.


PErhaps

[wareami]

The Kidz have yet to explore the acidic side(H3PO3), But they have used UP their share of RedP. In every stalled or incomplete rxn, they never had any success with bringing it to completion with the addition of extra RP. They alwayz wound UP with excess RP to be filtered off in the end. They don't know the mechanics behind it, but never stopped to question the success they had when bumping the rxn back on track with added I ₂. Usually in .5g increments.
The calculations involved in starting a reaction, ratio wise, should be sufficient to supply the P4 molecule and the only loss it seems is the generation of HI. If a good amount of H3PO3 was used at the proper concentration, the addition of I₂ should provide the needed boost for a recycling to take place and thereby create a re-generation of HI!
IMHO....
Peace of the REaction
Have FUN-Bee SAFE

[Worlock]

You enjoy this too much

If one is using P,P tanks and doing a fast reaction, it becomes apparent that most of the iodine escapes from the reactin flask and passes doen the hose to be trapped in the water.

This water after 4 oz or so is red and thick with iodine & HI
If one were to recover the Iodine after a reaction  most of it cones from the tanks ans a snaller amout is recovered from the wastes generated in the reaction  flask,
The hhose is always stained red on the inside  whether you do a fast or slow reaction. 
Even in a reflux reaction with a long condensor  Iodine is escaping, the proper technique is to bring a hose off the top of the condenser, attach it io a funnel and invert the funnel letting the wide end rest in a tray with a few inches of water, this  captures the gasses enitted from the reflux.

The way most reactions are rin from the stoicochemistry
Red P is added in abundance and a big mud pie of it sits in the flask.
It is common  knowledge that the I is producing the driving force of the reaction ,
It is the fuel  the Red P is like the spark plug ,
Now I doubt that Mnky boy has any H3PO4  available if he did he would have used it.
He wants to be certain ther the Ephed is reduced to the iodate and changed to meth m there is a theory out that hther pills are never cleaned well enough any more , the 10 grams of E is more like 6 grams of E and the rest is either a sulferouds compound ar a spiral macromolwcule like povidone that traps E and I. Silfer is a inhibitor of iodine in this reaction, it will shut a reactionm down

So because of the dirty E
THe certain losses of I
and  from prior experience ,

 adding a little I2 will refresh the reaction fuel and allow it to complete, immediately upon adding the Iw it begins to react it is visibly apparent.
Now I hsve no problem with H3PO4, but I have to stick with what I know so until I actually use the H3PO4 , I am not qualified to reccommend it, suppose something happened that Mnkboy did not understand, If I can't ecplain what he needs to do I look like a fool, and my credibility is ahot, and mnkyboy could lose the reaction or get hurt You see  I only advise bees on actual knowledge about, I don't speak from Knowledge of aquaintence.
that rule seems best  from emnarressment.
So although I will try new things on my own , I don't send a bee into the trenches first.
The additional I2 was needed in his reaction did you note the temperature he was running at 91 F 
not C but 91 F that is  around 33 C
The phospjourous is famose for changing its activity in quantum leaps
 and not in a linear fashion , but in sudden stages the next major excitation for the phos is around 110 F, then at 150 F, 189 F .

His chances of increasing the heat to any of those levels is slim
the theory is that heaty is ther culprit in creating toxins.
Six more hours on a dryside reaction is going to make it very dry m so a little water is efficatious, and a litrtlw \e iodine will definately drive the reaction to the right.

Now C'mon you know this stuff, don't you? Making me explain it when I need to get on with the Cosmic Power thing. It is difficult to type that much, for me it is anyway, two years ago I had a reaction , to some bad antibiotics , and it left me almost totally blind Stevens Johmsons Syndrome  you can look it uo on the web

Molecule:

meth ("c1(ccccc1)CCC[NH2+]C ")

[Worlock]

Nards of steel,
Nonads 
 the phosphorous is not very active at these cold temperatures, and the amount of exothermic reactionm heat is not noticable, some external heat source is needed, infra red at the frequencies that show up on the spectroscope would be a wild way to control a reaction, just at the freqwuency needed and the side reactions would be minor in some  systems

Molecule:

meth ("c1(ccccc1)CCC[NH2+]C ")

[alphacentauri]

There is something I don't understand in the theoric aspects of the reduction: wether you start from elemental P4 or H₃PO₂ (hypo) or H₃PO₃ (phosphorous) you start from a substance in a reduced oxidation state, and what remains of that at the end of the reaction should be a form of P at a higher oxidation state, if not the highest. At the end you should have H₃PO₄ (phosphoric). In the meantime elemental I₂ is reduced to its negative anion in acidic environment. So you have the species HI not all dissociated, maybe, due to the high concentration in very little solvent (water). So the first of this reduction chain is the redox reaction between reduced P and elemental Iodine. Then appears pfed, E-OH, let's call it. We have as intermediate product iodo-ephedrine ( synonimous of iodo-methamphetamine). What does this mean? That the first reaction in which E-OH is involved is the substitution of the alcoholic group with an iodine. But OH is a very bad leaving group by itself, and the species I^- (anionic) is not enough nucleophilic to replace OH, so chemical theory teaches us that the neutral H₂O molecule is a good leaving group and iodide can kick it off, so we have first the protonation of the alcohol with the formation of the species E-OH₂⁺, from which H₂O goes away leaving E⁺, the carbocation of ephedrine, a secondary carbocation, and this reaction is favoured in acidic medium (the H⁺) of HI and H₃PO_{2 or 3}. It is a SN1 reaction, a monomolecular nucleophilic substitution. I think this should be settled. What I don't understand is how the species E-I reaches the full hydrogen state (E-H, that is to say the honey, the cake, the fabulous meth). Can someone give me some light and drive away the darkness of ignorance ruling in my drug-fucked brain, please?
Then Organikum says that he can use H₃PO₄, that is to say phosphorous at the maximum oxidation state (+5). At that state P can't give away any more electrons, since it has the same electronic configuration of Neon. So how is chemically possible that the P(+5) species reduces I(0) to I(-1)? I can't understand that, or there's something I miss. It would be too nice if we could use phosphoric, since it's freely available even in industrial quantities!
Worlock, I'm still on the Ac..t..ed extraction phase. I'm pretty long because this is a period in which I work a lot, when I'm ready with the cold reaction I'll post or PM. Bee patient, I work over 48 hrs a week!

[alphacentauri]

Another consideration before going in bed: an SN1 reaction has by its own nature a kinetic law of the first order. I don't know how the reaction proceeds, but I know that since I deal with a SN1 I'll have only one species determining velocity. We have an intermediate state costituted by a secondary carbocation, a pretty unstable particle. So, since the carbocation reacts rapidly as soon as it is formed, the reaction will proceed with considerable speed (the carbocation is craving for something negative, a base, a nucleophileto saturate its positive charge). So if the carbocation reacts rapidly as soon as it is formed, that will not be the decisive stage for the reaction speed. But the formation of the carbocation is indeed; in other words everything speeds up the formation of the carbocation speeds up the whole reaction. And what promotes the formation of the carbocation? It's definitely not the species I(-1), otherwise we had an SN2 and not an SN1 reaction - as a matter of fact I(-1) is involved after, when the protonated alcohol has already expelled the neutral H₂O from the molecule (this process is favoured by a polar solvent, water) -. The carbocation promoting species can't be other than H⁺. Where does this H⁺ come from? Or from HI or from H-H₂PO₂ (monoprotic acid), or H₃PO₃. If we use redP, obviously the only source of H⁺ will be HI. If we accept this becomes understandable why we can use H₃PO₄ as Organikum states: if, instead of using I₂ we pour in the flask directly HI solution, then phosphoric acid enhances the formation of the carbocation, because it works as a simple generator of acidity, not as a reducing agent to generate the anions out of iodine. Am I wrong? Then if my considerations are verified, we can increase the conversion of pfed to meth only adding more hypo or phosphorous or redP, and not iodine; the concentration of iodine above a certain quantity would be ininfluent, isn't it? Folks, while there is early evening, here is late night. I'm going to rest my tired whitish bones between the sheets. Good night, or better, good evening for you.

[Organikum]

Please read my posts on this, I did say that H3PO4, phosphoric acid can be added to function as an inert aka in no way participating in the reaction, solvent. It has the function to work as "heat spreader" eliminating "hot spots" and mediating the "dry" reaction for being less violent. Water forces the the reaction and is not   useful to add. (of course you can add so much water or H3PO4 to drown the reaction complete but thats another story).
H3PO4 is no reactand.
It can in no way substitute red phosphorus, H3PO2 or H3PO3.

I thought halogenations are direct radical substitutions? I maybe wrong.

Worlock! I know Mnkyboy so I was sure he would have some H3PO4 handy and his pseudo will be clean. So I thought I overlooked something as my practical experiences with RP/I2 are limited. It was never my favored reaction - to violent to messy - but the new developments regarding HI in special made me change my mind and walk the HI path as a HI only reaction is for sure after my taste. (thanks to Ballz as he never gave up on this)

Worlock? I didn´t tweak your name as you will have mentioned....
And I didn´t post the questions for teasing you as I regard thus as very bad style.

FREEDOM
ORG

[MnkyBoy78]

In my post above (

Post 401913

(MnkyBoy78: "Hold On, bite down on this", Stimulants)) the term H2PO3 is wrong.  It needs to be H3PO2 or hypophosphorous acid.

I edited your post for you /Rhodium

[Daphuk_up]

Could some bee please tell me where in the hell you guys are getting this stuff.  I have hit the blue roof, orange roof, the marts, and a couple of local nurseries, none of which have anything resembling a phosphorus based acid anywhere in the entire nursery department.  Sure, they got triple super-phosphate, but I dont think that is going to help any.  

Someone, just please PM me with a brand name, description of container, or something which will clue me in.  I would be forever and ever grateful for this help.  

[alphacentauri]

I thought halogenations are direct radical substitutions?

Halogenations can be radical or ionic, and can be substitution or addition reactions. I would like to explain in details, it would be very enjoying for me, but I don't know if a post is the right place for a beautiful chemistry discussion like this. Anyway if you irradiate chlorine with UV you obtain free radicals. Those are single atoms with an unpaired electron each, 17 electrons, practically the neutral chlorine atom. But why neutral, and so unstable? Because the valence shell has two sublevels, one full, and one should have 6 electrons to be full, but chlorine has only 5 electrons. So, in order to get the 6th electrons the chlorine radical snatches an electron from anything, anybody. If sir chlorine finds mr methane, well, mr methane remains without an hydrogen. Yes, because the only way sir chlorine can take one of mr methane's electrons, is "convincing" one of mr methane's hydrogen to divorce from him and marry sir chlorine. In doing so mr methane becomes a radical and when he finds a happy couple of chlorine gemini, he says:"hey, chlorine, now gimme one of your atoms, and tell the other gemini to fuck off", so mr methane radical becomes mr methyl chloride. And the other gemini? He is a radical now (once runs the hound, and once the hare) and he repeats the cycle. But in chemistry there are two ways of breaking a bond: the omolisis (the bond breaks right in the middle like sir chlorine and UV rays before) or the eherolysis (one of the gemini is greedy and keeps all the two electrons). In this last case we have an ionic couple, or better, a positive part and a negative one. Imagine sir chlorine gemini approaching mr ethylene. Mr ethylene has a negative crowd between the two carbons, and the electrons of the chlorine single bond are not very friendly with this negative crowd, and run by the opposite side of the two chlorine gemini, leaving one of the gemini uncovered, and the other supercovered, so one part of the chlorine molecule becomes partially positive, and the opposite side partially negative. The positive part is oriented towards the ethylene double bond, till one of the double bond, the easier to move (pai bond is called), leaves the carbon-carbon axis and joins the positive part of the chlorine molecule. Doing so the other carbon becomes positive, because it has one bond less: he is a carbocation now, he has two electrons less than what he should have, and so he wants to fill this gap, searching for something negative. And what does he find? The negative chlorine anion, the gemini left apart by the positively polarized chlorine of before. When he left his brother, he remained near, so now he can join the carbocation and says to his former gemini:"fuck off, my old mate, I'm better with this carbon now than before with you", so we have 1,2di-chloro-ethane out of ethylene and chlorine. This is an addition reaction, of ionic type. And there are susbtitution reactions as well, in which the particle is something thirsty of nucleus: they want protons, big massibe bulky protons. Usually nucleophiles (nucleus lovers) are conceptually associated with bases, at least with coniugate bases after acidic ionization, but not necessarily a strong base is a good nucleophile. Too many parameters in between: temperature, solvent, the leaving group. Yes, a nucleophile to behave as a nucleophile needs necessarily something he can kick off from a neutral molecule: in the case of ephedrine and iodidric acid the leaving group is water, it can't be the alcoholic oxydril, since it has a negative charge, and it is more basic than the iodide anion, but the oxydrilic oxygen of the aminoalcohol ephedrine has two couples of electrons not shared with anybody, and an hydrogen ion is attracted by one of those couples, doing so the OH turns out to be water, and the positive charge of the original hydrogenion is distributed to all the "acid coniugate" ephedrine. Now the alcoholic group is a good leaving group, as water, and can be kicked off even by weakly basic and pretty lazy iodide anion. In this case the iodide anion is a nucleophile and water the leaving group.
Organikum, we could go on all the night, I'm enjoing too much, I love chemistry, I reach climax if I keep going like this........I leave all now. If you need other explanations I can manage to give you, you're welcome, I'm at your complete disposal.    

[MnkyBoy78]

Thanx alphacentauri, a understand was obtained through your explanation with the "drama" in your post.

Rhodium: Thanx cheif

Daphuk_up:  doubt the "anything -ous" would be found at Wally-World, (any) Depot, or what not.  Well, unless your local chemshop has a yellow or orange roof.