Author Topic: Super High Potentcy Push/Pull dope  (Read 60157 times)

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BenWiFFen

  • Guest
Urea looks promising
« Reply #140 on: January 31, 2003, 01:44:00 PM »
Urea looks promising. Found at good garder or farm supply. How about using starch with a little bit of your post rxn liquid. Won't it detect any traces of I2? Freebase will heat up right now. Hcl will take longer to heat up on it's own.


Worlock

  • Guest
Organikum
« Reply #141 on: January 31, 2003, 08:57:00 PM »
Gottcha on the phospohoric acid,
I was making it more difficult than it is.
Just woke up in the middle of a dream, about this prospector
in the Alaskan North slope, was cooking some chili in the cold, damp
reaction flask method thing. The old timer was real wxcited about adding phosphoric acid to the solution,  Wondered if it would be beneficial to add it in the begining or just at the final cook off?
It was a dream that was on going, so I had no answer for an imaginary old woman (who says prospectors can't be female),



Worlock? I didn´t tweak your name as you will have mentioned....


Don't know what you mean.



And I didn´t post the questions for teasing you as I regard thus as very bad style.



I know that, it is how you ASK IT HHEHEHE, just kidding. You said nothing out of line, sometimes I just need some sleep, and I make odd posts, there is nothing amiss

Hey ,tease me go ahead! I have had worse,
Don't worry about me, if I could not take the heat, I would not stand in the fire, all the time.
The story about blindness is true, except that several months ago, thanks to God, and a good doctor. and my own effort, the vision has improved, I am a bit of road hazard but I can drive during the day.
A little sad because when I could not see, I was really into playing music (guitar), noW i CAN SEE PRETTY WELL, HE MUSIC LOST ITS IMPORTANCE.





zibarium

  • Guest
perhaps you could close your eyes?
« Reply #142 on: February 01, 2003, 12:45:00 AM »
and stay with the music?

;)

glad you're with us!

Organikum

  • Guest
Worlock: H3PO4 is added at start
« Reply #143 on: February 01, 2003, 06:42:00 AM »
This is the way I can testify it was done in countries far away by the local wizards.

ORG


Worlock

  • Guest
Zib
« Reply #144 on: February 01, 2003, 07:54:00 AM »
Good one,
That's pretty much the way I had it figured, too.
But I open my eyes and 'ur still here


Maybe I can get a gig every night,
playing in that Old Man's piss bar, The one you hang at
Wit the rest of the pud slappers 


Oh take my Cadiiillllllllac, and take my heaaaaarrrrrt
Soooooooeein U'll leave me\,  must we paaarrrrt 
before I drink your beeeeer and you smell my fart.

Yeah its The Space Soldier of Texas,
fomerly known as The Cyber Corn
all the way from across the road
music to slap a pud by
Hey Space why did you change your name from Cyber Corn

{"Bug off")

It was cause they kept calling you "Cyber Corn holer"
Hey I heared they called him the "Cyber Bunge holer," cause his music stinks.

("Man, lemme outta here")

That's what your chili tastes like, "Bunge hole"

("Flap")

Yeah how come it always tastes like dog bunge hole?

("Suzy, Come up here, show these stub legged, short horned cowboys, your modified sudiferous glands, while I get the pony warmed up, this is gettin' ugly")

yeah thats what we wanna know

("It tastes like dog bunge hole cause it IS doggy bunge and you been byn' it for a month, 'bout time you fingered it out")

Get him boys

("Lets blow , you drive babe")
(" Ha Ha  Smokers of dog Bunge, is what you are")

Zooooommm
MMuuuaaaahahahahahahasha


Worlock

  • Guest
Hey Alph
« Reply #145 on: February 01, 2003, 08:18:00 AM »
Easy for you with your Sn1 reaction,
Sn1 - dehydration
                      H+
E-OH  +  HI  ------> E-OH2+   +   I-
E-OH2+    -----> H2O  +  E+(carbocation)
E+   +    I-  ------>   E-I



But what about the Sn2 reaction,
                   P4I3
E-OH  +  I2 ----->  E-I  +  OH-  +  I-



Both Sn1 and Sn2 have the intermdiate carbonium(+)ion
or Carbocation(+)
How do we get the I off the E ?
Does it fall off when it hits   H I ?
Does it stay on until the solution is made basic?
What about forming double bonds, or reforming the alcohol?

Seriously I would like to know what you think about the final step.



I am positive(+) it is a CAT  ion
NA its an ANion(-)



ZingoBingo

  • Guest
Dang , Wor . . .
« Reply #146 on: February 01, 2003, 06:30:00 PM »

Worlock

  • Guest
Zingo Bingo Knows good tunes
« Reply #147 on: February 01, 2003, 08:12:00 PM »
I see you are a Maestro and appreciate the subtleties of a love ballad, THe way a good man rakes and takes, so she may give.
but he keeps takimg from a bad woman,

You are also  Able to see between the lines, his years of sacrifice and his displeasure at her disloyalty .
Who is realy the Sadist,  who is at fault. It matters little any more, drunk, bad digestion,

 works for me.

The Tabs I'll get 'em right out, same story , same song, that's been repeated over and over with a variation  so slight.
It is classic 12 bars of country blues, Chuck Berry style , As written byt Claptin but played by the Ventures with  the drive of Axels  full flanger &  gain maxed for optimal feedbach.


http://www.geocities.com/zabba_alba/tab2.htm



Where the hell is my Do-bro
The Sultan of Korn
Or maybe you would prefer  Cool edit pro 4.0


alphacentauri

  • Guest
I think this, Worlock.
« Reply #148 on: February 02, 2003, 05:18:00 PM »
I think this, Worlock. E is a secndary substrate, since the leaving group is a secondary alcohol and leaves a secondary carbocation. Apart from any consideration, a secondary substrate could react with both the mechanism, SN1 or SN2. You know, SN1 is favoured with tertiary substrates (a tertiary carbocation is much more stable than a primary one), and SN2 is favoured with primary substrates. Now let's consider two parametres: the steric hindrance and the effect of the solvent. The OH bearing carbon atom is between a massive lipophilic aromatic 6term ring at one side, and a carbon bearing a methylaminic group at the other side. All these groups are pretty voluminous and they don't allow easily the formation of a penta-cohordinated transition state, as would be in a SN2 in one single stage. So, if we consider the steric effect, we could say that a SN2 would be not favoured in the formation of iodomethamphetamine. Now the solvent: SN1 proceeds via the formation of a carbocation, so it requires a medium that can solvate charged particles, a polar solvent, in a word. SN2 not, is favoured in apolar solvents, particularly if a solvent can solvate anions less than cations, so the leaving group (less nucleophile than the attacking nucleophile, so less basic) can be solvated a bit better than the nucleophile. Such solvents are only slightly polar, and usually don't contain hydrogens that can be removed as acidic hydrogens, the so-called aprotic solvents (DMF, DMSO, crown ethers). Now what we've got in our rection flask? Water, HI, hypo or phosphorous acids, polar medium anyway, even if we use RP. For these reasons I think SN1 way is more probable. But it's only my opinion. Which reactions could compete with a SN reaction? Elimination ones, thus forming an alkene unsaturated on C1-C2, but....who knows? Anyway when eliminations compete with substitutions, eliminations are favoured by a high temperature (other parameter to consider), and the low polarity of the medium (an alkene needs apolar medium to get solvated). Obviously in acidic water and low temperature elimination is not favoured and substitution should be the main reaction. Ah, I was forgetting this: the more ramified is the forming alkene, the higher probability you have of elimination instead of substitution (Saytzev's rule, if I remember). Now a secondary substrate is liable to give elimination more a primary and less than a tertiary. In the end the reaction will be a compromise between all these factors, for this reason I think that your way, the cold reaction, can give better results than a long initial reflux. I think the best is some days with the cold reaction, and then finishing it with some reflux. I liked your way from the beginning and frankly I can tell you that I believe in it.
About the second stage (the full H state) I'm still thinking about it.
8)

alphacentauri

  • Guest
Shit, I forgot the stereochemical aspect.
« Reply #149 on: February 02, 2003, 05:28:00 PM »
Shit, I forgot the stereochemical aspect.
If we had a SN2 we should have inversion of configuration, so from a R reagent we should obtain a S product, so from L-ephedrine we should obtain d-meth, and from d-ephedrine (pseudoephedrine) we should get L-meth. But this doesn't happen, so this could be another evidence against an SN2 mechanism. With SN1 we should have racemization, but we have a product of optical purity (d-meth, wether we start from ephedrine or pseudo). Shit, this is a maze, what's the answer to this? ::)

Rhodium

  • Guest
not an issue
« Reply #150 on: February 02, 2003, 09:10:00 PM »
The stereochemical aspect is not an issue here, as the only chiral center left after the reduction (the amine-bearing carbon) is not touched by HI, only the OH, which is nixed anyway in the process.

Worlock

  • Guest
You are good at this
« Reply #151 on: February 02, 2003, 09:37:00 PM »
SWIA-C has got to be careful , all reports in the past approached from the point of view that it all was the same reaction.
However depending upon the environment I /RP /E /water  seem to be able to take a variety of paths , all of them going on at once but with some mechanisms predominating in different enviromnments at different times Phosphorous behaving  strangely at different temperatures, Iodine never sure if it wants to be a metal, or a salt or just evaporate into thin air , only to show up somewhere else red again


So, if we consider the steric effect, we could say that a SN2 would be not favoured in the formation of iodomethamphetamine.



hmmm  a very good point, SwiW always wondered how the I- could get in so close to the Alpha carbon. SwiA-C answer is , general it  won't


Water, HI, hypo or phosphorous acids, polar medium anyway, even if we use RP. For these reasons I think SN1 way is more probable.



hmmm this would lead one to think that the Sn2 mechanism is a hold over from earlier liturature and SwiA-C is right again,

The splitting of I2 into 2I- in the presence of P4 is a farce without a source of easy to get protons, supplied by water or an acid.

But the SN1 reaction is initiated with a simple acids  addition of H+ and not a redox reaction.


Obviously in acidic water and low temperature elimination is not favoured and substitution should be the main reaction.



Your logic is outstanding, makes ,much sense,

SwiW called the cold reaction "Cold Lazy Arse method"
SwiW knows that when they used a 1 hour cook, it always would get a bitterness, when vaporized in a glass tube, that requiires re-crystalization to remove,  With the Cold Lazy Arse method, under 110F SwiW has not had to recrystaslize , since it was not bitter. The bitterness was present in samples taken before the Cold L A method had finished, Indicating that some chemical has not finished cooking, the obviouis choice would be epherdrine, but E has a distinktive taste anmd it must be that  Ephedrine and pseudoephedrine have different taste and reaction rates.. In fact they are seeming to be much less similar than we have been led to believe. It is no wonder extractions are harder, they add things that will block one or both from being extracted , complicating a once simple process. .

Swi-Us may have to extract ephedrine ,then go back and extract pseudo with different conditions.

formation of Ephedrine from pseudo would be apparent in much higher yelds  20-25% improved , but AwiW has not done enough to yet be certain what is going on there. It is possible and it may prove out  in the end, somewhere near the 48 hour point.
Unfortunately, precursors are becoming scarce and another approach may need  to replace this.

Thanks for your insight, It has cleared up much and has left SwiW with a new set of problems, Always gaining ground, the playing field keeps changing.





Worlock

  • Guest
Stereo
« Reply #152 on: February 02, 2003, 10:11:00 PM »
The SN1 May Be occuring on the surface of an I- charged Phosphorous particle, The RP does not dissolve. This fact forces the reaction to o be a surface effect since the P4 loads up with Iodide.

The E also has to apptoach with the proper bonds exposed,
and in the proper configuration.
I have been able to model only two stable forms of the Ephed / pseudo  molecules

1. One is relatively flat and long
2. The other is L shaped with the ring at a 90 degree offset from the rest of the molecule , in the L shaped form m the stable isomer has the oxygen sticking outward and ro one side at the bend of the L. leaving an opening at the alpha carbon
If the reaction occured at this opening , although the bonds can rotate, they won't, it is too crowded for them to begin spinning and flipping.
Heat may force more I- to leave the RP particle, making it easier to reconfigure abd also making the molecules more energetic and able to reconfigure.
This would explain the violence in the 1 hour reaction at 190 F
I- ions and not HI  would suddenly become abundant, and most of the reaction occurs away feom yje phos particles damaging a potion rotating and not rotating achieving equilibriom based  mainly on stereo chemical probability with a minor influence  from the precurserd configuration.

i'LL TOSS UP PICTURES IN A WHILE
  they are revealing about the nature of the isomers of E



Rhodium

  • Guest
more pathways
« Reply #153 on: February 02, 2003, 10:12:00 PM »
The splitting of I2 into 2I- in the presence of P4 is a farce without a source of easy to get protons, supplied by water or an acid.

I wouldn't say so, 2 P + 3 I2 2 PI3 happens easily with a little heat, and the hydroxyl group of (pseudo)ephedrine then performs a nucleophilic attack on PI3 and with a little I- shuffling and elimination of HPOI2 we get iodoephedrine. The HPOI2 can then turn another two (pseudo)ephedrines into iodoephedrine with relative ease.

All this at the same time as more direct iodinations take place, so there are many reaction mechanisms doing their thing simultaneously...

wareami

  • Guest
Hey Now....
« Reply #154 on: February 02, 2003, 11:50:00 PM »
Don't think ya'll are gonna ditch the Kidz that easy with all this ChemSpeak!
This HIGHly informative rally has stamped a hugh on Ibee's farhead replacing that Big "L" normally stamped there!(silly bee's...."L" = Looped nØt Lunatic! ;) )
Ibee's question is: Can anybee identify the governing factor that creates the racemic variety of Meth (50% l-meth/50% d-meth)in the rxn. Does this occur with ephedrine as well as pseudoephedrine?
•Is it the iodo factor?
•Is it heat?
•Is it ratio of reactants?
•Is it length of time exposed to HI or
•Is it concentration of HI?
Ibee doesn't want to derail the current discussion, he's just searching for a clearer understanding if one is known.
The correct answer may help direct us all toward a consistant successful rxn method thereby directing bees toward more d-meth and less l-meth.
Peace of the REaction
Have FUN-Bee SAFE



Rhodium

  • Guest
Why racemate?
« Reply #155 on: February 03, 2003, 12:10:00 AM »
Can anybee identify the governing factor that creates the racemic variety of Meth (50% l-meth/50% d-meth) in the rxn.

Where is the evidence that would be the case? In my opinion racemic meth could only be the product if the starting material used was either racemic ephedrine or racemic pseudoephedrine, which is not likely if commercial cold medication was used as a source for the starting material.

wareami

  • Guest
Loose Terminology
« Reply #156 on: February 03, 2003, 01:09:00 AM »
Where is the evidence that would be the case?
Ibee sees that term used way too often and just assumed that most HI/RP reductions of pseudoeph resulted in the racemic mixture...incorrectly I suppose!
Upon referencing the Merck listing you posted on pseudo(thanx! 8) ), I see where Ibee was mistaken!
So would Ibee be correct in assuming that l-meth is excluded from the picture if the starting precursor is d-pseudo?
An unfinished reduction would then include a mixture of d-pseudo and d-meth only....correct?



Worlock

  • Guest
2d representation of 3d simulation of isomers...
« Reply #157 on: February 03, 2003, 01:19:00 AM »
2d representation of 3d simulation of isomers


L shaped configuration


Flat configuration



Rhodium

  • Guest
Dynamic structures
« Reply #158 on: February 03, 2003, 02:05:00 AM »
Here you have all the interesting molecules, a thick bond means the bond is pointing towards us, a dashed that it is pointing away. You can rotate the molecules individually by holding down the left mousebutton on a molecule and drag it around. Right-click and choose ball-and-stick models, animate it and whatever you want. This must make the stereochemistry clear to most.













Molecule:

d-Methamphetamine ("C[C@H](NC)CC1=CC=CC=C1")


d-Methamphetamine













Molecule:

l-Ephedrine ("C[C@H](NC)[C@H](O)C1=CC=CC=C1")


l-Ephedrine













Molecule:

d-Pseudoephedrine ("C[C@H](NC)[C@@H](O)C1=CC=CC=C1")


d-Pseudoephedrine

wareami

  • Guest
PharmOut Rhodium :•þ !!!!!!
« Reply #159 on: February 03, 2003, 02:18:00 AM »
It's a good thing Ibee has TunnelVision to keep UP with all them Moles! It's funny how when the fork in tunnel bares to the Right....Ibee Gets HIGHer! :P
IWare