Author Topic: Chloroephedrine electroreduction at a lead cathode  (Read 6423 times)

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moo

  • Guest
Chloroephedrine electroreduction at a lead cathode
« on: June 26, 2004, 07:45:00 PM »
Methyl(2-phenylisopropyl)amine.
Fujisawa, Shunro. (Fujisawa Drug Manufg. Co.).

Patent JP173746

  Publication date 19460927  CAN 46:13963.

Abstract
The cathode soln. contains 5 g. 1-chloro-2-methylamino-1-phenylpropane dissolved in 200 ml. 20% H2SO4 and the Pb cathode is 60 sq. cm.  The anode soln. consists of 2000 ml. 50% H2SO4 surrounding the Pb anode.  Electrolysis takes place at 10 Deg, 15 amp., and 3.5 v. for 4 hrs. with
an unglazed porcelain cylinder as a diaphragm.  The soln. is neutralized with NaOH, steam distd., the distillate is extd. with ether, dry HCl gas is passed in after removal of the ether, and the crystals are sepd. by filtration and recrystd. from acetone, m. 173 Deg, [a]D20 16.1; yield 3.8 g (75%).


WizardX

  • Guest
Document or database inaccessible?
« Reply #1 on: June 27, 2004, 10:44:00 PM »
Does that link

http://l2.espacenet.com/dips/viewer?PN=JP173746&CY=gb&LG=en&DB=EPD

work for others? I get the following message: Document or database inaccessible

moo

  • Guest
It appears that is one of the patents they...
« Reply #2 on: June 28, 2004, 06:24:00 AM »
It appears that is one of the patents they haven't got in their database...


fierceness

  • Guest
Wow!
« Reply #3 on: June 28, 2004, 10:11:00 AM »
I'm confused as to why this thread isn't getting more hits?  WizX has an easy way to make chloroephedrine on his site (CaCl2 or ZnCl2).  I think most car batteries are made of lead anodes but the cathodes are just coated with PbO2.  3.5 volts is  less than 2 cells of a car battery.

I'm obviously missing something here, so don't go opening up any batteries yet!

ChemoSabe

  • Guest
Clean Up of Final Product
« Reply #4 on: June 28, 2004, 11:37:00 AM »
So is it pretty easy to be certain that no lead will make it into the final product?

I know that plenty of the substances used in nearly every process here involve chemicals that are in some way poisonous so it's really nothing new to have to deal with cleaning up final products proper but I just thought I'd ask.

I know that recrystsallizing is always a supreme idea no matter what process is used.


moo

  • Guest
Lead sulfate isn't soluble in the electrolyte...
« Reply #5 on: June 28, 2004, 02:59:00 PM »
Lead sulfate isn't soluble in the electrolyte nor the ether and even if it was, those impurities would be eliminated by the steam distillation or a standard A/B procedure, which is a good idea anyways.


fierceness

  • Guest
So, what are the reaction conditions exactly?
« Reply #6 on: June 28, 2004, 04:16:00 PM »
So, what are the reaction conditions exactly?  Do you just fill a nonreactive container with 200mL 20% H2SO4 and 5g of chloroephedrine on one side of unglazed porcelain and then 2L of 50% H2SO4 on the other and slowly dip the correctly measured lead rod ends (connected by a wire, a voltmeter, et al) into both sides at the same time?

Of course there is the danger of the hydrogen and oxygen gasses igniting, but if you keep the two sides sealed this shouldn't be a problem, no?

WizardX

  • Guest
Cell Construction
« Reply #7 on: June 28, 2004, 06:54:00 PM »

Post 431203 (missing)

(WizardX: "110V @ 15 Amps", Stimulants)


Have a look at these cells.

http://www.orgsyn.org/orgsyn/prep.asp?prep=cv3p0060


http://www.orgsyn.org/orgsyn/prep.asp?prep=cv1p0485




A Ceramic or Alundum porous cup

The reduction is carried out in four cells of the type shown in Fig. 2. Each cell consists of a 1-l. beaker (B), a porous cup (P), a mechanical stirrer, and sheet lead electrodes (E1 and E2) each having a total surface area of 100 sq. cm. (Note 1). In the cathode space of each cell are placed 25 g. (0.18 mole) of anthranilic acid (Note 2) and 400 ml. of 15% sulfuric acid. In each porous cup is placed 200 ml. of 15% sulfuric acid. The cells are connected in series as shown in Fig. 5 with an ammeter (A) and suitable resistance (R) (Note 3) also in the circuit.

http://www.orgsyn.org/orgsyn/prep.asp?prep=cv3p0060



I don't think the condom will survive the reaction because of the highly acidic solution, but it's worth a try? That why they use an unglazed ceramic/porcelain or alundum (aluminium oxide) porous cups.

moo

  • Guest
Electrolytic cell
« Reply #8 on: June 28, 2004, 07:33:00 PM »
Also see the mescaline electrosynthesis-related post

Post 47461

(uemura: "Electrochemical Reduction of Nitrostyrenes", Novel Discourse)
which contains a translation of the description of the cell from the article in

https://www.thevespiary.org/rhodium/Rhodium/pdf/mescaline.slotta-szyszka.pdf

which contains the picture missing from that post. :)

It appears that a hyperlab bee used layers of fiberglass cloth as his diaphragm. See the thread starting with

Post 381288

(Xicori: "Pyridine to Piperidine (electrochemical)", Chemistry Discourse)
.

The current, voltage, dimensions of the cell, type of diaphragm, concentrations of the electrolyte and reactants, and the cell resistance are all related to each other, so there is no point in blindly following the numbers. Strong evolution of gas is a side-reaction and most likely an indication of too high a voltage but that is not to say there could be no gas evolution at all at the correct voltage. Current consumption is indicative of and proportional to the main reaction going forward and there should be a notable increase in the current when that voltage is reached - if you plot the current as a function of the voltage you should see a "wave" rising. The only way to reliably and reproducibly measure the potential at which the reaction takes place is to use a suitable reference electrode to measure the potential of the working electrode and in many cases the potential is even then pH-dependent.


ChemoSabe

  • Guest
No Iodine or Red Phos
« Reply #9 on: June 28, 2004, 08:33:00 PM »
I hope a new wave of electro experimenting comes through.

It's nice to see these procedures that use no iodine or red phos.

Whoever starts a decent trend of starting from homegrown e or p2p and finishing with something like this will win some sort of prize from me.

Now that's the future.


fierceness

  • Guest
Wiz, I'm having problems understanding why the
« Reply #10 on: July 02, 2004, 04:13:00 PM »
Wiz, I'm having problems understanding why the diagrams are presented in the way they are in the links you provide.  I've been reading other sites and it seems like all one would need for this reduction is a beaker with two sides separated by a porcelin barrier.  Putting the right electrolyte solution on each side, and running a current through it for the right amount of time should yield meth.  Building the apparatus should be the only difficult thing to do here.  Whats the best device to use to get a current to run for this specific reduction?

ChemoSabe

  • Guest
DC power supplies
« Reply #11 on: July 02, 2004, 04:20:00 PM »
The coolest looking ones are the Hewlett Packard DC power supplies. Woould not look out of place in your stereo rack.

I've got an old "atomic era" kepco (subsidiary of general electric) that looks like it's brothers with the old HAM radio.

As usual for hardware check your favorite auction sites

You get two in one with this particular unit. $5.00 starting bid. They are heavy though. Expect to pay decently more than that for shipping.



the rackmountable HP6275b - Prices do vary - starting bid $499.00

PS: Palladiums still damn cheap. Get it while it's hot.



WizardX

  • Guest
Electrochemical
« Reply #12 on: July 02, 2004, 07:49:00 PM »
fierceness: Your right in saying...


I've been reading other sites and it seems like all one would need for this reduction is a beaker with two sides separated by a porcelin barrier.  Putting the right electrolyte solution on each side, and running a current through it for the right amount of time should yield meth.




...however, it is a little more involved.

To get the correct voltage and amps, you need a scientific regulated power supply like ChemoSabe posted.

Keeping the electrolyte solutions to the 10 oC as required, is not easy, as 15 Amps of current is passing through the cell.

General rules: The more amps that pass through a cell at a given cell resistance will generate heat. The higher the cell resistance, the greater the heat generated.

It is for this reason that the electrolyte is stirred well as all times with ice bath cooling. The surface area contact between the cell wall (the glass beaker) and the ice bath is another reason why some electrochemical cells are constructed as they are?


Organikum

  • Guest
problems
« Reply #13 on: July 03, 2004, 10:53:00 AM »
The main problem I see on the procedure is related to the overall yields. As chlorination of (pseudo)ephedrine by Zn/HCl will yield not much more than 75% best and the electroreduction also 75% the overall yields will be about 56% not more. Thats comparable low to a well-done HI reduction and I see no way to boost these yields without the use of hard-to-get chemicals like thionylchloride or PCl3.

The second problem is the separation of remaining chlorinated ephedrine from the finished reaction. This is IMHO much more a matter of concern than lead. Steamdistillation and recrystallization are a must.

To the electrolytic procedure:
The amount of electrolyte in the anode chamber seems to be very large to me, 2000ml. I guess this could be reduced drastically what would also allow to place the electrodes nearer together what would reduce heating up. (please correct me Wiz if I overlooked here something).
What counts is voltage and current density on the cathode, 3,5V and 0,25Amp/cm2. This says always a large area cathode is to be used. Lead-battery plates should work fine.

To form the sulfate-ester of the ephedrine in-situ and to reduce would be favorable IMHO. The sulfate-ester is told to be easier reducible as I remember and will be easier to separate from the product I guess. But I dont know exactly how this could actually be done, but there should be a way.

Chemosabe, as long there are pills, matchboxes and tincture nobody will be able to trigger a new wave and when the pills run out ppl will be to tired for anything. The new wave will come when a no-pills-avail generation has grown up and this generation will solve the problem to fullfill their needs with new synthesises. Or just buy the shit from the local frontend of some cartel.


ORG


ChemoSabe

  • Guest
Me, Myself & Iodine (& RP)
« Reply #14 on: July 03, 2004, 11:46:00 AM »
I'm sure you're correct about the pills Orgy.

When I first made my comment I realized as I was submitting it that if I ever wish to really have the satisfaction of supplying myself with some supreme product which has no connection whatsoever to extracted pills or matchbooks or to IOtinc that it's going to have to me, myself that starts my own trend.

A trend that most likely would bee followed by nobody else.

And all of this in a purely theoretical context of course.  8)


WizardX

  • Guest
inconclusive?
« Reply #15 on: July 04, 2004, 09:45:00 PM »
Fellow bees. I'm trying to get the entire reference so I can examine why 15 Amps @ 4 hours was used? I have run this in my electrochemical programs and the results are inconclusive? Moo, can you upload the entire reference if you have it?



The cathode soln. contains 5 g. 1-chloro-2-methylamino-1-phenylpropane dissolved in 200 ml. 20% H2SO4 and the Pb cathode is 60 sq. cm.  The anode soln. consists of 2000 ml. 50% H2SO4 surrounding the Pb anode.  Electrolysis takes place at 10 Deg, 15 amp., and 3.5 v. for 4 hrs. with an unglazed porcelain cylinder as a diaphragm.  The soln. is neutralized with NaOH, steam distd., the distillate is extd. with ether, dry HCl gas is passed in after removal of the ether, and the crystals are sepd. by filtration and recrystd. from acetone, m. 173 Deg, [a]D20 16.1; yield 3.8 g (75%).





Current Density = (Pb cathode is 60 sq. cm / 15 amps) = 0.25 amps/cm2


COUNTING MOLES OF ELECTRONS

C = A x s       C = Coulombs   A = Amperes   s = seconds

C = (15 x (3600 x 4) = 216000 Coulombs


Since, 1 mole of electrons = 96500 C

(216000 / 96500) = 2.238 moles of electrons for only 5 g. 1-chloro-2-methylamino-1-phenylpropane.

1 mole of CHLOROEPHEDRINE freebase, C6H5-CHCl-CH(NHCH3)-CH3, (1-chloro-2-methylamino-1-phenylpropane) is 183.68 g/mol

(5 / 183.68) = 0.027 moles.


moo

  • Guest
The patent
« Reply #16 on: July 05, 2004, 05:34:00 AM »
I only found the abstract but then I looked a bit around and found it here:

http://www4.ipdl.jpo.go.jp/NSAPITMP/web222/20040705212907443475.htm#DL000

.
If that link doesn't work (never trust an url with the string tmp ;) ), enter the patent number without the country code to the number field and the letter c to the kind code field in

http://www4.ipdl.jpo.go.jp/Tokujitu/tjsogodben.ipdl?N0000=115

.


WizardX

  • Guest
Thanks.
« Reply #17 on: July 05, 2004, 08:22:00 PM »
Thanks Moo, I got it. Now, we need a Japanese speaking/reading bee to translate it?  :o  Anyone?

slowmind

  • Guest
Anyone?
« Reply #18 on: July 07, 2004, 06:03:00 AM »
Anyone tried a nano of this yet?

Shane_Warne

  • Guest
I'm struggling to see why you'd want to ...
« Reply #19 on: July 07, 2004, 08:05:00 AM »
I'm struggling to see why you'd want to develope a procedure using chloroephedrine.

If you can obtain enough ephedrine to prepare enough chloroephedrine, to waste on this pursuit...you may aswell have used a HI reduction or NH3 reduction, for racemic meth, which your already versed on.

this synth is about 8-10yrs too late, maybe more. some parts of history don't need to bee written, or just don't get written, this is one of them.

P2P or the biosynth is the way of the future, my money is on the good old P2P routes, but it'd be a shame to see the biosynth go to waste... :(