Beginners Long Wet Reflux

Elementary · Sat Jul 16, 2005 4:33 pm

I was referering to wizard x's procedure 2(note6) where the hydriodic acid is prepared in situ, see paste below.

Alternatively, 150 mls of hot distilled water, 15 grams (0.48 moles) of red phosphorus, and 33.50 grams (0.132 moles) of iodine in small 3-5 gram portions is added. The mixture is allowed to react for 15-20 min, until all of the iodine has reacted. Then add 8.07 g (0.04 mole) of EPHEDRINE HCl and follow Procedure 1. The yield is 80-88%

If one was to do the curbshot for the same amount of E; (E=1.5g,I=3g,RP=2g) times this by 5.33 to give E=8g, I=16g, RP=10.7 (decimals rounded up)

Now lets convert this "beginners wet reflux" to 8 grams of E; (E=10g,I=10g,RP=10g) divide this by 1.25 to give E=8g, I=8g, RP=8g

If this wet reflux being described uses a 1:1:1 ratio then it is indeed better than Wiz X's as far as the amount of RP and I used, infact it even beats the curbshot.

So on conclusion the reason the reaction takes so long is because of the lack of an excess of I and RP.

If want the reaction complete in 2.5 hours then use Wiz X's amounts, for 10g of E you would need 41.88g of I, 18.75g of RP and 187.5 mls of distilled water.

At the end of the day is comes down to whos procedure you trust as RP is like gold dust for some, what evidence do we have that this "beginners long reflux" has good yields, even a small return of goodies mixed with (pseudo)E can speed you out if you've never tried it before.
There have been hundreds of procedures written, how many made it onto rhodiums archive though ? does this not say something ?

geezmeister · Busy Bee

What it boils down to is having done it repeatedly with excellent results.

Of course you can reduce pseudo with an excess of HI present. If we had an excess of precursors to waste everyone would be interested in doing the reaction the way WizardX's writeup describes. The iodine and red phosphorous used in the writeup you cite from Rhodium's page is sufficient to reduce twenty eight grams of pseudo HCl if refluxed for thirty six hours. The phosphorous and iodine in your calculation for ten grams of E is enough to reduce forty grams of pseudo HCl if refluxed for thirty six hours.

If everyone had kilos of red phosphorous and could source iodine in quantity they might choose to take less time and use more precursor to do so. Reality indicates that few cooks have large stocks of red phosphorous or iodine, and all of them have time available to do the reaction.

Your approach focuses on the time it takes for the redox.

SWIG had been involved with this reaction for a number of years before he finally took the time to do a long wet reflux of pseudo HCl. SWIG had good results with other approaches to this reaction, but none were as consistently complete and successful as the long wet reflux. He experimented with this reflux-- varied the ratios of precursors, varied time, varied temperature, varied water quantity. He came to the ratio he advocates based on the results of those reactions.

WizardX's post is in Rhodium's archive because it contains good information. The fact that it is in Rhodium's archive does not automatically give the information a goodhousekeeping stamp of approval. Curbshot's nano was in the archive, and it results in notoriously incomplete reduction, inconsistent yields, and inconsistent quality. No big writeup of the long wet reflux was needed. Its that simple, that effective. The window of success is so wide that a completely inexperienced individual can do the method with success on his first try.

As to what is in the pile, lets just say the recrystallized product compares very favorably with Desoxyn. Complete reduction means just that.

WizardX's writeup is one way to insure success with this reaction. It is not the only way. If you find your red phosphorous and iodine are very dear to you, the long wet reflux using a 1:1:0.5:0.8 Psuedo:I2:RP:dH2O ratio is just what you need to try.

Elementary · Mon Aug 01, 2005 7:33 pm

Ok fair enough.

EDIT 02/08/2005

It seems there are some wrong calculations on wizard X's procedure 2-note 6