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                   Ephedrine Reduction with HI/P by Wizard X
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Disclaimer:

The following information is for informational purposes or academic study only.
Check your local, state and federal laws, and procure the necessary permits
before undertaking any of the reactions described below. Wizard X  shall not
be held liable and indemnified from impeachment for the use, misuse, injury,
death, imprisionment or fellation due to the application of this information.

57% HI Solution (Hydriodic Acid):

Boiling point 125.5-126.5 deg C
Density 1.70 gr/ml
55-57%  w/w HI is 0.936 to 0.99  grams of HI per ml.

Reaction: H2S + I2 ===>> 2HI + S

In a fume cupboard a 1.5 three neck flask is added a mixture of 480 grams of 
iodine and 600 mls of distilled water. The centre neck is added a sealed
stirring unit which leads almost to the bottom of the flask. The left neck is
fitted and sealed with a glass tube which extends almost to the bottom of the
flask but does not touch the stirrer. This is connected to the hydrogen sulphide
(H2S) generator.  The right neck is fitted and sealed with a short glass tube
connected with a plastic tube to the bottom of an inverted funnel in a 5%
sodium hydroxide solution.

The mixture is vigorously stirred and a stream of hydrogen sulphide (H2S) gas is
passed rapidly in the iodine/water mixture as rapidly as it reacts with the
iodine. After several hours, 2-3 hours, the iodine disappears and the liquid
assumes a yellow colour (sometimes almost colourless) and most of the
precipitated sulphur sticks together in the form of a hard lump.[NOTE 1] This
liquid contains a mixture of HI, H2S and S. The liquid is now filtered through a
large funnel plugged with glass wool to  remove the sulphur S. No need removing
the dissolved H2S as this enhances the reductive power. Add a few crystals of
iodine to this HI/H2S solution and store at 0-5 deg C.

[NOTE 1] The sulphur lump in the flask can be removed by refluxing with
         concentrated nitric acid.


Reduction of Ephedrine HCl to Methamphetamine
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Ephedrine HCl, molecular weight  MW = 201.73 grams/mole
Ephedrine Freebase, molecular weight  MW = 165.23 grams/mole


PROCEDURE 1:
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A mixture of 80.7 grams (0.4 mole) of EPHEDRINE HCl, 20 grams of red phosphorus
and 170 mls of 57% hydriodic acid is refluxed for 25 hours. Stopping refluxing
after 25 hours, and allowing to stand at room temperature for another 12 hours.
A total of 37 hours. After this time a large amount of crystalline material has
seperated out from the solution.

The reaction mixture was diluted with 700 mls of distilled water and filtered to
remove the red phosphorus.[NOTE 2] The yellow filtrate was treated with a few
crystals of sodium thiosulphate to remove any free iodine.[NOTE 3] It is then
made basic with 40 % sodium hydroxide solution.

The liberated freebase methamphetamine which seperates is solvent extrated with
two, 75-100 ml ether portions and the ether/amine solution is first washed with
50 mls of distilled water and the ether/amine solution dried with anhydrous
sodium carbonate.[NOTE 4] After removal of the ether, the oil was vacuum
distilled at a vacuum  of 15 mmHg at 93 degC.[NOTE 5] The yield is 80 - 88%

[NOTE 2] Rinse the red phosphorus with a little distilled water and collect with
         other filtrate. Use suction filtration.

[NOTE 3] Add one or two crystals at a time and mix. Then add more sodium
         thiosulphate in one or two crystal portions until the filtrate shows
         no red or purple iodine.

[NOTE 4] Anhydrous magnesium sulphate can be used and rinse with a little ether
         to extract all the freebase out.

[NOTE 5] The oil is a mixture of methamphetamine and iodo-ephedrine.
         It is very difficult to vacuum distil the methamphetamine out alone.
         A better method is to reduce the iodo-ephedrine/methamphetamine oil
         residue with lithium aluminium hydride in ether. The iodo-ephedrine is
         easy reduced further to methamphetamine.


PROCEDURE 2:
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In a 500 mls round-bottom flask is added 150 mls of glacial acetic acid,
15 grams of red phosphorus, and 33.50 grams of iodine.[NOTE 6] The mixture is
allowed to react for 15 - 20 minutes, until all of the iodine has reacted.
Then add 10 mls of distilled water and mix, then add 8.07 grams (0.04 mole) of
EPHEDRINE HCl, and mix. A reflux condenser is added to the 500 mls round-bottom
flask and the mixture is gently refluxed (gently boiled continuously) for 2 and
1/2 hours. The hot mixture is suction filtered while still hot to remove the
excess red phosphorus.[NOTE 7] The hot filtrate is added slowly by pouring into
a solution of 20 grams of sodium bisulphite in 1 liter of distilled water. The
solution is made basic and solvent extracted as above using two 100 ml portions
of ether, washed and dry.[NOTE 4] The oily residue extracted (after removing the
ether) is a mixture of methamphetamine, iodo-ephedrine and minor amount of
acetic ester ephedrine.[NOTE 5]

[NOTE 6] 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%

[NOTE 7] Have some hot glacial acetic acid on hand to rinse the flask and the
         filtered red phosphorus.


Calculation of HI to Ephedrine Ratio
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In Procedure 1 and 2, two moles of HI react with one mole of Ephedrine giving
Methamphamine and I2.

C6H5-(CHOH)-CH(NHCH3)-CH3  + 2HI ==>>  C6H5-(CH2)-CH(NHCH3)-CH3  + I2 + H2O

Therefore the Ephedrine:HI ratio is theoretically 1:2, but to assure that
the reaction goes to completion, the actual ratios used above are 1:3.3

This ratios has been calculated as follows:

Procedure 1:

1 ml of 57% HI = 0.99 grams of HI
170 mls of 57% HI = (170 x 0.99) = 168.3 grams of HI = 1.32 moles HI
Since we use 0.4 moles of Eph then (1.32/0.4) = 3.3

Since the reaction of P + I + H2O is 2P + 3I2 + 3H2O ==>>  3HI + H3PO3
Therefore the ratio of HI:I2 is 3:3

Procedure 2

For every 0.04 moles of Eph, then (0.04 x 3.3) = 0.132 moles of HI,
since the ratio of HI:I2 is 3 : 3 = 1, then we need 0.132 moles of
I2 = 33.50 grams of I2.

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