Ephedrine Reduction with HI/P

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

Ephedrine HCl, molecular weight MW = 201.73 grams/mole

Ephedrine Freebase, molecular weight MW = 165.23 grams/mole

PROCEDURE 1:

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

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

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

Ephedrine + 2HI ==>> Methamphetamine + 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 phosphorus, iodine and water is 2 P + 3 I2 + 6 H2O --> 6 HI + 2 H3PO3 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.