Preparation of Hypophosphorus AcidP

Preparation of Hypophosphorus Acid

Preparation of Sodium Hypophosphite (Brit. Pat. 803179)

115g of white phosphorous is emulsified and reacted with aqueous NaOH to obtain 270g NaH2PO2.H2O (70%).

Preparation of aqueous sodium hypophosphite (Jap. Pat. 58185412)

White phosphorus is dispersed together with slaked lime used as an assistant in an aqueous medium by stirring in an inert gas atmosphere under heating at or above the melting point of white phosphorus. The dispersion is mixed with a solution of caustic soda under heating and stirring to effect the reaction of the components. The product is separated into solid and liquid, and the mother liquor is added with a phosphate donor such as phosphoric acid. When the phosphate donor is phosphoric acid, its amount is about 0.6mol per 1mol of dissolved calcium, and the pH of the system is maintained usually to about 8-11. The reaction is carried out at about 50-100°C, and the reaction product is aged for about 30min. The Ca(II) dissolved in the mother liquid is precipitated in the form of calcium apatite, and the high-purity aqueous solution of sodium hypophosphite can be recovered by this process.

Hypophosphorous acid purification

Free hypophosphorous acid, H3PO2, is prepared by acidifying aqueous solutions of hypophosphite ions, H2PO2-. For example, the solution remaining when phosphine is prepared from the reaction of white phosphorus and a base contains the H2PO2- ion. If barium hydroxide, Ba(OH)2, is used as the base and the solution is acidified with sulfuric acid, barium sulfate, precipitates and an aqueous solution of hypophosphorous acid results.

Ba2+ + 2 H2PO2- + 2 H3O+ + SO42- => BaSO4 + 2 H3PO2 + 2 H2O

The pure acid cannot be isolated merely by evaporating the water, however, because of the easy oxidation of the hypophosphorous acid to phosphoric acids (and elemental phosphorus) and its disproportionation to phosphine and phosphorous acid. The pure acid can be obtained by extraction of its aqueous solution by diethyl ether. Pure hypophosphorous acid forms white crystals that melt at 26.5° C. The electronic structure of hypophosphorous acid is such that it has only one hydrogen atom bound to oxygen, and thus it is a monoprotic oxyacid. It is a weak acid and forms only one series of salts, the hypophosphites. Hydrated sodium hypophosphite, NaH2PO2×H2O, is used as an industrial reducing agent, particularly for the electroless plating of nickel onto metals and nonmetals.

Hypophosphorous acid from sodium hypophosphite

To a stirred solution of 717.8 g of a 32% hydrochloric acid solution in a 3-necked 2 liter flask was added 615.42 g of powdered sodium hypophosphite. The temperature of the solution rose about 2° C. Water was removed from the stirred reaction mixture by reduced pressure distillation at a temperature of about 55°C ±7° C at a pressure of 44-72 mmHg until a hypophosphorous acid concentration of about 80 wt% was obtained. After cooling to room temperature, sodium chloride that had precipitated was filtered from the reaction mixture. The filter cake was washed twice with 32 wt % hydrochloric acid.

The recovered product contained 355.7 g of hypophosphorous acid. The analysis showed that the product contained 0.9 wt% sodium, 3.2 wt% chloride, and 80.96 wt% hypophosphorus acid. Chloride ion was removed from hypophosphorous acid using an ion-exchange column (height 221/4", diameter 11/8"). The column was packed with Rohm and Haas Resin IRA-410 in the chloride form and was regenerated using 5% NaOH.

The results from using this column at different H3PO2 and Cl- concentrations are shown below.
Influent Composition

Effluent Composition

H3PO2 %

Chloride %
46

3.0

49.6

0.06
46

3.0

46.7

0.09
71

5.9

52.6

0.02


Ephedrine Reduction to Methamphetamine with Hypophosphorus Acid and Iodine

By Wizard X

I would recommend a large excess of reducing agent for quick reduction. Charge flask with 100 ml (0.1Lt) of 50% H3PO2 (0.965 mole H3PO2 per 100 ml) fit reflux condenser, add 3x33 grams (99 grams, 0.39 mole) portions of I2 while cooling in ice bath down the reflux condenser[1]. After addition of I2, gently heat so HI gas is evolving from the condenser, add 5 mls potions of H2O down the condenser till HI gas stops evolving and hence maximum amount of HI saturation kept in solution is acheived. Now add 50.4 g (0.25 mole) of ephedrine hydrochloride and boil under reflux for at least 2 hours, let cool and then made basic with 20 % sodium hydroxide solution (20 grams NaOH in 100 mls H2O) in ice bath to liberate the free base. Set-up glassware for steam distillation and steam distil until the distillate is almost neutral to litmus.[2]

The libertated freebase methamphetamine which seperates is solvent extracted with three, 50-75 ml ether (or toluene) 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.[3] After removal of the ether (or toluene), the oil was vacuum distilled at a vacuum of 15 mmHg at 93°C. The yield is 80 - 82%[4].

Notes:

If I2 sticks to the condenser wall, wash down with distilled water.
Have the distillate receiving flask in ice. And cool down to 4-5 degC.
Anhydrous magnesium sulphate can be used. Rinse the anhydrous magnesium sulphate with a little ether (or toluene) after drying the main ether/amine solution.

Ephedrine Reduction

Major reduction reaction:

C6H5-(CHOH)-CH(NHCH3)-CH3 + HI =====> C6H5-(CHI)-CH(NHCH3)-CH3 + H2O

C6H5-(CHI)-CH(NHCH3)-CH3 + HI =====> C6H5-CH2-CH(NHCH3)-CH3 + I2

Minor reduction reaction:

C6H5-(CHI)-CH(NHCH3)-CH3 + H3PO2 + H2O =====> C6H5-CH2-CH(NHCH3)-CH3 + H3PO3 + HI

Ratio of ephedrine to HI is theoretically 1:2, however a 1:3 is used for better reduction and yeild.
Hypophosphorus to HI Calculations

H3PO2 + H2O + I2 ==>> H3PO3 + 2HI

Hypophosphorous 50% w/w. F.W = 66 g/mol. Density = 1.274 g/ml.

100mls (0.1Lt) of Hypophosphorous 50% w/w contains:

(1.274 / 50)/100 = 0.637 g/ml H3PO2 = 0.00965 mol/ml H3PO2.

0.00965 mol/ml H3PO2 x 100 = 0.965 mol/100ml H3PO2.

OR

(0.637/66) x 1000 = 9.65 moles H3PO2 per 1000 mls. (mol/Lt)

Since we use 100mls (0.1Lt), then 9.65 x 0.1 = 0.965 mol/100ml H3PO2. Now since the ratio of Ephedrine : HI is 1:3 = (3/1), we require 0.75 moles of HI for every 0.25 moles of ephedrine hydrochloride. Since we have 0.965 mol of H3PO2 and 0.39 moles of I2 (99/253.8 = 0.39), then the ratio of I2:HI is 1:2 = (2/1); so 0.39 moles of I2 reacts with the Hypophosphorous acid to form 0.39 x 2 = 0.78 moles of HI.

Finally, the excess Hypophosphorous acid, H3PO2 is 0.965-0.39 = 0.575 moles of H3PO2 is excess. The ratio for H3PO2 : I2 is 1:1, so only 0.39 moles of H3PO2 is needed to react with 0.39 moles of I2 to form 0.78 moles of HI. Not only do we have enough HI; 0.78 moles to reduce 0.25 moles of ephedrine hydrochloride, but a large excess of 0.575 moles of H3PO2.

Alternatively, Charge a 1000ml flask with 100 ml (0.1Lt) of 50% H3PO2 (0.965 mole H3PO2 per 100 ml), 100 mls of distilled water, fit reflux condenser, add 4x49.5 grams (198 grams, 0.78 mole) portions of I2 while cooling in ice bath down the reflux condenser[1]. After addition of I2, gently heat so HI gas is evolving from the condenser, add 10 mls potions of H2O down the condenser till HI gas stops evolving and hence maximum amount of HI saturation kept in solution is acheived. Now add 100.8 g (0.5 mole) of ephedrine hydrochloride and boil under reflux for at least 2 hours, let cool and then made basic with 20 % sodium hydroxide solution (20 grams NaOH in 100 mls H2O) in ice bath to liberate the free base. Set-up glassware for steam distillation and steam distil until the distillate is almost neutral to litmus.[2]

Since we have 0.965 mol of H3PO2 and 0.78 moles of I2 (198/253.8 = 0.78), then the ratio of I2 : HI is 1: 2 = (2/1); so 0.78 moles of I2 reacts with the Hypophosphorous acid to form 0.78 x 2 = 1.56 moles of HI. Finally, the excess hypophosphorus acid, H3PO2 is 0.965-0.78 = 0.185 moles of H3PO2 is excess. The ratio for H3PO2 : I2 is 1:1, so only 0.78 moles of H3PO2 is needed to react with 0.78 moles of I2 to form 1.56 moles of HI. Not only do we have enough HI; 1.56 moles to reduce 0.5 moles of ephedrine hydrochloride, but an excess of 0.185 moles of H3PO2.
P-fed Reduction to Methamphetamine with Hypophosphorus Acid and Iodine

By Pebble

Assuming that you have extracted P-fed from your pills, you are now ready to convert the p-fed to methamphetamine. The structure of P-fed is so closly related to methamphetamine, that people have learned to alter it's structure in order to convert it to methamphetamine.

What one is actually doing is reducing a benzylic alcohol. In our case, it appears as OH in P-fed. So, we strip this off completly and throw on an additional H (Hydrogen) to the p-fed. So, by removing the OH and adding a H we have thus altered the structure and have created methamphetamine.

This is for small-scale reduction of P-fed to methamphetamine. If one wanted to make meth in the range of ounces, etc. I would not reccomend one cook in an open container. Fire or explosion are at a much higher risk when cooking anything above 10 grams using this procedure. This is why the Push/Pull apparatus is so popular. No, fumes, smell, and IT CAN reduce the risk of injury if something does go wrong. Notice I said IT CAN. There is no fool-proof method of safety when working with chemicals of this nature, unless proper lab equipment, etc are used; and this still does not elimanate the dangers, just reduces the probability that an accident will happen.

If one wanted to have 5 grams as their finished peoduct, than 6-8 grams of P-fed. Some can get 90% yield where others may only get 35% yield. This takes time to perfect, but results will vary WITH YOU.

Take your 8 grams of p-fed and put this into a 250ml. flask or pyrex meassuring cup. Slowing add Iodine crystals unto the p-fed is mixed well with the

Hasil (Bahasa Indonesia) 1:

Persiapan Hypophosphorus asamPersiapan Hypophosphite natrium (Brit. Pat. 803179)115g putih fosfor berformulasi dan bereaksi dengan larutan NaOH untuk mendapatkan 270g NaH2PO2.H2O (70%).Persiapan larutan natrium hypophosphite (Jap. Pat. 58185412)Fosfor putih tersebar dengan kapur digunakan sebagai asisten dalam media berair oleh aduk dalam atmosfir inert gas di bawah Penghangat Ruangan pada atau di atas titik lebur fosfor putih. Dispersi dicampur dengan larutan soda kaustik di bawah Penghangat Ruangan dan diaduk untuk efek reaksi komponen. Produk dipisahkan menjadi padat dan cair, dan minuman keras ibu ditambahkan dengan fosfat donor seperti asam fosfat. Ketika donor fosfat asam fosfat, yang besarnya adalah sekitar 0.6mol per 1mol terlarut kalsium, dan pH sistem dijaga biasanya sekitar 8-11. Reaksi dilaksanakan di sekitar 50-100° C, dan produk reaksi ini berusia selama sekitar 30 menit. Ca(II) yang terlarut dalam cairan ibu yang dipercepat dalam bentuk apatit kalsium, dan kemurnian tinggi larutan natrium hypophosphite dapat dipulihkan dengan proses ini.Pemurnian asam hypophosphorousGratis hypophosphorous asam, H3PO2, disiapkan oleh acidifying larutan hypophosphite ion, H2PO2-. Sebagai contoh, solusi yang tersisa ketika fosfin disiapkan dari reaksi fosfor putih dan basis berisi H2PO2-ion. Jika barium hidroksida, Ba (OH) 2, digunakan sebagai dasar dan solusi Diasamkan dengan asam sulfat, barium sulfat, endapan dan larutan asam hypophosphorous hasil.BA2 + + 2 H2PO2 - 2 H3O + + SO42 - = > BaSO4 2 H3PO2 + 2 H2OAsam murni tidak terisolasi hanya oleh evaporasi air, namun, karena mudah oksidasi asam hypophosphorous untuk asam fosfat (dan fosfor elemental) dan yang disproportionation untuk fosfin dan asam fosfor. Asam murni dapat diperoleh dengan ekstraksi dengan larutan dengan dietil eter. Kristal bentuk putih murni hypophosphorous asam yang mencair pada 26.5° C. Struktur elektronik hypophosphorous asam adalah sedemikian rupa sehingga memiliki hanya satu atom hidrogen yang terikat oksigen, dan dengan demikian itu adalah monoprotic oxyacid. Ini adalah asam lemah dan membentuk satu siri garam, hypophosphites. Terhidrasi natrium hypophosphite, NaH2PO2 × H2O, digunakan sebagai agen pereduksi industri, terutama untuk mekanika plating nikel ke logam dan nonlogam.Asam hypophosphorous dari hypophosphite natriumSolusi yang bergerak 717.8 g larutan asam klorida 32% dalam 2 berleher 3 liter labu ditambahkan 615.42 g bubuk natrium hypophosphite. Suhu solusi mawar sekitar 2° C. air telah dihapus dari campuran reaksi yang bergerak oleh mengurangi tekanan distilasi pada suhu 55 ° C ±7 ° C pada tekanan mmHg 44-72 sampai konsentrasi asam hypophosphorous sekitar 80 wt % diperoleh. Setelah pendinginan ke suhu kamar, natrium klorida yang telah diendapkan disaring dari campuran reaksi. Kue penyaring dicuci dua kali dengan 32 wt % asam klorida.Produk pulih berisi 355.7 g asam hypophosphorous. Analisis menunjukkan bahwa produk terkandung 0.9 wt % natrium, 3.2 wt % klorida dan 80.96 wt % hypophosphorus asam. Ion klorida dihapus dari asam hypophosphorous menggunakan kolom pertukaran ion (tinggi 221/4", diameter 11/8"). Kolom dikemas dengan Rohm dan Haas Resin IRA-410 berupa klorida dan adalah diregenerasi menggunakan 5% NaOH.Hasil dari menggunakan kolom ini di berbeda H3PO2 dan Cl-konsentrasi yang ditunjukkan di bawah.Komposisi banget Komposisi limbah H3PO2% Klorida %46 3.0 49,6 0.0646 3.0 46. 7 0.0971 5.9 52,6 0.02Pengurangan efedrin shabu-shabu dengan asam Hypophosphorus dan yodiumOleh Wizard XSaya akan merekomendasikan berlebih agen pereduksi untuk cepat pengurangan. Biaya labu dengan 100 ml (0.1Lt) 50% H3PO2 (0.965 mol H3PO2 per 100 ml) cocok pendingin refluk, tambahkan bagian-bagian 3 x 33 gram (99 gram, 0.39 mol) dari I2 sedangkan pendinginan dalam penangas es turun pendingin refluk [1]. Setelah penambahan I2, lembut panas sehingga HI gas berkembang dari kondensor, tambahkan 5 mls ramuan dari H2O turun kondensor sampai HI gas berhenti berkembang dan karenanya jumlah maksimum HI saturasi disimpan dalam solusi dicapai. Sekarang menambahkan 50,4 g (0.25 mol) efedrin hidroklorida dan mendidih refluk untuk setidaknya 2 jam, biarkan dingin, dan kemudian membuat dasar dengan larutan 20% natrium hidroksida (20 gram NaOH 100 Rummenigge H2O) dalam penangas es untuk membebaskan dasar gratis. Gelas set-up untuk distilasi uap dan uap menyaring sampai distilat hampir netral ke lakmus. [2]Libertated freebase shabu-shabu yang memisahkannya pelarut diambil dengan tiga, 50-75 ml eter (atau toluena) bagian dan Eter amina solusi pertama dicuci dengan 50 ml air suling dan solusi eter amina dikeringkan dengan anhidrat natrium karbonat. [3] setelah penghapusan eter (atau toluena), minyak adalah vakum didistilasi pada vakum 15 mmHg pada 93° C. Hasil adalah 80-82% [4].Catatan: Jika I2 tongkat ke dinding kondensor, mencuci dengan air suling. Memiliki distilat menerima labu es. Dan mendinginkan untuk 4-5 degC. Anhidrat magnesium sulfat dapat digunakan. Bilas anhidrat magnesium sulfat dengan sedikit eter (atau toluena) setelah pengeringan solusi eter/amina utama. Pengurangan efedrinReaksi pengurangan besar:C6H5 - CH3 (CHOH) - CH (NHCH3)- + HI === > C6H5 - CH3 (CHI) - CH (NHCH3)- + H2OC6H5 - CH3 (CHI) - CH (NHCH3)- + HI === > C6H5-CH2-CH (NHCH3)-CH3 + I2Reaksi pengurangan kecil:C6H5 - CH3 (CHI) - CH (NHCH3)- + H3PO2 + H2O === > C6H5-CH2-CH (NHCH3)-CH3 H3PO3 + HIRasio efedrin untuk HI secara teoritis adalah 1:2, namun 1:3 digunakan untuk pengurangan dan yeild yang lebih baik.Hypophosphorus untuk HI perhitunganH3PO2 H2O + I2 == >> H3PO3 + 2HIHypophosphorous 50% w/w. F.W = 66 g mol. kepadatan = 1.274 g/ml.100mls (0.1Lt) Hypophosphorous 50% w/w berisi:(1.274 / 50) / 100 = 0.637 g / ml H3PO2 = 0.00965 mol/ml H3PO2.0.00965 mol/ml H3PO2 x 100 = 0.965 mol / 100ml H3PO2.ATAU(0.637/66) x 1000 = 9,65 mol H3PO2 per 1000 mls. (mol/Lt)Karena kita menggunakan 100mls (0.1Lt), kemudian 9,65 x 0,1 = 0.965 mol/100 ml H3PO2. Sekarang sejak rasio efedrin: HI adalah 1:3 = (3/1), kami memerlukan 0.75 mol HI untuk mol 0,25 setiap efedrin hydrochloride. Karena kita memiliki mol 0.965 dari H3PO2 dan 0.39 mol I2 (99 253.8 = 0.39), maka rasio I2:HI 1:2 = (2/1); Jadi 0.39 mol I2 bereaksi dengan asam Hypophosphorous untuk bentuk 0.39 x 2 = 0.78 mol HI.Akhirnya, kelebihan asam Hypophosphorous, H3PO2 adalah 0.965-0.39 = 0.575 mol H3PO2 kelebihan. Rasio untuk H3PO2: I2 adalah 1:1, sehingga hanya 0.39 mol H3PO2 diperlukan untuk bereaksi dengan mol 0.39 I2 untuk membentuk 0,78 mol HI. Tidak hanya kita memiliki cukup HI; 0,78 mol untuk mengurangi 0.25 mol efedrin hidroklorida, tapi berlebih mol 0.575 H3PO2.Alternatively, Charge a 1000ml flask with 100 ml (0.1Lt) of 50% H3PO2 (0.965 mole H3PO2 per 100 ml), 100 mls of distilled water, fit reflux condenser, add 4x49.5 grams (198 grams, 0.78 mole) portions of I2 while cooling in ice bath down the reflux condenser[1]. After addition of I2, gently heat so HI gas is evolving from the condenser, add 10 mls potions of H2O down the condenser till HI gas stops evolving and hence maximum amount of HI saturation kept in solution is acheived. Now add 100.8 g (0.5 mole) of ephedrine hydrochloride and boil under reflux for at least 2 hours, let cool and then made basic with 20 % sodium hydroxide solution (20 grams NaOH in 100 mls H2O) in ice bath to liberate the free base. Set-up glassware for steam distillation and steam distil until the distillate is almost neutral to litmus.[2]Since we have 0.965 mol of H3PO2 and 0.78 moles of I2 (198/253.8 = 0.78), then the ratio of I2 : HI is 1: 2 = (2/1); so 0.78 moles of I2 reacts with the Hypophosphorous acid to form 0.78 x 2 = 1.56 moles of HI. Finally, the excess hypophosphorus acid, H3PO2 is 0.965-0.78 = 0.185 moles of H3PO2 is excess. The ratio for H3PO2 : I2 is 1:1, so only 0.78 moles of H3PO2 is needed to react with 0.78 moles of I2 to form 1.56 moles of HI. Not only do we have enough HI; 1.56 moles to reduce 0.5 moles of ephedrine hydrochloride, but an excess of 0.185 moles of H3PO2.P-fed Reduction to Methamphetamine with Hypophosphorus Acid and IodineBy PebbleAssuming that you have extracted P-fed from your pills, you are now ready to convert the p-fed to methamphetamine. The structure of P-fed is so closly related to methamphetamine, that people have learned to alter it's structure in order to convert it to methamphetamine.What one is actually doing is reducing a benzylic alcohol. In our case, it appears as OH in P-fed. So, we strip this off completly and throw on an additional H (Hydrogen) to the p-fed. So, by removing the OH and adding a H we have thus altered the structure and have created methamphetamine.This is for small-scale reduction of P-fed to methamphetamine. If one wanted to make meth in the range of ounces, etc. I would not reccomend one cook in an open container. Fire or explosion are at a much higher risk when cooking anything above 10 grams using this procedure. This is why the Push/Pull apparatus is so popular. No, fumes, smell, and IT CAN reduce the risk of injury if something does go wrong. Notice I said IT CAN. There is no fool-proof method of safety when working with chemicals of this nature, unless proper lab equipment, etc are used; and this still does not elimanate the dangers, just reduces the probability that an accident will happen.If one wanted to have 5 grams as their finished peoduct, than 6-8 grams of P-fed. Some can get 90% yield where others may only get 35% yield. This takes time to perfect, but results will vary WITH YOU.Take your 8 grams of p-fed and put this into a 250ml. flask or pyrex meassuring cup. Slowing add Iodine crystals unto the p-fed is mixed well with the