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

النتائج (الكشف التلقائي) 1:

丙酸酸的製備次亞磷酸鈉 （英，麥琪 803179） 的製備白磷 115g 是乳化和反應與水溶液的氫氧化鈉獲得 270 g NaH2PO2.H2O （70%）。編寫的次亞磷酸鈉水溶液 （日本。派特。) 58185412白磷是加熟石灰攪拌下加熱白磷的熔點以上惰性氣體氣氛中被用作助理在水介質中分散的。分散混合在一起的下加熱並攪拌，影響元件的反應的苛性鈉溶液。這種產品分離成固體和液體，和母液添加磷酸捐贈者如磷酸等。磷酸捐助時磷酸，其量是大約 0.6mol 每 1mol 溶解的鈣，和系統 ph 值通常保持約 8-11。約 50-100 ° C，在進行反應，反應產物歲約 30 分鐘。Ca(II) 在母液中溶解析出鈣磷灰石的形式和次亞磷酸鈉的純度高水溶液可以回收利用這一進程。次膦酸淨化游離磷酸，： h3po2），備酸化次磷酸鈉離子，H2PO2-水溶液。例如，剩餘的磷化氫準備從白磷彈和一個基地的反應時該解決方案包含 H2PO2 離子。如果氫氧化鋇，廣管局 （OH） 2，使用作為基礎和解決辦法是用硫酸、 硫酸鋇、 沉澱和磷酸結果水溶液酸化。Ba2 + 2 的 H2PO2-2 H3O + + SO42-= > BaSO4 + 2： h3po2） + 2 H2O純酸不能僅僅通過蒸發的水，但是，由於容易氧化的有機膦酸 （和單質磷） 磷酸和磷酸和磷化氫對其歧孤立。純酸能求得其水溶液提取，乙醚。純磷酸形式白色晶體，融化在 26.5 ° c。次膦酸的電子結構是這樣它已綁定到氧氣，只有一個氫原子，因此它是一元弱酸的含氧酸。它是一種弱酸和形式的鹽類，hypophosphites 只有一個系列。次磷酸水合的鈉，NaH2PO2 × H2O，用作工業的還原劑，特別是對化學鍍鎳上金屬和非金屬。次膦酸從次亞磷酸鈉為 717.8 g 3 長頸的 2 升 32%鹽酸溶液攪拌溶液瓶增加了次磷酸鈉粉 615.42 克。解決方案玫瑰約 2 ° C.水已經被移除了由攪拌的反應混合物的溫度降低減壓蒸餾溫度約 55 ° C ± 7 ° C 在 44 72 毫米汞柱壓力直到獲得了大約 80 ％ 磷酸濃度。冷卻至室溫後, 從反應混合物過濾了沉澱的氯化鈉。濾餅是兩次用 32 wt %鹽酸酸洗的。回收的產品載 355.7 g 次膦酸。分析表明，該產品含有 0.9 wt %鈉，3.2 wt %氯化和 80.96 wt %丙酸酸。氯離子被調離次膦酸使用離子交換柱 （高度 221/4 英寸，直徑 11/8 英寸）。列擠滿 Rohm and Haas 樹脂 IRA 410 氯型和再生後使用 5%氫氧化鈉。使用此列在不同： h3po2） 和 Cl 濃度結果如下所示：進水組成 污水成分 ： H3PO2） % 氯化 %46 3.0 49.6 0.0646 3.0 46.7 0.0971 5.9 52.6 0.02麻黃堿減少甲基苯丙胺碘與丙酸酸堿由嚮導 X我會推薦大量過剩的還原劑進行快速減少。控告瓶 100 毫升 （0.1Lt） 的 50%： h3po2） （0.965 摩爾為： h3po2），每 100 毫升） 適合的回流冷凝器、 冷卻在冰浴下回流冷凝器 [1] 時添加 I2 3 x 33 克 0.39 鼴鼠 99 克） 部分。I2 的加法以後, 輕輕熱所以 HI 氣體正在從凝汽器，添加 5 毫升藥水的 H2O 下凝汽器，直到喜氣體停止不斷變化的因此取得了 HI 飽和度保持在溶液中的最大量。現在添加 50.4 g （0.25 摩爾） 鹽酸麻黃堿和沸騰回流為至少 2 個小時、 放涼，然後在冰浴解放自由基地基本用 20%氫氧化鈉溶液 （20 克氫氧化鈉在 100 毫升 H2O）。設置玻璃器皿為水蒸氣蒸餾法和蒸汽蒸餾直到餾出物是近乎中立的試金石。[] 2三、 50-75 毫升醚 （或甲苯） 部分提取哪些難是溶劑 libertated 化學甲基安非他明和醚/胺溶液先洗手用蒸餾水和醚/胺溶液用無水碳酸鈉幹 50 毫升。[3] 後醚 （或甲苯） 的去除，石油是真空蒸餾真空 15 毫米汞柱，在 93 ° c。收益率是 80-82%[4]。注釋：如果 I2 堅持凝汽器壁，用蒸餾水沖洗。有接收瓶在冰餾出物。和冷靜下來到 4-5 攝氏度。可以用無水硫酸鎂。後乾燥主要醚/胺溶液沖洗無水硫酸鎂與一個小醚 （或甲苯）。麻黃堿減少主要還原反應：C6H5-(祖)-CH （NHCH3）-CH3 + HI = = = > C6H5-（志）-CH （NHCH3）-CH3 + H2OC6H5-（志）-CH （NHCH3）-CH3 + HI = = = > C6H5-CH2-CH （NHCH3）-CH3 + I2輕微的還原反應：C6H5-（志）-CH （NHCH3）-CH3 +： h3po2） + H2O = = = > C6H5-CH2-CH （NHCH3）-CH3 + H3PO3 + HI麻黃堿與 HI 比理論上是 1： 2，然而 1： 3 用於更好的減少和產量。丙酸喜計算： h3po2） + H2O + I2 = = >> H3PO3 + 2HI次磷酸 50 %w / w.日用淡水 = 66 g/分子密度 = 1.274 g/ml。100mls (0.1Lt) 的次磷酸 50 %w / w 包含：(1.274 / 50) / 100 = 0.637 g/ml： h3po2） = 0.00965 條每毫升 mol： h3po2）。0.00965 mol/ml： h3po2） x 100 = 0.965 mol 100 毫升： h3po2）。或(0.637/66） x 1000 = 96.5 摩爾： h3po2） 每 1000年毫升。（mol/Lt）由於我們使用 100mls （0.1Lt），然後 96.5 × 0.1 = 0.965 mol 100 毫升： h3po2）。現在由於麻黃堿的比例： 喜是 1： 3 = (3/1)，0.75 摩爾數喜，我們需要為每個 0.25 摩爾的鹽酸麻黃堿。因為我們有： h3po2） 的 0.965 mol 和 I2 0.39 摩爾 (99/2538 = 0.39），然後 I2:HI 的比例是 1： 2 = （2/1） ；所以 0.39 摩爾數 I2 反應與磷酸形式 0.39 x 2 = 0.78 痣的喜。最後，過剩的次膦酸： h3po2） 是 0.965 0.39 = 0.575 痣： h3po2） 是多餘。： h3po2） 的比率： I2 是 1： 1，所以只有 0.39 0.39 摩爾數 I2 形成 0.78 摩爾的喜與反應所需的痣： h3po2）。我們不僅有足夠的喜 ；0.78 摩爾以減少 0.25 摩爾的鹽酸麻黃堿，但大量過剩的 0.575 摩爾數： h3po2）。或者，用蒸餾水 50%： h3po2） （0.965 摩爾： h3po2） 每 100 毫升），100 毫升 100 毫升 （0.1Lt） 收取 1000 毫升瓶，適合回流冷凝器，而冷卻在冰浴下回流冷凝器 [1] 添加 I2 4x49.5 克 0.78 鼴鼠 198 克） 部分。I2 的加法以後, 輕輕熱所以 HI 氣體正在從凝汽器，添加 10 毫升藥水的 H2O 下凝汽器，直到喜氣體停止不斷變化的因此取得了 HI 飽和度保持在溶液中的最大量。現在添加 100.8 g （0.5 摩爾） 鹽酸麻黃堿和沸騰回流為至少 2 個小時、 放涼，然後在冰浴解放自由基地基本用 20%氫氧化鈉溶液 （20 克氫氧化鈉在 100 毫升 H2O）。設置玻璃器皿為水蒸氣蒸餾法和蒸汽蒸餾直到餾出物是近乎中立的試金石。[] 2因為我們有： h3po2） 的 0.965 mol 和 I2 0.78 摩爾 (198/2538 = 0.78），然後 I2 的比率： 喜為 1： 2 = （2/1） ；所以 0.78 摩爾數 I2 反應與磷酸形式 0.78 x 2 = 1.56 痣的喜。最後，多餘的丙酸酸，： h3po2） 是 0.965 0.78 = 0.185 痣： h3po2） 是多餘。： h3po2） 的比率： I2 是 1： 1，所以只有 0.78 0.78 摩爾數 I2 形成 1.56 摩爾的喜與反應所需的痣： h3po2）。我們不僅有足夠的喜 ；1.56 摩爾以減少 0.5 摩爾的鹽酸麻黃堿，但過量的 0.185 摩爾數： h3po2）。P 餵養減少甲基苯丙胺碘與丙酸酸堿By 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

النتائج (الكشف التلقائي) 2:

次磷酸的製備的次磷酸鈉的製備（英國人。專利803179）的白磷115克乳化並用NaOH水溶液反應，得到270克NaH2PO2.H2O（70％）。製備水性次磷酸鈉（日語。專利58185412）白磷與由在加熱下，在惰性氣體氣氛下攪拌在或高於白磷熔點用作助劑在水性介質中熟石灰一起分散。將分散液混合，用苛性鈉中在加熱和攪拌以進行組分的反應溶液。該產物被分離成固體和液體，並母液中加入磷酸供體，例如磷酸。當磷酸鹽供體是磷酸，其量為約每溶解的鈣的1mol的0.6摩爾，並且系統的pH維持通常約8-11。該反應是在約50-100℃，和反應產物老化約30分鐘。溶解在母液中的Ca（Ⅱ）被沉澱的鈣磷灰石的形式，和次磷酸鈉的高純度的水溶液可以用這種方法來回收。次磷酸純化免次磷酸，H 3 PO 2，是通過酸化含水中，次磷酸根離子H2PO2-解決方案。例如，當磷化氫從白磷的反應製備鹼包含H2PO2-離子的溶液剩餘。如果氫氧化鋇，鋇（OH）2，被用作鹼，將溶液用硫酸酸化，硫酸鋇，沉澱物和次磷酸的結果的水溶液。鋇+ 2 H2PO2- + 2 H 3 O + SO 4 2- =>硫酸鋇+2 H3PO2 + 2 H 2 O 的純的酸不能被然而，蒸發由於次磷酸的容易氧化的水，以磷酸（和元素磷）僅僅分離和其歧化膦和亞磷酸。所述純的酸可以通過萃取其由乙醚水溶液來獲得。純次磷酸形式的熔化在26.5℃。次磷酸的電子結構是這樣的，它只有一個氫原子結合到氧，因此它是一個單質子含氧白色晶體。它是一種弱酸和形式只有一個系列的鹽，該次磷酸鹽。水合次磷酸鈉，次磷酸鈉×H 2 O，被用作工業還原劑，特別是用於化學鍍鎳的上金屬和非金屬。次磷酸從次磷酸鈉向717.8克的32％的鹽酸溶液的攪拌的溶液中3 -necked2升燒瓶中加入615.42克粉狀次磷酸鈉。將溶液的溫度升至約2℃。將水從通過減壓蒸餾將攪拌的反應混合物中除去在約55℃±7℃，在44-72毫米汞柱的壓力下的溫度直到大約一次磷酸濃度得到80％（重量）。在冷卻至室溫後，氯化鈉該已沉澱從反應混合物中過濾。濾餅用水洗滌兩次，用32％（重量）鹽酸。將回收的產物含有355.7克次磷酸。的分析表明，該產品含有0.9％（重量）鈉，3.2氯化物重量％，80.96重量％次磷酸。氯離子是從次磷酸用離子交換柱（高度四分之二百二十一“，直徑11/8”）除去。該柱用羅門哈斯樹脂IRA-410裝填在氯化物形式並用5％的NaOH再生。從使用該柱在不同H3PO2和氯離子濃度，結果如下所示。進水組成廢水組成H3PO2％氯化物％46 3.0 49.6 0.06 46 3.0 46.7 0.09 71 5.9 52.6 0.02 麻黃鹼還原成甲基苯丙胺與次磷酸和碘通過嚮導X 我會建議大量過剩的還原劑快速還原。電荷燒瓶用50％H 3 PO 2為100毫升（0.1Lt）（每100毫升0.965摩爾H 3 PO 2）配合回流冷凝器，添加3x33克（99克，0.39摩爾）部分I2的同時冷卻在冰浴下的回流冷凝器[1] 。此外I2後，輕輕熱量，使HI氣體演變從冷凝器，加水5毫升藥水倒在冷凝器到HI燃氣停止進化，從而HI飽和度保持在溶液中的最大數量來達到的。現在（在100毫升水20克氫氧化鈉）在冰浴中添加50.4克鹽酸麻黃鹼和燒開（0.25摩爾）回流至少2個小時，晾涼後再做基本的用20％氫氧化鈉溶液以釋放游離鹼。設置玻璃器皿進行水蒸氣蒸餾和蒸汽DISTIL直至餾出液呈中性至石蕊。[2] 的libertated游離鹼苯丙胺其中seperates是溶劑有三個，50-75毫升乙醚（或甲苯）部分和醚/胺溶液中提取首先用50毫升的蒸餾水，並用無水碳酸鈉乾燥醚/胺溶液中。[3]後除去乙醚（或甲苯）中，將油進行真空蒸餾在15mmHg的真空下於93℃。的產率是80 - 82％的[4]。注意：如果I2附著在冷凝器壁，沖洗用蒸餾水。有無餾出接收燒瓶在冰中。並冷卻到4-5攝氏度。無水硫酸鎂都可以使用。乾燥的主要醚/胺溶液沖洗後，在無水硫酸鎂一點點醚（或甲苯）。麻黃素減少主要還原反應：C6H5-（CHOH）-CH（NHCH3）-CH 3 + HI =====> C6H5- （CHI）-CH（NHCH 3）-CH 3 + H 2 O C 6 H 5（CHI）-CH（NHCH 3）-CH 3 + HI =====> C6H5-CH2-CH（NHCH 3）-CH 3 + I 2 次要還原反應：C6H5- （CHI）-CH（NHCH 3）-CH 3 + H 3 PO 2 + H 2 O =====> C6H5-CH2-CH（NHCH 3）-CH 3 + H 3 PO 3 + HI 麻黃素到HI的比率在理論上是1：2，然而1:3用於更好還原和yeild。Hypophosphorus到HI計算H3PO2 + H 2 O + I 2 == >> 3 PO 3 + 2HI 次磷酸50％重量/瓦特 FW = 66克/摩爾。。密度=1.274克/ ml的100mls的次磷酸50％（0.1Lt）w / w的包括：（1.274 / 50）/ 100 =0.637克/ ml的H3PO2 = 0.00965摩爾/毫升H3PO2。0.00965摩爾/毫升H3PO2×100 = 0.965摩爾/ 100ml的H3PO2。OR （0.637 / 66）×1000 =每1000毫升9.65摩爾的H3PO2。（摩爾/升）由於我們使用100mls（0.1Lt），則9.65×0.1 = 0.965摩爾/ 100ml的H3PO2。現在，因為麻黃鹼的比例：HI為1：3 =（3/1），我們需要0.75摩爾的HI的鹽酸麻黃鹼每0.25摩爾。因為我們有0.965摩爾H 3 PO 2的與0.39摩爾I2的（99 / 253.8 = 0.39），I2的則比率：HI為1：2 =（2/1）; 所以I 2為0.39摩爾發生反應的次磷酸以形成HI 0.39×2 = 0.78摩爾。最後，過量次磷酸，H 3 PO 2是0.965-0.39 = H3PO2的0.575摩爾過剩。的比率為H3PO2：I2為1：1，因此H3PO2只有0.39摩爾是需要與I2的0.39摩爾反應以形成HI 0.78摩爾。我們不僅有足夠的HI; 0.78摩爾，以減少0.25摩爾鹽酸麻黃鹼，但大量過量的H 3 PO 2的0.575摩爾。可替代地，充電一千毫升燒瓶用100ml（0.1Lt）50％H 3 PO 2（每100毫升0.965摩爾H 3 PO 2）的，蒸餾水100毫升水，適當的回流冷凝器，添加4x49.5克（198克，0.78摩爾）12的部分，同時冷卻在冰浴下的回流冷凝器[1]。此外I2後，輕輕熱量，使HI氣體演變從冷凝器，加水10毫升藥水順著冷凝器，直到HI燃氣停止進化，從而HI飽和度保持在溶液中的最大數量來達到的。現在（在100毫升水20克氫氧化鈉）在冰浴中添加100.8克鹽酸麻黃鹼和沸騰（0.5摩爾）回流至少2個小時，晾涼後再做基本的用20％氫氧化鈉溶液以釋放游離鹼。設置玻璃器皿進行水蒸氣蒸餾和蒸汽DISTIL直至餾出液呈中性至石蕊[2]。由於我們有0.965摩爾H 3 PO 2和I 2的（198 / 253.8 = 0.78）0.78摩爾，I2的則比率：HI是1：2 =（2/1）; 所以I 2為0.78摩爾發生反應的次磷酸以形成HI 0.78×2 = 1.56摩爾。最後，將過量的次磷酸，H 3 PO 2是H 3 PO 2的0.965-0.78 = 0.185摩爾過剩。的比率為H3PO2：I2為1：1，因此H3PO2只有0.78摩爾是需要與I2的0.78摩爾反應以形成HI 1.56摩爾。我們不僅有足夠的HI; 1.56痣減少0.5摩爾鹽酸麻黃鹼，但過量的H3PO2 0.185摩爾。P-減少美聯儲對甲基苯丙胺與次磷酸和碘通過卵石假設您已經提取的P-從餵藥丸你，你現在準備轉換在p送入甲基苯丙胺。的P-供給的結構被如此closly涉及甲基苯丙胺，人們已經了解到，以改變它的結構，以便將其轉換為甲基苯丙胺。什麼人真正做的是降低芐醇。在我們的例子中，它顯示為OH P中餵養。所以，我們剝離這一關完全地，扔在一個額外的H（氫）與P-餵養。這樣，通過除去OH和添加ħ因而我們改變了結構，並已創建甲基苯丙胺。這是為小規模削減P-送入甲基苯丙胺。如果一個人想使甲基盎司的範圍，等等。我不會在一個開放的容器reccomend 1名廚師。火災或爆炸是在一個更高的風險煮東西上面使用此過程10克時。這就是為什麼推/拉設備是如此受歡迎。不，煙霧，氣味，它可以減少受傷的風險，如果確實存在錯誤。請注意，我說可以。與有這種性質的化學品進行工作時，除非有適當的實驗室設備等均採用的是安全無防呆法; 而這仍然沒有elimanate的危險，只是減少了事故會發生的概率。如果有人想有5個公克為他們完成？產品，超過6-8克的P-餵養。有些可以得到90％的收益率在別人可能只能拿到35％的收益率。這需要時間來完善，但結果將隨你。把你的8克對美聯儲並把這個於250mL。瓶或耐熱玻璃meassuring杯。減緩補充碘對晶體的p美聯儲與混合均勻

النتائج (الكشف التلقائي) 3:

編寫hypophosphorus酸 鈉

編寫 次(布裡。 帕特。 115 803179)

白磷 的反應是乳化與水汽NaOH獲得2 270G.H2O( 70%).

編寫 水汽鈉(眨次。 帕特。 58185412)

白磷是分散連同仔石灰作為一個助理,一個水汽中等的攪拌在一個惰性氣體的氣氛中加熱的熔化點或以上的白磷。 分散的混合的一個解決辦法是根據苛性鈉加熱和震撼,效果的反應的組成部分。 該產品是分成固體和液體,和母親酒是由同一種磷酸鹽捐助如磷酸。 當磷酸鹽捐助是磷酸,其金額約0.6mol每1,解散鈣、酸鹼值的制度是維持一般約8-11。 的反應是在進行了約50-100,和反應產品是年齡介乎約30分鐘。(ii)解散的CA的母親在液體是倉卒的形式在鈣磷灰石,和高純度的鈉水溶液次可以收回的這一 進程.

hypophosphorous 酸酸、h3po2 hypophosphorous 淨化

免費 ,准備的酸化水溶液番瓜素次離子、h2po2至今)。 舉例來說,