PowerPoint Presentation: ASSAYS OF ALKALOIDS AND AMINE DRUGS CHAPTER 15 PowerPoint Presentation: Alkaloids are obtained from plant, animal, or synthetic sources. contain organic nitrogen(s) within their chemical structure. usually possess physiological activity. The type of formula R 3 N will be used to represent amines and amine-type alkaloid. PowerPoint Presentation: Alkaloidal Drugs & Preparations employed frequently in modern therapy. atropine morphine quinine reserpine PowerPoint Presentation: Assay of alkaloids is generally performed for purposes of: standardization proof of purity commercial evaluation pharmacological purposes PowerPoint Presentation: Methods for Quantitative Estimation: gravimetric volumetric spectometric electrometric physiological PowerPoint Presentation: The amount of alkaloids that occur in crude drugs are subject to considerable variation in different samples of the same drug. Factors: The age of the plant when it is collected. The season of the year when the drug is harvested. The soil and climate in which the drug is grown. The conditions under which the drug is collected, dried, and stored. PowerPoint Presentation: The quantity of alkaloid present in galenical preparations is also subject to variation. Factors: The quality of drug employed The menstruum used in the extraction of the alkaloid The amount of decomposition of the alkaloid during the process of extraction and of storage. PowerPoint Presentation: The rate of deterioration of alkaloidal preparations are being markedly affected by: nature of the alkaloid pH value of the preparation heat light PowerPoint Presentation: The principles employed in the quantitaive determination of alkaloids by chemical methods are based upon certain characteristics: PHYSICAL PROPERTIES CHEMICAL PROPERTIES PowerPoint Presentation: PHYSICAL PROPERTIES Solubility Alkaloids are sparingly soluble in water but readily soluble in most organic solvents that are immiscible in water. Alkaloidal salts are readily soluble in water and sparingly soluble in immiscible solvents. PowerPoint Presentation: CHEMICAL PROPERTIES Reaction in acids and bases Alkaloids combine directly w/ acids to form salts that are usually soluble in water but insoluble on certain organic solvents. Alkaloids are liberated from aqueous solutions of their salts by alkanes . Alkaloid form highly insoluble precipitates w/ a considerable number of reagents, especially w/ the salts of some heavy metals. PowerPoint Presentation: The assay are conducted largely through the use of immiscible solvents: chloroform ether ether-chloroform amyl alcohol PowerPoint Presentation: Separatory Funnel is used to separate immiscible solvents PowerPoint Presentation: Sources of Errors Errors are encountered in alkaloidal assaying other than those common to all analytical work, such as those cause by:: Errors are encountered in alkaloidal assaying other than those common to all analytical work, such as those cause by: PowerPoint Presentation: Inaccurate weighing Faulty calibration and reading of measuring apparatus Incorrect strength volumetric solutions Dirty apparatus Some of more common errors are: : Some of more common errors are: Failure to secure complete extractions of alkaloids from the drug or from solution Loss of volatile solvents during maceration of the drug before the aliquot portion is taken Imperfect separation of the immiscible liquids PowerPoint Presentation: Failure to wash the stems of separators and funnels, graduates, and other apparatus with solvent to recapture small amounts of alkaloidal residues Decomposition of the alkaloids, especially those which have an ester structure, upon prolonged contact with strong acids and alkalies Use of the wrong indicator PowerPoint Presentation: Variation in the moisture content of the sample is another sources of error, frequently cause different analysts to obtain noncorcordant results in the assay of same drug. PowerPoint Presentation: Choice of Indicators for Alkaloidal Titrations Methyl Red Solution: Methyl Red Solution -is recommended in all the titration in alkaloidal assays by volumetric methods PowerPoint Presentation: Alkaloidal Test Solution Mercuric iodine TS: Mercuric iodine TS Commonly known as Valser’s reagent Prepared by adding slowly a 10% solution of potassium iodine to red mercuric iodide until almostt all the red mercuric iodide is dissolved. Forms white precipitates Iodine TS: Iodine TS Commonly known as Wagner’s reagent A solution containing approximately 1.275 g of iodine and 3.6 g of potassium iodide dissolved in sufficient distilled water. Yeilds reddish or red-brown precipitates Mercuric Potassium Iodide TS: Mercuric Potassium Iodide TS Commonly known as Mayer’s reagent A solution of 1.358 g of mercuric chloride in 60 ml of mixed water with 5 g of potassium iodide in 10 ml of water and the mixture diluted to 100 ml water Yeilds white or slightly yellow precipitates Extraction Theory: Extraction Theory Hansel Jalalon PowerPoint Presentation: Liquid-liquid extraction is a useful method to separate components (compounds) of a mixture PowerPoint Presentation: Let's see an example. Suppose that you have a mixture of sugar in vegetable oil (it tastes sweet!) and you want to separate the sugar from the oil .
You observe that the sugar particles are too tiny to filter and you suspect that the sugar is partially dissolved in the vegetable oil. What will you do? PowerPoint Presentation: How about shaking the mixture with water Will it separate the sugar from the oil? Sugar is much more soluble in water than in vegetable oil , and, as you know, water is immiscible (=not soluble) with oil. Did you see the result?The water phase is the bottom layer andthe oil phase is the top layer , because water is denser than oil. *You have not shaken the mixture yet, so sugar is still in the oil phase. PowerPoint Presentation: By shaking the layers (phases) well, you increase the contact area between the two phases. The sugar will move to the phase in which it is most soluble: the water layer Now the water phase tastes sweet,
because the sugar is moved to the water phase upon shaking.** You extracted sugar from the oil with water .**In this example,water was the extraction solvent ;the original oil-sugar mixture was the solution to be extracted; and sugar was the compound extracted from one phase to another. Separating the two layers accomplishes the separation of the sugar from the vegetable oil PowerPoint Presentation: Did you get it? .....the concept of liquid-liquid extraction? Liquid-liquid extraction is based on the transfer of a solute substance from one liquid phase into another liquid phase according to the solubility.Extraction becomes a very useful tool if you choose a suitable extraction solvent.
You can use extraction to separate a substance selectively from a mixture, or to remove unwanted impurities from a solution.In the practical use, usually one phase is a water or water-based (aqueous) solution and the other an organic solvent which is immiscible with water. The success of this method depends upon the difference in solubility of a compound in various solvents. For a given compound, solubility differences between solvents is quantified as the " distribution coefficient " PowerPoint Presentation: Partition Coefficient Kp (Distribution Coefficient Kd) When a compound is shaken in a separatory funnel with two immiscible solvents, the compound will distribute itself between the two solvents. Normally one solvent is water and the other solvent is a water-immiscible organic solvent . Most organic compounds are more soluble in organic solvents, while some organic compounds are more soluble in water. PowerPoint Presentation: Here is the universal rule: At a certain temperature, the ratio of concentrations of a solute in each solvent is always constant. ﾊAnd this ratio is called the distribution coefficient, K. (when solvent 1 and solvent 2 are immiscible liquids For example, Suppose the compound has a distribution coefficient K = 2 between solvent 1 and solvent 2 By convention the organic solvent is (1) and waater is (2) PowerPoint Presentation: If there are 30 particles of compound , these are distributed between equal volumes of solvent 1 and solvent 2 .. (2) If there are 300 particles of compound , the same distribution ratio is observed in solvents 1 and 2 (3) When you double the volume of solvent 2 ( i.e., 200 mL of solvent 2 and 100 mL of solvent 1 ),
the 300 particles of compound distribute as shown If you use a larger amount of extraction solvent , more solute is extracted PowerPoint Presentation: What happens if you extract twice with 100 mL of solvent 2 ? In this case, the amount of extraction solvent is the same volume as was used in Figure 3, but the total volume is divided into two portions and you extract with each. As seen previously, with 200 mL of solvent 2 you extracted 240 particles of compound . One extraction with 200 mL gave a TOTAL of 240 particles You still have 100 mL of solvent 1 , containing 100 particles. Now you add a second 100 mL volume of fresh solvent 2 . According to the distribution coefficient K=2, you can extract 67 more particles from the remaining solution PowerPoint Presentation: An additional 67 particles are extracted with the second portion of extraction solvent (solvent 2 ).The total number of particles extracted from the first (200 particles) and second (67 particles) volumes of extraction solvent is 267.This is a greater number of particles than the single extraction ( 240 particles) using one 200 mL portion of solvent 2 ! It is more efficient to carry out two extractions with 1/2 volume of extraction solvent than one large volume! PowerPoint Presentation: If you extract twice with 1/2 the volume, the extraction is more efficient than if you extract once with a full volume. Likewise, extraction three times with 1/3 the volume is even more efficient…. four times with 1/4 the volume is more efficient….five times with 1/5 the volume is more efficient… ad infinitum The greater the number of small extractions, the greater the quantity of solute removed. However for maximum efficiency the rule of thumb is to extract three times with 1/3 volume PowerPoint Presentation: Chemically active (acid-base) extraction Can you change the solubility property of a compound? How? Most organic compounds are more soluble in organic solvents than in water ,usually by the distribution coefficient K > 4 However, specific classes of organic compounds can be reversibly altered chemically to become more water-soluble . This is a powerful technique and allows you to separate organic compounds from a mixture -- if they belong to different solubility classes PowerPoint Presentation: What type of organic compounds can be made water-soluble? Compounds belonging to the following solubility classes can be converted to their water-soluble salt form Organic acids include carboxylic acids (strong organic acids) and phenols (weak organic acids) . Organic bases includes amines PowerPoint Presentation: How can organic acids or bases be converted to a water-solubleform? Organic Acids can be converted to their salt form when treated with an aqueous solution of inorganic base ( e.g., NaOH (sodium hydroxide) and NaHCO 3 (sodium bicarbonate)).Salts are ionic , and in general, ions are soluble in water but not soluble in water-immiscible organic solvents .Remember: water is a very polar solvent thus salts ( i.e., ionic species) are well dissolved in it. Carboxylic Acids are converted to the salt form with 5% NaOH aqueous solution. NaOH is a strong inorganic base. Carboxylic acids are strong organic acids (pKa = 3 to 4), so they can also be ionized with weak inorganic bases ( e.g., NaHCO 3 (sodium bicarbonate) ) aqueous solution. PowerPoint Presentation: Let's try a sample problem. Here is a mixture of naphthalene and benzoic acid , dissolved in dichloromethane . You want to separate these two compounds. What will you do? You may use an aqueous solution of either 5% NaOH or sat. NaHCO 3 , to extract benzoic acid as a salt form PowerPoint Presentation: B. Phenols are considered to be weak organic acids. Phenol, the parent compound, is partially water-soluble (1 g will dissolve in 15 mL of water), whereas substituted phenols are not.Sodium bicarbonate (NaHCO 3 ) aqueous solution, a weak inorganic base, will not deprotonate phenols to make it ionic , because it is not strong enough.However, treatment with NaOH, a strong inorganic base, can change phenol to its ionic (salt) form. PowerPoint Presentation: Let's try a another sample problem. Here is a mixture of benzoic acid and p-methoxyphenol , dissolved in dichloromethane . You want to separate these two compounds. What will you do? You cannot use 5% NaOH to separate these two compounds. NaOH will react with both benzoic acid and p -methoxyphenol , thus both compounds will be extracted into the aqueous layer. PowerPoint Presentation: Let's try this problem again. Here is another mixture of benzoic acid and p-methoxyphenol , dissolved in dichloromethane . Strong organic acids such as benzoic acid would be deprotonated and ionized , while weak organic acids such as phenols would NOT be deprotonated NaOH was too strong a base, thus it does not differentiate the strong and weak organic acids. Use of weak inorganic base such as NaHCO 3 will differentiate between the compounds PowerPoint Presentation: 2. Organic Bases (amines) can be converted to their salt form when treated with an aqueous solution of an inorganic acid such as HCl (hydrochloric acid). Recall that salts are ionic and generally soluble in water but not soluble in water-immiscible organic solvents . PowerPoint Presentation: Let's try a third sample problem. Here is a mixture of benzoic acid and p-chloroaniline , dissolved in dichloromethane . You want to separate these two compounds. What will you do? You may use an aqueous solution of either 5% HCl, to extract the amine as a salt form and benzoic acid has remained in the organic layer PowerPoint Presentation: You can separate four different classes of compounds from a mixture based on differing solubility properties. The four classes are: 1.Amines (organic base) 2.Carboxylic acids (strong acid) 3.Phenols (weak acid) 4.Neutral compounds . PowerPoint Presentation: After the separation of the mixture of four components, we will have four solutions: each solution contains one component. The first three compounds are chemically altered, existing in their salt form dissolved in aqueous solution . The fourth compound is not chemically altered, but it is dissolved in an organic solvent .We now want to recover each compound in its original state ( i.e., in the non-ionic form) to complete the experiment. We call this step isolation or recovery . Let's see, one by one, how to recover each compound obtained from the separation process PowerPoint Presentation: Isolation (Recovery) of amines An amine is a basic compound. It is protonated in the presence of excess HCl forming a salt that is soluble in aqueous solution. This is how you separated the amine from the original mixture containing it. An amine is soluble in acidic aqueous solution because it forms a salt, an ionic form. However, if you change the pH of the solution to basic the amine can no longer stay dissolved because it is no longer ionic! This process is called basification . Basification is done by carefully adding concentrated NaOH solution to the solution containing the amine salt until it becomes basic. PowerPoint Presentation: ･ In the basification step, you use concentrated NaOH solution to minimize the volume of the final solution. Recall that a dilute solution of HCl was used to extract the amine as its water-soluble salt (see the picture on the right side). ･ Basification must be done carefully, portion by portion, with swirling each time because the acid-base neutralization reaction is exothermic. ･ Check the pH of the solution to ensure that it is basic. (~pH 10) PowerPoint Presentation: Isolation (Recovery) of Acids There are two different groups of organic acids: carboxylic acids (strong acids) and phenols (weak acids).In the separation procedure, acids were extracted using (weak or strong) basic aqueous solutions Both acids can be returned to the original form in the same manner! Organic acids are currently dissolved in a basic aqueous solution, because the acid forms a salt, an ionic form. When you make the aqueous solution acidic , the organic acids no longer remain dissolved because they are no longer ionic and usually precipitate out of solution. This process is called acidification . Acidification is done by carefully adding concentrated HCl solution until the mixture becomes acidic, When the weak base, NaHCO 3 , was the extracting solution, CO 2 gas will evolve during acidification. PowerPoint Presentation: ･ The recovery of organic acids requires acidification with concentrated HCl solution. Recall that in the extraction step for the separation of an organic acid either dilute NaHCO 3 or NaOH was used. Concentrated HCl will now help minimize the volume of the final aqueous solution ･ Acidification must be done carefully, portion by portion, with swirling each time because the acid-base neutralization reaction is exothermic. ･ Check the pH of the solution to ensure that it is acidic. (~pH 3) PowerPoint Presentation: Separatory Funnel Extraction Procedure Separatory funnels are designed to facilitate the mixing of immiscible liquids PowerPoint Presentation: Separatory Funnel Extraction Procedure 1. Support the separatory funnel in a ring on a ringstand. Make sure stopcock is closed 2. Pour in liquid to be extracted 3. Add extraction solvent 4. Add ground glass Stopper (well greased) PowerPoint Presentation: Separatory Funnel Extraction Procedure Pick up the separatory funnel with the stopper in palce and the stopcock closed, and rock it once gently. Then, point the stem up and slowly open the stopcock to release excess pressure. Close the stopcock. Repeat this procedure until only a small amount of pressure is released when it is vented Shake the separatory funnel . PowerPoint Presentation: Separatory Funnel Extraction Procedure Shake the separatory funnel vigorously . Now, shake the funnel vigorously for a few seconds. Release the pressure, then again shake vigorously. About 30 sec total vigorous shaking is usually sufficient to allow solutes to come to equilibrium between the two solvents. Vent frequently to prevent pressure buildup, which can cause the stopcock and perhaps hazardous chemicals from blowing out. Take special care when washing acidic solutions with bicarbonate or carbonate since this produces a large volume of CO 2 gas PowerPoint Presentation: Separatory Funnel Extraction Procedure Separate the layers. Let the funnel rest undisturbed until the layers are clearly separated While waiting, remove the stopper and place a beaker or flask under the sep funnel. Carefully open the stopcock and allow the lower layer to drain into the flask. Drain just to the point that the upper liquid barely reaches the stopcock PowerPoint Presentation: QUALITY CONTROL II Continuation of the …… ASSAY OF ALKALOIDS AND AMINE DRUGS ( CHAPTER 15) QUILANTANG, ROSELYN S. submitted by: MRS. MARY JEAN TUBA submitted to: PowerPoint Presentation: An assay is an investigative (analytic) procedure in laboratory medicine, pharmacology, environmental biology, and molecular biology for qualitatively assessing or quantitatively measuring the presence or amount or the functional activity of a target entity (the analyte ) which can be a drug or biochemical substance or a cell in an organism or organic sample. The measured entity is generally called the analyte , or the measurand or the target of the assay. The assay usually aims to measure an intensive property of the analyte and express it in the relevant measurement unit ( eg molarity,density , functional activity in enzyme international units, degree of some effect in comparison to a standard, etc). PowerPoint Presentation: Fluid extract is a type of fluid-solid extraction, that usually employ Soxhlet Apparatus to extract certain compound with known solubility in a solvent. Such as extracting lipids from a plant(solid) using ethanol (liquid). In recent years this has expanded to include specialized extraction methodologies and equipment of a proprietary nature to ratio-intact extract multiple groups of compounds with discovered solubility in a solvent. Such as extracting polysaccharides , resins and organic acids from a plant(solid) using glycerol(liquid). PowerPoint Presentation: A tincture is typically an alcoholic extract of plant or animal material or solution of such or of a low volatility substance (such as iodine and mercurochrome). To qualify as an alcoholic tincture, the extract should have an ethanol percentage of at least 40-60% or 80-120 proof. Sometimes even a 90% or 180 proof tincture is achieved. In herbal medicine, alcoholic tinctures are made with various concentrations of ethanol, 25% being the most common. Other concentrations include 45% and 90%. PowerPoint Presentation: Herbal tinctures are not always made using ethanol as the solvent, though this is most commonly the case. Other solvents include vinegar , glycerol , ether and propylene glycol , not all of which can be used for internal consumption. Ethanol has the advantage of being an excellent solvent for both acidic and basic (alkaline) constituents PowerPoint Presentation: QUESTIONS : 1.) What isomeric change does the chief alkaloid of belladona undergo in the assay process? Answer: Hyoscyamine , the chief alkaloid of belladona , and small quantities of other alkaloids are set free from the acids with which they are combined in the crude drug by the ammonia, and dissolve in the organic solvent. The hyoscyamine is largely converted into its racemic isomer, atropine, in the ammonial solution. PowerPoint Presentation: 2.) Would the error in the assay result be considerable if the ether soluble alkaloids of ipecac consisted almost entirely of emetine? Explain.. Answer: The error in the assay result would be considerable if the ether soluble alkaloids of ipecac consisted almost entirely of emetine because ipecac is an ether and that emetine is soluble in ether. So, ipecac is used in medicine as an emetic. PowerPoint Presentation: 3.) The alkaloids of belladona have the structure of esters. What type of decomposition might they undergo if allowed to stand in contact for along time with a strong alkali? Answer: The alkaloids of belladona have the structure of esters. So , HYDROLYSIS is the type of decomposition might they undergo if allowed to stand in contact for along time with a strong alkali.