Separation by Solvent Extraction

Transcription

1 Experiment 3 Separation by Solvent Extraction Objectives To separate a mixture consisting of a carboxylic acid and a neutral compound by using solvent extraction techniques. Introduction Frequently, organic chemists must separate an organic compound from a mixture of compounds, often derived from natural sources or products of synthetic reactions. One technique used to separate the mixture compounds is called extraction. Extraction is a process that selectively dissolves one or more of the mixture components into an appropriate solvent. The solution of these dissolved compounds is often referred to as the extract. Extraction processes include removal of soluble compounds from a solid matrix, such as in the brewing of coffee or tea, or in decaffeinating coffee with liquid carbon dioxide. In the organic chemistry laboratory, however, extraction almost always refers to the transfer of compounds from one liquid solvent to another. A compound can be separated from impurities in a solution by extracting the compound from the original (or first) solvent into a second solvent. For the process to be selective, the compound must be more soluble in the second solvent than in the first solvent, and the impurities must be insoluble in the second solvent. Additionally, the two selected solvents must be immiscible, or not soluble in one another, so that they produce two separate solvent layers. After dissolving the mixture in the first solvent, the solution is added to a second solvent. The two layers are vigorously mixed to maximize the surface area between them. This mixing facilitates the transfer of a dissolved compound from one solvent layer to another. Once the transfer process is complete, the layers are again allowed to form, as shown in Figure 1. Separation of the two layers then completes the separation of the desired compound from the impurities. 16

2 Washing is the reverse process, in which the impurities are removed to the second solvent, leaving the desired compound in the original solvent, as shown in Figure 2. Extractions using large quantities of solvents, tens or hundreds of milliliters, require a separatory funnel, as shown in Figure 3. The solvent layers are mixed by shaking the separatory funnel. Then the layers are allowed to reform. The bottom layer is drained through the stopcock; the top layer is poured from the top of the separatory funnel. Figure 3. Separatory funnel for extractions 17

3 The first requirement in the extraction process is to select two immiscible solvents. One solvent, usually aqueous (water based), should be polar in nature. The second solvent should be nonpolar and might be hydrocarbon, ether, or chlorinated solvent, such as dichloromethane. When the two immiscible solvents are placed into a container, two liquid layers are formed. The more dense solvent is always the bottom layer. It is important to identify the solvent in each layer. Hydrocarbons and ethers are less dense than water or the dilute aqueous solutions used in extractions. When one of these nonpolar solvents is used, the water layer is the bottom layer, as shown in Figure 4. However, dichloromethane is denser than water. When dichloromethane is used as the nonpolar solvent, the water layer will be the top layer, as shown in Figure 5. Although the identity of each layer can be established from the density of each solvent, their identities should be confirmed. To confirm the identities of the layers, one or two drops of water are added just below the surface of the top layer. If the drops of water fall through the top layer to the layer below, then the water layer is the bottom one. It is a good practice to save all layers in labeled containers until the experiment is complete and the desired product is isolated. Often the two solvents will not completely separate after shaking, due to the formation of an emulsion at the interface between them. An emulsion is a suspension of small droplets of one liquid in another liquid. Emulsions are generally opaque or cloudy in appearance and are often mistaken as a third layer. The small size of the droplets in an emulsion causes the separation of the two solvents to take place very slowly. Several procedures may be helpful to facilitate this separation. For example, gentle swirling of the container, addition of a few drops of saturated aqueous sodium chloride or ethanol, or addition of more solvent to dilute the solutions may help. In particularly 18

4 difficult cases, it may be necessary to filter the mixture to remove small solid particles that promote emulsion formation. A simple, but useful, guide to solubility is like dissolves like. That is, nonpolar compounds, including organic compounds, are more soluble in nonpolar solvents than in polar solvent. On the other hand, ionic and polar compounds are more soluble in polar solvents, such as water. These solubility differences can be exploited to separate nonpolar compounds from ionic or polar compounds. For example, synthetic reactions often produce ionic, inorganic salts as by products of the desired nonpolar organic product. In such cases, these salts are removed by washing the nonpolar solvent with water. The organic compound remains dissolved in the nonpolar solvent. Some organic compounds are sufficiently polar to be quite soluble in water. Extraction of such polar compounds into a nonpolar solvent is often difficult. The process can be facilitated by using the technique called salting out. Inorganic salts, such as NaCl, are dissolved in water to reduce the solubility of the organic compound in the aqueous layer. Under these conditions, the organic compound preferentially dissolves in the nonpolar layer. Extraction is a particularly effective means of separating organic compounds if one compound in the mixture can be chemically converted to an ionic form. The ionic form is soluble in an aqueous layer and can be extracted into it. Other non ionized organic compounds in the mixture will remain dissolved in the nonpolar solvent layer. Separation of the two layers results in the separation of the dissolved compounds. Ionic forms of some organic compounds can be produced by reacting them with aqueous acids or bases (Figure 6). Treatment of organic acids with bases such as sodium hydroxide (NaOH) converts these acids to water soluble anions. Reacting basic amines with dilute aqueous acid solutions such as hydrochloric acid (HCl) converts the amines to watersoluble cations. Figure 6. Conversion of organic compounds to ionic forms by reaction with base or acid 19

5 The extent to which an acid base reaction proceeds to completion depends upon the relative acidity and basicity of the reactants and products. Reactions occur so that stronger acids and bases react to produce weaker conjugate bases and acids. Recall that the pka is a measure of the acidity of an acid, as shown in Equation 1. Stronger acids have smaller pka s and their conjugate bases are inherently weaker. The position of an acid base equilibrium can then be predicted from a knowledge of the pka s of the acids involved. Stronger acids, those with a smaller pka, will react with the conjugate bases of weaker acids, those with a larger pka. An analysis of pka s indicates that aqueous NaOH can be used to extract both p toluic acid and p tert butylphenol from a nonpolar solvent, as shown in Equation 2 and 3. The stronger base, OH, removes a hydrogen ion, H +, from p toluic acid to form the salt, p toluate. The polar salt is soluble in aqueous solution. Both OH and p toluate are bases. The pka of 16 indicates that OH is a stronger base than p toluate, with a pka of 4.2. The stronger base takes H + from the weaker base. Similarly, OH is a stronger base than p tert butylphenoxide ion, with a pka of Therefore, OH takes H + from p tert butylphenol to form the water soluble p tertbutylphenoxide ion. Sodium hydrogen carbonate (NaHCO 3 ), with a pka of 6.4, is a weaker base than p tertbutylphenoxide ion, so HCO 3 will not take H + from p tert butylphenol, as shown in Equation 4. As a results p tert butylphenol is not converted to a salt in aqueous sodium hydrogen carbonate and does not become water soluble. 20

6 Although aqueous NaHCO 3 is not sufficiently basic to react with p tert butylphenol, it will react with p toluic acid to form the water soluble p toluate, as shown in Equation 5. The p toluic acid and p tert butylphenol can be recovered by adding HCl to the aqueous solutions. The p toluate and p tert butylphenoxide ions are stronger bases than is Cl, so each one takes H + from HCl. The acid forms are not water soluble and, therefore, precipitate from solution. The procedure you will use in this experiment exploits the differences in these reactions to separate salicylic acid and acetanilide from the nonpolar solvent in which they are dissolved. First, you will extract only salicylic acid into NaHCO 3 solution. Since acetanilide is neutral and so does not react with NaHCO 3, it remains dissolved in the nonpolar solvent. Then, you will add HCl to the aqueous layer to precipitate the water insoluble salicylic acid. You will isolate the precipitates from the solutions by vacuum filtration, then air dry them. To recover acetanilide, you will dry the nonpolar layer with anhydrous sodium sulfate (Na 2 SO 4 ) and evaporate the solvent in a fume hood. Finally, you will recrystallize the acetanilide. To facilitate the understanding of the experimental concept, each student should prepare a flowchart by filling in the blanks with the names or structures of the compounds prior to your lab session (next page). 21

7 Laboratory flowchart Experimental Procedure 1) Perform a leak test on your separatory funnel by adding some water, put on the cap, and close the stopcock. Shake and place the funnel on a support ring. If there is a leak, consult your instructor. 2) Obtain a 1:1 mixture of salicylic acid and acetanilide from your instructor. 3) Weigh the bag containing this mixture and transfer the mixture into the separatory funnel. Re weigh the empty bag and calculate the weight of the mixture. 4) Place 10 ml of ethyl acetate in the separatory funnel. Swirl the funnel until all of the solid compounds dissolve. 5) Add 5 ml of 10% NaHCO 3 solution. Put on the cap and shake the funnel carefully for a few seconds as demonstrated by your instructor. (*Remember to hold the separatory funnel with both hands and vent it frequently with the stopcock pointed upward and away from other people.) Settle for a few minutes and drain the aqueous layer into a beaker. 22

8 6) Extract the ethyl acetate layer with another portion of 5 ml of 10% NaHCO 3 solution. Combine the aqueous layers into the same beaker. Leave the ethyl acetate layer in the separatory funnel. 7) Chill the combined aqueous layer in an ice water bath. Acidify it by slow addition of concentrated HCl solution. Notice that foaming and precipitation occur. Continue to add HCl until no more solid is produced. 8) Filter the precipitate, transfer onto a pre weighed watch glass, and dry it on a steam bath. (Which compound should be obtained at this stage?) 9) Add 10 ml of saturated NaCl solution into the ethyl acetate layer (in the separatory funnel) to wash out polar impurities. Shake the funnel and settle for a few minutes. Drain and discard the aqueous layer. Transfer the ethyl acetate layer into an Erlenmeyer flask. 10) Dry the ethyl acetate layer with anhydrous sodium sulfate. Decant the solution into a preweighed ceramic evaporating dish. 11) Evaporate the ethyl acetate on a steam bath. (Which compound should be obtained at this stage?) 12) Calculate the recovery percentages of each compound. 13) Determine the melting points of the separated compounds. Laboratory Safety Precaution 1) Wear safety goggles and lab coat at all times while working in the laboratory. 2) Acetanilide is toxic and irritating. Concentrated hydrochloric acid is toxic and corrosive. Magnesium sulfate is irritating and hygroscopic. 3) Wash your hands thoroughly with soap or detergent before leaving the laboratory. 23