Lecture Notes on Extraction - Organic Laboratory I | CHEM 2130, Lab Reports of Organic Chemistry

Download Lecture Notes on Extraction - Organic Laboratory I | CHEM 2130 and more Lab Reports Organic Chemistry in PDF only on Docsity! 1 Extraction W. H. Bunnelle, L. A. Meyer, R. E. Glaser (Version 4) Introduction Chances are, everyone in this class has done an extraction, probably several of them. Ever make a cup of tea or a pot of coffee? Anything but instant, and you've done extraction. Simply put, extraction involves mixing of mutually insoluble materials, where a component of one of the phases moves into the other. To make a cup of tea, hot water is mixed with dried tea leaves. The leaves are not soluble in the water, so there are two immiscible phases, the solid and the water. However, the water dissolves some of the compounds originally present in the tea leaves - these materials are extracted into the water. On the other hand, the preparation of a cup of instant coffee is not an extraction at all. Instant coffee powder is completely soluble in the water, and so there is no second phase. This is simply a process of solution. Extractions are often classified according to the nature of the phases involved. The water-tea leaves system is an example of liquid-solid extraction, and one can think of many such examples where components migrate from a solid phase to a liquid phase and vice versa. In this experiment, we will be concerned with liquid-liquid extraction, a very common laboratory operation which can be used as a separation or purification technique. Consider the following reaction: This is a fairly standard organic reaction, involving the conversion of an alcohol to an alkyl halide. For now, imagine that we have the crude reaction mixture, which contains the organic product benzyl bromide (1), and phosphorous acid in dichloromethane solvent. How do we separate these to get the pure product? It turns out that a very simple way to separate the phosphorous acid from the organic material is to add water. The dichloromethane solvent is immiscible with water, and separates as a second layer, the organic phase. Now the molecules of benzyl bromide can stay in the organic phase, or migrate to the water layer, the aqueous phase. What property will determine where these molecules go? The tendency for a compound to reside in a solution is, of course, measured by its solubility. Benzyl bromide has a very much greater solubility in dichloromethane than in water, and so essentially all of this compound remains in the organic layer. For the phosphorous acid, on the other hand, the situation is just the opposite. As a very polar, ionized material, H3PO4, is much more soluble in water than in dichloromethane. Consequently, the phosphorous acid molecules will migrate out of the dichloromethane phase into the aqueous layer, much like those water-soluble constituents of tea leaves. Quite obviously, extraction will only work for separation of materials which have very different solubilities. It is a technique used most often for rudimentary purification, or what we call work-up, of crude reaction mixtures. Now, if the dichloromethane layer can be removed and the aqueous layer left behind, we will achieve, in a very simple way, a separation of the products of the reaction. The device used to carry out this separation is called a separatory funnel (see Figure 1). It does resemble a funnel, if you use your imagination a bit. The glass body has a conical shape like a funnel, but this one closes at the top to a ground-glass joint, which can be shut with a stopper. The stem of the separatory funnel incorporates a stopcock, so that flow out of the bottom of the funnel can be interrupted. A separatory funnel is well-designed for its function. The larger volume at the top of the funnel is helpful for mixing the phases when the funnel is inverted, and the tapering lower portion of the funnel permits clean separation of the phases. CH2OH + PBr3 CH2Br + H3PO3 CH2Cl2 1 2 Figure 1. The separatory funnel in the ring stand (left), and venting the funnel (right). How to Use a Separatory Funnel Operation of a separatory funnel is not difficult, but there are several potential pitfalls which must be avoided. The first order of business is to get the two liquid phases in the funnel. Make sure the stopcock is closed! This may be obvious, but every year, without fail, some students forget or don't pay attention, and wind up with a mess on the benchtop, and precious little product. Just in case, double check. The funnel is suspended in an iron ring -- choose one that allows free passage of the stopcock, and catches the separatory funnel about 1/2 - 1 inch below its widest part. The ring should be clamped high enough on a ring stand so that a collection vessel (usually an Erlenmeyer flask) will fit beneath the lower tip of the funnel when it is in place in the ring. Use a funnel to fill the sep funnel. It is important not to contaminate the ground joint at the top. If you get some grit or insoluble material on the joint, it will not seal tightly when you stopper it, and will leak all over your hands when you invert the funnel for mixing. By the way, you should make sure that your stopper and stopcock don't leak anyway. Before you try a 'real' extraction, take some water into the sep funnel, shake it around, and look carefully for leaks. If it does, check with your TA for help - you may have a misfit stopper or stopcock. Now, with both phases together, stopper the funnel and remove from the iron ring. With one hand cupped over the stopper end of the funnel (and holding the stopper in place between the fingers), and the other hand cradling the stopcock, gently invert the separatory funnel so that the contents run back to the fat end of the funnel, leaving the stopcock end empty. Now, with the sep funnel held at about a 45° incline, and with the stopcock end facing away from you and away from your lab neighbors, open the stopcock. You will hear a pfffft, as excess pressure vents from the sep funnel. Where does this pressure come from? There are many possibilities, depending on the extraction. In some cases, a gas (CO2) is evolved, in others the warmth of your hands is enough to vaporize some volatile solvent, causing a pressure increase. Often there is a small heat of mixing which causes the same thing. This can go both ways, in fact, sometimes a vacuum is created. Regardless, it is very important to vent the separatory funnel repeatedly during use. Since the pressure increase can be substantial, especially when CO2 is generated in the mixing, you should always begin cautiously - one gentle inversion of the sep funnel and vent. Next, close the stopcock, and gently rock the funnel back and forth one or two times. Vent again. Continue this process for several cycles, until you are sure that the phases have mixed thoroughly. The mixing may gradually become more vigorous, from rocking to sloshing to shaking, but never for more than a few seconds at a time, and always with frequent and repeated venting. Of course, you must make certain that the narrow neck of the funnel is completely drained before opening the stopcock, the pressure will squirt any remaining liquid out through the stopcock. 5 we assume that water is one of the solvents, and a generic organic solvent is the other, the distribution coefficient is given by: Korg/water = [A]org/[A]water The distribution coefficient is a measure of the tendency for the solute to reside in one phase versus the other and, to a very good approximation, is equal to the ratio of solubilities for A in the respective solvents. Thus, if A is much more soluble in water than in the organic solvent, the distribution coefficient will be small (<1), while a large distribution coefficient implies that A is much more soluble in the organic than in water. You should note that it is the concentration in each phase which is important, so that the quantity of A which is found in each phase will also depend on the volumes of solvent involved. Those compounds with very large (say, >20) or very small (<0.05) distribution coefficients will be found nearly totally in one of the phases. In contrast, for compounds with intermediate values for the distribution coefficient, the distinction between phases is not so clear-cut. Significant amounts of material will be found in each phase. This raises the question of extraction efficiency. Suppose we begin with an aqueous solution of material we want to extract into ether. What fraction of the material can be extracted for a given volume of ether? Will more compound be extracted if all of the ether is used in one single extraction, or is it better to divide the ether into smaller portions and do multiple extractions of the aqueous phase? In fact, the latter course of action will provide significantly better recovery of the compound, as the following calculations will illustrate. Let's assume, for A, that Kether/water = 4. Now, if we have an aqueous solution of A (10 g) in 100 ml of water, how much A will be recovered by extraction with (a) 100 ml of ether in one extraction, or (b) 2 extractions of 50 ml each. Begin with case (a): Kether/water = [A]org/[A]water g A in ether / ml of ether Kether/water =  = 4 g A in water / ml of water (Notice that the particular concentration units used are unimportant, as long as the same units are used in the numerator as in the denominator.) Now the volume of ether and the volume of water are both 100 ml, and substituting, we have: g A in ether  = 4 g A in water Since: (g A in ether) + (g A in water) = 10 g, we have two equations and two unknowns, and this can be solved: g A in ether = 8 g, or 80% of the compound is extracted into the organic phase with a single 100 ml extraction. For case (b), involving extraction of the same aqueous solution with two-50 ml portions of ether, we carry out the same kind of calculation for each extraction. For the first portion of ether: g A in ether / ml of ether  = 4 g A in water / ml of water 6 For this example, (ml of ether) = 50 ml, and (ml of water) = 100 ml, so: g A in ether  = 2 g A in water Again, the total amount of A is 10 g, and solving the simultaneous equations yields: g A in ether = 6.67 g Now, if this much A was extracted into the ether phase, 3.33 g must remain in the aqueous phase. For another extraction with the 2nd 50 ml portion of ether, we have again: g A in ether  = 2 g A in water but for this second extraction the total amount of A is 3.33 g, and the equations can be solved to give: g A in ether (2nd extract) = 2.22 g and the total extracted by the 2 - 50 ml portions of ether is 6.67 + 2.22 = 8.89 g, or 89 % of the available A. Comparison with the results obtained for case (a) clearly demonstrates that a more efficient extraction is obtained by the multiple extraction. In fact, one can carry out similar calculations for three 33.3 ml extractions, or four 25 ml extractions, and the efficiency would be still greater. Of course, some practical limit is reached beyond which the increased efficiency is not worth the extra effort required to carry out so many extractions. Somewhere between two and four extractions seems to be a reasonable compromise. Extraction With Acids and Bases Acids and bases can be easily separated when you take advantage of their “solubility” in aqueous solutions of various pHs. In fact, acids and bases dissolved in organic solvents, will react during extraction with aqueous basic or acidic solutions respectively, to form the corresponding salt: RCO2H + aq. NaOH → RCO2- Na+ + H2O RNH2 + aq. HCl → RNH3+ Cl- The salt, being an ionic compound, is much more soluble in aqueous solution than it is in the relatively non-polar organic solvent. Some fine-tuning of the technique allows for separation of compounds with closer acidity or basicity by controlling the pH of the aqueous solution. For example, benzenols (phenols) are weaker acids than are carboxylic acids, and are only converted to OH CO2H phenol pKa = 10 benzoic acid pKa = 5 salts with relatively strong bases (NaOH). Carboxylic acids, on the other hand, are converted to the salt even with weaker bases such as NaHCO3. Thus, the benzoic acid can be selectively removed by extraction into an aqueous solution of bicarbonate; after the stronger acid has been removed, the phenol can be extracted by use of NaOH solution. Extractions with NaHCO3 (and Na2CO3) are often used to 7 remove acidic materials, but particular care must be taken to vent the separatory funnel early and often during the extraction. The reaction of acid with these bases generates CO2 gas, and the pressure buildup in a closed separatory funnel can be dangerous. H-A + NaHCO3 → Na+ A- + H2O + CO2 (g) (acid) These points will be illustrated in the laboratory by the use of extraction to separate a mixture of three compounds. The solid mixture consists of varying amounts of p-nitroaniline, biphenyl, and benzoic acid. NH2O2N CO2H p-nitroaniline biphenyl benzoic acid The first of these is an amine derivative, and therefore basic. Biphenyl is a typical organic hydrocarbon, and is neither acidic or basic, while benzoic acid, as already discussed, is an acid. The mixture will be separated according to the scheme outlined below: NH3 +Cl -O2N NH2O2N CO2 - Na+ CO2H p-nitroaniline + benzoic acid + biphenyl in CH2Cl2 extract with 3N HCl benzoic acid + biphenyl neutrilize with NaOH aqueous aqueous organic organic extract with 3N NaOH neutrilize with HCl dry, r emove solvent by distillation The mixture is dissolved in an organic solvent, and extracted first with aqueous HCl. The HCl combines with p-nitroaniline to form the amine hydrochloride salt, which moves to the aqueous phase. Separation of the layers provides an aqueous solution of the amine salt, and an organic solution of benzoic acid and biphenyl. The p-nitroaniline can be recovered from the aqueous solution by neutralizing the acid. This converts the amine salt back to the amine, which is not water-soluble, and precipitates. The amine is then collected by vacuum filtration (see Expt. 1). Meanwhile, the next step of the separation is to extract the organic phase with aqueous NaOH. The benzoic acid is converted to its sodium salt, and moves to the aqueous layer. On separation of the phases, one obtains an aqueous solution of the sodium benzoate, and the organic solution still containing biphenyl. The benzoic acid is recovered by acidifying the aqueous solution, which causes the benzoic acid to precipitate, whereupon it is isolated by filtration. The biphenyl is isolated by drying the aqueous solution to remove dissolved water, and finally distillation of the solvent - the biphenyl will remain as a non-volatile residue. The whole operation is accomplished simply in a short period of time using these principles of extraction. The three components are isolated in nearly pure form; minor contaminants could be removed, if desired, by recrystallization.