Chemical Concepts eaction Efficiency Mass, molecular weight, the mole, balanced equations, stoichiometry, density. Green Concepts eaction efficiency atom economy and other metrics; safer products; paradigm shifts asking new types of questions. Introduction In Green Chemistry, our emphasis is on changing the thought processes about chemistry thinking about the implications of what we do as chemists instead of just the immediate outcomes. We teach students to ask questions like the following. Are we designing and using safe chemicals? What is safe with respect to human health and the environment? Are we making our chemical products wisely, avoiding dangerous chemicals in their manufacture and avoiding the generation of hazardous waste? By the very nature of these thought processes, we address many things that are important to both teachers and their students. Most importantly, we address chemistry from a real life perspective, and, by worrying about these real life issues, we explore chemical experiments that can be carried out without the need for specialized equipment or laboratories. ow do we determine how good a chemical reaction is? This question can form the basis for both valuable experience with the fundamentals of chemical calculations and essential thought processes regarding green/sustainable chemistry. In this lesson, we will look at several ways to assess reaction efficiency the quality of a reaction in terms of its production of a desired product. Through this analysis, new and more exciting ways to introduce old (boring?) chemical concepts (e.g., mass, molecular weight, the mole, balanced equations, stoichiometry, density) may be revealed. In this lesson, we will work through a set of exercises. We will look at a seemingly simple chemical reaction in order to develop our thinking about reaction efficiency, guided by worksheets. This sort of work would be ideal for adaptation to the peer led small group approach to problem solving that is being widely adopted throughout the United States, and elsewhere, and the following description identifies places in the lesson where it would be an excellent idea to stop talking, arrange for students to break into smaller groups, and discuss the questions being raised, then report back to the full group. The following discussion provides references to a worksheet that is provided as the last few pages of this packet. Copyright Kenneth M. Doxsee, University of regon Page 69 of 84
ere is an experimental procedure for a simple inorganic reaction, reported in a standard US laboratory textbook used by college undergraduate students, but not atypical of experiments that might be performed by high school students as well. (Note: I do not recommend that students actually perform this experiment. We will simply analyze it using pencil and paper.) (N 3 ) 2 PtI 2 + Ag 2 S 4 + 2 KCl (N 3 ) 2 PtCl 2 + 2 AgI + K 2 S 4 Experimental Procedure Prepare a solution of 63 mg (0.202 mmol) of silver sulfate in 10 ml of water in a 25-mL beaker containing a magnetic stirring bar. Add 100 mg (0.207 mmol) of the cis-diiodo derivative, in small portions, to this Ag + solution. eat the suspension, with stirring, on a sand bath (70-80 ºC) for 10-12 min. Filter the mixture to separate the precipitate of AgI. Isolation of Product Concentrate the filtrate to a volume of about 2.0 ml. Treat this solution with 330 mg (4.43 mmol, a large excess) of KCl. eat the mixture on a sand bath at 70-80 ºC for 2-3 min. Bright yellow crystals of cis-diammine dichloroplatinum(ii) should precipitate out. The heating is continued for an additional 5-8 min. Cool the mixture to ) ºC in an ice-water bath. Filter the product using a irsch funnel. Wash the crystals with 500 ml of ethanol followed by 1 ml of ether and dry them under suction in air. Determine the percentage yield. Let s say a student has run this reaction and has isolated 50 mg of the desired product. Since we started with 100 mg and ended up with 50 mg, a simple attempt to quantify the efficiency of the reaction might be simply to compare the two amounts, calculating a mass yield (part 1 on the cisplatin worksheet): (50 mg/100 mg)*100% = 50%. For a student just beginning studies of chemistry, this would represent a good and perhaps the only way to attempt to quantify how good the reaction was. As such, it can serve as a good starting point for discussion, guided by questions from the teacher or, even better, questions and comments from the students. Is this a good measure of reaction efficiency? If the reaction went perfectly, with every molecule of the starting compound converted into a molecule of the desired product, would this measure give a perfect score for the reaction (=100%)? Does a molecule of the product weigh the same as a molecule of the starting material? Copyright Kenneth M. Doxsee, University of regon Page 70 of 84
The end result of these discussions, of course, should be the recognition that we need to consider both the amount of product obtained and its molecular weight relative to that of the starting material. This leads to the concept of a theoretical yield the maximum amount of product that could be obtained given the amount of starting material used. As appropriate, this could serve as the point of introduction of basic chemical concepts such as molecular weight and the mole, but realistically it may be better to have students already acquainted with these sorts of concepts, as otherwise this lesson may be slowed too much by the need to discuss all these other issues. 1 molecule diiodide = 482.960 amu 1 molecule dichloride = 300.057 amu so 100 mg of diiodide could theoretically provide 100 mg * (300.057/482.960) = 62.1 mg of the dichloride. Suddenly our actual yield of 50 mg looks a lot better than we thought, since it is (50 mg/62.1 mg)*100% = 80.5% of the theoretically possible amount of product. Part 2 on the cisplatin worksheet: corrected mass yield = 50 mg / [100 mg * (MW product / MW starting material )] = mass yield * (MW starting material / MW product ) = 50% * (482.960/300.057) = 80.5% This, now, again represents an opportunity for student discussion about the merits of the calculation and any issues that it may present. Perhaps the most obvious thing that has been missed is consideration of the stoichiometry of the reaction and the amounts of all the reactants used, not just the amount of the starting material forming the focus of our transformation. In short, we need to consider two other key chemical concepts we need to ensure that we are looking at a balanced chemical equation, and we need to examine the amounts of each reagent used to determine which is serving as the limiting reagent. The full chemical reaction we are considering involves the treatment of the diiodide with silver (I) sulfate and potassium chloride. Students can be presented with the opportunity to write a balanced equation, showing all starting materials and products and the relative amounts of each needed. Small group discussion is always a good possibility, and can help weaker students get up to speed. The end result (part 3 on the worksheet): (N 3 ) 2 PtI 2 + Ag 2 S 4 + 2 KCl (N 3 ) 2 PtCl 2 + K 2 S 4 + 2 AgI Now, to determine the limiting reagent, we need to convert the quantities used (in grams) to the number of moles used. Setting up a simple table (as in the worksheet) is a nice, organized way to do this. Copyright Kenneth M. Doxsee, University of regon Page 71 of 84
Compound molecular wt milligrams millimoles (N 3 ) 2 PtI 2 482.960 100 0.207 Ag 2 S 4 311.794 63 0.202 KCl 74.551 330 4.43 (N 3 ) 2 PtCl 2 300.057 Examining the calculations, we see that we used a large excess of KCl (only two equivalents were required by the reaction stoichiometry), but that we actually should have used a bit more silver sulfate. The silver sulfate thus serves as the limiting reagent, and our theoretical yield of cisplatin was actually only 0.202 millimoles, or 60.6 mg, and our actual yield then is (50 mg/60.6 mg)*100% = 82.5% of the theoretically possible amount of product (worksheet parts 4 and 5). This is typically where we end with such discussions the chemical yield of a reaction, measured by comparing how much of the desired product was obtained to how much could theoretically have been obtained given the reaction stoichiometry and the amounts of reactants used. But does this number really capture the quality of the chemical reaction? Does 82.5% of theoretical yield represent a reasonably efficient reaction, perhaps worthy of a grade of B? Professor Barry Trost, a chemist at Stanford University, felt that reliance on yield as a measure of reaction efficiency represented an inappropriate measure, suggesting that we are better at carrying out chemical transformations than we in fact are. e developed a concept called atom economy, looking at a chemical reaction from the perspective of how many input atoms are incorporated in the desired product, vs. how many are discarded as waste. Let s set up another table, this time considering this issue (worksheet part 6): Compound atoms used in wt used atoms wt discarded pdt discarded (N 3 ) 2 Pt I 2 (N 3 ) 2 Pt 229.151 I 2 253.809 Ag 2 S 4 - - Ag 2 S 4 311.794 2 KCl Cl 2 70.906 K 2 78.197 Totals (N 3 ) 2 PtCl 2 300.057 I 2 K 2 Ag 2 S 4 643.800 Looking at the reaction this way, it doesn t look like such an efficient process any more, does it? Less than one third of the total mass of the reactants ends up as our desired product; the rest is converted to waste. Atom Economy = 300.057/(300.057 + 643.800)*100% = 31.8% ow is this as a measure of the efficiency of our reaction? [Another good point for discussion, questions, etc.] It iss getting better, but it ignores at least two really important things, both arising because this calculation is based only on the balanced chemical equation, without consideration of the actual experiment. Copyright Kenneth M. Doxsee, University of regon Page 72 of 84
First, it does not take into account the actual quantities of reagents we used. For example, we used a large excess of KCl, presumably in an effort to increase the yield of the reaction. In so doing, all the excess KCl becomes waste. Second, it does not take into account the amount of product we actually obtained. Let s inject this experimental reality into the calculation: Experimental atom economy (worksheet part 7) = (theoretical yield/mass of all reactants)*100% = [60.6 mg/(100 mg + 63 mg + 330 mg)]*100% = 12.3%. And if we look at how much we actually obtained rather than the theoretical amount: Actual atom economy (worksheet part 8) = (actual yield/mass of all reactants)*100% = [50 mg/(100 mg + 63 mg + 330 mg)]*100% = 10.1%. What does this number 10.1% - really mean? It means that in carrying out this reaction, we converted 89.9% of all of our starting materials into waste! Put another way (worksheet part 9), this number means that for every gram of desired product (cisplatin) that we produce, we produce (89.9%/10.1%)*1g = 8.9 g of waste. We paid for those starting materials! And we will pay to dispose of the waste! So we pay twice for the inefficiency of our reaction, and get that much less desired compound that could be sold. This is beginning to look like a respectable measure of our reaction efficiency. I find it very interesting that this is what a student would calculate if you neglected to tell them anything about structure, stoichiometry, balancing equations, moles, etc.! owever, we have still neglected to consider one additional important factor the reaction is not carried out as written in the balanced equation we also use a solvent, and when we isolate the product by filtration, we use more solvent. (And if we were to recrystallize our product, still more solvent.) In many cases, solvent usage is the single largest source of waste in chemical production. ur solvent usage, should we actually follow the written procedure exactly as written, is 10 ml of water for the reaction, and 0.5 ml of ethanol and 1 ml of diethyl ether in the isolation step. ow much waste is generated from this solvent use (worksheet part 10)? 10 ml 2 * 1g/mL = 10 g 2 0.5 ml C 3 C 2 * 0.794 g/ml = 0.397 g C 3 C 2 1.0 ml (C 3 C 2 ) 2 * 0.715 g/ml = 0.715 g (C 3 C 2 ) 2 Total solvent waste = 11.112 g Let s now, as the last stage of our analysis, fold this information into our calculation, and simply look at how much desired product is obtained from how much chemical input (worksheet part 11). Copyright Kenneth M. Doxsee, University of regon Page 73 of 84
[50 mg/(100 mg + 63 mg + 330 mg + 11,112 mg)]*100% = (50/11,605)*100% = 0.43% Les than half a percent of all of our chemical inputs ended up as a desired product!! Put another way, known as the environmental factor, a measure of how much waste a reaction produces relative to the amount of desired product (worksheet part 12): E factor = waste/product = (11,605 mg 50 mg)/50 mg = 231 We ve come a long way from our initial observation of an 82.5% yield, haven t we? A final, related measure that may be catching on, courtesy of Green Chemistry Institute s Pharmaceutical Sciences oundtable: Process Mass Intensity Metric (worksheet part 13) = quantity of raw materials input/quantity of bulk active pharmaceutical ingredient (API) out, where process = all steps from commonly available materials, raw materials = everything used, including water, and bulk API out is the final salt form, dried to specifications. In our simple case, this is = (11,605)/50 = 232, and in general one might imagine it will be very similar to the E factor. Before we move on, let s spend just a moment more thinking about this experiment and what it teaches. n the positive side, it gives experience with balancing equations, calculating mole ratios, and classical inorganic coordination chemistry, as well as introducing the concept of chemotherapy. It provides laboratory experience with running reactions at elevated temperatures and isolation by filtration. n the negative side, it presents a reaction that requires elevated temperatures (energy consumption), and that generates large amounts of waste relative to the amount of desired product (232 grams of waste per gram of product!) Finally, the product is an active anticancer drug! It has dramatic biochemical activity at low dosage. Why are we having beginning students risk potential exposure to a compound with proven biological activity??? This discussion of the experiment represents some of the key concepts in green chemistry in short, we learn to ask questions at every stage of an experiment, synthesis, or process. Why are we making this compound? What are its properties particularly in terms of health and environmental impact? Can we accomplish the same goals by making a different and safer compound? Can we make the compound from safer starting materials, or using safer solvents, or using less energy input (either heating or cooling)? Can we reduce the amount of waste that is generated, or eliminate it entirely? Given time to think about the 12 principles and our discussions today, you would probably conclude that we have illustrated principles 1, 2, 3, 7, 8, and 9. Copyright Kenneth M. Doxsee, University of regon Page 74 of 84
Cisplatin worksheet (N 3 ) 2 PtI 2 + Ag 2 S 4 + KCl (N 3 ) 2 PtCl 2 + K 2 S 4 + AgI compound molecular weight mass (milligrams) millimoles (N 3 ) 2 PtI 2 482.960 100 0.207 Ag 2 S 4 311.794 63.0 0.202 KCl 74.551 330 4.43 (N 3 ) 2 PtCl 2 300.057 50.0 * Experimental result 1. Mass yield = (mass of product/mass of starting material) * 100% = ( mg/ mg) * 100 % = % 2. Corrected mass yield = mass yield * (MW of starting material/mw of product) = % * ( g-mol -1 / g-mol -1 ) = % 3. Balanced equation (N 3 ) 2 PtI 2 + Ag 2 S 4 + KCl (N 3 ) 2 PtCl 2 + K 2 S 4 + AgI 4. Theoretical yield = (millimoles limiting reagent) * MW of product = millimoles * mg-mmol -1 = mg 5. Actual yield = (mass of product/theoretical yield) * 100 % = ( mg/ mg) * 100 % = % 6. Atom economy compound atoms used in pdt formula wt used atoms discarded formula wt discarded (N 3 ) 2 Pt I 2 (N 3 ) 2 Pt 229.151 I 2 253.809 Ag 2 S 4 - - Ag 2 S 4 311.794 2 KCl Cl 2 70.906 K 2 78.197 Totals (N 3 ) 2 PtCl 2 300.057 I 2 K 2 Ag 2 S 4 643.800 Copyright Kenneth M. Doxsee, University of regon Page 75 of 84
Atom economy = [formula weight of product/(formula weight of product + formula weight discarded)] * 100 % = ( / + ) * 100 % = % 7. Experimental atom economy = (theoretical yield/mass of all reactants) * 100 % = [ mg/( mg + mg + mg)] * 100 % = % 8. Actual atom economy = (actual yield/theoretical yield) * experimental atom economy = (actual yield/mass of all reactants) * 100 % = [ mg/( mg + mg + mg)] * 100 % = % 9. Waste generated for every gram of desired product = [(100 % - actual atom economy)/actual atom economy] * 1 g = [(100 % - %)/ %] * 1 g = g 10. Solvent usage in this reaction: Water: 10 ml * density = 10 ml * 1.00 g-ml -1 = g = mg Ethanol: 0.5 ml * density = 0.5 ml * 0.794 g-ml -1 = g = mg Ether: 1.0 ml * density = 1.0 ml * 0.715 g-ml -1 = g = mg Total solvent usage = mg + mg + mg = mg 11. Yield of product based on all chemical inputs = [mass of product/(mass of all reagents + mass of all solvents)] * 100 % = [ mg/( mg reagents + mg solvents)] * 100 % = % Copyright Kenneth M. Doxsee, University of regon Page 76 of 84
12. E-factor (Environmental factor) = mass of waste/mass of product = (mass of all reagents and solvents mass of product)/mass of product = ( mg - mg)/ mg = 13. Process mass intensity = mass of raw materials input/mass of product output = mg/ mg = Copyright Kenneth M. Doxsee, University of regon Page 77 of 84
B. Aniline oxidation An example of an organic reaction note that organic chemists are notorious for focusing only on the reagents and products of interest, ignoring other inputs and byproducts and not always even balancing equations. xidation of aniline is a classical method for the production of hydroquinone: 2 S 4 Fe N 2 + Mn 2 The balanced equations: 2 PhN 2 + 4 Mn 2 + 5 2 S 4 2 quinone + (N 4 ) 2 S 4 + 4 MnS4 + 4 2 And the overall equation: 2 quinone + 2 Fe + 2 2 2 hydroquinone + 2 Fe C 6 5 N 2 + 2 Mn 2 + 2.5 2 S 4 + Fe C 6 4 () 2 + Fe + 0.5 (N 4 ) 2 S 4 + 2 MnS4 + 2 I don t have full experimental details for this reaction, so let s just use it for practice in the calculation of atom economy. Calculation of the atom economy for this one may seem more complicated, but as before, all we need to do is balance the equation, then look at how many input atoms are used and how many are discarded. (efer to the worksheet on the following pages.) Imagine how this one would look upon full analysis, given that the best it could be is 19.4%, and certainly there are large amounts of solvents involved in the process. Experimental details are required to calculate the other measures of reaction efficiency Experimental Atom Economy, Actual atom economy, Waste generated for every gram of desired product, Solvent and other chemical usage, Yield of product based on all chemical inputs, E-factor, and Process Mass Intensity. You may wish to have students try to find such experimental information through the Internet. I have provided on the following page two abbreviated reports of the reaction that include most (but not all) of the information needed. aving students discuss and identify what essential information is not provided, and having them make educated guesses, would be valuable exercises. (The first procedure does not report how much steam is needed, and the second procedure omits the type and amount of base used to adjust the p of the reaction mixture to 5-6. Neither procedure says how much iron was needed, and the second procedure does not provide the final yield of hydroquinone.) Copyright Kenneth M. Doxsee, University of regon Page 78 of 84
Abstract (Quinones. Gibbs, C. F. (The B. F. Goodrich Co.). (1944), US Patent 2,343,768; Chemical Abstracts 38:22768). A mixture of Mn 2 (11.6 g) and 50% 2 S 4 (36 g) is placed in a suitable vessel and maintained at 50-60 over a water bath. A mixture of Mn 2 (58.8 g), 2 S 4 (50 g) and 2 (40 g) is added over a period of 1 hour concurrently with a mixture of aniline (24 g), 50% 2 S 4 (140 g) and 2 (165 g). The pressure is maintained at 40 mm and a constant stream of water vapor is passed into the bottom of the vessel. The quinone collects in a water-cooled receiver immediately after the start of the addition of aniline sulfate; the yield is 73% or higher. This process makes it possible to carry out the oxidation at a higher temperature without the formation of large amounts of by-products. It is also speedier. Abstract (Quinone production from aniline and its reduction to hydroquinone. (Instytut Przemyslu rganicznego). Brit. Patent 1,151,673 (1969), 3 pp.; Chemical Abstracts 71:61012). In the preparation of quinone (I) by the oxidation of Ph-N 2 (II), unreacted Mn 2 is reduced to water-soluble manganous salts as an alternative to steam distillation to remove the I formed. Thus, 400 L water, 115 kg concentrated 2 S 4 and 15 kg Mn 2 (85% pure) is stirred, cooled to 5 and 30 kg II added over 3 hours at 5-8. Simultaneously, an additional 60 kg Mn 2 is added in 15 kg-portions at 30-minute intervals, then the mixture stirred 7 hours at 5-10 ; 3.13 kg 2 2 (calculated as 100% 2 2 ) is added over 30 minutes and the mixture neutralized to p 5-6, then reduced with Fe filings at 65-75 until one half the I present was reduced to hydroquinone (III). Cooling to 10 and filtration yields a precipitate containing 30 kg quinhydrone which is then further reduced to III. NaS 3 may be used in place of 2 2. Copyright Kenneth M. Doxsee, University of regon Page 79 of 84
Synthesis of ydroquinone Worksheet 2 S 4 Fe N 2 + Mn 2 Balanced equations C 6 5 N 2 + 2 Mn 2 + 2.5 2 S 4 C 6 4 2 + 0.5 (N 4 ) 2 S 4 + 2 MnS4 + 2 2 C 6 4 2 + Fe + 2 C 6 4 () 2 + Fe verall equation C 6 5 N 2 + 2 Mn 2 + 2.5 2 S 4 + Fe C 6 4 () 2 + Fe + 0.5 (N 4 ) 2 S 4 + 2 MnS4 + 2 compound atoms used in pdt wt used atoms discarded wt discarded C 6 5 N 2 C 6 4 76.098 N 3 17.030 2 Mn 2 2 31.999 Mn 2 2 141.875 2.5 2 S 4 2 2.016 1.5 2 + 2.5 S 4 243.169 Fe - - Fe 55.847 Totals C 6 6 2 110.112 N 3 Mn 2 2 Fe + 2.5 S 4 + 1.5 2 457.921 Atom economy = (formula weight of product/formula weight of all reactants) * 100 % = ( / ) * 100 % = % Experimental details are required to calculate other measures of reaction efficiency, as in the cis-platin section. Consider having students try to find such information through the Internet. n the previous page are short reports of the reaction that include most (but not all) of the information needed. Ask your students to discuss and identify what essential information is not provided, and having them make educated guesses. (The first procedure does not report how much steam is needed, and the second procedure omits the type and amount of base used to adjust the p of the reaction mixture. Neither procedure says how much iron was needed, and the second procedure does not provide the final yield of hydroquinone.) Copyright Kenneth M. Doxsee, University of regon Page 80 of 84
Chemical Concepts Solventless Aldol Condensation Carbonyl chemistry; the aldol reaction; phase behavior; melting points of solids and mixtures; recrystallization. Green Concepts Solventless reactions; atom economy. (Consider Green Principles 1, 3, 4, 5, 7, 8, and 9.) Introduction The aldol condensation represents a powerful general method for the construction of carboncarbon bonds, one of the central themes of synthetic organic chemistry. In the base-catalyzed aldol condensation reaction, deprotonation alpha (adjacent) to a carbonyl group affords a resonance-stabilized anion called an enolate, which then carries out nucleophilic attack at the carbonyl group of another molecule of the reactant. (Analogous acid-catalyzed reactions are also well-known.) The product, a beta-hydroxy carbonyl compound, often undergoes facile elimination of water (dehydration), affording an alpha, beta-unsaturated carbonyl compound as the final product. - + - enolate - - + + - 2 Mechanism of the base-catalyzed aldol condensation Aldol condensation reactions between two different carbonyl compounds can lead to complex product mixtures, due to the possibility of enolate formation from either reactant and to the possibility of competing homo coupling rather than the desired cross coupling. + ' ' + + ' ' ' crossed aldol products homo aldo products Crossed aldol condensation can afford complex mixtures If, however, only one of the carbonyl compounds has alpha hydrogens available for deprotonation and enolate formation, the crossed aldol reaction can provide synthetically Copyright Kenneth M. Doxsee, University of regon Page 81 of 84
useful yields of products. Thus, for example, benzaldehyde cannot be converted to an enolate, yet reacts readily with enolates of other carbonyl compounds, including acetone. + 3 C C 3 C 3 C 3 A successful crossed aldol condensation omo coupling of acetone (or other ketones) is generally not a problem in such reactions, as the aldol condensation of ketones is generally not a very efficient reaction. (More specifically, each step of the aldol condensation is reversible under the reaction conditions at least until the dehydration step and thus equilibrium is established. With aldehydes, equilibrium favors the aldol product, but with ketones, primarily for steric reasons, very little aldol condensation product is present at equilibrium.) In this experiment, you will explore the aldol condensation reaction of 3,4- dimethoxybenzaldehyde and 1-indanone. 3 C 3 C 3,4-dimethoxybenzaldehyde + 1-indanone Na C 3 C 3 In contrast to typical experimental procedures for aldol condensation reactions, this reaction will be carried out without solvent. ngoing research is revealing a number of reactions that proceed nicely in the absence of solvent, representing the best possible solution to choice of a benign solvent. Although these reactions are frequently referred to as solid-state reactions, it has been noted [1] that in many cases, mixture of the solid reactants results in melting, so that the reactions actually occur in the liquid, albeit solvent-free state. This melting phenomenon is interesting and actually represents one of the key points of this experiment. You have learned that impurities lead to lower melting points. ere, you will experience this in a vivid way as you mix the two solid reactants, they will melt. In addition to providing a memorable demonstration of the impact of impurities on melting points and illustrating the possibility of carrying out organic reactions in the absence of solvents, this experiment highlights another key green concept the design of efficient, atom-economical reactions. The aldol condensation, if effected without dehydration, has an atom economy of 100% and requires only a catalytic amount of acid or base, and even with dehydration, the atom economy remains quite high. 1. G. othenberg, A. P. Downie, C. L. aston, and J. L. Scott, Understanding Solid/Solid rganic eactions," J. Am. Chem. Soc. 2001, 123, 8701-8708. Copyright Kenneth M. Doxsee, University of regon Page 82 of 84
Laboratory eaction SAFETY PECAUTINS: Use care to avoid contact with solid sodium hydroxide or the reaction mixture. 1. Place 0.25 g of 3,4-dimethoxybenzaldehyde and 0.20 g of 1-indanone in a test tube. Using a metal spatula, scrape and crush the two solids together until they become a brown oil. Use care to avoid breaking the test tube. 2. Add 0.05 g of finely ground (using a mortar and pestle) solid Na to the reaction mixture and continue scraping until the mixture becomes solid. Workup and purification 4. Allow the mixture to stand for 15 minutes, then add about 2 ml of 10% aqueous Cl solution. Scrape well in order to dislodge the product from the walls of the test tube. Check the p of the solution to make sure it is acidic. 5. Isolate the solid product by vacuum filtration, continuing to pull air through the solid to facilitate drying. Determine the mass of the crude product. 6. ecrystallize the product from 90% ethanol/10% water, using the hot solvent first to rinse any remaining product from the test tube. You should not require more than 20 ml of solvent to effect this recrystallization. Characterization 7. Determine the mass and melting point of the recrystallized product. (A typical melting point range is 178 181 C.) Questions 1. Describe the physical properties (color and state) of your crude product. eport the mass and percent of theoretical yield of the crude product. 2. eport the color and melting point range of your recrystallized product. eport the mass and percent of theoretical yield of the recrystallized product. 3. Calculate the atom economy for the reaction. 4. Perform an economic analysis for the preparation of your product. esearch Questions 1. What is a eutectic mixture? A eutectic liquid? Are there practical applications of eutectic mixtures? If so, give an example, including an explanation of why the properties of a eutectic mixture are helpful to the particular application. Copyright Kenneth M. Doxsee, University of regon Page 83 of 84
2. The introduction to this experiment comments on the difference between a true solidstate reaction and a reaction involving two solids that in fact proceeds through formation of an intermediate liquid state. Find other reported examples of solid-state reactions, solventless reactions, and related concepts and attempt to analyze them with regards to whether they have been described correctly. (For example, a solidstate reaction between two low-melting solids might well be expected to proceed through formation of a liquid, as did our experiment.) eferences This experiment was adapted from the primary literature and has been presented in Green rganic Chemistry: Strategies, Tools, and Laboratory Experiments, Kenneth M. Doxsee and James E. utchison, Brooks-Cole, 2003. Any number of solventless aldol reactions are possible, and it may be attractive to allow students some latitude in choosing their reactants, taking care to avoid unexpectedly hazardous reagents or products. The reactants reported here were chosen deliberately to highlight the melting point depression phenomenon; other pairs of reagents may or may not visibly melt upon mixing. Copyright Kenneth M. Doxsee, University of regon Page 84 of 84