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Melting Point and Re-crystallization:Is My Stash Is Pure? If Not, Can I Make It Pure?IntroductionIn the relevant scene, the chemist loads a small sample of the produit into a capillary tube which he attached to a thermometer. The assemblage was then placed in a Thiele tube (see Figure 1A on the right) containing clear, colorless mineral oil which was slowly heated with an alcohol burner as the chemist raptly observed the behavior of the heroin in the small tube. After a time, he looked up and exultantly declared the product the purest he’d ever seen. Magnifique ! So how did he know this? Devereaux’s chemistIn the 1971, Academy Award winning movie, The French Connection, the technique you are about to explore shows up prominently. In order to check the purity of his 32 million dollar heroin shipment, Devereaux, the French drug king pin gets a chemist.Figure 1A: Thiele Tube Melting Point Apparatus ApparatusTwo ways. First, the sample probably began to melt just below the expected melting point of pure heroin (173 oC). (See Figure 1B) The chemist would have observed this when the originally crystalline white powder begin to look glassy and not entirely solid, perhaps near 172o. Second, he would have seen the sample completely liquefy by 173oCThe fact that the range of melting is narrow, e.g. < 1o, and, that this range is very close to the accepted melting point of pure heroin means the sample is, as they say in France, good shit. 172 oC 173oCFigure 1B: What the Chemist Saw Of course, street heroin is rarely sold pure. Dealers commonly cut the pure product with all kinds of things including acetaminophen (Tylenol),?tranquilizers, baking soda, powdered milk, starch, talcum powder and even rat poison and laundry detergent. So what does diluting the pure product do to the melting point behavior? First, the cut sample would melt well below the expected 173 oC for pure heroin. Second, and even more telling, the range of the melting would be wide, often 4-8 oC. So, if you were to test the toot your local dealer sold you via melting point analysis, you might see something like a 164-170o C melting range. Bad shit.Why Melting Point Ranges Move Around: A Qualitative ModelThe reason for this change in behavior is that the impurities tend to interrupt and disconnect the crystalline bonds holding the heroin together. This is illustrated in Figure 2 below. (See also, page 18 of your lab manual.)Figure 2: Sketch of Pure vs. Impure Solid Structure Pure heroin Heroin cut with rat poison The pure heroin’s structure on the left is tightly packed and in a regular array. This means that the same energy is needed to loosen the crystalline bonds that hold the solid together. A narrowly defined melting point is thus expected since the melting temperature reflects the amount of energy necessary to cause the loosening of the crystalline bonds. In contrast, the cut heroin containing, let say rat poison impurities, (the smaller and randomly distributed blue balls in the structure on the right) produces a splotchy appearance in the once regular array of heroin molecules. Like termites inside a piece of wood, the impurities disrupt and disorganize the bonding between the heroin molecules, thereby weakening the structure. Also, since the impurities will be distributed irregularly throughout, some parts of the structure are seriously weakened, while other parts are not so weakened. Put another way, if the impurities are like termites, sometimes the termites do a lot of damage to one section ofwood “impurity” the wood, while they leave other parts less damaged. The net effect of this random weakening of the structure is that some parts fall apart sooner than other parts, which –in melting point terms-means that a wide range of (lower) temperatures can make the cut heroin melt. Exterminating Termites, Removing Rat Poison: Re-crystallizationNow if you are taking heroin, and you know the dealer cuts it with rat poison, you probably want to clean up your stash before shooting up, particularly if you’re a rat. A classic method to do so is re-crystallization.As discussed in your lab manual, this means finding a solvent that can dissolve both your desired compound and any impurity (or impurities) present when the solvent is heated to some elevated temperature. Once re-cooled, say to room temperature or ice bath temperatures, though, the right solvent will only spit back the desired compound as a solid and exclude (hint: see questions on pg. 19 of Lab manual) the impurities, leaving them still dissolved in the cooled solution. The process is captured in Figure 3. Figure 3: Re-crystallization in Picture FormThe trick to making it work as shown above is to find the right solvent. The whole point of the lab exercise on page 20 of your lab manual is to sniff out that solvent.So what do you look for in a good re-crystallization solvent? As already noted, your target compound should be relatively easy to dissolve at hot water bath temperatures but not very soluble at lower temperatures, like room or ice bath temperatures. If you examine the table for the four solvents chosen for the exercise on page 20, you see the choices (soluble, partially soluble, insoluble) for solvent and for each temperature (hot or cold.) The outcome devoutly to be wished is that in the room temperature column, acetanilide drops out of solution as a white solid, but at hot water bath temperatures it dissolves entirely. Mimicking the table on pg. 20 of the manual, you ideally want the behavior of solvent Y:Solubility of Acetanilide in Solvents X, Y and ZSolventResult at Room Temp.Result upon heating(if applicable)XSolubleSolubleYInsolubleSolubleZPartially SolublePartially SolubleOne other caveat, not easily tested, is that the impurities in your chosen solvent stay in solution, excluded from the purified solid at both elevated and room temperature. This condition is usually assumed. That’s ok. Impurities are normally in low concentrations compared to the target material, (too much rat poison kills the paying customers after all), and nearly everything dissolves at least a little bit in any solvent, regardless of type.still more stuff down this a waySome Practical Notes on Re-CrystallizationEven when you’ve found the right solvent, you’re not out of the woods. Often, rookies at re-crystallization find that their sample passes through the filter paper and into the filter flask. It can be a bummer for beginners. Some of this is technique. Make sure to thoroughly wet the paper with the solvent and apply vacuum to suck it down onto the filter flask surface. Also (duh), make sure the filter paper circle completely covers the holes of the flask. Quickly pour your slurry of purified crystals down the middle of the filter paper. Don’t let the slurry run under the paper. And avoid over washing which can re-dissolve major amounts of your product.It’s also worth noting that at a practical level going from B C in Figure 3 should be done slowly. This is because the process of crystal formation has two basic stages: nucleation followed by growth. (Hint: check questions on page 19 of your lab manual).Nucleation means an initial formation of a site where a crystal can start to grow. This can be a itty bitty micro seed crystal of the compound itself (=> homogeneous nucleation), or, some foreign surface (like a scratch in the beaker=> heterogeneous nucleation) that spells home to the compound. But cool your hot solution too quickly, or agitate the solution, and you form many nucleation sites at once. Growth is then spread out and diluted, leading to a finely divided final solid which can slip past the filter paper. Too many quarterbacks not enough linebackers. You want large, well-formed, robust crystals that flopdown and go right to sleep in the filtration funnelduring filtration. This is encouraged by letting the`growthh’Once the nucleating site appears, molecules of the compound began to flock to it, packing and stacking on it and each other like a band of angry linebackers piling on top of an opposing quarterback. This is the growth piece of the crystallization. nucleation site solution cool very slowly, maybe at the hotplate cooling rate, and, without stirring so that only a few nucleating sites can gather up most of the dissolved compound to its’ bosom(s).A good example of such crystalline growth is seen in Figure 4. Note the quarter on the bottom right hand corner of the picture, which gives you a sense of the size of these crystals. They’re the cat’s meow for desirable crystals. Figure 4: Re-crystallized o-Acetylsalicylic Acid in Water Large crystals are not only easy to filter they are easy to wash since their surface area/volume is low. Small crystals require much more washing. They have higher surface area/volume, which can lead to partial re-dissolution of the small crystals. So go slow and stay chill on the final cooling step. Let it stand at room temperature until you get a bunch of nice big crystals, then, and only then ice the sucker.As noted, the final step to finish the re-crystallization process is to filter the cold solution achieved in (C) above. This not only puts the desired crystals in the filtration funnel, it also leaves the offending impurities, including any rat poison, (hopefully excluded and still in solution), in the filtration flask for easy disposal.Figure 5 shows a Before vs. After’ case of re-crystallization of one of your practice compounds, phthalic acid. FYI, the NIST melting point reported for the stuff is 207oC 1.Awesome shit.Figure 5: Before vs. After Re-crystallization of Impure Phthalic Acid in WaterBefore at 40X: mp 202-206 oCAfter at 40X: mp 206-207oC5. Binary Eutectic Mixtures and Melting RangesThe Post-Lab questions on pages 26-29 of the lab manual assume you’ve been exposed not only to the intuitive ideas above, but to basic ideas concerning something called a binary eutectic phase diagram (see page 21 of the lab manual) , which is a plot of melting point versus the composition of an A+B mixture. I’m not crazy about this. Most explanations you’ll find on this topic are:found in scary physical chemistry or geology texts (and this is Organic Chem, right?)focused on cooling a liquid mixture of A+ B. We need to heat an initially solid mixture since we interested in understanding melting point behavior.But your lab manual insists on doing the eutectic boogie, so I’ll try to explain the basics. First, let’s describe what a eutectic mixture is. In Figure 5 it’s where the two solid blue curves intersect at E. At that composition (55 wt %A/45 wt % B) two things are true:Figure 5: A Simple Binary Eutectic Melting Diagram Undergoing Cooling Start herecooling (only A solidifying) Shifts back to 80%A/20%B and falls below TE only after all A and B are solidB starts to solidify here with A in 55 %A/45% B ratioend hereA and B both solidCwt %the melting point of the A+B mixture reaches its lowest point, TE.If you started with the 55/45 mixture you can’t purify it into either purer A or purer B by heating (or cooling) since the same composition occurs in both liquid (above point E) and solid phase (below point E). You’re stuck there.Fortunately, at least for a simple binary mixture, there is just one point along the composition axis where this happens. If we take any other point (say M in Figure 5) and cool 100 grams of a mixture containing 80 grams A and 20 grams B from liquid, only A will start to solidify since it’s in much greater abundance (and melts lower). But as soon as a little A solidifies, the remaining liquid becomes enriched in B and the melt composition (symbolized by the blue ball in Figure 5) rolls down the curve to a lower temperature say, point C.You can see where this is going; the liquid composition continually enriches in B as cooling continues and more solid A forms2. Eventually, the blue ball will roll all the way down to bottom of the curve at E. Now B starts to solidify, dropping out of the liquid with the remaining liquid A at a fixed eutectic ratio of 55% A/45% B and at the fixed temperature TE . (if you want to know why the ratio’s constant, look up `phase rule.’) The temperature stops falling because B is undergoing a phase change. It’s like what happens when liquid water become ice as the temperature outside falls below zero - the temperature of the ice-water mixture pegs at 0o C until all the liquid water is gone. Similarly, until all the B is solid and the solid mixture returns to the original 80:20 ratio the solid mix’s temperature remains pinned at TE. Temperature only starts falling again when both A and B are completely solidified. The pathway for the blue ball then jogs left back to the original 80:20 line before the temperature drops further. Now, how does all of the above explain what happens as we heat an initially solid mixture? As already noted, this is what we here in Organic land are interested since we want to understand melting point behavior. We’ll start with the same A:B composition, 80:20 , but at a temperature way lower than TE. (See Figure 6, below). Stays at E until all of B melts with A at 55 %A/45% B ratio leaving only solid AB all melted, only A continuing to meltheating (only As seen in Figure 6, when we heat up to TE, we essentially reverse what occurs when we cool. No matter what starting solid composition is used, the initial liquid formed at TE will end at the 55% A/45% B, eutectic ratio. More liquid at this ratio is formed until all the B is liquefied. Since B is only 20% of the sample, the remaining solid is all A. The liquid composition (symbolized by the red ball) will then roll up the curve all the way to M as the liquid now continually enriches in A and ends up all liquid at point M at the original solid composition (80:20 =A/B) .The only times we don’t end up travelling through the eutectic point when we heat a solid up are:when the solid is in fact pure.if the binary mixture doesn’t have a eutectic.Otherwise, using these binary diagrams, the predicted melting point range for any simple binary composition should be between TE and TM.3 (Check out questions h and ion page 22 of your lab manual.)FYI, If you are crazy glutton for punishment and want even more details, check out the websites cited in footnotes 4 and 5. Both are geology articles but pretty easy to read.end hereboth A and B meltedstart herewt. %Figure 6: A Simple Binary Eutectic Melting Diagram Undergoing HeatingFootnotes1 Note that a variety of different reported melting points are cited in variousreferences, ranging from 191-230oC. Your Lab Book uses the CRC value of 210-211 OC. I didn’t find that to be the case in my own experiments.WSA = grams solid A=100* y ( x+y)WL= grams liquid = 100* x ( x+y)Note: Until TE is reached, B is all in the liquid phase and thus has the mass it started with in the initial composition.at the start of cooling. 2You can compute the mass of A, solid WSA, and the mass of liquid, WL, at any point on the curve using the recipes below (and assuming 100 gram mixture)4:yxsolid Aliquid A+B3At first blush this may seem contradictory since it means even a solid that is only 0.2% impure will have a wide melting point range. However, practically speaking, at such a low % impurity you would probably never see the liquidus formation of the eutectic before all of the B impurity had liquefied and the liquid composition had rapidly risen near the liquid temperature of the 99.8/0.2 composition, TM, close to A’s pure melting point TA. If, however, you have substantial impurity B, then intermediate compositions above theeutectic but below TM would start to be observed since more B means the net volume of liquid (A+B) will be more visible as the temperature is increased.4 ................
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