CRYSTALLIZATION: PURIFICATION OF SOLIDS ANSWERS TO PROBLEMS: 1. (a) (b) (c) (d) A plot similar to line A in Figure 5.1 on page 559 will be obtained. The line will be slightly curved. All of the substance A would dissolve at 80 C. A solubility of 17.0 g in 100 ml of water is equivalent to a solubility of 0.17 g in 1 ml of water. This is a greater solubility than is required. Crystals of A should appear around 56 OC. The recovery of A would amount to 0.085 g. A solubility of 1.5g in 100 ml of water at 0 OC is equivalent to a solubility of 0.015 g in 1 ml of water. Therefore, 0.015 g of A would remain dissolved in the water, with the remainder being formed as crystals. 2. If a saturated hot solution was filtered by vacuum filtration, the cooling which occurred as the solvent was drawn through the filter paper would cause the solute to precipitate in the form of crystals. The result would be that the filter paper would become clogged with crystals, and impurities would not be removed successfully from the solution being filtered. PHYSICAL CONSTANTS: MELTING POINTS, BOILING POINTS, DENSITY ANSWERS TO PROBLEMS: 1. A mixed-melting-point could be used. Equal quantities of A and B are intimately mixed using a mortar and pestle. The mixture is placed in a melting point capillary tube, and the melting point determined. If the melting point is identical to that of pure A (and pure B) without depression or expansion, then A and B are identical. 2. Curve 2 would be an ideal heating rate. 8. Perform a mixture-melting-point with the unknown and each of the other substances independently. The mixture that gives no depression or expansion of the melting point reveals the identity of the compound.
EXTRACTIONS, SEPARATIONS, AND DRYING AGENTS ANSWERS TO PROBLEMS: 4. Following the method given in Section 7.2, we obtain: (50-x mg) K = 1.0 = C 2 C 1 = (0.25 ml ether) (x mg) (1.0 ml H 2 O) 1.0 = (50-x)(1.0) 0.25x 0.25x = 40-x 1.25x = 40 x = 32 mg remaining in aqueous phase 50-x = 18 mg extracted into ether phase For one 0.50 ml extraction: (50-x mg) K = 1.0 = (0.50 ml ether) (x mg) (1.0 ml H 2 O) 1.0 = (50-x)(1.0) 0.50x 0.50x = 50-x 1.5x = 50 x = 33.3 mg remaining in aqueous phase 50-x = 16.7 mg extracted into ether phase
2. RCOO - Na + + HCl Ar-O - Na + + HC1 R-NH + 3 CI - + NaOH RCOOH + NaCl Ar-OH + NaC1 R-NH 2 + H 2 O + NaC1 3. One could extract the mixture with hydrochloric acid at any point in the separation procedure. However the sodium becarbonate extraction must be done before the sodium hydroxide extraction. 4.
8. The washing procedure involves adding 1 ml of aqueous sodium bicarbonate to an organic mixture and shaking the mixture vigorously. When this procedure is followed, any acidic substances are removed into the sodium bicarbonate as their sodium salts. Extracting an aqueous layer three times with methylene chloride involves adding a volume of methylene chloride to an aqueous layer, shaking it vigorously, allowing the layers to separate and removing the lower (methylene chloride) layer while leaving the aqueous layer behind. Save the methylene chloride layer in another container. Then add another portion of methylene chloride to the aqueous layer and repeat the above procedure. Each time save the methylene chloride layer and leave the aqueous layer in the original container. The procedure involves a total of three extractions with methylene chloride. 9. You should not add drying agent at this point. First, transfer the organic layer with a dry Pasteur pipet to a dry container and then add drying agent to the organic layer. SIMPLE DISTILLATION 1. (a) Approximately 92% A, 8% B. (b) Approximately 18% A, 82% B. 3 The line connects the compositions of the boiling liquid (lower curve) and its equilibrium vapor (upper curve) at the same temperature. Using line xy in Figure 8.3, if the boiling liquid has composition W, its vapor has composition Z. 4. It would boil at about 110 C. 7. A good separation can be achieved in a simple distillation only if thereis a large ( >100 C) difference in the boiling points of the liquids to be separated. FRACTIONAL DISTILLATION, AZEOTROPES 1. (a) 3.9 g = 0.05 mole benzene; 4.6 g = 0.05 mole toluene N benzene = 0.05 0.05 + 0.05 = 0.5; N toluene = 0.5 (b) (e) Partial vapor pressure of benzene = (270 mm)(0.5) = 135 mm At 90, P total = (1010 mm)(o.5) + (405 mm)(o.5) = 707 mm at 100, P total = (1340 mm)(o.5) + (560 mm)(o.5) = 950 mm The boiling point is greater than 90, but less than 100. Assume a linear relationship between the vapor pressure of each substance and the temperature from 90 to 100.
2 For benzene; (1340-1010) = 10 66 mm change in vapor10 pressure for each 2 change in temperature 2 For toluene; (560-405) = 10 31 mm change in vapor pressure for each 2 change in temperature The following approximate vapor pressures are obtained at certain temperatures: benzene toluene 90 1010 405 92 1076 436 94 1142 467 96 1208 498 98 1274 529 100 1340 560 At 92 ; P total = (1076)(0.5) + (436)(0.5) = 756mm Thus, the boiling point is approximately 92. (d) Partial vapor pressure of benzene at 92 = (1076)(0.5) = 538 mm Partial vapor pressure of toluene at 92 = (436)(0.5) = 218 mm 538 538 Vapor composition: = 0.71 benzene, 0.29 toluene 760 760 (e) 0.71 mole = 55.4 g benzene; 0.29 mole = 26.7 g toluene 55.4 g 55.4 + 26.7 g X 100 = 67.5% benzene; 32.5% toluene
COLUMN CHROMATOGRAPHY 1. If the components of the mixture all passed through the column, the solvent must have been too polar. In that case, repeat the chromatography using a less polar solvent, such as petroleum ether. If all of the mixture stayed on the column too long, the solvent was probably not polar enough. Switch to a more polar solvent, such as methanol or ethanol. Other parameters which might be adjusted include the length of the column and the column packing. 3. Order of elution: Biphenyl (elutes first) > benzyl alcohol > benzoic acid (elutes last). 4 The person forgot to allow the column to drain until the surface of the liquid containing the orange compound had passed the upper surface of the adsorbent. Then a small amount of solvent should have been added, and that solvent should have been allowed to drain below the upper surface of the adsorbent. Only then would it be permissible to fill the solvent reservoir. The solvent reservoir should not be filled until the sample has been completely applied to the adsorbent. 5 Clearly, petroleum ether is not sufficiently polar to elute this sample. A more polar solvent, such as methanol or ethanol, should be selected.
THIN-LAYER CHROMATOGRAPHY 1. The presence of only one, highly-mobile spot does not necessarily indicatethat the unknown material contains one pure compound. It is possible that the material is a mixture, but that all of the components travel all the way up the TLC plate because the solvent is too polar. The experiment should be repeated with a less polar solvent, such as petroleum ether or cyclohexane. If only one spot appears in this experiment, it is safer to conclude that the unknown is a single, pure compound. 2. Owing to differences in adsorbent thickness, it is not safe to conclude that the two samples are identical. The experiment should be repeated with both samples applied to the same TLC plate. If they both have the same Rf value on the same plate, one may conclude that they are identical. 3. Relative Rf values: biphenyl > benzyl alcohol > benzoic acid. 5. This is the reverse situation of that described in Problem 1. In this case, the solvent is not sufficiently polar to move the components down the plate. The experiment should be repeated in a more polar solvent. If only one spot is observed in this second experiment, one may conclude that the sample is pure. GAS CHROMATOGRAPHY 1. (a) I-Chloroproane would have the shorter retention time (elutes first). The elution order is according to boiling point. (b) The retention times would not be identical because it is impossibleto duplicate exactly all the factors affecting retention times. However, the order of elution would be the same. 2. Area Peak A = 66 x 8 = 528 mm 2 Area Peak B = 43 x 8 = 344 mm2 Total area= 872 mm2 528 %A= x 100 = 60.6% 872 344 %B = x 100 = 39.4% 872 4. (a) Retention time would increase. (b) Retention time would decrease. (c) Retention time would increase.