1Separations and purifications test w solutions

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ORGANIC CHEMISTRY TOPICAL: Separations and Purifications Test Time: 22 Minutes* Number of Questions: 17 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Separations and Purifications Test Passage I (Questions 1–7) A second student carried out the same reaction and obtained the TLC chromatogram shown in Figure A student attempted to synthesize Nphenylphthalimide according to the following reaction: solvent front O C OH + H2N C spotting point OH O Phthalic acid Figure Analine O C N + 2H2O C The solubility of the product in several solvents at two different temperatures was also determined These results are presented in Table O N - Phenylphthalimide Table Solvent Reaction l Solubility (g/100mL) 0¡C 50¡C 0.002 0.003 6.3 6.8 0.1 0.3 0.3 7.5 Phthalic acid and aniline were heated to reflux for one hour Upon cooling, the mixture was dissolved in ether and placed in a separatory funnel The solution was then washed with 5% aqueous Na2CO3 followed by 5% aqueous HCl and allowed to stand over solid anhydrous Na2SO4 for 15 minutes The ether was then evaporated and a solid product obtained Dichloromethane 1-Propanol Water Toluene The crude product was dissolved in an appropriate hot solvent and after slowly cooling the solution, crystals were obtained, which after filtering and drying gave a melting point of 205¡ to 207¡C The literature value for the melting point of the expected product is 208¡C In the experiment, which of the following would have been removed by washing the solution with sodium carbonate? The recrystallized product was analyzed by thin layer chromatography (TLC) and the chromatogram in Figure was obtained A B C D Phthalic acid N-Phenylphthalimide Aniline Water solvent front spotting point Figure GO ON TO THE NEXT PAGE KAPLAN MCAT What was the purpose of allowing the washed product mixture to stand over solid anhydrous sodium sulfate? A B C D To drive the reaction to completion To remove the acidic starting material To remove water from the solution To increase the purity of the product What conclusion can be drawn from the melting point data given in the passage? I The reaction produced a high yield II The product was fairly pure III The expected product was obtained A B C D Based on the two TLC chromatograms in Figure and Figure 2, what can be said about the product obtained by the first student compared to that obtained by the second student? A The first student obtained a better yield of product B The first student obtained a 100% pure product while the second student obtained a 93% pure product C The first studentÕsproduct is more pure than the second studentÕs D The first student used a different recrystallizing solvent I only II only II and III only I, II, and III From the solubility data given, which of the four solvents would be most suitable for recrystallization of the product? A B C D Water Dichloromethane 1-Propanol Toluene What is the IUPAC name for aniline? A B C D Benzenamine Cyclohexanamine Benzylamine Benzoic acid A proton NMR spectrum of the final product was also taken How would the student be able to confirm the presence of N-phenylphthalimide from this spectrum? A B C D Absorption peaks in the region 10-12 ppm Absorption peaks in the region 7-8 ppm Absorption peaks in the region 3-4 ppm Absorption peaks in the region 1-2 ppm GO ON TO THE NEXT PAGE as developed by Separations and Purifications Test Passage II (Questions 8–13) A student carried out the synthesis of heptanonitrile by reacting 1-chlorohexane with sodium cyanide (Reaction 1) Cl + NaCN CN + NaCl -Chlorohexane Heptanonitrile Reaction The student placed sodium cyanide (32g) in a flask along with a suitable aprotic solvent 1-Chlorohexane (72g) was then added with continuous stirring and the resulting mixture was heated for 30 minutes Upon cooling, distilled water was added and the solution extracted into ether After washing and drying this layer, the ether was evaporated, leaving behind a liquid residue This residue was then distilled under reduced pressure (50 torr) The fraction that boiled between 70¡C and 80¡C was collected since the desired product has a boiling point of 75¡C at 50 torr The product weighed 53g, and analysis by gas chromatography (GC) revealed the following chromatogram: B In the chromatogram shown in Figure 1, the area under peak A is units and the area under peak B is 180 units If peak B is due to the presence of heptanonitrile, what is the purity of the product? A B C D 97.0% 80.0% 5.0% 2.7% Based on the information given in the passage, what would you expect the boiling point of the product to be at atmospheric pressure? A B C D 60¡C 75¡C 80¡C 190¡C Assuming 100% purity, what is the yield of heptanonitrile in the experiment? A 72 / 111 × 100 53 / 120.5 B 53 / 120.5 × 100 72 / 111 C 53 / 111 × 100 72 / 120.5 D 72 / 120.5 × 100 53 / 111 A l How can the fact that heptanonitrile has a longer retention time in the gas chromatograph be explained? 10 15 20 I It has a higher boiling point II It is more polar than the other components in the mixture and thus more attracted to the polar stationary phase of the chromatography column III It is more soluble in ether during extraction and therefore moves more slowly during the chromatographic separation IV It is optically active and will move more slowly through the chromatographic separation based on its rotation of plane polarized light 25 Minutes A B C D I and II only I, II, and III only II and III only I, II, III, and IV GO ON TO THE NEXT PAGE KAPLAN MCAT 12 Reaction most likely proceeds by which of the following mechanisms? A B C D Bimolecular elimination Bimolecular nucleophilic substitution Unimolecular nucleophilic substitution Hydrolysis In the experiment, why was ether chosen as the extraction solvent for heptanonitrile? A The NaCl by-product was not soluble in ether and would therefore remain in the water layer while the nitrile would be extracted into the ether layer B All nitriles are more soluble in water than in ether C Ether and water are completely soluble in each other D The sodium salt of the nitrite would remain dissolved in the water GO ON TO THE NEXT PAGE as developed by Separations and Purifications Test 1 Which of the following factors usually increases the solubility of a compound in a given solvent? Questions 14 through 17 are NOT related to a descriptive passage I II III IV Which of these statements correctly describes the separation of the compounds below by extraction into dichloromethane and water? Acidic Solution O N Caffeine O N NH H 3C N O CH Octanoic acid O CH + H 3C Basic Solution CH3 N N N CH3 O OH CH 3(CH2) 6C I, II, III, and IV I, II, and IV only I, II, and III only I and II only Separation by thin-layer chromatography is based primarily on: N O CH 3(CH2) C A B C D Higher temperature Similar polarities Higher molecular weight of the compound Lower density of the solvent OÐ A If the solution were acidic, caffeine would be more soluble in dichloromethane than in water B If the solution were basic, caffeine would be more soluble in dichloromethane than in water C If the solution were basic, octanoic acid would be more soluble in dichloromethane than in water D If the solution were acidic, octanoic acid would be more soluble in water than in dichloromethane A molecular weight B relative attraction of the components towards the mobile and stationary phases C relative refractive indices of the solvent and the components D relative specific rotations of the various components Which of the following techniques would be most suitable in the separation and analysis of two miscible liquids with boiling points of 80¡C and 150¡C? A Recrystallization followed by gas chromatography B Recrystallization followed by electrophoresis C Vacuum distillation followed by electophoresis D Simple distillation followed by gas chromatography END OF TEST KAPLAN MCAT ANSWER KEY: A C C D A 8 10 B C A D C 11 12 13 14 15 A B A B D 16 17 D B as developed by Separations and Purifications Test EXPLANATIONS Passage I A Sodium carbonate (Na2CO3) is commonly employed in many experiments It is the salt of a weak diprotic acid (carbonic acid H2CO3) and a strong base, formed, for example, by completely reacting carbonic acid with two molar equivalents of NaOH It reacts with acidic compounds by converting them into their corresponding sodium salts: or, in ionic form, Na2CO3 + HA → NaHCO3 + NaA 2Na+ + CO32Ð + HA → 2Na+ + HCO3Ð + AÐ These sodium salts are water soluble and so can be extracted into the aqueous layer Choice A phthalic acid is the only acidic compound out of all of the answer choices This molecule is a dicarboxylic acid and conversion to its corresponding sodium salt involves loss of a proton from each carboxyl functionality resulting in the formation of a COO−Na+ group As a result, any phthalic acid that contaminates the final product will be dissolved in water and consequently removed All the other answer choices are wrong since they're not acidic N-phenylphthalimide will not form a sodium salt and so will not dissolve in the aqueous layer, making choice B incorrect Aniline choice C is basic and so definitely will not form a sodium salt It is in fact treatment with 5% hydrochloric acid, not sodium carbonate, that will result in the removal of any basic contaminants such as aniline Finally, choice D is wrong since it is the addition of anhydrous sodium sulfate that results in the removal of water, not the addition of sodium carbonate C The word ÒanhydrousÓ meansÒwithout water,Ó and a substance that is anhydrous, an anhydride, will typically react with water, thus removing it in the process Therefore, even if one is unaware of this experimental procedure, one can still arrive at the correct answer After extracting or washing an organic material in an aqueous solution, the organic layer is wet it contains a small amount of water Several anhydrous salts such as sodium sulfate or calcium chloride can be used as drying agents and so the addition of sodium sulfate will result in removal of water, making choice C the correct response Choice A is incorrect since the addition of sodium sulfate has nothing to with driving the reaction to completion The acidic starting material is removed by sodium carbonate not sodium sulfate and so choice B is also wrong Finally, the addition of sodium sulfate has nothing to with the purity of the product The purity of the product comes into play in when the crude product is dissolved in a hot solvent otherwise known as recrystallization, so choice D is wrong C The melting point of a compound is a physical property of that compound; the expected product must have been obtained since the experimental melting point is quite close to the literature value roman numeral III must be in the correct answer Answer choices A and B can be eliminated The presence of impurities acts to lower the melting point of a substance and also to broaden the range of temperatures over which melting occurs (i.e the melting point is no longer just a point) The melting point of the crystals is 205¡ to 207¡, which indicates the presence of some impurities, but not too much, making roman numeral II correct However, the melting point is not related to the yield of the compound so statement I is wrong Choice C is the correct answer then, since statements II and III are true D This question tests our understanding of the fundamental principles behind recrystallization, but it also incorporates table reading into it In recrystallization, we dissolve an impure product in a solvent at high temperature, then let it cool: as the system cools, the pure product precipitates out, while leaving the impurities behind dissolved in the solvent In choosing a solvent for recrystallization, then, the solute (product) should be relatively soluble in it at elevated temperatures, and insoluble in it at cooler temperatures The greater the difference between these two solubilities, the greater the yield will be Toluene choice D shows the greatest difference in solubilities, so answer D is correct You can see from the data in the table that the product is much more soluble at 50 degrees Celsius than at degrees Celsius The rest of the answer choices show very little difference between the solubilities in hot and cold solvent In addition, the solute has a low solubility in both choices A water and B dichloromethane, even at high temperature A The accepted IUPAC name for aniline is benzenamine This describes an benzene ring with a primary amine substituent attached to it Choice B is wrong as cyclohexanamine is a six membered carbon ring with an amine substituent on one of its carbons There is no aromatic ring in this compound Benzylamine choice C would be a benzene ring substituted at one carbon with a CH2NH2 group Therefore, there would be one too many CH2 groups to use this as an alternative name KAPLAN MCAT Finally, benzoic acid choice D is a benzene ring with a carboxyl substituent There isn't even an amine functionality (or even an nitrogen atom) in this molecule, so this answer choice is also wrong NH cyclohexanamine H2 C NH2 benzylamine COOH benzoic acid B If you look at the product in Reaction 1, you can see that the only types of proton that are present are aromatic ones There are no protons attached to the nitrogen or carbonyl functionalities, so in the spectrum of the pure product, only this one type of signal should be observed From your knowledge of NMR, you should know that aromatic protons give rise to signals in the region of to parts per million, making choice B the correct response Choices C and D are wrong since these signals are too far upfield to be characteristic of aromatic protons Remember that these protons are deshielded, hence they absorb in the 7-8 ppm region On the other hand, the signal in choice A would be too far downfield and is characteristic of highly deshielded protons such as carboxylic acid protons C TLC analysis gives a qualitative indication of purity You can tell how many components are in the mixture as they separate out on the chromatogram to give individual spots So, the product obtained by student must be more pure than that obtained by student 2, as the chromatogram in Figure two indicates the presence of an impurity, whereas that in Figure only shows the presence of one compound Although you can see how many compounds are present, you can't determine their amounts, so choice A which talks about yield is wrong The purity of the product cannot be determined quantitatively either, so choice B is also wrong Finally, the nature of the recrystallization solvent does not determine the nature of the TLC chromatogram, so choice D is also wrong Passage II A Before we dive into the explanation to this question, let's just briefly discuss the principal behind gas chromatography Basically, the technique is employed so that different components in a mixture can separated The sample to be analyzed is injected into the GC column and vaporized; an inert carrier gasÑusually helium or nitrogenÑtransports this gaseous sample through the column which contains an adsorbant Depending on the nature of the components, they are attracted to different extents to the adsorbant (also known as the stationary phase) and so are eluted from (swept out from) the column at different times From the chromatogram in Figure 1, it should be evident that there are two components eluted from the column In addition, you should know that the areas under these peaks are proportional to the amount of eluted material In this case, the units of heptanonitrile is 180 the area of peak B divided by the sum of areas A and B which is 185 This figure then has to be multiplied by 100 to get a percentage Without even bothering to carry out the calculation, we should be able to see that 180/185 is really close to 100%, making choice A the correct response D For this question, you need to know that at higher pressures, boiling points are also higher: this is simply a consequence of the definition of boiling Despite our everyday experience telling us that boiling implies a high temperature (100¡C), the actual scientific definition of boiling does not involve any mention of hotness or coldness Boiling occurs when the vapor pressure of the liquid equals the pressure of the environment If this ambient pressure is low (as at high altitudes), then a low vapor pressure would suffice for boiling to occur Conversely, if the ambient pressure is high (as in a pressure cooker), a high vapor pressure is needed for boiling to occur Temperature comes into play because the higher the temperature, the higher the vapor pressure of the liquid The higher the ambient pressure, then, the higher the temperature needs to be in order for the vapor pressure to reach that same valueÑi.e., the higher the boiling point In this case we have been given the boiling point of the product at 50 Torr If the boiling point of the product at 50 torr is 75°C, then the boiling point at 760 torr or atmospheric pressure would be significantly higher Answer choice D at 190°C is the only one that is much higher than 75°C Choice B is incorrect, because the boiling point is not expected to remain the same when the pressure is increased, and choice A is incorrect as this boiling point is lower not higher Finally, choice C is incorrect as this is not a significant increase in the boiling point As the pressure has drastically increased, you would expect the boiling point to so as well 10 as developed by Separations and Purifications Test 10 C This question is a pretty simple calculation to determine the yield of product The yield of product is defined as the actual number of moles produced divided by the expected number of moles multiplied by 100 to get this figure into a percentage As you can see from Reaction 1, mole of 1-chlorohexane goes on to form mole of heptanonitrile The expected yield should therefore be equivalent to the number of moles of 1-chlorohexane which is 72 divided by 120.5 The actual yield of heptanonitrile is equivalent to its mass 53g divided by its molar mass which is 111g/mol Putting these figures together, you should see that the yield is 53/111 which in turn is divided by 72/120.5 This is then multiplied by 100 and corresponds to choice C 11 A The components of mixtures separate as they are carried along the GC column by virtue of their relative attractions to the mobile (gas) and stationary (column packing or coating) phases Therefore, those components which are more attracted to the column have a longer retention time Statement II is correct as heptanonitrile is more polar than the other components in the mixture so will be attracted to the column more, making its retention time longer Also, compounds with a higher boiling point have longer retention time Therefore, I and II are valid reasons making choice A the correct response Past treatment and optical activity have nothing to with retention time In addition, heptanonitrile isn't even optically active and so answer choices which contain statements III and IV can be eliminated 12 B This question tests your ability to recognize fundamental organic reactions Since the alkyl halide 1-chlorohexane is primary and the nucleophile cyanide is strong, the mechanism is bimolecular nucleophilic substitution The cyanide ion nucleophilically attacks 1-chlorohexane and displaces the leaving group, Cl − By contrast, tertiary alkyl halides and weak nucleophiles tend to undergo unimolecular nucleophilic substitution, so choice C can be discarded All elimination reactions produce double or triple bonds, which is not the case here, making choice A incorrect Finally, hydrolysis involves the splitting of a molecule and adding -H to one part and -OH to the other Again, this doesn't happen a chloride is simply replaced by a cyanide, so again the correct response is B 13 A The correct answer here is choice A All alkali metal salts are soluble in water, so sodium chloride will dissolve while heptanonitrile remains in the ether layer Heptanonitrile contains a pretty long nonpolar hydrocarbon portion and if you remember the phrase Òlike dissolves like,Ó this is why it dissolves in ether Therefore,choice B, which states that nitriles are more soluble in water than in ether is completely wrong Also, there is no such thing as a sodium salt of a nitrile, so choice D is wrong You should know from your fundamental practical knowledge that ether and water are immiscible, so choice C is also incorrect Independent Questions 14 B The correct answer here is choice B Under basic conditions, the caffeine molecule is deprotonated and so is neutral As a result, it would tend to dissolve in organic solvents rather than in water Under the same conditions, octanoic acid is in its charged or polar form and would therefore be soluble in water Under basic conditions then, caffeine would be found in the organic layer, not the aqueous layer making choice B the correct answer Under acidic conditions, caffeine is charged and the octanoic acid is not Therefore, caffeine would be more soluble in water if the solution was acidic, not in dichloromethane, which is what choice A states Choice C is incorrect since octanoic acid would be more soluble in water under basic conditions and finally, choice D is wrong as octanoic acid would be more soluble in dichloromethane under acidic conditions 15 D Two liquids which boil below 150¡C and at least 25¡ apart can be separated by simple distillation Gas chromatography can be used to analyze volatile materials such as these since they boil at relatively low temperatures Hence, these two techniques would be ideal in order to separate and analyze the two liquids Choices A and B are wrong since recrystallization is commonly used purify solids, not liquids Choice C is incorrect for two reasons Firstly, vacuum distillation is necessary only when boiling points are high and are not far apart As there is a difference of 70¡C between the boiling points of the two liquids, and they boil at 150¡C and 80¡C, there would be no need to call upon this technique Secondly, electrophoresis is used to analyze charged species such as amino acids, so its inclusion in choices B and C make these incorrect responses 16 D Most substances tend to become more soluble at higher temperatures and so statement I is correct You only need to look back at Table in the first passage in this topical to confirm this In doing so, you can see that the compound is always more soluble at 50°C than at 0°C In fact, the whole process of recrystallization is based on the fact that a compound is more soluble at higher temperature and so again, the solubility does increase as the temperature is raised Statement II is also KAPLAN 11 MCAT correct Again, recall the phrase Òlike dissolves likeÓ so compounds thatare polar or ionic will dissolve in solvents that are also polar, whereas compounds that are non-polar will dissolve in non-polar solvents Therefore, similar polarities will increase the solubility Statement III molecular weight and statement IV density of the solvent not determine solubilities, so I and II are the only true statements, and choice D is the correct response 17 B All types of chromatography rely on the principal that components in a mobile phase flow through a stationary phase and are removed at different stages depending on the attraction towards these phases In simple terms then, if a component is attracted to the stationary phase (which in the case of thin-layer chromatography is the surface of the TLC plate) it will be held immobile and produce a spot If a component is weakly attracted to the stationary phase it will remain in the mobile phase (which constitutes the solvent moving up the TLC plate) and produce a spot further up All the other answer choices not influence the separation of the components on the TLC column 12 as developed by ... ON TO THE NEXT PAGE as developed by Separations and Purifications Test Passage I (Questions 1–7) A second student carried out the same reaction and obtained the TLC chromatogram shown in Figure... optically active and will move more slowly through the chromatographic separation based on its rotation of plane polarized light 25 Minutes A B C D I and II only I, II, and III only II and III only... dichloromethane and water? Acidic Solution O N Caffeine O N NH H 3C N O CH Octanoic acid O CH + H 3C Basic Solution CH3 N N N CH3 O OH CH 3(CH2) 6C I, II, III, and IV I, II, and IV only I, II, and III
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