Final S09 Key

Final S09 Key - Chem 6AL Russak 18 March 2009 7:30‐10:30...

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Unformatted text preview: Chem 6AL Russak 18 March 2009 7:30‐10:30 pm Final Exam The following questions are all multiple choice. Please fill out the score sheet with your name and perm number. Your perm number MUST be both written and the bubbles filled. There are equations and spectral charts on the back sheet that you may need to calculate or solve your answers. You may use a calculator, ruler, and pencil or pen. You may NOT keep the exam. It will be posted online once all grading has been completed. Experiment 1: Density, melting point, refractive index 1. The refractive index of compound X is 1.4. What is the speed of light through a pure sample of compound X? a. 1.16 x 10‐8 m/s b. 2.14 x 10‐8 m/s c. 4.28 x 10 8 m/s d. 2.14 x 10 8 m/s e. None of the above 2. How many moles of KOH (MW = 56.11) are needed to make a 10 wt % solution of KOH in water, starting with 12 mL of water? a. 67.3 mol b. 0.023 mol c. 0.090 mol d. 74.62 mol e. None of the above 3. What is the weight percent of cinnamic acid in a mixture that is 28.8 mol percent cinnamic acid in urea? (Urea = 60.06 g/mol, cinnamic acid = 148.16 g/mol) a. 71 % b. 50 % c. 29 % d. 21 % e. None of the above 4. In general, how do impurities affect the melting point and range? a. They increase both b. They decrease both c. Melting point increases but the range decreases d. Melting point decreases but the range increases e. None of the above 5. The tared weight of an empty 10 mL syringe is 4.24 g. Once filled it weighs 16.16 g. What is the density of the liquid? a. 1.192 g/mL b. 0.965 g/mL c. 1.887 g/mL d. 1.354 g/mL e. None of the above Experiment 2: Distillation, Boiling points, Rault’s law 6. A solution of 2 grams of ethanol (46.0 g/mol) and 3 grams of methanol (32.04 g/mol) was prepared. Assuming the pure vapor pressure of methanol is 81 kPa and the total vapor pressure for the system is 69.5 kPa, what is the pure vapor pressure of ethanol? a. 11.5 kPa b. 22.1 kPa c. 35.7 kPa d. 45.0 kPa e. None of the above 7. Consider a 20 mol % mixture of hexane (P°=72 kPa) in toluene (P°=35 kPa). What is the combined vapor pressure? a. 7 kPa b. 14.4 kPa c. 28 kPa d. 42.4 kPa e. None of the above 8. Water is a much smaller molecule than benzene, yet it has a higher boiling point. What factor could be causing this? a. It has much less surface area b. The pressure exerted by the atmosphere is much more in the case of water c. The intermolecular forces are greater for water d. Water is more dense, therefore would have a higher boiling point e. None of the above 9. What feature is responsible for influencing the boiling point of a liquid? a. Hydrogen Bonding b. Surface area c. Intermolecular forces d. Overall size of molecule e. All of the above 10. A fractional distillation, when compared to a simple distillation, will usually give: a. more volume of a more pure substance b. less volume of a more pure substance c. more volume of a less pure substance d. less volume of a less pure substance e. None of the above 11. Use the graph below to answer the question. What does V1 stand for in the graph? a. b. c. d. e. 12. When an azeotrope occurs in a solution of liquids a. It is easy to distill the mixture to purity b. Many times column hold‐up is the cause for such an effect c. The forces of boiling the impure substance out are greater than the boiling point ofthe solution d. The forces of boiling the impure substance out are far less than the boiling point of the solutions e. None of the above 13. What is a way to prevent the distillation solution from “bumping”? a. Wrap the column in foil b. Do the distillation outside of the hood to prevent air from cooling the apparatus c. Add a boiling stone d. Heat the solution rapidly e. None of the above Experiment 3: Acid/Base chemistry, Extraction, Drying, Recrystallization 14. Methanol‐water cannot be used as a pair of liquid‐liquid extraction solvents because: a. they have different boiling points b. water is more polar than ethanol c. they are immiscible d. they are miscible e. None of the above The vapor composition at L2 The liquid composition of V2 The vapor composition of L1 The liquid composition of V1 None of the above 15. Choose an extraction procedure that would separate the structures below from each other considering they are in a separatory funnel with diethyl ether and water. a. Add acid and remove the water layer containing cyclohexanol. Then add water and base and remove the water layer containing methyl m‐nitrobenzoate. b. Add acid and remove the water layer containing methyl benzoate. Then add water and base and remove the water layer containing methyl m‐nitrobenzoate. c. Add base and remove the water layer containing cyclohexanol. Then add water and acid and remove the water layer containing methyl m‐nitrobenzoate. d. Add base and remove the water layer containing methyl benzoate. Then add water and acid and remove the water layer containing cyclohexanol. e. None of the above 16. Consider the following equilibrium Given that the pKa for benzoic acid is around 4 and the pKa for protonated aniline is 5, how many molecules of benzoate are there for every molecule of benzoic acid? a. 0.1 b. 1 c. 10 d. 1.0 E4 e. None of the above 17. Which organic solvent would constitute the bottom layer when performing an extraction with water? a. Ethyl Acetate b. Hexanes c. Chloroform d. Toluene e. None of the above 18. The reason recrystallization works to purify a compound is because a. Impurities are highly concentrated in the solvent b. The solute is lowly concentrated in the solvent c. Impurities are lowly concentrated in the solvent d. The concentration has no effect on the ability to crystallize the compound e. None of the above Experiment 4: Extraction coefficients, Sublimation 19. In experiment 4, the tea bag was soaked in hot basic water. What was the purpose of this? a. For the acidic tannins to fully protonate the caffeine and make it water soluble. b. For the base to neutralize and tannins and prevent protonation of caffeine c. Because caffeine is not soluble in water itself d. The tea bag was soaked in hot acidic water, not base e. None of the above 20. If dichloromethane and water are the extraction solvents, vigorous shaking can lead to: a. Evaporation b. Separation c. Byproducts d. Emulsion e. None of the above 21. Consider the phase diagram above. The compound is heated from ‐40 °C to 30 °C at 20 atm. What process describes its change in matter a. Melting b. Freezing c. Sublimation d. Boiling e. None of the above 22. What is the total % recovered after performing an extraction 2 times with equal volumes of each phase. The partition coefficient of the compound is 3. a. 3.13% b. 6.25% c. 25.0 % d. 52.3 % e. None of the above 23. How many extractions will it take to recover greater than 98% of a compound with partition coefficient of 5. You are extracting with 3 mL of ethyl acetate each time from a 5 mL volume of water. a. 1 b. 2 c. 3 d. 4 e. More than 4 Experiment 5: Photoisomerization, Thin Layer Chromatography 24. An example of photoisomerization in the body is a. The Krebs cycle b. The oxidation of Vitamin A1 to Vitamin A c. Retinal d. Hemoglobin exchange mechanism e. Skin Cancer 25. Why is the above reaction carried out with light and not heat? a. The heat required would probably decompose the compound before isomerization b. Heat would give the thermodynamically more stable product..which is trans c. Heat is unable to polarize electrons d. Heat would cause dimerization e. None of the above 26. The reaction in experiment 5 was done in homogeneous solution to start with. Why? a. Molecules in solution allow space between themselves so reactions can take place b. In a solid form, most of the trans is protected from light by the crystal lattice c. The solution provided a medium that was soluble for the starting material and insoluble for the product d. A and B e. A, B, and C 27. Look at the TLC below and answer the question that follows: A B The TLC of reaction A B tells you that: a. Starting material is still present in the reaction b. The reaction is complete c. Compound B is less polar that compound A d. Compound A is more polar than compound B e. None of the above 28. Regarding the TLC from above, let’s say that a mixture of 5:1 hexanes:ethyl acetate was used to develop the plate. What would happen to the spots if a 4:1 mixture of hexanes:ethyl acetate was used? a. The solvent mixture is more polar so all spots would travel further b. The solvent mixture is less polar so all spots will travel more c. Compound A will travel further while compound B will be retained d. Compound B will travel further while compound A will be retained e. All spots will travel less Experiment 6 and Gas Chromatography 29. The acid catalyzed dehydration of 2‐butanol to butene gave a mixture of products. Another possible product resulted from a methyl shift. This product would have been a. Isopropanol b. 2‐methylpropene c. Isopropene d. 2‐methyl butene e. None of the above 30. Of the products shown in the reaction for experiment 6, which product predominates? a. 1‐butene b. trans‐2‐butene c. cis‐2‐butene d. 2‐methylpropene e. None of the above 31. In Gas Chromatography, compounds generally separate based on: a. Affinity for stationary phase b. Temperature of sample c. Flow rate of the carrier gas d. Boiling point e. All of the above 32. The products in the above reaction from experiment 6 eluted on the GC in order of a. Polarity b. Boiling point c. Unsaturation d. B and C e. All of the above 33. Gas Chromatography uses a carrier gas to transport the sample through. Which is a typical carrier gas used? a. Helium b. Oxygen c. Hydrogen d. Butene e. None of the above Experiment 7 O MeOH NaBH4, NaOMe t-Bu 4-tert-butylcyclohexanone t-Bu H trans 80:20 H OH + t-Bu H cis OH H 4-tert-butylcyclohexanol 34. Which of the following are carbonyl reducing agents: a. NaBH4 b. Metal hydride agents c. LiAlH4 d. All of the above e. None of the above 35. The starting material for experiment 7 is uniquely absent of indicated chirality. This is because a. The tbutyl locks the ring in one conformer b. The starting material is racemic c. The structure is symmetrical and therefore does not have chirality d. All of the above e. None of the above 36. For Experiment 7, the product Rf will be ________ as compared to the starting material Rf in TLC analysis. a. Larger b. Smaller c. the same d. non‐existent e. None of the above 37. When a reaction proceeds with only 1 isomer as its results, the reaction is said to be a. Stereoelectronic b. Stereogenic c. Stereospecific d. Stereoselective e. None of the above 38. In experiment 7, methanol was used as the solvent because a. It is polar enough to dissolve the NaBH4 salts b. It participates in the transition state of the reaction c. It’s conjugate base is more basic than nucleophilic d. All of the above e. None of the above Experiment 8 H N O OH EtI K2CO3 O H N O 39. Which compound is known as Tylenol? a. The intermediate formed in the reaction b. The starting material c. The product d. Phenacetin e. None of the above 40. The solvent methyl ethyl ketone (MEK) was removed prior to the extraction process. This was because: a. It is soluble in water b. It has a high boiling point c. It has a polarity similar to acetone d. It is an excellent solvent for SN2 processes e. All of the above 41. After the MEK was removed, an extraction using NaOH as the aqueous phase was used. Why? a. To remove any excess MEK from the product b. To dissolve the product into the aqueous layer c. To dissolve any remaining starting material in the aqueous layer d. To generate the suitable phenoxide needed for ether formation e. None of the above 42. Assuming the pKa of the phenol in the starting material is 9 and NaOH was added to generate the conjugate base. Assuming the pKa of water is 15, the equilibrium constant should be a. 106 in favor of the phenoxide b. 106 in favor of the phenol c. 10‐6 in favor of the phenoxide d. The reaction is in perfect equilibrium e. None of the above Miscellaneous Spectroscopy Questions 43. Which of the following spectroscopic techniques uses the lowest energy of the electromagnetic radiation spectrum? a. UV b. visible c. IR d. X‐ray e. NMR 44. Which of the following molecular changes is necessary for mass spectrometry to occur? a. excitation of an electron from the ground state to higher energy state b. change of alignment of an electron in a magnetic field c. change of alignment of a proton in a magnetic field d. loss of an electron e. molecular vibration 45. In the electromagnetic spectrum, __________ frequencies, __________ wavenumbers, and __________ wavelengths are associated with high energy. a. high, small, long b. low, large, short c. low, small, short d. high, large, short e. high, small, short 46. Which of the infrared regions is considered to be the fingerprint region? a. 4000cm‐1 ‐ 1000cm‐1 b. 1500μm ‐ 400μm c. 4000μm ‐ 1000μm d. 2200μm ‐ 1000μm e. 1500cm‐1 ‐ 400cm‐1 47. 1H nuclei located near electronegative atoms tend to be __________ relative to 1H nuclei which are not. a. shielded b. deshielded c. resonanced d. split e. none of the above 48. How many signals would you expect to see in the 1H NMR spectrum of the following compound? 2 3 4 5 6 49. How many signals would you expect to see in the 1H NMR spectrum of the following compound? a. b. c. d. e. a. b. c. d. e. 6 3 5 4 2 50. How many signals would you expect to see in the 1H NMR spectrum of the following compound? a. b. c. d. e. 1 2 3 4 5 51. What splitting pattern is observed in the proton NMR spectrum for the indicated hydrogens? Singlet doublet triplet quartet septet 52. There are no splitting patterns in 13C NMR spectroscopy. The reason is because: a. Carbon nuclei are not interfered with by neighboring carbon nuclei like in H NMR b. Carbon NMR obeys the N=0 rule c. The abundance of 13C is too high for splitting to be accurate d. The abundance of 13C is too low for splitting to be accurate e. None of the above 53. How many different types of carbon are in the following structure? a. b. c. d. e. a. b. c. d. e. 3 4 5 6 7 54. What is responsible for an alkene proton appearing so far downfield at 5‐7 ppm? a. Hydrogen and carbon are electron withdrawing b. Anisotropy c. Magnetic field d. Polarity e. None of the above Infrared Spectroscopy Match the following structures to their IR spectra 55. D 56. B 57. C 58. E 59. A 1H NMR Spectroscopy Match the following structures to their 1H NMR Spectra: s=singlet, d=doublet, t=triplet, q=quartet, qt= quintet, sx=sextet, st=septet, ot=octet, m=multiplet (signals are overlapped and difficult to determine splitting). The integrations are giving in parentheses next to the splitting pattern. Match the following structures to their 1H NMR spectra: O H O H O O CH3 H O O A B C D E 60. s(2H), multiplet (4H) Answer =B 61. multiplet (5H), s(3H) Answer: E 62. s(1H), q(1H),q(2H),d(3H),t(3H) =C 63. q(2H), t(3H) Answer: A 64. C13 NMR Spectroscopy Match the following structures to their 13C NMR spectra: t(4H), quintet(4H) Answer: D 65. 4 signals [E] 66. 5 signals [C] 67. 5 signals [D] 68. 3 signals (2@64) [A] 69. Mass Spectrometry Match the following structures to their Mass Spectra: 5 signals [B] 70. [C] 71. [E] 72. [A] 73. [D] 74. [B] Combined Spectra Use the combined spectral data to solve the structures for the following problems 75. Which compound matches the following information: 13 Molecular formula: C6H11N C NMR: 78.5 75.9 41.2 32.5 16.1 4.1 NH H NMR: 2.80, t, J=6.8 Hz, 2H 2.20, t, J=7.1 Hz, 2H 1.78, s, 3H 1.63, m, 2H 1.45, broad s, 2H d. NH2 1 a. H2N b. NH2 HN c. e. 76. IR: 2921 2853 1678 1657 1615 1356 1260 13 Molecular formula: C8H12O C NMR: 198.3 154.1 135.6 41.0 34.2 30.2 21.3 16.7 d. H NMR: 2.66, t, J=7.2 Hz, 2H 2.50, t, J=7.6 Hz, 2H 2.25, s, 3H 2.09, s, 3H 1.82, m, 2H 1 a. O O b. c. O O e. O Equations: N=c/ʋ E=hc/λ P=P°X PTotal=PA+PB Υ=PA/PTotal K = [A]org / [A]aq [A]aq final / [A]aq initial = (V2/(V2+V1K)n pKa = ‐log Ka weight % = mass ratio of solute to solution mol % = mol ratio of solute to solution Tables: ...
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