Unformatted text preview: Chem 6AL M09 Russak 24 August 2009 2pm ‐ 4pm 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 velocity of light through a compound is 1.16 x 108 m/s. What is the refractive index? a. 0.0533 b. 0.533 c. 1.785 d. 1.875 e. None of the above 2. What is the weight percent of compound A in a mixture that is 25 mol percent compound A in compound B? (A= 73.2 g/mol, B = 110.0 g/mol) a. 11.8 % b. 18.1 % c. 33.3 % d. 50 % e. None of the above Experiment 2: Distillation, Boiling points, Rault’s law 3. 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 4. Consider a 40 mol % mixture of hexane (P°=72 kPa) in toluene (P°=35 kPa). What is the combined vapor pressure? a. 21 kPa b. 28.8 kPa c. 49.8 kPa d. 57.2 kPa e. None of the above 5. 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 6. Use the graph below to answer the question. What does V2 stand for in the graph? a. b. c. d. e. 7. 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 8. Acetone‐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 acetone c. they are immiscible d. they are miscible e. None of the above The liquid composition at L2 The liquid composition of L3 The vapor composition of V1 The liquid composition of V1 None of the above 9. 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 base and remove the water layer containing cyclohexanol. Then add water and acid and remove the water layer containing methyl m‐nitrobenzoate. b. Add acid and remove the water layer containing cyclohexanol. Then add water and base and remove the water layer containing methyl m‐nitrobenzoate. c. Add acid and remove the water layer containing methyl benzoate. Then add water and base 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 containingcyclohexanol. e. None of the above 10. Consider the following equilibrium Given that the pKa for benzoic acid is around 4 and the pKa for protonated aniline is 5, what direction and by how much does the equilibrium lie? a. Right, 1 E‐1 b. Right, 1 E+1 c. Left, 1 E‐1 d. Left, 1 E+1 e. A and D 11. 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: Partition coefficients, Sublimation 12. 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 13. Consider the phase diagram above. The compound is cooled from ‐30 °C to ‐80 °C at 20 atm. What process describes its change in matter a. Melting b. Freezing c. Sublimation d. Boiling e. None of the above 14. How many extractions will it take to recover greater than 95% of a compound with partition coefficient of 5. You are extracting with 4 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: Liquid CO2 Extraction 15. What type of extraction did we use to get the limonene from the orange peel? a. Liquid/liquid extraction b. Liquid extraction c. Solid extraction d. Solid/liquid extraction e. None of the above 16. What must be required on the system in order for solid carbon dioxide to turn into a liquid? a. The temperature should be higher b. The pressure should be higher c. The pressure should be lower d. The volume must be larger e. None of the above Experiment 6: Photoisomerization 17. 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 18. 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 Chromatography 19. 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 20. 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 21. 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 22. 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 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 23. Why could the product spot have been visualized on the TLC as higher than the starting material? a. The methanol in the spotted sample is making it travel further b. The products are more non‐polar than the starting material c. The reaction was not to completion d. Not enough sodium borohydride was used e. None of the above 24. When a reaction proceeds with more of one isomer than the other as its products, the reaction is said to be a. Stereoelectronic b. Stereogenic c. Stereospecific d. Stereoselective e. None of the above Experiment 8 H N O OH EtI K2CO3 O H N O 25. If potassium carbonate is not a strong enough base to fully deprotonate the phenol, how does this reaction even work? a. The loss of hydrogen drives the reaction forward b. The loss of carbon dioxide drives the reaction forward c. The solvent must be assisting somehow d. It is not necessary for the proton to be removed before substitution e. All of the above 26. After the MEK was removed, an extraction using NaOH as the aqueous phase was performed. 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 Spectroscopy Theory 27. 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 28. 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 29. 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 30. 1H nuclei located near electronegative atoms tend to be __________ relative to 1H nuclei which are not. a. b. c. d. e. shielded deshielded resonanced split none of the above 31. How many signals would you expect to see in the 1H NMR spectrum of the following compound? a. b. c. d. e. 6 3 5 4 2 32. What splitting pattern is observed in the proton NMR spectrum for the indicated hydrogens? Singlet doublet triplet quartet septet 33. 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 34. How many different types of carbon are in the following structure? a. b. c. d. e. 3 4 5 6 7 35. 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 Optical Activity 36. The %ee of (S)‐2‐bromobutane was calculated to be 95. Based on this number, and that the theoretical rotation is +15, what was the observed rotation? a. ‐14.25 b. +14.25 c. ‐0.75 d. +0.75 e. None of the above 37. What does the D stand for in the optical rotation? [αD20] a. Deuterium b. Sodium D‐line c. Decomposition d. Diastereomer e. None of the above Infrared Spectroscopy Match the following structures to their Infrared Spectra: a. b. c. d. e. 38. C 39. A 40. E 41. B 42. D 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: 43. s(2H), multiplet (4H) Answer =B 44. multiplet (5H), s(3H) Answer: E 45. s(1H), q(1H),q(2H),d(3H),t(3H) =C 46. q(2H), t(3H) Answer: A 47. t(4H), quintet(4H) Answer: D 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. 48. s(8H) C 49. t(2H),m(1H),s(1H),m(2H),d(6H) E 50. m(4H),qt(2H),t(3H),t(3H) A 51. d(1H),d(1H),m(2H),s(1H),s(3H) B 52. Mass Spectrometry Match the following structures to their Mass Spectra: d(2H),d(2H),s(1H),s(3H) D 53. [C] 54. [E] 55. [A] 56. [D] 57. Combined Spectra Use the combined spectral data to solve the structures for the following problems 58. Which compound matches the following information: 13 [B] Molecular formula: C6H11N C NMR: 78.5 75.9 41.2 H NMR: 2.80, t, J=6.8 Hz, 2H 2.20, t, J=7.1 Hz, 2H 1.78, s, 3H 1 32.5 16.1 4.1 NH 1.63, m, 2H 1.45, broad s, 2H d. NH2 a. H2N b. NH2 HN e. c. 59. IR: 2921 2853 1678 1657 1615 1356 1260 Molecular formula: C8H12O C NMR: 198.3 154.1 135.6 41.0 34.2 30.2 21.3 16.7 d. O 13 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 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 % ee = actual/observed weight % = mass ratio of solute to solution mol % = mol ratio of solute to solution Tables: ...
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This note was uploaded on 05/17/2011 for the course CHEM 6al taught by Professor Pettus during the Spring '08 term at UCSB.
- Spring '08