Final F08 Key

Final F08 Key - Chem 6AL Russak 10 December 2008...

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Unformatted text preview: Chem 6AL Russak 10 December 2008 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 on the back sheet that you may need to calculate your answers. You may use a calculator….not your phone or your iPod/iPhone. You may NOT keep the exam. It will be posted online once grading is finished. Experiment 1 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. What is the mol percent of cinnamic acid in a mixture that is 50 weight percent urea in cinnamic acid? (Urea = 60.06 g/mol, cinnamic acid = 148.16 g/mol) a. 71 % b. 50 % c. 29 % d. 21 % e. None of the above 3. The eutectic point is defined as: a. a suspension of droplets in another liquid b. a ratio of compounds expressing a small melting range c. a solution of two liquids which cannot be separated by distillation d. a mixture of two liquids e. a solution with a precise boiling point 4. 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 5. Colligative properties are those properties that depend on a. The mass of the particles b. The quantity of particles 6. 7. 8. 9. c. The melting point of particles d. The boiling point of particles e. None of the above How much volume of water added to 58.4 g NaCl will make a 1.0 M solution. a. 2 L b. 100 mL c. 10 mL d. 1000 mL e. None of the above 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 The conventional units for mass are Less than 10 percent of the class answered this question correctly. Most said b, probably because those are the units we used in lab…but please be aware that these are not the S.I. or conventional units for mass. You received credit for answer b or c. a. Kg/L b. g c. Kg d. lb e. None of the above 2.4 mL acetyl chloride (MW=78.5 g/mol, d=1.11 g/mL) is allowed to react with 1.0 mL ethanol (MW=46.06 g/mol, d=0.789 g/mL) to produce ethyl acetate (MW=88.10 g/mol, d=0.897 g/mL). What is the limiting reagent? Most people chose B here...I got 0.0275 mol for each one, which the answer would be D, however, what did you guys do? Carry it out 4 decimal places? Pretty hard if you put either you got it right. a. Acetyl chloride b. Ethanol c. Ethyl acetate d. Both ethanol and acetyl chloride are in equal amounts so they are both limiting e. None of the above 10. Which of the following is NOT a main route for chemicals to enter the body I threw this one out because it did not get a good response. The book explains that the eyes are not a main route for chemicals. The circulation to the eye is very minimal. In fact, oxygen gets to the eye through the atmosphere and not the circulatory system. a. Eyes b. Mouth c. Lungs d. Skin e. None of the above Experiment 2 11. Suppose you have a mixture of 9 moles of methanol and 4 moles of ethanol at a particular temperature. The vapor pressure of pure methanol at this temperature is 81 kPa, and the vapor pressure of pure ethanol is 45 kPa. What is the mole fraction of ethanol in solution? a. 2.25 b. 0.12 c. 0.18 d. 0.31 e. 0.44 12. Which of the following equations is Dalton’s Law? a. PTotal = PA + PB b. P = XAPA + XBPB c. P = nRT/V d. P = XBPA + XAPB e. pH = pKa + log(B/A) 13. 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 14. A chemist performs two successive simple distillations on an 88/12 mol % solution of hexane/pentane. What approximately is the mol % of hexane in the distillation? (ignore L1, L2, L3, V1, V2 in the graph) a. b. c. d. e. <1 % 15 % 45 % 86 % 95% 15. How could one obtain more theoretical plates when performing a fractional distillation? a. add more boiling chips b. add more solvent c. heat the mixture rapidly d. add more surface area to the condenser e. shorten the condenser 16. The chemical shift of 0 ppm (d) is arbitrarily assigned to a. Chloroform b. Benzene c. Trimethylsilane d. Methylene chloride e. Tetramethylsilane 17. Extended conjugation has which effect on a carbonyl stretch? a. Decreases intensity b. Increases intensity c. Decreases wavenumber d. Increases wavenumer e. None of the above 18. A relatively polar solvent often used for normal phase TLC: a. Hexanes b. Water c. Petroleum ether d. Ethyl Acetate e. Dioxane 19. The vapor pressure of Compound A is 100 kPa and for Compound B it is 200 kPa. a. Compound A is more volatile than Compound B b. Compound B is more volatile than Compound A c. Compound A has a higher melting point than Compound B d. Compound B has a higher melting point than Compound A e. A and C 20. 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 Experiment 3 21. The density of water is 1.0 g/mL; the density of methylene chloride is 1.326 g/mL; the density of diethyl ether is 0.713 g/mL; and the density of ethanol is 0.789 g/mL. Only one of the following statements is correct. a. when mixed with water, diethyl ether is the bottom layer b. when mixed with water, methylene chloride is the bottom layer c. when mixed with water, ethanol will be the top layer d. when mixed with water, ethanol will be the bottom layer e. methylene chloride and water are miscible. 22. 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. ethanol is too toxic d. they are miscible e. they are immiscible 23. Compound A has a partition coefficient (K) of 4.0. You dissolve 1.0 g of Compound A in 15 mL of water. How many grams will be in the aqueous layer after one extraction with 15 mL of ether? a. 0.53 g b. 0.20 g c. 0.10 g d. 0.15 g e. 0.30 g 24. Given that the partition coefficient value for a compound in dichloromethane and water is 4.0, how many extractions would it take to get over 99% of it back assuming you used equal volumes of solvent? a. 1 b. 2 c. 3 d. 4 e. 5 25. Recrystallization assumes that: a. The impurities are more soluble that the product b. The solvent used must dissolve the compound at room temperature c. The melting ensures adequate separation of the impurities from the product d. Solvents with similar functionalities are required e. All of the above 26. Which of the following bonds would show the strongest absorption in the IR? a. Acid chloride carbonyl b. C=C stretch c. C‐O stretch d. Aldehyde carbonyl e. A and C 27. 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 28. What would be the best way to report the following: A sample of urea (25 g, 60.06 g/mol) was recrystallized from hexanes/ethyl acetate to give 18.6 g of pure urea. a. % yield b. Theoretical yield c. % recovery d. % conversion e. None of the above 29. Considering what you know about normal phase TLC, what could you assume about reverse phase? a. Compounds elute in opposite order b. Polar compounds tend to be much less retained than non‐polar ones c. The form of silica is non‐polar d. Different solvent systems would need to be used e. All of the above Experiment 4 30. Caffeine belongs to which family of molecules? a. Taxol b. Caffeinoids c. Serines d. Purines e. Coffee 31. The solid/liquid extraction performed in the caffeine experiment (lab 4) was the direct mimic of: a. A dean‐stark extractor b. A Soxhlet extractor c. Centrifugation 32. 33. 34. 35. d. The phase change of melting e. None of the above If caffeine is mixed with water, what is the equilibrium constant assuming protonated caffeine has a pKa of 14?Everyone gets this right. I messed up the answer choices. Disregard this one. a. For every 1 molecule of water, there are 100‐200 molecules of caffeine b. For every 1 molecule of protonated caffeine, there are 2 molecules of water c. The equilibrium lies to the left by 20 times d. The equilibrium lies to the right by 20 times e. None of the above If acidic tannins are present in tea leaves, what would be the best extraction method to remove them?I gave everyone this one because almost all of you chose D. Base (Na2CO3) would ensure DE‐protonated caffeine stays in the organic layer. If caffeine was protonated then it would go into the water layer because it would be positively charged (salt). a. Add HCl so that the protonated caffeine would stay in the water and the tannins would remain in the organic layer b. Add HCl so that the protonated caffeine would stay in the organic layer and the tannins would remain in the water layer c. Add Na2CO3 so that the protonated caffeine would stay in the water and the tannins would remain in the organic layer d. Add Na2CO3 so that the protonated caffeine would stay in the organic layer and the tannins would remain in the water layer e. None of the above Sublimation relies on what principle? a. That the compound does not decompose b. That the compound goes from a liquid to a gas c. That lowering the pressure increases the temperature of sublimation d. That all impurities sublime as well e. All of the above The K value for extraction was very poor in experiment 4. Considering what you know now about this, what could be a possible solution? a. Increase the volumes of each layer when performing extractions b. Allow the centrifuge to spin longer c. Find another solvent with a better K value d. Steep the tea bag until dry e. All of the above are possible solutions Experiment 5 36. 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 37. When two liquids are miscible with each other, the solution is said to be a. Extracted b. Homogeneous c. Heterogeneous d. Intractable e. Eutectic 38. Consider the reaction scheme of Experiment 5 Notice that this reaction is in equilibrium. It was not necessarily proven in lab for the reaction to progress beyond halfway, but nonetheless it does go nearly all the way when subjected to longer times under the lamp or in the sun (gonna have to trust me). Since the reaction is product favored, something must be driving it towards that product. A possible explanation is: The cis product is very good at absorbing light The crystallization of the product allows for protection to reconversion The solubility of the cis product is much greater in the solution than the trans The cis product is more stable chemically than the trans None of the above 39. The solvent used for this conversion was an ethanol/water solution. This was important because: Everyone gets credit. Perhaps too difficult a question to ask you. The solvent mixture was chosen for crystallization purposes. This is a solvent that dissolves the starting material well, but not the product (the cis crashes out of solution). a. Polar solvent accelerated the reaction stabilizing polar intermediates b. Strongly hydrogen bonding solvents were necessary to stabilize intermediates c. The solvent was assisting in the breaking of the pi bond allowing for rotation d. All of the above e. None of the above 40. The mother liquor of a recrystallization refers to a. The solvent system used b. The crystals collected c. The solution of the crystals and the solvent d. The filtrate collected e. The solid collected from filtration Experiment 6 and Gas Chromatography a. b. c. d. e. 41. 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 42. 43. 44. 45. 46. 47. Isopropanol 2‐methyl propene Isopropene 2‐methyl butene None of the above The mechanism of action for this transformation is solely dependent upon the starting material. This means that the process was a. SN1 b. SN2 c. E1 d. E2 e. None of the above 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 The major component to packed columns in GC is silica gel. This is because a. The inherent insolubility in many solvents b. The high surface area c. The ability to absorb water d. All of the above e. None of the above Which of the following is not a type of GC column: a. Chiralcel OD‐H b. WCOT c. SCOT d. Diatomaceous earth packed e. All are types of GC columns The products in the above reaction from experiment 6 eluted in order of a. Polarity b. Boiling point c. Unsaturation d. B and C e. All of the above Consider the following diagram a. b. c. d. e. The gas delivery tube was positioned under the water. What could happen if the reaction cooled and the gas delivery tube was still submerged?Everyone gets credit. There was some trickery in this, but nonetheless fundamental. Water added to concentrated acid = highly exothermic reaction. This situation was warned to you in your experiment handout. D cannot be the right answer because the acid was not “dilute.” There would not be a violent exotherm should water and dilute acid come into contact. a. Absolutely nothing b. The product could be sucked back into the vial causing it to convert back to starting material c. The water could suck back in the tube and explode upon contact with the acid d. An explosion could occur due to violent reaction of water with dilute acid e. C and D 48. Polysiloxanes and polyethylene glycol are examples of a. Liquid solid phases for Gas Chromatography b. Extremely non‐polar solvents used for liquid chromatography c. Dessicants (reagents that remove water) d. All of the above 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 49. 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 50. The starting material for experiment 7 is uniquely absent of indicated chirality. This is because a. The t‐butyl locks the ring in one conformer b. The starting material is racemic 51. 52. 53. 54. c. The structure is symmetrical and therefore does not have chirality d. All of the above e. None of the above 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 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 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 Sodium methoxide (NaOMe) was also added to this reaction. This is because a. The reaction should be performed under basic conditions b. The reaction should be performed under acidic conditions because H‐ is a strong acid c. It would remove any excess water that could interfere with the NaBH4 d. A and C e. B and C Experiment 8 H N O OH EtI K2CO3 O H N O 55. The potassium carbonate used in the experiment ensures a. The reaction is basic so Iodide can attack effectively b. Alleviates the need for solvent in the reaction c. Allows generation of an ethoxide ion d. Ensures phenoxide formation e. None of the above 56. Which compound is known as Tylenol? The intermediate formed in the reaction The starting material The product Phenacetin None of the above 57. The reaction above is missing the solvent, methyl ethyl ketone a. True b. False 58. To ensure no starting material was isolated along with the final product a. A TLC was conducted to ensure completion b. A brine wash was conducted to remove any water c. Any phenoxides were regenerated with basic sodium hydroxide and extracted d. Excess starting material was used e. None of the above 59. 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 Experiment 9 NaBr OH H2SO4 Br a. b. c. d. e. 60. The reaction is said to be unusual because a. Bromide rarely acts as an electrophile in these situations b. Acid concentrated reactions of alcohols in the presence of bromide usually lead to elimination products c. SN1 mechanisms should be favored d. All of the above e. None of the above 61. Water is added to the above reaction. This is a. To dilute the concentrated acid b. To ensure protonation of the alcohol c. To allow for greater solubility of the sodium bromide d. All of the above e. None of the above 62. What is the purpose of reflux? a. To boil the solvent until full evaporation b. To maintain the integrity of the concentration of the reaction c. To limit the loss of solvent through evaporation d. B and C e. None of the above Thin Layer Chromatography 63. Thin Layer Chromatography (TLC) can give valuable information regarding a. The progress of the reaction b. The quantity of starting material vs. product c. The relative polarities of the compounds d. A and C e. A, B, and C 64. 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 65. Regarding the TLC from question 7, 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 66. Consider the two TLC plates shown below (run on silica gel). Both plates were used to analyze the same mixture. The plate on the left was run using acetone as the solvent. Which solvent do you think was used to run the plate on the right? a. toluene b. diethyl ether c. hexane d. ethanol e. any of the above 67. The alkaloids listed on the table were analyzed by TLC on silica gel using three different solvents. The Rf are given below: If you were running a forensic laboratory, which solvent would you choose to analyze a sample seized by the police in a drug smuggling case and suspected to contain morphine, cocaine and aconitine?Not a good response for this question from students, so I threw it out. The purpose of the question was for you to find a developing solvent (1,2 or 3) that gave the best separation of each compound to be analyzed. Solvent 3 gives markedly differing Rf values, whereas solvents 1 and 2 do not hardly give separation at all. a. solvent 1 b. solvent 2 c. solvent 3 d. either solvent 1 or solvent 2 e. none of the above are useful solvent systems for detection Infrared Spectroscopy 68. Wavenumber is represented as a. v° 1/ʋ 1/λ 1/K None of the above 69. Stretching vibrations of covalent bonds takes more energy than bending vibrations a. True b. False b. c. d. e. 70. Presence of OH stretching is concentration dependent in the IR a. True b. False 71. Conjugation, in general, lowers the wavenumber in the IR by about a. 5 b. 10 c. 15 d. 20 e. None of the above Match the following structures to their IR spectra 72. D 73. B 74. C 75. E 76. A 1H NMR Spectroscopy 77. Which compound matches the spectra? (integration and splitting given) I will accept answer b or d here. This is because you weren’t given the signal for a formate hydrogen as in answer b. C was not an option because the quartet next to an oxygen atom should be closer to 4 ppm. 1s 3t 2d 2d 2q 10 8 6 PPM 4 2 0 O O H a. H Br c. O Br O O H O H b. d. 78. The integration of a signal refers to a. The number of hydrogens on the adjacent carbon b. The chemical shift of the signal c. The number of protons the signal corresponds to d. All of the above e. None of the above 79. What allows for the detection of a proton in a molecule?Everyone was split on this question, which means that 20% chose answer a, 20% chose b, etc…which means it is a bad question. Everyone gets credit. FYI, it is the angular momentum of the precession of a nucleus that gives rise to spin. a. The presence of even numbers of protons, neutrons, and electrons b. Precession c. A north pole and a south pole d. Isotopes e. None of the above 80. In comparison, A hydrogen atom of an aldehyde is _______ to a hydrogen atom on an aromatic ring with respect to chemical shift. a. Upfield, Deshielded b. Downfield, Shielded c. Electron poor, upfield d. Electron rich, downfield e. None of the above Below is the 1H‐NMR spectrum of n‐butyl propionate. Assign the peaks (capital A‐D) in the spectrum to the protons (lowercase a‐e). Integration is given. 81. 82. 83. 84. Assign the peak A in the spectrum to the protons (a‐e).A Assign the peak B in the spectrum to the protons (a‐e).C Assign the peak C in the spectrum to the protons (a‐e).D Assign the peak D in the spectrum to the protons (a‐e).B 85. What is the observed multiplicity of HA in the compound below which was not made in lab? Ha O singlet doublet quartet multiplet broad singlet a. b. c. d. e. 86. Considering J value of 18 Hz in the 1H‐NMR, which is the proper substitution? Cl Cl a. Br Cl Cl c. Br b. Br d. Br Cl C13 NMR Spectroscopy 87. Which carbon appears most downfield in this compound? D a b c O d O O 88. How many unequivalent carbon signals would you expect for a. 6 b. 7 c. 8 d. 9 89. Which compound matches the spectra? ? O OH a. b. c. d. OH 90. How many 13C‐NMR chemical peaks would you predict for cyclohexane a. 6 b. 5 c. 4 d. 3 91. 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 e. c. 92. 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 O b. c. O e. O Mass Spectrometry 93. Which of the following best explains why cyclohexanone does not undergo a McLafferty Rearrangement. a. It is cyclic and it cannot align in the required 6‐membered transition state b. It is a ketone, which is a functional group that does not undergo McLafferty Rearrangements c. Cyclohexanone does not have the required gamma hydrogens d. Cyclohexanone is much too stable to fragment. e. Cyclohexanone does undergo a McLafferty Rearrangement 94. Which compound has a molecular ion at m/z = 58, an IR absorption at 1715 cm‐1 and just one singlet in its 1H NMR spectrum? a. butane b. CH3COCH3 c. CH3CH2CHO d. 2‐methylpropane e. CH3CH2CH2OCH3 Match the Mass Spectra with the following compounds: 95. D 96. B 97. E 98. C 99. A Equations: N=c/ʋ E=hc/λ P=P°X P=PA+PB 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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This note was uploaded on 05/17/2011 for the course CHEM 6al taught by Professor Pettus during the Spring '08 term at UCSB.

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