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Page 1 EXAM ONE CHEMISTRY 0310 PROF. WILCOX October 4, 2012 NAME ________________________________ PeopleSoft Number_______________________ Signature_______________________________ Instructions. 1. Write your name above and or initials on the second and last pages of the exam. 2. Read all questions carefully. There are questions and there are pages to this exam. Check to be sure that you have all pages. 3. Check the blackboard at the front from time to time to see any "late bulletins" about the exam. Stay calm and stay in your seat. Raise your hand if you have a question. Ask for help if you feel there is an error in the exam, or if a question is unclear. 1.________/12 8.________/8 2.________/8 9.________/10 3.________/4 10.________/5 4.________/4 11._______/5 5.________/6 12._______/8 6.________/8 13._______/8 7._______/8 14._______/6 TOTAL: _________/100Page 2 1. Use the table of bond dissociation energies to calculate the .H (enthalpy of reaction) for the following processes. Show your work. In c one � �of the needed bond energies is provided in the figure. CH3 Br CH2 H + Br2 H + H-Br CH3 CH3(a) CH3 CH3 101 46 -70 -87 Total: 146-157 = -11 kcal/mol H + H-OH + �OH (b) 98.5 -119 Total 98.5-119 = -20.5 89 kcal/mol CH3 HH H CH3 CH3 CH3(c) CH3�HH Break C-CH3 88 kcal/mol Break Primary C-H 101 kcal/mol Total +1.5 kcal/mol Make C-CH3 -89 kcal/mol Make Secondary C-H -98.5 kcalPage 3 2. For each pair of structures shown, indicate whether the two species are Enantiomers, Diastereomers, or Identical molecules. You may use the abbreviations E, D, or I, respectively. OH OH CH3CH3a) E H Br HBr b) HBr HBr I CH3 F HHF H3C Cl c) Cl E CH3 F HH d) F Cl Cl CH3 I 3. Circle each chiral molecule. Put a star (*) next to each stereocenter and label it as R or S. H3CCHCl2CCl3CH2BrHOHHHHClH3CCHCl2CCl3CH2ClHOHHHH3CClSSR***Page 4 4. Show how you would calculate the equilibrium constant for the following process. Include the appropriate data (numbers) required for the calculation, and set up the equation. HBr + H2O Br-+ H3O+ pKa -4.7 -1.7 Keq = 103 (difference in pKa = 3; equilibriumfavors weak acid side.) 5. Use Newman projections to illustrate three different staggered Cl conformations of 2-chlorobutane, , as seen looking down the C2-C3 bond. Cl Cl Cl CH3H3C CH3 CH3 CH3CH3 6. Use the table of Change in Free Energy on Flipping from the �CyclohexaneConformer with the Indicated Substituent Equatorial to the Conformer with the SubstituentAxial (page 2) to calculate the free energy change in each �of the following reactions. Show your work. a) H3C +1.7 kcal/mol CH3 Cl H3C b) Cl +1.7 for CH3-0.52 for Cl Total: +1.18 kcal/mol CH3Page 5 7. Redraw the following molecules as Fischer projections. Br H C2H5H H H Br H3CF Cl F C2H5 Br HH3C Cl CH3 Br H CH3 8. For each of the following sentences, circle the word or item that best completes atrue statement. a) When comparing two acids, an acid with the (higher/lower) pKa is the stronger acid. b) Bond angles around sp2 hybridized atoms will be close to (90, 109, 120) degrees. c) It is (true/false) that water (H2O) can act as a base. d) The transition state for the first propagation step in radical bromination is (earlier/later) than the corresponding step in chlorination. e) A molecule that has a plane of symmetry is (chiral/not chiral). 9. Give a correct equation that describes the relationship between the standard freeenergy change in a reaction (DG ) and the reaction equilibrium constant (Keq). G/RT) .Keq = e-(.G = -� �RTlnKeqPage 6 10. Give the mechanism for the free radical chlorination of methane. Showinitiation, propagation, and at least one termination step. CH4 + Cl2 CH3Cl + HCl Cl + Cl Initiation Cl2 Two propagation steps CH4 + Cl CH3 + HCl ClCH3 + � � � �Cl CH3 + Cl2 Examples of termination Cl + Cl Cl2 ClCH3 CH3 + Cl 11. Fill in� � � � � �the blank: Show your work. Set up the equations no need to use a calculator. �a. If the specific rotation of a mixture of enantiomers is +12.0, and the pure R isomer has a specific rotation of 20.0, the optical purity (also called � � �enantiomeric excess) of the mixture is___60%___, and the __S___(R or S?) isomer is the major component. b. A solution of substance C, with a concentration of 0.2 g/mL, in a polarimeter with a path length of 10 cm, rotates 589 nm polarizedlight -10.0 degreesclockwise. The specific rotation [a] of C is ___-50____.Page 7 i. 12. Monochlorination of isobutane (2-methylbutane) affords fourdifferent products (ignore stereochemistry). The relative reactivity of primary, secondary, and tertiary C-H bonds toward achlorine radical is 1:4:5 (1 :2 :3 ) Write the four possible products of monochlorination of 2-� � �methylbutane. Cl ii. A B C D Calculate the relative amount (ratio) of the four products expected in this chlorination. Show your work. No calculator is needed.Whole number ratios are fine. (No need to convert to percentages.) If you cannotdo thenecessary math by hand, set up the equations with all necessary numbersincluded. ClClClA31 B24 C15 = = = D61 D61 D61 � � �Page 8 CHALLENGE 13) The heats of combustion of the following molecules are different. The difference is 0.8 kcal/mol. 1 2 a) Which molecule has the larger heat of combustion? NUMBER 1 b) Briefly explain why the difference is 0.8 kcal/mol. A correct answer can be based on your knowledge of cyclohexane derivatives and butane conformations. The axial methyl, when all alone on the ring and not near another group, will be 1.7 kcal lessstable than the equatorial. In this case, in 2 the equatorial methyl is not alone . It is � � � �gauche to the other methyl group that is always axial. This gauche interaction causes strain in the 2 . Not as much strain as in 1 but significant. Gauche � � � �methyl-methyl interactions give 0.9 kcal/mole of strain. (Butane number) So there is 1.7 kcal/mol strain in 1 and 0.9 kcal/mol strain in 2 . The � � � �difference is 0.8


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Pitt CHEM 0310 - EXAM ONE

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