CHM 171Q - Final - 2004

CHM 171Q - Final - 2004 - Name

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Unformatted text preview: Name _______________________________________________________________ Chemistry 171Q, Final Exam December 18, 2004 8:30 - 11:30 a.m. There are eight (8) questions to this exam. Before you begin, please make sure that your exam copy contains all eight. If not, please notify the instructor immediately. Note that the last four pages are deliberately left blank for use as scratch paper. You may carefully remove them to use as scratch paper. Good Luck!!! Problem 1. 2. 3. 4. 5. 6. 7. 8. TOTAL Score /40 points /20 points /45 points /30 points /20 points /15 points /65 points /45 points ____________________ /280 points -1- 1. (40 Points) Illustrated below are similar reactions of similar pairs of compounds. In each case, circle the one that reacts more rapidly. Then provide a brief explanation for the difference in rate. Br CH3OH OCH3 (a) Br CH3OH CH3O (b) + HCl -40 Et2O -40 Et2O H Cl Cl H + HCl -2- (c) CH3 Cl Me + NaSCH3 CH3 O S Me SCH3 + Cl NaSCH3 Me O S Me SCH3 (d) (CH3)3C (CH3)3C H H + + I Cl (CH3)3C (CH3)3C + + H-I H-Cl -3- (2) (20 points) Alcohols typically show broad characteristic infrared (IR) absorptions at 3400 cm-1 due to the O-H stretching vibration. However, if the OH group is intramolecularly hydrogen bonded to a Lewis basic group within the molecule, the IR absorption is typically shifted to slightly higher energy. The IR spectra of the two diastereoisomers of butane-2,3-diol both show two IR absorptions in the OH stretching region. Interestingly, the intensity of IR resonance assigned to the intramolecular, hydrogen-bonded OH group is significantly larger in one of the stereoisomers, indicating a greater fraction of the isomer exists in the hydrogen-bonded state. (a) Using Newman projections, draw the most stable conformers of the two diastereoisomers of butane-2,3-diol that would be expected to show an intramolecular, hydrogen bonded OH group. It may be helpful to show the hydrogen bond in each structure with a dashed line (i.e. O-----H). 1 (b) 2 Which of the two isomeric conformers drawn above (1 or 2) would you expect to be more stable? (Hint: A CC/CC gauche interaction is more destabilizing than a CC/CO gauche interaction.) Explain your reasoning. (c) Which diastereoisomer of butane-2,3-diol (1 or 2) would you expect to show the greater fraction of the IR resonance due to the intramolecular, hydrogen-bonded OH group? Briefly explain your reasoning. -4- 3. (45 points) Complete each of the following: (a) (10 points) Free radical bromination of 1,1-diphenylpropene (3) or 3,3-diphenyl-1-propene (4) with N-bromosuccinimide (NBS) in CCl4 give identical ratios of monobromination products. Ph Ph 3 Ph Ph 4 Clearly explain why 3 and 4 give identical product ratios with NBS. The enthalpies of formation of 3 and 4 are 52 and 62 kcal/mol, respectively. If both isomers were present in solution at equal concentrations, which would you expect to react more rapidly with NBS? Explain your reasoning. -5- (b) (5 points) The reaction of n-chlorobutane with sodium hydroxide to give n-butanol in aqueous THF is found to be significantly accelerated by the addition of a small amount of sodium iodide. Careful analysis shows that sodium iodide is not consumed in the reaction, therefore, it acts as a reaction catalyst. n-C4H9Cl + NaOH THF H2O n-C4H9OH + NaCl Provide an explanation for how sodium iodide catalyzes the conversion of n-chlorobutane to n-butanol. (c) (30 points) For each of the three following interconversions, circle the compounds that would predominate at equilibrium. Then provide a brief explanation for your choice. CH3OH + HNEt3 K CH3OH2 + NEt3 -6- CH3Br + OH K CH3OH + Br CH3O + O K CH3OCH2CH2O -7- 4. (30 points) Supply the missing starting material, reagent(s), or product(s) for each of the following reactions. Be careful to show the appropriate stereochemistry where relevant. (a) CH3CH2 C C CH3 Na NH3 (b) O (c) CH3 CH2OH HCl (d) H (S)-pent-4-en-2-ol Cl -8- (e) 1) Hg(OAc)2 H2O, THF CH3 CH3 2) NaBH4, NaOH (f) Br Br2 CH3OH CH3O D CH3 CH2CH3 H 5. (20 points) As discussed in class, one procedure for the conversion of alkyl halides (R-X) to hydrocarbons (R-H) is to transform the halide into a Grignard reagent and then treat it with water. This procedure is not useful, however, if the starting halide contains substituents that react with Grignard reagents. A useful alternative procedure is to react the halide with tri-n-butyl tin hydride (HSnBu3 ), e.g. Br + H-SnBu3 ButOOBut H + Br-SnBu3 Bond Dissociation Energies (kcal/mol) Br 71 Bu3Sn H 74 ButOO ButO H 100 Bu3Sn Br 85 ButO ButO But OBut H Br 66 38 105 55 -9- (a) Estimate H for the reaction of bromocyclohexane with HSnBu3 . Show how you arrived at your estimate. The direct reaction of cyclohexane with HSnBu3 proceeds very slowly. The reaction is greatly accelerated, however, by the addition of small amounts of di-tert-butyl peroxide (ButO-OBut). The di-tert-butyl peroxide is ultimately converted to tert-butanol during the reaction. Mechanistic studies have demonstrated that the reaction proceeds via a free radical chain mechanism. (b) Write a plausible mechanism (with single-barbed arrows to show electron movement) for the reaction of bromocyclohexane with HSnBu3 . Include all relevant initiation and propagation steps. Label the initiation steps as I1, I2, I3, ... and the propagation steps as P1, P2, P3, ... (You need not show termination steps.) - 10 - 6. (15 points) The reaction of cyclohexene with HCl in acetic acid as solvent leads to the formation of two products, chlorocyclohexane (5) and cyclohexyl acetate (6). HCl CH3CO2H 5 (25%) Cl + 6 (75%) OAc (a) It is found that the product percentages change when tetramethylammonium chloride (TMAC) is added to the solution (note: TMAC dissolves in CH3 CO2 H). For example, when [TMAC] = 0.1 M, the reaction produces 50% 5 and 50% 6. When [TMAC] = 0.9 M, the 5:6 product ratio is 80:20. Control experiments show that 5 and 6 are stable under the reaction conditions. Explain how adding TMAC changes the product ratio. (b) What dependency, if any, would you expect on the product ratio if [HCl] was changed? (note: Ka for HCl in CH3 CO2 H is 3 x 10-9 M) Explain your answer. - 11 - (c) The reaction with cyclohexene has also been run with DCl in CH3 CO2 D. When the reaction was stopped at partial conversion and the cyclohexene reactant recovered, it was found the cyclohexene contained no deuterium. What is the mechanistic implication of this observation? 7. (65 points) Specify the reagents that would be needed to carry out each of the following synthetic transformations. You may use any organic building block with 3 carbons for your syntheses. However, the reagents that you use that are not incorporated into the products may contain more than 3 carbon atoms. The structure of the reactant and the product should be shown for each chemical transformation. Retrosynthetic analysis will be considered for partial credit. (a) H OH CH3C CH H3C HO H OH - 12 - (b) CH3 O CH3 (c) HC CH CH3 OH HO CH3 - 13 - (d) OH OH - 14 - 8. (a) (45 points) Compound A and B have formula C5 H10. A and B react with H2 /Pt to give a single compound, C5 H12. A reacts with BH3 THF followed by treatment with a basic solution of hydrogen peroxide to give C (C5 H12O). B reacts with BH3 THF followed by treatment with a basic solution of hydrogen peroxide to give D (C5 H12O). When C or D is treated with chromic acid the solutions turn blue-green. Preparative reaction of C with chromic acid gives E (C5 H10O2 ). E has a broad IR absorption at 3000 cm-1 and a strong, sharp absorption at 1710 cm-1 , and the following 1 H NMR spectrum: d 11.9 (1H, singlet), 2.23 (2H, doublet), 2.12 (1H, multiplet), and 0.99 (6H, doublet). Based on these observations, propose structures for A-D. Your observation/deduction reasoning will be considered for partial credit. Observation Deduction A fl B fl C fl D fl E fl - 15 - (b) Compound F has formula C6 H12 and an IR absorption at 1640 cm-1. Reaction of F with either Hg(OAc)2 /aqueous THF followed by NaBH4 , or with BH3 THF followed by treatment with a basic solution of hydrogen peroxide, gives the same compound G, C6 H14O. Reaction of F with H2 O2 in the presence of a catalytic amount of OsO4 gives H, C6 H14O2 , which can not be resolved into a pair of enantiomers. Reaction of F with O3 followed by reduction with (CH3 )2 S gives only one product I, C3 H6 O. Based on these observations, propose structures for F-I. Your observation/deduction reasoning will be considered for partial credit. Observation Deduction F fl G fl H fl I fl - 16 - - 17 - - 18 - - 19 - - 20 - ...
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