CHEMISTRY 5309 Test Number 2 Fall, 2003 ANSWER KEY

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1 EMISTRY 5309 Test Number 2 Fall, 2003 ANSWER KEY 1. Explain briefly why the dipole moment of 2,3-dichloro-2,3-dimethylbutane changes with temperature, as shown in the table below. (The solvent is heptane.) (10 points) Dipole Moment (m, D) Temperature ( ) m > 0 m = There is an equilibrium between these two conformations. Since the dipole moment increases with temperature, there must be more of the one with the gauche chlorines at higher temperature. This means that that conformation is the higher energy one. 2. List the following in order of stability. (10 points) Anomeric effect Most Stable B A B A D Least Stable D 3. Given the following mechanism for the formation of a product from A and B. A B K k 2 D A E E D k 3 Product Derive the rate expression for the formation of the product ( d[product] dt ) using the steady state approximation for [E] in terms of [A], [B] and. (13 points) rate = d[product] dt = k 3 [E] Steady state for E k 2 [][A] = [E] [E] k [E] = 2 [A][] rate = k k [A][] 2 3 K = [] [] = K [A][B] [A][B]

2 rate = k k 2 3 K[A]2 [B] ( ) = k k 2 3 K[A]2 [B] 4. Explain the following trend in activation entropies (DS ), including why in 2 DS is positive while in 90% acetone it is negative. The reaction is the solvolysis (S N 1) of tertbutyl chloride. (10 points) ( 3 ) 3 ( 3 ) 3 products Solvent DS e.u. 5 e.u. 90% acetone 10% 2 17 e.u. Water is a good solvent so you need less solvent reorganization to solvate the polar transition state (this is a negative DS ) d d DS is positive without solvent. As the solvent gets less polar more solvent reorganization is required to solvate the polar transition state so there is a larger negative contribution to DS which becomes dominant giving rise to a negative DS. 5. Explain briefly why when the solvent is changed from pure water to DMS the ammett r value for benzoic acid ionization goes from 1.00 to 2.6. (10 points) R K A R The anion is not solvated very well in DMS 3 S 3 compared to 2 so the R group has to do more toward stabilizing the benzoate anion than when the solvent is water. 6. The reaction of ethylene oxide in water to produce ethylene glycol is believed to follow the following mechanism: (14 points) 2 2 K S N k

3 If the reaction is studied in buffered aqueous solution, how would you expect the observed pseudo-first order rate constants to vary with p? The pseudo-first order rate constant is the rate constant (for the reaction which is first order in ethylene oxide) at a given p. rate = k [ ] K = [ ] rate = kk [ ][ ] rate = kk K W = kkk W This is K W So rate is independent of p. 7. Given that k / k D for the free radical chlorination of toluene (rate determining step given below) is 1.3 (at 77 ) and that the Ph 2 bond strength is 85 kcal/mol and the bond strength is 103 kcal/mol. Also, k / k D for the free radical chlorination of methane (rate determining step given below) is 7.1 (at 71 ). Discuss the significance of these kinetic isotope effects and whether these data are consistent with the bond strength in methane being 102 kcal/mol. Use energy profile diagrams to illustrate your answer. (14 points) Ph 2 Ph 2 D 3 3 D Ph 2 Ph 2 D 3 3 D Since k /k D is a maximum the transition state for methane is symmetric. d d For toluene the transition state is unsymmetric. The bond strength in methane (102 kcal/mol) is about the same as the bond strength (103 kcal/mol) so the transition state can easily be symmetric. The bond strength in toluene (85 kcal/mol) is considerably smaller than that of so an unsymmetric transition state is reasonable. According to the ammond postulate, since this is an exothermic reaction the transition state will look like the reactant so the transition state for this reaction will look like: d d Symmetric trans. state Trans. state looks like reactants Ph D = 1 kcal/mol Exothermic Ph 2 D = 18 kcal/mol Extent of reaction Extent of reaction

4 8. Which of the following pairs of reactions will occur er? ircle the Roman number (I or. (8 points) a) LiAl 4 3 LiAl 4 3 I ydride is small, axial attack II b) 3 2 Br 2 3 Br 2 Br 2 Br 2 Br 2 Br vs 3 Axial attack favors II, less hindrance. 9. Suggest an experiment, or experiments, which would distinguish between Mechanism I and Mechanism II for the aisen rearrangement: (15 points) Be sure to explain how your experiment(s) would rule out one mechanism and would be consistent with the other.

5 Label a carbon atom in the allyl group (but not the middle carbon). The red dot is a carbon-13 ( 13 ). Mechanism I This is the only product. Mechanism II The anion can attack the allyl cation at either end. ere you will get two products. 13 NMR spectroscopy will show the label present at either one carbon (Mechanism or at two carbons (Mechanism.