1 Chemistry 5011/8011 Monday, November 14, 2005 Midterm Exam 2 Please do not open or sign this packet until you are instructed to do so. Please write all of your answers for this exam in this exam packet. Although you may use as many blue books for scratch work as you would like, the blue books will not be collected at the end of the exam or graded. Answer each question in the space provided if you can, but feel free to continue your answer on the back of the page if you need more room. (Please write a note by your answer pointing us to the continuation if you do this.) You will be given 2 hours total to finish the exam. This exam contains two problems, which are split into parts. Do not get stuck on one part and then assume that you will be unable to answer the rest of the question—move on. In addition, partial credit will be given for incorrect but plausible or consistent answers, so guess on problems you cannot answer perfectly. At the end of the 2-hour exam period you will be asked to return your exam to the proctor. (You may, of course, also turn the packet in earlier if you choose.) This exam is open-resource—you may use any books, notes, calculator, etc. you have brought with you to the exam. However, you are not allowed to communicate with anyone during the exam, or to bring any materials in or out of the room while you are taking the exam. You are also not allowed to use any devices that could be used to communicate with anyone (laptop computers, cellphones, etc.). Please do not take any part of the exam packet with you when you are done; everything will be returned to you after the exams are graded. This packet should contain 12 pages, including this one. Please check to make sure that your packet contains 12 pages before beginning your exam. Name: __________________________________________________ Signature: __________________________________________________ Helpful constants to know for this exam: Boltzmann’s constant: kB = 2.94 × 10-24 cal K-1 = 1.38 × 10-23 J K-1 Gas constant: R = 1.99 cal K-1 mol-1 = 8.314 J K-1 mol-1 Planck’s constant: h = 1.58 × 10-34 cal sec = 6.626 × 10-34 J sec e = 2.7182 1. Driver and Woerpel found that the dialkylsilyl fragment of silacyclopropane 1 spontaneously migrated to allylbenzene 2 to form the new benzylsilacyclopropane 3.1 Based on kinetic observations, these investigators proposed a two-step mechanism in which SiR2 dissociates from 1 as a silylene (SiR2). Overall reaction: PhPhSiSi++12 3cyclohexene Proposed Mechanism: SiSi1k1k-1+cyclohexeneSiR2SiSiR2Ph+2k2PhSi3 The proposed SiR2 intermediate, however, was never observed. Because step 2 and the reverse of step 1 both represent the addition of SiR2 to a double bond, the authors assumed that k2 and k-1 were similar in magnitude. a. (15 pts) On the next page, derive an appropriate rate law for the appearance of product 3 (∂[3]/∂t) in terms of measurable quantities. If you use any assumptions or approximations in your derivation, name them and briefly (in one sentence or less) justify them. 1 Driver, T. G.; Woerpel, K. A. J. Am. Chem. Soc. 2003, 125, 10659-10663. (step 1) (step 2)3 answer to 1(a):4 b. (6 pts) Driver and Woerpel performed kinetic experiments under the assumption that the overall reaction was first-order in 1, such that ][][obs13kt=∂∂; tkteobs][][−=011. Using the method of initial rates, the authors determined kobs for different starting concentrations of allylbenzene (2). Using the data shown below, calculate k1 for the reaction. Make sure to include units in your answer. If you answer incorrectly, we will try to use any calculations you show on the bottom half of the page to assign partial credit. k1 = calculations:5 c. (5 pts) Driver and Woerpel also studied the effect of substituents on rate for a series of substituted allylbenzenes: SiSi++XX The results of these experiments are shown at right. Calculate a linear substituent effect ρ from the plot. d. (10 pts) These experiments were designed to test for transient, charged intermediates that might lie in the middle of step 1 or 2 but be kinetically indistinguishable from the overall steps. Potential alternative explanations for each of the two steps are shown in boxes on the next page. So, what is really happening in “step 1” and “step 2”? Circle one answer—either one of the alternatives, or the direct step without intermediates—in each box. (answer on the next page) ρ =6 Step 2 (circle one row): Sik2SifastSik2SifastSiSiSik2SifastSiSiSik2SifastSik2SifastSi Step 1 (circle one row): SiSik1+SiSik1+SiSik1+SiSifastfastfastSi7 e. (10 pts) The investigators also constructed an Eyring plot, with the goal of determining activation parameters for the reaction: Calculate ΔH‡ and ΔS‡ for the reaction from this plot. ΔH‡ = . ΔS‡ = f. (24 pts) Based on your calculated substituent effects and activation parameters, which step do you think is the rate-determining step? On the next two pages, draw the transition state for each pathway you chose in part (d). (If you chose a two-step alternative for either step, draw the higher-energy transition state for the pair of steps.) Then, explain whether or not ρ and ΔS‡ are consistent with each transition state being rate-limiting. All answers will be graded independently and need not make collective sense; you may argue that each piece of data is consistent with both, either, or neither transition state being rate-limiting.8 Is ρ consistent with step 1 being rate-limiting? (Circle one.) yes or no Explain: Is ΔS‡ consistent with step 1 being rate-limiting? (Circle one.) yes or no Explain: Step 1 transition state9 Is ρ consistent with step 2 being rate-limiting? (Circle one.) yes or no Explain: Is ΔS‡ consistent with step 2 being rate-limiting? (Circle one.) yes or no Explain: Step 2 transition state10 2. Amyloid diseases, such as Alzheimer’s and Parkinson’s diseases, are caused by the misfolding and misassembly of particular proteins in the brain. Transerythretin (TTR) is one such protein and has been implicated in a number neurodegenerative disorders. Normally folded TTR forms tetrameric structures, but dissociation of these tetramers into monomers can lead to partial denaturation of the protein into a misfolded, “amyloidogenic monomer” which readily