Chemistry 14C Summer 2017 Second Midterm Exam Page 1

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1 Chemistry 14C Summer 2017 Second Midterm Exam Page 1 Begin this exam by gently removing the last two pages of this exam. Nothing o n these pages will be graded. They will be discarded before grading. Please use the back of the data table and information pages for scratch space. Please do not use the exam margins for this purpose. When a pharmaceutical is taken, only rarely does the body absorb the entire dose. Some portion of the drug is excreted. The excreted portion can then swirl down the drain, evade filters in wastewater treatment facilities, and merge with the general water supply. This is of concern because these drug traces can alter animal behavior. For example, traces of oxazepam, a commonly used benzodiazopine anti-anxiety drug, cause perch to eat more and socialize less than normal. Mass spectrometry is a useful tool to evaluate the presence of trace pharmaceutical pollutants in water. Here is some information about a few common benzodiazopines: Drug name: xazepam (Molecule ) Alprazolam (Molecule A) Bromazepam (Molecule B) Diazepam (Molecule D) Flunitrazepam (Molecule F) Lorazepam (Molecule L) Formula: C 15 H 11 N 2 2 Cl C 17 H 13 N 4 Cl C 14 H 10 BrN 3 C 16 H 13 ClN 2 C 16 H 12 FN 3 2 C 15 H 10 Cl 2 N 2 2 m/z for M: (4) Write a drug letter in each blank. If there is a tie, write two or more letters. If the answer cannot be determined from the information given, write "C" (for 'cannot determine') in the blank. (a) Which benzodiazopine(s) has/have the most intense (i.e., tallest) M+1 peak (when M = 100%)? (b) Which benzodiazopine(s) give(s) a mass spectrum in which the intensity of the M+2 is at least 5% (when M = 100%)? 2. (2) Complete this statement by writing the letter(s) of one or more answer choice(s) in the blank. In a test sample containing oxazepam, the mass spectrum included a peak at m/z = 251. (ther peaks were present as well). This peak could be due to. Answer choices: (a) M; (b) M+1; (c) M+2; (d) molecular ion; (e) fragment ion; or (f) none of these. 3. (2) Show how you would calculate the best (i.e., the most precise) value for the intensity of the M+1 peak of oxazepam (molecule ) when M = 100%. Clearly include the result of the calculation. 4. (2) In the blank write the best estimate (rounded to the nearest 5; i.e., 0, 5, 10, 25, etc.) for the intensity of the M+2 peak of oxazepam (molecule ) when M = 100%: % 5. (4) Write a drug letter in each blank. If there is a tie, write two or more letters. If the answer is 'none of these' write 'N' in the answer blank. If the answer cannot be determined from the information given, write "C" (for 'cannot determine') in the blank. (a) Which benzodiazepine(s) violate(s) the nitrogen rule? (b) Which benzodiazepine(s) violate(s) the hydrogen/halogen rule? Page 1 score =

2 Chemistry 14C Summer 2017 Second Midterm Exam Page 2 Molecular structures for questions 6 12 can be found on the information page (page 9). For questions 6 12 we need only be concerned with the molecular structures. We don't have to know how the reactions occur, or why. 6. (4) Complete each statement by writing one or more molecule numbers in each blank. If none of the answer molecules meets the requirement, write '0' in the blank. (a) Write the molecule(s) whose IR spectrum is/are predicted to contain any peaks in the 1750 cm -1 to 1450 cm -1 range: (b) Write the molecule(s) whose IR spectrum is/are to contain predicted only one peak in the 1750 cm -1 to 1450 cm -1 range: 7. (7) Using the scale shown below, sketch the IR spectrum of molecule 1. Neatness and accuracy count. Label each peak with the corresponding bond and name of functional group. Also include stretching frequencies (or frequency ranges) if you think it will help clarify your answer. Hint: Practice in the work area below before sketching your final answer here Work area for question 7. Nothing below this line of this page will be graded Page 2 score =

3 Chemistry 14C Summer 2017 Second Midterm Exam Page 3 8. (7) The following IR spectrum may belongs to either molecule 1, molecule 2, or molecule cm Complete each statement below by writing 'can' or 'cannot', in the blank. If you write 'can' you are done for this part of the question. If you write 'cannot' complete the explanation by adding no more then fifteen words in each case. Be precise and specific. Make your explanations for all parts of this problem distinctly different. (a) This IR be from molecule 1 because... (b) This IR be from molecule 2 because... (c) This IR be from molecule 3 because (2) Complete the statement by writing one or more molecule numbers in the blank. If there is a tie write two or more molecule numbers. Which molecule(s)' 1 H-NMR spectrum/spectra have/has the most signals between 8.0 and 6.5 ppm. Assume each proton attached to the benzene ring is resolved (i.e., gives a distinct signal and is not buried in a multiplet). Answer: Page 3 score =

4 Chemistry 14C Summer 2017 Second Midterm Exam Page 4 Here is a drawing of molecule 3 with protons labeled, useful for questions 10 12: A B C D E F Molecule (5) Complete this statement by writing either 'shielded' or 'deshielded' in the first blank, then add no more than six words in second and third blanks. Be as specific as you can within the six-word limit. Proton B of molecule 3 is highly due to and. 11. (6) Assign 1 H-NMR chemical shifts to protons A F of molecule 3 by writing one chemical shift in each blank. Each chemical shift will be used only once. Chemical shift choices: 7.4 ppm, 6.1 ppm, 3.0 ppm, 1.4 ppm, 1.3 ppm, and 0.9 ppm. There are no ties. Proton A = ppm Proton B = ppm Proton C = ppm Proton D = ppm Proton E = ppm Proton F = ppm 12. (12) The conversion of molecule 3 into molecule 4 might be monitored by 1 H-NMR. At the start of the reaction, the spectrum would contain signals due to molecule 3 only. In the middle of the reaction the spectrum would contain signals due to molecule 3 and molecule 4. At the end of the reaction the spectrum would contain signals due to molecule 4 only. Fill in the blanks below to describe one signal that would disappear as molecule 3 is consumed, and another signal that would appear as molecule 4 is formed. Write only one choice per answer blank. For all of the 1 H-NMR spectra in this experiment the smallest integral = 1. Assume the benzene ring protons are not resolved (i.e., all the benzene ring proton signals are lumped together as one multiplet.) Write a letter (A F) in the blank. The proton I am using for this question is proton For the signal that disappears......its chemical shift is. Answer choices: ppm, 6.1 ppm, 3.0 ppm, 1.4 ppm, 1.3 ppm, or 0.9 ppm....the splitting is. Answer choices: Multiplet, singlet, doublet, triplet, quartet, pentet, or sextet....the integral is. Answer choices: 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, or 5.0. For the signal that appears......its chemical shift is. Answer choices: ppm, 6.1 ppm, 4.9 ppm, 1.4 ppm, 1.3 ppm, or 0.9 ppm....the splitting is. Answer choices: Multiplet, singlet, doublet, triplet, quartet, pentet, or sextet....the integral is. Answer choices: 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, or 5.0. Page 4 score =

5 Chemistry 14C Summer 2017 Second Midterm Exam Page (9) Each set of spectral information below corresponds to one of the molecules shown on the information page (page 9). This molecule is not necessarily one of the numbered molecules. In the answer blank write the number (in the case of a numbered molecule) or the letter (in the case of a molecule which does not have a number) of the molecule that best fits the given data. If none of the given molecule choices is correct write 'N' (for none) in the blank. Hint: This problem does not require a complete spectral analysis like a combined spectroscopy problem such as question 14 below. Answer choices: Molecule 1, molecule 2, molecule 3, molecule 4, molecule 5, molecule 6, (a) CH 3 CH 2 CH 2 CH 2 MgBr, (b) H 3 +, (c) CH 3 CH 2 CH 2 CH 3, (d) Cr 3, (e) H 2 S 4, (f) CHCl 3, or (g) Br 2. (a) IR has no peak cm -1, but there are peaks elsewhere in the spectrum. 1 H-NMR: 3.47 ppm (quartet; integral = 1.0) and 1.21 ppm (triplet, integral = 1.5). Answer: (b) IR has a strong peak at 1697 cm -1, as well as other peaks elsewhere in the spectrum. 1 H-NMR: 9.61 ppm (singlet; integral = 1), ppm (multiplet; integral = 3), and 6.07 ppm (singlet; integral = 2). Answer: (c) The IR spectrum contains strong peaks at 3050 cm -1, 2950 cm -1, and 1695 cm -1, as well as elsewhere in the spectrum. The 1 H-NMR spectrum contains ten signals. Answer: Work space for problem 13. Nothing below this line on this page will be graded. Page 5 score =

6 Chemistry 14C Summer 2017 Second Midterm Exam Page (34) Deduce the structure that corresponds to the spectral data on pages 6 8. Write your final answer in the box. A correct answer is worth full credit. If the answer is incorrect, your analysis of the spectra can be worth significant partial credit, so show your work clearly in the space below each set of data only. Answers outside of these places will be ignored. Final Structure Box (4) Mass spectrum: m/z = 210 (M; 100%), m/z = 211 (13.82%), and m/z = 212 (0.75%). No fluorine or iodine. Write in the box the one formula that is consistent with the mass spectrum, and is not rejected due to other reasons. Page 6 score =

7 Chemistry 14C Summer 2017 Second Midterm Exam Page 7 (12) IR: IR workspace: Page 7 score =

8 Chemistry 14C Summer 2017 Second Midterm Exam Page 8 Anything written outside the boxes on this page will be ignored. Write only 1 H-NMR implications in these boxes. (12) 1 H-NMR: Chemical shift Splitting Integral # H Implications ppm multiplet ppm triplet ppm singlet ppm triplet ppm pentet ppm singlet 1 Page 8 score =

9 Chemistry 14C Summer 2017 Second Midterm Exam Page 9 This is an information sheet. Nothing that you write on this sheet will be graded. It will be discarded prior to grading. So-called "bath salts" are a complex mixture of designer drugs. Despite the harm caused by these drugs (and especially the toxic impurities they contain), they remain legal in many states. The Internet has a wealth of procedures to manufacture bath salts components such as methylenedioxypyrovalerone. Although IR and NMR are beyond the means of most bath salt makers, let's explore how spectroscopy might be useful in this process. Here is one set of reactions that might be used to convert piperonal (molecule 1) into methylenedioxypyrovalerone (molecule 6): H Molecule 1 (C 8 H 6 3 ) H 1. CH 3 CH 2 CH 2 CH 2 MgBr 2. H 3 + Molecule 2 (C 12 H 16 3 ) Cr 3, aq. H 2 S 4 CHCl 3 Br 2 NH Molecule 5 (C 4 H 9 N) Molecule 3 (C 12 H 14 3 ) CHCl 3 Br Molecule 4 (C 12 H 13 Br 3 ) CH 3 CH 2 CH 2 CH 3 N Molecule 6 (C 16 H 21 N 3 ) Page 9 score =