Chemistry 14C Spring 2018 Second Midterm Exam Page 1

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1 Chemistry 14C Spring 2018 Second Midterm Exam Page 1 Begin this exam by gently removing the last two pages. Nothing on these pages will be graded. These pages will be discarded prior to grading. Please use the backs of the data tables page and other exam pages for scratch space. Please do not use the exam margins for this purpose. 1. (13) Imagine you have a sample of a clear, colorless liquid, smelling vaguely like gasoline, that might be either octane, CH 3 (CH 2 ) 6 CH 3, or nonane, CH 3 (CH 2 ) 7 CH 3. How might you use spectroscopy to determine the correct identity? In the first blank for each part write 'Y' (for yes) or 'N' (for no). If you write 'N' you are done this part of question 1. For parts (a) and (c) if you write 'Y' complete the explanation by writing no more than ten words in each answer space. For part (b) write one or more zone numbers (1 through 5) and/or 6 (for the fingerprint region) or 0 (no zones) in the blank. (a) Can mass spectrometry be used to differentiate octane and nonane (Y/N)? Explanation: The mass spectrum of octane has......whereas the mass spectrum of nonane has... (b) Can infrared spectroscopy be used to differentiate octane and nonane (Y/N)? In which zone(s) do the IR spectra of these molecules have significant, interpretable differences? Zone(s) (0-6): (c) Can 1 H-NMR be used to differentiate octane and nonane (Y/N)? Explanation: The 1 H-NMR spectrum of octane has......whereas the 1 H-NMR spectrum of nonane has... In lecture 6 we learned there are 328 C 6 H 6 isomers. Kekulé's proposal for the structure of benzene came to him in a dream ("the snake seized its own tail and the form whirled mockingly before my eyes...") What if Kekulé had access to spectroscopy? Would he have proposed a different structure for benzene? (Kekulé didn't have spectroscopy, because the oldest Chem 14C spectroscopic technique -- mass spectrometry -- wasn't invented until 1913, after Kekulé's death in 1896.) Benzene Fulvene 2. (6) Let's predict the molecular ion region for the mass spectrum of benzene with some calculations. Write the requested number with the requested number of decimal places in each blank. Don't round until the very last step of your calculations. m/z for M = No decimal places. Relative intensity of M+1 (when M = 100%) = One decimal place. Relative intensity of M+2 (when M = 100%) = No decimal places. Page 1 score =

2 Chemistry 14C Spring 2018 Second Midterm Exam Page 2 3. (1) Complete this statement by writing 'Y' (for yes) or 'N' (for no) in the blank. Are the molecular ion regions in the mass spectra of benzene and fulvene identical (Y/N)? 4. (6) Now let's consider the 1 H-NMR spectra of some C 6 H 6 isomers. In each blank write a number (0-100): Benzene: Total number of 1 H-NMR signals = Total number of 1 H-NMR singlets = Fulvene: Total number of 1 H-NMR signals = Total number of 1 H-NMR triplets = Prismane: Total number of 1 H-NMR signals = Total number of 1 H-NMR doublets = 5. (6) The molecular ion region for the mass spectra of epinephrine (adrenaline) and norepinephrine (a biochemical precursor to adrenaline) are shown below. Epinephrine: m/z = 183 (M; 100%), m/z = 184 (10.64%), and m/z = 185 (0.84%). Norepinephrine: m/z = 169 (M; 100%), m/z = 170 (9.48%), and m/z = 171 (0.59%). Based on these mass spectra, what is the difference in the molecular formulae of these molecules? Write a number (0-100) in the blank after each element: Carbon =, hydrogen =, and oxygen =. 6. (4) In the blank after each 1 H-NMR spectrum below, write the molecule letter (A E) that is the best fit. Write 'N' (for none) fit none of the given choices is correct. Molecule A Molecule B Molecule C Molecule D Molecule E (a) 1 H-NMR: 3.4 ppm (triplet; integral = 1.0), 1.8 ppm (pentet; integral = 1.0), 1.5 ppm (sextet; integral = 1.0), and 0.9 ppm (triplet; integral = 1.5). Best fit = molecule (A E or N): (b) 1 H-NMR: 4.3 ppm (pentet; integral = 1.0), 2.2 ppm (pentet; integral = 2.0), 2.0 ppm (pentet; integral = 2.0), 1.7 ppm (sextet; integral = 1.0), and 1.6 ppm (sextet; integral = 1.0). Best fit = molecule (A E or N): Page 2 score =

3 Chemistry 14C Spring 2018 Second Midterm Exam Page 3 7. (6) We've seen this graph several times in lecture: Complete each statement by writing no more than three words in each blank. Be precise. The correct label for the graph's x-axis is. The correct label for the graph's y-axis is. On this graph ΔE refers to (hint: not just 'difference in energy'). Questions 8 13 concern phenylbutazone, a nonsteroidal anti-inflammatory drug (NSAID) used for short-term treatment of pain and fever in animals (especially race horses). It is not longer approved for human use because it can cause severe adverse effects. Despite is banishment for human use phenylbutazone has been detected in horsemeat intended for human consumption. Knowing the correct molecular structure is critical to understanding a drug's biology, as well as when using spectroscopy to detect the drug's presence. Phenylbutazone might have one of two structures shown below: Phenylbutazone (keto tautomer) Phenylbutazone (enol tautomer) 8. (4) Complete this statement by writing 'K' (for keto) or 'E' (for enol) in the first blank, a bond and functional group name in the second blank (such as alkyl bromide C Br), followed by a number (1-5) in the third blank: The IR spectrum shown below agrees with the (K or E) phenylbutazone structure, because this spectrum lacks a peak caused by (name of functional group and bond) in zone (1-5). Page 3 score =

4 Chemistry 14C Spring 2018 Second Midterm Exam Page 4 9. (4) What aspect(s) of molecular structure correlate with each axis on an IR spectrum? Write no more than five words in each space. Hint: Energy is not an aspect of molecular structure. "Stretching frequency" and "photon energy" are not acceptable answers. IR spectrum x-axis: IR spectrum y-axis: 10. (4) We could also study the phenylbutazone structure by 1 H-NMR. Complete the following statement by writing number (0-100) in the first blank, and 'M' (for more), 'S' (for the same number), or 'L' (for less) in the second blank: Assuming all nonequivalent proton signals for benzene ring protons are resolved (separated), we predict (0-100) signals in the 1 H-NMR spectrum for the phenylbutazone keto form. The 1 H-NMR spectrum for the phenylbutazone enol form has signals than the 1 H-NMR spectrum for the keto form. 11. (4) In the previous question, what causes protons to be equivalent? Complete the following statement by writing no more than three words in each blank. Make your answers distinctly different. Equivalent protons have the same and the same. 12. (4) The chemical shift of the phenylbutazone keto form methine group proton is 3.4 ppm. Using no more than three words in each case, list two features or factors of the molecule's carbonyl groups that cause this deshielding. Feature #1: Feature #2: 13. (4) The laboratory synthesis of phenylbutazone might involve a reaction with 1- bromobutane. In each blank write the letter (F I) of a 1-bromobutane proton that meets the given requirement. Most deshielded proton (F I): Most complex 1 H-NMR signal (F I): Page 4 score =

5 Chemistry 14C Spring 2018 Second Midterm Exam Page (34) Deduce the structure that corresponds to the spectral data on pages 5 7. 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 = 220 (M; 100%), m/z = 221 (16.2%), and m/z = 222 (0.54%). No fluorine or iodine. Write in the box the one (and only one) formula that is consistent with the mass spectrum, and is not rejected due to other reasons. Page 5 score =

6 Chemistry 14C Spring 2018 Second Midterm Exam Page 6 (13) IR: IR analysis: Page 6 score =

7 Chemistry 14C Spring 2018 Second Midterm Exam Page 7 Anything written outside the boxes on this page will be ignored. Write only 1 H-NMR implications in the 1 H-NMR implications boxes. Write only 13 C-NMR conclusions in the 13 C-NMR conclusions box. (12) 1 H-NMR: Chemical shift Splitting Integral # H Implications ppm multiplet ppm singlet ppm triplet ppm pentet ppm triplet ppm singlet 3.0 (2) 13 C-NMR: 173 ppm (singlet), 150 ppm (singlet), 128 ppm (doublet), 126 ppm (doublet), 125 ppm (doublet), 52 ppm (quartet), 44 ppm (triplet), 37 ppm (singlet), 34 ppm (triplet), 29 ppm (quartet), and 20 ppm (triplet). 13 C-NMR conclusions: Page 7 score =