Chemistry 14C Fall 2015 Second Midterm Exam Page 1

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1 Chemistry 14C Fall 2015 Second Midterm Exam Page 1 1. (6) Provide brief yet precise definitions. Use no more than fifteen words for each definition. (a) Molecular ion: (b) Spin-spin coupling: 2. (2) Which of the following spectroscopy techniques do not depend on a magnetic field? Circle one or more: Mass spectrometry Infrared spectroscopy 1 -NMR 13 C-NMR MRI 3. (6) Consider the functional groups that may or may not be present in the molecule that produced the following IR spectrum. The molecular formula for this molecule is C cm Wavenumber (cm -1 ) In the blank after each group or functional group name, write "Absent", "Present", or "Cannot be certain", as appropriate. Alcohol: Ether: Methyl group: Carboxylic acid: Ketone: Terminal alkyne: 4. (2) Write the name of a functional group that is not listed in question 3, but based on the IR spectrum, clearly absent from the molecule. Page 1 score =

2 Chemistry 14C Fall 2015 Second Midterm Exam Page 2 5. (7) The following 1 -NMR spectrum was produced by one of the molecules listed at the right. A: B: C: N C 3 For each molecule choice circle "does" or "does not", as appropriate. If you circle "does" you are done with that part. If you circle "does not" complete the explanation by adding no more than fifteen words in each case. Your "does not" explanations must be distinctly different. Be very specific. (a) The spectrum does / does not correspond to molecule A because... (b) The spectrum does / does not correspond to molecule B because... (c) The spectrum does / does not correspond to molecule C because (8) In the blank after each set of spectral data, circle the letter(s) of one or more molecules that fit the given data. "None of these" is also an acceptable answer. In this problem, each molecule might be used only once, or more than once, or not at all. Note that the spectral data are incomplete. Each spectrum may contain peaks that are not specified. Br Br Cl D E F G I (a) Mass spectrum: m/z = 150 (M; 100%) and m/z = 152 (97.4%): D E F G None of these (b) IR: Peaks at 2962 and 2862 cm -1 ; no other peaks in the cm -1 range: D E F G None of these (c) 1 -NMR spectrum has exactly five signals: D E F G None of these (d) 13 C-NMR spectrum contains at least one doublet: D E F G None of these Page 2 score =

3 Chemistry 14C Fall 2015 Second Midterm Exam Page 3 ere's a real-world application of spectroscopy techniques, taken from the field of archeology. Excavations at the Neolithic city of Jiahu in enan Province, China uncovered shards of pottery that may have been used for fermented beverages 9,000 years ago. Material extracted from these shards was found to contain long-chain alkanes, β-amyrin, oleanolic acid, and tartaric acid. The combination of these substances suggests the pots were used to produce, store, and/or serve a mixed fermented beverage of rice, honey, and fruit. Questions 7 13 are based on this shard analysis. 3 C C 3 3 C C 3 3 C C 3 C 3 C 3 3 C C 3 3 C C 3 3 C C 3 C 3 C 3 C 3 3 C C 3 3 C C 3 3 C C 3 β-amyrin (C ) α-amyrin (C ) leanolic acid (C ) 7. (4) (a) The mass spectrum of one alkane extracted from the shards has m/z = 352 (M; 100%) and m/z = 353 (28.07%). This alkane is (circle one or more): C C C C C (b) In this same mass spectrum, what is the best estimate for the intensity of the peak at m/z = 354? Circle one: 0 2% 2 4% 4 8% 8 16% 16 32% 32 64% % 8. (4) Circle "can be" or "cannot be", then complete the statement by adding no more than fifteen words: The alkanes C and C can be / cannot be differentiated by their infrared spectra because (2) Write a number in the blank. The structure of β-amyrin contains DBE. 10. (2) Circle all the peaks that might have different m/z in the mass spectra of β-amyrin and its constitutional isomer α-amyrin. M M+1 M+2 Fragments Base peak No differences Cannot determine Page 3 score =

4 Chemistry 14C Fall 2015 Second Midterm Exam Page 4 Fruit fermentation yields various carboxylic acids, depending upon the fruit used. Some of these acids are shown below. That the Jiahu fragments contained tartaric acid suggests that grapes were used in this case. Questions refer to these carboxylic acids. Assume that the protons do not exchange. (If you don't know what this means, it probably won't change your answers.) Mesotartaric acid (d,l)-tartaric acid Malic acid xalic acid Question (4) In the empty box below each carboxylic acid, write the number of signals in its 1 -NMR spectrum. 12. (2) Circle the molecule(s) whose 1 -NMR spectrum contains a proton with a chemical shift of at least 6.5 ppm. Mesotartaric acid (d,l)-tartaric acid Malic acid xalic acid None of these 13. (2) Write the integrals for the 1 -NMR spectrum of mesotartaric acid. Assume the smallest integral = 1. Write one number per blank, arranged from smallest to largest. Note that one integral is already written. There may be more blanks than integrals. (Smallest integral) 1,,,,, (Largest integral) Page 4 score =

5 Chemistry 14C Fall 2015 Second Midterm Exam Page 5 Spearmint oil is used as an odorant and flavorant in many consumer products. The oil contains at least 28 different organic molecules, including carveol, carvone, dihydrocarveol, and limonene. A half of the oil is carvone. Carvonic acid is produced by human metabolism of carvone. The questions on this exam page concern the molecules shown below. Carveol Carvone Carvonic acid Dihydrocarveol Limonene 14. (2) Complete this sentence with the name of one spearmint oil molecule listed above. The molecule most probably produced the IR spectrum shown below is cm Wavenumber (cm -1 ) 15. (10) In the space below each question write the name(s) of one or more (as indicated) spearmint oil molecule(s) with the indicated property. (a) Its 1 -NMR spectrum has the most deshielded proton (write one or more than if there is a tie): (b) Its 1 -NMR spectrum has the most signals (write one, or more than if there is a tie): (c) Its 1 -NMR spectrum has no singlets (write one or more): (d) The chemical shifts of its 1 -NMR spectrum are influenced by magnetic induction (write one or more): (e) as the least quartets in its 13 C-NMR spectrum (write one or more than one if there is a tie): Page 5 score =

6 Chemistry 14C Fall 2015 Second Midterm Exam Page (3) Draw in the box a molecule that is consistent with the given molecular formula, IR, and NMR data. Use the back of the spectroscopy data page, or back of the previous exam page, for scratch paper if needed. Formula: C 4 10 N 2 IR: Peaks at 3271 cm -1 and 2961 cm -1, but nothing else cm NMR: 2.67 ppm (singlet; integral = 4) and 1.91 ppm (singlet; integral = 1). 13 C-NMR: 53.3 ppm (triplet). 17. (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 = 149 (M; 100%), m/z = 150 (7.08%) and m/z = 151 (32.4%). 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 (13) IR: Chemistry 14C Fall 2015 Second Midterm Exam Page 7 IR workspace: Page 7 score =

8 Chemistry 14C Fall 2015 Second Midterm Exam Page 8 Anything written outside the boxes on this page will be ignored. Write only 1 -NMR implications and 13 C-NMR conclusions in these boxes. (12) 1 -NMR: Chemical shift Splitting Integral # Implications 4.54 ppm singlet ppm triplet ppm quartet ppm triplet ppm singlet ppm triplet 3 (2) 13 C-NMR: ppm (singlet), 48.0 ppm (triplet), 46.3 ppm (triplet), 44.1 ppm (triplet), 40.7 ppm (triplet), and 15.5 ppm (quartet). 13 C-NMR conclusions: Page 8 score =