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Unformatted text preview: Chemistry 14C Fall 2009 Exam 2 Page 1 Page 1 score = Whether wet or dry, cardboard has a distinctive odor. If used in food packaging, absorption of the odorant molecules from the cardboard can significantly change the flavor of the food. Studies revealed a complex mixture of no less than 36 molecules cause this distinct odor. ( E ) and ( Z )-non-2-enal are among the more potent odorant molecules. ( E )-non-2-enal (C 9 H 16 O) ( Z )-non-2-enal (C 9 H 16 O) Questions 14 refer to the mass spectrum of ( E )-non-2-enal. The relative abundance of M is 100%. 1. (2) Is M the base peak in this mass spectrum? Circle one: Yes No Cannot determine 2. (2) Write a number in the blank. M has m/z = _________________. 3. (2) The best estimate of the relative abundance of the M+1 peak is (circle one): Less than 9.6 9.6 9.9 10.2 10.5 10.8 More than 10.8 4. (2) Write in the blank a number which is a whole-number multiple of five (i.e., 0, 15, 165, etc.). The best estimate for the relative abundance of the M+2 peak is _________________. 5. (2) Circle the way(s) in which the mass spectra of ( E )-non-2-enal and ( Z )-non-2-enal differ. m/z for M m/z for M+1 m/z for M+2 Fragmentation Relative abundance of M Relative abundance of M+1 Relative abundance of M+2 None of these 6. (3) ( E )-non-2-enal reacts with atmospheric oxygen to form a new molecule of molecular formula C 9 H 16 O 2 . Comparing the IR spectra of ( E )-non-2-enal with this new molecule, we see these changes: peaks at 1700 cm-1 and 2710 cm-1 disappear, zone 2 becomes broader, and a new peak appears at 1682 cm-1 . Write a reasonable structure for this new molecule....
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