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Unformatted text preview: and 58 look like two carbons next to oxygen and 41 is a carbon not next to oxygen but not far away. As the second Solutions for Chapter 3 – Determining Organic Structures oxygen doesn’t show up in the IR, it must be an ether. As there is only one double bond, the compound must be cyclic. The suggests just one structure. (b) IR: 3300 cm–1 (broad) must be an OH; 13C NMR 62 and 79 show a symmetrical molecule and no C=O so it must have a triple bond and a saturated carbon next to oxygen. This again gives only one structure. (c) IR: 1770 cm–1 must be some sort of carbonyl group; 13C NMR 178 suggests an acid derivative, 86 is a saturated carbon next to oxygen, 40 and 27 are saturated carbons not next to oxygen. There is only one double bond so it must be a ring. Looks like a close relative of (a). (d) IR 1720 and 1650 cm–1 (strong) must be C=C and C=O; 13C NMR 165 is an acid derivative, 131 and 133 must be an alkene, and 54 is a saturated carbon next to oxygen. That defines all the carbon atoms. It is not significant that we cannot say which alkene carbon is which. (a) (b) 82 O O 58 41 (c) 62 HO 214 79 (d) O 86 165 O OH 27 40 178 O 133/131 O Me 54 PROBLEM 8 You have dissolved tert- butanol in MeCN with an acid catalyst, left the solution overnight, and found crystals in the morning with the following characteristics. What are the crystals? OH H MeCN ? IR: 3435 and 1686 cm–1; 13C NMR: 169, 50, 29, and 25 ppm. Mass spectrum (%): 115 (7), 100 (10), 64 (5), 60 (21), 59 (17), 58 (10...
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