There is only one double bond so it must be a ring

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Unformatted text preview: case. ■ You will meet other ways of distinguishing these compounds in chapters 13 and 18. 6 Solutions Manual to accompany Organic Chemistry 2e NH2 H2N O H O (c) One strong broad band above 3000 cm–1 must be an OH group and a band at about 2200 cm–1 must be a triple bond, presumably CN as otherwise we have nowhere to put the nitrogen atom. This means structures of this sort. OH HO N N PROBLEM 7 Four compounds having the formula C4H6O2 have the IR and NMR data given below. How many DBEs (double bond equivalents—see p. 75 in the textbook) are there in C4H6O2? What are the structures of the four compounds? You might again find it useful to draw a few structures to start with. (a) IR: 1745 cm–1; 13C NMR 214, 82, 58, and 41 ppm (b) IR: 3300 cm–1 (broad); 13C NMR 62 and 79 ppm. (c) IR: 1770 cm–1; 13C NMR 178, 86, 40, and 27 ppm. (d) IR: 1720 and 1650 cm–1 (strong); 13C NMR 165, 133, 131, and 54 ppm. Purpose of the problem First steps in identifying compound from two sets of data. Because the molecules are so small (only four carbon atoms) drawing out a few trial structures is a good way to start. Suggested solution Here are some possible structures for C4H6O2. It is clear that there are two double bond equivalents and that double bonds and rings are likely to feature. Functional groups are likely to include alcohol, aldehyde, ketone and carboxylic acid. O OH H HO CO2H O Me O O (a) IR: 1745 cm–1 must be a carbonyl group; 13C NMR 214 must be an aldehyde or ketone, 82...
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