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12 Organic Chemistry Solutions Manual (h) Two sharp bands above 3000 cm must be an group and two bands between 1600 and 1700 cm¹ suggests two double bonds, presumably a and a C=C. This leaves just three hydrogen atoms and gives us structures such as b(i) or b(ii). NMR would help because the shift would show whether the compound was an amide or an aldehyde and the alkene shifts would reveal the presence of the NH: group in b(ii). NH₂ H2N H 0 0 (c) One strong broad band above 3000 cm must be an OH and a band at about 2200 cm must be a triple bond, presumably CN as there would otherwise be NHs as well. That leaves again but we do not need a ring and we have structures like these. OH C(ii) HO N N Problem 6 Four compounds having the molecular formula C₄H₆O₂ have the IR and NMR spectra given below. How many DBEs (Double Bond Equivalents) are there in C₄H₆O₂? What are the structures of the four compounds? You might again find it helpful to draw out some or all possibilities before you start. (a) IR: 1745 cm⁻¹: NMR: 214, 82, 58, and 41 p.p.m. (b) IR: 3300 (broad) cm BC NMR: 62 and 79 p.p.m. (c) IR: 1770 cm⁻¹: "C NMR: 178, 86. 40, and 27 p.p.m. (d) IR: 1720 and 1650 (strong) cm NMR: 165, 131. 133, and 54 p.p.m. Purpose of the problem First steps in real identification using two different methods. Because the molecules are small (only four carbon atoms) drawing out a few trial structures gives you some ideas as to the types of compounds likely to be found. Suggested solution Here are some possible structures for It is clear that two double bonds or one double bond and a ring are likely to feature. The double bonds have got to be C=O or C=C (or both if there is no ring). Functional groups are likely to include alcohol, aldehyde, acid, and ketone. 0 0 0 HO CO2H H 0 OH Me (a) IR: 1745 cm must be a group; BC NMR: 214 p.p.m. taldehyde or ketone), 82 and 58 82 p.p.m. (two saturated carbons next to oxygen?), and 41 p.p.m. (one saturated carbon not next 0 to oxygen but near some electron-withdrawing group). The second oxygen does not show up in 0 58 214 the IR it must be an ether. As there is only one double bond. there must be a ring. This 41 suggests one structure. (b) IR: 3300 (broad) must he an OH; NMR: 62 and 79 must be a symmetrical molecule with no alkenes and no it must be a bond usually 70-80 p.p.m.) and a HO 79 saturated carbon next to oxygen. This again gives one structure. that the alkyne does not OH show up in the IR because it is symmetrical. 62