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15.26 The DU of E and F is 3. Both produce G, with DU = 1, upon catalytic hydrogenation. G cannot have a pi bond because it is produced by hydrogenation, so its DU must be due to a ring. Therefore E and F must have two pi bonds and a ring. Because G shows only a single peak in its spectrum, all of its C's are identical. It must be cyclohexane. Thus E and F are isomers of cyclohexadiene. The at 259 nm in the UV spectrum of compound F suggests that its two double bonds are conjugated. The absence of any absorption maximum above 200 nm in the UV spectrum of E is consistent with non-conjugated double bonds. E F 15.27 The peaks of equal intensity at m/z 122 and 124 in the mass spectrum indicate the presence of a bromine atom in the unknown. Subtracting 79 from 122 leaves 43 as the mass of the rest of the molecule. The spectrum shows the presence of three different types of hydrogens in 2:2:3 ratio. The two H's at 3.4 δ appear as a triplet, so they must be coupled to two H's, the ones appearing at 1.9 δ. The three H's at 1.0 δ also appear as a triplet, so they are also coupled to the two H's at 1.9 δ. Therefore the two H's at 1.9 δ are coupled to five H's and appear as a sextet. This shows the presence of a propyl group, CH₃CH₂CH₂, in the compound. (The mass of a propyl group is 43.) Thus, the unknown is 1-bromopropane. CH₃CH₂CH₂Br 15.28 To have an absorption at 280 nm, the amino acid must have conjugated double bonds in its side chain. Both phenylalanine and tryptophan have absorptions at 280 nm. Leucine, proline, and serine do not have any chromophores that absorb in this region. 15.29 As can be seen in the models, the double bonds of the compound on the left do not quite lie in the same plane, so its conjugation is decreased and its occurs at higher energy (shorter wavelength). The double bonds of the compound on the right lie in the same plane, so it is more conjugated and its occurs at lower energy (longer wavelength). 15.30 As can be seen in the models, the compound on the left (biphenyl) has its two benzene rings in the same plane. The methyl group in the ortho position of the compound on the right causes its benzene rings to lie in slightly different planes, decreasing their conjugation, and shifting the for this compound to higher energy (shorter wavelength). 235