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14.28 A (DU = 3) gives B (DU = 1) upon catalytic hydrogenation, so A must have two pi bonds and one ring and B has only the ring remaining. Because all of the carbons of B are identical, it must be cycloheptane. So A must be a cycloheptadiene. It only remains to establish the positions of the double bonds. Because one of the ozonolysis fragments contains only two C's, the double bonds must be adjacent. A is 1,3-cycloheptadiene. H H O ozonolysis O + O O H A H C₂H₂O₂ C₅H₈O₂ Pt H₂ B CH₃ 14.29 Because the product has a C with no H's and a it CH₃-C-CH₂CH₃ must be the alcohol formed by rearrangement of the initial carbocation. OH 14.30 There will be three absorptions in the spectrum of this compound. The hydrogens on the carbons bonded to the chlorines are enantiotopic and will produce one signal. The hydrogens of CH₂ group are diastereotopic, and will give two signals. 14.31 The triplet (2 H's) near 3.6 δ must be split by two H's, as is also the case for the triplet (2 H's) near 3.0 δ, so they must result from a group. The singlet at 2.2 δ (3 H's) results from a CH₃ group with no nearby H's. The product is This is the product of anti- Markovnikov addition of HCI to the reactant. The inductive electron withdrawing effect of the carbonyl group must destabilize the secondary 224