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324 Organic Chemistry Solutions Manual use oxonium ion intermediates and not do SN2 reactions (Chapter 14). Loss of BF₃ and rotation the last intermediate though 180° gives the product. 1 1 0 H 2 BF₃ 0 2 2 1 3 3 3 0 OH 4 0 0 OH 4 4 0 0 BF₃ OBF₃ OBF₃ 0 OH 0 0 0 E. J. Corey and M. G. Bock, Tetrahedron Lett., 1975, 2643. The third reaction involves a cyclization. Atoms 1 and 7 clearly make the new bond and the rest of the atoms fit into place without change except that the Br is gone and the alkene has moved from 7/8 to Br 9 H Zn 2 5 6 8 4 6 This new reaction paved the way 1 7 4 0 for the synthesis of the seven- 8 CO₂Me 1 3 5 7 3 2 0 membered ring antitumour H compounds (M. F. Semmelhack and E. C. S. Wu, J. Am. Chem. Soc., Zinc will insert oxidatively into the C-Br bond and the mechanism follows from the 1976, 98, 3384). nature of the organometallic compound. ZnBr H H OMe Zn starting material CO₂Me 0 H H H Problem 2 Explain this series of reactions. 0 0 1. H N 2. pyridine Purpose of the problem Working out the stereochemistry and mechanism of the Beckmann rearrangement. This example comes from a Suggested solution general investigation into the The first reaction forms the oxime by the usual mechanism (Chapter 14). This reaction is under Beckmann and the related Schmidt thermodynamic control so the OH group will bend away from the aryl substituent. Then we rearrangements by R. H. Prager et al., Aust. J. Chem., 1978, 31, the Beckmann rearrangement (pp. 997-9). The group anti to the OH group migrates from C 1989. and that gives the product after rehydration and adjustment of protons. HO 0 N N 1. 2. pyridine C to N migration more stable oxime H o H N N