1 (148)

Prévia do material em texto

Chapter 19 Suggested solutions for Chapter 19 145 Problem 5 Only one of the diastereomeric bromides shown here eliminates to give alkene A. Why? Neither bromide gives alkene B. Why not? 0 0 0 0 0 0 0 0 base base alkene A alkene A alkene B Br Purpose of the problem Helping to appreciate that cage molecules often have restricted opportunities for eliminations. Suggested solution The first molecule has one anti-periplanar H and Br so elimination can occur. The second has no drogens in the right place. Alkene B is a bridgehead alkene and could not exist. 0 0 0 B H 0 0 0 0 Br Br an impossible alkene alkene A Problem 6 Suggest mechanisms for these reactions, paying particular attention to the elimination steps. R Br Br Br 1. NaOH, H2O 1. NaOH N R 0 0 R N 2. HCI, 2. Br R 0 H Purpose of the problem Exploiting multiple eliminations. Suggested solution The first compound is an amide and hydrolyses in base to a nitrogen anion that can lose one Read about this in B. S. Furniss A second elimination then expels the remaining bromide and a molecule of nitrogen. et al., Vogel's textbook of organic chemistry (5th edn), Longmans, R Br R Br R Br Harlow, 1989, p. 744. Br Br R N N CO OH 2 N N NH N H Ho The second example includes a cyclization and an elimination. They could occur in either order Read about this in B. S. Furniss but starting with the cyclization removes any difficulty about the geometry of the alkene. In et al., Vogel's textbook of organic chemistry (5th edn), Longmans, whichever order you do the reactions, the elimination should be by the mechanism. Harlow, 1989, p. 805. e Br H OH Br 0,0 HO₂C 0 R R R R