Answers to problems

Summary

Answers to problems from Chapter 1

1. Methylation of pentan-2-one using a base (to form the enolate) and iodomethane can give either hexan-3-one or 3-methylpentan-2-one. The less-substituted enolate is formed under kinetic conditions (by using a strong base for irreversible proton abstraction) with LDA in THF at low temperature. On addition of iodomethane, this gives hexan-3-one. The more-substituted enolate can be formed under thermodynamic conditions (reversible reaction) with t-BuOK, t-BuOH. However, these conditions tend to result in polyalkylation. More satisfactory results can be obtained by treating the silyl enol ether (formed from pentan-2-one with Me₃SiCl, Et₃N) with methyllithium or benzyltrimethylammonium fluoride (to form the specific enolate) and iodomethane.

2. Compounds (a)–(c) can be formed using the Michael reaction. See (a) H. Feuer, A. Hirschfield and E. D. Bergmann, Tetrahedron, 24 (1968), 1187; (b) J. Cason, Org. Synth. IV (1963), 630; (c) E. D. Bergmann, D. Ginsburg and R. Pappo, Org. Reactions, 10 (1959), 179.

3. The iodide 1 is formed by a Baylis–Hillman reaction (Scheme 1.61). See Z. Han, S. Uehira, H. Shinokubo and K. Oshima, J. Org. Chem., 66 (2001), 7854. Note that the syn aldol product is formed from the intermediate Z-titanium enolate.

4. The synthesis of aldol product 2 and cytovaricin is described by D. A. Evans, S. W. Kaldor, T. K. Jones, J. Clardy and T. J. Stout, J. Am. Chem. Soc., 112 (1990), 7001.