SolutionsDirected aldol addition
Exercise 1
The lithium enolate of 2,2-dimethylpentan-3-one (ethyl tert-butyl ketone), prepared with LDA, is almost exclusively Z. Which aldol will form with benzaldehyde? Justify it with the Zimmerman–Traxler model.
The syn aldol, with high selectivity. In the chair transition state, the phenyl group of the aldehyde occupies an equatorial position to avoid the 1,3-diaxial interaction with the tert-butyl group on C1 of the enolate. With the Z enolate, the methyl on Cα is pseudoaxial, and the resulting relationship between methyl and hydroxyl in the product is syn. The very bulky tert-butyl group makes the energy difference between the possible chairs very large.
Exercise 2
Which reagents would you use to obtain, from 3-pentanone and propanal, (a) the syn aldol and (b) the anti aldol? Why are boron enolates more selective than lithium enolates?
(a) Bu2BOTf and i-Pr2NEt, which give the Z boron enolate, then propanal at −78 °C. (b) (c-C6H11)2BCl and Et3N, which give the E enolate. O–B bonds are shorter than O–Li bonds, the chair transition state is more compact and the steric interactions that distinguish the two chairs are magnified; moreover, boron enolates do not aggregate and always react through the cyclic transition state.