SolutionsFacial control by the aldehyde
Exercise 1
Addition of methylmagnesium bromide to 2-phenylpropanal gives mainly one of the two diastereomers of 3-phenylbutan-2-ol. Identify the L, M and S groups and draw the favoured conformation according to the Felkin–Anh model.
L = phenyl, M = methyl, S = H. In the favoured conformation the phenyl group is perpendicular to the plane of the carbonyl, and the nucleophile enters from the opposite side, along the Bürgi–Dunitz trajectory, passing next to the hydrogen (S) rather than the methyl (M). The major product is the Felkin diastereomer (also predicted by Cram’s rule).
Exercise 2
2-(Benzyloxy)propanal gives one diastereomer with a silyl enol ether and TiCl4, and mainly the opposite one with BF3·OEt2. Explain.
TiCl4 can coordinate to both the carbonyl oxygen and the benzyloxy oxygen and forms a five-membered chelate that locks the conformation; the nucleophile enters from the less hindered face and gives the chelation (anti-Felkin) product. BF3 can bind only one oxygen, there is no chelate, and the polar Felkin–Anh model applies, with the alkoxy group perpendicular to the carbonyl: the Felkin product is obtained.