Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B4 a Solution 2026-10-20
A is the predicted major diastereomer. At the carbon adjacent to the ketone, take phenyl as the large group, vinyl as medium, and hydrogen as small. There is no adjacent heteroatom donor that would force a five-membered chelate. The appropriate first model is therefore the non-chelating Felkin-Anh model.
View along the carbonyl-carbon–stereocentre bond. Rotate to place the large alpha phenyl group approximately perpendicular to the carbonyl plane. Of the two resulting approaches opposite that large group, the favourable Bürgi-Dunitz trajectory passes nearer hydrogen rather than vinyl. Propyl transfer to the carbonyl and protonation of its oxygen give the configuration labelled A: in the printed projection the new OH projects forward and the propyl bond backward, while the original vinyl bond remains backward. Rotation used to construct a Newman projection must not be mistaken for inversion of the original stereocentre.
This is a relative-selectivity prediction, not a numerical product ratio. Simple attack on the face that happens to look less crowded in the initial flat drawing is unreliable because the relevant C–C single bond can rotate. B is the competing product from the less favourable approach beside vinyl. The model and its preference for the trajectory beside the small group are described in the institutional treatment.[ref]
Felkin-Anh and Cram Chelate by University of Windsor. pp. 1–2, section 1, nonpolar alpha substituents and the small-group trajectory
Original Newman construction of the preferred Felkin–Anh approach and the local stereochemistry corresponding to product A
. Details. Aqueous base selectively converts the methyl ester to a carboxylate, and brief acidification gives the corresponding monoacid while retaining the tert-Butyl ester. At the methyl-ester end, a hydroxide lone pair attacks the acyl carbon and the carbonyl pi pair moves to oxygen. The tetrahedral intermediate collapses as the oxygen anion reforms C=O and the acyl C–OMe bond breaks to methoxide. proton transfer then gives the carboxylate and methanol. carboxylate formation makes the net saponification effectively irreversible under the basic conditions; protonation protonates that carboxylate.
The tert-Butyl ester reacts more slowly under these conditions because its bulky alkoxy group hinders acyl attack, and tert-butoxide is a stronger base and poorer leaving group than methoxide. The selectivity is kinetic, not an assertion that tert-butyl esters can never undergo base hydrolysis. The work-up should be sufficiently mild and brief to avoid the separate acid-labile pathway.
Anhydrous trifluoroacetic acid instead removes the tert-butyl group, leaving the methyl ester. protonation activates the ester; in a useful reactive protonation form the alkoxy oxygen bears the proton. The O–C(tert-butyl) bond then cleaves with its electron pair going to oxygen, producing the carboxylic acid and a tert-butyl-cation-like fragment. Loss of a methyl proton gives 2-methylpropene and regenerates the acid catalyst. Carbonyl-oxygen protonation and charge delocalization can precede cleavage; drawing alkoxy-oxygen protonation is a transparent way of tracking the bonds broken. The key distinction is alkyl–oxygen cleavage here versus acyl–oxygen cleavage in the base reaction. Methyl-cation formation is strongly unfavourable, and without water ordinary acid-catalysed methyl-ester hydrolysis is not available. No water is required to account for the tert-butyl deprotection or its alkene by-product.

