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Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B4 b ii Solution 2026-10-20
Write the reacting terminus as , where includes the pre-existing stereocentres and the intact substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as , the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments and formaldehyde.
Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes . The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.
The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.
Reductive ozonolysis cuts the unsaturated backbone into cyclopentane-1,3-dicarbaldehyde units, apart from end fragments. Both alkene carbons carry a hydrogen, so the carbonyl products are aldehydes rather than ketones. Ring substituent stereochemistry is inherited from the polymer repeat; backbone E/Z geometry does not survive cleavage.
Ozone adds across a backbone alkene by cycloaddition, giving a primary ozonide. Its fragmentation produces a carbonyl group and a carbonyl oxide, which recombine to a secondary ozonide. Dimethyl sulfide reduces the peroxide-containing intermediate, forming dimethyl sulfoxide and the two carbonyl termini. Repetition at every backbone double bond liberates the dialdehyde. An oxidizing work-up would instead risk oxidation to carboxylic acids, so the specified reductive work-up is essential.