The lactol is a cyclic hemiacetal and can open to , retaining its pre-existing alkoxy-bearing stereocentre. Its two anomers lose their anomeric configuration on opening to the planar aldehyde, so both can feed the same addition pathway. A first equivalent of decylmagnesium bromide can be consumed by the free OH, producing decane and a magnesium alkoxide; the excess reagent supplies the carbon nucleophile for addition.
magnesium can coordinate both the aldehyde oxygen and the adjacent benzyloxy oxygen. This produces a five-membered chelate, with ring sequence Mg–O(carbonyl)–C(aldehyde)–C(alpha)–O(benzyloxy), that restricts rotation about the aldehyde–alpha-carbon bond. Decyl attack on the less crowded face, away from the alpha-carbon's hydroxyethyl chain in the chelated arrangement, forms the new C–C bond and a magnesium alkoxide at the former aldehyde carbon. Acidic workup protonates both alkoxides to give G. The OBn bond is retained; it is not displaced by the Grignard reagent.
The product configuration is the one with OBn and the newly formed OH both projecting forward in the specified zigzag chain, while the opposite configuration at the new alcohol carbon is disfavoured. This is the Cram chelation model for an alpha-alkoxy aldehyde, rather than stereochemical memory of the starting lactol OH. Coordination is a mechanistic explanation consistent with MgBr2 and the product; the scheme does not quantify individual organomagnesium aggregates.[ref]
Felkin-Anh and Cram Chelate by University of Windsor. p. 3, section 2, Chelation and Addition–Cram Chelate Rule
The relevant primary synthetic sequence is documented independently.[ref]Highly Stereoselective Asymmetric Total Synthesis of (−)-Jimenezin via Sequential Intramolecular Amide Enolate Alkylation by Thien Nhan Lu and others. 2022, synthetic sequence leading through the protected oxygen heterocycle
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.
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