In boiling benzene containing a trace of iodine, methyl benzylpenicillinate (1) isomerises to methyl benzylpenillonate (10a). The rearrangement is triggered by a 4,7-bond cleavage of the penicillinate (1a) and probably involves the formation of methyl (4S)-2-[(5R)-2-benzyl-4-oxo-Δ²-oxazolin-5-yl]-5,5-dimethylthiazolidine-4carboxylate (9a) as the primary intermediate. This intermediate undergoes an irreversible fragmentation to methyl (4S)-5,5-dimethyl-Δ²-thiazoline-4-carboxylate (8a) and 2-benzyl-Δ²-oxazolin-4-one (5), which then combine to give the penillonate (10a), probably by way of the species (24).

The reaction pathway is, however, complicated by the preferential reaction of the primary intermediate (9a) with the thiazoline (8a) to give dimethyl (3R,4R,7S,12S)-2-oxo-3-phenylacetamido-6,6,11,11-tetramethyl-5,10-dithia-1,8-diazatricyclo[7.3.0^4.8]dodecane-7,12-dicarboxylate (20a). On the basis of deuterium-labelling experiments, it is shown that the aforementioned addition reaction is reversed in the presence of hydrogen iodide. Therefore the tricyclododecane (20a) is not a direct intermediate in the penicillinate→penillonate transformation but it is formed in a reversible side-reaction.