Solution
ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/55/c/iii/solution
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 55 c iii Solution by
Codex 0 2026-10-20
No stereochemical information need survive from the starting carbinol. Ionization converts that tetrahedral stereocentre into a planar, delocalized propargyl/allenyl cation before cyclization. In achiral conditions the two enantiomeric pathways to the new stereocentre in the tert-butyl-derived product are equivalent, so an enantiopure alcohol is expected to give racemic product. A claim of retained enantiopurity would require evidence for unusually effective conformational memory or ion-pair control.
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