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Question

The major product formed in the following reaction sequence is

Bicyclic ketone treated with KHMDS then a Davis oxaziridine

This question was previously asked in
MH SET 2026 Paper 1 Question Paper (25-Oct-2026)
The correct answer is
Endo alpha-hydroxy ketone

KHMDS removes an α-proton to give the potassium enolate, and an N-sulfonyloxaziridine (Davis reagent) transfers its oxygen to the enolate carbon, giving an α-hydroxy ketone after work-up.

The substrate is a 7,7-dimethylbicyclo[2.2.1]heptan-2-one. Only C-3 can enolise (C-1 is a bridgehead, so Bredt’s rule forbids an enolate there). The exo face of the C-2/C-3 enolate is shielded by the syn methyl group on the one-carbon bridge, so the bulky oxaziridine approaches from the endo face, as in the Davis hydroxylation of camphor. The new OH therefore occupies the endo position and the hydrogen at C-3 is exo.

A bridgehead hydroxy ketone would need enolisation towards the bridgehead, which is not possible. An amino ketone would require nitrogen transfer, whereas N-sulfonyloxaziridines transfer oxygen. The exo alcohol is the diastereomer formed by attack from the hindered face.

endo-3-hydroxy-7,7-dimethylbicyclo[2.2.1]heptan-2-one option is correct.

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