Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 2 / B10 / b / iii

Hydride is delivered to steroid C5 and a proton to C4; the C3 ketone is retained. Use the conjugated fragment $\mathrm{O}=C3-C4=C5$. Carbonyl hydrogen bonds from Tyr91–OH and protonated Glu57 lower the barrier to conjugate reduction. Draw a curved arrow from the nicotinamide C4–H bond of NADPH to steroid C5, shift the C4=C5 pi pair toward C3–C4, and shift C3=O pi electrons onto oxygen. The nicotinamide ring rearomatizes to NADP+; the steroid becomes a bound enolate, or its protonated enol equivalent. electron shifts across the nicotinamide ring must accompany C–H bond cleavage rather than leaving an unexplained neutral carbon radical.

Next, reform C3=O and protonate steroid C4 from an appropriately positioned acid, directly or through an enol/proton-relay route. This regenerates the catalytic acid after solvent reprotonation. The enzyme fixes the hydride-accepting face so that the new C5–H has the 5α configuration, giving dihydrotestosterone; delivery to the opposite face would give a different ring-junction configuration. This is not simple 1,2 reduction of the C3 carbonyl to an alcohol. The net local bookkeeping is one hydride plus one proton across C4=C5, with two electrons supplied by NADPH. The original reactive-fragment diagram shows the enolate-forming arrows and the separate protonation step.