= Solution
\b[B is 4-hydroxy-1-methylpyrrolidin-2-one; C is 1-methylpyrrolidin-3-ol.] Their ring connectivities are shown below. Each ring remains five membered; the attached N-methyl group is retained.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025-reduction-products.png]
{title=Original cyclic product structures and <hydride> <electron-pair arrows> for ketone and lactam reduction}
For B, a B–H bond of <sodium borohydride> supplies a <hydride> equivalent to the <ketone> carbonyl carbon, and the carbonyl $\pi$ pair moves to <oxygen>. This gives an <alkoxide>, whose <protonation> during the reaction or workup gives the secondary <alcohol>. The <lactam> carbonyl remains. Converting one $\mathrm{C{=}O}$ to $\mathrm{CH(OH)}$ adds two H atoms and gives $\mathrm{C_5H_9NO_2}$.
For C, <lithium aluminium hydride> reduces that <ketone> in the same net way and also reduces the <lactam>. Coordination of the amide <oxygen> to <aluminium> assists <hydride> attack at its carbonyl carbon, giving a tetrahedral intermediate. The N <lone pair> then forms a C=N bond while the C–O bond breaks to an aluminium-bound oxygen group. A resulting <iminium ion> accepts a second <hydride> at carbon, yielding the cyclic tertiary <amine>. Aqueous workup protonates the ketone-derived <alkoxide>.
The <lactam> conversion replaces $\mathrm{N-C({=}O)}$ by $\mathrm{N-CH_2}$; its <oxygen> is removed rather than retained as another <alcohol>. Together with <ketone> reduction, the net change is $+4\mathrm H-\mathrm O$, giving $\mathrm{C_5H_{11}NO}$. No stereochemistry is specified: the new <alcohol> stereocentre is formed as a <racemic mixture> in an achiral environment.
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