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Fluoride favors geminal disubstitution: after the first substitution, the second F replaces the other chloride at that same phosphorus, giving a $\mathrm{P}{\mathrm{F}}_2$ unit and two unchanged $\mathrm{PC}{\mathrm{l}}_2$ units. The first highly electronegative F withdraws electron density inductively, maintaining or increasing the electrophilicity of its phosphorus and activating the neighboring P–Cl bond toward another nucleophile. The small fluoride ion also has little steric hindrance to attack there. In this context geminal substitution means both incoming substituents on one phosphorus; it does not describe the relative positions of different ring nitrogen atoms. Electronic activation at the substituted phosphorus competes with effects elsewhere in the ring, so the argument predicts the preferred pathway rather than an unconditional exclusive product.

Sodium borohydride is chemoselective for the ketone; lithium aluminium hydride reduces both the ketone and the lactam. In an amide, donation of the N lone pair into the carbonyl $\pi$ system reduces the carbonyl carbon's electrophilicity. It is therefore substantially less susceptible to the milder hydride donor.

Lithium aluminium hydride supplies more strongly reducing hydride equivalents, and the strong affinity of aluminium for oxygen assists removal of the amide oxygen through the iminium pathway. This difference explains both the unchanged lactam carbonyl in B and the absence of that oxygen in C. Both reactions are interpreted after the usual protonating workup.