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Write $R$ for the current growing chain. Propene first binds at the available site of $[\mathrm{C}{\mathrm{p}}_2\mathrm{ZrR}{]}^{+}$, giving a coordinated alkene adjacent to the metal-bound chain. In the Cossee-Arlman mechanism, the chain migrates to the substituted alkene carbon while the terminal carbon forms the new Zr–C bond:The electron-pair arrows for insertion take the old Zr–R bond toward the substituted alkene carbon and the alkene pi bond toward the new Zr–terminal-carbon bond. A four-center arrangement permits both bonds to change in one migratory insertion; Zr remains Zr(IV). The product again has an accessible site, so coordination and insertion can repeat.

Predominant 1,2 insertion leaves a primary Zr–CH${}_2$ chain end and generates head-to-tail polypropylene. It avoids the more crowded alternative secondary Zr–CH(Me) end and unfavorable methyl contacts with the cyclopentadienyl ligands in the insertion geometry. This explains regioselectivity. An unbridged, unsubstituted zirconocene does not by itself establish a fixed tacticity; the schematic alignment of methyl groups in a polymer drawing is not evidence for an isotactic product.

The product is carbonic acid, $\mathrm{HO}-\mathrm{C}(=\mathrm{O})-\mathrm{OH}$. The oxygen lone pair Homo of water donates into the carbon-centred component of the carbon dioxide ${\pi }^{\ast }$ LUMO. Draw an electron-pair arrow from the water oxygen to carbon and a second from one C=O pi bond onto its oxygen:The first step bends the original linear electrophile and gives carbon three oxygen neighbours; carbon is approximately trigonal planar in the resulting carbonate framework. Proton transfer from the oxonium oxygen to the negatively charged oxygen produces the neutral acid. In real water, solvent molecules relay that proton rather than requiring a strained isolated intramolecular transfer. The forward mechanism does not assert that most dissolved carbon dioxide is present as carbonic acid at equilibrium.

The products are hexanenitrile, $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}\equiv \mathrm{N}$, and ethyl triphenylmethyl ether, $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The nitrile carbon comes from cyanide, so replacing chloride in the five-carbon chain gives a six-carbon product.

For the primary chloride, the carbon-end lone pair of $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted nucleophilic substitution step forms the C–C bond and releases $\mathrm{C}{\mathrm{l}}^{-}$. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$. An ethanol oxygen lone pair then attacks the carbocation, giving $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-{\mathrm{O}}^{+}(\mathrm{H})\mathrm{Et}$. Finally pyridine accepts that O–H proton, with the O–H bond pair returning to oxygen, giving the neutral ether and pyridinium chloride.

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.

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 $\mathrm{C}=\mathrm{O}$ to $\mathrm{CH}(\mathrm{OH})$ adds two H atoms and gives ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{N}{\mathrm{O}}_2$.

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 $\mathrm{N}-\mathrm{C}(=\mathrm{O})$ by $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2$; its oxygen is removed rather than retained as another alcohol. Together with ketone reduction, the net change is $+4\mathrm{H}-\mathrm{O}$, giving ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{NO}$. No stereochemistry is specified: the new alcohol stereocentre is formed as a racemic mixture in an achiral environment.