Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 2 / B9 / c / Solution

$B={\mathrm{N}}_3{\mathrm{P}}_3(\mathrm{NHCy}{)}_6$, with two cyclohexylamino substituents on each original phosphorus. All six P–Cl bonds have been replaced; the six N–H bonds of those substituents remain. Each equivalent of triethylamine traps one equivalent of HCl.

$C$ contains three four-membered ${\mathrm{P}}^{\mathrm{V}}-\mathrm{N}(\mathrm{Cy})-{\mathrm{P}}^{\mathrm{III}}(\mathrm{Cl})-\mathrm{N}(\mathrm{Cy})$ spiro rings, one at each original ring phosphorus. Each PCl${}_3$ joins the two geminal NHCy groups: replacement of two of its chlorides forms two new P–N bonds and releases two HCl, consuming both N–H protons. The new phosphorus retains one chloride and a lone pair, making it three-coordinate P(III); the original six-membered ${\mathrm{P}}_3{\mathrm{N}}_3$ ring retains P(V). The product formula is ${\mathrm{N}}_3{\mathrm{P}}_3\{(\mathrm{NCy}{)}_2\mathrm{PCl}{\}}_3$. For each spiro ring, the original phosphorus is bonded to both exocyclic nitrogens, and each of those nitrogens is bonded to the same new PCl center; the braces do not indicate a simple linear substituent.

The three new P–Cl directions permit an all-cis isomer and a cis,trans isomer. Relative to the central ring, write these as $(+,+,+)$ and $(+,+,-)$, with cyclic relabeling giving equivalent choices. Reversing all signs does not create a third relative configuration. This describes the configuration pattern; puckering and substituent conformations are additional structural details.