 Solution

C and the dimer of D have the same composition, $\mathrm{C}{\mathrm{u}}_2\mathrm{L}{\mathrm{i}}_2(\mathrm{TMP}{)}_4$, but different metal ordering around an eight-membered metal–nitrogen ring. C has adjacent Cu sites and adjacent Li sites; in the conventional Gilman dimer D${}_2$, Cu and Li alternate around the four metal positions. D is represented by $\mathrm{Li}[\mathrm{Cu}(\mathrm{TMP}{)}_2]$ per formula unit. The salt by-product is LiCl.

A TMP anion has formula ${\mathrm{C}}_9{\mathrm{H}}_{18}\mathrm{N}$ and mass about 140.25. Therefore the mass per empirical unit is $2(140.25)+63.55+6.94\simeq 350.99$, giving $M_r\simeq 702$ for the dimer, consistent with $699\pm 5$. A monomer would be near 351 and is excluded. In C, inequivalent Cu/Cu, Cu/Li and Li/Li amide surroundings, combined with the constrained TMP orientations, permit the three methyl environments in the observed 1:2:1 integral ratio. D${}_2$ has the higher exchange-averaged methyl equivalence needed for one methyl signal. One signal need not imply that all instantaneous solid-state methyl environments are identical.

The remaining formulae follow from atom and charge balance:E can itself aggregate; its formula here specifies the copper-amide unit. F is a mixed organyl/amido Gilman cuprate, and G combines one F unit with one D unit. The distinct metal-ordering isomers and these mixed aggregates have experimental precedent.^[ref]