 Solution

The simplest assignment is two phenylethynyl bridges and four equivalent terminal methyl groups, represented as $\mathrm{A}{\mathrm{l}}_2\mathrm{M}{\mathrm{e}}_4(\mu -\mathrm{C}\equiv \mathrm{CPh}{)}_2$. All 12 methyl protons then occupy the same class of environment, accounting for the single nonaromatic singlet without requiring methyl bridge/terminal averaging. The phenyl protons account for the aromatic-region signals; the metal-bound acetylide has no terminal acetylenic hydrogen.

An alkynyl bridge has electronic possibilities unavailable to a saturated methyl: in addition to sigma attachment from its carbon end to one Lewis acid centre, its occupied C≡C pi system can donate to the other aluminium. A delocalized sigma/pi description stabilizes the phenylethynyl-bridged arrangement while keeping the methyl groups terminal. One must not draw an unchanged localized triple bond plus two independent full ordinary Al–C bonds at the same carbon and count all of them as separate two-electron bonds. The exact distribution of sigma and pi donation requires structural or orbital evidence; this NMR observation alone does not supply a numerical bridge bond order.

The absence of concentration or temperature dependence is consistent with this preferred arrangement throughout the measured range. It is not proof that the molecule has no motion: exchange between symmetry-equivalent terminal positions, or exchange remaining fast throughout that range, could be spectroscopically invisible. Phenylethynyl bridging in this particular dimethylaluminium dimer has independent primary NMR precedent.^[ref]