[o alquimista]

R.G. Amiet, The Preparation of Lucigenin, Journal of Chemical Education, Volume 59 no. 2, February 1982

"To a refluxing solution of N-methylacridone (1g) in ethanolic HCl (50 ml ethanol, 10 ml conc. HCl) is added zinc dust (3.2 g) portionwise over 15 min. The mixture is refluxed for a further 30 min, then added to water (100 ml) to give a green precipate of the crude bis-acridinium compound."

Does anyone know the mechanism of this reductive coupling reaction? Attached is my guess.

[Albert]

Do you mean Clemmensen reduction?

[o alquimista]

Due to the reagents I already figured out that it is a Clemmensen reduction. I was wondering how the coupling of the two aromatic ketons takes place (mechanism). As in the above N-methylacridone --> bis-acridinium example.

[movies]

For a homoogeneous mechanism (the one Albert posted is a heterogeneous mechanism) of the Clemmenson reduction you form a benzyllic radical (in this case doubley benzyllic) so you could have a radical-radical coupling to make the first carbon-carbon bond.  Alternatively, you could draw a mechanism with an doubley benzyllic carbanion and an extended iminium ion in order to make the carbon-carbon bond.  I think I favor the radical coupling, myself.

[o alquimista]

In this example the product has one extra double bond, namely the coupling bond. If it's a reaction between two radicals, there is one double bond too little.

[movies]

You would couple two ketyl type radicals and then eliminate water.