 Solution

Each imidazole already has one nitrogen substituted. Its remaining pyridine-like nitrogen attacks a benzylic carbon bearing chlorine in an $S_{\mathrm{N}}2$ reaction, expelling chloride and producing an imidazolium center. Repetition at the other end closes the ring. The expected product is therefore a dicationic bis-imidazolium macrocycle, in which two meta-xylylene bridges connect the two imidazolium rings. Both nitrogens of each imidazolium are alkylated; the aryl rings remain intact.

A compact way to specify its connectivity is to traverse the ring:Here each indicated imidazolium connects its two substituents through its two different ring nitrogen atoms, rather than through a single nitrogen.

chloride acts as an anion template: association with the positively charged intermediates and the inward-facing imidazolium CH donors brings the reacting ends together, favoring cyclization over intermolecular oligomerization. Tetrabutylammonium is a spectator counterion, not a covalent part of the product. The two substitutions produce two chloride counterions for the dication; template chloride can also occupy the cavity, with the remaining ions in solution. The schematic below specifies connectivity and charge without reproducing the source drawing.