= Solution
Each <imidazole> already has one nitrogen substituted. Its remaining pyridine-like nitrogen attacks a benzylic <carbon> bearing chlorine in an $S_\mathrm N2$ reaction, expelling <chloride> and producing an <imidazolium> center. Repetition at the other end closes the ring. The expected product is therefore \b[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:
$$
\mathrm{m\!\!-!C_6H_4-CH_2-N(imidazolium)-CH_2-
m\!\!-!C_6H_4-CH_2-N(imidazolium)-CH_2-}\quad\text{back to the first aryl group}.
$$
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 \b[<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.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2021/paper-1-macrocycle.png]
{title=Connectivity of the chloride-templated bis-imidazolium macrocycle}
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