 Solution

Label the enamine carbon bonded to nitrogen C1 and its alkene partner C2; C6 is the other carbon next to C1. B is the C-alkylated iminium ion bromide. Its new substituent at C2 is $-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}(\mathrm{C}{\mathrm{O}}_2\mathrm{Et})=\mathrm{C}{\mathrm{H}}_2$, while C1 is double-bonded to the pyrrolidine nitrogen, which bears positive charge. bromide is the counterion, accounting for Br in the empirical formula without requiring a retained C–Br bond.

The C2 end of the enamine attacks the bromomethyl carbon by $S_{\mathrm{N}}2$ displacement. Simultaneously, donation of the nitrogen lone pair creates the C1=N bond. This is ordinary enamine C-alkylation, not N-alkylation.

proton transfer removes an alpha proton at C6, converting the iminium ion into the enamine regioisomer C1=C6. That enamine attacks the terminal $\mathrm{C}{\mathrm{H}}_2$ of the tethered acrylate in an intramolecular Michael addition. nitrogen donation restores the iminium ion, and protonation of the ester enolate completes the sequence.

The new bridge is $C2-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{C}{\mathrm{O}}_2\mathrm{Et})-\mathrm{C}{\mathrm{H}}_2-C6$. Together with the original ring and its C1 bridge this produces the bicyclo$[\mathrm{3.3.1}]$nonane iminium ion skeleton in C. It retains the pyrrolidine group and the ester; hydrolysis to a ketone is not part of the displayed reaction. The second panel of the original annulation figure shows the chemical atom connectivity and distinguishes ionic bromide in B from a covalent bromine substituent.