 Solution

Let $\mathrm{Ar}$ denote the 2,5-dimethoxyphenyl group. A is the doubly C-alkylated beta-ketoesterThe central carbon has no hydrogen after the second step. The alkene, ester and tert-butyl ether remain intact.

carbonate reversibly removes a proton from the methylene group between the two carbonyl groups. The resulting resonance-stabilized enolate attacks the primary iodoethyl carbon of the aryl-containing reagent by an $S_{\mathrm{N}}2$ displacement, releasing iodide. The monoalkylated intermediate still has one proton at the same activated carbon. A second deprotonation followed by $S_{\mathrm{N}}2$ attack on the primary bromide introduces the but-3-en-1-yl group.

The beta-dicarbonyl position is much more acidic than the methylene group next to the ketone and ether at the other end, explaining the alkylation site. Counting the retained carbon skeleton gives ${\mathrm{C}}_{24}{\mathrm{H}}_{36}{\mathrm{O}}_6$, including the two carbonyl groups and one terminal alkene. The second panel of the original product drawing specifies this connectivity.