 Solution

First the tertiary amine attacks the primary allylic bromide carbon by SN2 reaction, giving the quaternary ammonium salt. The less hindered nitrogen face is favored by the pre-existing fused-ring stereochemistry.

DBU removes the proton at the carbon alpha to both the ammonium nitrogen and the ester carbonyl. The resulting ammonium ylide undergoes a concerted 2,3-sigmatropic rearrangement: the N–CH${}_2$(allyl) bond breaks, a bond forms between the ylide carbon and the aryl-bearing terminal allyl carbon, and the double bond moves to give $-\mathrm{CH}(\mathrm{Ar})-\mathrm{C}(\mathrm{Me})=\mathrm{C}{\mathrm{H}}_2$. Nitrogen becomes neutral again. An ordinary alkyl substitution would attach the wrong end of the allyl unit.

An envelope-like five-membered transition structure keeps Ar outside the crowded bicyclic framework. With the pre-existing OMe dash and ring-junction geometry fixed, place the newly installed side chain on the solid-wedge face at the former ylide carbon and choose the benzylic configuration that places Ar outward, as in the displayed intermediate. This is diastereoselective, suprafacial allyl transfer.

Finally methanesulfonic acid protonates the methyl acetal. Methanol leaves to give an oxocarbenium/iminium-stabilized electrophilic centre, which is attacked intramolecularly by the electron-rich arene at its available ortho position. Deprotonation restores aromaticity and closes the bridge. The terminal isopropenyl group survives; it is not the bond consumed in this final Intramolecular Friedel-Crafts reaction. The constrained tether selects the displayed bridged diastereomer.