 Solution

Dibutylboron triflate coordinates the imide carbonyl and triethylamine removes the propionyl $\alpha$ proton, giving predominantly the Z dibutylboron enolate. Its connectivity is $\mathrm{Aux}-\mathrm{N}-\mathrm{C}(\mathrm{O}-\mathrm{BB}{\mathrm{u}}_2)=\mathrm{CHMe}$; the imide auxiliary remains attached, rather than being displaced during enolization. Coordination of cinnamaldehyde's oxygen to boron organizes a six-membered Zimmerman–Traxler transition state.

A chair with the aldehyde's styryl group pseudoequatorial avoids the auxiliary's bulky benzyl side. Formation of the C–C bond and transfer of the aldehyde oxygen into the boron coordination sphere give the boron aldolate, which hydrolyses to D. A Z enolate gives the syn aldol relationship. In the printed product zigzag, the labelled $\alpha$-methyl bond should therefore be a solid wedge, on the same stereochemical face as the neighbouring solid-wedge hydroxyl bond. With the stated connectivity, the labelled carbon is R: priority is imide carbonyl, then $\mathrm{CH}(\mathrm{OH})\mathrm{CH}=\mathrm{CHPh}$, then methyl, then hydrogen. The adjacent hydroxyl centre is S. These assignments refer to the actual auxiliary and product orientation, not an arbitrary choice of an otherwise equivalent chair.