= Solution
Dibutylboron triflate coordinates the imide carbonyl and triethylamine removes the propionyl $\alpha$ proton, giving predominantly the \b[Z dibutylboron <enolate>]. Its connectivity is $\mathrm{Aux-N-C(O-BBu_2)=CHMe}$; 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. \b[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 \b[R]: priority is imide carbonyl, then $\mathrm{CH(OH)CH=CHPh}$, 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.
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