 Solution

Label the ynone carbons $\mathrm{Ar}(\mathrm{OH})-\mathrm{CO}-{\mathrm{C}}_{\alpha }\equiv {\mathrm{C}}_{\beta }-\mathrm{Ph}$. The N-oxide oxygen adds to the electrophilic $C_{\beta }$, giving an O-vinylated quinolinium species with an enolate-like nucleophilic $C_{\alpha }$.

The chain $C_{\alpha }-C_{\beta }-O-N-C{2}_{\mathrm{quinoline}}$ is now arranged for rearrangement: form a bond between $C_{\alpha }$ and quinoline C2 while cleaving N–O and forming the $C_{\beta }=O$ bond. This can be represented through a cyclic addition/rearrangement intermediate, followed by proton transfer and rearomatization. The resulting carbon skeleton is the heteroarylated 1,3-dicarbonyl systemFinally, the ortho phenolic oxygen attacks the distal carbonyl. Cyclodehydration forms the chromone ring and eliminates water. The product is 2-phenyl-3-(quinolin-2-yl)chromen-4-one. The original ynone carbonyl becomes the chromone 4-one, the phenyl-bearing alkyne carbon becomes C2, and the other alkyne carbon becomes C3 carrying the heteroaryl substituent.

This atom mapping explains both the new C–C bond and the loss of the N-oxide oxygen from the heteroarene; it is not a direct substitution on the phenyl ring.