Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 9

A is 1-propyl-1,2,4-triazole, with propyl on the nitrogen that is adjacent to one N and one C in the triazole ring, rather than on the nitrogen between the two carbon atoms.

DBU removes the ring NH proton, producing a resonance-stabilized triazolide anion. Nitrogen attacks the primary bromide carbon by an SN2 reaction, displacing bromide and giving the N-propyl product. Alkylation at the equivalent tautomer-related N1/N2 positions gives the same constitutional product, while N4 alkylation would give a different isomer. Charge distribution in the triazolide and the accessibility of its nitrogen lone pairs favor the N1 product under the stated conditions. This is kinetic regioselectivity, not an assertion that the unalkylated tautomer has a permanently localized negative charge.

The nonnucleophilic magnesium amide metalates the triazole carbon between the two ring nitrogen atoms, conventionally C5 of 1-propyl-1,2,4-triazole. That proton is more readily removed because the resulting heteroaryl anion is stabilized by the neighboring nitrogen atoms; the reagent therefore acts as a base rather than adding a tetramethylpiperidinyl group.

The heteroaryl organomagnesium intermediate attacks the allylic bromide, forming a new C–C bond. The product is 5-(cyclohex-2-en-1-yl)-1-propyl-1,2,4-triazole. Ordinary allylic substitution and an $S_{\mathrm{N}}{2}^{\prime }$ description give the same unlabelled cyclohexenyl connectivity for this unsubstituted allylic ring, so the structural assignment does not distinguish those trajectories.

The molecular formula checks the assignment: N-propyltriazole is ${\mathrm{C}}_5{\mathrm{H}}_9{\mathrm{N}}_3$; replacing a ring H by ${\mathrm{C}}_6{\mathrm{H}}_9$ gives ${\mathrm{C}}_{11}{\mathrm{H}}_{17}{\mathrm{N}}_3$, as printed.

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.

The analogous N-oxide rearrangement uses the carbon next to nitrogen that most readily supports addition and rearomatization. In isoquinoline this is C1. Thus replace the quinolin-2-yl group of D by isoquinolin-1-yl:The ynone-to-chromone atom mapping is unchanged. The change is attachment at isoquinoline C1, rather than copying quinoline numbering and assigning an isoquinolin-2-yl carbon substituent at the nitrogen position.

TODO: user-deferred biology. Enzymatic drug metabolism and conjugation is deferred under the current nonbiology scope.