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

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.