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A diazine is an aromatic six-membered heterocyclic compound whose parent ring contains four carbon atoms and two nitrogen atoms. The three unsubstituted parent arrangements share formula ${\mathrm{C}}_4{\mathrm{H}}_4{\mathrm{N}}_2$ but differ in connectivity: pyridazine is the $1,2$ arrangement, pyrimidine the $1,3$ arrangement, and pyrazine the $1,4$ arrangement.^[ref]

Why only three? Fix one nitrogen at position $1$ on a six-membered ring. The other can be adjacent, separated by one intervening ring atom, or opposite. Reflection identifies the remaining apparent alternatives: positions $6$ and $2$ describe the same adjacent arrangement, while $5$ and $3$ describe the same next-nearest arrangement. Thus the distinct shortest ring separations are one, two and three bonds.

The three diazines are constitutional isomers, not different orientations of one molecule. Rotating a drawing changes its appearance without changing its atom-to-atom connections. Moving a ring nitrogen to another connectivity position instead changes which chemical compound the drawing represents. This makes the diazine family a compact example of why a molecular formula alone does not specify chemical identity.

Pyrazine belongs to $D_{2h}$, the D2h point group. With the printed axes, its three twofold axes are $x,y,z$; it also has an inversion centre and the three coordinate mirror planes. The nitrogen atoms occupy sites 1 and 4 on the $y$ axis. Interchanging carbon and nitrogen is not a molecular symmetry operation, so the sixfold benzene symmetry is lost.

The six $\pi$ electrons give the Hückel method energyThe appropriate localized reference contains two C=N $\pi$ bonds and one C=C $\pi$ bond. Three C=C bonds would incorrectly replace the nitrogen one-centre parameters. For one isolated C=N pair, the dimensionless matrix iswhose eigenvalues are $(1\pm \sqrt{5})/2$. Since $\beta <0$, the occupied bonding molecular orbital has energy $\alpha +(1+\sqrt{5})\beta /2$. Two doubly occupied C=N bonds and one doubly occupied C=C bond consequently giveTherefore the signed delocalization energy isThe positive stabilization is $1.82039|\beta |$, smaller than the benzene value $2|\beta |$ in the stated model but still stabilizing.

Pyrazine is aromatic: its planar cyclic conjugated system contains six $\pi$ electrons and has a filled set of three bonding $\pi$ molecular orbitals. Each ring nitrogen contributes one $p_z$ electron; its in-plane lone pair is outside this $\pi$ system. The heteroatoms change orbital energies and coefficients without interrupting cyclic conjugation or the $4n+2$ count. Aromaticity does not require the stabilization to equal benzene's.

The cyclic amino sugar equilibrates with its open-chain $\alpha$-amino aldehyde. Two such units condense head-to-tail: each amino group adds to the other unit's aldehyde, followed by dehydration to the cyclic bis-imine/dihydropyrazine system. Formula accounting identifies the observed intermediate:This corresponds to ${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{N}}_2{\mathrm{O}}_8$, so the accumulating intermediate has lost two waters but has not yet acquired the fully aromatic pyrazine oxidation state. The unchanged portions of both carbon chains become the substituents on opposite ring carbons.

Dehydrogenative aromatization gives B, retaining both hydroxylated side chains and releasing H${}_2$: the intermediate loses two hydrogen atoms to reach ${\mathrm{C}}_{12}{\mathrm{H}}_{20}{\mathrm{N}}_2{\mathrm{O}}_8$, mass 320. A competing elimination of the hydroxyl group immediately adjacent to the ring in one side chain, with loss of a ring-adjacent proton, gives an exocyclic unsaturated intermediate. Proton transfers shift that unsaturation into the ring and leave a deoxygenated side-chain methylene. This dehydrative aromatization gives A, formally $322-18=304$, rather than requiring a separate unexplained reducing agent. Both pathways gain aromatic stabilization, while only the dehydrogenative branch requires H${}_2$ release. The acid/base proton-transfer steps can proceed through several tautomeric representations; mass balance distinguishes the two net branches.

Pyridazine is the $1,2$-diazine: an aromatic six-membered heterocyclic compound with adjacent ring nitrogen atoms and formula ${\mathrm{C}}_4{\mathrm{H}}_4{\mathrm{N}}_2$. The systematic ring position labels refer to connectivity, not to two added substituents on an unchanged hydrocarbon ring.^[ref]

Tracing once around the ring gives the atom sequence $\mathrm{N}-\mathrm{N}-\mathrm{C}-\mathrm{C}-\mathrm{C}-\mathrm{C}$. In pyrimidine, the shortest path between the two nitrogen atoms includes one carbon; in pyrazine, it includes two. Those are different constitutional isomers, although all three have the same elemental formula.^[ref]

For a structure-reading exercise, start with a hexagon marked with two adjacent $\mathrm{N}$ labels and rotate the entire page. It remains pyridazine. Move one $\mathrm{N}$ label across a ring vertex instead, while keeping the ring connectivity interpretation, and the identity can become pyrimidine. The useful distinction is between moving a picture and changing a molecule.

Pyrimidine is the aromatic organic compound ${\mathrm{C}}_4{\mathrm{H}}_4{\mathrm{N}}_2$ with a six-membered ring containing nitrogen at positions $1$ and $3$. It is therefore a diazine, specifically 1,3-diazine.^[ref]

Changing the nitrogen positions gives different molecules: pyridazine has the $1,2$ arrangement, while pyrazine has the $1,4$ arrangement. The same elemental formula therefore does not uniquely specify ring connectivity. In the $1,3$ arrangement, the two nitrogen atoms have one ring carbon between them along the shorter path; in the $1,2$ arrangement they are adjacent.^[ref]

Cytosine, thymine and uracil contain a pyrimidine-derived ring framework, but their added groups and bonding make them chemically distinct from unsubstituted pyrimidine. This distinguishes a parent ring structure from the biological nucleobases named by association with that structure.