hand_formulas_appendix_aromatic.json

{
  "_README": [
    "APPENDIX: TEN MORE FORMULAS COMPUTED BY HAND, COVERING THE LOWERCASE AROMATIC",
    "NITROGEN CASE THAT THE ORIGINAL THIRTY MISSED.",
    "",
    "WHY THIS FILE EXISTS, AND WHY IT IS A SEPARATE FILE. The original hand set of thirty",
    "was frozen first and must stay frozen; appending to it after the fact would destroy the",
    "one property that makes it worth anything. So the gap is recorded here instead, with the",
    "history intact.",
    "",
    "THE GAP. The first thirty contain pyrrole (written [nH], a bracket atom), pyridine (a",
    "bare two-connected aromatic n) and adenine. None contains a bare lowercase aromatic",
    "nitrogen with THREE connections, which is what an N-substituted azole has: N-methyl",
    "pyrrole, N-methylimidazole, caffeine, carbazole, indolizine. The parser's original rule 3",
    "gave every aromatic atom one extra unit of bond order for its formal double bond, which",
    "for a three-connected nitrogen came to 3 + 1 = 4 against a normal valence of 3. The parser",
    "therefore REFUSED rather than answering, which is the designed safe behaviour and is why",
    "no wrong number was ever produced. But refusing is not free: 197 of the 4,610 comparable",
    "structures in the aromatic measurement set (4.3 percent) fall in this class, and they are",
    "chemically central rather than exotic.",
    "",
    "HOW THE GAP WAS FOUND. By the scorer's own selftest, on caffeine, which is one of its",
    "three fixed reference molecules. It was NOT found by comparing against rdkit and was not",
    "fixed by consulting rdkit's answer. That distinction is the whole discipline here.",
    "",
    "THE RULE, DERIVED FROM CHEMISTRY RATHER THAN FROM RDKIT. An aromatic ring needs six pi",
    "electrons. An atom supplies them either by contributing one electron through a formal",
    "double bond, or by donating a lone pair. It cannot do both. The original rule 4 already",
    "said this for aromatic o and s, which is why furan and thiophene were right from the",
    "start; the omission was that it was never extended to nitrogen. A bare lowercase aromatic",
    "n with two connections is pyridine-type and carries the formal double bond: 2 + 1 = 3,",
    "which fills nitrogen exactly. A bare lowercase aromatic n with three connections has spent",
    "all three sigma bonds and must donate its lone pair instead, so it carries no formal double",
    "bond: 3 + 0 = 3, which also fills nitrogen exactly. Both cases reach valence 3 with no room",
    "left, so a bare lowercase aromatic nitrogen NEVER takes an implicit hydrogen. The pyrrole",
    "NH is written [nH] and gets its hydrogen from the bracket, as it always did.",
    "",
    "Each formula below was walked atom by atom on paper before that rule was written into the",
    "parser, exactly as the original thirty were."
  ],
  "frozen_utc": "2026-07-28",
  "supersedes_nothing": "hand_formulas.json is unchanged and still governs the original 30",
  "n": 10,
  "set": [
    {
      "id": "a01", "name": "N-methylpyrrole", "smiles": "Cn1cccc1",
      "formula": "C5H7N", "charge": 0,
      "element_counts": {"C": 5, "H": 7, "N": 1},
      "heavy_atom_count": 6, "bond_count": 6,
      "hand": "The aromatic n has three neighbours (the methyl and two ring carbons) so it donates its lone pair and takes 0 H. Four aromatic c with two neighbours each take 1 H = 4. Methyl 3 H. H=7. bonds = 6-1+1 ring = 6."
    },
    {
      "id": "a02", "name": "1-methylimidazole", "smiles": "Cn1ccnc1",
      "formula": "C4H6N2", "charge": 0,
      "element_counts": {"C": 4, "H": 6, "N": 2},
      "heavy_atom_count": 6, "bond_count": 6,
      "hand": "Atoms 0 methyl, 1 n (neighbours 0, 2 and the ring closure to 5, so three: 0 H), 2 c and 3 c with two neighbours each (1 H each), 4 n with two neighbours (pyridine-type, 0 H), 5 c with two neighbours (1 H). H = 3 + 1 + 1 + 1 = 6. bonds = 6-1+1 = 6."
    },
    {
      "id": "a03", "name": "caffeine", "smiles": "Cn1cnc2c1c(=O)n(C)c(=O)n2C",
      "formula": "C8H10N4O2", "charge": 0,
      "element_counts": {"C": 8, "H": 10, "N": 4, "O": 2},
      "heavy_atom_count": 14, "bond_count": 15,
      "hand": "Fourteen heavy atoms: 8 C, 4 N, 2 O. All three aromatic nitrogens that carry a methyl have three neighbours and take 0 H; the one bare n has two neighbours and is pyridine-type, also 0 H. Of the aromatic carbons only atom 2 has two neighbours and takes 1 H; the ring-fusion carbons and the two carbonyl carbons all have three and take 0. Three methyls give 9 H. H = 9 + 1 = 10. Fifteen bonds: 13 chain plus 2 ring closures, and 14-1+2 rings = 15 confirms it."
    },
    {
      "id": "a04", "name": "1-methylpyrazole", "smiles": "Cn1nccc1",
      "formula": "C4H6N2", "charge": 0,
      "element_counts": {"C": 4, "H": 6, "N": 2},
      "heavy_atom_count": 6, "bond_count": 6,
      "hand": "The substituted n has three neighbours so 0 H. The adjacent bare n has two so it is pyridine-type, also 0 H. Three aromatic c with two neighbours give 1 H each = 3. Methyl 3 H. H=6. bonds = 6-1+1 = 6. Note this is the two-adjacent-nitrogen case, which the original thirty had no example of."
    },
    {
      "id": "a05", "name": "indole", "smiles": "c1ccc2[nH]ccc2c1",
      "formula": "C8H7N", "charge": 0,
      "element_counts": {"C": 8, "H": 7, "N": 1},
      "heavy_atom_count": 9, "bond_count": 10,
      "hand": "Nine heavy atoms, 8 C and 1 N. The two ring-fusion carbons have three neighbours and take 0 H. The other six aromatic carbons have two neighbours and take 1 H each = 6. The bracket [nH] states 1 H. H=7. bonds = 9-1+2 rings = 10."
    },
    {
      "id": "a06", "name": "N-methylpyridinium", "smiles": "C[n+]1ccccc1",
      "formula": "C6H8N+", "charge": 1,
      "element_counts": {"C": 6, "H": 8, "N": 1},
      "heavy_atom_count": 7, "bond_count": 7,
      "hand": "[n+] is a bracket atom with no written H so 0 H, and the bracket rule settles it without any aromatic reasoning at all. Five aromatic c with two neighbours give 1 H each = 5. Methyl 3 H. H=8. Net charge +1 so the formula takes a plus suffix. bonds = 7-1+1 ring = 7."
    },
    {
      "id": "a07", "name": "3-methylthiophene", "smiles": "Cc1ccsc1",
      "formula": "C5H6S", "charge": 0,
      "element_counts": {"C": 5, "H": 6, "S": 1},
      "heavy_atom_count": 6, "bond_count": 6,
      "hand": "The substituted ring carbon has three neighbours and takes 0 H. Three aromatic c with two neighbours give 1 H each = 3. Aromatic s donates its lone pair and takes 0 H. Methyl 3 H. H=6. bonds = 6-1+1 = 6."
    },
    {
      "id": "a08", "name": "2-pyridone", "smiles": "O=c1cccc[nH]1",
      "formula": "C5H5NO", "charge": 0,
      "element_counts": {"C": 5, "H": 5, "N": 1, "O": 1},
      "heavy_atom_count": 7, "bond_count": 7,
      "hand": "The aromatic carbon bearing the exocyclic double-bonded O has three neighbours and takes 0 H. Four aromatic c with two neighbours give 1 H each = 4. [nH] states 1 H. The doubly bonded O reaches valence 2 and takes 0 H. H=5. bonds = 7-1+1 ring = 7. This is the aromatic-carbon-with-an-exocyclic-double-bond case, which the original thirty had only in uppercase form."
    },
    {
      "id": "a09", "name": "indolizine", "smiles": "c1ccn2cccc2c1",
      "formula": "C8H7N", "charge": 0,
      "element_counts": {"C": 8, "H": 7, "N": 1},
      "heavy_atom_count": 9, "bond_count": 10,
      "hand": "A bridgehead nitrogen shared by both rings, so it has three neighbours and takes 0 H. One aromatic carbon is the other fusion atom with three neighbours, 0 H. The remaining seven aromatic carbons have two neighbours each and give 1 H each = 7. bonds = 9-1+2 rings = 10."
    },
    {
      "id": "a10", "name": "2-methylbenzimidazole", "smiles": "Cc1nc2ccccc2[nH]1",
      "formula": "C8H8N2", "charge": 0,
      "element_counts": {"C": 8, "H": 8, "N": 2},
      "heavy_atom_count": 10, "bond_count": 11,
      "hand": "Ten heavy atoms, 8 C and 2 N. The methyl-bearing ring carbon and both ring-fusion carbons have three neighbours and take 0 H. Four benzo carbons have two neighbours and give 1 H each = 4. The bare n has two neighbours, pyridine-type, 0 H. [nH] states 1 H. Methyl 3 H. H = 3+4+1 = 8. bonds = 10-1+2 rings = 11."
    }
  ]
}