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RDKit bond type - query #7167

Answered by ptosco
Suganesh-repo asked this question in Q&A
Feb 15, 2024 · 1 comments · 1 reply
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  1. The second molecule is chemically incorrect as carbon 0 is pentavalent - you cannot have an exo double bond on an aromatic ring.
  2. The bond does not show up as EITHERDOUBLE because it would not make sense: your system is symmetric, and therefore there is no possibility for geometric isomerism. If you introduce an asymmetry, the bond will show up as EITHERDOUBLE. However, the carbon is still pentavalent and hence that molecule is chemically incorrect.
m = Chem.MolFromSmiles("CC=C1=CC=CC=C1", sanitize=False)
m.GetBondWithIdx(1).SetBondDir(Chem.BondDir.EITHERDOUBLE)
m

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