In this question, the compound below was named 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one.

enter image description here

I assume that we can split this into three parts.

  1. The propyl group attached to the nitrogen, namely: 1-(propan-2-yl)- (basic part of nomenclature)
  2. The pyrazole derivative like ring, namely 1,2,3,5-tetrahydro-6H-pyrazolo (For this refer to this answer)
  3. The pyridine ring, namely [3,4-b]pyridin-6-one.

The third part is where I am confused, what does the [3,4-b] stand for?

Another example for said nomenclature is in Acdlabs

enter image description here

This leads me to assume the b stands for the fusing of the two bicyclo rings. But then, why [3,2-b] and why does the numbering not count the bridgehead carbons?

  • $\begingroup$ The bracketed letters gives you "which one is the face ip". When you merge the rings the second one can be fused head or cross up with respect to the other one. I hope it helps else I can better formulate it....also acdlabs.com/iupac/nomenclature/79/…. $\endgroup$
    – Alchimista
    Commented May 24, 2021 at 11:47
  • 2
    $\begingroup$ look also here link.springer.com/article/10.1007/s40828-016-0035-3. a, b... refers to the bond fused $\endgroup$
    – Alchimista
    Commented May 24, 2021 at 11:54
  • $\begingroup$ And the number in bracket refers to the original numbering. The entire molecule is afterwards renumbered. $\endgroup$
    – Alchimista
    Commented May 24, 2021 at 12:56
  • $\begingroup$ @Alchimista, according to the paper you linked "The first face described by the numbered positions 1 and 2 is lettered ‘a’, with the letters continuing alphabetically clockwise around the exterior of the parent", then how do we get the second example? shouldn't it be [3,2-d]? $\endgroup$ Commented May 24, 2021 at 13:25
  • $\begingroup$ given the molecule there is no need to go to d. Thieno[2.3-b]furan would be that with the heteroatoms "on the same side". $\endgroup$
    – Alchimista
    Commented May 24, 2021 at 14:48


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