I've just come across a figure indicating the positions of conjugated bonds in indigo and leucoindigo, and I'm quite confused about why both molecules aren't fully conjugated. enter image description here

My reasoning is as follows:

  1. Both carbons above and below the bond are sp2 hybridized and have π electrons.
  2. The carbon to the left is sp2 hybridized and has π electrons, and the nitrogen to the right has a lone pair.
  3. Same as 2.
  4. Same as 1.

I also want to ask about why leuco-indigo isn't planar. According to Wikipedia:

In indigo white, the conjugation is interrupted because the molecule is nonplanar.

  • $\begingroup$ If a carbon atom is sp3 hybridized, it doesn't have any pi electrons. Other than that, you are mostly right, and the book tends to oversimplify things. Besides, they have their leucoindigo formula downright wrong. $\endgroup$ Commented Oct 20, 2017 at 4:38

2 Answers 2


Let's just start with stating the very obvious: Figure 2.23 has more errors than there should be in an entire chapter.


With the exception of bonds to hydrogen, all other bonds are nicely conjugated in indigo. The π molecular orbital show this very clearly.[1] The molecule has a fantastic C2h symmetry, which also means it is in one plane. Intramolecular hydrogen bonds certainly further stabilise this conformation. They can be visualised with the help of the Quantum Theory of Atoms in Molecules (QTAIM).[2]

pi orbitals of indigo (click for full image)

qtaim of indigo

Interestingly eleven out of twenty π orbitals are occupied, which leaves the HOMO being anti-bonding with respect to the central carbon-carbon bond, which is 138 pm long.

Going further I will first add electrons, then add protons according to the following scheme:

redox of indigo


I am happy to report, that the molecule stays flat. The central carbon-carbon bond elongates a little to 141 pm. It basically still has comparable orbitals, although the location of the nodal planes already accommodate the divide of the central bond. I am guessing that the intramolecular hydrogen bonds are accountable for this since the molecule is highly charged.

pi mo of indigo(2-)

qtaim of indigo(2-)

Now the basic question is how do the protons behave, where do they go, will they break the symmetry.


A bit to my surprise, the flat version of this compound is also most stable.

pi mo of leucoindigo

qtaim of leucoindigo

Preliminary conclusions

There are several points, where this project could have derailed. Level of theory is only one of them. The evidence presented here is not conclusive enough. In due time I might be able to do some more, but for now, at least the textbook picture is debunked.

  1. Calculated at the DF-BP86/def2-SVP level of theory. Occupied orbitals are blue and orange, virtual orbitals (unoccupied) are red and yellow.
  2. Analysis with MultiWFN 3.3.8, see http://sobereva.com/multiwfn
    Nuclei are brown dots, bond critical points are blue dots, ring critical bonds are orange dots. Blue lines are zero-flux surfaces that seoarate the molecule into atomic basins.

Concerning this part of your question:

I also want to ask about why leucoindigo isn't planar. According to Wikipedia, "In indigo white, the conjugation is interrupted because the molecule is nonplanar

If I combine the images on the Wikipedia page for leuco dye we get:

enter image description here

which shows that the leuco- 'form' has a single (rotatable) central bond, not a double bond. Also the -OH and -NH look like they might clash - but would need more detailed 3D geometry for that, rather than just a 2D image.

Alternatively, there could be intra-molecular hydrogen bonds in indigo, that would not be present in the reduced leucoindigo. These hbonds would tend to stabilise the flat structure : for example see this study ("TD-DFT study on electron transfer mobility and intramolecular hydrogen bond of substituted indigo derivatives" C. Ma et al, Chemical Physics Letters, 2015).

  • 2
    $\begingroup$ It is not that simple. Clash is equally possible in both structures; so is hydrogen bond. $\endgroup$ Commented Oct 20, 2017 at 13:05
  • $\begingroup$ @IvanNeretin : Yes, I see you are right. I wonder what the optimal geometry of these bonds are. Perhaps the lone pairs on the carbonyl oxygen are in the plane, while those of the hydroxyl oxygen are not? $\endgroup$
    – gilleain
    Commented Oct 20, 2017 at 13:39
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    $\begingroup$ Lone pairs would readily adjust if that was the problem. No, it's not that simple. $\endgroup$ Commented Oct 20, 2017 at 14:04
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    $\begingroup$ Indigo: nice and planar, see crystal structure. $\endgroup$ Commented Oct 20, 2017 at 15:14
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    $\begingroup$ The hydrogen bonding argument is supported by the fact that leucoindigo is soluble in water while the purplish blue stuff is not (en.wikipedia.org/wiki/Leuco_dye). Indigo forming internal hydrogen bonds instead of external ones can do that. $\endgroup$ Commented Oct 21, 2017 at 0:13

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