Let's take Hematoxylin, which is a pH indicator and a stain for histology, and therefore a chromophore, but does not fluoresce:

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Now let's take Eosin, which stains tissue pinkish, but is also fluorescent:

enter image description here

I would like to know which features of the molecules decide if it is a chromophore or a fluorophore. As far as I can see it (I'm a biologist with some training in organic chemistry), both have conjugated systems, although the system in Eosin spans a lot of more bonds. Is it really just the size of the conjugated system that decides this, e.g., if Hematoxylin had a larger conjugated system, would it also be fluorescent?

  • $\begingroup$ A molecule "is" not a chromophore, it may contain a chromophore! Second one cannot say some dye "does not" chromophore in general. It may or may not fluorecse in a certain environment (solvent etc). Whether or not fluorescence orcurres, is a question of electronic and vibrational levels of dyes. Look for some literature on dye lasers. $\endgroup$
    – Georg
    May 31, 2013 at 21:36

2 Answers 2


The physical chemistry of fluorescence is a little bit complicated, and so you should consult a text book for "the reason" that things fluoresce.

One of the things that I have noticed is that fluorescent compounds often have several "heavy halogens" Br and I attached to some poly-aromatic hydrocarbon.

Disclaimer: this is not to say that the halogens are the fluorescing portion of the molecule (it's probably almost the whole molecule, conjugation + halogen), and it does not mean that all compounds that contain Br and I will be fluorescent. It also does not mean that Br and I are required for fluorescence.

I think that a lot of fluorescent molecules were discovered by chance, or even brute-force changing substituents until the molecule exhibits the desired fluorescence. (It's really too hard to predict with qualitative theory, and electronic structure theory descriptions of fluorescence are also very difficult.)


The key to fluorescence is an extended Pi system. In hematoxylin, the two aromatic rings are far apart, while in Eosin, the conjuated Pi system is uninterrupted and continuous. You need at least four conjugated groups to absorb in the visible light region. Aromatic systems absorb strongly in the UV region. In addition, the attachment of halogens (i.e. Br) reduces fluorescence.

Other things that reduce fluorescence besides halogens are $NO_2^-$ (nitro groups) and flexibility of the molecules. Rigidity increases fluorescence.

  • $\begingroup$ Other things that reduce fluorescence besides halogens are NO2 (nitro groups) and flexibility of the molecules. Rigidity increases fluorescence. $\endgroup$
    – user2471
    Oct 13, 2013 at 17:02
  • $\begingroup$ Why do these groups reduce fluorescence? Is it due to their high electronegativity drawing electrons and thus somehow disrupting the conjugation? $\endgroup$
    – Eekhoorn
    Oct 13, 2013 at 18:49
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    $\begingroup$ I incorporated your comment into the answer. Feel free to edit if you don't think it belongs. Also, can do you have a reference for the 'need at least four conjugated groups..' statement. $\endgroup$ Oct 14, 2013 at 1:35
  • 1
    $\begingroup$ @bobthechemist if you do a rough partice-in-a-box calculation, you find that the box needs to be about 10 angstroms long in order to absorb at the electron-volt order of magnitude (where visible light is). With a C-C bond at roughly 150 pm and a C-atom at roughly 50pm, that's 6 carbons to get to 1000pm. I'll buy four conjugated groups to absorb light in the visible, although the particle-in-a-ring calculations may yield something quite different. $\endgroup$
    – chipbuster
    Oct 14, 2013 at 6:46
  • $\begingroup$ While I agree that an extended pi system helps fluorescence, I do not agree that the reason is that it needs to absorb in the visible portion of the spectrum. $\endgroup$
    – buckminst
    Oct 15, 2013 at 15:14

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