A paper I'm looking at1 mentions a few electron deficient heterocycles:

electron deficient heterocyclic compounds such as di(azines), benzothiazole and imidazole...

What are the criteria to determine whether a heterocycle is electron deficient or not? (Why is imidazole considered electron deficient, for example?)


(1) Bureš, F. Fundamental aspects of property tuning in push–pull molecules. RSC Adv. 2014, 4 (102), 58826–58851. DOI: 10.1039/C4RA11264D.

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    $\begingroup$ I'd point out that these aren't just heterocycles but heteroaromatic systems. The conjugation is important to the discussion. $\endgroup$ Feb 24, 2016 at 23:20

1 Answer 1


As indicated in the comment below, a rule of thumb for these particular compounds, is that aromatic nitrogens with double bonds (e.g., not pyrrole) are "electron deficient."

The term is a bit ill-defined. It does not mean the aromatic system (which usually still has 4n+2 electrons) actually has a different electron count.

More broadly, there can be several ways of declaring "electron rich" or "electron deficient," for organic compounds, but in the context of a "push-pull" conjugated or aromatic system (in the paper you link), these refer to rings or functional groups that are electron donors or electron acceptors.


  • Electron deficient - tends to gain negative charge as an electron accepting group or ring
  • Electron rich - tends to gain positive charge as a donor group or ring

The charge transfer need not be a full electron - it is almost always much less (e.g., a small fraction).

The second part of your question is how you can determine this.

Conjugated systems are typically imagined as a particle-in-a-box.. that the electrons are highly delocalized. The "electron donor" is a region with higher potential energy and the electron acceptor is a region with lower potential energy. Thus, the electrons have slightly greater probability on the electron acceptor.

Therefore, you determine "electron rich" or "electron deficient" in comparison with the rest of the conjugated system.

Typically, electron deficient rings have more negative HOMO and LUMO energies - making them more difficult to oxidize (i.e., they are "deficient" in electrons and it's hard to give up another) and easier to reduce (i.e., adding an electron makes them less "deficient").

Electron rich rings have more positive HOMO and LUMO energies - making them easier to oxidize and give up an electron.

Again, this is a rule of thumb and it's almost always partial charge transfer.

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    $\begingroup$ This answer is quite correct but fairly generic. A rule of thumb is that heteroarmoatic compounds with double bonded nitrogens (pyridine for example) are generally electron-deficient. $\endgroup$
    – Lighthart
    Feb 25, 2016 at 0:06
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    $\begingroup$ Absolutely true for the specific case, but there are plenty of push-pull conjugated systems without double-bonded nitrogens. So I went on the second (broader) question - when is it an acceptor? Truthfully, it depends on the rest of the conjugated system (i.e., electron deficient compared to what?) $\endgroup$ Feb 25, 2016 at 3:39
  • $\begingroup$ Surely the electron deficiency refers to neighbouring aromatic atoms as opposed to the whole ring I.e. It should be clarified that the aromatic rings still have the same number of electrons (2n+2). $\endgroup$
    – Beerhunter
    Feb 25, 2016 at 7:46
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    $\begingroup$ @Beerhunter You mean 4n+2 electrons. Yes, the aromatic rings still have 4n+2 electrons, but it does not refer to any neighboring atoms. The "electron deficiency" refers to the tendency for the ring to pull charge density from other conjugated systems. The charge transfer is rarely a full electron. $\endgroup$ Feb 25, 2016 at 14:26
  • $\begingroup$ thank you all very much for your answers and comments! do you know the rationale behind that rule of thumb ("aromatic nitrogens with double bonds")? $\endgroup$ Feb 25, 2016 at 15:12

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