Lone Pairs and Aromaticity

One of the things my students find most challenging about aromaticity is whether to include lone pairs as part of a cyclic π system. If a lone pair is included, then the number of π electrons increases by two, and a student’s prediction about whether a species is aromatic will also change. What I think makes this challenging is that the rules appear to change depending on the nature of the ring and the nature of the atom that has the lone pair. This issue is apparent when we compare pyridine, pyrrole, and furan:

karty16_imgA

Pyridine and pyrrole both contain a nitrogen atom with a lone pair of electrons, but, whereas the lone pair in pyridine is not included in the π system, the one in pyrrole is. In furan, one of the lone pairs on oxygen is included but not the other.

Students can certainly memorize these specific results for pyridine, pyrrole, and furan, but the true problem manifests itself when students are asked to make predictions about the aromaticity of unfamiliar molecules that contain atoms with lone pairs. For many years my students struggled with this, but with an additional tool I now teach, they have become quite good at making these kinds of predictions.

Years ago I would explain the results of pyridine, pyrrole, and furan just as textbooks do, using only arguments that invoke orbitals, as follows:

karty16_imgB

Over the years, however, I saw that this explanation benefited only my best students, leaving the others behind. I think one reason is that, to use these orbital arguments, students must envision orbitals in the Lewis structures or line structures they are typically given in a problem, something that many students have difficulty doing. Second, it requires students to consider various hybridizations for the atom with the lone pair—the hybridization that students would normally predict via VSEPR theory and one that might benefit the molecule by resulting in aromaticity.

I still teach the orbital arguments as presented above, but now I also show students how to arrive at the same answers using only resonance theory, something that students are already very comfortable with by this point in the course. The idea is that the atoms over which electrons can be shifted via resonance are the ones that contribute p orbitals to the same π system, and the number of curved arrows that are used to shift the electrons corresponds to the number of pairs of electrons in that π system.

karty16_imgC

In pyrrole, six electrons can be shifted via resonance, involving all five atoms of the ring, so there is a single, cyclic π system containing six π electrons. In pyridine, the double bonds are shifted via resonance without involving the lone pair (as shown above on the left), so the lone pair is not part of the same π system as the double bonds. (On the right above, I show that an attempt can be made to involve that lone pair in resonance, but doing so results in a resonance structure in which the lone pair remains on N.) Finally, for furan, just one of the lone pairs on oxygen is involved in resonance with the double bonds, so the second lone pair is not part of that cyclic π system. (Simultaneously including the second lone pair on oxygen in resonance would exceed an octet on one of the atoms.)

Now that I teach both methods of predicting whether a lone pair is part of a π system—one that invokes arguments of orbitals and one that circumvents orbitals—I find that I am reaching many more students. For this reason, I have included both methods in my textbook in the chapter on conjugation and aromaticity (Chapter 14). My better students still appreciate the depth of understanding they receive with an orbital explanation, while all of my students are better equipped to solve problems on aromaticity.

I would be interested in hearing from others about how they and their students deal with these types of problems.

— Joel Karty

17 comments on “Lone Pairs and Aromaticity

  1. Amit Khanna
    August 1, 2014 at 12:50 am #

    That helps me………….Thanks

  2. Harsh Barua
    October 9, 2014 at 11:40 am #

    Thank You Sir….the description u provide helped

  3. Amit
    October 27, 2014 at 12:16 am #

    This is what my teacher taught me long ago..but i forgot as the time passed..thanks to you which helped me to remember..

  4. shuhail
    January 25, 2015 at 4:41 am #

    “In Pyridine , six electrons can be shifted via resonance, involving all five atoms of the ring, so there is a single, cyclic π system containing six π electrons. In pyridine, the double bonds are shifted via resonance without involving the lone pair (as shown above on the left), so the lone pair is not part of the same π system as the double bonds. (On the right above, I show that an attempt can be made to involve that lone pair in resonance, but doing so results in a resonance structure in which the lone pair remains on N.) Finally, for furan, just one of the lone pairs on oxygen is involved in resonance with the double bonds, so the second lone pair is not part of that cyclic π system. (Simultaneously including the second lone pair on oxygen in resonance would exceed an octet on one of the atoms.)”

    I think you wanted to say “In pyrrole” rather “In pyridine”? In the first line of the quotation
    And Thanks so much for the explanation, it helped me as well.

  5. shuhail
    February 1, 2015 at 2:00 am #

    Do you know why true scientists are great people? One of the reasons is that they keep accepting small notices or remarks from even a student with little knowledge of chemistry (like me). I think you are a true scientist and a great person, Thanks.

  6. Atieqah Asma
    May 11, 2015 at 12:08 am #

    thanks a lot!!! 😘

  7. Anik
    June 18, 2015 at 12:45 am #

    Best concept….. It will surely help me to solve this type of problems

  8. J. Manuel Recio
    October 30, 2015 at 3:27 am #

    I like the way you explain this point. My question is close to this issue. As a teacher of General Chemistry, I have problems to explain why In pyrrole and furano, if the stochiometric formula according to Lewis’ theory is AX3E (N) and AX2E2 (O), the electronic geometry is not tetrahedral but planar trigonal. The concept of aromaticity should not be invoked at this very moment. I hope you understand my worry.
    Thanks in advance.
    J. Manuel Recio
    University of Oviedo
    Oviedo, Spain

  9. Jyotinder Singh
    February 2, 2016 at 6:22 am #

    Lovely explanation! Helped me out!

    • kamal singh
      October 4, 2016 at 4:22 am #

      Very nice concept.

  10. Akansha Singh
    February 27, 2016 at 1:55 am #

    Which one has higher resonance energy

  11. abhas saxena
    April 9, 2016 at 5:08 am #

    nice explanation

  12. Ashley Sexton-Radney
    October 10, 2016 at 3:29 pm #

    This was awesome! I am teaching this tomorrow! Thanks!

  13. Chathuri
    November 4, 2016 at 12:47 am #

    Thank you sir…the explanation you provide it helped to me to solve this kind of questions..

  14. anonymous
    January 13, 2017 at 2:39 am #

    Its indeed a very nice method.

  15. anonymous
    January 13, 2017 at 2:40 am #

    Its indeed a very nice method. Thankyou

  16. Prasteena Suresh
    February 16, 2017 at 12:55 am #

    this makes sense but it did not clear my doubt. I mean, by what you have said we can make out how many lone pairs get into the ring provided the arrow are given. What if only the bond structure is given?? And as it is an unfamiliar molecue we do not know in which orbitals the lone pair is in??
    then what will we do? how will we know?? i hope you understood my doubt.

    Thankyou

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