Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Aromaticity of 1-methyl-1H-pyridin-2-one?

  1. Jun 3, 2015 #1
    Hi everyone,
    I've recently had a little discussion at work concerning the concept of aromaticity applied to some molecules.
    To cut a long story short, while I see why 2-pyridone is considered aromatic:

    2pyridone.PNG

    I don't understand why 1-methyl-2-pyridone is also considered aromatic:

    1Me2pyridone.PNG

    The two structures on the right do respect the rules for aromaticity (6 pi electrons, cyclic delocalisation, etc).
    However, if a molecule must be seen as the average of its resonance structures, 'weighted' on their relative stability, then a molecule whose only aromatic resonance structures are very unstable (e.g. by charge separation as above), will be represented more closely by non-aromatic structures, and on the whole it won't 'be' aromatic. Or am I wrong?

    I had a look at this nice review:

    http://www.ark.chem.ufl.edu/published_papers/pdf/1074.pdf [Broken]

    which however didn't clarify the situation in my mind.

    To add to the confusion, in the past I did a chlorination of a fused pyridine-2,4-dione, which gave this:

    dichlorination.PNG

    We expected the chlorination to go only once on either carbonyl, yielding 2 isomeric mono-Cl products, and instead we got the dichloro compound with a positive charge. The justification was that the latter was more stable than expected due its aromatic character. It was so stable that it took a rather long time to hydrolyse off one Cl at pH 10 (large excess of aqueous carbonate). And we could even isolate it as a solid salt.

    This fact is used as an argument to say that charged structures are perfectly valid.
    But for me this is a different case, as here there is no *separation* of charge, and all resonance structures have one positive charge on one atom, there is no possible structure with no charges anywhere.

    So... not clear at all, as far as I'm concerned.

    I would like to know your opinion on the subject, please.

    Thanks!
    L
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jun 4, 2015 #2

    Ygggdrasil

    User Avatar
    Science Advisor

    The top figure doesn't present a proper resonance structure as it involves the motion of a proton from the nitrogen to the oxygen (this is called a tautomerization). If you were to draw the resonance structures without the tautomerization, you would have resonance contributors with a separated positive and negative charge. Aromaticity does not rely on having a structure with alternating double bonds like benzene. The resonance structure of pyridone with a double bond on the oxygen is still aromatic. All it requires is a series of atoms bonded in a (planar) cycle where each of the atoms in the cycle have a p-orbital that it can overlap with the p-orbital from its neighbor (along with the 4n+2 electron rule). Another good example of a molecule that exhibits aromaticity without looking anything like benzene is thymine.
     
  4. Jun 6, 2015 #3
    Hi Ygggdrasil, thanks for your reply.
    I did tautomerise 2-pyridone to 2-hydroxypyridine intentionally (as you see I used reaction arrows for the first transformation, and the electron shift arrow + square brackets notation for the other two), because I thought the ability to tautomerise to structures with fully sp2 cycles was part of what made a chemical entity aromatic.
    So from what you say I understand that instead I should keep the original pyridone structure and only shift electrons around, when trying to assess aromaticity.
    If this is the case, then one would say that 2-pyridone and 1-methyl-2-pyridone have a similar degree of aromaticity, whereas 2-hydroxypyridine is more aromatic because all its predominant resonance structures are already aromatic and don't require charge separation.
    Would you agree with this interpretation?

    I looked at thymine as well. http://en.wikipedia.org/wiki/Thymine
    For me, in the molecule as it's drawn there are 4 sp2 carbons (the carbonyl carbons and the double bond carbons) and 2 sp3 nitrogens (the NH's). So there are 4 p orbitals that can overlap (perpendicular to the ring). The p orbitals on the nitrogen are not aligned properly for that. To align them I would have to move both nitrogen doublets into the ring (thus making the nitrogens positively charged) and place negative charges on both oxygens. A double separation of charge.
    How abundant would a resonance structure like that be? I would say very little, so I would say this molecule is very poorly aromatic. Its 2,4-dihydroxy-pyrimidine tautomer, on the other hand, would be much more aromatic, according to the rules.

    If you think I'm wrong in my reasoning, please let me know.

    Thanks
    L
     
  5. Jun 6, 2015 #4

    Ygggdrasil

    User Avatar
    Science Advisor

    I would not say that 2-pyridone is less aromatic than 2-hydroxypyridine. Because they differ by an intramolecular proton shift, any solution of 2-pyridone will exist in an equilibrium between 2-pyridone and 2-hydroxypyridine. This equilibrium has been measured, and it suggests that 2-pyridone is 0.3 kcal/mol more stable than 2-hydroxypyridine (http://www.chem.wayne.edu/schlegel/Pub_folder/54.pdf). If 2-hydroxypyridine really experienced much more resonance stabilization than 2-pyridone, we would expect 2-hydroxypyridine to be much more stable than 2-pyridone, but this is not the case. Rather, the relative basicity of the nitrogen vs the oxygen determines which tautomer is more stable, not their relative aromaticity.

    In the case of sp3 atoms with lone pairs, they can change their hybridization to sp2 in order to place the lone pair into a p-orbital that can form a conjugated pi-system with neighboring p-orbitals. See:
    http://teachthemechanism.com/2013/03/12/lone-pairs-and-aromaticity/
    http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf
     
  6. Jun 7, 2015 #5

    DrDu

    User Avatar
    Science Advisor

    I wonder whether the methylpyrrolidone is not also considerably stabilized by an intramolecular hydrogen bond from the methyl group to the oxygen. What solvent do we talk about? Maybe it is even protonated?
    Personally, I don't consider "aromaticity" to be a very useful concept in terms of making predictions, for reasons that are obvious in this case. It is hard to estimate the importance of various resonance structures and anyhow it is not straight forward to discuss aromaticity in terms of valence bond theory.
     
  7. Jun 9, 2015 #6
    Thank you both for your replies.
    I think I understand what you mean. It was the ucla tutorial that made it clearer.
    Basically, what I considered to be separation of charge is actually just a way the molecule has to put its electrons in the orbitals that make more sense energetically.
    I was well aware of the sp3 to sp2 transition for the heteroatoms of pyrrole, furan, thiophene, which allows two 'usually' sp3 electrons to become part of the aromatic cycle, and I saw this process as 'normal', because it didn't require any bizarre charges to appear.
    Turning the pyridone sp3 NH into sp2 was also 'OK' (as if it were a pyrrole NH), but then I was still counting one electron for each C=O group, the one on the carbon, which meant the cycle couldn't be aromatic. The only way I could get that electron out of the way was to place it on the oxygen, thus going from C=O to C+-O-, i.e. a separation of charge.
    I now seem to understand that this is not necessary for Huckel's rule. The tutorial specifically mentions that such electrons must not be counted. It doesn't really explain why, but OK, I suppose people have thought about this.

    Based on the above, the next question would be, if the C=O electron that should be on the carbon actually isn't in the aromatic ring, it must still be somewhere (closer to the oxygen?), so does that mean that the carbonyl oxygen is more electron-rich than it would be in the corresponding lactam (i.e. in valerolactam?). And would the ring then be just as electron-rich as in a pyridine?
    That would be strange, because we know very well that 2- and 4-pyridones are more reactive than the corresponding pyridines in electrophylic substitutions.

    As DrDu seems to suggest, this may just be a byproduct of the fact that I was taught organic chemistry using mostly the VB theory. Molecular orbitals were considered witchcraft back then :O)
    So perhaps all these electrons are going just wherever they want, enriching one atom or the other in ways that are inscrutable to the non-quantum-physicist, and there's no point in learning silly arithmetic rules or playing with doublets and charges anymore. :O(
     
  8. Jun 9, 2015 #7

    DrDu

    User Avatar
    Science Advisor

    Well, this is indeed a widespread oppinion nowadays. In fact, it is not difficult to do some quite accurate density functional calculation for these molecules on a personal computer, and some of the software is even freely available.
    But on the other hand I think you are on the right track with your considerations about the stability of that molecule.
    However, in applied chemistry, it is extremely important to formulate your question precisely. What are you really interested in? In some specific chemical reaction, like chlorination and its possible products?
    Or in spectroscopic properties of your molecule? This makes an enormous difference, also in the concept of aromaticity which is appropriate.
    As a last point, it is always important to remember that these molecules usually react in a solvent which may have huge influence on reactivity. What solvent are you considering?
     
  9. Jun 10, 2015 #8
    I think you're right, the answer strongly depends on the exact question asked.
    In this case it was just me saying (about a molecule much like 1-Me-2-pyridone): that's not aromatic, because it can't make a 6 electrons cycle without separation of charge; and a colleague replying: yes it is, because N-methyl-pyridinium salts are still aromatic.

    I'm pretty sure students are still asked the question: 'is' this molecule aromatic, as if a yes or no answer were possible.
    We know from this discussion, from the review I attached and from several books, that aromaticity is not a binary property, but more a parameter that can be measured.
    Furan is said to be the 'least' aromatic of 5-membered-ring aromatic heterocycles, based on the fact that it doesn't differ that much in certain properties from a theoretical oxacyclopenta-2,4-diene, or whatever it's called.
    Benzene on the other hand is considered a prototypical 'strongly' aromatic compound, because it has very different properties from a theoretical 1,3,5-cyclohexatriene. Just to name one, hydrogenating one double bond in benzene to go to 1,3-cyclohexadiene is a very thermodynamically unfavourable reaction, in contrast with what would be expected if each double bond behaved independently from the other two.

    So the real question should be something like: to what extent the properties of this molecule differ from the properties that would be expected in the absence of aromatic character? And then again, one may have to be quite specific on which properties we're talking about.

    In the case of my 1-Me-2-pyridone-like molecule, should I expect it to be much more stable than the analogue where the double bond in 5,6 is hydrogenated, i.e. this:

    pyridone_hydr.PNG

    By looking at the molecule on the right, I would call that a Michael acceptor, whereas I would not expect that kind of behaviour from the one on the left, simply because of the electron-donating ability of the nitrogen.

    Thanks
    L
     
  10. Jun 12, 2015 #9

    At first glance, I'd agree with your assumptions about those two molecules. Of course, it again boils down to what exactly you would try to react them with though. I can imagine the molecule on the right being difficult to actually act as a Michael acceptor unless the solvent and nucleophile were chosen carefully. You'd still probably have a decent chance of simply deprotonating the carbon at #5 as shown below:
    upload_2015-6-12_20-13-58.png

    You end up with a wonderfully aromatic product.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Aromaticity of 1-methyl-1H-pyridin-2-one?
  1. 1H NMR spectroscopy (Replies: 1)

Loading...