So, apparently pyridine is more stable than benzene. My lecture notes are saying that pyridine will undergo electrophilic aromatic substitution much more slowly than benzene will, which got me thinking - slower means the compound is more stable (higher activation energy). But isn't benzene more aromatic than pyridine?

I checked the heats of hydrogenation and it checks out - pyridine is indeed more stable than benzene. I'm a little bit lost though, because I don't get where this unexpected stability is coming from. Surely benzene is an ultra-stable compound because of the superb resonance it has (higher resonance energy than pyridine), electron density is (AFAIK) equally distributed around the ring, whereas this isn't the case in pyridine.

Also, the reason I call this a 'paradox' (in my head, at least) is because I've been learning about how benzene is a really stubborn compound when it comes to getting it to react - it doesn't undergo normal alkene addition reactions and all that - yet pyridine is quite a bit more stable than benzene and it has a million different weird reactions going on (acid/base, nucleophilic, EAS, N-oxide formation, metal complexes, etc. etc.)

Apparently pyridine is more stable, yet more reactive due to the lone pair electrons on the nitrogen (which allow lots of crazy stuff). In general, is it true that more stable = less reactive? That's the basic assumption I've been using until now - based on thermodynamics (?) - but I've seen a couple of things that make me think it's not as simple as that. Sorry for the rambling, just blurting my questions out!

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    $\begingroup$ Stable does mean less reactive, but it's a relative term. A compound can be stable to one set of reactions and highly reactive towards another set. $\endgroup$
    – Tyberius
    Nov 20, 2017 at 6:13
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    $\begingroup$ I think the ‘paradox’ boils down to you wanting a simple black/white description for a much more complex observation. Note that benzene is not ‘ultra-stable’ or it could not get hydrogenated to cyclohexane. $\endgroup$
    – Jan
    Nov 20, 2017 at 10:10
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    $\begingroup$ To build on Tyberius' point: reactivity (reaction rate, determined by activation energy) is determined not only by the energy of the starting material, but also the energy of the transition state (or intermediate, by the Hammond postulate). Obviously, this depends on the exact reaction pathway under consideration. There are some cases where benzene is more reactive than pyridine and some where benzene is less reactive. $\endgroup$ Nov 21, 2017 at 11:48
  • $\begingroup$ I don't think that this is primarily opinion-based; if anything it should be unclear, but what would be most helpful would be to have an answer explaining what the difference is. $\endgroup$ Nov 21, 2017 at 11:49
  • $\begingroup$ I fully agree with @ortho that this is not opinion-based I personally have a problem understanding the question at all. However, it seems some have found it out. I urge these to edit the question and — more importantly — write a corresponding answer. Thus, I am voting to reopen. $\endgroup$
    – Jan
    Nov 21, 2017 at 14:25

2 Answers 2


When you think about stability, think of it as a relative value, meaning in comparison to atoms of other elements. Therefore, pyridine is more stable than the benzene but less stable than other elements, in general.

You are correct when you say stability is inversely proportional to reactivity. But because stability is relative, it is very difficult to comment on the relative reactivity.


Heat of hydrogenation of an alkene in which the double bonds are conjugated is not additive because such an alkene is resonance stabilized, and, therefore, its Lewis structure, which the calculation would be based on, is a poor representation of the molecule.

I have found 2 aspects :

  1. Pyridine is less reactive, than benzene toward electrophilic aromatic substitution, because nitrogen is more electronegative, than carbon and acts like an electron withdrawing group. As a result, the meta hydrogen is substituted.

  2. Pyridine is more reactive than benzene because the presence of nitrogen enables pyridine to react with nucleophiles. Nitrogen is an electron withdrawing substituents enables the aromatic ring to participate in such reactions and the substitution happens in the ortho and meta positions.

  • $\begingroup$ Surely it's reactivity that's relative rather than stability? There's an objective measurement for stability (correct me if I'm wrong) - heats of hydrogenation, right? Reactivity is relative seeing as one molecule may be more prone to react than another due to factors like lone pairs, etc. $\endgroup$ Nov 21, 2017 at 14:59
  • $\begingroup$ @Malnurturedㅤㅤㅤㅤ stability is also relative. You may call an aluminum less reactive than chorine ,but not compared to oxygen $\endgroup$
    – Vedant
    Nov 21, 2017 at 15:02
  • $\begingroup$ Fair, that's true in most cases maybe, but for C=C bonds you can measure the energy stored within them using heats of hydrogenation. I'm really just focused on organic compounds (and in this case two resonant, aromatic compounds), and for them I thought stability could be objectively ranked based on the energy content of the bonds? $\endgroup$ Nov 21, 2017 at 15:14
  • $\begingroup$ Please see edit $\endgroup$
    – Vedant
    Nov 21, 2017 at 15:17
  • $\begingroup$ @Malnurturedㅤㅤㅤㅤ like always, happy to help :) $\endgroup$
    – Vedant
    Nov 21, 2017 at 16:33

Six membered heterocycles with an electronegative heteroatom are generally electron deficient compared to benzene. Such compounds are classified as p-deficient. Electron-withdrawing heteroatoms decrease the p-electron density at the carbon atoms and are thus p-deficient relative to benzene


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