While studying Molecular Orbital Theory, I ran into a comparison of basicity between pyridine and piperidine. The latter was concluded to be more basic because of the sp3 hybridization of the positively charged nitrogen opposed to the sp2 hybridization of the pyridine nitrogen. The p orbitals are higher in energy, hence, the higher the p influence on the final hybrid orbital the bigger it will be in energy. Higher energy orbitals result in higher stability for positive charges.


On another moment, when reading about carbocation stability, it was stated that tertiary carbocations are more stable due to the delocalization of the positive charge.

Both statements regarding the positive charge stabilization makes sense, but here's where I started getting confused:

  • If positive charge delocalization plays a big role on carbocation stability, wouldn't the π-system of the pyridine help delocalize the positive charge through resonance? Wouldn't that compensate the effect of the orbitals lower energy (due to higher s orbital character)?

  • Considering the answer to the above question was that the orbital hybridization actually plays a bigger factor in positive charge stabilization than delocalization, then why alkynes like methylacetylene are attacked in the secondary carbon (which are sp hybridized in the carbocation form?) and not in the sp3 hybridized carbon? (Consider R = CH3 in the image below)

Alkyne chlorination

Could someone give me a light on those matters? I assume I might be making a confusion regarding the carbocation hybridization of methylacetylene, would the secondary carbon have a sp geometry when it's positively charged?


1 Answer 1


Basicity is the ability to donate a pair of electrons. In the case of pyridine, the lone pair is in a sp2 orbital. This orbital does not participate in any resonance, and is smaller than the sp3. This means the electrons will be closer to the nucleus due to stronger attraction, making the pyridine a weaker base, since it's harder for it to donate these electrons.

Tertiary carbocations are more stable, that is correct, but you should look closely on what that means. A tertiary carbon is carbon atom that is to 3 other alkyl groups , not just any other atom. Alkyl groups are weak electron-donating groups, therefore stabilizing the carbocation, because they increase the electron density in the electron-poor center. That is why according to Markovnikov's rule, the electrophile, in this case the proton, will bond to the less substituted carbon - in order to create a more stable secondary carbocation, which the nucelophile (the chloride) will attack.

  • $\begingroup$ So the stabilization of charges require the orbitals interacting to be the ones actually holding the charge? I thought maybe other orbitals could help stabilize the charge if, for example, they held pi-system electrons for more/less time, even if they were not the actual orbitals responsible for the charge. But thinking better it does make sense that the orbitals with the excess/deficit of electrons are the ones that really matter on stability through resonance. $\endgroup$
    – IanC
    May 29, 2019 at 14:55
  • $\begingroup$ By the way, just confirming, the positive carbon on the carbocation that has a double bond, would still have sp hybridization right? $\endgroup$
    – IanC
    May 29, 2019 at 14:58
  • $\begingroup$ Yes, as it still has only two "hands" (That's an easy way to remember hybridization.) $\endgroup$
    – Argento
    May 29, 2019 at 15:18

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