So, whenever there is abstraction of H+ by an acid-base reaction, I see that the proton on the sp3-hybridized carbon is taken. Why is this if you take into account that vinylic hydrogens are more acidic?

Also, on a sidenote, what is the hybridization state of a carbon in a C=C double bond where the vinylic proton was removed by a base?

  • $\begingroup$ Are you sure you don't mean transfer of hydride, i.e. oxidation to the tropylium cation? $\endgroup$ Commented Oct 25, 2014 at 1:22
  • $\begingroup$ Actually Abel, the tropylium cation is drawn. But the anion with the abstraction of the proton only on the sp3 carbon is also shown (Cycloheptatrienyl anion). It was an exercise where we had to say whether the compounds were aromatic or not. Could we make such compound from acid-base reactions? $\endgroup$
    – yolo123
    Commented Oct 25, 2014 at 1:31

1 Answer 1


The pKa for a vinylic proton is ~ 43, the pKa for propene is ~ 40, and the pKa for 1,4-pentadiene is ~ 35. Consequently it's not the vinylic protons that are the most acidic, it's the allylic ones. Now the question whether tropylium anion is aromatic/antiaromatic is complicated.

  • $\begingroup$ Abel, are you saying that an acid would rather rip apart a hydrogen from the sp3 carbon in the propene? $\endgroup$
    – yolo123
    Commented Oct 25, 2014 at 1:57
  • $\begingroup$ If you think about it, is CH3:- not a much stronger base than CH2=CH:-? So, it is the CH2=CH2 that is the stronger acid. $\endgroup$
    – yolo123
    Commented Oct 25, 2014 at 1:59
  • $\begingroup$ In the allyl anion the charge is delocalized over all three carbons. In saturated hydrocarbons it's confined to a specific carbon. $\endgroup$ Commented Oct 25, 2014 at 2:02
  • $\begingroup$ Okay, so in this particular molecule (Cycloheptatrienyl anion), the abstracted hydrogen will always come from the sp3 carbon. $\endgroup$
    – yolo123
    Commented Oct 25, 2014 at 2:04
  • 2
    $\begingroup$ I think you are correct in saying this, $\endgroup$ Commented Oct 25, 2014 at 2:08

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