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From what I can tell, deprotonation of the central hydrogen leads to the more thermodynamically stable product (most substituted double bond. i.e. most stabilization by hyperconjugation). Therefore, if butanone is allowed to equilibriate in the gas phase, I would expect the deprotonated form A to be present in greater amounts than B. Therefore, I would conclude that the pKa of the central hydrogen is lower than that of the terminal one, in the gas phase.

My questions:

  1. Is this correct? (please point out any mistakes or omissions from my argument)
  2. How does this change with solvation, and why?

EDIT: To clarify questions: Any data on the acidity of the different sites in solution is greatly appreciated. The gas-phase specification above is simply there to be able to treat the problem solely thermodynamically (I appreciate that data on gas phase equilibration of acids might be hard to come by)

I've seen many other posts about ketone deprotonation, but browsing them I have been unable to answer my questions. Refer: "Which enolate of butanone is more stable?", "Regioselective enolate formation"

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    $\begingroup$ I take it the opposite is true (B is in greater amounts) that's why you asked the question. If so could you cite the source? $\endgroup$ – Avyansh Katiyar Mar 5 '18 at 0:25
  • $\begingroup$ Equilibrate in gas phase? Why are you considering that? $\endgroup$ – Zhe Mar 5 '18 at 22:53
  • $\begingroup$ @Zhe The gas-phase part is simply to rule out any solvation effects which may (or may not?, i don't know) have an effect. $\endgroup$ – Adroit Mar 6 '18 at 4:46
  • $\begingroup$ @AvnishKabaj I am not sure what is correct, thats why i asked the question, and would like the answer to include a credible source. $\endgroup$ – Adroit Mar 6 '18 at 4:48
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    $\begingroup$ pKa is directly related to thermodynamic stability, so... $\endgroup$ – Zhe Mar 7 '18 at 22:51

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