I need to find out the one having highest acidic hydrogen.

As per my observations, there are two equivalent resonating structures possible in case of conjugate base of $\ce{CH2=C=CH2}.$ However, the correct answer is given as $\ce{HC≡CH}$ due to inductive effect of sp hybridization of carbon.

But the priority order in determining stability of a conjugate base is:

  1. Aromaticity
  2. Number of equivalent resonating structures
  3. Resonance
  4. Hyperconjugation
  5. Inductive effect

As such, the correct answer should have been the allene. Where did I go wrong?

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    $\begingroup$ Related: Relative acidities of alkanes, alkenes, and alkynes. $\endgroup$ – andselisk Jun 21 '20 at 13:44
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    $\begingroup$ I did not downvote, but likely some people will if they seem to see a homework problem with no/too little work input from the asker. You might want to look at the link given be andselisk for some ideas, as well. $\endgroup$ – Oscar Lanzi Jun 21 '20 at 13:58
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    $\begingroup$ This "priority order" is hardly a law of nature. In my opinion, instead of trying to force everything into a fixed order, it's a better idea to get more experience with various functional group pKa's so that you have a better idea of how much each factor affects the pKa. That takes time. You might think that memorising some order is a quick shortcut to the top, and maybe it does work for quite a few cases, but I fear that this sort of approach comes with the risk of having to unlearn wrong things later on. $\endgroup$ – orthocresol Jun 21 '20 at 14:04
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    $\begingroup$ But that's just me. If you have to learn this sort of thing, it's ok, just make sure to take it with a pinch of salt. Especially because these sort of orders were worked out precisely by observing the pKa's of various compounds, so using them to predict pKa's is a little bit like circular logic. TLDR: Models only work in the contexts in which they were developed. $\endgroup$ – orthocresol Jun 21 '20 at 14:14
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    $\begingroup$ I agree with @orthocresol. The stability order here would be $\mathrm{sp \gt sp^2 \gt sp^3}$. In allene, $\ce{H}$ is on $\mathrm{sp^2}$ carbon, so most stable carbanion would be that from acetylene. $\endgroup$ – Mathew Mahindaratne Jun 21 '20 at 14:26

The inductive effect will always make the stronger acidic compound, especially with an sp hybridization as it will ultimately stabilize the electrophile best in solution. Though resonance will also contribute to the stability, a structure similar to allene will have more resonance as it will have more bonds focused in one area so close to the electrophilic site.

I think you over resonance, its an easy mistake, but my notes from Orgo I show that the difference in pKa from an sp2 to an sp hybridization are roughly a factor of 2.

TL:DR, Induction beats resonance.

  • $\begingroup$ But in most of the cases that I've seen in these types of problem, resonance is prioritized over induction :( $\endgroup$ – Sanu_012 Jun 23 '20 at 14:38

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