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I am trying to understand acid-base reactions and I am trying to figure out which conjugate base is more stable.

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The molecule on the top with dashed lines around the $\ce{H}$ is the original molecule and then below it I remove $\ce{H}$ and see what the resultant conjugate base looks like. I don't understand why the bottom left conjugate base is less stable than the one on the right. Why doesn't nitrogen create a more stable conjugate base than the triple bond?

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  • $\begingroup$ Maybe it's due to geometry? The triple bonded C has a single lone pair to donate, while the N has two. So that would allow a wider range of acceptable orientations when considering N-/H+ collisions? $\endgroup$ – Dan Burden Apr 1 '17 at 23:51
  • $\begingroup$ Hey, just so you know, the homework tag is being phased out, so please don't tag questions with it. Thanks! $\endgroup$ – heather Apr 1 '17 at 23:58
  • $\begingroup$ @heather thanks! I won't do that from now on. $\endgroup$ – ahat Apr 2 '17 at 0:06
  • $\begingroup$ @DanBurden could you elaborate some more? $\endgroup$ – ahat Apr 2 '17 at 0:08
  • $\begingroup$ Well I think the crux of this lies in the fact that ammonia has pKa 33 while ethyne has 25, which you can apply here. Why is it so, because I guess a sp carbon is more electronegative than a sp3 nitrogen perhaps. $\endgroup$ – Sawarnik Apr 2 '17 at 13:21

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