Here,1,2-dibromobutane is reacted with
i) alc. $\ce{KOH}$
ii) $\ce{NaNH2}$ in liq. $\ce{NH3}$

According to me, since alc. $\ce{KOH}$ causes dehydrohalogenation, using Saytzeff's rule the product obtained after reaction (i) should be one of the either:
enter image description here ![enter image description here

And after reaction (ii), since $\ce{NH2-}$ is a strong base, an E2 reaction proceeds and again using Saytzeff's rule the final product should be:
![enter image description here

However, according to the answer key the final product obtained is given as:
![enter image description here

Can someone help by explaining where and how I went wrong in my attempt to solve the question?

  • $\begingroup$ An allylic bromide will react with KOH. What happens to terminal alkynes in the presence of NaNH2? Check pKa's! $\endgroup$ – user55119 Apr 7 at 18:38
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    $\begingroup$ Hey, all of your compounds except the first one are wrong as they have 5 carbons. $\endgroup$ – Anubhab Das Apr 7 at 20:03
  • $\begingroup$ @AnubhabDas Yeah sorry, I've corrected them now. $\endgroup$ – Asterix Apr 8 at 3:28

When you form the allene, the methylene hydrogen has enhanced acidity because deprotonating it leads to a carbanion in which the negative charge is delocalized:

$\ce{CH3-CH=C=CH2} - \ce{H^+ -> CH3-CH=C=CH^- <-> CH3-C^-H-C#CH}$

The proton then reattaches to the opposite end of the allene chain to form a more stable product. In effect this is a (strong) base catalyzed tautomerization.

  • $\begingroup$ Thanks @zhe. How to get my resonance structures lined up properly? $\endgroup$ – Oscar Lanzi Apr 7 at 22:45
  • $\begingroup$ Honestly, I'm not the best person to ask. I liked you what did you with $\equiv$, and that might be the best approach here. @OscarLanzi $\endgroup$ – Zhe Apr 7 at 23:46
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    $\begingroup$ @OscarLanzi For a triple bond mhchem uses # symbol, and yes, it looks somewhat misaligned and I guess one should just accept it. I also tried to improve the location of the negative charge on carbons with \overset{...}{...} so that it doesn't distance hydrogens away. $\endgroup$ – andselisk Apr 8 at 0:19
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    $\begingroup$ Thanks @andselisk. $\endgroup$ – Oscar Lanzi Apr 8 at 1:20
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    $\begingroup$ Thanks @OscarLanzi, but since the methylene H of the allene is acidic, shouldn't it react with NaNH2 instead of undergoing tautomerisation? And another thing, are alkynes, in general, more stable compared to allenes? $\endgroup$ – Asterix Apr 8 at 3:26

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