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 '19 at 18:38
  • 2
    $\begingroup$ Hey, all of your compounds except the first one are wrong as they have 5 carbons. $\endgroup$ – Anubhab Das Apr 7 '19 at 20:03
  • $\begingroup$ @AnubhabDas Yeah sorry, I've corrected them now. $\endgroup$ – Asterix Apr 8 '19 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.

| improve this answer | |
  • $\begingroup$ Thanks @zhe. How to get my resonance structures lined up properly? $\endgroup$ – Oscar Lanzi Apr 7 '19 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 '19 at 23:46
  • 1
    $\begingroup$ @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 '19 at 0:19
  • 1
    $\begingroup$ Thanks @andselisk. $\endgroup$ – Oscar Lanzi Apr 8 '19 at 1:20
  • 1
    $\begingroup$ 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 '19 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.