The major product of the following reaction 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:




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:



However, according to the answer key the final product obtained is given as:



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

$\ce{CH3-CH=C=CH2} - \ce{H^+ -&gt; CH3-CH=C=CH^- &lt;-&gt; CH3-C^-H-C#CH}$

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.