# 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?

• An allylic bromide will react with KOH. What happens to terminal alkynes in the presence of NaNH2? Check pKa's! – user55119 Apr 7 '19 at 18:38
• Hey, all of your compounds except the first one are wrong as they have 5 carbons. – Anubhab Das Apr 7 '19 at 20:03
• @AnubhabDas Yeah sorry, I've corrected them now. – Asterix Apr 8 '19 at 3:28

$$\ce{CH3-CH=C=CH2} - \ce{H^+ -> CH3-CH=C=CH^- <-> CH3-C^-H-C#CH}$$
• 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 – Zhe Apr 7 '19 at 23:46
• @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. – andselisk Apr 8 '19 at 0:19