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Gem-dihalides react with $\ce{NaNH2}$ to give terminal alkynes. Why is a terminal alkyne prefered?

For example, enter image description here

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I had thought that 2-pentyne would be preferred over 1-pentyne as it would be more stable due to hyperconjugation, but it says the final product is terminal alkyne.

Why?

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  • $\begingroup$ It's all about how the reaction is run. 2-Pentyne can form under milder conditions. More vigorous conditions induce the "zipper reaction" producing the anion of 1-pentyne. 1-Alkyne (pKa ~ 25; NH3, pKa ~ 35. $\endgroup$ – user55119 May 4 '18 at 12:02
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this once was my doubt...terminal alkynes are preferred because the acidic hydrogen at the terminal carbon is acidic(sp hybrid of Carbon) and which when taken up by the base gives a species with a negative charge on sp hybridized carbon which is indeed stable...so this provides the incentive for forming a terminal alkyne as the final product.

Trusted source: Jerry March Organic Chemistry 6th edition.

When the base is NaNH2 1-alkynes predominate (where possible), because this base is strong enough to form the salt of the alkyne, shifting any equilibrium between 1- and 2-alkynes. When the base is $\ce{OH-}$ or $\ce{OR-}$, the equilibrium tends to be shifted to the internal alkyne, which is thermodynamically more stable.

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