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Terminal alkynes are unusual for simple hydrocarbons in that they can be deprotonated (pKa = 26) using an appropriate base (typically NaNH2, pKa = 36) to generate a carbanion which can function as a C centered nucleophile and so allow for the formation of new C-C bonds by reacting with C centered electrophiles (such as alkyl halides).

In order to appreciate what makes the terminal alkyne more acidic than most other hydrocarbons, we should look at the stability of the conjugate base (i.e. the carboanion).

For each type of carbanion shown, the nature of the hybrid orbital containing
the electron pair is important. Increased s character (sp
= 50%, sp2 = 33% and sp3 = 25%) implies
that the alkyne sp orbital is closer to the nucleus and so there
is greater electrostatic stabilization of the electron pair. Therefore
the conjugate base of the alkyne is the most stable and the most readily
formed.
However the terminal alkyne C-H bond is not strongly acidic and a strong
base, such as the amide ion, NH2-, is required to
form the carbanion.

Could you use a base such as NaOH or NaOEt for this reaction ?
Related reactions