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Interconversion of terminal to internal alkyne and vice versa takes place by the following reagents (A) and (B).
Reagents (A) and (B) are:

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${NaNH}_{2} and {NaNH}_{2}$

$alc. KOH and {NaNH}_{2}$

$alc. KOH and P - 2 catalyst$

${NaNH}_{2} and Lindlar's catalyst$

answer is A.

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Detailed Solution

Sodamide is a very strong base and it can remove hydrogen atom from the sp carbon of the triple bond. But Alc KOH cant, hence it can be used for sp3 carbons preferbly.

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