Why Is a Terminal Alkyne Favored When Sodium Amide (NaNH₂)

Why is a terminal alkyne favored when sodium amide (NaNH₂) is used in an elimination reaction with 2,3-dichlorohexane?

A) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
B) The resonance favors the formation of the terminal rather than internal alkyne.
C) The positions of the Cl atoms induce the net formation of the terminal alkyne.
D) The terminal alkyne is more stable than the internal alkyne and is naturally the favored product.
E) The terminal alkyne is not favored in this reaction.