I\'m stuck between answer A and E. Leaning more towards A. Can you please explai

Question

I'm stuck between answer A and E. Leaning more towards A. Can you please explain why which is correct and the other is wrong. I will "like" your answer.

4. Why is sodium amide (NaNH2) a good choice for double dehydrohaiogenation of a dihalide to a terminal alkyne? A. The pKa of ammonia (NH3) is much larger than the pKa of a terminal alkene (H-CEC-R) B. NH2e allows the product to equilibrate to the most-stable triple bond. C. Only 1 equivalent of NaNH2 is needed D. The reaction can be carried out in water, a much safer, cheaper, and easier solvent to use than ammonia. E. The terminai alkyne formed initially will be irreversibly deprotonated to an acetylide anion.

Explanation / Answer

Option E is the correct answer, in which the formed alkyne gets deprotonated to acetylide anion because the NH2- i.e. sonjugate base of a week acid i.e. NH3 and therefore, it is very strong base and alkyne proton is weekly acidic

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