When preparing terminal alkynes by an elimination reaction, sodium amide (NaNH₂) dissolved in liquid ammonia (NH₃) is used most frequently. Which of the following statements offers the best explanation for the above statement? A) The above statement is false; terminal alkynes are not produced under the given conditions. B) Sodium amide deprotonates the terminal alkyne, driving formation of the alkynide ion. C) Only sodium amide is a strong enough base to deprotonate a carbon. D) Terminal alkynes are more stable than the internal alkynes, and are always the favored product. E) Steric hindrance favors preparation of the less substituted terminal alkyne.

Question

Quizplus · Accepted Answer

Sodium amide is a strong enough base to deprotonate the acidic hydrogen on the terminal carbon of an alkyne. This creates an alkynide ion, which can then react with a variety of electrophiles to form terminal alkynes. Therefore, sodium amide dissolved in liquid ammonia is commonly used to prepare terminal alkynes via elimination reactions.

When Preparing Terminal Alkynes by an Elimination Reaction, Sodium Amide