Final answer:
The major product for the reaction with excess NaNH₂ followed by H₂O is an alkyne. The strong base NaNH₂ deprotonates the alkyne to form an acetylide anion, and subsequent addition of water protonates the anion back to a neutral alkyne, with ammonia being a side product.
Step-by-step explanation:
The major product formed when an amide anion (provided by a compound such as sodium amide, NaNH₂) reacts with an alkyne is typically an alkyne. This is because the amide anion is a strong enough base to deprotonate the terminal hydrogen of an alkyne, generating a negatively charged acetylide anion. When water (H₂O) is then added, it acts as an acid donating a proton back to the acetylide anion, resulting in a neutral alkyne molecule and ammonia (NH₃) as a side product.
Given the reactants in the student's question (excess NaNH₂ and H₂O), the major product would most likely be:
The balanced chemical reaction can be represented as R-C≡CH + NaNH₂ → R-C≡C⁻ Na⁻ + NH₃ followed by R-C≡C⁻ Na⁻ + H₂O → R-C≡CH + NaOH + NH₃.