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When ethyne (H–C≡CH) is treated first with (1) NaNH2 in liquid NH3 and then with (2) CH3CH2Br, the product X is formed. If X is again subjected to the same two-step sequence, the product Y obtained is:

1. X = 1-Butyne; Y = 3-Hexyne
2. X = 2-Butyne; Y = 3-Hexyne
3. X = 2-Butyne; Y = 2-Hexyne
4. X = 1-Butyne; Y = 2-Hexyne

Correct answer: X = 1-Butyne; Y = 3-Hexyne

Solution

The correct option is right because treating ethyne with NaNH2 generates a terminal alkyne, which in this case is 1-butyne. When 1-butyne is further treated with CH3CH2Br, a nucleophilic substitution occurs, leading to the formation of 3-hexyne.

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