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Question:

In the reaction HC≡CH (1) NaNH2/liq.NH3 → (2) CH3CH2Br X (1) NaNH2/liq.NH3 → (2) CH3CH2Br Y, X and Y are:

X=2−Butyne;Y=2−Hexyne

X=1−Butyne;Y=3−Hexyne

X=1−Butyne;Y=2−Hexyne

X=2−Butyne;Y=3−Hexyne

Solution:

NaNH2 is a strong base which on reacting with terminal alkyne removes terminal acidic hydrogen leaving alkynyl carbanion, which on reaction with alkyl halide produces longer alkyne by nucleophilic addition. Thus, X is 1-Butyne and Y is 3-Hexyne.