In the reaction H-Cequiv CH underset -2-CH-3-CH-2-Br-xrightarrow -1-NaNH-2-liq-NH-3- X underset -2-CH-3-CH-2-Br-xrightarrow -1-NaNH-2-liq-NH-3- Y- X and Y are-X - 1 - Butyne- Y - 3-HexyneX - 2 - Butyne- Y - 3-HexyneX - 1 - Butyne- Y - 2-HexyneX - 2 - Butyne- Y - 2-Hexyne

Question

Toppr Admin · Accepted Answer

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-xA0-

In the reaction H−C≡CH(1)NaNH2/liq.NH3−−−−−−−−−−−−→(2)CH3CH2BrX(1)NaNH2/liq.NH3−−−−−−−−−−−−→(2)CH3CH2BrY, X and Y are:X=1−Butyne;Y=3−HexyneX=2−Butyne;Y=3−HexyneX=1−Butyne;Y=2−HexyneX=2−Butyne;Y=2−Hexyne

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.

$N a N H_{2}$ 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.