Question

In the reaction $H-C\equiv CH\underset{\left(2\right)C{H}_{3}C{H}_{2}Br}{\overset{\left(1\right)NaN{H}_2/liq.N{H}_3}{\to }}X\underset{\left(2\right)C{H}_{3}C{H}_{2}Br}{\overset{\left(1\right)NaN{H}_2/liq.N{H}_3}{\to }}Y,$ Xand Y are:

• A. X = 1 - Butyne; Y = 3 - Hexyne
• B. X = 2 - Butyne; Y = 3 - Hexyane
• C. X = 2 - Butyne; Y = 2 - Hexyane
• D. X = 1 - Butyne; Y = 2 - Hexyane

Answer 1

The correct option is B X = 2 - Butyne; Y = 3 - Hexyane

Higher alkynes are prepared from terminal alkynes like acetylene by treating it with sodium metal at high temperature or with sodamide$\left(NaN{H}_2\right)$.
$NaN{H}_2$ 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 substitution.
Here $N{H}_2^{-}$ attacks the acetylene and removes the acidic proton and the carbanion formed attack the ethyl bromide to form 1-Butyne (higher alkyne).
Again, excess $N{H}_2^{-}$ attacks the terminal alkyne of 1-Butyne to form carbanion and the process repeats to give 3-Hexyne.
Thus, X is 1-Butyne and Y is 3-Hexyne.