In the reaction H - C - CH 1 NaN H 2 - liq - NH 3-2 CH 3 CH 2 Br X 1 NaNH 2 - liq - NH 3-2 CH 3 CH 2 Br - Y- Xand Y are-A- X -1 - Butyne- Y -3 - HexyneB- X-2 - Butyne- Y-3 - HexyaneC- X-2 - Butyne- Y-2 - HexyaneD- X -1 - Butyne- Y -2 - Hexyane

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Higher Alkynes from Alkynes

In the reacti...

Question

In the reaction $H - C \equiv C H (1) N a N H_{2} / l i q . N H_{3} - ----------- \to (2) C H_{3} C H_{2} B r X (1) N a N H_{2} / l i q . N H_{3} - ----------- \to (2) C H_{3} C H_{2} B r Y,$ Xand Y are:

X = 1 - Butyne; Y = 3 - Hexyne

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X = 2 - Butyne; Y = 3 - Hexyane

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X = 2 - Butyne; Y = 2 - Hexyane

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X = 1 - Butyne; Y = 2 - Hexyane

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Solution

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 $(N a N H_{2})$ .
$N a N H_{2}$ 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 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.

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Higher Alkynes from Alkynes

Standard XII Chemistry

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In the reaction H−C≡CH(1) NaNH2/liq.NH3−−−−−−−−−−−−→(2) CH3CH2BrX(1) NaNH2/liq.NH3−−−−−−−−−−−−→(2) CH3CH2BrY, Xand Y are:

In the reaction $H - C \equiv C H (1) N a N H_{2} / l i q . N H_{3} - ----------- \to (2) C H_{3} C H_{2} B r X (1) N a N H_{2} / l i q . N H_{3} - ----------- \to (2) C H_{3} C H_{2} B r Y,$ Xand Y are: