In the reaction H-C- CH 2CH3CH2Br X 2CH3CH2Br Y- X and Y are-

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Dehydrohalogenation of Dihalides

In the reacti...

Question

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

X = 1 - B u t y n e; Y = 3 - H e x y n e

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X = 2 - B u t y n e; Y = 3 - H e x y n e

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X = 2 - B u t y n e; Y = 2 - H e x y n e

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X = 1 - B u t y n e; Y = 2 - H e x y n e

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Solution

The correct option is A $X = 1 - B u t y n e; Y = 3 - H e x y n e$
$N a N H_{2}$ 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.

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$H - C \equiv C H \frac{(1) N a N H_{2}}{l i q . N H_{3}} - -------- \to (2) C H_{3} C H_{2} B r X \frac{(1) N a N H_{2}}{l i q . N H_{3}} - -------- \to (2) C H_{3} C H_{2} B r Y$

X and Y are

H - C \equiv C - H (i) N a N H_{2} / l i q N H_{3} - ---------- \to (i i) C H_{5} - I X (i) N a N H_{2} / l i q N H_{3} - ---------- \to (i i) C H_{3} C H_{2} - I

Q. 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:

Q. 3,3-dimethyl-1-butyne and 1-hexyne show position isomerism.

\begin{matrix} Column - I & Column - II (a) 3,3,4,4-tetramethyl-1-pentyne & (p) Has only three collinear carbons (b) 2-butyne & (q) Has four collinear carbons (c) 1-butyne & (r) Substitution of H by Cl gives single product (d) 2,2,5,5-tetramethyl-3-hexyne & (s) Substitution of H by Cl gives three different product (excluding stereoisomers) \end{matrix}

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

The correct option is A X=1−Butyne;Y=3−HexyneNaNH2 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.

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