In the reaction H - Cl - CH X Y-

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Hydrohalogenation of Alkenes

In the reacti...

Question

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

X = 1 -

Butyne;

y = 2 -

Hexyne

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X = 1 -

Butyne;

y = 3 -

Hexyne

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X = 2 -

Butyne;

y = 3 -

Hexyne

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X = 2 -

Butyne;

y = 2 -

Hexyne

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Solution

The correct option is A $X = 1 -$ Butyne; $y = 2 -$ Hexyne
On addition of $N a N H_{2}$ , $⊖ N H_{2}$ abstract proton of alkyne and carbanion formation takes place. Then carbanion attacks on alkyl halide and addition of alkyl takes place.

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Similar questions

In the reaction

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

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

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:

\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−Cl=CH(1)NaNH2/liq.NH3−−−−−−−−−−−−→(2)CH3CH2BrX(1)NaNH2/liq.NH3−−−−−−−−−−−−→(2)CH3CH2BrY.

The correct option is A X=1−Butyne; y=2− HexyneOn addition of NaNH2, ⊖NH2 abstract proton of alkyne and carbanion formation takes place. Then carbanion attacks on alkyl halide and addition of alkyl takes place.

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

The correct option is A $X = 1 -$ Butyne; $y = 2 -$ Hexyne
On addition of $N a N H_{2}$ , $⊖ N H_{2}$ abstract proton of alkyne and carbanion formation takes place. Then carbanion attacks on alkyl halide and addition of alkyl takes place.