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Physical Properties of Alkynes

B 2-butyne A....

Question

$B Y \leftarrow 2 - b u t y n e X - \to$ A. A and B are geometrically isomers. 'A' is more symmetrically than 'B'. 'B' has higher heat of hydrogenation than 'A'. Then 'X' and 'Y' are respectively:

L i / L i q N H_{3}, H_{2} /

Lindlar's catalyst

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L i / L i q . N H_{3}, N a / L i q . N H_{3}

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H_{2} /

Lindlar's catalyst,

N a / L i q . N H_{3}

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H_{2} / P t, H_{2} /

Lindlar's catalyst

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Solution

The correct option is A $L i / L i q N H_{3}, H_{2} /$ Lindlar's catalyst
'X' and 'Y' are $L i / L i q N H_{3}, H_{2} /$ Lindlar's catalyst respectively.
$L i / L i q N H_{3}$ reduces $2 - b u t y n e$ to $t r a n s - 2 - b u t e n e$ (A).
$H_{2}$ Lindlar's catalyst reduces $2 - b u t y n e$ to $c i s - 2 - b u t e n e$ (B).
$t r a n s - 2 - b u t e n e$ is more symmetrical than $c i s - 2 - b u t e n e$ as $t r a n s - 2 - b u t e n e$ has inversion centre which is not present in $c i s - 2 - b u t e n e$ .
$c i s - 2 - b u t e n e$ has higher heat of hydrogenation than $t r a n s - 2 - b u t e n e$ .

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Similar questions

$B Y \leftarrow 2 - B u t y n e + H_{2} X - \to A .$ A and B are geometrical isomers, A is more symmetrical than B. Then X and Y are

H_{2}

X

N a / l i q . N H_{3}

Y

X

Y

Y

X

B L i n d l a r^{'} s c a t a l y s t / H_{2} \leftarrow ------------ - R C \equiv C R N a / L i q . N H_{3} - ------- \to A

\equiv

H_{2} / Lindlar's

N H_{3}

S t a t e m e n t

H_{2} / L i n d l a r^{'} s

B r_{2} / C C l_{4}

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