The balance chemical equations for the following :
The preparation of ethyne from 1, 2 - dibromoethane.

C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r

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C H_{2} = C H B r N a N H_{2} - ---- \to C H_{2} = C H_{2}

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C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C B r_{2} N a N H_{2} - ---- \to C H \equiv C H

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C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r N a N H_{2} - ---- \to C H \equiv C H

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Solution

The correct option is D $C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r N a N H_{2} - ---- \to C H \equiv C H$
Reaction:
$C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r N a N H_{2} - ---- \to C H \equiv C H$ (ethyne)

In the reaction first $β$ elimination reaction of the alkyl halide takes place to form the corresponding alkene. This alkene on reaction with sodamide forms corresponding alkyne.(As $N a N H_{2}$ is strong base)

Option D is correct.

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