$C_{2} H_{2} N a N H_{2} - ---- \to C H_{3} B r - --- \to N a N H_{2} - ---- \to C_{2} H_{5} B r - ---- \to N a - --- \to N H_{3} (l)$ trans-2-pentene.

If the given reaction is correct then enter 1 else enter 0.

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Solution

Acetylene reacts with sodamide and methyl bromide to form prop-1-yne.
It then reacts with sodamide and ethyl bromide to form pent-2-yne.
The carbon carbon triple bond is then selectively reduced to double bond by sodium in liq ammonia to obtain trans-2-pentene.

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