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The hydrocarbon which can react with sodium in liquid ammonia is
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Verified Answer
The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}$
$$
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{Na} / \mathrm{Liq} 9 \mathrm{NH}}{\Delta} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \stackrel{\ominus}{\mathrm{C}} \mathrm{Na}^{\oplus}
$$
It is a terminal alkyne, having acidic hydrogen.
Note: Solve it as a case of terminal alkynes, otherwise all alkynes react with $\mathrm{Na}$ in liq. $\mathrm{NH}_3$.
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH} \stackrel{\mathrm{Na} / \mathrm{Liq} 9 \mathrm{NH}}{\Delta} \longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \stackrel{\ominus}{\mathrm{C}} \mathrm{Na}^{\oplus}
$$
It is a terminal alkyne, having acidic hydrogen.
Note: Solve it as a case of terminal alkynes, otherwise all alkynes react with $\mathrm{Na}$ in liq. $\mathrm{NH}_3$.
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