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The hydrocarbon which can react with sodium in liquid ammonia is

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\begin{aligned} {CH}_{3} {CH}_{2} C \equiv {CCH}_{2} {CH}_{3} \end{aligned}

\begin{aligned} {CH}_{3} {CH}_{2} {CH}_{2} C \equiv {CCH}_{2} {CH}_{2} {CH}_{3} \end{aligned}

\begin{aligned} {CH}_{3} {CH}_{2} C \equiv CH \end{aligned}

\begin{aligned} {CH}_{3} CH \equiv {CHCH}_{3} \end{aligned}

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detailed solution

Correct option is C

The compound ${CH}_{3} {CH}_{2} C \equiv CH$ contains hydrogen atom attached to triply bonded $sp$ hybridized carbon which acquires acidic nature due to high electronegativity of carbon. This acetylinic hydrogen is replaced by ${Na}^{+}$ .

${CH}_{3} {CH}_{2} C \equiv CH + Na \to {CH}_{3} {CH}_{2} C = \bar{C} : N \overset{+}{a} + \frac{1}{2} H_{2} O .$

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