JEE Main 2025 — Hydrocarbons Question with Solution

Statement I

“One mole of propyne reacts with excess of sodium to liberate half a mole of ${\mathrm{H}}_2$ gas.”

Reaction and Stoichiometry

Propyne (terminal alkyne): $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{CH}$.

Reaction with sodium:

$2\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{CH}+2\mathrm{Na}⟶2(\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv {\mathrm{C}}^{-}\mathrm{N}{\mathrm{a}}^{+})+{\mathrm{H}}_2.$

From this balanced equation, 2 moles of propyne produce 1 mole of ${\mathrm{H}}_2$.

Hence, 1 mole of propyne will produce $\frac{1}{2}$ mole of ${\mathrm{H}}_2$.

$\boxed{\text{Statement I is correct.}}$

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Statement II

“Four grams of propyne reacts with $\mathrm{NaN}{\mathrm{H}}_2$ to liberate $\mathrm{N}{\mathrm{H}}_3$ gas which occupies 224 mL at STP.”

Analysis

Moles of propyne

Molecular mass of propyne (${\mathrm{C}}_3{\mathrm{H}}_4$):

$3\times 12+4\times 1=36+4=40g/mol.$

Four grams of propyne is:

$\frac{4\mathrm{g}}{40g/mol}=0.1\mathrm{mol}.$

Reaction with $\mathrm{NaN}{\mathrm{H}}_2$

For a terminal alkyne:

$\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{CH}+\mathrm{NaN}{\mathrm{H}}_2⟶\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv {\mathrm{C}}^{-}\mathrm{N}{\mathrm{a}}^{+}+\mathrm{N}{\mathrm{H}}_3.$

1 mole of propyne produces 1 mole of $\mathrm{N}{\mathrm{H}}_3$.

Moles of $\mathrm{N}{\mathrm{H}}_3$ produced

With $0.1\mathrm{mol}$ of propyne, we get $0.1\mathrm{mol}$ of $\mathrm{N}{\mathrm{H}}_3$.

Volume of $\mathrm{N}{\mathrm{H}}_3$ at STP

1 mole of any ideal gas at STP $\approx 22.4\mathrm{L}=22400\mathrm{mL}.$

$0.1\mathrm{mol}$ of $\mathrm{N}{\mathrm{H}}_3$ occupies $0.1\times 22.4\mathrm{L}=2.24\mathrm{L}=2240\mathrm{mL}.$

However, Statement II says the liberated $\mathrm{N}{\mathrm{H}}_3$ occupies only 224 mL at STP, which corresponds to $0.01\mathrm{mol}$ of $\mathrm{N}{\mathrm{H}}_3$, not $0.1\mathrm{mol}$. Therefore, the statement’s volume is off by a factor of 10 and is thus incorrect if the reaction goes to completion in a typical way.

$\boxed{\text{Statement II is incorrect.}}$

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Conclusion

Statement I is correct.

Statement II is incorrect.

Hence, the best choice is:

$\boxed{\text{Option D: Statement I is correct but Statement II is incorrect.}}$