Given below are two statements : Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of $\mathrm{H}_{2}$ gas. Statement II : Four g of propyne reacts with $\mathrm{NaNH}_{2}$ to liberate $\mathrm{NH}_{3}$ gas which occupies 224 mL at STP. In the light of the above statements, choose the most appropriate answer from the options given below:

(A) Statement I is correct but Statement II is incorrect.

(B) Both Statement I and Statement II are incorrect

(C) Statement I is incorrect but Statement II is correct

(D) Both Statement I and Statement II are correct.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Hydrocarbons

Given below are two statements :

Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of

H_{2}

gas.

Statement II : Four g of propyne reacts with

NaNH_{2}

to liberate

NH_{3}

gas which occupies 224 mL at STP.

In the light of the above statements, choose the most appropriate answer from the options given below:

Answer: (A)

Propyne,

C H_{3} - C \equiv C H

, has one acidic terminal alkyne hydrogen. With excess sodium metal it forms the sodium acetylide:

C H_{3} - C \equiv C H + N a \to C H_{3} - C \equiv C^{-} N a^{+} + \frac{1}{2} H_{2}

One mole of propyne liberates exactly half a mole of

H_{2}

, so Statement I holds.

For Statement II, propyne reacts the same way with sodamide, releasing ammonia instead of hydrogen:

C H_{3} - C \equiv C H + N a N H_{2} \to C H_{3} - C \equiv C^{-} N a^{+} + N H_{3}

The molar mass of propyne is 40 g/mol, so 4 g corresponds to 0.1 mol, giving 0.1 mol of

N H_{3}

by 1:1 stoichiometry. At STP this occupies

0.1 \times 22400 = 2240

mL, not the 224 mL claimed.

Statement I is correct and Statement II is incorrect.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option