Given below are two statements : Statement I : One mole of propyne reacts with excess of sodium to liberate…

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:
  1. Statement I is incorrect but Statement II is correct
  2. Both Statement I and Statement II are correct
  3. Statement I is correct but Statement II is incorrect
  4. Both Statement I and Statement II are incorrect

Solution

$\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}+\mathrm{Na} \rightarrow \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}+\frac{1}{2} \mathrm{H}_2$
Statement-l is correct.
Moles of $\mathrm{C}_3 \mathrm{H}_4=\frac{4}{40}=0.1$ mole
$\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \rightarrow \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}+ \mathrm{NH}_3$
0.1 mole
0.1 mole
Volume of $\mathrm{NH}_3=(0.1)(22.4)=2.24 \mathrm{~L}$
Statement-II is incorrect.

Asked in: JEE Main 2025 (22 Jan Shift 1)

Practice more Hydrocarbons questions on Aicharya