Given below are two statements :
Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas.
Statement II : Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Propyne has only one acidic hydrogen, so count moles carefully: it is 0.1 mol, not 0.01 mol.

1. One acidic hydrogen. In propyne, CH₃-C≡C-H, only the hydrogen on the triple-bonded carbon is acidic. The sp carbon holds the negative charge well, so this H can be removed. The three CH₃ hydrogens are not acidic.
2. Statement I. Each propyne gives one H atom to sodium, and two H atoms make one H₂. So 1 mol propyne gives 1/2 mol H₂. Statement I is correct.

   2CH₃C≡CH + 2Na → 2CH₃C≡CNa + H₂

3. Statement II. Molar mass of C₃H₄ is 36 + 4 = 40 g mol⁻¹, so 4 g is 0.1 mol. The reaction CH₃C≡CH + NaNH₂ → CH₃C≡CNa + NH₃ is 1 : 1, so it gives 0.1 mol NH₃, which is 0.1 × 22400 = 2240 mL at STP, not 224 mL. Statement II is incorrect.
4. So. Statement I is correct and Statement II is incorrect, which is option (d).

The idea to keep. A terminal alkyne gives one acidic H per C≡C-H group. With Na, two of them make one H₂. With NaNH₂, each one makes one NH₃.

Answer from the NTA final answer key. Solution by PrepWiser.