What was the NEET 2025 assertion-reason question on propyne reactions?
One mole of propyne with excess sodium produces exactly half mole of hydrogen gas. Four grams of propyne with NaNH2 produces ammonia gas occupying 224 mL at STP. The four evaluation choices are I correct II wrong, both wrong, I wrong II correct, both correct.
The official solution treats both statements as independent facts and checks them against balanced equations. Statement I matches the stoichiometry of the sodium reaction. Statement II fails on the volume calculation at STP.
How do you solve this NEET 2025 assertion-reason question on propyne reactions step by step?
Statement I is correct but Statement II is incorrect, so the answer is choice (1). Molar mass of propyne C3H4 = 40 g mol⁻¹ therefore 4 g = 0.1 mol.
For Statement I the balanced equation is
so 1 mol alkyne liberates exactly 0.5 mol H2. This matches the claim.
For Statement II the reaction is
therefore 0.1 mol alkyne must liberate 0.1 mol NH3. 0.1 mol NH3 at STP occupies 0.1 × 22.4 L = 2.24 L = 2240 mL not 224 mL. The factor-of-10 mismatch makes Statement II wrong.
What exact calculation error produces the wrong option in this propyne question?
Students pick both correct when they treat 4 g propyne as 0.01 mol instead of 0.1 mol by mistakenly using molar mass 400 or shifting the decimal in 4/40. They then use 22.4 mL per millimole instead of 22.4 L per mole when calculating STP volume, producing exactly 224 mL for 0.01 mol.
Forgetting to balance the sodium reaction as 2Na per terminal hydrogen leads them to expect 1 mol H2 per mole alkyne and wrongly reject Statement I. Write the balanced equation and mole ratio first, convert mass to moles once, then apply the mole-to-gas ratio before any volume number. This order makes the factor-of-10 error in Statement II automatic to spot.
What numerical relations govern acidity and gas evolution from terminal alkynes?
Terminal alkynes liberate gas with sodium or sodamide because the sp-hybridised C–H bond has pKa ≈ 25. This value is acidic enough for Na and NaNH2 but not for NaHCO3.
With sodium metal the balanced equation is
so 1 mol alkyne produces 0.5 mol H2. With sodamide the equation is
giving a strict 1 : 1 mol ratio. Alkenes and internal alkynes produce zero gas with either reagent because they lack the acidic terminal hydrogen.
What practice questions on terminal alkynes match the NEET 2025 difficulty?
Assertion: One mole of 1-butyne on treatment with excess Na liberates 11.2 L H2 at STP. Reason: Terminal alkynes are monobasic acids. Both are true and the reason correctly explains the assertion because one acidic hydrogen produces exactly 0.5 mol H2 which occupies 11.2 L at STP.
Numerical: 2.0 g of propyne is reacted with NaNH2. Volume of NH3 evolved at STP is? Options: 1.12 L, 2.24 L, 0.56 L, 22.4 L. Answer is 1.12 L. Moles of propyne equal 2/40 which is 0.05. The 1 : 1 ratio gives 0.05 mol NH3 which occupies 0.05 × 22.4 L equals 1.12 L.
Assertion-Reason: Ethyne reacts with ammoniacal AgNO3 but not with Na metal to give H2. The assertion is false because ethyne is HC≡CH and does react with sodium metal to liberate H2. The “but not” clause makes the whole claim incorrect.
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What strategy solves assertion-reason questions on alkynes in NEET Chemistry 2025?
Read both statements independently. Ignore the label “Statement I/II” and treat them as two separate true/false facts. Write the balanced equation and mole ratio before looking at any numerical value given.
Convert every mass to moles and every mole to volume using 22.4 L = 22400 mL exactly once. If numbers match within 1 % the statement is correct. A factor-of-10 error always makes it incorrect. Students who apply this checklist can correctly evaluate any assertion-reason on terminal alkyne acidity and gas volumes in under 90 seconds because it forces them to catch the volume mismatch and never confuse 224 mL with 2240 mL at STP.
What quick revision table for terminal alkyne reactions can you memorise in 2 minutes?
- Reagent Na (excess): gas liberated H2, moles of gas per mole alkyne = 0.5, volume at STP = 11.2 L
- Reagent NaNH2: gas liberated NH3, moles of gas per mole alkyne = 1.0, volume at STP = 22.4 L
- Reagent NaHCO3 or Na2CO3: no reaction, 0 mL gas
- 4 g propyne (0.1 mol) → 2.24 L NH3 or 1.12 L H2 at STP
You can expect to clear all doubts on stoichiometry in two minutes by reviewing this table before solving questions from the past-paper archive.
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Read next: Polarisation and Malus' Law NEET 2025: Three Polaroids at 22.5°.
Frequently asked questions
How much H2 is liberated from one mole of propyne with excess sodium?
One mole of propyne liberates exactly 0.5 moles of H2. The balanced equation is CH3-C≡CH + 2Na → CH3-C≡CNa + 1/2 H2, so the stoichiometry matches Statement I of the NEET 2025 question.
What volume of NH3 is produced from 4g propyne with NaNH2 at STP?
4 g propyne equals 0.1 mol. The 1:1 reaction with NaNH2 produces 0.1 mol NH3 which occupies 2.24 L (2240 mL) at STP. The claim of 224 mL in the NEET question is therefore incorrect.
Why is Statement II wrong in the NEET 2025 propyne assertion-reason question?
Statement II claims 4 g propyne with NaNH2 produces 224 mL NH3 at STP. Calculation shows 0.1 mol propyne yields 2240 mL, a factor-of-10 error. Students often mistake molar mass or use 22.4 mL instead of 22.4 L.
Do internal alkynes give H2 gas with sodium metal?
No. Only terminal alkynes with acidic sp C–H (pKa ≈ 25) react with Na or NaNH2 to evolve gas. Internal alkynes and alkenes lack this acidic hydrogen and produce zero gas with either reagent.
What is the common calculation mistake in the NEET 2025 propyne question?
Students treat 4 g propyne as 0.01 mol instead of 0.1 mol by wrongly using molar mass 400 or shifting decimals. They then calculate 0.01 × 22.4 L as 224 mL, matching the wrong statement instead of the correct 2240 mL.