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Chemistry · Hydrocarbons

JEE Main 2025 — 22 January, Morning Shift — Question 35

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II :

Statement I : One mole of propyne reacts with excess of sodium to liberate half a mole of H2\mathrm{H}_{2} gas.

Statement II : Four g of propyne reacts with NaNH2\mathrm{NaNH}_{2} to liberate NH3\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. Option A:

    Statement I is correct but Statement II is incorrect.

    Correct
  2. Option B:

    Both Statement I and Statement II are incorrect

  3. Option C:

    Statement I is incorrect but Statement II is correct

  4. Option D:

    Both Statement I and Statement II are correct.

Answer: A

Step-by-step solution

figure

CH3−C≡CH+NaNH2→CH3C≡C‾Na++NH3\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{C} \equiv \overline{\mathrm{C}} \stackrel{+}{\mathrm{Na}}+\mathrm{NH}_{3}

4gm4 gm

$\frac{4}{40}=0.1 \mathrm{~mole} \quad \frac{0.1 \mathrm{~mole}}{2240 \mathrm{~mole}} $$

Statement I is correct but Statement II is incorrect

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
22 January, Morning Shift
Subject
Chemistry
Chapter
Hydrocarbons
Topic
Properties & Uses of Alkynes
Given below are two statements : one is labelled as Statement I and… | JEE Main 2025 PYQ with Solution · DhiX AI