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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Hydrocarbons

JEE Main 2025 · 22 January · Morning Shift · Chemistry · Hydrocarbons

Given below are two statements : Statement I : One mole of propyne...

Given below are two statements : 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:

Options

  1. AStatement I is incorrect but Statement II is correct
  2. BBoth Statement I and Statement II are correct
  3. CStatement I is correct but Statement II is incorrect
  4. DBoth Statement I and Statement II are incorrect

Correct answer

C. Statement I is correct but Statement II is incorrect

Step-by-step solution

CH3−C≡CH+Na→CH3−C≡C−Na++12H2\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 C3H4=440=0.1\mathrm{C}_3 \mathrm{H}_4=\frac{4}{40}=0.1 mole CH3−C≡CH+NaNH2→CH3−C≡C−Na++NH3\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 NH3=(0.1)(22.4)=2.24 L\mathrm{NH}_3=(0.1)(22.4)=2.24 \mathrm{~L} Statement-II is incorrect.
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