MCQEasyJEE 2025Alkynes (Acidic Nature, Reactions)

JEE Chemistry 2025 Question with Solution

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 224mL224 \, \text{mL} at STP.

In the light of the above statements, choose the most appropriate answer from the options given below:

  • A

    Statement I is correct but Statement II is incorrect.

  • B

    Both Statement I and Statement II are incorrect.

  • C

    Statement I is incorrect but Statement II is correct.

  • D

    Both Statement I and Statement II are correct.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Two statements about the reactions of propyne with sodium and NaNH2\mathrm{NaNH_2} are to be checked.

Find: Which statement is correct.

For Statement I, propyne is a terminal alkyne and contains one acidic hydrogen atom. It reacts with sodium as:

C3H3H+NaC3H3Na+12H2\mathrm{C_3H_3-H + Na \rightarrow C_3H_3Na + \frac{1}{2}H_2}

So, one mole of propyne liberates half a mole of H2\mathrm{H_2} gas. Therefore, Statement I is correct.

For Statement II, molar mass of propyne (C3H4)\mathrm{(C_3H_4)} is:

3×12+4×1=40g mol13 \times 12 + 4 \times 1 = 40 \, \text{g mol}^{-1}

Hence, moles of propyne in 4g4 \, \text{g} are:

440=0.1mol\frac{4}{40} = 0.1 \, \text{mol}

Reaction with NaNH2\mathrm{NaNH_2} is:

C3H4+NaNH2C3H3Na+NH3\mathrm{C_3H_4 + NaNH_2 \rightarrow C_3H_3Na + NH_3}

Thus, 0.1mol0.1 \, \text{mol} propyne gives 0.1mol0.1 \, \text{mol} NH3\mathrm{NH_3}.

At STP, 1mol1 \, \text{mol} gas occupies 22.4L22.4 \, \text{L}. Therefore, volume of 0.1mol0.1 \, \text{mol} NH3\mathrm{NH_3} is:

0.1×22.4=2.24L0.1 \times 22.4 = 2.24 \, \text{L}

This is 2240mL2240 \, \text{mL}, not 224mL224 \, \text{mL}. Therefore, Statement II is incorrect.

Hence, the correct option is A.

Check each statement separately

Given: Two independent statements are provided.

Find: Truth value of each statement.

  1. Statement I depends on the acidic hydrogen of a terminal alkyne. One hydrogen atom from two molecules combines to form one molecule of H2\mathrm{H_2}, so per mole of propyne the amount of hydrogen released is:
12mol of H2\frac{1}{2} \, \text{mol of } \mathrm{H_2}

Hence, Statement I is true.

  1. Statement II requires stoichiometry and gas volume calculation.

Mass of propyne = 4g4 \, \text{g}

Moles of propyne=440=0.1\text{Moles of propyne} = \frac{4}{40} = 0.1

From the reaction:

C3H4+NaNH2C3H3Na+NH3\mathrm{C_3H_4 + NaNH_2 \rightarrow C_3H_3Na + NH_3}

Mole ratio of propyne to NH3\mathrm{NH_3} is 1:11:1, so moles of NH3\mathrm{NH_3} formed = 0.1mol0.1 \, \text{mol}.

Now,

Volume at STP=0.1×22.4=2.24L\text{Volume at STP} = 0.1 \times 22.4 = 2.24 \, \text{L}

Given statement says 224mL224 \, \text{mL}, which is incorrect.

Therefore, Statement I is correct but Statement II is incorrect, so the correct option is A.

Common mistakes

  • Students often forget that only the terminal alkyne hydrogen is acidic. Propyne has one such hydrogen, so it can liberate hydrogen with sodium. Do not treat all hydrogens in propyne as equally reactive.

  • A common mistake is calculating the gas volume for 0.1mol0.1 \, \text{mol} at STP as 224mL224 \, \text{mL} instead of 2.24L=2240mL2.24 \, \text{L} = 2240 \, \text{mL}. Always convert 22.4L22.4 \, \text{L} carefully.

  • Some students use the wrong molar mass of propyne. For C3H4\mathrm{C_3H_4}, the molar mass is 40g mol140 \, \text{g mol}^{-1}, not any other value. Compute moles first before applying stoichiometry.

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