NVAEasyJEE 2023Alkynes (Acidic Nature, Reactions)

JEE Chemistry 2023 Question with Solution

Molar mass of the hydrocarbon (X)\left(X\right) which on ozonolysis consumes one mole of O3O_3 per mole of (X)\left(X\right) and gives one mole each of ethanol and propanone is:

Answer

Correct answer:70

Step-by-step solution

Standard Method

Given: The hydrocarbon (X)\left(X\right) consumes one mole of O3O_3 per mole of (X)\left(X\right) on ozonolysis and gives one mole each of ethanol and propanone.

Find: The molar mass of (X)\left(X\right).

From the given reaction in the solution:

CH3CCHCH3+O3CH3CHO+CH3COCH3\text{CH}_3-\text{C}\equiv\text{CH}-\text{CH}_3 + \text{O}_3 \to \text{CH}_3\text{CHO} + \text{CH}_3\text{COCH}_3

Thus, the hydrocarbon (X)\left(X\right) is pent-2-yne with molecular formula C5H8C_5H_8.

Its molecular mass is:

5×12+8×1=70g mol15 \times 12 + 8 \times 1 = 70 \, \text{g mol}^{-1}

Therefore, the molar mass of the hydrocarbon is 70g mol170 \, \text{g mol}^{-1}.

Common mistakes

  • Identifying the hydrocarbon incorrectly from the ozonolysis products is a common mistake. This is wrong because the products must be matched to the correct alkyne skeleton. Use the given ethanol and propanone fragments to reconstruct the parent hydrocarbon carefully.

  • Using the wrong molecular formula for pent-2-yne can lead to an incorrect molar mass. This is wrong because alkynes follow the formula CnH2n2C_nH_{2n-2}. For n=5n=5, use C5H8C_5H_8 before calculating the mass.

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