Question:

17 mg of a hydrocarbon (M.F. C10H16) takes up 8.40 mL of the H2 gas measured at \(0\degree C\) and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields
Ozonolysis of the hydrocarbon
The number of double bond/s present in the hydrocarbon is_______.

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For determining double bonds:
• Use the degree of unsaturation formula to find the total number of double bonds and rings.
• Analyze ozonolysis fragments to confirm the number and position of double bonds.

Updated On: Jan 9, 2025
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Correct Answer: 3

Solution and Explanation

1. Molecular Formula Analysis:
The given molecular formula is \(\text{C}_{10}\text{H}_{16}\). The degree of unsaturation is:
\[\text{Degree of unsaturation} = \frac{2C + 2 - H}{2} = \frac{2(10) + 2 - 16}{2} = 3.\]
This indicates 3 double bonds or rings.
2. Hydrogenation Data:
8.40 mL of \(\text{H}_2\) gas is absorbed. At STP, 1 mol of gas occupies 22.4 L. Moles of \(\text{H}_2\) absorbed:
\[n = \frac{8.40}{22400} = 3.75 \times 10^{-4}~\text{mol}.\]
3. Ozonolysis Data:
Ozonolysis fragments indicate three distinct double bonds in the hydrocarbon.
4. Conclusion
The hydrocarbon contains 3 double bonds.
Final Answer: \(3\).

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