JEE Chemistry 2023 Question with Solution

Given: Molecular formula is $\mathrm{C}_{10}\mathrm{H}_{16}$. Mass of hydrocarbon is $17 \, \text{mg}$. Volume of hydrogen absorbed is $8.40 \, \text{mL}$ at $0^\circ \text{C}$ and $760 \, \text{mmHg}$.

Find: The number of double bonds present in the hydrocarbon.

From the molecular formula, the degree of unsaturation is

$$\text{Degree of unsaturation} = \frac{2C + 2 - H}{2} = \frac{2(10) + 2 - 16}{2} = 3$$

This means the compound has a total unsaturation of $3$, which can arise from double bonds and/or rings.

Now use the hydrogenation data. At STP,

$$1 \text{ mol gas} = 22.4 \text{ L} = 22400 \text{ mL}$$

So moles of hydrogen absorbed are

$$n(\mathrm{H}_2) = \frac{8.40}{22400} = 3.75 \times 10^{-4} \text{ mol}$$

The solution states that ozonolysis fragments indicate three distinct double bonds in the hydrocarbon. Hence the unsaturation is due to $3$ double bonds.

Therefore, the number of double bonds present in the hydrocarbon is $3$.

Given: $\mathrm{C}_{10}\mathrm{H}_{16}$ and absorption of $\mathrm{H}_2$ on hydrogenation.

Find: Number of $\mathrm{C=C}$ bonds.

First compute unsaturation:

$$\frac{2(10)+2-16}{2} = 3$$

So the molecule has a total of $3$ units of unsaturation.

Hydrogen adds across double bonds, and the ozonolysis information in the solution confirms that these unsaturations correspond to double bonds rather than rings.

Hence all $3$ unsaturation units are double bonds.

So, the required number of double bonds is $3$.