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Hydrocarbons: JEE Main Chemistry Question with Solution

17\text{mg}

of a hydrocarbon (M.F.

\mathrm{C_{10}H_{16}}

) takes up

8.40\text{ mL}

of the

\mathrm{H_2}

gas measured at

0^\circ\mathrm{C}

and

760\text{ mm}

\mathrm{Hg}

. Ozonolysis of the same hydrocarbon yields The number of double bond/s present in the hydrocarbon is

\text{\quad\quad\quad\quad}

Chemical structure formulas of the ozonolysis products: acetone, formaldehyde, and 3-oxopentanedial.

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Step-by-step solutionView

Correct answer

The hydrocarbon contains 3 double bonds, as 1 mole of the hydrocarbon consumes 3 moles of H₂ gas.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Mass of hydrocarbon $\mathrm{C_{10}H_{16}} = 17\text{ mg} = 1.7 \times 10^{-2}\text{ g}$ . Molar mass of $\mathrm{C_{10}H_{16}} = 10 \times 12 + 16 \times 1 = 136\text{ g/mol}$ . Volume of $\mathrm{H_2}$ gas at STP ( $0^\circ\mathrm{C}$ , $760\text{ mm Hg}$ ) $= 8.40\text{ mL}$ .

02find

The number of double bonds present in one molecule of the hydrocarbon.

03strategise

Each double bond consumes one mole of $\mathrm{H_2}$ upon complete catalytic hydrogenation. The number of double bonds equals the mole ratio $n_{\mathrm{H_2}} / n_{\text{hydrocarbon}}$ . Calculate moles of hydrocarbon from mass/molar mass, moles of $\mathrm{H_2}$ from $V / 22400\text{ mL}$ (or ideal gas law at STP), and find their ratio.

04execute

Moles of $\mathrm{C_{10}H_{16}} = \frac{17 \times 10^{-3}\text{ g}}{136\text{ g/mol}} = 1.25 \times 10^{-4}\text{ mol}$ . Moles of $\mathrm{H_2} = \frac{8.40\text{ mL}}{22400\text{ mL/mol}} = 3.75 \times 10^{-4}\text{ mol}$ . Number of double bonds $= \frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}} = 3$ .

✓verify

Degree of unsaturation (DU) for $\mathrm{C_{10}H_{16}}$ is $10 + 1 - 16/2 = 3$ . The number of double bonds is 3. Therefore, all 3 degrees of unsaturation come from aliphatic double bonds. This corresponds to an acyclic triene. It matches the ozonolysis fragments that sum to 10 carbons. The result is fully consistent.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we not need to interpret the ozonolysis product structure given in the figure?

The hydrogenation data alone uniquely determines the number of double bonds (1 mole H₂ per double bond). The ozonolysis diagram is supplementary structural info confirming the structure, but quantitative hydrogenation data directly yields 3.

Answer

The hydrocarbon contains 3 double bonds, as 1 mole of the hydrocarbon consumes 3 moles of H₂ gas.

Why each option works or fails

Step-by-step solution

given: Mass of hydrocarbon $\mathrm{C_{10}H_{16}} = 17\text{ mg} = 1.7 \times 10^{-2}\text{ g}$ . Molar mass of $\mathrm{C_{10}H_{16}} = 10 \times 12 + 16 \times 1 = 136\text{ g/mol}$ . Volume of $\mathrm{H_2}$ gas at STP ( $0^\circ\mathrm{C}$ , $760\text{ mm Hg}$ ) $= 8.40\text{ mL}$ .
find: The number of double bonds present in one molecule of the hydrocarbon.
strategise: Each double bond consumes one mole of $\mathrm{H_2}$ upon complete catalytic hydrogenation. The number of double bonds equals the mole ratio $n_{\mathrm{H_2}} / n_{\text{hydrocarbon}}$ . Calculate moles of hydrocarbon from mass/molar mass, moles of $\mathrm{H_2}$ from $V / 22400\text{ mL}$ (or ideal gas law at STP), and find their ratio.
execute: Moles of $\mathrm{C_{10}H_{16}} = \frac{17 \times 10^{-3}\text{ g}}{136\text{ g/mol}} = 1.25 \times 10^{-4}\text{ mol}$ . Moles of $\mathrm{H_2} = \frac{8.40\text{ mL}}{22400\text{ mL/mol}} = 3.75 \times 10^{-4}\text{ mol}$ . Number of double bonds $= \frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}} = 3$ .
verify: Degree of unsaturation (DU) for $\mathrm{C_{10}H_{16}}$ is $10 + 1 - 16/2 = 3$ . The number of double bonds is 3. Therefore, all 3 degrees of unsaturation come from aliphatic double bonds. This corresponds to an acyclic triene. It matches the ozonolysis fragments that sum to 10 carbons. The result is fully consistent.

Shortcut:

When to use it: When the molecular formula has very low degree of unsaturation, providing an immediate upper bound.

given: Molecular formula of hydrocarbon is $\mathrm{C_{10}H_{16}}$ .

find: Degree of unsaturation (Double Bond Equivalent, DBE).

strategise: Calculate DBE: $\text{DBE} = C + 1 - \frac{H}{2} = 10 + 1 - 8 = 3$ . Since each double bond corresponds to 1 unit of unsaturation, the maximum possible double bonds is 3. Since hydrogenation absorbs $\mathrm{H_2}$ , the ratio $n(\mathrm{H_2})/n(\text{hydrocarbon})$ confirms whether rings are present. $8.40/22.4 = 0.375\text{ mmol}$ , and $17/136 = 0.125\text{ mmol}$ . Ratio is exactly 3, matching total DBE.

execute: $\text{DBE} = 10 + 1 - 8 = 3$ . Ratio $= \frac{0.375\text{ mmol}}{0.125\text{ mmol}} = 3$ .

verify: All unsaturation sites take up hydrogen, so there are 0 rings and exactly 3 double bonds.

Hydrocarbons: JEE Main Chemistry Question with Solution

What feels right?

The hydrocarbon contains 3 double bonds, as 1 mole of the hydrocarbon consumes 3 moles of H₂ gas.

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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