Hydrocarbons — JEE Main practice

38 questions

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Sample questions with solutions

Q1 · 2026

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:

A.
$\mathrm{C}_2\mathrm{H}_2$
B.
$\mathrm{C}_2\mathrm{H}_4$
C.
$\mathrm{C}_4\mathrm{H}_{10}$
D.
$\mathrm{C}_2\mathrm{H}_6$

Answer: A

Set up the general combustion equation. Since a hydrocarbon $\mathrm{C}_x\mathrm{H}_y$ combusts as

\mathrm{C}_x\mathrm{H}_y + \left(x+\frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O}

and at 273 K water is liquid (negligible volume), the residual gas after combustion consists of unreacted $\mathrm{O}_2$ plus $\mathrm{CO}_2$ produced.

Use the total residual gas volume to find y. Given 80 mL of hydrocarbon reacts with 264 mL $\mathrm{O}_2$ , and the residual gas volume (unreacted $\mathrm{O}_2$ + $\mathrm{CO}_2$ ) is 224 mL,

264 - 80\left(x+\frac{y}{4}\right) + 80x = 224

Simplifying,

264 - \frac{80y}{4} = 224 \quad \Rightarrow \quad 40 = 20y \quad \Rightarrow \quad y = 2

Use the KOH absorption step to find x. Since KOH absorbs $\mathrm{CO}_2$ , leaving only unreacted $\mathrm{O}_2$ (64 mL) behind,

264 - 80\left(x+\frac{y}{4}\right) = 64

Substituting $y=2$ ,

264 - 80\left(x+\frac{1}{2}\right) = 64 \quad \Rightarrow \quad 264 - 80x - 40 = 64 \quad \Rightarrow \quad x = 2

Combine the results. Since $x=2$ and $y=2$ , the hydrocarbon formula is $\mathrm{C}_2\mathrm{H}_2$ .

Hence, the answer is option A — $\mathrm{C}_2\mathrm{H}_2$ .

Q2 · 2026

Given below are two statements:

Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with $\mathrm{CH}_3\mathrm{COCl}/\mathrm{AlCl}_3$ will give

Statement II: $-\mathrm{NO}_2$ group is a m-directing, and deactivating group.

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

A.
Statement I is incorrect but Statement II is correct
B.
Both Statement I and Statement II are correct
C.
Both Statement I and Statement II are incorrect
D.
Statement I is correct but Statement II is incorrect

Answer: A

Consider what nitration does. Since nitrating benzene installs a $-\mathrm{NO}_2$ group, forming nitrobenzene.
Consider whether Friedel-Crafts acylation can then proceed. Given the $-\mathrm{NO}_2$ group is a strongly deactivating substituent (it withdraws electron density from the ring), the ring becomes far too electron-poor to undergo Friedel-Crafts acylation with $\mathrm{CH}_3\mathrm{COCl}/\mathrm{AlCl}_3$ — this reaction does not proceed on nitrobenzene. So Statement I's claim that this reaction gives the shown product is incorrect.
Evaluate Statement II. Since $-\mathrm{NO}_2$ is indeed both a meta-directing and deactivating group (due to its strong electron-withdrawing resonance and inductive effects), Statement II is correct.

Hence, Statement I is incorrect but Statement II is correct — the answer is option A.

Q3 · 2026

The dibromo compound [P] (molecular formula: $\mathrm{C}_9\mathrm{H}_{10}\mathrm{Br}_2$ ) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yield [R] which gives positive Iodoform test but negative Tollen's test. The compound [P] is:

Answer: C

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