Hydrocarbons Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Hydrocarbons with step-by-step solutions covering alkane combustion, monochlorination, Friedel-Crafts alkylation, alkene stability, ozonolysis, Wurtz and Grignard reactions.
A curated set of JEE Main 2026 previous-year questions on Hydrocarbons, each solved step by step so you can verify both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Trace the sequence:
- (i) Friedel-Crafts methylation gives toluene ($\text{C}_6\text{H}_5\text{CH}_3$).
- (ii) Ring chlorination (the $-\text{CH}_3$ is o/p-directing) gives chlorotoluene.
- (iii) $\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$ oxidises the $-\text{CH}_3$ side chain to $-\text{COOH}$.
So $(X)$ is chlorobenzoic acid, $\text{Cl-C}_6\text{H}_4\text{-COOH}$ (formula $\text{C}_7\text{H}_5\text{ClO}_2$).
Only the $-\text{COOH}$ group reacts with $\text{NaHCO}_3$, releasing $\text{CO}_2$:
$$\text{Ar-COOH} + \text{NaHCO}_3 \rightarrow \text{Ar-COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$Moles of gas:
$$n_{\text{CO}_2} = \dfrac{11.2}{22.4} = 0.5\ \text{mol}$$Since $1\ \text{mol}$ of acid $\to 1\ \text{mol}\ \text{CO}_2$, moles of $(X) = 0.5$.
Molar mass of $\text{C}_7\text{H}_5\text{ClO}_2$:
$$7(12) + 5(1) + 35.5 + 2(16) = 84 + 5 + 35.5 + 32 = 156.5\ \text{g mol}^{-1}$$$$P = 0.5 \times 156.5 = 78.25\ \text{g}$$Answer: 78.25
Solution
Check Statement I:
- Decarboxylation (soda-lime): $\text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3$ — gives methane. ✓
- $\text{CH}_3\text{MgBr} + \text{H}_2\text{O} \rightarrow \text{CH}_4 + \text{Mg(OH)Br}$ — gives methane. ✓
- Kolbe’s electrolysis of sodium acetate: $2\,\text{CH}_3\text{COO}^- \rightarrow \text{CH}_3\text{-CH}_3 + 2\text{CO}_2 + 2e^-$ — gives ethane, not methane. ✗
So Statement I is false.
Check Statement II:
- Wurtz reaction couples two alkyl halides, so the smallest possible product is ethane; it cannot give methane.
- Hydrogenation of unsaturated hydrocarbons (e.g. $\text{CH}_2\text{=CH}_2 \to \text{C}_2\text{H}_6$) also cannot yield methane.
So Statement II is true.
Statement I false, Statement II true.
Answer: D
Solution
Isomers of $\text{C}_5\text{H}_{10}$ that do not decolourise $\text{KMnO}_4$ are the saturated cycloalkanes (no C=C):
- Cyclopentane — all 10 H equivalent $\Rightarrow$ 1 product.
- Methylcyclobutane — distinct positions: $-\text{CH}_3$, C1 (ring C bearing methyl), C2/C4 (equivalent), C3 $\Rightarrow$ 4 products.
- Ethylcyclopropane — distinct positions: terminal $\text{CH}_3$, ethyl $\text{CH}_2$, ring C1, ring C2/C3 (equivalent) $\Rightarrow$ 4 products.
- 1,1-Dimethylcyclopropane — distinct positions: $\text{CH}_3$ (both equivalent), ring $\text{CH}_2$ (C2/C3 equivalent) $\Rightarrow$ 2 products.
- 1,2-Dimethylcyclopropane (one structural isomer; cis/trans are stereoisomers) — distinct positions: $\text{CH}_3$ (equivalent), ring CH bearing methyl (C1/C2 equivalent), ring $\text{CH}_2$ (C3) $\Rightarrow$ 3 products.
Total:
$$1 + 4 + 4 + 2 + 3 = 14$$Answer: 14
Solution
General combustion of an alkane $\text{C}_n\text{H}_{2n+2}$:
$$\text{C}_n\text{H}_{2n+2} + \left(\dfrac{3n+1}{2}\right)\text{O}_2 \rightarrow n\,\text{CO}_2 + (n+1)\,\text{H}_2\text{O}$$Set the $\text{O}_2$ requirement equal to 8:
$$\dfrac{3n+1}{2} = 8 \;\Rightarrow\; 3n + 1 = 16 \;\Rightarrow\; n = 5$$So the alkane is $\text{C}_5\text{H}_{12}$ (pentane).
$$\text{Carbon atoms} + \text{Hydrogen atoms} = 5 + 12 = 17$$Answer: 17
Solution
Track the transformation:
- (i) $\text{Br}_2$ adds across the double bond: $\text{C}_6\text{H}_5\text{-CHBr-CH}_2\text{Br}$ (a vicinal dibromide).
- (ii) Excess $\text{NaNH}_2$ performs double dehydrohalogenation to give the terminal alkyne phenylacetylene, $\text{C}_6\text{H}_5\text{-C}{\equiv}\text{CH}$ (and deprotonates it to the acetylide).
- (iii) The acetylide is alkylated by $\text{CH}_3\text{I}$: $\text{C}_6\text{H}_5\text{-C}{\equiv}\text{C-CH}_3$ (1-phenylpropyne, $\text{C}_9\text{H}_8$).
- (iv) $\text{Na}/\text{NH}_3(l)$ (dissolving-metal reduction) reduces the internal alkyne to the trans-alkene: $\text{C}_6\text{H}_5\text{-CH=CH-CH}_3$ ($\text{C}_9\text{H}_{10}$).
Molar mass of $(Y) = \text{C}_9\text{H}_{10}$:
$$9(12) + 10(1) = 108 + 10 = 118\ \text{g mol}^{-1}$$Answer: B
Solution
All three products contain 7 carbon atoms, so $(X)$ is a $\text{C}_7$ hydrocarbon:
- 2,4-dimethylpentane is $\text{C}_7\text{H}_{16}$ — a skeletal isomer of heptane (formed by $\text{AlCl}_3$-catalysed isomerisation of an alkane).
- Aromatization (dehydrogenation) to toluene ($\text{C}_7\text{H}_8$) and cyclisation to methylcyclohexane ($\text{C}_7\text{H}_{14}$) are the classic catalytic reforming reactions of a straight-chain alkane.
Only a saturated straight-chain $\text{C}_7$ alkane, n-heptane, undergoes all three (isomerisation, aromatization, cyclisation). The triene/alkene options are unsaturated and would not give the saturated 2,4-dimethylpentane by mere $\text{AlCl}_3$ isomerisation.
Answer: C
Solution
Find the molecular formula. From $\dfrac{3n+1}{2} = 8$, we get $n = 5$, so $(Y)$ is $\text{C}_5\text{H}_{12}$.
Use the “only one monochloro product” clue. Among the three $\text{C}_5\text{H}_{12}$ isomers:
- n-pentane $\to$ 3 monochloro products,
- 2-methylbutane $\to$ 4 monochloro products,
- 2,2-dimethylpropane (neopentane), $\text{C(CH}_3)_4$ $\to$ all 12 H are equivalent $\to$ only 1 monochloro product. ✓
So $(Y)$ is neopentane, $\text{C(CH}_3)_4$. Its four $\text{CH}_3$ groups are all primary carbons (the central carbon is quaternary).
$$\text{Primary carbon atoms} = 4$$Answer: 4
Solution
Alkene stability increases with the number of alkyl groups on the double-bonded carbons (hyperconjugation and $+I$ effect):
- C. 2,3-Dimethylbut-2-ene, $(\text{CH}_3)_2\text{C=C(CH}_3)_2$ — tetrasubstituted (most stable).
- A. 2-Methylbut-2-ene, $(\text{CH}_3)_2\text{C=CHCH}_3$ — trisubstituted.
- B. cis-But-2-ene, $\text{CH}_3\text{CH=CHCH}_3$ — disubstituted.
- D. Prop-1-ene, $\text{CH}_3\text{CH=CH}_2$ — monosubstituted (least stable).
Order of stability:
$$\text{C} > \text{A} > \text{B} > \text{D}$$Answer: A
Solution
Molar mass filter. $M = 72\ \text{g mol}^{-1}$ corresponds to $\text{C}_5\text{H}_{12}$ ($5\times12 + 12 = 72$).
- 1,1-Dimethylcyclopropane is $\text{C}_5\text{H}_{10}$ ($M = 70$) — rejected.
Primary-carbon filter (a primary carbon is a $\text{CH}_3$ bonded to only one other carbon):
- 2,2-Dimethylpropane, $\text{C(CH}_3)_4$ — 4 primary carbons. ✗
- 2-Methylbutane, $(\text{CH}_3)_2\text{CH-CH}_2\text{-CH}_3$ — the two methyls on C2 plus the terminal $\text{CH}_3$ give 3 primary carbons. ✓
- n-pentane, $\text{CH}_3(\text{CH}_2)_3\text{CH}_3$ — only 2 primary carbons. ✗
Both conditions are satisfied only by 2-methylbutane.
Answer: C
Solution
Identify the gas from its density at STP. Molar volume at STP is $22.4\ \text{dm}^3\text{/mol}$:
$$M = \dfrac{22.4\ \text{dm}^3/\text{mol}}{1.4\ \text{dm}^3/\text{g}} = 16\ \text{g mol}^{-1} \;\Rightarrow\; \text{gas is } \text{CH}_4$$So $\text{R} = \text{CH}_3$ and $\text{RMgI} + \text{H}_2\text{O} \rightarrow \text{CH}_4 + \text{Mg(OH)I}$.
Iodination. $\text{CH}_4 + \text{I}_2 \rightleftharpoons \text{CH}_3\text{I} + \text{HI}$. $\text{HIO}_3$ oxidises the $\text{HI}$ back to $\text{I}_2$, driving the reaction forward, so $(X) = \text{CH}_3\text{I}$.
Wurtz reaction. $2\,\text{CH}_3\text{I} + 2\,\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{-CH}_3 + 2\,\text{NaI}$, so $(Y)$ is ethane, $\text{C}_2\text{H}_6$.
Molar mass of $(Y)$:
$$2(12) + 6(1) = 30\ \text{g mol}^{-1}$$Answer: 30
Solution
Examine each pathway for a carbocation rearrangement:
- A. Benzene + n-propyl chloride/$\text{AlCl}_3$: the 1° n-propyl cation rearranges by a 1,2-hydride shift to the more stable 2° isopropyl cation, giving cumene. Rearrangement occurs.
- B. tert-Butyl alcohol + $\text{H}_3\text{O}^+$: protonation and loss of water gives the 3° tert-butyl cation directly, which simply eliminates a $\beta$-H to form 2-methylprop-1-ene. No rearrangement — the initial cation is already the most stable one.
- C. n-Hexane + $\text{AlCl}_3/\text{HCl}$: isomerisation to 2-methylpentane proceeds through hydride/methyl shifts. Rearrangement occurs.
- D. Neopentyl alcohol + $\text{H}_3\text{O}^+$: the 1° neopentyl cation undergoes a 1,2-methyl shift to a 3° cation, giving 2-methylbut-2-ene. Rearrangement occurs.
Only B gives its product without any rearrangement.
Answer: B