Hydrocarbons: JEE Main 2026 PYQs

Answer: 14

Key idea: Free radical halogenation of cycloalkanes and identification of unique monochloro products based on molecular symmetry

1. Identify that $\text{C}_5\text{H}_{10}$ with degree of unsaturation = 1 and no $\text{KMnO}_4$ decolorization means cycloalkanes (not alkenes)
2. List all cycloalkane isomers: cyclopentane, methylcyclobutane, ethylcyclopropane, 1,1-dimethylcyclopropane, and 1,2-dimethylcyclopropane
3. For cyclopentane: all H equivalent → 1 monochloro product
4. For methylcyclobutane: 4 types of non-equivalent H → 4 monochloro products
5. For ethylcyclopropane: 4 types of non-equivalent H → 4 monochloro products
6. For 1,1-dimethylcyclopropane: 2 types of non-equivalent H → 2 monochloro products
7. For 1,2-dimethylcyclopropane: 3 types of non-equivalent H (by symmetry) → 3 monochloro products. Total = 1+4+4+2+3 = 14

Answer: 17

Key idea: Complete combustion of alkanes and stoichiometry

1. For an alkane $C_nH_{2n+2}$, write the combustion equation: $C_nH_{2n+2} + O_2 \to CO_2 + H_2O$
2. Balance the equation: $C_nH_{2n+2} + \frac{3n+1}{2}O_2 \to nCO_2 + (n+1)H_2O$
3. Given that moles of $O_2$ required = 8, so $\frac{3n+1}{2} = 8$
4. Solve: $3n+1 = 16 \Rightarrow n = 5$
5. The alkane is $C_5H_{12}$ with 5 carbon atoms and 12 hydrogen atoms
6. Sum of carbon and hydrogen atoms = $5 + 12 = 17$

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Key idea: Reactions of conjugated polyenes: electrocyclic reactions, aromatization, and acid-catalyzed rearrangements

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Key idea: Ozonolysis of alkenes - cleavage of C=C bonds to form carbonyl compounds

Answer: Option (3)

Key idea: Ozonolysis, Aldol Condensation, Iodoform Test, Retrosynthetic Analysis

1. Work backwards from product $S$: cyclopentene with $CH_3$ on double bond and $COOH$ adjacent
2. Before acidification, $R$ has $COO^-$ group. The iodoform test ($I_2/NaOH$) on $R$ indicates presence of $CH_3CO-$ group that converts to $COOH$ after haloform reaction
3. Compound $Q$ undergoes intramolecular aldol condensation under basic conditions to form cyclopentene ring in $R$
4. Ozonolysis of $P$ (C₈H₁₄) cleaves a double bond to give dicarbonyl compound $Q$
5. Working backwards: $P$ must be a cyclopentene with $CH_3$ on double bond carbon and $CH_2CH_3$ substituent positioned to give correct dicarbonyl upon ozonolysis
6. The structure matches option (3): Cyclopentene ring with $CH_3$ group on double bond carbon and $CH_2-CH_3$ side chain

Answer: Option (2)

Key idea: Friedel-Crafts Alkylation: Carbocation Rearrangement and Polyalkylation

1. In Friedel-Crafts alkylation, $n$-propyl chloride reacts with $\text{AlCl}_3$ to form $n$-propyl carbocation ($\text{CH}_3\text{CH}_2\text{CH}_2^+$)
2. Primary carbocations are unstable and undergo 1,2-hydride shift: $\text{CH}_3\text{CH}_2\text{CH}_2^+ \rightarrow \text{CH}_3\text{CH}^+\text{CH}_3$ (isopropyl carbocation)
3. The major product is isopropyl benzene (cumene), not $n$-propyl benzene. Statement A is incorrect, Statement B is correct.
4. Alkyl groups are electron-donating and activating, so once attached, they make the ring more reactive, leading to polyalkylation. Statement C is correct.
5. Electron-donating groups generally activate the ring (except those with lone pairs that coordinate with $\text{AlCl}_3$). Statement D is incorrect.
6. Correct statements are B and C only.

Answer: Option (1)

Key idea: Alkene stability based on degree of substitution and hyperconjugation

1. Identify the degree of substitution for each alkene by counting alkyl groups attached to the double bond carbons
2. **C (2,3-Dimethylbut-2-ene)**: $(CH_3)_2C=C(CH_3)_2$ - tetrasubstituted (4 alkyl groups)
3. **A (2-Methylbut-2-ene)**: $CH_3-C(CH_3)=CH-CH_3$ - trisubstituted (3 alkyl groups)
4. **B (cis-But-2-ene)**: $CH_3-CH=CH-CH_3$ (cis) - disubstituted (2 alkyl groups)
5. **D (Prop-1-ene)**: $CH_2=CH-CH_3$ - monosubstituted (1 alkyl group)
6. Apply stability rule: More substituted alkenes are more stable due to hyperconjugation and inductive effects
7. Stability order: $C > A > B > D$ (tetra > tri > di > mono)

Answer: Option (3)

Key idea: Primary carbon identification and molecular formula determination from molar mass

1. Determine molecular formula: Molar mass = 72 g/mol. For alkane $C_nH_{2n+2}$: $12n + 2n + 2 = 72$ gives $n = 5$, so $C_5H_{12}$
2. Recall: Primary carbon ($1°$) is bonded to only ONE other carbon atom
3. Check option (1): 1,1-Dimethylcyclopropane is $C_5H_{10}$ (mass = 70 g/mol) ✗
4. Check option (2): 2,2-Dimethylpropane has 4 primary carbons (all four $CH_3$ groups) ✗
5. Check option (3): 2-Methylbutane: $CH_3-CH(CH_3)-CH_2-CH_3$ has exactly 3 primary carbons (three $CH_3$ groups) ✓
6. Check option (4): n-Pentane has only 2 primary carbons (two terminal $CH_3$ groups) ✗
7. Answer: 2-Methylbutane satisfies both conditions (mass = 72 g/mol and 3 primary carbons)

Answer: Option (4)

Key idea: Methods of preparation of methane - decarboxylation, Kolbe's electrolysis, Grignard reagent hydrolysis, Wurtz reaction, and hydrogenation of unsaturated hydrocarbons

1. **Statement I Analysis:** Check three methods - (a) Decarboxylation: $\text{CH}_3\text{COONa} + \text{NaOH} \rightarrow \text{CH}_4 + \text{Na}_2\text{CO}_3$ ✓ gives methane
2. (b) Kolbe's electrolysis: $2\text{CH}_3\text{COO}^- \rightarrow \text{CH}_3\text{CH}_3 + 2\text{CO}_2 + 2e^-$ ✗ gives **ethane**, not methane
3. (c) Grignard + water: $\text{CH}_3\text{MgBr} + \text{H}_2\text{O} \rightarrow \text{CH}_4 + \text{Mg(OH)Br}$ ✓ gives methane
4. Since Kolbe's electrolysis produces ethane (not methane), **Statement I is FALSE**
5. **Statement II Analysis:** (a) Unsaturated hydrocarbons need at least 2 carbons for C=C or C≡C bonds, so cannot give $\text{CH}_4$ ✓
6. (b) Wurtz reaction: $2\text{R-X} + 2\text{Na} \rightarrow \text{R-R} + 2\text{NaX}$ couples alkyl groups, giving ethane from $\text{CH}_3\text{X}$, not methane ✓
7. Both parts of Statement II are correct, so **Statement II is TRUE**. Answer: Option (4)

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Hückel's Rule for Aromaticity - (4n+2) π electrons

Answer: 16

Key idea: Gattermann-Koch reaction, Friedel-Crafts acylation, Claisen-Schmidt condensation (aldol condensation), and counting π electrons in conjugated systems

1. Step 1: Identify $x$: Benzene + CO + HCl with CuCl gives benzaldehyde (C$_6$H$_5$CHO) via Gattermann-Koch reaction
2. Step 2: Identify $y$: Benzene + CH$_3$COCl with AlCl$_3$ gives acetophenone (C$_6$H$_5$COCH$_3$) via Friedel-Crafts acylation
3. Step 3: Identify $z$: Benzaldehyde + acetophenone with alkali undergoes Claisen-Schmidt condensation to form chalcone: C$_6$H$_5$-CH=CH-CO-C$_6$H$_5$
4. Step 4: Count $\pi$ electrons in chalcone: First benzene ring (6) + C=C bond (2) + C=O bond (2) + Second benzene ring (6) = 16 $\pi$ electrons