Take timed practice tests on Hydrocarbons for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.
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1. The IUPAC name of CH₃−CH(CH₃)−CH₃ is:
Explanation: The longest chain has 3 carbons (propane), with a methyl branch at C2: 2-methylpropane. (Isobutane is the common name.)
2. Markovnikov's rule predicts that in HX addition to CH₂=CHCH₃:
Explanation: Markovnikov's rule: H adds to the carbon with more H atoms (less substituted). The more stable (more substituted) carbocation intermediate forms: CH₃−CH⁺−CH₃ (2°) rather than CH₃−CH−CH₂⁺ (1°).
3. Terminal alkynes (RC≡CH) are more acidic than alkanes because:
Explanation: sp carbon has 50% s-character (vs sp² = 33%, sp³ = 25%). More s-character → electrons held closer to nucleus → more electronegative C → more stable carbanion → more acidic C−H bond. pKa: alkyne (25)
4. Benzene undergoes electrophilic substitution (NOT addition) because:
Explanation: Benzene's aromatic stabilisation energy (~150 kJ/mol) would be lost in addition reactions. Substitution restores aromaticity while addition would not. This thermodynamic preference drives EAS over addition.
5. In free radical chlorination of propane, the major product is:
Explanation: Selectivity: 2° H atoms (on C2) are more readily abstracted than 1° H atoms (on C1 and C3). 2° radical is more stable. Propane has 2 × 2° H (on C2) and 6 × 1° H (on C1/C3). Even accounting for numbers, 2-chloro product is major.
6. Ozonolysis of CH₃−CH=CH−CH₃ (but-2-ene) gives:
Explanation: Ozonolysis breaks the C=C double bond. But-2-ene: CH₃−CH=CH−CH₃ → CH₃CHO + CH₃CHO. Each carbon of the double bond gets an oxygen (=O), giving two ethanal molecules.
7. In nitration of benzene, the electrophile is:
Explanation: Mixed acid (conc. HNO₃ + conc. H₂SO₄): HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. The nitronium ion (NO₂⁺) is the electrophile that attacks the benzene ring.
8. In EAS, −OH group is an ortho/para director because:
Explanation: −OH group shows +M effect: lone pair on O donates into the ring via resonance, increasing electron density at ortho and para positions. Electrophiles attack electron-rich positions → o/p products predominate.
9. The product of treating propene with HBr in presence of peroxides is:
Explanation: Peroxides initiate a free radical chain mechanism. Br• adds to the less substituted end of propene (more stable radical forms at C2). The result is anti-Markovnikov addition: Br goes to C1 → 1-bromopropane. This is the Kharasch effect (peroxide effect).
10. The stability order of alkenes is:
Explanation: Hyperconjugation and inductive effects from alkyl groups stabilise alkenes. Order: tetrasubstituted > trisubstituted > disubstituted > monosubstituted > ethylene. Heats of hydrogenation confirm this: more stable alkene releases less heat when hydrogenated.
11. Cyclopentadienyl anion (C₅H₅⁻) is aromatic because:
Explanation: C₅H₅⁻ has 5 carbons, 4 π electrons from double bonds + 1 lone pair from the carbanion = 6 π electrons total. Hückel's rule: 4n+2 = 6 (n=1). Cyclic, planar, fully conjugated + 6π electrons → aromatic.
12. The general formula of alkanes is:
Explanation: Alkanes (saturated hydrocarbons): CₙH₂ₙ₊₂. Alkenes (one double bond): CₙH₂ₙ. Alkynes (one triple bond): CₙH₂ₙ₋₂.
13. The product of combustion of CH₄ in excess O₂ is:
Explanation: Complete combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O. In excess O₂, complete oxidation to CO₂ and H₂O occurs.
14. Lindlar's catalyst reduces alkynes to:
Explanation: Lindlar's catalyst (Pd on CaCO₃, poisoned with lead acetate and quinoline) selectively reduces alkynes to cis (Z) alkenes via syn addition of H₂. Sodium in liquid NH₃ gives trans alkenes.