Take timed practice tests on Haloalkanes and Haloarenes 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. CH₃Cl is classified as a:
Explanation: Classification is based on the degree of the carbon bearing the halogen: 1° = attached to one alkyl group, 2° = two, 3° = three. In CH₃Cl, the C has no alkyl groups → primary.
2. SN2 reactions proceed with:
Explanation: SN2 is a concerted, one-step mechanism. The nucleophile attacks from the back of the leaving group (backside attack) → umbrella flip → inversion of configuration at the stereogenic centre (Walden inversion).
3. SN1 reactions are favoured by:
Explanation: SN1: unimolecular, two-step (carbocation intermediate). Favoured by: 3° > 2° > 1° substrates (stable carbocations), polar protic solvents (stabilise ions), weak/neutral nucleophiles.
4. The order of reactivity of alkyl halides toward SN2 is:
Explanation: SN2 is bimolecular — steric hindrance at the reactive carbon is the key factor. Less hindered substrates react faster: primary (1°) > secondary (2°) > tertiary (3°). Methyl halide reacts fastest of all.
5. In polar protic solvents, which is a better nucleophile for SN2?
Explanation: In polar protic solvents, small anions (F⁻, OH⁻) are heavily solvated → poor nucleophiles. Large, polarisable anions (I⁻, RS⁻) are less solvated → better nucleophiles. Order: I⁻ > Br⁻ > Cl⁻ > F⁻ in protic solvents.
6. Dehydrohalogenation of 2-bromobutane with KOH/ethanol gives the major product:
Explanation: E2 elimination follows Zaitsev's rule: the major product is the more substituted (more stable) alkene. From 2-bromobutane: but-2-ene (disubstituted, more stable) > but-1-ene (monosubstituted). Alcoholic KOH provides the base for E2.
7. Chlorobenzene is less reactive toward nucleophilic substitution than chloroalkanes because:
Explanation: In chlorobenzene, the lone pair on Cl delocalises into the ring (p-π overlap) → C−Cl bond has partial double bond character → shorter, stronger bond → harder to break by nucleophiles. Alkyl C−Cl has no such resonance.
8. The reaction of (R)-2-bromobutane with KOH/H₂O predominantly gives:
Explanation: KOH/H₂O = hydroxide nucleophile in protic solvent + primary substrate (actually secondary here). The conditions favour SN2 for secondary halide (OH⁻ is a strong nucleophile). SN2 → Walden inversion → (S)-butan-2-ol from (R)-2-bromobutane.
9. Chlorobenzene reacts with NaOH at 350°C and high pressure to give phenol via:
Explanation: Chlorobenzene + NaOH (300-350°C, 200 atm) → sodium phenoxide → phenol. This is a nucleophilic aromatic substitution via an addition-elimination mechanism (Meisenheimer complex intermediate), not SN2 which cannot occur at sp² carbon.
10. The Finkelstein reaction converts:
Explanation: Finkelstein reaction: alkyl chloride/bromide + NaI in dry acetone → alkyl iodide. NaCl and NaBr are insoluble in acetone and precipitate, driving equilibrium to the right. This is an SN2 exchange reaction.
11. The leaving group ability in SN reactions follows the order:
Explanation: Leaving group ability = ability to leave as a stable anion. Better leaving groups have weaker C−X bonds and more stable anions. I⁻ is the most stable (large, polarisable) → best leaving group. F⁻ is the worst leaving group despite F being most electronegative (C−F bond is very strong).
12. Grignard reagent (RMgX) from alkyl halide reacts with water to give:
Explanation: Grignard reagents (organometallic) are strong bases and react with any acidic hydrogen: RMgX + H₂O → RH + Mg(OH)X. This is why Grignard reactions must be conducted under anhydrous conditions.
13. The order of boiling points of alkyl halides with same alkyl group is:
Explanation: Boiling point depends on van der Waals forces (London dispersion). Larger halogen atoms have higher polarisability → stronger London forces → higher boiling point. RI > RBr > RCl > RF for the same alkyl group.