Haloalkanes and Haloarenes: JEE Main PYQ, 2020–2025

Answer: (A)

Principle: persistent organic pollutants such as PCBs are non-polar and lipophilic (fat-soluble). Because they are not readily broken down or excreted, they are stored in the fatty tissue of organisms. When a predator eats many contaminated prey, the pollutant accumulates in its body, so its concentration rises at each higher trophic level. This is biomagnification. The key property is lipid solubility, not volatility, water solubility or ionic character. The question is answered by recalling that the compounds are non-polar and fat-soluble.

Answer: (D)

Chlorine withdraws electrons by its −I effect, which deactivates the ring, so substitution is slower than in benzene. But its lone pairs, by resonance, raise the electron density at the ortho and para positions, so the incoming group goes there.

Answer: (A)

Dehydrohalogenation removes HBr, with H lost from a β-carbon. Here there are two β-carbons. The CH₃ carbon gives C₆H₅CH₂CH=CH₂, which is non-conjugated and monosubstituted. The benzylic CH₂ carbon gives C₆H₅CH=CHCH₃, which is disubstituted and conjugated with the aromatic ring. By Zaitsev's rule, with extra stabilisation from conjugation, the second alkene is the major product (mainly the trans isomer). Alcoholic KOH is not a bulky base, so the Hofmann product is not favoured, and the medium favours elimination over substitution.

Answer: (D)

Bond length depends on the sum of the covalent radii of carbon and the halogen. Because the halogen radius increases going down the group from F to I, the bond length increases in the order C-F < C-Cl < C-Br < C-I. Hence the C-I bond is the longest.

Answer: (A)

Dipole moment μ = charge × distance = δe × d. Convert μ to SI units: 1.86 D = 1.86 × 3.336 × 10⁻³⁰ C m = 6.21 × 10⁻³⁰ C m. Then e·d = 1.602 × 10⁻¹⁹ C × 1.78 × 10⁻¹⁰ m = 2.85 × 10⁻²⁹ C m. So δ = 6.21 × 10⁻³⁰ / 2.85 × 10⁻²⁹ ≈ 0.22. The C–Cl bond is therefore polar covalent and not ionic, with about 0.22 e of charge displaced from C to Cl.

Answer: (A)

Principle: use the balanced equation 2CHCl3 + O2 → 2COCl2 + 2HCl and convert mass to moles and back with the right molar mass. Application: n(CHCl3) = 11.95/119.5 = 0.100 mol, and 30% of it is 0.030 mol. The ratio CHCl3 : COCl2 is 1:1, so n(COCl2) = 0.030 mol and its mass is 0.030 × 99 = 2.97 g. The ratio CHCl3 : O2 is 2:1, so n(O2) = 0.015 mol. This slow air oxidation in light, which also releases HCl, is why chloroform is stored in dark, completely filled bottles with no air space.

Answer: (A)

Non-biodegradable chemicals are not broken down or excreted quickly, so they are retained in body fat. Each predator eats many prey, so the chemical's concentration grows at each higher trophic level. This process is called biomagnification. Students may confuse it with eutrophication, which concerns nutrient enrichment of water rather than food-chain transfer.

Answer: (A)

Principle: chloroform is slowly oxidised by atmospheric oxygen in the presence of light to carbonyl chloride (COCl2, phosgene), 2CHCl3 + O2 → 2COCl2 + 2HCl. Phosgene is a highly poisonous gas. Application: to stop this reaction, both light and oxygen must be removed. Dark brown bottles cut off light, and filling the bottle to the brim leaves almost no air. The other options either confuse chloroform with CCl4, treat the practice as purely physical, or propose a hydrolysis reaction that is not the hazard.

Answer: (A)

A bond dipole moment is μ = q × d, where q is the charge on each end and d is the separation between the charges. Here q = δe. Convert the units first: μ = 1.86 × 3.34 × 10⁻³⁰ = 6.21 × 10⁻³⁰ C m, and d = 178 pm = 1.78 × 10⁻¹⁰ m. Then q = 6.21 × 10⁻³⁰ / 1.78 × 10⁻¹⁰ = 3.49 × 10⁻²⁰ C, so δ = 3.49 × 10⁻²⁰ / 1.60 × 10⁻¹⁹ ≈ 0.22. The C–Cl bond is therefore about 22% ionic, polar covalent and nowhere near fully ionic. The chlorine carries a partial negative charge and the carbon a partial positive charge. The common errors are halving the charge, halving the distance, and slipping a power of ten in the picometre conversion.

Answer: (C)

Haloarenes (C–X bond on an sp2 carbon of a benzene ring) and vinylic halides do not undergo SN1 or SN2 readily. In SN2 the nucleophile must approach the carbon from the side opposite the leaving group, which is blocked by the ring plane. In SN1 the phenyl cation that would form is very unstable. Alkyl halides, whether primary, secondary, tertiary, or methyl, do undergo substitution, so chlorobenzene is the only choice that fits the rule.