Hydrocarbons: JEE Advanced PYQ, 2007–2026

Answer: (C)

The number of products depends on how many non-equivalent ring C–H positions there are. In o-xylene there are two types (positions 3 and 4). In m-xylene there are three types (2, 4 and 5). In toluene there are three types (ortho, meta, para). In p-xylene, by symmetry, all four ring hydrogens are equivalent, so only one mononitro product forms. The answer is p-xylene.

Answer: (A)

Conformers are arrangements that interconvert by rotation about a single bond without breaking any bond. The ethane C–C bond is a σ bond formed by head-on overlap of sp³ orbitals, and this overlap is symmetrical about the bond axis. Rotation therefore leaves the overlap unchanged, and the hybridisation, bond lengths and bond angles stay the same. The only change is the dihedral angle between C–H bonds on the two carbons. In the eclipsed form the electron clouds of these bonds are closest, so repulsion (torsional strain) is highest and stability is lowest.

Answer: (A)

In nitrobenzene the –NO₂ group withdraws electrons by the −R effect (delocalisation of ring π electrons into the N=O bond) and by the −I effect. The resonance structures show positive charge developing at the ortho and para carbons, but not at the meta carbons. The ortho and para positions are therefore more strongly deactivated towards an electrophile. The meta position is relatively the most electron-rich, although still less reactive than benzene, so substitution occurs there.

Answer: (A)

Two conditions control the site of reaction. Cl₂ with heat and light is a free-radical process and substitutes the benzylic hydrogens of toluene successively, so excess Cl₂ gives benzotrichloride, C₆H₅CCl₃. Cl₂ with FeCl₃ in the dark is electrophilic aromatic substitution on the ring. The substituent now on the ring is –CCl₃, whose three electronegative Cl atoms withdraw electrons by the –I effect. It is therefore a deactivating, meta-directing group, and the major product is 1-chloro-3-(trichloromethyl)benzene. Treating –CCl₃ as an o/p-directing alkyl group, stopping at –CH₂Cl, or calling the ring unreactive are the errors in the other options.

Answer: (D)

The order of introducing groups decides the isomer formed. To get a meta relationship, the meta-directing NO₂ group must be introduced first. Nitrobenzene is deactivated but undergoes ring chlorination with Cl₂/FeCl₃ at the meta position. Introducing Cl first would give o/p products. Cl₂ in sunlight does not give ring substitution.

Answer: (A)

The reaction is 3 C₂H₂ → C₆H₆. Moles of ethyne = 26.0/26 = 1.00 mol. Moles of benzene at 100% = 1.00/3 = 0.333 mol, which is 0.333 × 78 = 26 g. At 75% conversion the mass is 0.75 × 26 = 19.5 g.

Answer: (A)

Principle: aromatic nitration needs a strong electrophile, the nitronium ion NO₂⁺. In the mixed acid, H₂SO₄ is the stronger acid and protonates HNO₃: HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄⁻. The protonated nitric acid then loses water: H₂NO₃⁺ → NO₂⁺ + H₂O. The benzene π-electrons attack NO₂⁺ to give the arenium ion, which loses H⁺ to give nitrobenzene. Reversing the acid–base roles, calling H₂SO₄ an oxidant, or treating it as a mere solvent each ignores this electrophile-generating step.

Answer: (A)

The amount of each product is proportional to (number of equivalent H) × (relative reactivity per H). 2-Methylbutane, (CH₃)₂CH–CH₂–CH₃, has six primary H on the two equivalent CH₃ groups on C-2, which give 1-chloro-2-methylbutane: 6 × 1 = 6. It has one tertiary H on C-2, which gives 2-chloro-2-methylbutane: 1 × 5 = 5. It has two secondary H on C-3, which give 2-chloro-3-methylbutane: 2 × 3.8 = 7.6. It has three primary H on C-4, which give 1-chloro-3-methylbutane: 3 × 1 = 3. The total is 21.6, so the tertiary chloride is 5/21.6 × 100 ≈ 23%.

Answer: (A)

The staggered conformation is the energy minimum and the eclipsed conformation is the maximum, about 12.5 kJ mol⁻¹ higher. Collisions at room temperature easily supply this energy, so molecules rotate rapidly and the different conformations cannot be separated or isolated. The lower-energy arrangement is still favoured, so at any instant more molecules are near the staggered form. The eclipsed form is only a brief arrangement passed through during rotation.

Answer: (A)

Work through the clues in turn. (1) Hydration with HgSO₄/H₂SO₄ gives an enol that tautomerises to a ketone. Two different ketones need a triple bond with different groups on its two carbons, and that bond must not be terminal. A terminal alkyne gives a methyl ketone only, and one product. (2) Na/liq. NH3 reduces internal alkynes to trans-alkenes. The product must have two different groups on each alkene carbon to show cis–trans isomerism. Among C₅H₈ alkynes (pent-1-yne, pent-2-yne, 3-methylbut-1-yne), only pent-2-yne is internal. CH₃–C≡C–CH₂CH₃ gives CH₃COCH₂CH₂CH₃ and CH₃CH₂COCH₂CH₃ on hydration. Na/liq. NH3 turns it into trans-CH₃CH=CHCH₂CH₃, which has a cis isomer. Cyclopentene fails both tests.