JEE Chemistry Hydrocarbons MCQs Set 35

Answer: (d)

Explanation:
1. The primary cation (or polarised complex) from isobutyl chloride rearranges by a 1,2-hydride shift to the tertiary butyl cation.
2. This attacks benzene to give tert-butylbenzene.

Teacher's Note:
1. Primary alkyl halides rearrange in Friedel-Crafts alkylation when a tertiary cation is accessible.
2. The product is the more stable-cation product.

Answer: (c)

Explanation:
1. The ring carrying the -NH- group is activated (the N lone pair is delocalised into the ring), the ring carrying C=O is deactivated.
2. The positions ortho to N are blocked (ring fusion and the methyl group), so bromine enters para to the N atom, giving the product shown.

Teacher's Note:
1. Attack occurs on the ring with the strongest donor.
2. When ortho is blocked, para is used.

Answer: (b)

Explanation:
1. HOCl adds as Cl⁺ (electrophile) and OH^-. Cl⁺ forms the chloronium (bridged) ion, and water/OH attacks the carbon that can best carry positive charge, the benzylic carbon.
2. So OH goes to the carbon bearing phenyl and Cl to the carbon bearing CH₃: Ph-CH(OH)-CHCl-CH₃.

Teacher's Note:
1. The nucleophile attacks the carbon with more positive character (benzylic).
2. HOCl gives a chlorohydrin, not ring substitution.

Answer: (d)

Explanation:
1. H⁺ adds to the carbon farther from phenyl, because that gives the benzylic cation, which is stabilised by the ring.
2. Br^- then attacks the benzylic carbon: Ph-CHBr-CH₂-CH₂Br (1,3-dibromo-1-phenylpropane).

Teacher's Note:
1. Benzylic cation formation controls the orientation.
2. The CH₂Br group slightly disfavours the other cation (-I effect).

Answer: (c)

Explanation:
1. The Lewis acid generates I⁺ from ICl (I^δ+-Cl^δ-).
2. CH₃ is an ortho-para director; the para position is favoured, giving 4-iodotoluene.

Teacher's Note:
1. No light, so no side-chain radical substitution.
2. Interhalogen: less electronegative halogen is the electrophile.

Answer: (d) vinylcyclopentane

Explanation:
1. Protonation of vinylcyclopentane gives a secondary carbocation next to a cyclopentane ring. Ring expansion (relief of strain) gives a six-membered secondary cation, and a 1,2-hydride shift gives the tertiary 1-methylcyclohexyl cation, which is captured by Cl^-.
2. Cycloheptene would give chlorocycloheptane, 1-methylcyclopentene gives 1-chloro-1-methylcyclopentane (C₆H₁₀), and cyclohexene is C₆H₁₀ giving chlorocyclohexane.

Teacher's Note:
1. Carbocations rearrange (shift or ring expansion) to a more stable cation before the nucleophile attacks.
2. Always check the carbon count of the starting material.

Answer: (b) all four (1, 2, 3 and 4)

Explanation:
1. (1) gives a secondary cation, then a methyl shift gives the tertiary benzylic cation Ph-C⁺(CH₃)-CH(CH₃)₂.
2. (2) gives a tertiary cation, then a hydride shift gives the same benzylic cation. (3) and (4) protonate directly to this tertiary benzylic cation.
3. All four give Ph-C(OH)(CH₃)-CH(CH₃)₂ (3-methyl-2-phenylbutan-2-ol).

Teacher's Note:
1. Rearrangements funnel different alkenes to the most stable common cation.
2. Tertiary benzylic is the most stable cation here.

Answer: (a)

Explanation:
1. Hot alkaline KMnO₄ oxidises any alkyl side chain that has a benzylic hydrogen to -COOH, so M is benzoic acid.
2. The COOH group is a meta director and deactivator, so nitration gives 3-nitrobenzoic acid.

Teacher's Note:
1. Side-chain oxidation shortens any alkyl chain to COOH.
2. Meta directors: COOH, NO₂, CHO, CN.

Answer: (b)

Explanation:
1. -OCH₃ is a much stronger activator than -CH₃ (lone pair +M effect), so it decides the position.
2. Its para position is blocked by CH₃, so the nitro group enters ortho to OCH₃: 4-methyl-2-nitroanisole.

Teacher's Note:
1. Lone-pair (+M) donors beat hyperconjugative donors.
2. If para is blocked, go ortho.

Answer: (c) X is the product of an ionic reaction and Y is the product of a radical reaction

Explanation:
1. The peroxide effect works only for HBr. For HCl the chain propagation step (Cl• adding to the alkene is fine but H abstraction from HCl by the carbon radical) is endothermic, so HCl adds by the ionic (Markovnikov) mechanism giving 2-chlorobutane.
2. HBr with peroxide adds by the radical chain mechanism and gives the anti-Markovnikov 1-bromobutane.

Teacher's Note:
1. Peroxide effect: HBr only (not HCl, not HI).
2. Anti-Markovnikov product signals a radical mechanism.

Answer: (d)

Explanation:
1. Chlorine adds across the C=C to give the 3,4-dichlorobutyl compound ArCH₂CH₂CHCl-CH₂Cl.
2. AlCl₃ ionises the secondary chloride (more easily than the primary), giving ArCH₂CH₂CH⁺-CH₂Cl, which closes a five-membered ring para to OCH₃: 1-(chloromethyl)indane with OCH₃ on the carbon para to the ring-fusion carbon bearing the new bond.

Teacher's Note:
1. Secondary cations form more easily than primary ones.
2. Ring closure para to OCH₃ is preferred over the hindered ortho position.

Answer: (b)

Explanation:
1. Protonation of the terminal CH₂ gives the secondary cation CH₃-CH⁺-CH(CH₃)Ph.
2. A 1,2-hydride shift gives the more stable tertiary benzylic cation CH₃CH₂-C⁺(CH₃)Ph, which adds water to give 2-phenylbutan-2-ol.

Teacher's Note:
1. Look for a hydride shift that creates a benzylic tertiary cation.
2. Acid hydration follows Markovnikov addition plus rearrangement.

Answer: (b)

Explanation:
1. Protonation of the =CH₂ carbon (Markovnikov) gives the tertiary carbocation (CH₃)₂C⁺- attached to the ring.
2. Chloride captures this cation to give the tertiary chloride; no rearrangement is needed because a tertiary cation is already formed.

Teacher's Note:
1. Markovnikov: H goes to the carbon with more hydrogens.
2. A tertiary cation has no driving force to shift.

Answer: (d)

Explanation:
1. Bromine radicals are highly selective and abstract the hydrogen that gives the most stable radical.
2. The tertiary hydrogen at C-2 gives a tertiary radical, so the major product is 2-bromo-2-methylpentane.

Teacher's Note:
1. Br• is far more selective than Cl•.
2. Order of H abstraction: tertiary > secondary > primary.

Answer: (a) (II) (iii) (P)

Explanation:
1. An arenium ion forms only when an electrophile attacks an aromatic ring in electrophilic substitution.
2. Benzene with the protonated aldehyde (electrophile) fits (II)(iii)(P); the other combinations involve addition to an alkyne or alkene, or a radical.

Teacher's Note:
1. Arenium ion = intermediate of electrophilic aromatic substitution.
2. Radical and addition pathways do not form it.

Answer: (a)

Explanation:
1. AlCl₃ polarises the C-Cl bond; the incipient primary cation undergoes a 1,2-hydride shift to the more stable secondary cation.
2. Benzene attacks this cation and gives isopropylbenzene (cumene) as the major product.

Teacher's Note:
1. Friedel-Crafts alkylation can rearrange the alkyl group.
2. Benzene does not undergo ring chlorination with an alkyl chloride.

Answer: (a) Ph-CO-CD₂-CH₂-CH₃

Explanation:
1. Hydration of the alkyne puts the oxygen on the carbon attached to phenyl (the carbon better able to carry positive charge), giving the ketone Ph-CO-CH₂-CH₂-CH₃ via the enol.
2. In D₂O/D⁺ the enol is formed as Ph-C(OD)=CD-CH₂CH₃ and the keto form has both alpha hydrogens exchanged: Ph-CO-CD₂-CH₂-CH₃.

Teacher's Note:
1. Acid-catalysed enolisation exchanges alpha hydrogens with deuterium.
2. Alkyne hydration gives ketones.

Answer: (d)

Explanation:
1. Br₂ forms a bridged bromonium ion on one face; Br^- attacks from the opposite face (anti addition).
2. Anti addition to cyclohexene gives trans-1,2-dibromocyclohexane (as a racemic pair), not the cis isomer.

Teacher's Note:
1. Anti addition: the two Br atoms end up on opposite faces.
2. 1,2-Product, not 1,3 or 1,4.

Answer: (a)

Explanation:
1. The ring double bond is trisubstituted and more electron-rich than the monosubstituted vinyl group, and it gives a tertiary carbocation on protonation (the vinyl group would give only a secondary cation).
2. H⁺ adds to the ring CH carbon and Br^- to the tertiary carbon (Markovnikov): the vinyl group is retained.

Teacher's Note:
1. More substituted alkenes protonate faster (more stable cation).
2. With 1 mole of HBr only the more reactive double bond reacts.

Answer: (b)

Explanation:
1. H⁺ adds to the CH₂ carbon, giving the secondary cation (CH₃)₃C-CH⁺-CH₃.
2. A methyl group migrates (1,2-methyl shift) to give the more stable tertiary cation (CH₃)₂C⁺-CH(CH₃)₂; Cl^- captures it to give 2-chloro-2,3-dimethylbutane.

Teacher's Note:
1. A secondary cation next to a quaternary carbon rearranges by methyl shift.
2. The rearranged product is the major one.