JEE Chemistry Hydrocarbons MCQs Set 36

Answer: (b) (I) (i) (Q)

Explanation:
1. Phenylacetylene with Hg²⁺/H₃O⁺ adds water (electrophilic addition, Q) to give the enol, which tautomerises to acetophenone, a ketone.
2. Styrene with H⁺/H₂O gives an alcohol; the other combinations do not give a ketone.

Teacher's Note:
1. Hydration of a terminal aryl alkyne by Hg²⁺ gives a methyl ketone.
2. Alkene hydration gives alcohols.

Answer: (c) N-bromosuccinimide (NBS)

Explanation:
1. The CH₃ group of propene is allylic. Substitution of an allylic hydrogen by bromine needs a low, steady concentration of Br₂ and radical conditions, which NBS provides.
2. Br₂/H₂O gives a bromohydrin, HBr gives 2-bromopropane and HBr/peroxide gives 1-bromopropane (both additions).

Teacher's Note:
1. NBS = allylic (and benzylic) bromination.
2. Additions to the C=C bond do not give allyl bromide.

Answer: (c)

Explanation:
1. HCl adds by Markovnikov addition to give the 3-(3-chlorobutyl)phenol (secondary chloride).
2. AlCl₃ forms the secondary cation, which alkylates the ring intramolecularly (five-membered ring) para to the OH group, the less hindered activated position. The OH group stays; chlorine is lost as AlCl₄^-.

Teacher's Note:
1. The ring OH survives; the alkyl chloride is the leaving group.
2. Cyclisation para to the donor avoids steric crowding.

Answer: (c)

Explanation:
1. In the presence of peroxide HBr adds by a free radical chain mechanism. Br• adds first to the less substituted alkene carbon (C-2), giving the more stable tertiary radical at C-1.
2. The radical abstracts H from HBr, so Br ends up at C-2: 1-bromo-2-methylcyclohexane (anti-Markovnikov product).

Teacher's Note:
1. Peroxide effect applies to HBr only.
2. Br• adds to give the more stable radical.

Answer: (b)

Explanation:
1. Friedel-Crafts acylation with succinic anhydride puts the acyl group para to OCH₃: A = 4-(4-methoxyphenyl)-4-oxobutanoic acid.
2. Clemmensen reduction converts only the aryl ketone C=O to CH₂; the COOH group remains. X = 4-(4-methoxyphenyl)butanoic acid.

Teacher's Note:
1. Clemmensen reduces C=O of aldehydes and ketones to CH₂ and does not touch COOH.
2. Acylation gives no rearrangement.

Answer: (b)

Explanation:
1. The ring attached to nitrogen is activated by the N lone pair (the -NHCOR group is an ortho-para director). The ring attached to C=O is deactivated.
2. Bromination takes place on the N-bearing ring at the para position (less hindered).

Teacher's Note:
1. The amide N donates into its own ring; the carbonyl ring is deactivated.
2. Only monobromination is asked.

Answer: (c)

Explanation:
1. The ring is deactivated by the acyl group and bears only a weak donor (acetate oxygen), so the alkene reacts with Br₂ rather than the ring.
2. Br₂ forms a bromonium ion; the solvent methanol (present in large excess) opens it. The nucleophile attacks the carbon that is farther from the electron-withdrawing C=O group (the beta carbon, which carries CH₃), so OCH₃ goes to the beta carbon and Br to the alpha carbon.

Teacher's Note:
1. In a halo-ether formation the solvent acts as the nucleophile.
2. The carbon next to C=O is less able to carry positive charge, so the nucleophile adds at the other carbon.

Answer: (d) 4

Explanation:
1. D⁺ adds to C-2 to give the planar tertiary cation. Cl^- can attack from either face of the cation.
2. The product has two stereocentres (C-1 and C-2): 2² = 4 stereoisomers, cis and trans (each racemic), all formed since the cation is planar (no stereospecificity).

Teacher's Note:
1. Planar carbocation: attack from both faces gives both diastereomers.
2. Count stereocentres: 2 centres give 4 isomers.

Answer: (a) 21

Explanation:
1. Br₂ adds anti through a bromonium ion. Anti addition to the trans alkene gives only the meso-2,3-dibromobutane, so X = 1.
2. HBr gives 2-bromobutane, which has one stereocentre and forms as both R and S enantiomers, so Y = 2.
3. YX = 21.

Teacher's Note:
1. Anti addition to a trans alkene gives meso; to a cis alkene it gives the racemic pair.
2. A new stereocentre formed from a planar cation gives a racemic mixture.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Assertion: bromine adds anti through a cyclic bromonium ion; anti addition to the symmetrical trans alkene gives the meso compound. True.
2. Reason: Br⁺ (the electrophile) attacks the pi bond first, so the addition is electrophilic. True.
3. The electrophilic nature does not by itself explain the meso product; the stereochemistry follows from the anti (bromonium) opening, so R is not the correct explanation.

Teacher's Note:
1. The stereochemical outcome comes from the bromonium ion and anti attack.
2. Be careful to separate the mechanism type from the stereochemical result.

Answer: (b) (a) < (b) < (d) < (c)

Explanation:
1. In the acid mixture aniline is converted to the anilinium ion, -NH₃⁺, which is strongly deactivating, so it is the slowest.
2. Chlorobenzene is deactivated (-I), toluene is activated by CH₃ (+I, hyperconjugation), and anisole is the most activated (+M of OCH₃).

Teacher's Note:
1. Check what a basic group becomes in strongly acidic medium.
2. +M donors beat +I donors.

Answer: (b)

Explanation:
1. IBr is polarised I^δ+-Br^δ-, so the electrophile is I⁺ (iodine is the less electronegative atom).
2. The CH₂F group is an ortho-para director; substitution at the para position gives 4-iodobenzyl fluoride.

Teacher's Note:
1. In an interhalogen compound, the less electronegative halogen is the electrophile.
2. Para is less hindered than ortho.

Answer: (a)

Explanation:
1. The ring joined to oxygen is activated (oxygen lone pair), though less than in an ether because the lone pair is shared with the C=O group. The ring joined to C=O is deactivated.
2. Nitration therefore occurs in the O-bearing ring, at the less hindered para position.

Teacher's Note:
1. Compare the two rings: the one attached through O is richer in electrons.
2. Para is preferred over ortho for steric reasons.

Answer: (d) 4

Explanation:
1. H⁺ adds to the carbon away from oxygen, giving the oxygen-stabilised cation CH₃CH₂CH(CH₃)-CH=O⁺-CH(CH₃)₂; Br^- then adds to give CH₃CH₂CH(CH₃)-CHBr-OCH(CH₃)₂.
2. The product has two different stereocentres, so there are 2² = 4 stereoisomers, all optically active.

Teacher's Note:
1. Oxygen lone pairs stabilise the adjacent cation strongly.
2. n different stereocentres give 2ⁿ stereoisomers.

Answer: (a)

Explanation:
1. HBr adds to the alkene by Markovnikov addition through the benzylic cation, giving Ar-CHBr-CH₂CH₃.
2. Excess hot HBr also cleaves the aryl methyl ether (Ar-O-CH₃ to Ar-OH + CH₃Br); the aryl C-O bond is not broken.

Teacher's Note:
1. Aryl alkyl ethers give phenol and alkyl halide with HBr/HI.
2. The aromatic C-O bond is strong and is not cleaved.

Answer: (a)

Explanation:
1. AlCl₃ ionises the primary alkyl chloride; the primary cation undergoes a 1,2-hydride shift to the tertiary cation ArCH₂CH₂-C⁺(CH₃)₂.
2. Intramolecular alkylation closes a five-membered ring at the ring position para to OCH₃ (activated and less hindered), giving the 1,1-dimethylindane in which the gem-dimethyl carbon is bonded to the ring carbon para to OCH₃.

Teacher's Note:
1. Rearrangement to the tertiary cation comes before ring closure.
2. Five-membered rings close easily in these Friedel-Crafts cyclisations.

Answer: (a) 1-methylcyclohexene + HBr in the presence of R₂O₂

Explanation:
1. Br₂/CCl₄, acid-catalysed hydration and oxymercuration all begin with attack of an electrophile (Br⁺, H⁺, Hg²⁺) on the pi bond.
2. HBr with peroxide adds by a free radical chain mechanism (Br• attacks first), so it is a radical addition.

Teacher's Note:
1. Peroxide effect = radical addition (anti-Markovnikov).
2. Identify the first attacking species.

Answer: (d)

Explanation:
1. AlCl₃ ionises the C-Cl bond and the cation attacks the benzene ring of the same molecule (intramolecular Friedel-Crafts alkylation).
2. The six-membered ring closes easily, giving tetralin (1,2,3,4-tetrahydronaphthalene). In contrast, benzoic acid (deactivated) and 3-nitro-toluene (deactivated) do not undergo Friedel-Crafts reactions, and toluene with AlCl₃ alone has no electrophile.

Teacher's Note:
1. Friedel-Crafts reactions fail on strongly deactivated rings.
2. An intramolecular reaction forming a five- or six-membered ring is favourable.