JEE Chemistry Haloalkanes and Haloarenes MCQs Set 28

Answer: (c) B < A < D < C

Explanation:
1. S_N1 rate depends on the stability of the carbocation formed.
2. Ethyl chloride gives an unstable primary cation. Isobutyl chloride gives a primary cation that can rearrange by a hydride shift to a tertiary cation, so it is faster than ethyl chloride.
3. Benzyl chloride gives a delocalisation-stabilised benzyl cation, and the +R methoxy group in the para position stabilises it even more.
4. Order: ethyl chloride < isobutyl chloride < benzyl chloride < 4-methoxybenzyl chloride, that is B < A < D < C.

Teacher's Note:
1. Faster S_N1 means a more stable carbocation.
2. Electron-donating groups on the ring stabilise a benzylic cation.

Answer: (b) B < A < C < D

Explanation:
1. All four form a secondary benzylic carbocation; the rate depends on the substituent on the ring.
2. A methoxy group at the meta position acts only through its –I effect, so it destabilises the cation (B is slowest).
3. A para methyl group stabilises the cation by hyperconjugation (+I), so C is faster than the unsubstituted A.
4. A para methoxy group stabilises it by a strong +R effect, so D is fastest. Order: B < A < C < D.

Teacher's Note:
1. Meta substituents cannot donate by delocalisation; para substituents can.
2. +R of OCH₃ is stronger than hyperconjugation of CH₃.

Answer: (c) 2-methoxy-2-methylbutane

Explanation:
1. A secondary bromide in a polar protic solvent with a weak nucleophile reacts by S_N1.
2. The secondary carbocation rearranges by a 1,2-hydride shift to the more stable tertiary carbocation (CH₃)₂C⁺CH₂CH₃.
3. Methanol attacks the tertiary cation to give 2-methoxy-2-methylbutane (after loss of H⁺).

Teacher's Note:
1. Carbocation rearrangement always leads to the most stable cation.
2. Solvent attack (solvolysis) gives ethers in methanol.

Answer: (b) 2-bromo-2-methylbutane

Explanation:
1. Strong base and heat give E₂ elimination following Saytzeff's rule: X = 2-methylbut-2-ene (more substituted alkene).
2. HBr adds to X by Markovnikov's rule: H goes to the less substituted carbon, Br to the tertiary carbon.
3. Y = 2-bromo-2-methylbutane.

Teacher's Note:
1. Small alkoxide bases give the Saytzeff (more substituted) alkene.
2. Markovnikov: the halogen goes to the carbon that can better hold the positive charge.

Answer: (a) 7-bromocyclohepta-1,3,5-triene

Explanation:
1. A precipitate of AgBr forms only if the C–Br bond ionises readily in water to give Br^–.
2. 7-Bromocycloheptatriene ionises to the tropylium cation, which is aromatic (6 π electrons) and very stable.
3. Bromobenzene and 3-bromopyridine are aryl halides (partial double-bond character), and the cyclohexyl cation is an unstable secondary cation.

Teacher's Note:
1. The tropylium ion C₇H₇⁺ is aromatic.
2. Aryl and simple secondary halides do not ionise easily at room temperature.

Answer: (d) 4-(1-bromopropyl)phenol

Explanation:
1. Excess hot HBr does two things: it adds across the double bond (Markovnikov) and it cleaves the aryl alkyl ether.
2. H adds to the carbon farther from the ring, so Br goes to the benzylic carbon (stable benzylic cation): ArCH(Br)CH₂CH₃.
3. The ether ArO–CH₃ is cleaved by S_N2 (Zeisel's method), giving the phenol and CH₃Br, since the aryl C–O bond cannot break.
4. Product: 4-(1-bromopropyl)phenol.

Teacher's Note:
1. Aryl alkyl ethers always cleave to a phenol and an alkyl halide.
2. The benzylic carbocation decides the position of Br.

Answer: (b) ethyl 2-methylbut-2-enoate (CH₃CH=C(CH₃)COOEt)

Explanation:
1. The substrate is a tertiary halide and the reagent is a strong base on heating, so E₂ elimination dominates over substitution.
2. Two β-hydrogens are available (on the CH₃ and on the CH₂). By Saytzeff's rule the more substituted alkene, which is also conjugated with the ester, is the major product.
3. Product: CH₃CH=C(CH₃)COOEt.

Teacher's Note:
1. Tertiary halide + strong base gives elimination.
2. The more substituted, conjugated alkene is the major product.

Answer: (b) 3,3-dimethylcyclohexene

Explanation:
1. NaOMe is a strong base, so the secondary halide undergoes E₂ elimination.
2. The adjacent carbon bearing the two methyl groups has no hydrogen, so the only β-hydrogen is on the other neighbouring ring carbon.
3. Loss of HBr gives a double bond between C-1 and C-6, which is 3,3-dimethylcyclohexene.

Teacher's Note:
1. E₂ needs a β-hydrogen anti to the leaving group.
2. No rearrangement occurs in E2, unlike E1 or S_N1.

Answer: (b) II < I < III

Explanation:
1. S_N1 reactivity follows the stability of the carbocation.
2. 1-Chloropropane gives an unstable primary cation (least reactive). 2-Chlorobutane gives a secondary cation.
3. 4-Methoxybenzyl chloride gives a benzylic cation strongly stabilised by delocalisation and by the +R methoxy group, so it is the most reactive.
4. Order: II < I < III.

Teacher's Note:
1. A delocalisation-stabilised benzylic cation beats a simple secondary cation.
2. Rate of S_N1 depends only on the concentration of the alkyl halide.

Answer: (d) I, II and III

Explanation:
1. A tertiary halide with a strong nucleophile/base (OMe^–) in a polar protic solvent gives mostly elimination products, but substitution (ether) also occurs in smaller yield.
2. Elimination can occur towards the CH₃ (giving 2-methylpent-1-ene) or towards the CH₂ (giving 2-methylpent-2-ene, the Saytzeff product).
3. So all three are formed, in different yields.

Teacher's Note:
1. Substitution and elimination compete; all products are possible.
2. The more substituted alkene is the major elimination product.

Answer: (a) CH₃CH⁺CH₂Cl

Explanation:
1. Cl⁺ (the electrophile from HOCl) adds to the double bond first, giving a chloronium ion that opens to the more stable secondary carbocation CH₃CH⁺CH₂Cl.
2. Water then attacks the secondary carbon, giving CH₃CH(OH)CH₂Cl (a chlorohydrin).

Teacher's Note:
1. The electrophile is Cl⁺, not OH⁺.
2. Water attacks the more substituted carbon.

Answer: (d) Swarts reaction

Explanation:
1. Swarts reaction: alkyl chloride or bromide is heated with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃ to give the alkyl fluoride.
2. Free radical fluorination is explosive, Sandmeyer gives aryl halides and Finkelstein gives alkyl iodides.

Teacher's Note:
1. Finkelstein: NaI in acetone gives R–I; Swarts: metal fluoride gives R–F.
2. Remember the pair of names with their products.

Answer: (c) 2-butyne

Explanation:
1. Silver abstracts the three chlorine atoms of two molecules: 2CH₃CCl₃ + 6Ag gives CH₃C≡CCH₃ + 6AgCl.
2. The product is 2-butyne.

Teacher's Note:
1. gem-Trihalides with Ag (Wurtz-type coupling) give alkynes.
2. Each pair of CCl₃ carbons forms a triple bond.

Answer: (d) 2-methylbenzyl bromide

Explanation:
1. A gives a precipitate with alcoholic AgNO₃, so the bromine is in the side chain (a benzylic bromide), not on the ring.
2. Oxidation gives a dicarboxylic acid C₈H₆O₄, so there are two side chains on the ring.
3. B forms an anhydride on heating, which is possible only for a phthalic acid (the two –COOH groups ortho to each other: phthalic anhydride).
4. So A is 2-methylbenzyl bromide.

Teacher's Note:
1. A cyclic anhydride needs the two acid groups on adjacent carbons.
2. Halogen on the ring is not removed by AgNO₃ under these conditions.

Answer: (c) carbocation

Explanation:
1. SbCl₅, a Lewis acid, removes chloride to give the planar benzylic carbocation C₆H₅CH⁺CH₃.
2. Chloride can attack from either face with equal ease, giving a 50:50 mixture of enantiomers, which is racemisation.

Teacher's Note:
1. Planar intermediates cause racemisation.
2. This is the same idea as an S_N1 reaction.

Answer: (d) CH₃Cl

Explanation:
1. S_N2 is a one-step backside attack, fastest at the least hindered carbon (CH₃ > 1° > 2° > 3°).
2. Methyl chloride reacts by S_N2 almost completely, giving complete inversion of configuration.

Teacher's Note:
1. More alkyl groups mean more steric hindrance to backside attack.
2. Tertiary halides prefer S_N1 or E1.