CBSE Class 12 Chemistry Haloalkanes and Haloarenes MCQs Set 11

Answer: (c) a < c < b

Explanation:
1. A –NO₂ group at the ortho position withdraws electrons by both −I and −R effects and stabilises the carbanion intermediate, so (b) reacts fastest.
2. A meta-NO₂ group acts only by its −I effect, so (c) is faster than chlorobenzene but slower than (b). Order: a < c < b.

Teacher's Note:
1. Only ortho and para nitro groups can carry the negative charge of the intermediate by resonance; a meta nitro group helps only through the inductive effect.

Answer: (a) i < ii < iii

Explanation:
1. –CH₃ is electron releasing and deactivates the ring towards nucleophilic attack, while each o/p –NO₂ group activates it.
2. More nitro groups at ortho/para positions mean faster substitution: i < ii < iii.

Teacher's Note:
1. Electron-withdrawing groups speed up nucleophilic aromatic substitution; electron-releasing groups slow it down – the opposite of electrophilic substitution.

Answer: (b) 2, 3

Explanation:
1. C-2 carries H, OH, COOH and CH(OH)CHO; C-3 carries H, OH, CHO and CH(OH)COOH – four different groups on each.
2. C-1 (COOH) and C-4 (CHO) are not attached to four different groups, so only C-2 and C-3 are asymmetric.

Teacher's Note:
1. A carbon with a double bond (C=O) can never be a chiral centre, since it has only three attached groups.

Answer: (d) iv

Explanation:
1. Racemisation needs the halogen to be on a chiral carbon; only in (iv) is the C–Cl carbon attached to four different groups (H, Cl, CH₃, C₂H₅).
2. In (iii) the chiral carbon is not the carbon that bears Cl, so its configuration is not affected. (i) and (ii) are achiral.

Teacher's Note:
1. Check whether the carbon that loses the halogen is itself chiral; a chiral centre elsewhere in the molecule is not touched by substitution.

Answer: (d) inversion more than retention, leading to partial racemisation

Explanation:
1. The planar carbocation can be attacked from either face, but the departing halide ion partly shields the front face.
2. Hence attack from the back side is slightly favoured: inversion is somewhat more than retention, giving partial racemisation.

Teacher's Note:
1. For board answers, "S_N1 gives racemisation" is accepted, but the exact picture is a racemic mixture with slight excess of the inverted product.

Answer: (c) C₆H₅–CH(Br)–CH₂–CH₃

Explanation:
1. H⁺ adds to the carbon farther from the ring so that the benzylic carbocation C₆H₅–C⁺H–CH₂CH₃ is formed; it is stabilised by resonance with the ring.
2. Br⁻ then attacks this carbon, giving C₆H₅CH(Br)CH₂CH₃.

Teacher's Note:
1. When both alkene carbons carry one substituent, decide by carbocation stability – a benzylic cation wins.

Answer: (c) i > iv > iii > ii

Explanation:
1. S_N2 reactivity is decided by steric hindrance at the carbon under attack: methyl > 1° > 2° > 3°.
2. So CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl, i.e. i > iv > iii > ii.

Teacher's Note:
1. S_N2 and S_N1 orders are exact opposites for simple alkyl halides – remember one and reverse it for the other.

Answer: (b) ii < i < iii

Explanation:
1. S_N1 rate depends on carbocation stability. (ii) gives a 1° cation, (i) a 2° cation and (iii) a benzylic cation further stabilised by the electron-donating p-OCH₃ group.
2. Hence ii < i < iii.

Teacher's Note:
1. A p-methoxy group pushes electrons into the ring by resonance and makes a benzylic carbocation even more stable.

Answer: (a) p-CH₃C₆H₄CN

Explanation:
1. Diazotisation gives D = p-CH₃C₆H₄N₂⁺Cl⁻.
2. Sandmeyer reaction with CuCN/KCN replaces –N₂⁺ by –CN: E = p-CH₃C₆H₄CN, with loss of N₂.

Teacher's Note:
1. The diazonium group is replaced at the same ring position, so the methyl group stays para to the new group.

Answer: (a) H₃C–C(Br)(CH₃)–CH₂–CH₃

Explanation:
1. Markovnikov addition of H⁺ gives the 2° carbocation (CH₃)₂CH–C⁺H–CH₃.
2. A 1,2-hydride shift converts it into the more stable 3° carbocation (CH₃)₂C⁺–CH₂CH₃, which takes up Br⁻ to give 2-bromo-2-methylbutane.

Teacher's Note:
1. Whenever a carbocation is formed next to a carbon carrying more alkyl groups, look for a hydride or methyl shift to a more stable cation.

Answer: (d) X = p-chlorotoluene and Y = benzyl chloride

Explanation:
1. With FeCl₃, electrophilic substitution occurs on the ring at o/p positions (X = p-chlorotoluene, major).
2. In light, free radical substitution occurs in the side chain, giving Y = C₆H₅CH₂Cl (benzyl chloride).

Teacher's Note:
1. Remember the "CCC/SSS" rule: Cold, Catalyst, Core (ring) versus Sunlight, Side chain, high temperature.

Answer: (b) resonance stabilization

Explanation:
1. In aryl halides the lone pair of the halogen is in resonance with the ring, giving the C–X bond partial double bond character.
2. This makes the C–X bond shorter and stronger and difficult to break.

Teacher's Note:
1. The sp² carbon of the ring (not sp³) is the second reason – it holds the bonding electrons more tightly and shortens the C–X bond.

Answer: (a) C₆H₆ + CH₃Cl –[anhy. AlCl₃]→

Explanation:
1. Friedel–Crafts alkylation gives C₆H₅–CH₃: a new C–C bond between the ring and CH₃.
2. (b) forms a C–N bond, (c) a C–O bond (Williamson synthesis) and (d) a C–O bond (ethanol).

Teacher's Note:
1. Reactions with CN⁻, Grignard reagents, Wurtz/Fittig coupling and Friedel–Crafts alkylation are the main ways to form new C–C bonds in this chapter.

Answer: (a) Benzyl halides are more reactive than vinyl and aryl halides

Explanation:
1. The benzyl carbocation (or the S_N2 transition state) is stabilised by the ring, so benzyl halides are very reactive.
2. Vinyl and aryl halides have a C–X bond with partial double bond character and are much less reactive than alkyl or benzyl halides.

Teacher's Note:
1. Benzyl and allyl halides are the "reactive pair"; vinyl and aryl halides are the "unreactive pair" in nucleophilic substitution.

Answer: (a) C₆H₅–CH₂X

Explanation:
1. In benzyl halide the halogen is on the CH₂ carbon, which is sp³ hybridised.
2. In (b) X is on a vinylic carbon and in (c) and (d) on a ring carbon; all three are sp².

Teacher's Note:
1. Benzylic and allylic halides have X on an sp³ carbon next to a C=C or ring; vinylic and aryl halides have X directly on an sp² carbon.

Answer: (a) 1-chloro-4-methylbenzene

Explanation:
1. Substituents are numbered and cited in alphabetical order: chloro before methyl, so Cl gets position 1.
2. Name: 1-chloro-4-methylbenzene (common name 4-chlorotoluene or p-chlorotoluene).

Teacher's Note:
1. When two substituents on benzene would get the same locants either way, the one cited first alphabetically gets the lower number.

Answer: (b) A–S, B–R, C–Q, D–P

Explanation:
1. CCl₄ is sold as the fire extinguisher "Pyrene"; (p-ClC₆H₄)₂CHCCl₃ is DDT, an insecticide; CHI₃ (iodoform) is an antiseptic.
2. This leaves CH₂Cl₂ with refrigerant: A–S, B–R, C–Q, D–P.

Teacher's Note:
1. In matching questions, fix the items you are sure of (DDT, iodoform) first; the rest often follows by elimination.

Answer: (c) A–Q, B–S, C–P, D–R

Explanation:
1. (A) X on an sp³ carbon of an alkane – alkyl halide; (B) X on the carbon next to C=C – allyl halide.
2. (C) X on a benzene ring carbon – aryl halide; (D) X on a C=C carbon – vinyl halide. So A–Q, B–S, C–P, D–R.

Teacher's Note:
1. Allyl = X one carbon away from C=C; vinyl = X directly on C=C. Mixing these two is the most common slip.

Answer: (d) Vinyl halide

Explanation:
1. In vinyl halide, CH₂=CH–X, the halogen is attached to an sp² carbon of the double bond.
2. Vinyl halide contains C_sp²–X bond.

Teacher's Note:
1. In allyl and benzyl halides the halogen sits on an sp³ carbon next to the double bond or ring; only vinyl and aryl halides have X on an sp² carbon.

Answer: (a) CH₃CHBr₂

Explanation:
1. In a gem-dihalide both halogen atoms are on the same carbon atom.
2. In CH₃CHBr₂ both Br atoms are on C-1, so it is a gem-dihalide (ethylidene bromide).

Teacher's Note:
1. gem (geminal) = same carbon; vic (vicinal) = adjacent carbons, as in BrCH₂CH₂Br.

Answer: (c) 3-bromo-1-chlorocyclohexene

Explanation:
1. The double-bond carbons get numbers 1 and 2; numbering starts at the carbon carrying Cl so that the substituents get the lowest locants (1, 3).
2. Substituents are written alphabetically: 3-bromo-1-chlorocyclohexene.

Teacher's Note:
1. In cycloalkenes the C=C must get locants 1 and 2 first; only then apply lowest locants to the substituents.

Answer: (b) C₂H₆ (excess) + Cl₂ → (UV light)

Explanation:
1. With excess ethane, a Cl atom is far more likely to meet a C₂H₆ molecule than a C₂H₅Cl molecule, so polychlorination is minimised and monochloroethane is the main product.

Teacher's Note:
1. Excess of the hydrocarbon favours monohalogenation; excess of halogen gives polyhalogenated products.

Answer: (c) anhydrous ZnCl₂

Explanation:
1. R–OH + HCl → R–Cl + H₂O in presence of anhydrous ZnCl₂ (Groove’s process).
2. ZnCl₂ is a Lewis acid that helps break the C–O bond.

Teacher's Note:
1. Conc. HCl + anhydrous ZnCl₂ is the Lucas reagent; it is also used to distinguish 1°, 2° and 3° alcohols.

Answer: (c) SOCl₂ in presence of pyridine

Explanation:
1. R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑
2. Both by-products are gases that escape, leaving pure alkyl chloride; pyridine absorbs HCl.

Teacher's Note:
1. Remember why thionyl chloride is preferred: the by-products SO₂ and HCl are gases, so no separation is needed.

Answer: (d) 2-bromopentane + 3-bromopentane

Explanation:
1. CH₃–CH=CH–CH₂CH₃ (2-pentene) + H⁺ → CH₃–C⁺H–CH₂–CH₂CH₃ (I) + CH₃CH₂–C⁺H–CH₂CH₃ (II)
2. With Br⁻: (I) → CH₃–CHBr–CH₂CH₂CH₃ (2-bromopentane); (II) → CH₃CH₂–CHBr–CH₂CH₃ (3-bromopentane)
3. Addition of a proton to 2-pentene gives two carbocations. Whereas carbocation (I) is stabilized by five hyperconjugation structures while carbocation (II) is stabilized by four hyperconjugation structures. Since both these carbocations are almost equally stable, therefore, nucleophilic attack by Br⁻ on these two carbocations gives a mixture.

Teacher's Note:
1. Markovnikov’s rule gives one product only when the two alkene carbons carry different numbers of alkyl groups; here both are 2° carbons.

Answer: (a) 1-bromo-3-phenylpropane

Explanation:
1. CH₂=CH–CH₂Ph (3-phenyl-1-propene) + HBr/ROOR (anti-Markovnikov addition) → BrCH₂–CH₂–CH₂Ph (1-bromo-3-phenylpropane)

Teacher's Note:
1. Peroxide effect works only with HBr; Br attaches to the terminal CH₂ of the double bond.

Answer: (b) X = conc. alcoholic NaOH, 80°C; Y = HBr/acetic acid, 20°C

Explanation:
1. Conc. alc. NaOH at 80°C brings about dehydrohalogenation (giving CH₃CH=CH₂) while HBr/acetic acid at 20°C brings about Markovnikov’s addition (giving CH₃CHBrCH₃).

Teacher's Note:
1. Alcoholic alkali favours elimination, aqueous alkali favours substitution – a very common exam distinction.

Answer: (c) 1-Bromo-1-methylcyclopentane

Explanation:
1. Ease of abstraction of hydrogen is: 3° > 2° > 1°. The ring carbon carrying CH₃ has the only 3° H, so Br goes there. Therefore, option (c) is correct.

Teacher's Note:
1. Bromination is highly selective (much more than chlorination), so the 3° bromide is the major product.

Answer: (d) CH₃Br + AgF →

Explanation:
1. CH₃Br + AgF →(Swarts reaction) CH₃F + AgBr

Teacher's Note:
1. Alkyl fluorides are made by halogen exchange (Swarts reaction) with AgF, Hg₂F₂, CoF₂ or SbF₃; direct fluorination is explosive.

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

Explanation:
1. Both statements are true. AgCN is mainly covalent, so the carbon of CN is tied to Ag and only the nitrogen lone pair is free to attack, giving C₂H₅NC.
2. The ambident nature of CN⁻ alone does not explain why the isocyanide is the major product, so R is not the correct explanation.

Teacher's Note:
1. KCN (ionic) gives alkyl cyanide by attack through C; AgCN (covalent) gives isocyanide by attack through N.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion: At high temperature, addition of Cl₂ to the double bond is reversible but propene undergoes free radical substitution at 3-position to afford allyl chloride.
2. CH₃–CH=CH₂ →(Cl₂/773 K) ClCH₂–CH=CH₂
3. So A is false and R is true.

Teacher's Note:
1. At low temperature Cl₂ adds to propene; at 773 K allylic substitution takes over.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. C₆H₆ + I₂ ⇌ C₆H₅I + HI is reversible; HI being a strong reducing agent reduces C₆H₅I back to benzene.
2. HNO₃ oxidises HI to I₂, so the equilibrium shifts forward. Both A and R are true and R explains A.

Teacher's Note:
1. An oxidising agent (HNO₃ or HIO₃) is always needed in direct iodination to remove HI.

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

Explanation:
1. Correct explanation: Iodides decompose on standing to liberate I₂ which makes the solution dark.
2. Both statements are true, but R does not explain A.

Teacher's Note:
1. The weak C–I bond breaks easily in light, which is why alkyl iodides are stored in dark bottles.

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

Explanation:
1. Correct explanation: NaCN being ionic, nucleophilic attack by CN⁻ ion on the alkyl halide occurs through carbon giving CH₃CN as the major product.
2. Na^{+ −}:C≡N + CH₃–Br → CH₃–C≡N + NaBr

Teacher's Note:
1. Ambident nature only tells you that two products are possible; whether C or N attacks depends on whether the cyanide is ionic (KCN, NaCN) or covalent (AgCN).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. AgCN is covalent (Ag–C≡N:); the carbon is bonded to Ag, so only the lone pair on N is available for attack.
2. R–Br + AgCN → R–N⇄C (alkyl isocyanide/carbylamine) + AgBr. Both are true and R explains A.

Teacher's Note:
1. KCN → nitrile (R–CN); AgCN → isocyanide (R–NC).

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

Explanation:
1. Correct explanation: AgNO₂ being a covalent compound, nucleophilic attack occurs through lone pair of electrons on the N-atom giving CH₃CH₂NO₂ as the major product.
2. Ag–O–N=O + CH₃CH₂–Br → CH₃CH₂NO₂ + AgBr

Teacher's Note:
1. With ionic KNO₂ the same bromoethane gives ethyl nitrite (C₂H₅–O–N=O).

Answer: (d) A is false, but R is true.

Explanation:
1. Correct assertion: Isopropyl bromide on treatment with aq. NaOH undergoes nucleophilic substitution to give isopropyl alcohol.
2. So A is false and R is true.

Teacher's Note:
1. Aqueous KOH/NaOH → substitution (alcohol); alcoholic KOH → elimination (alkene).

Answer: (c) A is true, but R is false.

Explanation:
1. 1-Chlorobutane has only one kind of β-hydrogen (on C-2), so it can only give but-1-ene.
2. Correct reason: Due to presence of six hyperconjugation structures in but-2-ene as compared to only two in but-1-ene, but-2-ene is more stable than but-1-ene. So A is true but R is false.

Teacher's Note:
1. Saytzeff’s rule matters only when there is a choice of β-hydrogens; a 1-halobutane has no such choice.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In 1-bromo-2-methylbutane, CH₃CH₂CH(CH₃)CH₂Br, the methyl branch is on the β-carbon, close to the site of backside attack.
2. In 1-bromo-3-methylbutane the branch is on the γ-carbon, farther away, so it reacts faster. R explains A.

Teacher's Note:
1. The closer a branch is to the carbon bearing the halogen, the more it slows down S_N2.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Fluoride is a poor leaving group, so the elimination has carbanion character; the base removes the more acidic proton from the terminal CH₃ (1° β-H).
2. Hence the less substituted alkene, 1-butene (Hofmann product), is the major product. R explains A.

Teacher's Note:
1. Alkyl fluorides are an exception to Saytzeff’s rule: they give mainly the less substituted (Hofmann) alkene.