JEE Chemistry Haloalkanes and Haloarenes MCQs Set 14

Practice MCQs for JEE Chemistry Haloalkanes and Haloarenes

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Multiple Choice Questions

Question 1: Benzyl alcohol reacts with PCl₃ to give X and H₃PO₃. X is:
(a) 3-chlorotoluene
(b) benzyl chloride (chloromethylbenzene)
(c) 2-chlorotoluene
(d) (4-chlorophenyl)methanol
Show Answer & Explanation

Answer: (b) benzyl chloride (chloromethylbenzene)

Explanation:
1. PCl₃ replaces the alcoholic OH group by Cl: 3ROH + PCl₃ → 3RCl + H₃PO₃.
2. The ring is untouched, so C₆H₅CH₂OH gives C₆H₅CH₂Cl.

Teacher's Note:
1. Halogen enters at the carbon that carried OH.

Question 2: 2-Methylpropan-2-ol reacts with HBr to give Y and water. Y is:
(a) 1-bromo-2-methylpropane
(b) 1-bromobutane
(c) 1-bromo-2,2-dimethylpropane
(d) 2-bromo-2-methylpropane
Show Answer & Explanation

Answer: (d) 2-bromo-2-methylpropane

Explanation:
1. A tertiary alcohol reacts with HBr by an SN1 path.
2. Br⁻ attacks the tertiary carbocation, giving (CH₃)₃CBr.

Teacher's Note:
1. No skeletal change occurs.

Question 3: X reacts with PCl₅ to give 1-chloropropane, HCl and POCl₃. X is:
(a) propan-1-ol
(b) propan-2-ol
(c) propane
(d) 2-methylpropane
Show Answer & Explanation

Answer: (a) propan-1-ol

Explanation:
1. PCl₅ converts R–OH into R–Cl + POCl₃ + HCl.
2. 1-Chloropropane requires the alcohol with OH on the terminal carbon of a three-carbon chain.

Teacher's Note:
1. Propan-2-ol would give 2-chloropropane.

Question 4: Benzyl alcohol reacts with PCl₅ to give X, HCl and POCl₃. X is:
(a) benzyl bromide
(b) 2-chlorotoluene
(c) benzyl chloride
(d) (3-chlorophenyl)methanol
Show Answer & Explanation

Answer: (c) benzyl chloride

Explanation:
1. PCl₅ replaces the OH group of the alcohol with Cl.
2. C₆H₅CH₂OH + PCl₅ → C₆H₅CH₂Cl + POCl₃ + HCl.

Teacher's Note:
1. Only the side-chain OH is replaced, not ring hydrogens.

Question 5: 2,2-Dimethylpropane (neopentane) is treated with Cl₂ in light. The monochloro product X is:
(a) 2-chloro-2-methylbutane
(b) 2-chloro-3-methylbutane
(c) 3-chloropentane
(d) 1-chloro-2,2-dimethylpropane
Show Answer & Explanation

Answer: (d) 1-chloro-2,2-dimethylpropane

Explanation:
1. All twelve hydrogens of neopentane are equivalent (all on CH₃ groups).
2. Only one monochloro product is possible: (CH₃)₃CCH₂Cl.

Teacher's Note:
1. No rearrangement occurs in radical halogenation.

Question 6: The number of monochloro structural isomers formed when isobutane (2-methylpropane) is treated with Cl₂/hν is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Isobutane has two kinds of hydrogen: nine primary H on the methyls and one tertiary H.
2. Substituting each kind gives 1-chloro-2-methylpropane and 2-chloro-2-methylpropane.

Teacher's Note:
1. Count the different types of H.

Question 7: The number of monochloro structural isomers formed when isooctane (2,2,4-trimethylpentane) is treated with Cl₂/hν is:
(a) 7
(b) 2
(c) 4
(d) 3
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Isooctane is (CH₃)₃C–CH₂–CH(CH₃)₂. Group the hydrogens by type: the nine H of the tert-butyl methyls, the CH₂ hydrogens, the single CH hydrogen and the six H of the two equivalent isopropyl methyls.
2. That gives four distinct sets of hydrogen, hence four monochloro structural isomers.

Teacher's Note:
1. Structural isomers only; stereoisomers are not counted.

Question 8: The number of monochloro products obtained when cyclopentane reacts with Cl₂/hν is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. All ten hydrogens of cyclopentane are equivalent.
2. So only chlorocyclopentane forms.

Teacher's Note:
1. Symmetry decides the number of monosubstituted products.

Question 9: Propene reacts with HBr (no peroxide). The major product is:
(a) 1-bromopropane
(b) 2-bromopropane
(c) propane
(d) 1-bromo-2-methylpropane
Show Answer & Explanation

Answer: (b) 2-bromopropane

Explanation:
1. Markovnikov's rule: H adds to the carbon with more hydrogens.
2. Br goes to the central carbon, giving CH₃CHBrCH₃.

Teacher's Note:
1. With peroxide, 1-bromopropane would form.

Question 10: The alkene Z reacts with HCl to give chlorocyclopentane. Z is:
(a) cyclopentane
(b) cyclopenta-1,3-diene
(c) methylcyclopentane
(d) cyclopentene
Show Answer & Explanation

Answer: (d) cyclopentene

Explanation:
1. HCl adds across a C=C bond to form an alkyl chloride.
2. Cyclopentene gives chlorocyclopentane.

Teacher's Note:
1. Cyclopentane has no double bond to add to.

Question 11: Chloroethane is treated with NaI in dry acetone. The product is:
(a) iodoethane
(b) ethane
(c) iodomethane
(d) butane
Show Answer & Explanation

Answer: (a) iodoethane

Explanation:
1. This is the Finkelstein reaction: R–Cl + NaI → R–I + NaCl.
2. NaCl is insoluble in dry acetone and precipitates, driving the reaction forward.

Teacher's Note:
1. Used to prepare alkyl iodides.

Question 12: Z reacts with NaI in dry acetone to give benzyl iodide. Z is:
(a) 2-chlorotoluene
(b) 4-chlorotoluene
(c) benzyl chloride
(d) 3-bromotoluene
Show Answer & Explanation

Answer: (c) benzyl chloride

Explanation:
1. The Finkelstein reaction replaces a halogen on an sp³ carbon.
2. Benzyl chloride gives C₆H₅CH₂I; ring halides do not react.

Teacher's Note:
1. Aryl halides do not take part in SN2 exchange.

Question 13: Chloroethane is heated with AgF. The product is:
(a) ethane
(b) 1,1-difluoroethane
(c) fluoromethane
(d) fluoroethane
Show Answer & Explanation

Answer: (d) fluoroethane

Explanation:
1. This is the Swarts reaction: R–Cl + AgF → R–F + AgCl.
2. One chlorine is replaced by one fluorine.

Teacher's Note:
1. Swarts reagents: AgF, Hg₂F₂, CoF₂, SbF₃.

Question 14: Benzyl chloride is converted into benzyl fluoride by which reagent?
(a) any of AgF, Hg₂F₂ or SbF₃
(b) only AgF
(c) only Hg₂F₂
(d) only SbF₃
Show Answer & Explanation

Answer: (a) any of AgF, Hg₂F₂ or SbF₃

Explanation:
1. All the metallic fluorides AgF, Hg₂F₂, CoF₂ and SbF₃ are used in the Swarts reaction.
2. Any of them can replace Cl by F.

Teacher's Note:
1. Swarts reaction works for alkyl chlorides and bromides.

Question 15: Which compound reacts with CoF₂ to give CH₃CH=CHCH₂F (1-fluorobut-2-ene)?
(a) butane
(b) but-2-ene
(c) 1-chlorobut-2-ene
(d) buta-1,3-diene
Show Answer & Explanation

Answer: (c) 1-chlorobut-2-ene

Explanation:
1. The Swarts reaction exchanges the halogen on the sp³ carbon for fluorine.
2. The double bond is untouched, so the reactant is CH₃CH=CHCH₂Cl.

Teacher's Note:
1. Compare the carbon skeletons of reactant and product.

Question 16: 1-Chloropropane is treated with KNO₂. The main product is:
(a) propyl nitrite (CH₃CH₂CH₂ONO)
(b) 1-nitropropane
(c) propane
(d) propene
Show Answer & Explanation

Answer: (a) propyl nitrite (CH₃CH₂CH₂ONO)

Explanation:
1. KNO₂ is ionic; the oxygen of NO₂⁻ attacks the carbon.
2. The product is the alkyl nitrite R–O–N=O.

Teacher's Note:
1. KNO₂ gives nitrite; AgNO₂ gives the nitro compound.

Question 17: 1-Chloropropane is treated with AgNO₂. The main product is:
(a) propyl nitrite
(b) 1-nitropropane
(c) propane
(d) propene
Show Answer & Explanation

Answer: (b) 1-nitropropane

Explanation:
1. AgNO₂ is largely covalent, so nitrogen attacks as the nucleophile.
2. The product is the nitroalkane R–NO₂.

Teacher's Note:
1. KNO₂ gives nitrite; AgNO₂ gives the nitro compound.

Question 18: Potassium tert-butoxide reacts with CH₃Br. The product Y is:
(a) 2-methylpropene
(b) 2-methylbutane
(c) ethane
(d) tert-butyl methyl ether
Show Answer & Explanation

Answer: (d) tert-butyl methyl ether

Explanation:
1. This is a Williamson synthesis with a primary methyl halide, so substitution wins.
2. (CH₃)₃CO⁻ + CH₃Br → (CH₃)₃COCH₃ + Br⁻.

Teacher's Note:
1. Elimination would only dominate for a 3° halide with a strong bulky base.

Question 19: Benzyl bromide is heated with aqueous KOH. The product X is:
(a) benzyl alcohol
(b) toluene
(c) 3-(chloromethyl)phenol
(d) 4-methylphenol
Show Answer & Explanation

Answer: (a) benzyl alcohol

Explanation:
1. The benzylic C–Br bond is hydrolysed by OH⁻.
2. C₆H₅CH₂Br + KOH → C₆H₅CH₂OH + KBr.

Teacher's Note:
1. The ring is unaffected.

Question 20: In the Wurtz reaction, 2,3-dimethylbutane is prepared from:
(a) 1-bromopropane
(b) bromoethane
(c) 2-bromopropane
(d) 1-bromo-2,2-dimethylpropane
Show Answer & Explanation

Answer: (c) 2-bromopropane

Explanation:
1. Wurtz coupling joins two identical alkyl groups: 2 R–X + 2Na → R–R.
2. Two isopropyl groups give (CH₃)₂CH–CH(CH₃)₂, which is 2,3-dimethylbutane.

Teacher's Note:
1. Count carbons: 3 + 3 = 6.

Question 21: Bromobenzene, bromoethane and sodium in dry ether (Wurtz-Fittig reaction) mainly give:
(a) toluene
(b) butane
(c) biphenyl only
(d) ethylbenzene
Show Answer & Explanation

Answer: (d) ethylbenzene

Explanation:
1. The aryl halide and alkyl halide couple through their organosodium intermediates.
2. C₆H₅–Br + C₂H₅–Br + 2Na → C₆H₅C₂H₅ + 2NaBr.

Teacher's Note:
1. The cross product is the major one wanted.

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Multiple Choice Questions (MCQs) for JEE Chemistry Haloalkanes and Haloarenes

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