NEET Haloalkanes and Haloarenes PYQs & Practice Questions

10 free NEET Chemistry questions on Haloalkanes and Haloarenes, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Reaction Sequence - Anti-Markovnikov Addition, Nucleophilic Substitution, Reduction · Classification and Nomenclature · Methods of Preparation · SN1 and SN2 Reactions · Optical Activity and Stereochemistry · Reactions of Haloarenes · Reactions – Elimination vs Substitution · Monochlorination - Free Radical Mechanism and Isomers · Type of Reaction in Toluene's Halogenation with FeCl3 · Identifying a vic-Dihalide · Classifying Ethylidene Chloride · Preferred Mechanism for Primary Alkyl Halides

1. Q1NEET 2026Stereochemistry of Anti Addition (Halogenation) vs Syn Addition (Dihydroxylation) to Alkenes

   Given below are two statements: Statement-I: trans-But-2-ene upon treatment with Br2 in CCl4 gives the dl (chiral, racemic pair) product shown in a Newman projection with the two Br atoms anti to each other and the methyl/H groups arranged accordingly. Statement-II: cis-But-2-ene upon treatment with alkaline KMnO4 gives the meso product shown in a Newman projection with the two OH groups arranged accordingly. In the light of the above statements, choose the most appropriate answer from the options given below.

   • A. Both Statement I and Statement II are incorrect
   • B. Statement I is correct but Statement II is incorrect
   • C. Statement I is incorrect but Statement II is correct
   • D. Both Statement I and Statement II are correct
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   Answer: B — Statement I is correct but Statement II is incorrect

   Br2/CCl4 addition proceeds via anti addition (through a bromonium ion intermediate). For anti addition to a TRANS-alkene, the product is the MESO compound (not the dl/chiral pair), so Statement-I's depicted product is incorrect. Alkaline KMnO4 dihydroxylation proceeds via syn addition (both OH groups added to the same face). For syn addition to a CIS-alkene, the product is the MESO compound, so Statement-II correctly depicts the meso-2,3-butanediol product. Hence Statement-I is incorrect but Statement-II is correct.

2. Q2NEET 2026Stability of Benzylic Carbocation - Resonance and Hyperconjugation

   The following carbocation, Ph-C⁺(H)-CH3 (a benzylic carbocation with the positive charge on the carbon attached to the phenyl ring and bearing a methyl group), is stabilized by the interaction of the empty p orbital with:

   • A. empty σ and empty π* orbitals
   • B. empty σ* and filled π orbitals
   • C. empty σ* and empty π* orbitals
   • D. filled σ and filled π orbitals
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   Answer: C — empty σ* and empty π* orbitals

   The empty p orbital on the benzylic carbocation is stabilized through two simultaneous electron-donating interactions: resonance (conjugation) with the filled π system of the attached phenyl ring, and hyperconjugation with the filled σ (C-H) bonds of the adjacent methyl group. Both of these are filled orbitals donating electron density into the empty p orbital, stabilizing the positive charge.

3. Q3NEET 2025Reaction Sequence - Anti-Markovnikov Addition, Nucleophilic Substitution, Reduction

   Predict the major product 'P' in the following sequence of reactions: 1-methylcyclopentene treated with (i) HBr, benzoyl peroxide (ii) KCN (iii) Na(Hg)/C2H5OH

   • A. 2-methylcyclopentyl-CH2NH2 (1,2-disubstituted, methyl and CH2NH2 adjacent)
   • B. 1-methyl-1-(CH2NH2)cyclopentane (both substituents on same carbon)
   • C. 2-methylcyclopentyl-NC (methyl and NC adjacent)
   • D. 1-(NC)-1-(CH3)cyclopentane (both substituents on same carbon)
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   Answer: —

   HBr with benzoyl peroxide (peroxide effect) adds anti-Markovnikov, placing Br on the carbon away from the methyl-bearing carbon, giving 2-methylcyclopentyl bromide. KCN does SN2 substitution of Br by CN. Na(Hg)/C2H5OH reduces the nitrile (CN) to CH2NH2. Final product: 2-methylcyclopentyl-CH2NH2.

4. Q4NEET 2025Monochlorination - Free Radical Mechanism and Isomers

   How many products (including stereoisomers) are expected from monochlorination of 2-methylbutane (H3C)2CH-CH2-CH3?

   • A. 2
   • B. 3
   • C. 5
   • D. 6
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   Answer: C — 5

   Monochlorination of 2-methylbutane at different positions: C1/C1' (two equivalent terminal methyls on the isopropyl part, 1 product), C2 (chiral centre created, giving a racemic pair - but counted structurally as 1 position with a new stereocenter giving 2 stereoisomers since it becomes chiral), C3 (chiral centre, 2 stereoisomers), and C4 (1 product). Total structural+stereo count = 6.

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5. Q5NEET 2023Classification of Halides

   The given compound CH=CH-CH(X)-CH2-CH3 attached to a benzene ring (phenyl group on the terminal alkene carbon, with halogen X on the carbon adjacent to the double bond) is an example of ________.

   • A. aryl halide
   • B. allylic halide
   • C. vinylic halide
   • D. benzylic halide
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   Answer: A — aryl halide

   The halogen X is attached to the carbon directly adjacent to a carbon-carbon double bond (the allylic carbon), not to the double bond carbon itself or to the aromatic ring. Hence this compound is an example of an allylic halide.

6. Q6NEET 2023Reaction of Alcohols with HBr (SN1 with Rearrangement)

   Consider the following reaction and identify the product (P). 3-Methylbutan-2-ol (CH3-CH(CH3)-CH(OH)-CH3) reacts with HBr to give Product (P).

   • A. CH3-CH=CH-CH3
   • B. CH3-CH(CH3)-CH(Br)-CH3 (2-bromo-3-methylbutane)
   • C. CH3-C(CH3)2-CH2Br (1-bromo-2,2-dimethylpropane)
   • D. CH3-C(CH3)(Br)-CH2-CH3 (2-bromo-2-methylbutane)
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   Answer: C — CH3-C(CH3)2-CH2Br (1-bromo-2,2-dimethylpropane)

   HBr protonates the OH to form a secondary carbocation after loss of water. This secondary carbocation undergoes a hydride (H-) shift to form a more stable tertiary carbocation, which is then attacked by Br- to give 2-bromo-2-methylbutane as the major product (P), via the classic SN1 rearrangement pathway.

7. Q7NEET 2023Reactions via Diazonium Salts and Grignard Reagents

   Identify the product in the following reaction: Benzenediazonium chloride (C6H5-N2+Cl-) is treated with (i) Cu2Br2/HBr, (ii) Mg/dry ether, (iii) H2O, to give Product.

   • A. Benzene
   • B. Phenylmagnesium bromide (C6H5MgBr)
   • C. 4-Bromophenol
   • D. Phenol
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   Answer: —

   Step (i) converts the diazonium salt to bromobenzene via a Sandmeyer-type reaction with Cu2Br2/HBr. Step (ii) converts bromobenzene to phenylmagnesium bromide (a Grignard reagent) using Mg in dry ether. Step (iii) quenches the Grignard reagent with water, protonating the C-Mg bond to give benzene (along with Mg(OH)Br) as the final product.

8. Q8NEET 2022Preparation of Haloarenes

   Which of the following is suitable to synthesize chlorobenzene?

   • A. Phenol, NaNO2, HCl, CuCl
   • B. Benzene with HCl directly
   • C. Aniline (benzene with NH2 group) treated with HCl and heating
   • D. Benzene, Cl2, anhydrous FeCl3
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   Answer: C — Aniline (benzene with NH2 group) treated with HCl and heating

   Chlorobenzene is synthesized by electrophilic aromatic substitution of benzene with Cl2 in the presence of anhydrous FeCl3 as catalyst (Friedel-Crafts type halogenation).

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9. Q9NEET 2022Chirality and SN1/SN2 Reactions

   The incorrect statement regarding chirality is:

   • A. SN1 reaction yields 1:1 mixture of both enantiomers
   • B. The product obtained by SN2 reaction of haloalkane having chirality at the reactive site shows inversion of configuration
   • C. Enantiomers are superimposable mirror images on each other
   • D. A racemic mixture shows zero optical rotation
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   Answer: B — The product obtained by SN2 reaction of haloalkane having chirality at the reactive site shows inversion of configuration

   Enantiomers are actually non-superimposable mirror images of each other, not superimposable, making statement 3 incorrect. SN1 gives racemic (1:1) mixtures, SN2 gives inversion of configuration, and racemic mixtures show zero net optical rotation - all correct statements.

10. Q10NEET 2022Synthesis of Chlorobenzene

    Which of the following sequence of reactions is suitable to synthesize chlorobenzene?

    • A. Benzene, Cl2, anhydrous FeCl3
    • B. Phenol, NaNO2, HCl, CuCl
    • C. Benzene, HCl
    • D. Aniline, HCl, Heating
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    Answer: —

    Chlorobenzene is best synthesized by direct electrophilic aromatic substitution: benzene reacting with Cl2 in the presence of anhydrous FeCl3 (Lewis acid catalyst), which generates the Cl+ electrophile for substitution onto the ring.