Haloalkanes and Haloarenes NEET PYQs (2005–2026) with Solutions

10 solved Haloalkanes and Haloarenes PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumCyanides, isocyanides and carbylamine reaction

   The following two reactions give the same foul smelling product Z. X and Z, respectively, are:

   1. AOption A: X = AgCN; Z = C₂H₅NC
   2. BOption B: X = KCN; Z = C₂H₅CN
   3. COption C: X = AgCN; Z = C₂H₅CN
   4. DOption D: X = KCN; Z = C₂H₅NC

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   Correct answer: (A) X = AgCN; Z = C₂H₅NC

   Explanation

   AgCN with alkyl halides gives isocyanides (R–NC), which have foul smell. Carbylamine reaction also forms isocyanides. Thus Z=C₂H₅NC and X=AgCN. Hence option (A) is correct.

2. Question 2 (NEET 2026)

   NEET 2026MediumReaction mechanisms involving alcohols and haloalkanes

   In the following reaction sequence, X and Z respectively are:

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (B) Option 2

   Explanation

   CH₃CH₂CH₂OH + PCl₅ → CH₃CH₂CH₂Cl + POCl₃ + HCl, so X=POCl₃. Elimination with alcoholic KOH gives propene. Addition of HBr in peroxide follows anti-Markovnikov rule, giving CH₃CH₂CH₂Br. Hence option (B) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumGrignard reagents with acidic hydrogen compounds

   One of the products formed in the following reaction is $\mathrm{C_6H_{11}MgBr} + \mathrm{C_6H_{11}NH_2} \longrightarrow$ (cyclohexylmagnesium bromide + cyclohexylamine)

   1. AOption A: Structure A
   2. BOption B: Structure B
   3. COption C: Structure C
   4. DOption D: Structure D

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   Correct answer: (D) Structure D

   Explanation

   The options show: (A) dicyclohexylamine, $\mathrm{(C_6H_{11})_2NH}$; (B) 2-cyclohexylcyclohexan-1-amine (two linked cyclohexane rings with $\mathrm{NH_2}$ on the ring carbon next to the link); (C) bicyclohexyl (two cyclohexane rings joined directly); (D) cyclohexane. Grignard reagents are very strong bases. Any compound with an acidic hydrogen (water, alcohols, amines) converts them into the hydrocarbon: $$\mathrm{RMgX + R'NH_2 \rightarrow R{-}H + R'NHMgX}$$ Here the N–H hydrogen of cyclohexylamine protonates the cyclohexyl carbanion: $$\mathrm{C_6H_{11}MgBr + C_6H_{11}NH_2 \rightarrow C_6H_{12} + C_6H_{11}NHMgBr}$$ So one product is cyclohexane, Structure D, option (D). No C–N or C–C coupling products (A, B, C) form.

4. Question 4 (NEET 2026)

   NEET 2026MediumDiazotisation and reactions of amines

   The major product Z formed in the following sequence of reactions is:

   1. AOption A: C₂H₅NO₂
   2. BOption B: C₂H₅–N=N–OH
   3. COption C: C₂H₅OH
   4. DOption D: C₂H₅NH₂

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   Correct answer: (C) C₂H₅OH

   Explanation

   Ethane gives chloroethane under UV chlorination, which reacts with NH₃ to form ethylamine. Diazotisation followed by hydrolysis converts ethyl diazonium salt to ethanol. Hence Z=C₂H₅OH and option (C) is correct.

5. Question 5 (NEET 2025)

   NEET 2025MediumChemical Reactions & Mechanisms

   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). In the light of the above statements, choose the correct answer from the options given below :

   1. AOption A: Both A and R are true and R is the correct explanation of A
   2. BOption B: Both A and R are true but R is not the correct explanation of A
   3. COption C: A is true but R is false
   4. DOption D: A is false but R is true

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   Correct answer: (A) Both A and R are true and R is the correct explanation of A

   Explanation

   In SN2 reactions, the rate depends on the leaving group ability. Iodide ion is a better leaving group than chloride ion because of its larger size and weaker C-I bond. Therefore, iodoalkanes undergo SN2 reactions faster than chloroalkanes.

6. Question 6 (NEET 2024)

   NEET 2024MediumChemical Reactions & Mechanisms

   The compound that will undergo Sₙ1 reaction with the fastest rate is:

   1. AOption A: Compound (1)
   2. BOption B: Compound (2)
   3. COption C: Compound (3)
   4. DOption D: Compound (4)

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   Correct answer: (B) Compound (2)

   Explanation

   SN1 reactions proceed through carbocation formation. The more stable the carbocation, the faster the SN1 reaction. Compound (2) forms a benzylic secondary carbocation which is highly stabilized by resonance and hyperconjugation. Therefore it undergoes SN1 reaction fastest. Hence, option (B) is correct.

7. Question 7 (NEET 2023)

   NEET 2023MediumClassification & Nomenclature

   The given compound is an example of ________.

   1. AOption A: benzylic halide
   2. BOption B: aryl halide
   3. COption C: allylic halide
   4. DOption D: vinylic halide

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   Correct answer: (C) allylic halide

   Explanation

   The halogen atom X is attached to the carbon adjacent to a carbon-carbon double bond. Such compounds are called allylic halides.

8. Question 8 (NEET 2022)

   NEET 2022MediumChemical Reactions & Mechanisms

   The incorrect statement regarding chirality is:

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

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   Correct answer: (D) Enantiomers are superimposable mirror images on each other

   Explanation

   Enantiomers are non-superimposable mirror images, not superimposable. Therefore statement (4) is incorrect and is the correct answer.

9. Question 9 (NEET 2022)

   NEET 2022MediumPreparation Methods

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

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (B) Option 2

   Explanation

   Chlorobenzene is prepared by electrophilic aromatic substitution: C₆H₆ + Cl₂ → C₆H₅Cl using anhydrous FeCl₃ catalyst. Hence option (B) is correct.

10. Question 10 (NEET 2016)

    NEET 2016MediumFriedel-Crafts Reactions

    Which of the following can be used as the halide component for Friedel-Crafts reaction?

    1. AOption A: Chlorobenzene
    2. BOption B: Bromobenzene
    3. COption C: Chloroethene
    4. DOption D: Isopropyl chloride

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    Correct answer: (D) Isopropyl chloride

    Explanation

    Friedel-Crafts alkylation requires formation of a stable carbocation. Isopropyl chloride forms relatively stable secondary carbocation. Aryl and vinyl halides do not undergo Friedel-Crafts reactions easily. Hence option (D) is correct.