Haloalkanes and Haloarenes NEET PYQReal NEET & AIPMT 2007–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Haloalkanes and Haloarenes question from the NEET & AIPMT papers on this site (2007–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Haloalkanes and Haloarenes previous year questions
19 questions. Tap an option to check your answer.
Q1NEET 2021
The major product formed in dehydrohalogenation reaction of 2-Bromo pentane is Pent-2-ene. This product formation is based on ?
Not quite — the answer is A.
In base-induced dehydrohalogenation, the more substituted (more stable) alkene is the main product, which is Saytzeff's rule. Pent-2-ene is more substituted than pent-1-ene. Hofmann's rule predicts the less substituted alkene and applies to bulky bases or quaternary ammonium bases.
Q2NEET 2021
The correct sequence of bond enthalpy of 'C–X' bond is :
Not quite — the answer is B.
As the halogen gets larger (F < Cl < Br < I), the C–X bond becomes longer and overlap poorer, so bond enthalpy decreases: CH₃–F > CH₃–Cl > CH₃–Br > CH₃–I. This is why C–I bonds break most easily.
Q3NEET 2020 · Q95
Identify compound X in the following sequence of reactions : Toluene (C₆H₅CH₃) —Cl₂/hν→ X —H₂O, 373 K→ Benzaldehyde (C₆H₅CHO)
Not quite — the answer is C.
Side-chain chlorination of toluene in light gives benzal chloride, C₆H₅CHCl₂; hydrolysis at 373 K gives the gem-diol C₆H₅CH(OH)₂, which loses water to form benzaldehyde. Benzyl chloride would hydrolyse to benzyl alcohol and benzotrichloride to benzoic acid.
Q4NEET 2020 · Q114
Elimination reaction of 2-Bromo-pentane to form pent-2-ene is : (a) β-Elimination reaction (b) Follows Zaitsev rule (c) Dehydrohalogenation reaction (d) Dehydration reaction
Not quite — the answer is A.
HBr is removed from the α-carbon and a β-hydrogen, so it is a β-elimination and a dehydrohalogenation, and the more substituted pent-2-ene is the major product as Zaitsev's rule predicts. No water is lost, so calling it dehydration is wrong.
Q5NEET 2019 · Q65
Among the following, the reaction that proceeds through an electrophilic substitution, is:
Not quite — the answer is B.
AlCl₃ polarises Cl₂ to generate the electrophile Cl⁺, which attacks the benzene ring: electrophilic aromatic substitution. The Sandmeyer reaction is a radical process, UV chlorination is free-radical addition, and benzyl alcohol with HCl is nucleophilic substitution.
Q6NEET 2018 · Q61
The compound C₇H₈ undergoes the following reactions: C₇H₈ —(3Cl₂/Δ)→ A —(Br₂/Fe)→ B —(Zn/HCl)→ C. The product 'C' is
Not quite — the answer is C.
Toluene with Cl₂ under heat (side-chain chlorination) gives C₆H₅CCl₃. The –CCl₃ group is electron-withdrawing and meta-directing, so Br₂/Fe gives m-bromobenzotrichloride; Zn/HCl reduces –CCl₃ back to –CH₃, giving m-bromotoluene. Direct bromination of toluene would give the ortho/para products.
Q7NEET 2017 (Gujarati paper) · Q40
Which of the following will react faster through Sɴ1 mechanism?
Not quite — the answer is A.
Sɴ1 rate depends on carbocation stability. Allyl chloride ionises to the resonance-stabilised allyl cation, so it reacts fastest. Ethyl chloride would give an unstable primary cation, and vinyl and aryl chlorides have a C–Cl bond with partial double-bond character and do not ionise easily.
Q8NEET 2016 Phase 2 · Q76
Which of the following can be used as the halide component for Friedel-Crafts reaction?
Not quite — the answer is D.
Friedel–Crafts alkylation needs a halide that forms a carbocation with AlCl₃, and isopropyl chloride readily gives the isopropyl cation. In chlorobenzene, bromobenzene and chloroethene the C–X bond has partial double-bond character (sp² carbon), so aryl and vinyl cations are not formed.
Q9NEET 2016 Phase 2 · Q86
Consider the reaction CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr. This reaction will be the fastest in
Not quite — the answer is C.
This is an S_N2 reaction of a primary bromide. Polar aprotic solvents such as DMF solvate Na⁺ but leave CN⁻ free and highly nucleophilic, so the rate is greatest. Protic solvents (water, methanol, ethanol) hydrogen-bond to CN⁻ and slow the reaction.
Q10NEET 2016 Phase 1 · Q85
For the following reactions: (a) CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O (b) (CH₃)₂CHBr + KOH → (CH₃)₂CHOH + KBr (c) Cyclohexene + Br₂ → 1,2-dibromocyclohexane Which of the following statements is correct?
Not quite — the answer is B.
In (a) HBr is removed to form an alkene (elimination); in (b) –OH replaces –Br on the same carbon (substitution); in (c) Br₂ adds across the C=C double bond (addition). Calling the second reaction an elimination is wrong because no double bond is formed.
Q11AIPMT 2011 Mains
Which of the following compounds undergoes nucleophilic substitution reaction most easily?
Not quite — the answer is C.
Aryl halides undergo nucleophilic substitution only when electron-withdrawing groups at ortho/para positions stabilise the negatively charged intermediate. A para –NO₂ does this strongly, so p-nitrochlorobenzene reacts most easily. –OCH₃ and –CH₃ donate electrons and make substitution harder than in chlorobenzene.
Q12AIPMT 2011 Prelims
Which one is a nucleophilic substitution reaction among the following?
Not quite — the answer is D.
In the reaction of the primary alkyl bromide with NH₃, the nucleophile NH₃ replaces Br⁻, which is nucleophilic substitution. The cyanohydrin and Grignard reactions are nucleophilic additions to C=O, and acid-catalysed hydration of propene is electrophilic addition.
Q13AIPMT 2010 Mains · Q120
In the following reaction: C₆H₅CH₂Br —(1. Mg, ether; 2. H₃O⁺)→ X the product 'X' is
Not quite — the answer is C.
Benzyl bromide with Mg in dry ether forms the Grignard reagent C₆H₅CH₂MgBr, and hydrolysis with H₃O⁺ replaces MgBr by H, giving toluene C₆H₅CH₃. Benzyl alcohol would need reaction with O₂ or formaldehyde, not plain hydrolysis; bibenzyl is a Wurtz-type side product.
Q14AIPMT 2010 Prelims · Q90
Which is most reactive towards SN1 reaction?
Not quite — the answer is D.
SN1 rate depends on carbocation stability. C₆H₅C(CH₃)(C₆H₅)Br gives a tertiary carbocation stabilised by two phenyl rings and a methyl group, the most stable of the four. Benzyl bromide gives only a primary benzylic cation.
Q15AIPMT 2009 Prelims · Q154
Which of the following reactions is an example of nucleophilic substitution reaction?
Not quite — the answer is D.
In RX + KOH(aq) → ROH + KX the nucleophile OH⁻ replaces the halide ion, a nucleophilic substitution. Reaction with Na is the Wurtz coupling, with H₂ is a reduction, and with Mg forms a Grignard reagent.
Q16AIPMT 2009 Prelims · Q198
Trichloroacetaldehyde, CCl₃CHO reacts with chlorobenzene in presence of sulphuric acid and produces:
Not quite — the answer is C.
In conc. H₂SO₄, chloral condenses with two molecules of chlorobenzene at their para positions, losing water, to give DDT: (p-Cl–C₆H₄)₂CH–CCl₃. The CCl₃ group stays intact; it is not replaced by a third ring.
Q17AIPMT 2008 Prelims · Q96
In a SN2 substitution reaction of the type R–Br + Cl⁻ —(DMF)→ R–Cl + Br⁻ Which one of the following has the highest relative rate?
Not quite — the answer is A.
SN2 needs backside attack at the carbon bearing Br, so it slows as branching near that carbon increases steric hindrance. Ethyl bromide is the least hindered, so it reacts fastest. The neopentyl-type (CH₃)₃C–CH₂Br, though primary, is extremely hindered and reacts slowest.
Q18AIPMT 2007 Prelims · Q69
If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it may mean that
Not quite — the answer is C.
A compound believed to be chiral can still show no rotation if it is present as a racemic mixture, where equal amounts of the two enantiomers cancel each other. 'Certainly meso' is too strong: a meso compound is not chiral at all, and a racemate explains the observation equally well.
Q19AIPMT 2007 Prelims · Q88
For (i) I⁻, (ii) Cl⁻, (iii) Br⁻, the increasing order of nucleophilicity would be
Not quite — the answer is A.
In protic solvents the larger, more polarisable iodide is least solvated and the best nucleophile, so nucleophilicity increases Cl⁻ < Br⁻ < I⁻. The reverse order applies to basicity, not nucleophilicity.
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