Organic Compounds Containing Halogens — NEET Chemistry MCQs with answers

Organic Compounds Containing Halogens is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 612 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Preparation and properties of haloalkanes · Nucleophilic substitution SN1/SN2 · Elimination reactions · Haloarenes · Haloalkanes & Haloarenes · Haloalkanes and haloarenes · Polyhalogen compounds and applications · Nucleophilic Substitution · Reactions with Metals · Optical Isomerism · Classification · Nomenclature · Nature of C-X bond · Haloarenes · SN2 mechanism · SN1 mechanism · SN1 vs SN2 · Preparation of haloalkanes · Preparation of haloarenes · Named reactions · Polyhalogen compounds · Elimination vs substitution · Ambident nucleophiles · Physical properties · Optical activity and stereochemistry · Haloalkanes · SN2 reaction · Preparation · Grignard reagent · Elimination reactions · Optical activity · Nucleophilic Substitution · Grignard Reagent · Polyhalogen Compounds · Physical Properties · Preparation of Haloalkanes · Optical Activity · SN1/SN2 kinetics · Optical rotation · SN2 kinetics

8 free Organic Compounds Containing Halogens practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Haloarenes

   Chlorobenzene resists nucleophilic substitution under conditions that readily convert chloroethane to ethanol. The main reason is that in chlorobenzene:

   • A.chlorobenzene is non-polar and therefore nucleophiles cannot approach it
   • B.the C–Cl bond has partial double-bond character due to resonance with the ring, making it shorter and stronger
   • C.chlorine is too electronegative to leave as a chloride ion under any conditions
   • D.the benzene ring sterically blocks nucleophilic attack at the carbon bearing chlorine
   Show answer and explanation

   Answer: B. the C–Cl bond has partial double-bond character due to resonance with the ring, making it shorter and stronger

   Lone pair delocalisation from Cl into the ring gives partial double bond character to C–Cl, shortening and strengthening the bond and disfavouring both SN1 (destabilised cation via resonance loss) and SN2 (sp2 carbon, back-side attack blocked by ring).

   In chlorobenzene, one lone pair on chlorine conjugates with the aromatic π system, giving resonance structures with C=Cl⁺ character. This (i) shortens the C–Cl bond (169 pm vs 177 pm in haloalkanes) and increases its bond dissociation energy, (ii) makes the ring carbon partially sp2-hybridised with less s-character available for a good leaving group departure, and (iii) makes the carbon less electrophilic since the ring donates electron density into it, repelling nucleophiles. SN2 is also geometrically impossible since backside attack would require approaching through the aromatic ring plane. Hence ordinary aqueous or alcoholic nucleophiles cannot displace Cl from chlorobenzene, unlike in haloalkanes.

   Common mistake: Attributing low reactivity purely to steric hindrance from the ring rather than resonance-based C–Cl bond strengthening

2. Question 2 · medium · Haloarenes

   Chlorobenzene is treated with Br₂ in the presence of anhydrous FeBr₃. The major product formed is:

   • A.no reaction occurs because chlorine deactivates the ring completely toward electrophilic attack
   • B.1-bromo-2-chlorobenzene and 1-bromo-4-chlorobenzene, with the para isomer predominating
   • C.1-bromo-3-chlorobenzene as the major product, since Cl is deactivating and hence meta-directing
   • D.a mixture of all three (ortho, meta, para) isomers in roughly equal amounts, since Cl has no directing preference
   Show answer and explanation

   Answer: B. 1-bromo-2-chlorobenzene and 1-bromo-4-chlorobenzene, with the para isomer predominating

   Chlorine is deactivating by induction (withdraws σ electron density, slowing the reaction overall) but ortho/para-directing by resonance (lone pair donation stabilises the arenium ion at ortho/para positions), so bromination gives mainly ortho and para dichloro-bromo products with para favoured due to less steric strain.

   Halogens on benzene are unique among substituents: they deactivate the ring (net electron withdrawal, inductive effect dominates overall reactivity) yet direct incoming electrophiles ortho/para (resonance donation of a lone pair into the ring stabilises the arenium ion intermediate specifically when the positive charge falls on the carbon bearing halogen's neighbour). In FeBr3-catalysed bromination of chlorobenzene, Br⁺ (generated from Br2 and FeBr3) attacks predominantly at ortho and para positions relative to Cl. Para product is typically the major isomer due to lower steric hindrance compared to ortho. Reaction data: substrate chlorobenzene, reagent Br2, conditions anhydrous FeBr3 catalyst, product mixture of o- and p-bromochlorobenzene (para major).

   Common mistake: Assuming a deactivating group must also be meta-directing

3. Question 3 · easy · Haloarenes

   Why does chlorobenzene undergo electrophilic substitution more slowly than benzene, even though the halogen directs incoming electrophiles to specific ring positions?

   • A.The inductive electron withdrawal by chlorine lowers electron density on the ring overall, outweighing the resonance donation at ortho/para positions
   • B.Chlorine's large size sterically blocks electrophiles from approaching any ring position
   • C.Chlorine withdraws electrons only inductively and does not donate any electron density by resonance at all
   • D.The C–Cl bond dissociates spontaneously in the presence of electrophiles, forming a less reactive intermediate
   Show answer and explanation

   Answer: A. The inductive electron withdrawal by chlorine lowers electron density on the ring overall, outweighing the resonance donation at ortho/para positions

   Chlorine simultaneously withdraws electron density inductively (through the sigma bond) and donates electron density by resonance (lone pair into the ring); the inductive effect dominates in magnitude, giving overall ring deactivation, while the resonance effect still directs substitution to ortho/para positions where it is felt most.

   Halogens present a classic dual electronic effect. The strong electronegativity of Cl withdraws sigma-bond electron density from the ring (inductive, -I effect), which lowers the overall electron density available to stabilise the transition state/arenium ion for any electrophilic attack, making the whole ring less reactive than benzene. Simultaneously, one of chlorine's lone pairs conjugates into the pi system (+M/resonance effect), donating electron density that is felt specifically at the ortho and para positions, stabilising an arenium ion generated by attack there more than at meta. Since the inductive withdrawal outweighs the resonance donation in net magnitude, the overall rate is slower than benzene (deactivation), but the positions that are attacked are still ortho/para (directing effect governed by where resonance donation helps most).

   Common mistake: Believing resonance donation alone should make chlorobenzene more reactive than benzene

4. Question 4 · medium · Haloarenes

   A student argues: "Since bromobenzene has a weaker C–Br bond than the C–Cl bond in chlorobenzene, bromobenzene should undergo SN1 hydrolysis with aqueous NaOH just like tert-butyl bromide does." This reasoning is flawed because:

   • A.an aryl cation formed by loss of Br⁻ from an sp2 carbon would be highly unstable (empty orbital in the sp2 plane, not conjugated with the ring pi system), so SN1 at an aromatic carbon is not feasible regardless of bond strength
   • B.bromobenzene actually does undergo SN1 hydrolysis readily, so the student's conclusion, though for the wrong reason, is correct
   • C.SN1 reactions never occur on any halogen-substituted carbon, whether aromatic or aliphatic
   • D.bromobenzene's C–Br bond is actually stronger than chlorobenzene's C–Cl bond, so the premise itself is wrong
   Show answer and explanation

   Answer: A. an aryl cation formed by loss of Br⁻ from an sp2 carbon would be highly unstable (empty orbital in the sp2 plane, not conjugated with the ring pi system), so SN1 at an aromatic carbon is not feasible regardless of bond strength

   Although weaker bond dissociation energy generally favours easier heterolysis, an aryl carbocation would form in an sp2 orbital lying in the plane of the ring, unable to conjugate with the aromatic pi system — this makes it far too unstable to form, so bond strength comparisons from alkyl systems do not transfer to aryl halides.

   The student correctly notes C–Br bonds are weaker than C–Cl bonds on average, but incorrectly extends alkyl-halide SN1 reasoning to aryl halides. In tert-butyl bromide, ionisation gives a planar sp2 carbocation stabilised by hyperconjugation/adjacent alkyl groups. In bromobenzene, loss of Br⁻ would require forming a cation at an aromatic ring carbon; the resulting empty orbital would lie in the plane of the ring (sp2, in-plane), completely orthogonal to and unconjugated with the delocalised aromatic pi system. Such a phenyl cation is extremely high in energy and does not form under normal conditions. Therefore bromobenzene, like chlorobenzene, resists both SN1 and SN2 substitution under ordinary aqueous NaOH conditions; only strong ortho/para-activating groups (via SNAr) or forcing conditions (high T/P with Cu catalyst, or via benzyne mechanism with very strong base) enable substitution.

   Common mistake: Extrapolating SN1 feasibility from alkyl to aryl halides based on bond dissociation energy alone

5. Question 5 · easy · Haloarenes

   Which statement correctly compares the C–X bond length in chlorobenzene versus 1-chlorobutane?

   • A.The C–Cl bond in chlorobenzene is longer than in 1-chlorobutane because the ring pulls electron density away
   • B.Both bonds have identical length since both involve an sp3 carbon bonded to chlorine
   • C.Bond length cannot be compared because chlorobenzene has no discrete C–Cl bond due to full delocalisation of chlorine's electrons into the ring
   • D.The C–Cl bond in chlorobenzene is shorter than in 1-chlorobutane
   Show answer and explanation

   Answer: D. The C–Cl bond in chlorobenzene is shorter than in 1-chlorobutane

   The C–Cl bond in chlorobenzene (sp2 ring carbon, resonance-assisted partial double bond character) is shorter (about 169 pm) than the C–Cl bond in a haloalkane like 1-chlorobutane (sp3 carbon, pure single bond, about 177-179 pm).

   In chlorobenzene, the carbon bonded to chlorine is sp2 hybridised (part of the aromatic ring), and resonance delocalisation of a chlorine lone pair into the ring pi system imparts partial double-bond character to the C–Cl bond. Both factors — greater s-character in sp2 hybrid orbitals and resonance double-bond character — act to shorten and strengthen this bond relative to the C–Cl bond in an alkyl chloride such as 1-chlorobutane, where the carbon is sp3 and there is no resonance donation. Experimentally, C(sp2)-Cl in chlorobenzene is about 169 pm, whereas C(sp3)-Cl in a typical alkyl chloride is about 177-179 pm.

   Common mistake: Assuming ring electron withdrawal by induction would lengthen rather than shorten the bond

6. Question 6 · medium · Haloarenes

   A chemist wants to convert p-nitrochlorobenzene into p-nitroaniline in a single practical step. Which reagent/condition should be used, and why does it work despite chlorobenzene itself being inert to the same reagent?

   • A.NH₃ under pressure/heat; the para-NO₂ group activates the ring toward SNAr by stabilising the Meisenheimer intermediate formed on NH₃ addition
   • B.NaNH₂ in liquid NH₃ via a benzyne intermediate, exactly as with plain chlorobenzene
   • C.Zn/HCl reduction directly converts the C–Cl bond into a C–NH₂ bond
   • D.Concentrated H₂SO₄ followed by NH₃ workup substitutes Cl for NH₂ via a sulfonation intermediate
   Show answer and explanation

   Answer: A. NH₃ under pressure/heat; the para-NO₂ group activates the ring toward SNAr by stabilising the Meisenheimer intermediate formed on NH₃ addition

   Because the p-NO2 group can delocalise the negative charge of the Meisenheimer intermediate formed when NH3 adds to the ring carbon bearing Cl, p-nitrochlorobenzene undergoes direct nucleophilic aromatic substitution with NH3 (heat/pressure) to give p-nitroaniline, while unactivated chlorobenzene requires much harsher conditions (very high T/P with Cu catalyst) for the same conversion.

   p-Nitrochlorobenzene reacts with excess ammonia under moderate heat and pressure via the addition-elimination (SNAr) mechanism: NH3 attacks the ring carbon bearing Cl, the resulting carbanion intermediate is significantly stabilised because negative charge can delocalise onto the oxygens of the para-NO2 group, and Cl⁻ is then eliminated to restore aromaticity, giving p-nitroaniline. Plain chlorobenzene lacks this stabilisation, so ammonolysis under similar mild conditions does not proceed (industrially requires ~473 K, high pressure, Cu2O catalyst for chlorobenzene, versus mild reflux conditions sufficing for the nitro-activated substrate). Reaction data: substrate p-nitrochlorobenzene, reagent NH3, conditions heat/pressure (SNAr), product p-nitroaniline.

   Common mistake: Assuming the benzyne mechanism (used for unactivated aryl halides with strong base) also applies here

7. Question 7 · medium · Haloarenes

   Chlorobenzene reacts with CH₃COCl in the presence of anhydrous AlCl₃ (Friedel-Crafts acylation). The major product is:

   • A.1-chloro-4-acetylbenzene (p-chloroacetophenone) as the predominant product
   • B.1-chloro-3-acetylbenzene (m-chloroacetophenone) as the predominant product
   • C.the reaction does not proceed at all since AlCl₃ decomposes chlorobenzene before acylation can occur
   • D.a mixture where meta product dominates because Friedel-Crafts acylation always favours the least hindered, most electron-poor position
   Show answer and explanation

   Answer: A. 1-chloro-4-acetylbenzene (p-chloroacetophenone) as the predominant product

   Friedel-Crafts acylation is a standard electrophilic aromatic substitution; Cl directs the acylium electrophile (CH3CO+) to ortho/para positions, and the bulky acyl group combined with steric hindrance from ortho-Cl favours the para product as major.

   AlCl3 coordinates with CH3COCl to generate the electrophilic acylium ion, CH3CO+, which attacks the chlorobenzene ring. Despite being deactivating overall (net electron withdrawal via induction), chlorine remains ortho/para-directing due to resonance donation of a lone pair stabilising the arenium ion at those positions. Given the bulk of the incoming acetyl group and existing steric hindrance near the ortho position (adjacent to Cl), the para product, p-chloroacetophenone (1-chloro-4-acetylbenzene), dominates over the ortho isomer, with negligible meta product. Reaction data: substrate chlorobenzene, reagent CH3COCl, conditions anhydrous AlCl3 catalyst, product predominantly p-chloroacetophenone.

   Common mistake: Assuming meta substitution because Cl is described as a 'deactivating' group

8. Question 8 · medium · Haloarenes

   Bromobenzene is nitrated using a mixture of concentrated HNO₃ and concentrated H₂SO₄. The major product formed is:

   • A.m-bromonitrobenzene, since bromine strongly deactivates ortho/para positions specifically
   • B.a 1:1:1 statistical mixture of ortho, meta and para bromonitrobenzene
   • C.p-bromonitrobenzene as the major product, along with a smaller amount of the ortho isomer
   • D.o-bromonitrobenzene exclusively, with no para isomer formed due to complete steric blocking by bromine
   Show answer and explanation

   Answer: C. p-bromonitrobenzene as the major product, along with a smaller amount of the ortho isomer

   Bromine, like other halogens, is deactivating overall (inductive withdrawal) but ortho/para-directing (resonance donation of a lone pair stabilises the arenium ion at those positions); nitration of bromobenzene therefore gives mainly p-bromonitrobenzene with some o-bromonitrobenzene, and only a trace of meta.

   The nitronium ion, NO2+, generated from HNO3/H2SO4, is the attacking electrophile. Bromine on the ring withdraws electron density inductively (deactivating the whole ring relative to benzene, so the reaction is slower) but donates a lone pair by resonance specifically stabilising the arenium ion intermediate when attack occurs ortho or para to it. Because the para position experiences less steric interference from the bulky bromine substituent than the ortho position, p-bromonitrobenzene is the major product, with o-bromonitrobenzene as a significant minor product and only trace amounts of the meta isomer. Reaction data: substrate bromobenzene, reagent conc. HNO3/conc. H2SO4, conditions controlled temperature (below 60°C) to avoid polynitration, product p-bromonitrobenzene (major) + o-bromonitrobenzene (minor).

   Common mistake: Assuming deactivating substituents must be meta-directors as a blanket rule

Questions about Organic Compounds Containing Halogens for NEET

More Chemistry chapters

← BiomoleculesAll Chemistry chaptersHydrocarbons →

Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (612 active practice questions in this chapter today).