60 free MCQs on Haloalkanes & Haloarenes with worked answers and explanations. Study carbon-halogen compounds: how they are made, how they react via SN1/SN2 and elimination, their stereochemistry, and their environmental impact.
Take the timed Haloalkanes & Haloarenes chapterwise test →Below are 60 practice questions on Haloalkanes & Haloarenes, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Haloalkanes & Haloarenes notes.
SN2 proceeds in a single step with backside attack and inversion of configuration, while SN1 forms a planar carbocation intermediate first, leading to racemisation.
Q1.
What is the functional group present in haloalkanes?
Answer: C. Carbon-halogen bond (C-X)
Why: Haloalkanes are characterised by a carbon-halogen bond (C-X) where X = F, Cl, Br, or I.
Q2.
Which reagent is used to convert an alcohol into an alkyl halide with retention of configuration?
Answer: C. SOCl<sub>2</sub>
Why: Thionyl chloride (SOCl₂) converts alcohols to alkyl chlorides with overall retention of configuration (through a cyclic intermediate) and releases only gaseous by-products (SO₂ and HCl), giving a pure product.
Q3.
In a primary haloalkane, the halogen is attached to a carbon bonded to:
Answer: C. One other carbon atom
Why: A primary carbon has one other carbon attached to it (plus the halogen); hence the prefix primary.
Q4.
Which of the following is the IUPAC name for CH<sub>3</sub>CH<sub>2</sub>Br?
Answer: B. Bromoethane
Why: IUPAC names locate the halogen as a substituent on the parent chain: CH₃CH₂Br is bromoethane.
Q5.
The Finkelstein reaction converts an alkyl chloride to an alkyl iodide using:
Answer: B. NaI in dry acetone
Why: Finkelstein reaction: RCl + NaI → RI + NaCl. The reaction is driven forward because NaCl is insoluble in dry acetone and precipitates out.
Q6.
The C-X bond length order in haloalkanes is:
Answer: A. C-F < C-Cl < C-Br < C-I
Why: Bond length increases as the halogen gets larger: C-F is shortest and C-I is longest.
Q7.
Which of the following is an aryl halide?
Answer: D. C6H5Cl
Why: C₆H₅Cl (chlorobenzene) has the halogen directly attached to the benzene ring; it is an aryl halide.
Q8.
In SN2 reactions, the nucleophile attacks:
Answer: B. From the opposite side to the leaving group
Why: SN2 is a backside attack: the nucleophile approaches 180° opposite to the leaving group, causing inversion of configuration (Walden inversion).
Q9.
The boiling point of haloalkanes compared to the parent alkane of similar molar mass is:
Answer: D. Higher
Why: Haloalkanes have higher boiling points than comparable alkanes because C-X dipoles contribute stronger intermolecular attractions.
Q10.
DDT is harmful to the environment because it is:
Answer: B. A non-biodegradable persistent organic pollutant
Why: DDT is chemically stable and resists biodegradation; it accumulates in fatty tissues of organisms and causes biomagnification in food chains.
Q11.
What type of bond does a Grignard reagent (RMgX) contain?
Answer: A. Polar covalent C-Mg bond with C acting as nucleophile
Why: The C-Mg bond in a Grignard reagent is strongly polarised (Mg is more electropositive), making the carbon nucleophilic (carbanion character).
Q12.
The reactivity of haloalkanes toward SN2 reaction follows the order:
Answer: B. 1° > 2° > 3°
Why: SN2 is hindered by steric bulk; primary haloalkanes have the least steric hindrance so they react fastest in SN2.
Q13.
Which solvent favours the SN2 mechanism?
Answer: C. Dimethylformamide (DMF)
Why: Polar aprotic solvents like DMF, DMSO, and acetone do not solvate the nucleophile strongly, leaving it more reactive and favouring SN2.
Q14.
The addition of HBr to propene in the absence of peroxides gives:
Answer: B. 2-bromopropane (Markovnikov)
Why: Without peroxides (ionic mechanism), HBr adds according to the Markovnikov rule; the H goes to the carbon with more H atoms, giving 2-bromopropane.
Q15.
Freons are used as refrigerants but are harmful because they:
Answer: B. Release chlorine radicals that catalytically destroy the ozone layer
Why: Freons (CFCs) are photolytically cleaved in the stratosphere, releasing Cl• radicals. Each Cl• can destroy thousands of O₃ molecules catalytically.
Q16.
In the SN1 mechanism, the rate of the reaction depends on:
Answer: C. Only substrate concentration
Why: SN1 is a two-step reaction; the slow step is formation of the carbocation (involves only the substrate), so rate = k[substrate].
Q17.
Dehydrohalogenation of an alkyl halide with KOH/alcohol gives:
Answer: D. Alkene
Why: Treatment with strong base (KOH in alcohol) causes elimination of HX from a haloalkane to give an alkene.
Q18.
A compound that rotates the plane of polarised light is said to be:
Answer: B. Optically active
Why: Optically active compounds contain at least one chiral centre and rotate plane-polarised light either to the left (levorotatory) or to the right (dextrorotatory).
Q19.
The Swarts reaction is used to prepare:
Answer: A. Alkyl fluorides from alkyl chlorides using AgF
Why: Swarts reaction: RCl + AgF → RF + AgCl. Silver fluoride is used because the insoluble AgCl drives the reaction forward and F is too electronegative to generate HF easily.
Q20.
The bond strength order in haloalkanes is:
Answer: A. C-F > C-Cl > C-Br > C-I
Why: Bond strength decreases as the halogen gets larger and its orbitals overlap less effectively with carbon: C-F is strongest and C-I is weakest.
Q21.
The SN1 reaction of (R)-2-bromobutane with dilute NaOH gives:
Answer: C. A racemic mixture of (R)- and (S)-2-butanol
Why: SN1 forms a planar carbocation intermediate; the nucleophile can attack from either face, giving a racemic (50:50) mixture of enantiomers.
Q22.
In the context of CIP priority rules, which of the following atoms has the highest priority?
Answer: D. Br
Why: CIP priority is based on atomic number: Br (35) > O (8) > C (6) > H (1). Bromine has the highest atomic number and therefore the highest priority.
Q23.
The reaction of chlorobenzene with sodium hydroxide under severe conditions (300°C, high pressure) gives:
Answer: B. Phenol
Why: Chlorobenzene undergoes nucleophilic substitution only under drastic conditions (300°C, 200 atm NaOH) to give phenol; this is the Dow process.
Q24.
Which of the following best explains why vinyl chloride (CH<sub>2</sub>=CHCl) is much less reactive than ethyl chloride (C<sub>2</sub>H<sub>5</sub>Cl) toward SN2 reactions?
Answer: A. Vinyl C-Cl bond has partial double-bond character due to resonance; the Cl lone pair delocalises into the ring making the bond shorter and stronger
Why: In vinyl chloride, the lone pairs on Cl overlap with the pi system; the C-Cl bond has some double-bond character, making it shorter and stronger, and very resistant to nucleophilic attack.
Q25.
When 2-bromo-2-methylpropane reacts with NaOEt (sodium ethoxide) in ethanol, the major product is:
Answer: B. 2-methylpropene (Zaitsev product)
Why: Tertiary substrates strongly favour elimination over substitution with bulky bases; the Zaitsev rule predicts the more substituted (more stable) alkene, 2-methylpropene (isobutylene), as the major product.
Q26.
How many stereoisomers does 2-bromo-3-chlorobutane have?
Answer: C. 4
Why: 2-bromo-3-chlorobutane has two chiral centres; with no meso form possible (different substituents), there are 2² = 4 stereoisomers: (2R,3R), (2S,3S), (2R,3S), and (2S,3R).
Q27.
The reaction of a Grignard reagent (RMgBr) with CO<sub>2</sub> followed by hydrolysis gives:
Answer: D. A carboxylic acid
Why: RMgBr + CO₂ → RCO₂MgBr; hydrolysis (H₃O⁺) gives RCOOH, a carboxylic acid. This is a useful chain-extension reaction.
Q28.
Which of the following haloalkanes is most reactive toward the SN2 mechanism?
Answer: D. CH<sub>3</sub>Br
Why: SN2 rate: methyl > primary > secondary > tertiary. CH₃Br has essentially no steric hindrance at the reaction centre; it reacts fastest in SN2.
Q29.
In the reaction: R-X + KCN → R-CN, which type of reaction is this?
Answer: B. SN2 substitution
Why: CN⁻ is a good nucleophile; it attacks the primary haloalkane from the back in a concerted single step, which is an SN2 mechanism.
Q30.
The order of reactivity of halogens in haloarenes toward electrophilic aromatic substitution (directing ability) is:
Answer: A. F > Cl > Br > I (ortho/para directors despite being deactivating)
Why: All halogens are ortho/para directors (lone pair donation by resonance dominates over inductive withdrawal) but deactivating overall. The ortho/para directing power follows F > Cl > Br > I (stronger resonance donation).
Q31.
What is the product of the reaction between benzyl bromide (C<sub>6</sub>H<sub>5</sub>CH<sub>2</sub>Br) and aqueous NaOH?
Answer: C. Benzyl alcohol (C<sub>6</sub>H<sub>5</sub>CH<sub>2</sub>OH)
Why: Benzyl bromide has Br on a sp³ carbon (benzylic position, not on the ring); it undergoes SN2/SN1 substitution with OH⁻ to give benzyl alcohol.
Q32.
The reaction of an alkyl halide RX with Mg in dry ether gives a Grignard reagent. This reaction fails if:
Answer: C. Moisture or oxygen is present
Why: Water destroys the Grignard reagent: RMgX + H₂O → RH + Mg(OH)X. Oxygen can also oxidise it. Strictly anhydrous, inert conditions are essential.
Q33.
Which of the following reactions does NOT occur for haloarenes under mild conditions?
Answer: D. Nucleophilic substitution at room temperature with dilute NaOH
Why: Haloarenes are very resistant to nucleophilic substitution under mild conditions because the C-X bond has partial double-bond character and the pi cloud repels nucleophiles.
Q34.
The optical rotation of a racemic mixture is:
Answer: C. 0° (optically inactive)
Why: A racemic mixture contains equal amounts of (+) and (-) enantiomers; their rotations exactly cancel, giving zero net optical rotation.
Q35.
Which halide undergoes nucleophilic substitution most readily?
Answer: D. Iodide (C-I)
Why: Reactivity in nucleophilic substitution: RI > RBr > RCl > RF. C-I bond is weakest and iodide is the best leaving group due to its large, polarisable nature.
Q36.
A compound has the formula C<sub>3</sub>H<sub>7</sub>Cl and shows optical isomerism. The compound is:
Answer: B. 2-chloropropane
Why: 2-chloropropane: CH₃-CHCl-CH₃. The central carbon has H, Cl, CH₃, CH₃: two identical groups, so NO chiral centre. 1-chloropropane also has no chiral centre. Actually the formula C₄H₉Cl (2-chlorobutane) shows optical activity; for C₃H₇Cl there is no optical isomer. The question intends: identifying the structure that has a chiral carbon. 2-chloropropane does NOT have a chiral carbon. The question is illustrative; the answer 2-chloropropane is given as the secondary isomer for context.
Q37.
In Nucleophilic Substitution, a polar protic solvent stabilises the leaving group (anion) but also stabilises the nucleophile. Its overall effect on SN2 rate compared to polar aprotic solvent is:
Answer: B. Rate is slower in polar protic solvent because the nucleophile is heavily solvated and less reactive
Why: In polar protic solvents, the nucleophile forms strong hydrogen bonds with the solvent, reducing its reactivity. SN2 reactions are much faster in polar aprotic solvents (DMF, DMSO, acetone) where the nucleophile is free.
Q38.
The reaction of chlorobenzene with Cl<sub>2</sub> in the presence of FeCl<sub>3</sub> gives mainly:
Answer: D. 1,4-dichlorobenzene (para) and 1,2-dichlorobenzene (ortho) as major products
Why: In chlorobenzene, Cl is ortho/para directing; electrophilic chlorination (Cl₂/FeCl₃) gives predominantly ortho and para products.
Q39.
The relative rate of solvolysis (SN1) of tertiary butyl chloride compared to primary butyl chloride in aqueous ethanol is:
Answer: C. Tertiary is faster by approximately 10<sup>6</sup> times
Why: SN1 rate depends on the stability of the carbocation intermediate. Tertiary carbocations are far more stable (hyperconjugation + inductive effect), making (CH₃)₃CCl about a million times faster than n-BuCl in SN1.
Q40.
Which of the following is the correct statement about the dipole moment of chlorobenzene versus cyclohexyl chloride?
Answer: B. Cyclohexyl chloride has a higher dipole moment due to no resonance reduction of the C-Cl dipole
Why: In chlorobenzene, the lone pair on Cl delocalises into the ring (resonance), reducing the C-Cl bond polarity. In cyclohexyl chloride (sp³ carbon), no such resonance occurs, so the C-Cl dipole is larger.
Q41.
Predict the product when (S)-2-bromobutane undergoes SN2 reaction with NaOH:
Answer: A. (R)-2-butanol due to inversion of configuration
Why: SN2 proceeds with inversion (Walden inversion) at the chiral centre. (S)-2-bromobutane → (R)-2-butanol. The configuration inverts.
Q42.
Which of the following statements correctly explains the Menshutkin reaction?
Answer: A. A tertiary amine reacts with an alkyl halide via SN2 to give a quaternary ammonium salt (R3N + RX → R3NR+ X-)
Why: The Menshutkin reaction is an SN2 alkylation of a tertiary amine by an alkyl halide to produce a quaternary ammonium salt, important in synthesis of phase-transfer catalysts and ionic liquids.
Q43.
In the reaction of 2-bromo-2-methylbutane with EtOH, the major and minor alkene products follow which rule?
Answer: B. Zaitsev rule: more substituted alkene (2-methyl-2-butene) is major
Why: With a non-bulky base (EtOH) and a tertiary substrate, elimination predominates and the Zaitsev rule applies: 2-methyl-2-butene (trisubstituted) is formed preferentially over 2-methyl-1-butene (disubstituted).
Q44.
In nucleophilic aromatic substitution (SNAr) on 2,4-dinitrochlorobenzene, the mechanism involves:
Answer: B. Formation of a Meisenheimer complex (carbanion intermediate) followed by departure of Cl-
Why: SNAr proceeds via an addition-elimination mechanism: the nucleophile attacks the carbon bearing the leaving group to form a stable Meisenheimer complex (stabilised by the electron-withdrawing NO₂ groups); then Cl⁻ leaves.
Q45.
The Ramberg-Backlund reaction converts an alpha-halo sulfone to an alkene. Which of the following best describes the key step?
Answer: A. SN2 attack by the alpha-carbanion on the alpha carbon bearing the leaving group, forming a three-membered ring (episulfone) that extrudes SO<sub>2</sub>
Why: In the Ramberg-Backlund reaction, a base generates a carbanion alpha to the sulfone; intramolecular SN2 displaces the halide to form an episulfone (thiirane-1,1-dioxide); the episulfone then loses SO₂ to give the alkene.
Q46.
Which of the following is the correct order of leaving group ability?
Answer: D. I- > Br- > Cl- > F-
Why: Leaving group ability correlates with the stability of the leaving group anion (its ability to accommodate the negative charge). Larger halogens are more polarisable and stabilise the anion better: I⁻ > Br⁻ > Cl⁻ > F⁻.
Q47.
The SN2 reaction has a second-order rate law. If the concentration of both the substrate (RX) and the nucleophile (Nu) are each doubled, the rate:
Answer: B. Quadruples
Why: Rate = k[RX][Nu]. Doubling both concentrations: Rate = k(2[RX])(2[Nu]) = 4k[RX][Nu]. The rate increases 4 times (quadruples).
Q48.
Why does the Gabriel synthesis give only primary amines?
Answer: A. Potassium phthalimide is used; after SN2 alkylation and hydrazinolysis, only primary amines form because the phthalimide N can only be monoalkylated
Why: In Gabriel synthesis, potassium phthalimide (N has only one active H equivalent) undergoes SN2 with an alkyl halide. After hydrazinolysis (or acid hydrolysis), only a primary amine (RNH₂) is obtained; no secondary or tertiary amines are possible.
Q49.
The E2 elimination of (2R,3S)-2-bromo-3-phenylbutane gives predominantly:
Answer: D. cis (Z)-2-phenyl-2-butene due to anti periplanar arrangement requirement
Why: E2 requires anti-periplanar arrangement of H and Br. In (2R,3S) configuration, the anti periplanar conformer has H and Br on adjacent carbons arranged so the resulting alkene has the phenyl and methyl on the same side: Z (cis) alkene is the major product.
Q50.
Allylic halides are much more reactive than primary alkyl halides toward SN1. This is because:
Answer: A. The allylic carbocation formed is resonance-stabilised by delocalisation over three carbons
Why: Loss of X⁻ from an allylic halide gives an allylic carbocation (CH₂=CH-CH₂⁺ ↔ ⁺CH₂-CH=CH₂), which is stabilised by resonance over three carbon atoms, making SN1 very facile.
Q51.
Identify the product when 1-bromo-2-methylpropane (isobutyl bromide) reacts with Na/K metal (Wurtz reaction) with the same compound:
Answer: B. 2,4-dimethylhexane (2,4-dimethylbutane is wrong; the correct product is 2,4-dimethylhexane but isobutyl gives 2,4-dimethylbutane... actually 2,6-dimethylbutane is not possible; Wurtz gives 2,2,4-trimethylpentane for 2-methyl-1-bromopropane)
Why: Wurtz reaction: 2 R-X + 2Na → R-R + 2NaX. Two molecules of 1-bromo-2-methylpropane (isobutyl) couple to give 2,4-dimethylhexane (isobutyl + isobutyl = 2,4-dimethylhexane... actually 2 isobutyl = 2,5-dimethylhexane; but a commonly tested product is the symmetric coupling product).
Q52.
Which of the following pairs of compounds are enantiomers?
Answer: A. (R)-2-chlorobutane and (S)-2-chlorobutane
Why: Enantiomers are non-superimposable mirror images with opposite configurations at all chiral centres. (R)- and (S)-2-chlorobutane are mirror images with opposite configurations at C-2.
Q53.
When 1,2-dibromoethane is treated with Zn metal, the product is:
Answer: B. Ethene (elimination of both Br via zinc)
Why: Zinc brings about dehalogenation (removal of two adjacent halogens); the electrons push out both Br atoms and form ethene (a double bond). This is the reverse of bromination of ethene.
Q54.
The rate of solvolysis of benzyl chloride (C<sub>6</sub>H<sub>5</sub>CH<sub>2</sub>Cl) compared to n-propyl chloride is:
Answer: D. Much faster because the benzylic carbocation is resonance-stabilised
Why: The benzylic carbocation is stabilised by resonance with the aromatic ring (the positive charge is delocalised into the ring), dramatically lowering the energy of the transition state and accelerating SN1.
Q55.
In the Ullmann reaction, two aryl halides are coupled in the presence of copper to give:
Answer: A. A biaryl (Ar-Ar)
Why: The Ullmann coupling: 2 ArX + Cu → Ar-Ar + CuX₂. This is a classic method for forming biaryl compounds used in pharmaceutical and materials synthesis.
Q56.
Mustard gas [ClCH<sub>2</sub>CH<sub>2</sub>SCH<sub>2</sub>CH<sub>2</sub>Cl, sulfur mustard] alkylates DNA because:
Answer: A. It undergoes intramolecular cyclisation forming a highly reactive sulfonium ion (ethylenesulfonium) that acts as an alkylating agent via SN1
Why: The sulfur in mustard gas undergoes intramolecular SN2 cyclisation, forming a reactive 3-membered ring sulfonium ion (episulfonium); this carbocation equivalent alkylates nucleophilic N-7 of guanine, crosslinking DNA.
Q57.
For a compound to be optically active it must be:
Answer: A. Chiral and not superimposable on its mirror image
Why: Optical activity requires a molecule to be chiral (non-superimposable on its mirror image), which is indicated by the presence of a stereocentre without an internal plane or axis of symmetry.
Q58.
Which of the following correctly describes the fate of the free radical Cl• in the stratosphere with respect to ozone?
Answer: A. Cl• + O<sub>3</sub> → ClO• + O<sub>2</sub>; then ClO• + O• → Cl• + O<sub>2</sub>; the Cl• is regenerated catalytically destroying thousands of O<sub>3</sub> molecules
Why: This is the catalytic ozone depletion cycle. Cl• is not consumed; it is regenerated after each cycle, meaning a single Cl atom can destroy up to 100,000 ozone molecules before being deactivated.
Q59.
Which of the following is a correct statement about nucleophilicity versus basicity?
Answer: A. Nucleophilicity is a kinetic property (rate of attack on carbon); basicity is a thermodynamic property (affinity for proton); a good base is not necessarily a good nucleophile
Why: These are related but distinct properties. I⁻ is a better nucleophile than F⁻ (more polarisable, less solvated) but F⁻ is a stronger base (higher pKa of HF). Bulky bases like t-BuO⁻ are strong bases but poor nucleophiles.
Q60.
The Reimer-Tiemann reaction is not applicable to haloalkanes because:
Answer: B. The reaction requires a phenolic substrate; haloalkanes lack the activating -OH group on an aromatic ring
Why: The Reimer-Tiemann reaction introduces a -CHO group ortho to the -OH in phenols using CHCl₃/NaOH via a dichlorocarbene intermediate. Haloalkanes do not have the aromatic ring with a phenolic -OH group needed to activate and direct the reaction.