Haloalkanes and HaloarenesNEET MCQs with solutions

Not quite — the answer is B.

The key factor is the leaving-group ability during C–X bond cleavage. C–I is the weakest bond and I⁻ is the best leaving group, whereas C–F is the strongest bond. Therefore dehydrohalogenation reactivity follows R–I > R–Br > R–Cl > R–F.

Not quite — the answer is C.

In vinyl chloride, chlorine is directly attached to an sp2-hybridised carbon of the C=C double bond. Allyl and benzyl chlorides have halogen on an sp3 carbon; ethyl chloride is a simple primary haloalkane with sp3 carbon.

Not quite — the answer is B.

Finkelstein reaction is a halide-exchange reaction using NaI in dry acetone. NaBr (or NaCl) precipitates out of acetone, driving the equilibrium forward to give the alkyl iodide.

Not quite — the answer is A.

Thionyl chloride (SOCl₂) reacts with alcohols to give alkyl chlorides; by-products SO₂ and HCl are gases that escape, driving the reaction to completion. PCl₅ also gives alkyl chloride but produces POCl₃ (liquid) as by-product, making purification harder. SOCl₂ is the preferred reagent precisely because of its gaseous by-products.

Not quite — the answer is A.

In CH₃CH₂Cl, the carbon bearing Cl is bonded to only one other carbon, making it primary. CH₃CHClCH₃ is secondary (two C neighbours) and (CH₃)₃CCl is tertiary (three C neighbours). C₆H₅Cl is an aryl halide, not a haloalkane at all.

Not quite — the answer is A.

The C–X bond involves unequal sharing of electrons due to the higher electronegativity of halogen compared to carbon, making it polar covalent — not ionic (no complete electron transfer). A pure or non-polar covalent bond would require equal electronegativities, which is never the case for C–X.

Not quite — the answer is A.

CH₃CH₂CH₂Cl has a higher molecular mass and stronger van der Waals + dipole–dipole forces than the corresponding alkane (propane) or ether (dimethyl ether). CH₃CH₂CH₂F is the key trap — same carbon chain but fluorine's weaker polarisability gives it lower dispersion forces than the chloro compound despite higher C–X polarity.

Not quite — the answer is A.

Benzene reacts with Cl₂ in the presence of FeCl₃ (Lewis acid) via electrophilic aromatic substitution — FeCl₃ activates Cl₂ by forming Cl⁺ (electrophile). Friedel-Crafts alkylation introduces alkyl groups, not halogen. Sulphonation introduces –SO₃H, not –Cl.

Not quite — the answer is A.

OH⁻ is a nucleophile — it attacks the electrophilic carbon bearing the halogen, displacing X⁻ via nucleophilic substitution. Electrophilic substitution involves attack by an electrophile on an electron-rich centre (e.g., benzene ring). Free radical substitution involves radical intermediates, not ionic nucleophiles.

Not quite — the answer is C.

Bleaching powder acts on ethanol (and acetone) to give chloroform via haloform reaction; ethanol is the most commonly cited NCERT substrate. Acetaldehyde is a trap — it also gives chloroform but is not the standard lab substrate listed. Propanone also works but ethanol is the primary NCERT answer.

Not quite — the answer is C.

Polar protic solvents stabilize the transition state of ionization via solvation of both ionic fragments — cation (by lone pairs) and anion (by H-bonding). Option D is a trap — H-bonding with β-H would actually reduce E2, not explain SN1 preference.

Not quite — the answer is B.

2,4-Dinitrochlorobenzene is the classic benchmark SNAr substrate: two EWGs at ortho and para positions both stabilize the Meisenheimer complex by resonance. Option C is the trap — methyl (EDG) actually deactivates the ring for SNAr. Option D (methoxy) is an EDG, also deactivating.

Not quite — the answer is A.

The benzene ring is electron-rich and readily attacked by electrophiles; EAS is the dominant reaction class for haloarenes. Nucleophilic substitution at the aryl carbon is extremely difficult unless strongly activating groups (e.g., –NO₂ at ortho/para) are present. Addition to the aromatic ring is disfavoured as it would destroy the delocalisation energy.

Not quite — the answer is B.

An allylic halide has the halogen on a carbon adjacent to (not part of) a C=C double bond; CH₂=CHCH₂Cl fits this exactly. CH₂=CHCl is vinylic (halogen on the double bond carbon). C₆H₅CH₂Cl is benzylic. CH₃CH₂CH₂Cl is a simple primary alkyl halide.

Not quite — the answer is A.

On ionisation, the benzyl carbocation delocalises positive charge over the ring via resonance, lowering activation energy for SN1. Ethyl chloride also gives a primary carbocation but without any resonance stabilisation. Bond polarity and molecular mass are irrelevant to SN1 rate here.

Not quite — the answer is D.

A vinylic halide has the halogen directly on a sp² carbon of a C=C double bond; CH₂=CHCl is the only such compound here. C₆H₅Cl is an aryl halide (halogen on aromatic sp² carbon, a common trap). CH₂=CHCH₂Cl is allylic, not vinylic.

Not quite — the answer is C.

A benzylic halide (e.g., C₆H₅CH₂Cl) has halogen on the –CH₂– carbon adjacent to the ring, not on the ring itself. Allylic halogen is adjacent to, not on, the double bond (eliminating A). Vinylic is on the C=C carbon, not adjacent (eliminating B). Aryl halide carbon is sp², not sp³ (eliminating D).

Not quite — the answer is B.

The longest chain has 3 carbons (propane); halogen is on C2 giving the lowest possible locant, so the name is 2-chloropropane. 1-chloropropane would place the halogen on the terminal carbon (that is CH₃CH₂CH₂Cl, a different compound). Propyl chloride is the common name, not IUPAC.

Not quite — the answer is B.

In aryl halides like C₆H₅Cl, the halogen is directly bonded to a ring carbon that is sp² hybridised. Option A is wrong because ring carbons are sp², not sp³. Option C describes benzylic halides. The halogen is bonded to a specific carbon, not the π system.

Not quite — the answer is D.

In (CH₃)₃CCl, the carbon bearing Cl is bonded to three other carbons — the definition of tertiary. CH₃CHClCH₃ is secondary (two C neighbours). CH₂ClCH₂CH₃ is primary (one C neighbour, a strong trap due to the three-carbon chain). CH₃CH₂Cl is also primary.

Not quite — the answer is A.

A benzyl halide has the halogen on the –CH₂– group directly attached to a benzene ring. C₆H₅Cl is an aryl halide (halogen on the ring itself), which is the key trap. CH₂=CHCl is vinylic and CH₃CH₂Cl is a simple primary alkyl halide.

Not quite — the answer is B.

Aralkyl (arene + alkyl) halides have the halogen on an alkyl side chain attached to an aromatic ring; C₆H₅CH₂Cl fits this exactly. C₆H₅Cl is an aryl halide (halogen on ring, not side chain). C₆H₅CH=CHCl is a vinylic-type halide on a side chain, not a saturated aralkyl halide.

Not quite — the answer is C.

Steric hindrance around the reaction site increases with the number of alkyl groups on the carbon bearing the halogen. A primary carbon has only one such alkyl group, leaving the carbon most accessible to nucleophiles or bases. Bond polarity and carbocation stability are unrelated to steric bulk.

Not quite — the answer is D.

In common nomenclature, alkyl halides are named as alkyl group + halide; the methyl group (CH₃–) + chloride gives methyl chloride. "Methane chloride" mimics IUPAC logic incorrectly. "Monochloro methane" confuses substitutive nomenclature. "Carbon monochloride" is a completely incorrect trivial name.