Haloalkanes and Haloarenes Class 12 Notes - CBSE Chemistry Chapter 10

Chapter summary

Haloalkanes and Haloarenes covers organic compounds where a halogen is bonded to an sp3 alkyl carbon or an sp2 aromatic carbon, including their classification, IUPAC naming, the polar C-X bond, and methods of preparation. The heart of the chapter is reaction mechanisms, namely nucleophilic substitution by the SN1 and SN2 routes, elimination giving alkenes by the Saytzeff rule, and why aryl halides resist substitution. It is a high-yield NEET chapter because mechanism, stereochemistry, and reactivity-order questions appear almost every year.

Chapter notes
🃏 Flash Cards: Haloalkanes and Haloarenes

Class 12 Chemistry · Chapter 23 – swipe through all 8 cards to understand the whole chapter.

🏷️Start here1/8

Classification & Nomenclature

Halogen on an sp3 carbon makes a haloalkane; on an sp2 ring carbon, a haloarene.

(CH3)3CBr = 2-bromo-2-methylpropane

Halogen is a ‘halo-‘ prefix; number the chain for the lowest locant set.

  • By carbon type: 1°, 2°, 3° (X on C bonded to 1, 2, 3 other carbons)
  • Allylic/benzylic = X next to C=C / benzene → very reactive
  • Vinylic/aryl = X on sp2 C → unreactive; this split drives the chapter
🧲Core idea2/8

The Polar C–X Bond

X is more electronegative than C, so carbon is δ+ and is the site nucleophiles attack.

C^δ⁺–X^δ⁻ · bond strength: C–F > C–Cl > C–Br > C–I

Down the group bond length increases and strength decreases; C–I breaks easiest.

  • C–I weakest → iodides are the most reactive
  • Insoluble in water (no strong H-bonds), soluble in organic solvents
  • Same R: b.p. R–I > R–Br > R–Cl > R–F; branching lowers b.p.
⚗️Preparation3/8

Making Haloalkanes

Alcohols are the most common starting point for haloalkanes.

ROH + SOCl2 → RCl + SO2↑ + HCl↑

SOCl2 is best — by-products escape as gases, giving pure RCl.

  • With HX: reactivity HI > HBr > HCl (HCl needs ZnCl2 / Lucas reagent)
  • Also PCl3 and PCl5 convert ROH to RCl
  • Alkene + HX follows Markovnikov; HBr + peroxide → anti-Markovnikov (Kharasch)
🔁Key reactions4/8

Finkelstein & Swarts

Halogen-exchange reactions swap one halide for another.

R–Cl + NaI →(dry acetone) R–I + NaCl↓

NaCl precipitates in acetone, pulling Finkelstein forward.

  • Finkelstein: R–Cl/R–Br + NaI in dry acetone → R–I
  • Swarts: R–Br + AgF → R–F + AgBr (route to fluorides)
  • Haloarenes: arene + Cl2/Br2 with anhydrous FeCl3 (Lewis acid)
🎯Core mechanism5/8

SN2 Substitution

A nucleophile attacks the back side in one concerted step as X leaves.

rate = k[RX][Nu] (second order)

Back-side attack flips the centre → inversion (Walden inversion).

  • Reactivity 1° > 2° > 3° (least steric hindrance wins)
  • Gives complete inversion of configuration
  • Favoured by strong nucleophiles and polar aprotic solvents
♻️Core mechanism6/8

SN1 Substitution

The C–X bond ionises first to a carbocation, then the nucleophile is captured.

rate = k[RX] (first order, Nu-independent)

Planar carbocation is attacked from both faces → racemic mixture.

  • Reactivity 3° > 2° > 1° (more stable carbocation wins)
  • Gives partial racemisation, not clean inversion
  • Favoured by polar protic solvents; allylic/benzylic react fast
✂️Key contrast7/8

Elimination (β-Elimination)

A strong base pulls off H and X from adjacent carbons to form an alkene.

alcoholic KOH → alkene · aqueous KOH → alcohol

The favourite NEET trap: ‘alcoholic = alkene’, ‘aqueous = substitution’.

  • Saytzeff (Zaitsev): more substituted, more stable alkene is the major product
  • 2-bromobutane + alc. KOH → mainly but-2-ene
  • Reactivity 3° > 2° > 1°; bulky base + 3° + heat favours elimination
🛡️Advanced8/8

Haloarenes: Low Reactivity

Aryl halides resist nucleophilic substitution far more than alkyl halides.

Resonance → partial C=X double-bond → short, strong C–X

Phenol from chlorobenzene needs ~623 K and 300 atm (forcing conditions).

  • Resonance + sp2 carbon hold the C–X bond tighter (stronger, shorter)
  • In EAS, halogen is deactivating but o,p-directing
  • –NO2 at ortho/para boosts reactivity (addition-elimination)
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📝 Practice Haloalkanes and Haloarenes — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Q1NEET 2021
Consider the reaction CH₃CH₂COO⁻Na⁺ + NaOH + ? xrightarrowHeat CH₃CH₃ + Na₂CO₃. The missing reagent is:
Correct answer: C. This is decarboxylation of a sodium carboxylate. Heating a sodium salt of a carboxylic acid with soda lime (NaOH + CaO) removes CO₂ and gives the alkane with one fewer carbon (ethane here). CaO keeps the NaOH dry and prevents it melting/attacking glass, so the missing reagent is CaO.
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Q2NEET 2021
The major product formed when 3-methylbut-1-ene, (CH₃)₂CH–CH=CH₂, reacts with HBr in the presence of organic peroxide (C₆H₅CO)₂O₂ is:
Correct answer: A. HBr with a peroxide adds by the anti-Markovnikov (Kharasch) free-radical route: Br• adds to the terminal CH₂ to give the more stable secondary radical, and the H caps the chain. So Br ends up on the terminal carbon: (CH₃)₂CH–CH₂–CH₂Br (1-bromo-3-methylbutane) is the major product.
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Q3NEET 2020
The dehydrohalogenation of 2-bromopentane to form pent-2-ene is: (1) a β-elimination reaction (2) follows Zaitsev rule (3) a dehydrohalogenation reaction (4) a dehydration reaction. Which statements are correct?
Correct answer: D. Losing H and Br from adjacent (α and β) carbons is a β-elimination and, specifically, a dehydrohalogenation (loss of HX). Pent-2-ene (more substituted) forms preferentially → Zaitsev rule. It is NOT a dehydration (no water/OH is lost). So statements 1, 2 and 3 are correct.
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Q4NEET 2020
Which of the following will NOT undergo an SN1 reaction with OH⁻?
Correct answer: C. SN1 needs a carbocation. Allyl chloride (allylic cation), tert-butyl chloride (3° cation) and the 1° chloride attached via a chain to a ring can all ionise. Chlorobenzene cannot: the Cl is on an sp² ring carbon with partial double-bond character (resonance), so the aryl C–Cl will not ionise to give a carbocation. Hence it does not undergo SN1.
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Q5NEET 2018
Toluene undergoes: C₇H₈ →[3Cl₂/Δ] A →[Br₂/Fe] B →[Zn/HCl] C. The product C is:
Correct answer: C. 3Cl₂/Δ (light/heat) chlorinates the side chain: toluene → benzotrichloride, C₆H₅CCl₃ (A). The –CCl₃ group is a strong electron-withdrawing (meta-directing) group, so Br₂/Fe substitutes meta → m-bromobenzotrichloride (B). Zn/HCl reduces –CCl₃ back to –CH₃, giving m-bromotoluene (C).
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Q6NEET 2016
Which of the following can be used as the halide component for a Friedel–Crafts alkylation reaction?
Correct answer: D. Friedel–Crafts needs a halide whose C–X bond can be polarised/ionised by AlCl₃ to give a carbocation (or its equivalent). In aryl halides (chloro-/bromobenzene) and vinyl halides the lone pair/π overlap gives partial double-bond character, so the C–X bond is too strong to ionise. Isopropyl chloride is an ordinary alkyl halide and works fine.
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Q7NEET 2016
For the reactions (i) CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O, (ii) (CH₃)₂CHBr + KOH → (CH₃)₂CHOH + KBr, (iii) cyclohexene + Br₂ → 1,2-dibromocyclohexane, the correct statement is:
Correct answer: A. (i) gives an alkene by loss of HBr → elimination. (ii) replaces Br by OH to give an alcohol → nucleophilic substitution. (iii) adds Br₂ across the C=C of cyclohexene → addition. So (i) elimination, (ii) substitution, (iii) addition.
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Q8NEET 1997
Replacement of the Cl of chlorobenzene to give phenol requires drastic conditions, but the chlorine of 2,4-dinitrochlorobenzene is readily replaced. This is because:
Correct answer: D. Nucleophilic aromatic substitution proceeds through a carbanion (Meisenheimer) intermediate bearing negative charge on the ring. –NO₂ groups ortho/para to the Cl withdraw electron density from exactly those positions, delocalising and stabilising that negative charge, so the chloride is displaced easily. Chlorobenzene lacks such groups, so it needs ~623 K/300 atm.
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Q9NEET 1991
In which compound are ALL the bond angles exactly 109°28′?
Correct answer: B. Only a molecule with four identical groups on a tetrahedral carbon has all angles equal to the regular tetrahedral angle 109°28′. CCl₄ is symmetrical (sp³, four equal C–Cl bonds), so every Cl–C–Cl angle is exactly 109°28′. CH₃Cl, CHCl₃ and CHI₃ have two different kinds of bonds, so their angles deviate slightly.
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Q10NEET 1988
The Cl–C–Cl bond angle in 1,1,2,2-tetrachloroethene (Cl₂C=CCl₂) and in tetrachloromethane (CCl₄) will be about, respectively:
Correct answer: A. In Cl₂C=CCl₂ each carbon is sp² hybridised (one C=C, two C–Cl), so the Cl–C–Cl angle is about 120°. In CCl₄ carbon is sp³ hybridised and tetrahedral, so the Cl–C–Cl angle is 109°28′ (≈109.5°).
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Frequently Asked Questions

What are haloalkanes and haloarenes?

Haloalkanes are compounds in which one or more halogen atoms are bonded to an sp3 hybridised carbon of an alkyl group, while haloarenes have a halogen bonded directly to an sp2 carbon of an aromatic ring. They are formed by replacing hydrogen atoms of hydrocarbons with halogen atoms.

What is the difference between SN1 and SN2 reactions?

SN2 is a single-step bimolecular reaction with rate proportional to both the halide and the nucleophile, and it gives inversion of configuration, favoured by primary halides. SN1 is a two-step unimolecular reaction with rate depending only on the halide, going through a carbocation and giving a racemic mixture, favoured by tertiary halides.

What product forms with alcoholic KOH versus aqueous KOH?

Alcoholic KOH causes beta-elimination (dehydrohalogenation) to give an alkene, while aqueous KOH causes nucleophilic substitution to give an alcohol. This contrast is a very common NEET trap and the Saytzeff rule decides the major alkene in elimination.

Why are haloarenes less reactive than haloalkanes toward nucleophilic substitution?

In haloarenes the halogen lone pair delocalises into the ring by resonance, giving the C-X bond partial double-bond character so it is shorter and stronger. The sp2 carbon is also more electronegative and holds the bonding electrons tightly, so aryl halides resist nucleophilic substitution and need very harsh conditions.

Is Haloalkanes and Haloarenes important for NEET?

Yes, it is part of the NEET Class 12 organic chemistry syllabus and is a reliably high-yield chapter. Questions on SN1 versus SN2, reactivity order of halides, Saytzeff elimination, and named reactions like Finkelstein and Swarts appear frequently.

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