Haloalkanes and Haloarenes Class 12 Notes - CBSE Chemistry Chapter 10

By /
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
πŸƒ Flash Cards: Haloalkanes and Haloarenes

Class 12 Chemistry Β· 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: 1Β°, 2Β°, 3Β° (X on C bonded to 1, 2, 3 other carbons)
  • Allylic/benzylic = X next to C=C / benzene β†’ very reactive
  • Vinylic/aryl = X on sp2 C β†’ unreactive; this split drives the chapter
🧲Core idea2/8

The Polar C–X Bond

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

C^δ⁺–X^δ⁻ Β· 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 β†’ 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.
βš—οΈPreparation3/8

Making Haloalkanes

Alcohols are the most common starting point for haloalkanes.

ROH + SOCl2 β†’ RCl + SO2↑ + HCl↑

SOCl2 is best β€” 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 β†’ anti-Markovnikov (Kharasch)
πŸ”Key reactions4/8

Finkelstein & Swarts

Halogen-exchange reactions swap one halide for another.

R–Cl + NaI β†’(dry acetone) R–I + NaCl↓

NaCl precipitates in acetone, pulling Finkelstein forward.

  • Finkelstein: R–Cl/R–Br + NaI in dry acetone β†’ R–I
  • Swarts: R–Br + AgF β†’ 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 β†’ inversion (Walden inversion).

  • Reactivity 1Β° > 2Β° > 3Β° (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 β†’ racemic mixture.

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

Elimination (Ξ²-Elimination)

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

alcoholic KOH β†’ alkene Β· aqueous KOH β†’ alcohol

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

  • Saytzeff (Zaitsev): more substituted, more stable alkene is the major product
  • 2-bromobutane + alc. KOH β†’ mainly but-2-ene
  • Reactivity 3Β° > 2Β° > 1Β°; bulky base + 3Β° + heat favours elimination
πŸ›‘οΈAdvanced8/8

Haloarenes: Low Reactivity

Aryl halides resist nucleophilic substitution far more than alkyl halides.

Resonance β†’ partial C=X double-bond β†’ 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)
Swipe β†’Click a card to focus β†’8 cards
πŸ“ Practice Haloalkanes and Haloarenes β€” 10 NEET PYQs
Real previous-year questions Β· with answers & solutions
Start β†’Close βœ•
Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
Consider the reaction CH₃CHβ‚‚COO⁻Na⁺ + NaOH + ? xrightarrowHeat CH₃CH₃ + Naβ‚‚CO₃. 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 COβ‚‚ 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.
πŸ”Ž See the full step-by-step solution β†’
Q2NEET 2021
The major product formed when 3-methylbut-1-ene, (CH₃)β‚‚CH–CH=CHβ‚‚, reacts with HBr in the presence of organic peroxide (C₆Hβ‚…CO)β‚‚Oβ‚‚ is:
Correct answer: A. HBr with a peroxide adds by the anti-Markovnikov (Kharasch) free-radical route: Brβ€’ adds to the terminal CHβ‚‚ to give the more stable secondary radical, and the H caps the chain. So Br ends up on the terminal carbon: (CH₃)β‚‚CH–CH₂–CHβ‚‚Br (1-bromo-3-methylbutane) is the major product.
πŸ”Ž See the full step-by-step solution β†’
Q3NEET 2020
The dehydrohalogenation of 2-bromopentane to form pent-2-ene is: (1) a Ξ²-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 (Ξ± and Ξ²) carbons is a Ξ²-elimination and, specifically, a dehydrohalogenation (loss of HX). Pent-2-ene (more substituted) forms preferentially β†’ Zaitsev rule. It is NOT a dehydration (no water/OH is lost). So statements 1, 2 and 3 are correct.
πŸ”Ž See the full step-by-step solution β†’
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 (3Β° cation) and the 1Β° chloride attached via a chain to a ring can all ionise. Chlorobenzene cannot: the Cl is on an spΒ² 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.
πŸ”Ž See the full step-by-step solution β†’
Q5NEET 2018
Toluene undergoes: C₇Hβ‚ˆ β†’[3Clβ‚‚/Ξ”] A β†’[Brβ‚‚/Fe] B β†’[Zn/HCl] C. The product C is:
Correct answer: C. 3Clβ‚‚/Ξ” (light/heat) chlorinates the side chain: toluene β†’ benzotrichloride, C₆Hβ‚…CCl₃ (A). The –CCl₃ group is a strong electron-withdrawing (meta-directing) group, so Brβ‚‚/Fe substitutes meta β†’ m-bromobenzotrichloride (B). Zn/HCl reduces –CCl₃ back to –CH₃, giving m-bromotoluene (C).
πŸ”Ž See the full step-by-step solution β†’
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/Ο€ 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.
πŸ”Ž See the full step-by-step solution β†’
Q7NEET 2016
For the reactions (i) CH₃CHβ‚‚CHβ‚‚Br + KOH β†’ CH₃CH=CHβ‚‚ + KBr + Hβ‚‚O, (ii) (CH₃)β‚‚CHBr + KOH β†’ (CH₃)β‚‚CHOH + KBr, (iii) cyclohexene + Brβ‚‚ β†’ 1,2-dibromocyclohexane, the correct statement is:
Correct answer: A. (i) gives an alkene by loss of HBr β†’ elimination. (ii) replaces Br by OH to give an alcohol β†’ nucleophilic substitution. (iii) adds Brβ‚‚ across the C=C of cyclohexene β†’ addition. So (i) elimination, (ii) substitution, (iii) addition.
πŸ”Ž See the full step-by-step solution β†’
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. –NOβ‚‚ 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.
πŸ”Ž See the full step-by-step solution β†’
Q9NEET 1991
In which compound are ALL the bond angles exactly 109Β°28β€²?
Correct answer: B. Only a molecule with four identical groups on a tetrahedral carbon has all angles equal to the regular tetrahedral angle 109Β°28β€². CClβ‚„ is symmetrical (spΒ³, four equal C–Cl bonds), so every Cl–C–Cl angle is exactly 109Β°28β€². CH₃Cl, CHCl₃ and CHI₃ have two different kinds of bonds, so their angles deviate slightly.
πŸ”Ž See the full step-by-step solution β†’
Q10NEET 1988
The Cl–C–Cl bond angle in 1,1,2,2-tetrachloroethene (Clβ‚‚C=CClβ‚‚) and in tetrachloromethane (CClβ‚„) will be about, respectively:
Correct answer: A. In Clβ‚‚C=CClβ‚‚ each carbon is spΒ² hybridised (one C=C, two C–Cl), so the Cl–C–Cl angle is about 120Β°. In CClβ‚„ carbon is spΒ³ hybridised and tetrahedral, so the Cl–C–Cl angle is 109Β°28β€² (β‰ˆ109.5Β°).
πŸ”Ž See the full step-by-step solution β†’
View 20+ more practice questions, gamified β†’
Free Β· no signup Β· works in your browser
Studying this chapter? Track it - saved on this device, no login.

Chapter Navigation

Previous: Coordination Compounds Class 12 Notes
Next: Alcohols, Phenols and Ethers Class 12 Notes

Related Chapters in Class 12 Chemistry

Practice What You Learned

Test yourself with our JEE Main Mock Test Set 1 to see how well you’ve mastered the concepts.

Explore More

Back to Class 12 Chemistry Notes Hub

You Might Also Like

Hand-picked cross-subject and entrance-exam resources to deepen your prep:

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.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top
CHEMISTRY · CH 10