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Haloalkanes & Haloarenes

Nucleophilic substitution (SN1, SN2), elimination (E1, E2), haloarenes reactivity β€” and why aryl halides don't undergo SN reactions easily. Expect 3–4 EAPCET questions.

3–4Questions in EAPCET
~3%Paper Weightage
8Key Reactions
4Mistake Traps

Concept Core

SN1 vs SN2, E2 elimination, and the unique chemistry of aryl halides.

SN1 vs SN2 β€” Comparison Table
FeatureSN1SN2
Mechanism2 steps (carbocation intermediate)1 step (concerted)
Rate-determining stepFormation of carbocationBackside attack by nucleophile
Rate lawRate = k[RX] (first order)Rate = k[RX][Nu] (second order)
Substrate preference3Β° > 2Β° (stable carbocation)1Β° > 2Β° (less steric hindrance)
StereochemistryRacemisation (mixture of enantiomers)Inversion (Walden inversion)
SolventPolar protic (stabilises carbocation)Polar aprotic (activates nucleophile)
Common Nucleophiles and Reactions
NucleophileProductType
OH⁻ (aq)Alcohol (Rβˆ’OH)Substitution
CN⁻Nitrile (Rβˆ’CN) β†’ carboxylic acidSubstitution (chain extension)
NH₃Amine (Rβˆ’NHβ‚‚)Substitution
AgNO₃ (aq)AgX precipitate + alcoholSubstitution + test
KCN (alc)Nitrile (Rβˆ’CN)Substitution
Elimination Reactions

E2 (Bimolecular elimination): Strong base (KOH in alcohol) removes Ξ²-H simultaneously with leaving group departure β†’ alkene (Saytzeff's rule: more substituted alkene is major product).

Saytzeff's rule: In elimination, the more substituted (stable) alkene is the major product.

CH₃CHβ‚‚CHBrCH₃ + KOH(alc) β†’ CH₃CH=CHCH₃ (major, more substituted) + CHβ‚‚=CHCHβ‚‚CH₃ (minor, less substituted)
Haloarenes β€” Reduced Reactivity to SN

Aryl halides (e.g., chlorobenzene, C₆Hβ‚…Cl) are very resistant to nucleophilic substitution because:

1. Resonance: Cβˆ’Cl bond acquires partial double bond character through resonance β†’ shorter, stronger Cβˆ’X bond.

2. Geometry: spΒ² carbon β€” backside attack (SN2) geometrically impossible (flat ring).

3. Carbocation: Aryl carbocations are highly unstable β€” SN1 not feasible either.

Important Reactions: DDT, Freon, Chloroform

DDT (dichlorodiphenyltrichloroethane): insecticide. Chlorobenzene + chloral β†’ DDT (Friedel-Crafts type).

Chloroform (CHCl₃): Prepared from acetone + bleaching powder (haloform reaction).

Carbon tetrachloride (CClβ‚„): Fire extinguisher. Prepared: CClβ‚„ from CSβ‚‚ + Clβ‚‚.

Freons (CFCs): Refrigerants (CHFβ‚‚Cl etc.). Cause ozone layer depletion (Clβ€’ radicals).

Substitution vs Elimination β€” When Each Occurs
Favours Substitution (SN): weak nucleophile, low temperature, 1Β° substrate Favours Elimination (E2): strong base, high temperature, 3Β° substrate, bulky base SN1: 3Β° substrate, polar protic solvent, weak nucleophile SN2: 1Β° substrate, polar aprotic solvent, strong nucleophile

Formula Vault

SN1/SN2 conditions, leaving group ability, and elimination rules.

Leaving Group Order
I⁻ > Br⁻ > Cl⁻ > F⁻
Weaker base = better leaving group
SN2 Rate Law
Rate = k[RX][Nu⁻]
Second order; bimolecular
SN1 Rate Law
Rate = k[RX]
First order; unimolecular
SN2 Substrate Order
CH₃X > 1Β° > 2Β° > 3Β°
Less steric hindrance = faster SN2
SN1 Substrate Order
3Β° > 2Β° > 1Β° > CH₃X
More stable carbocation = faster SN1
Saytzeff's Rule
More substituted alkene = major product
Applies to E2 elimination
SN2 Stereochemistry
Inversion of configuration (Walden)
Backside attack flips tetrahedral geometry
SN1 Stereochemistry
Racemisation
Planar carbocation attacked from either face

Worked Examples

5 problems β€” SN1 vs SN2 choice, nucleophile identification, elimination, haloarene, and leaving group.

Easy2-Bromo-2-methylpropane + NaOH(aq) β€” SN1 or SN2?β–Ύ
2-Bromo-2-methylpropane (tert-butyl bromide) reacts with dilute NaOH(aq). Predict SN1 or SN2 and the product.
1
Tert-butyl bromide is a 3Β° substrate β†’ favours SN1 (stable tertiary carbocation).
2
Dilute NaOH(aq) = weak nucleophile, polar protic solvent β†’ confirms SN1 conditions.
3
Product: (CH₃)₃Cβˆ’OH (tert-butyl alcohol) via carbocation intermediate.
βœ“  SN1; product = (CH₃)₃COH (tert-butanol)
EasyWhich is a better leaving group: F⁻ or I⁻?β–Ύ
Compare F⁻ and I⁻ as leaving groups in nucleophilic substitution reactions.
1
Leaving group ability increases with: weaker basicity, better ability to stabilise negative charge.
2
F⁻: small, highly electronegative, strong base β†’ poor leaving group.
3
I⁻: large, polarisable, weak base β†’ excellent leaving group.
4
Order: I⁻ > Br⁻ > Cl⁻ > F⁻
βœ“  I⁻ is the better leaving group (weaker base, more polarisable)
MediumPredict major elimination product: 2-bromobutane + KOH (alcohol)β–Ύ
2-Bromobutane + KOH in alcohol at high temperature. Predict the major elimination product (Saytzeff's rule).
1
Conditions: strong base (KOH), alcohol solvent, heat β†’ E2 elimination.
2
2-Bromobutane: CHβ‚ƒβˆ’CHBrβˆ’CHβ‚‚βˆ’CH₃ (Br on C2).
3
Two possible alkenes: but-1-ene (CHβ‚‚=CHβˆ’CHβ‚‚CH₃) and but-2-ene (CHβ‚ƒβˆ’CH=CHβˆ’CH₃)
4
Saytzeff's rule: more substituted alkene = major product.
5
But-2-ene has 2 substituents on double bond (more stable) vs but-1-ene (1 substituent).
6
Major product: but-2-ene (CH₃CH=CHCH₃)
βœ“  Major product: but-2-ene (Saytzeff's rule β€” more substituted alkene)
EAPCET LevelWhy does chlorobenzene not undergo SN2 with NaOH easily?β–Ύ
Explain why chlorobenzene is much less reactive toward NaOH(aq) compared to chloroethane.
1
1. Resonance stabilisation: Lone pair on Cl delocalises into benzene ring via resonance β†’ Cβˆ’Cl bond has partial double bond character β†’ stronger bond.
2
2. spΒ² carbon: Benzene ring carbons are spΒ² hybridised β†’ flat geometry β†’ backside attack (required for SN2) is geometrically impossible.
3
3. No SN1 either: Aryl carbocations (C₆H₅⁺) are extremely unstable β€” high energy species not formed.
4
Therefore chlorobenzene needs much more drastic conditions (high T, high P, Cu catalyst) for nucleophilic substitution.
βœ“  Chlorobenzene resists SN because: Cβˆ’Cl has partial Ο€ character (resonance), spΒ² geometry blocks backside attack, and aryl carbocations are unstable
Trap QuestionSN2 reaction gives inversion of configuration β€” this means the product and reactant are enantiomers, always.β–Ύ
In SN2 reaction of (R)-2-bromobutane with OH⁻. A student says the product is the (S) enantiomer. Is this always true?
1
SN2: backside attack by nucleophile β†’ Walden inversion (configuration at the chiral centre is inverted).
2
If reactant is (R)-2-bromobutane, the SN2 product with OH⁻ will have inverted configuration at C2.
3
The configuration at C2 changes from R to S (or vice versa) due to the Walden inversion.
4
However, whether the product is labelled (R) or (S) also depends on the CIP priority order of substituents.
5
In this specific case: (R)-2-bromobutane + OH⁻ β†’ (S)-2-butanol (inversion confirmed). This is correct.
βœ“  Correct for this case β€” SN2 always gives inversion at the reacting carbon (Walden inversion)

Mistake DNA

4 haloalkane errors from EAPCET distractor analysis.

πŸ”„
SN1 for Primary Halides in Polar Protic Solvent
Primary halides undergo SN1 because polar protic solvents are used β€” incorrect. Primary carbocations are too unstable for SN1.
❌ Wrong
1Β° RX in EtOH/Hβ‚‚O: SN1 mechanism βœ— (1Β° carbocation too unstable)
βœ“ Correct
1Β° RX: SN2 mechanism βœ“ (less steric hindrance) SN1 needs 3Β° or 2Β° stable carbocation βœ“
SN1 requires a stable carbocation intermediate. Only 3Β° (and some 2Β°) substrates can form stable carbocations. Primary substrates cannot β€” they undergo SN2 instead.
πŸ”’
E2 Product: Less Substituted Alkene is Major (Anti-Saytzeff)
Saytzeff's rule says MORE substituted alkene is the major E2 product. Students sometimes apply it backwards.
❌ Wrong
E2 of 2-bromobutane: but-1-ene is major βœ— (Saytzeff says more substituted)
βœ“ Correct
Saytzeff: more substituted βœ“ but-2-ene is major βœ“ (more stable, internal double bond)
Saytzeff's rule: thermodynamically more stable (more substituted) alkene is the major product. Exception: bulky base (like KOtBu) favours less substituted alkene (Hofmann product) due to steric reasons.
βš—οΈ
CN⁻ Attacks via Carbon in KCN vs Silver Cyanide
CN⁻ is an ambidentate nucleophile. KCN gives alkyl cyanide (Cβˆ’attack); AgCN gives isocyanide (Nβˆ’attack).
❌ Wrong
KCN + RX β†’ Alkyl isocyanide βœ— (KCN gives cyanide, not isocyanide)
βœ“ Correct
KCN β†’ RCN (nitrile) βœ“ (C attacks) AgCN β†’ RNC (isocyanide) βœ“ (N attacks) Ag⁺ deactivates C end
In KCN, CN⁻ is free and attacks through the softer carbon end β†’ alkyl nitrile (Rβˆ’CN). In AgCN, Ag⁺ coordinates to carbon, blocking it β†’ attack through N β†’ isocyanide (Rβˆ’NC).
πŸ”’
Haloarenes Undergo SN2 with Strong Nucleophiles
Haloarenes do NOT undergo SN2 under normal conditions because the aromatic ring's spΒ² geometry makes backside attack impossible.
❌ Wrong
C₆Hβ‚…Cl + NaOH(aq) β†’ C₆Hβ‚…OH βœ— (simple SN2 at aryl C not possible normally)
βœ“ Correct
C₆Hβ‚…Cl: very unreactive βœ“ Needs 300Β°C + NaOH(l) βœ“ (Dow process) or DDA conditions βœ“
Aryl halides require extreme conditions for nucleophilic substitution: (1) Dow process: 300Β°C, 200 atm, NaOH(l). (2) Meisenheimer complex via addition-elimination at high temperature with electron-withdrawing groups ortho/para.

Chapter Intelligence

Haloalkanes is the gateway to nucleophilic substitution β€” the mechanism framework applies throughout organic chemistry.

EAPCET Weightage (2019–2024)
SN1 vs SN2 identification
~8
Leaving group order (I>Br>Cl>F)
~7
Elimination (Saytzeff's rule)
~6
Nucleophile identification
~5
Haloarene reduced reactivity
~4
High-Yield PYQ Patterns
SN1 or SN2 for given substrateLeaving group ability orderMajor E2 product by SaytzeffKCN vs AgCN productsWhy chlorobenzene unreactiveSN2: inversion of configurationSN1: racemisation explanation
Exam Strategy
  • SN1 vs SN2: check substrate first. 3Β° β†’ SN1. 1Β° (or methyl) β†’ SN2. 2Β° β†’ depends on other conditions.
  • Leaving group ability: I⁻ > Br⁻ > Cl⁻ > F⁻ (bond strength order: Cβˆ’F strongest, Cβˆ’I weakest; weaker bond = better leaving).
  • Saytzeff's rule: E2 gives more substituted alkene as major product. Bulky base (KO-tBu) gives less substituted (Hofmann) product.
  • KCN β†’ nitrile (C attack, chain elongation by 1C). AgCN β†’ isocyanide (N attack, no chain elongation).
  • This chapter feeds directly into Alcohols (OH replaces X) and Amines (NHβ‚‚ replaces X).
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