Haloalkanes & Haloarenes — vidhyapath 2027

Concept Core

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

SN1 vs SN2 — Comparison Table

Feature	SN1	SN2
Mechanism	2 steps (carbocation intermediate)	1 step (concerted)
Rate-determining step	Formation of carbocation	Backside attack by nucleophile
Rate law	Rate = k[RX] (first order)	Rate = k[RX][Nu] (second order)
Substrate preference	3° > 2° (stable carbocation)	1° > 2° (less steric hindrance)
Stereochemistry	Racemisation (mixture of enantiomers)	Inversion (Walden inversion)
Solvent	Polar protic (stabilises carbocation)	Polar aprotic (activates nucleophile)

Common Nucleophiles and Reactions

Nucleophile	Product	Type
OH⁻ (aq)	Alcohol (R−OH)	Substitution
CN⁻	Nitrile (R−CN) → carboxylic acid	Substitution (chain extension)
NH₃	Amine (R−NH₂)	Substitution
AgNO₃ (aq)	AgX precipitate + alcohol	Substitution + test
KCN (alc)	Nitrile (R−CN)	Substitution

Elimination Reactions

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

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

CH₃CH₂CHBrCH₃ + KOH(alc) → CH₃CH=CHCH₃ (major, more substituted) + CH₂=CHCH₂CH₃ (minor, less substituted)

Haloarenes — Reduced Reactivity to SN

Aryl halides (e.g., chlorobenzene, C₆H₅Cl) are very resistant to nucleophilic substitution because:

1. Resonance: C−Cl bond acquires partial double bond character through resonance → shorter, stronger C−X bond.

2. Geometry: sp² carbon — backside attack (SN2) geometrically impossible (flat ring).

3. Carbocation: Aryl carbocations are highly unstable — SN1 not feasible either.

Important Reactions: DDT, Freon, Chloroform

DDT (dichlorodiphenyltrichloroethane): insecticide. Chlorobenzene + chloral → DDT (Friedel-Crafts type).

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

Carbon tetrachloride (CCl₄): Fire extinguisher. Prepared: CCl₄ from CS₂ + Cl₂.

Freons (CFCs): Refrigerants (CHF₂Cl etc.). Cause ozone layer depletion (Cl• radicals).

Substitution vs Elimination — When Each Occurs

Favours Substitution (SN): weak nucleophile, low temperature, 1° substrate Favours Elimination (E2): strong base, high temperature, 3° substrate, bulky base SN1: 3° substrate, polar protic solvent, weak nucleophile SN2: 1° 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 > 1° > 2° > 3°

Less steric hindrance = faster SN2

SN1 Substrate Order

3° > 2° > 1° > 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 — SN1 vs SN2 choice, nucleophile identification, elimination, haloarene, and leaving group.

Easy2-Bromo-2-methylpropane + NaOH(aq) — SN1 or SN2?▾

2-Bromo-2-methylpropane (tert-butyl bromide) reacts with dilute NaOH(aq). Predict SN1 or SN2 and the product.

Tert-butyl bromide is a 3° substrate → favours SN1 (stable tertiary carbocation).

Dilute NaOH(aq) = weak nucleophile, polar protic solvent → confirms SN1 conditions.

Product: (CH₃)₃C−OH (tert-butyl alcohol) via carbocation intermediate.

✓ SN1; product = (CH₃)₃COH (tert-butanol)

EasyWhich is a better leaving group: F⁻ or I⁻?▾

Compare F⁻ and I⁻ as leaving groups in nucleophilic substitution reactions.

Leaving group ability increases with: weaker basicity, better ability to stabilise negative charge.

F⁻: small, highly electronegative, strong base → poor leaving group.

I⁻: large, polarisable, weak base → excellent leaving group.

Order: I⁻ > Br⁻ > Cl⁻ > F⁻

✓ I⁻ is the better leaving group (weaker base, more polarisable)

MediumPredict major elimination product: 2-bromobutane + KOH (alcohol)▾

2-Bromobutane + KOH in alcohol at high temperature. Predict the major elimination product (Saytzeff's rule).

Conditions: strong base (KOH), alcohol solvent, heat → E2 elimination.

2-Bromobutane: CH₃−CHBr−CH₂−CH₃ (Br on C2).

Two possible alkenes: but-1-ene (CH₂=CH−CH₂CH₃) and but-2-ene (CH₃−CH=CH−CH₃)

Saytzeff's rule: more substituted alkene = major product.

But-2-ene has 2 substituents on double bond (more stable) vs but-1-ene (1 substituent).

Major product: but-2-ene (CH₃CH=CHCH₃)

✓ Major product: but-2-ene (Saytzeff's rule — more substituted alkene)

EAPCET LevelWhy does chlorobenzene not undergo SN2 with NaOH easily?▾

Explain why chlorobenzene is much less reactive toward NaOH(aq) compared to chloroethane.

1. Resonance stabilisation: Lone pair on Cl delocalises into benzene ring via resonance → C−Cl bond has partial double bond character → stronger bond.

2. sp² carbon: Benzene ring carbons are sp² hybridised → flat geometry → backside attack (required for SN2) is geometrically impossible.

3. No SN1 either: Aryl carbocations (C₆H₅⁺) are extremely unstable — high energy species not formed.

Therefore chlorobenzene needs much more drastic conditions (high T, high P, Cu catalyst) for nucleophilic substitution.

✓ Chlorobenzene resists SN because: C−Cl has partial π character (resonance), sp² geometry blocks backside attack, and aryl carbocations are unstable

Trap QuestionSN2 reaction gives inversion of configuration — this means the product and reactant are enantiomers, always.▾

In SN2 reaction of (R)-2-bromobutane with OH⁻. A student says the product is the (S) enantiomer. Is this always true?

SN2: backside attack by nucleophile → Walden inversion (configuration at the chiral centre is inverted).

If reactant is (R)-2-bromobutane, the SN2 product with OH⁻ will have inverted configuration at C2.

The configuration at C2 changes from R to S (or vice versa) due to the Walden inversion.

However, whether the product is labelled (R) or (S) also depends on the CIP priority order of substituents.

In this specific case: (R)-2-bromobutane + OH⁻ → (S)-2-butanol (inversion confirmed). This is correct.

✓ Correct for this case — SN2 always gives inversion at the reacting carbon (Walden inversion)

Mistake DNA

4 haloalkane errors from EAPCET distractor analysis.

🔄

SN1 for Primary Halides in Polar Protic Solvent

Primary halides undergo SN1 because polar protic solvents are used — incorrect. Primary carbocations are too unstable for SN1.

❌ Wrong

1° RX in EtOH/H₂O: SN1 mechanism ✗ (1° carbocation too unstable)

✓ Correct

1° RX: SN2 mechanism ✓ (less steric hindrance) SN1 needs 3° or 2° stable carbocation ✓

SN1 requires a stable carbocation intermediate. Only 3° (and some 2°) substrates can form stable carbocations. Primary substrates cannot — they undergo SN2 instead.

🔢

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 ✗ (Saytzeff says more substituted)

✓ Correct

Saytzeff: more substituted ✓ but-2-ene is major ✓ (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.

⚗️

CN⁻ Attacks via Carbon in KCN vs Silver Cyanide

CN⁻ is an ambidentate nucleophile. KCN gives alkyl cyanide (C−attack); AgCN gives isocyanide (N−attack).

❌ Wrong

KCN + RX → Alkyl isocyanide ✗ (KCN gives cyanide, not isocyanide)

✓ Correct

KCN → RCN (nitrile) ✓ (C attacks) AgCN → RNC (isocyanide) ✓ (N attacks) Ag⁺ deactivates C end

In KCN, CN⁻ is free and attacks through the softer carbon end → alkyl nitrile (R−CN). In AgCN, Ag⁺ coordinates to carbon, blocking it → attack through N → isocyanide (R−NC).

🔒

Haloarenes Undergo SN2 with Strong Nucleophiles

Haloarenes do NOT undergo SN2 under normal conditions because the aromatic ring's sp² geometry makes backside attack impossible.

❌ Wrong

C₆H₅Cl + NaOH(aq) → C₆H₅OH ✗ (simple SN2 at aryl C not possible normally)

✓ Correct

C₆H₅Cl: very unreactive ✓ Needs 300°C + NaOH(l) ✓ (Dow process) or DDA conditions ✓

Aryl halides require extreme conditions for nucleophilic substitution: (1) Dow process: 300°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 — the mechanism framework applies throughout organic chemistry.

EAPCET Weightage (2019–2024)

SN1 vs SN2 identification

Leaving group order (I>Br>Cl>F)

Elimination (Saytzeff's rule)

Nucleophile identification

Haloarene reduced reactivity

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. 3° → SN1. 1° (or methyl) → SN2. 2° → depends on other conditions.
Leaving group ability: I⁻ > Br⁻ > Cl⁻ > F⁻ (bond strength order: C−F strongest, C−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 → nitrile (C attack, chain elongation by 1C). AgCN → isocyanide (N attack, no chain elongation).
This chapter feeds directly into Alcohols (OH replaces X) and Amines (NH₂ replaces X).