Elimination Reactions: E1 and E2 Mechanisms

Master E1 and E2 elimination mechanisms, Saytzeff's rule, Hofmann elimination, and competition with substitution for JEE

15 min read

The Hook: How to Make Alkenes from Alkyl Halides

Connect: Real Life → Chemistry

Ever wondered how plastic polymers like polyethylene are made? It all starts with ethylene (H₂C=CH₂), which comes from elimination reactions! When you heat ethyl alcohol with concentrated H₂SO₄, water is eliminated to form ethylene - the building block of plastics worth billions of dollars.

Here’s the JEE question: Why does treating 2-bromobutane with alcoholic KOH give mainly 2-butene, not 1-butene? And why does the same reaction give different products with aqueous vs alcoholic KOH?

The Core Concept

What is Elimination?

Elimination is the removal of two atoms/groups from adjacent carbons to form a double bond.

General Reaction:

$$\boxed{\text{R-CH}_2\text{-CHX-R'} + \text{Base} \rightarrow \text{R-CH=CH-R'} + \text{HX}}$$

Also called: β-elimination or 1,2-elimination

Key terms:

  • α-carbon: Carbon bearing leaving group (X)
  • β-carbon: Adjacent carbon (has H to be removed)
  • β-hydrogen: Hydrogen on β-carbon
JEE Weightage
Elimination Reactions: 3-4 questions in JEE Main, 2-3 in JEE Advanced Focus areas: E1 vs E2 mechanisms, Saytzeff vs Hofmann products, competition with SN1/SN2 This topic is frequently combined with substitution reactions!

E2 Mechanism: Bimolecular Elimination

What is E2?

E2 = Elimination Bimolecular

“Bimolecular” means rate depends on BOTH alkyl halide and base.

The Mechanism (One Step)

Concerted process - all bonds break and form simultaneously

E2 Mechanism - Concerted EliminationAnti-periplanar geometry required, one-step processB:-CbH(beta-H)RCaBrR'Reactantconcertedone stepBHCCBr‡d-d-Transition StateCCRR'Alkene+ B-H + Br-Key Features:Concerted (one step)Strong base requiredAnti-periplanar geometryRate = k[R-X][Base]No rearrangement possible (no intermediate)

Three things happen at once:

  1. Base removes β-hydrogen
  2. C-H bond breaks
  3. C-X bond breaks
  4. π bond forms

Example: 2-bromobutane + KOH (alcoholic)

$$\text{CH}_3\text{CH}_2\text{CHBr-CH}_3 + \text{OH}^- \xrightarrow{\text{C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH=CH-CH}_3 + \text{H}_2\text{O} + \text{Br}^-$$

Rate Equation

$$\boxed{\text{Rate} = k[\text{R-X}][\text{Base}]}$$

Why bimolecular?

  • Both R-X and base involved in rate-determining step
  • Increasing either concentration increases rate

Stereochemistry: Anti-Periplanar Geometry

Most Important for JEE Advanced!

E2 requires specific geometry:

  • H and X must be anti-periplanar (180° dihedral angle)
  • Both must be in same plane
  • On opposite sides of C-C bond

Why this geometry?

  • Allows orbital overlap for π bond formation
  • Lower activation energy
  • Most stable transition state

Newman Projection:

Newman Projection - Anti-Periplanar Geometry for E2H and leaving group (X) must be 180 degrees apart for E2 eliminationAnti-Periplanar(180 dihedral angle)CaH(beta-H)RR'Br(leaving group)R''R'''180E2 Favored!Syn-Periplanar(0 dihedral angle)CaHRR'BrR''R'''0E2 DisfavoredE2 Requirement:Anti (180) = orbital alignment for pi bondIn cyclic systems, H and X must be diaxial (trans) for E2!Syn (0) = poor overlap
JEE Advanced: Stereochemistry Problem

Q: Why doesn’t the following compound undergo E2 elimination easily?

       Br
        |
    [Rigid cyclic structure where H and Br are cis]

Answer:

In a rigid ring system, if H and Br are cis (same side), they CANNOT achieve anti-periplanar geometry!

For E2, need:

  • H and Br on opposite sides (trans)
  • 180° dihedral angle

JEE Tip: When given cyclic compounds, check if H and leaving group can be anti!

Factors Affecting E2 Rate

Factor 1: Structure of Alkyl Halide

Reactivity order:

$$\boxed{3° > 2° > 1°}$$

Why?

  • More substituted alkene is more stable (product stability)
  • More alkyl groups on α-carbon help C-X bond breaking
  • Hyperconjugation stabilizes transition state

Factor 2: Base Strength

Strong bases favor E2

Good E2 bases:

  • OH⁻ (hydroxide)
  • OR⁻ (alkoxide) - e.g., C₂H₅O⁻
  • NH₂⁻ (amide)

Effect of base size:

  • Small strong base (OH⁻, OMe⁻): Can also do substitution
  • Bulky strong base (t-BuO⁻): Prefers elimination over substitution
Memory Trick: Base Selection

“Big Bases prefer Beta (β-elimination)”

For elimination:

  • Bulky bases (tert-butoxide): E2 predominates
  • High temperature: Favors elimination

For substitution:

  • Small bases (methoxide): SN2 can compete
  • Low temperature: Favors substitution

JEE Strategy:

  • t-BuO⁻ + 2° or 3° halide → E2 (almost exclusive)
  • OH⁻ + 1° halide → SN2 (competes)

Interactive Demo: Visualize E1 and E2 Elimination Mechanisms

Watch how β-hydrogen is removed and double bonds form in real-time.

Factor 3: Leaving Group

Better leaving group = faster E2

Same order as SN1/SN2:

$$\boxed{\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-}$$

Factor 4: Solvent

E2 works in both protic and aprotic solvents

Common E2 solvents:

  • Alcoholic KOH (C₂H₅OH + KOH)
  • Hot alcohol with base
  • DMSO with strong base

E1 Mechanism: Unimolecular Elimination

What is E1?

E1 = Elimination Unimolecular

“Unimolecular” means rate depends ONLY on alkyl halide concentration.

The Mechanism (Two Steps)

Step 1: Ionization (SLOW - Rate Determining)

$$\boxed{\text{R-X} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-}$$

Forms carbocation intermediate (same as SN1!)

Step 2: Deprotonation (FAST)

$$\boxed{\text{R}^+ \xrightarrow{\text{-H}^+, \text{ fast}} \text{Alkene}}$$

Base removes β-hydrogen, forming π bond.

Example: tert-butyl bromide + H₂O (heat)

Step 1:

$$(CH_3)_3C-Br \xrightarrow{\text{slow}} (CH_3)_3C^+ + Br^-$$

Step 2: Base removes β-hydrogen from adjacent carbon, electrons form π bond with carbocation.

E1 Mechanism - Two-Step EliminationVia carbocation intermediate, rate = k[R-X], rearrangement possibleStep 1: Ionization (SLOW)CCH3CH3CH3Br(CH3)3C-Brslow-Br-CarbocationC+CH3CH3CH2H(beta-H)+ Br-Step 2: Deprotonation (FAST)fastBase removes H+CCCH3CH3HH+ H+ + Br-(CH3)2C=CH2IsobutyleneEnergy ProfileR-XR+AlkeneTS1TS2Step 1 = RDSE1 Key Points:Carbocation intermediate (same as SN1)Rearrangement possible!Rate = k[R-X] onlyBest substrates: 3 degree > 2 degree >> 1 degreeFavored by: polar protic solvent, high temperature, weak base

Rate Equation

$$\boxed{\text{Rate} = k[\text{R-X}]}$$

Why unimolecular?

  • Rate depends ONLY on [R-X]
  • Base NOT involved in slow step
  • Same as SN1!

Same carbocation as SN1! → E1 and SN1 compete

Factors Affecting E1 Rate

Factor 1: Structure

Reactivity order:

$$\boxed{3° > 2° > 1°}$$

Why?

  • More stable carbocation → faster ionization
  • Same reason as SN1!

Factor 2: Solvent

Polar protic solvents favor E1

Best E1 solvents:

  • Water (H₂O)
  • Alcohols (heat)
  • Dilute acids

Why?

  • Stabilize carbocation intermediate
  • Same as SN1

Factor 3: Temperature

High temperature favors E1

Why?

  • Elimination is entropically favored (creates more molecules)
  • Heat provides energy for C-X bond breaking
E1 vs SN1 Competition

Both mechanisms:

  • Same first step (carbocation formation)
  • Same substrate preference (3° > 2°)
  • Same solvent preference (polar protic)

How to favor E1 over SN1:

  1. Higher temperature (E1 favored)
  2. Stronger base (removes H⁺ faster)
  3. Less nucleophilic solvent

Typical conditions:

  • SN1: Low temp, good nucleophile (H₂O, ROH)
  • E1: High temp, poor nucleophile or weak base

JEE Tip: If conditions favor carbocation formation BUT high temperature → expect E1!

Carbocation Rearrangements in E1

Just like SN1, E1 can have rearrangements!

Example:

    CH₃                      CH₃
     |                        |
CH₃—C—CH₂Br  →  CH₃—C—CH₂⁺  →  (rearrange)  →  CH₃—C⁺—CH₃
     |                        |                      |
    CH₃                      CH₃                    CH₃

                         →  CH₂=C(CH₃)₂

Saytzeff’s Rule: Predicting Major Product

The Rule

Saytzeff’s Rule (Zaitsev’s Rule):

In elimination, the more substituted alkene is the major product.

Also stated as:

Hydrogen is preferentially removed from the β-carbon with fewer hydrogens.

Why More Substituted is More Stable

Alkene stability order:

$$\boxed{\text{Tetrasubstituted} > \text{Trisubstituted} > \text{Disubstituted} > \text{Monosubstituted}}$$

Reasons:

  1. Hyperconjugation: More alkyl groups = more hyperconjugation
  2. Inductive effect: Alkyl groups donate electrons, stabilize π bond
Example: Saytzeff's Rule

Q: What are the products when 2-bromobutane undergoes elimination?

Structure: CH₃-CHBr-CH₂-CH₃

Possible products:

Product 1: 2-butene (more substituted)

$$\text{CH}_3\text{-CH=CH-CH}_3$$
  • Two alkyl groups on C=C
  • Disubstituted alkene
  • Major product (Saytzeff)

Product 2: 1-butene (less substituted)

$$\text{CH}_2\text{=CH-CH}_2\text{-CH}_3$$
  • One alkyl group on C=C
  • Monosubstituted alkene
  • Minor product

Ratio: ~80% 2-butene : 20% 1-butene

JEE Rule: “Saytzeff says: More Substituted!”

Counting β-Hydrogens

Method to predict major product:

  1. Identify α-carbon (has X)
  2. Identify all β-carbons (adjacent to α)
  3. Count hydrogens on each β-carbon
  4. Removal from β-carbon with fewer H gives more substituted alkene

Example:

       β₂  α   β₁
        |   |   |
   CH₃-CH-CHBr-CH₃
        |
       CH₃

   β₁: 3 hydrogens (CH₃ group)
   β₂: 1 hydrogen (CH group)

Removal from β₂ (fewer H) → more substituted alkene → major product

Memory Trick: Saytzeff

“S for Saytzeff, S for Substituted”

Saytzeff product:

  • More Substituted
  • More Stable
  • Standard conditions (alcoholic KOH)

To remember: “Zaitsev Zays: Zore Zubstituents!” (phonetically)

Hofmann Elimination: Exception to Saytzeff

When Does Hofmann Occur?

Hofmann elimination gives less substituted alkene as major product.

Conditions for Hofmann:

  1. Bulky base (difficult to reach more substituted position)
  2. Quaternary ammonium salts (R₄N⁺X⁻)
  3. Leaving group with extra positive charge

Hofmann’s Rule

With bulky bases or quaternary ammonium salts, less substituted alkene is major product.

Why?

  • Bulky base cannot easily approach more hindered β-carbon
  • Removes H from less hindered (more accessible) position
  • Steric factors override thermodynamic stability
Example: Hofmann Elimination

Q: What is the major product from Hofmann elimination of:

$$\text{(CH}_3)_2\text{CH-CH}_2\text{-N}^+(\text{CH}_3)_3 \text{ OH}^- \xrightarrow{\text{heat}} ?$$

Analysis:

Possible products:

Product 1: (CH₃)₂C=CH₂ (more substituted - Saytzeff) Product 2: CH₃-CH=CH-CH₃ (less substituted)

With quaternary ammonium + heat (Hofmann conditions):

Major product: CH₂=CH-CH(CH₃)₂ (less substituted)

Why?

  • Large N(CH₃)₃ group creates steric hindrance
  • Base preferentially removes H from less hindered position
  • Hofmann product predominates

JEE Tip: Hofmann elimination is exception to Saytzeff - look for bulky groups!

E1 vs E2: The Big Comparison

Side-by-Side Comparison

FeatureE2E1
Full NameElimination BimolecularElimination Unimolecular
Steps1 step (concerted)2 steps (via carbocation)
Rate EquationRate = k[R-X][Base]Rate = k[R-X]
IntermediateTransition state onlyCarbocation intermediate
Substrate3° ≥ 2° > 1°3° > 2° » 1°
BaseStrong base requiredWeak base OK
StereochemistryAnti-periplanar requiredNo restriction
RearrangementNo rearrangementCan rearrange
Major productSaytzeff (usually)Saytzeff
Competes withSN2SN1
TemperatureAny temperatureHigh temperature favors
JEE Decision Grid

The “BEAST” Rule for E2:

  • Bimolecular (rate depends on [R-X][Base])
  • Elimination in one step
  • Anti-periplanar geometry
  • Strong base needed
  • Transition state (no intermediate)

The “CHAIR” Rule for E1:

  • Carbocation intermediate
  • High temperature favors
  • All-around geometry (no restriction)
  • Ionization first (slow step)
  • Rearrangement possible

Master these and you’ll never confuse them!

Substitution vs Elimination: The Competition

When Do They Compete?

SN2 vs E2 - compete under similar conditions SN1 vs E1 - share same first step (carbocation)

Factors Determining Substitution vs Elimination

Factor 1: Structure of Substrate

Primary (1°):

  • SN2 strongly favored
  • E2 very slow (less stable alkene product)
  • Ratio: ~90% SN2, 10% E2

Secondary (2°):

  • Both compete
  • Small base → SN2
  • Bulky base → E2
  • High temp → E2

Tertiary (3°):

  • SN2 doesn’t occur (too hindered)
  • E2 strongly favored
  • Can also do E1/SN1 in polar protic solvents
Quick Decision Rule

For Primary (1°) Substrate:

  • Strong small base → SN2 (major)
  • Strong bulky base → E2 (major)

For Secondary (2°) Substrate:

  • Strong base, low temp → SN2
  • Strong base, high temp → E2
  • Bulky base → E2

For Tertiary (3°) Substrate:

  • Strong base → E2 (major)
  • Weak nucleophile, polar protic → SN1/E1 mixture
  • High temp → E1 over SN1

Memory: “1° loves SN2, 3° loves E2, 2° is confused!”

Factor 2: Base/Nucleophile

Strong nucleophile + small base:

  • Favors substitution (SN2)
  • Examples: I⁻, CN⁻, RS⁻

Strong base + bulky:

  • Favors elimination (E2)
  • Examples: t-BuO⁻, LDA

Weak nucleophile:

  • With 3°: SN1/E1 mixture
  • With 1°: Very slow reactions

Factor 3: Solvent

Polar aprotic (acetone, DMSO):

  • Favors SN2
  • Can also support E2

Polar protic (H₂O, ROH):

  • Favors SN1/E1 for 3° substrates
  • Heat: E1 predominates

Factor 4: Temperature

Low temperature:

  • Kinetic control
  • Favors substitution

High temperature:

  • Thermodynamic control
  • Favors elimination (ΔS positive)
JEE Advanced: Predict Products

Q: Predict major product and mechanism for each:

(a) CH₃CH₂Br + OH⁻ in ethanol (room temp) (b) (CH₃)₂CHBr + t-BuO⁻ in t-BuOH (c) (CH₃)₃CBr + H₂O (heat) (d) CH₃CH₂CH₂Br + CN⁻ in acetone

Solutions:

(a) SN2 major, some E2

  • 1° substrate
  • Strong base, moderate temp
  • Products: CH₃CH₂OH (major) + CH₂=CH₂ (minor)

(b) E2 exclusively

  • 2° substrate
  • Bulky strong base (t-BuO⁻)
  • Product: CH₃CH=CH₂

(c) E1 major, some SN1

  • 3° substrate
  • Weak nucleophile, heat
  • Products: (CH₃)₂C=CH₂ (major E1) + (CH₃)₃COH (minor SN1)

(d) SN2 exclusively

  • 1° substrate
  • Strong nucleophile (CN⁻), poor base
  • Polar aprotic solvent
  • Product: CH₃CH₂CH₂CN

JEE Strategy: Systematically check: substrate (1°/2°/3°), base/Nu strength, solvent, temperature!

Common Mistakes to Avoid

Mistake #1: Ignoring Saytzeff's Rule

Wrong: “All β-hydrogens removed equally”

Correct: Preferentially removes H from β-carbon with fewer H atoms

Example: 2-bromopentane gives mainly 2-pentene (trisubstituted), not 1-pentene (disubstituted)

JEE Tip: Always identify more substituted alkene - that’s Saytzeff product!

Mistake #2: Confusing E1 and E2 Conditions

Wrong: “Strong base means E1”

Correct:

  • Strong base → E2 (bimolecular)
  • Weak base + heat → E1 (via carbocation)

Memory: “Strong (2 syllables) goes with E2”

Mistake #3: Forgetting Rearrangements

Wrong: Predicting E1 product without considering rearrangement

Correct: E1 has carbocation intermediate → CAN rearrange

Example:

CH₃-CH(CH₃)-CH₂Br  →  (via E1)
    ↓ (ionization)
CH₃-CH(CH₃)-CH₂⁺  →  (rearrange)
    ↓ (1,2-H shift)
CH₃-C⁺(CH₃)-CH₃  →  (elimination)
    ↓
CH₂=C(CH₃)₂ (rearranged product!)

JEE Strategy: For E1, always check carbocation stability and rearrangement possibility!

Mistake #4: Wrong Competition Prediction

Wrong: “1° substrate with strong base gives only E2”

Correct: 1° substrate favors SN2, even with strong base

Competition depends on:

  • Substrate (1°/2°/3°)
  • Base size (small/bulky)
  • Temperature

JEE Rule:

  • 1° + any base → SN2 major (unless very bulky base + heat)
  • 3° + strong base → E2 major
  • 2° → depends on conditions!

Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Mechanism Identification

Q: Identify E1 or E2 mechanism:

(a) CH₃CH₂Br + KOH (alcoholic) → CH₂=CH₂ (b) (CH₃)₃CBr + H₂O (heat) → (CH₃)₂C=CH₂

Solutions:

(a) E2

  • Strong base (KOH)
  • Alcoholic solution
  • One-step concerted mechanism
  • Rate = k[R-X][OH⁻]

(b) E1

  • Weak base (H₂O)
  • 3° substrate forms stable carbocation
  • Heat provides energy
  • Two-step mechanism
  • Rate = k[(CH₃)₃CBr]
Problem 2: Saytzeff Product

Q: What is the major product from dehydrohalogenation of 2-bromobutane?

Solution:

Structure: CH₃-CHBr-CH₂-CH₃

Possible alkenes:

  1. CH₃-CH=CH-CH₃ (2-butene) - disubstituted
  2. CH₂=CH-CH₂-CH₃ (1-butene) - monosubstituted

Major product: 2-butene (CH₃-CH=CH-CH₃)

Reason: More substituted alkene (Saytzeff’s rule)

Additional: 2-butene exists as cis and trans isomers!

Level 2: JEE Main

Problem 3: E2 Stereochemistry

Q: Which stereoisomer of 1,2-dibromocyclohexane undergoes E2 elimination faster?

(a) Both Br trans (axial-equatorial) (b) Both Br cis

Solution:

Answer: (a) Trans isomer

Reason:

  • E2 requires anti-periplanar geometry
  • H and Br must be 180° apart
  • In chair conformation: both must be axial (diaxial)
  • Trans diaxial is easily achieved
  • Cis cannot achieve anti-periplanar

JEE Tip: For cyclic compounds in E2, check for diaxial arrangement!

Problem 4: Substitution vs Elimination

Q: Major product when 2-bromopropane reacts with: (a) KOH in ethanol at 25°C (b) KOH in ethanol at 80°C

Solution:

Structure: CH₃-CHBr-CH₃ (2° substrate)

(a) At 25°C: SN2 major

  • Product: CH₃-CHOH-CH₃ (2-propanol) - major
  • Some CH₃-CH=CH₂ (propene) - minor

(b) At 80°C: E2 major

  • Product: CH₃-CH=CH₂ (propene) - major
  • Some CH₃-CHOH-CH₃ - minor

Reason: High temperature favors elimination (entropically favored)

Level 3: JEE Advanced

Problem 5: Carbocation Rearrangement in E1

Q: What is the major product when 3,3-dimethyl-2-bromobutane undergoes E1 elimination?

Structure:

    CH₃  Br
     |   |
CH₃—C—CH—CH₃
     |
    CH₃

Solution:

Step 1: Ionization

    CH₃  Br              CH₃  ⁺
     |   |                |   |
CH₃—C—CH—CH₃  →  CH₃—C—CH—CH₃
     |                    |
    CH₃                  CH₃

  (2° carbocation - can rearrange)

Step 2: Rearrangement (1,2-CH₃ shift)

    CH₃  ⁺              CH₃  CH₃
     |   |               |   |
CH₃—C—CH—CH₃  →  CH₃—C⁺—C—CH₃
     |                   |
    CH₃                 CH₃

  2° carbocation      3° carbocation

Step 3: Elimination (remove H⁺)

       H
       |
   CH₂—C⁺(CH₃)—C(CH₃)₂  →  CH₂=C(CH₃)—C(CH₃)₂

                         2,3,3-trimethyl-1-butene

Answer: 2,3,3-trimethyl-1-butene (rearranged product)

JEE Tip: E1 can give unexpected products due to rearrangement!

Problem 6: Hofmann vs Saytzeff

Q: Predict major product:

(a) CH₃CH₂CH(Br)CH₃ + KOH/ethanol (b) CH₃CH₂CH(N⁺(CH₃)₃)CH₃ OH⁻ + heat

Solutions:

(a) Saytzeff product (E2 with normal base)

Products possible:

  • CH₃CH=CHCH₃ (2-butene) - trisubstituted
  • CH₃CH₂CH=CH₂ (1-butene) - disubstituted

Major: 2-butene (more substituted)

(b) Hofmann product (bulky leaving group)

Major: 1-butene (less substituted)

Reason:

  • Quaternary ammonium has bulky N(CH₃)₃ group
  • Steric hindrance prevents access to more substituted position
  • Base removes H from less hindered terminal position

Ratio: ~70% 1-butene : 30% 2-butene (reversed from Saytzeff!)

Problem 7: Complete Competition Analysis

Q: For each reaction, predict major mechanism (SN1/SN2/E1/E2) and major product:

(a) CH₃Br + CN⁻ in DMSO (b) (CH₃)₃CCl + CH₃OH (heat) (c) CH₃CH₂CH₂Br + t-BuO⁻ in t-BuOH (d) (CH₃)₂CHBr + I⁻ in acetone

Solutions:

(a) SN2 - CH₃CN

  • Methyl substrate (no hindrance)
  • Strong nucleophile
  • Polar aprotic solvent

(b) E1 major, SN1 minor

  • 3° substrate
  • Weak nucleophile
  • Protic solvent + heat
  • Major: (CH₃)₂C=CH₂, Minor: (CH₃)₃COCH₃

(c) E2 - CH₃CH=CH₂

  • 1° substrate BUT bulky base
  • t-BuO⁻ is very hindered → can’t do SN2
  • E2 predominates

(d) SN2 - (CH₃)₂CHI

  • 2° substrate
  • I⁻ is excellent nucleophile but poor base
  • Polar aprotic solvent
  • No heat → substitution favored

JEE Strategy: Consider ALL factors - substrate, base/Nu, solvent, temperature!

Quick Revision Box

TopicKey PointsJEE Rule
E2One step, anti-periplanarStrong base, concerted
E1Two steps, via carbocationWeak base + heat
Rate (E2)k[R-X][Base]Bimolecular
Rate (E1)k[R-X]Unimolecular
Substrate (E2)3° ≥ 2° > 1°Strong base needed
Substrate (E1)3° > 2° » 1°Stable carbocation
SaytzeffMore substituted alkeneStandard elimination
HofmannLess substituted alkeneBulky base or R₄N⁺
RearrangementE2: No, E1: YesSame as SN patterns
vs SN21° → SN2, 3° → E2Bulky base → E2
vs SN1Both via carbocationHeat → E1

Connection to Other Topics

Prerequisites:

Produces:

Related Mechanisms:

Applications:

Teacher’s Summary

Key Takeaways

1. Two Elimination Mechanisms:

  • E2: One step, needs strong base, anti-periplanar geometry
  • E1: Two steps, carbocation intermediate, heat favors

2. Product Prediction (Saytzeff vs Hofmann):

  • Saytzeff (normal): More substituted alkene (standard conditions)
  • Hofmann (exception): Less substituted (bulky base or R₄N⁺)

3. Stereochemistry:

  • E2: Requires anti-periplanar (H and X at 180°)
  • E1: No restriction (carbocation is planar)

4. Rearrangements:

  • E2: No rearrangement (concerted)
  • E1: Can rearrange (carbocation intermediate)

5. Competition with Substitution (MOST IMPORTANT):

Primary (1°) substrates:

  • Usually SN2 > E2 (unless bulky base)

Tertiary (3°) substrates:

  • Usually E2 > E1/SN1 (strong base → E2, weak + heat → E1)

Secondary (2°) substrates:

  • Depends on ALL factors (base, solvent, temperature)
  • Strong + small → SN2
  • Strong + bulky → E2
  • Weak + protic + heat → E1

6. JEE Strategy Matrix:

         1° Substrate    2° Substrate    3° Substrate
Small    SN2 major       SN2 or E2       SN1/E1
base

Bulky    E2 (some)       E2 major        E2 only
base

“Master the competition between substitution and elimination - it’s the heart of alkyl halide chemistry!”