Alkenes: Unsaturated Hydrocarbons

Complete guide to alkenes - preparation, electrophilic addition reactions, Markovnikov's rule, and mechanism analysis for JEE

13 min read

The Hook: The Chemistry Behind Ripening Fruits

Connect: Real Life → Chemistry

Ever noticed how raw bananas ripen faster when kept with ripe ones? The secret is ethylene (C₂H₄) - the simplest alkene! Fruits naturally produce this gas to trigger ripening. Commercial fruit vendors use ethylene gas chambers to ripen mangoes and bananas quickly.

But here’s the chemistry twist: Why is ethylene so reactive compared to ethane? Why do alkenes undergo addition reactions while alkanes don’t?

The answer lies in the π bond!

The Core Concept

What are Alkenes?

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C).

$$\boxed{\text{General formula: C}_n\text{H}_{2n}}$$

In simple terms: “Unsaturated” means the molecule can add more atoms across the double bond - like a partially filled container that has room for more!

The Double Bond: σ + π

A C=C double bond consists of:

  1. One σ (sigma) bond - strong, formed by head-on overlap (sp² - sp²)
  2. One π (pi) bond - weaker, formed by sideways overlap (p - p)

Bond energies:

  • C-C (single): 347 kJ/mol
  • C=C (double): 611 kJ/mol
  • π bond energy alone: 611 - 347 = 264 kJ/mol
Why Alkenes are Reactive

The π bond is weaker and above/below the plane → easily attacked by electrophiles!

This is why alkenes undergo addition reactions while alkanes undergo substitution.

Related: Chemical Bonding

JEE Weightage

Alkenes: 4-5 questions in JEE Main, 2-3 in JEE Advanced High-yield topics:

  • Markovnikov vs Anti-Markovnikov
  • Mechanism of electrophilic addition
  • Stability and heat of hydrogenation

Preparation of Alkenes

Method 1: Dehydrohalogenation of Alkyl Halides

β-Elimination (E2 mechanism)

$$\boxed{\text{R-CH}_2\text{-CH}_2\text{-X} + \text{KOH (alc.)} \xrightarrow{\Delta} \text{R-CH=CH}_2 + \text{KX} + \text{H}_2\text{O}}$$

Example:

$$\text{CH}_3\text{CH}_2\text{Br} + \text{KOH (alc.)} \xrightarrow{\Delta} \text{CH}_2\text{=CH}_2 + \text{KBr} + \text{H}_2\text{O}$$

Saytzeff Rule (Zaitsev’s Rule)

“In dehydrohalogenation, the more substituted alkene is the major product”

$$\boxed{\text{More substituted alkene = Major product}}$$

Example: 2-bromobutane elimination

     Br
     |
CH₃-CH-CH₂-CH₃  +  KOH (alc.)
                     ↓
        Two possible products:

1. CH₃-CH=CH-CH₃  (2-butene) ← Major (more substituted)
2. CH₂=CH-CH₂-CH₃  (1-butene) ← Minor (less substituted)

Why? More substituted alkenes are more stable due to:

  • Hyperconjugation (more α-H atoms)
  • Inductive effect (more alkyl groups donate electrons)
Memory Trick: Saytzeff Rule

“Say-tzeff says: Stable products Satisfy”

Stability order:

$$\text{Tetra} > \text{Tri} > \text{Di} > \text{Mono-substituted} > \text{Ethylene}$$

Example:

  • (CH₃)₂C=C(CH₃)₂ (tetra) - Most stable
  • CH₂=CH₂ (no substitution) - Least stable

JEE Tip: Count the total number of alkyl groups on both carbons of C=C!

Method 2: Dehydration of Alcohols

Elimination of H₂O using concentrated H₂SO₄

$$\boxed{\text{R-CH}_2\text{-CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 443\text{ K}} \text{R-CH=CH}_2 + \text{H}_2\text{O}}$$

Mechanism (E1):

Step 1: Protonation

$$\text{R-CH}_2\text{-CH}_2\text{-OH} + \text{H}^+ \rightarrow \text{R-CH}_2\text{-CH}_2\text{-OH}_2^+$$

Step 2: Carbocation formation (slow, rate-determining)

$$\text{R-CH}_2\text{-CH}_2\text{-OH}_2^+ \rightarrow \text{R-CH}_2\text{-CH}_2^+ + \text{H}_2\text{O}$$

Step 3: Deprotonation

$$\text{R-CH}_2\text{-CH}_2^+ \rightarrow \text{R-CH=CH}_2 + \text{H}^+$$

Ease of dehydration:

$$\boxed{3° > 2° > 1° \text{ alcohols}}$$

Why? Carbocation stability: 3° > 2° > 1°

JEE Trap: Carbocation Rearrangement

Common mistake: Ignoring carbocation rearrangements!

Example: Dehydration of 3,3-dimethyl-2-butanol

      OH
      |
CH₃-CH-C(CH₃)₃

      ↓ H₂SO₄

Expected: CH₃-CH=C(CH₃)₂
But also: CH₂=C(CH₃)-C(CH₃)₃  (after hydride shift!)

Rule: If a more stable carbocation can form by hydride shift or methyl shift, it will!

Related: Carbocation Rearrangements

Method 3: Dehalogenation of Vicinal Dihalides

Using Zinc metal

$$\boxed{\text{R-CHX-CHX-R} + \text{Zn} \rightarrow \text{R-CH=CH-R} + \text{ZnX}_2}$$

Example:

$$\text{CH}_2\text{Br-CH}_2\text{Br} + \text{Zn} \rightarrow \text{CH}_2\text{=CH}_2 + \text{ZnBr}_2$$

Using KI in acetone:

$$\boxed{\text{R-CHBr-CHBr-R} + 2\text{KI} \xrightarrow{\text{acetone}} \text{R-CH=CH-R} + 2\text{KBr} + \text{I}_2}$$

Method 4: Partial Reduction of Alkynes

Catalytic Hydrogenation (Lindlar’s Catalyst)

$$\boxed{\text{R-C≡C-R} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4 \text{ (Lindlar)}} \text{R-CH=CH-R}}$$
  • Gives cis-alkene (syn addition)
  • Catalyst is “poisoned” to prevent over-reduction

Reduction with Na/liq. NH₃ (Birch Reduction)

$$\boxed{\text{R-C≡C-R} + 2\text{Na/liq. NH}_3 \rightarrow \text{R-CH=CH-R}}$$
  • Gives trans-alkene (anti addition)
Memory Trick: Stereochemistry

“Lindlar is Like cis, Liquid ammonia is Like trans”

  • Lindlar → Like → cis (both have “l” sound!)
  • Sodium in liquid NH₃ → trans

JEE Application: Questions will ask about stereochemistry of products!

Related: Alkynes

Reactions of Alkenes: Electrophilic Addition

General Mechanism

Step 1: Electrophile attacks π bond → carbocation intermediate Step 2: Nucleophile attacks carbocation

   R-CH=CH₂  +  E⁺
       ↓
   R-CH⁺-CH₂-E   (carbocation)
       ↓  Nu⁻
   R-CH(Nu)-CH₂-E

Reaction 1: Hydrogenation

Addition of H₂ (Catalytic Reduction)

$$\boxed{\text{R-CH=CH}_2 + \text{H}_2 \xrightarrow{\text{Pt/Pd/Ni}} \text{R-CH}_2\text{-CH}_3}$$

Mechanism: Syn addition (both H atoms add from same face)

Heat of Hydrogenation: Useful to determine alkene stability

$$\boxed{\text{Stability} \propto \frac{1}{\text{Heat of hydrogenation}}}$$

Example:

  • 2-butene: ΔH = -120 kJ/mol (more stable)
  • 1-butene: ΔH = -127 kJ/mol (less stable)

Lower heat released → More stable alkene (already lower energy)

Reaction 2: Addition of Halogen (X₂)

Bromination (Br₂) - Test for Unsaturation

$$\boxed{\text{R-CH=CH}_2 + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{R-CHBr-CH}_2\text{Br}}$$

Observation: Red-brown Br₂ color disappears (decolorization test!)

Mechanism: Bromonium ion intermediate

Step 1: π electrons attack Br₂

   R-CH=CH₂  +  Br-Br
       ↓
       Br⁺
      / \
   R-CH—CH₂  (Bromonium ion - 3-membered ring)

Step 2: Br⁻ attacks from backside (anti addition)

       Br
      / \
   R-CH—CH₂  +  Br⁻
       ↓
   R-CHBr-CH₂Br  (trans addition)

Stereochemistry: Anti addition (opposite sides)

JEE Application: Stereochemistry

Q: What is the product when cis-2-butene reacts with Br₂?

Answer: Meso 2,3-dibromobutane (anti addition gives meso form from cis alkene)

From trans-2-butene → Racemic mixture (d + l forms)

JEE Advanced loves: Stereochemical outcomes of additions!

Related: Stereoisomerism

Reaction 3: Addition of Hydrogen Halides (HX)

Hydrohalogenation

$$\boxed{\text{R-CH=CH}_2 + \text{HX} \rightarrow \text{R-CHX-CH}_3}$$

Markovnikov’s Rule

“In the addition of HX to an unsymmetrical alkene, H goes to the carbon with more H atoms”

Or equivalently: “The halogen goes to the more substituted carbon”

$$\boxed{\text{Halogen → More substituted C}}$$

Example:

$$\text{CH}_3\text{-CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{-CHBr-CH}_3$$

Not → CH₃-CH₂-CH₂Br (this is minor product)

Mechanism:

Step 1: Protonation → more stable carbocation

CH₃-CH=CH₂  +  H⁺
      ↓
CH₃-CH⁺-CH₃  (2° carbocation - STABLE)

NOT: CH₃-CH₂-CH₂⁺  (1° carbocation - UNSTABLE)

Step 2: Bromide ion attacks

CH₃-CH⁺-CH₃  +  Br⁻
      ↓
CH₃-CHBr-CH₃
Memory Trick: Markovnikov's Rule

“Rich get Richer”

The carbon that’s already rich in H atoms gets another H atom!

Or think: “More Markovnikov = More substituted”

Modern statement (easier): “Positive charge goes to the more substituted carbon” (more stable carbocation)

Then nucleophile attacks there!

Reactivity of Hydrogen Halides:

$$\boxed{\text{HI} > \text{HBr} > \text{HCl} > \text{HF}}$$

Why?

  • HI has weakest bond (easiest to break)
  • HF has strongest bond (hardest to break)
  • Also, I⁻ is best leaving group

Anti-Markovnikov Addition (Peroxide Effect/Kharasch Effect)

Only works with HBr (not HCl or HI)

$$\boxed{\text{R-CH=CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \text{R-CH}_2\text{-CH}_2\text{Br}}$$

Mechanism: Free radical addition (NOT electrophilic!)

Step 1: Peroxide homolysis

$$\text{R-O-O-R} \xrightarrow{\Delta} 2\text{RO}^\bullet$$

Step 2: Radical abstracts H from HBr

$$\text{RO}^\bullet + \text{H-Br} \rightarrow \text{ROH} + \text{Br}^\bullet$$

Step 3: Br• adds to less substituted carbon

$$\text{CH}_3\text{-CH=CH}_2 + \text{Br}^\bullet \rightarrow \text{CH}_3\text{-CH}^\bullet\text{-CH}_2\text{Br}$$

Why? Forms more stable radical (2° radical)

Step 4: Chain propagation

$$\text{CH}_3\text{-CH}^\bullet\text{-CH}_2\text{Br} + \text{HBr} \rightarrow \text{CH}_3\text{-CH}_2\text{-CH}_2\text{Br} + \text{Br}^\bullet$$
JEE Trap: Why only HBr?

Q: Why doesn’t peroxide effect work with HCl or HI?

Answer:

For HCl:

  • H-Cl bond too strong (431 kJ/mol)
  • Cl• radical can’t form easily
  • First step (RO• + HCl) is endothermic

For HI:

  • H-I bond too weak (299 kJ/mol)
  • Reaction is too fast and uncontrolled
  • Forms iodine radicals that couple (I• + I• → I₂)

HBr is Goldilocks: Just right! (366 kJ/mol)

JEE One-liner: “Peroxide effect is the Best Bet for HBr only”

Reaction 4: Addition of Water (Hydration)

Acid-catalyzed hydration

$$\boxed{\text{R-CH=CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4} \text{R-CH(OH)-CH}_3}$$

Follows Markovnikov’s rule!

Mechanism:

Step 1: Protonation

$$\text{R-CH=CH}_2 + \text{H}^+ \rightarrow \text{R-CH}^+\text{-CH}_3$$

(2° carbocation)

Step 2: Water attacks

$$\text{R-CH}^+\text{-CH}_3 + \text{H}_2\text{O} \rightarrow \text{R-CH(OH}_2^+\text{)-CH}_3$$

Step 3: Deprotonation

$$\text{R-CH(OH}_2^+\text{)-CH}_3 \rightarrow \text{R-CH(OH)-CH}_3 + \text{H}^+$$

Example:

$$\text{CH}_3\text{-CH=CH}_2 \xrightarrow{\text{H}_2\text{SO}_4/\text{H}_2\text{O}} \text{CH}_3\text{-CH(OH)-CH}_3$$

(Propene → 2-propanol, NOT 1-propanol)

Related: Alcohols

Reaction 5: Oxymercuration-Demercuration

Alternative to acid-catalyzed hydration (no rearrangement!)

$$\boxed{\text{R-CH=CH}_2 \xrightarrow[\text{2. NaBH}_4]{\text{1. Hg(OAc)}_2/\text{H}_2\text{O}} \text{R-CH(OH)-CH}_3}$$

Advantages:

  • Follows Markovnikov’s rule
  • No carbocation rearrangement (mercurinium ion intermediate)
  • Better yields

Reaction 6: Hydroboration-Oxidation

Anti-Markovnikov hydration

$$\boxed{\text{R-CH=CH}_2 \xrightarrow[\text{2. H}_2\text{O}_2/\text{OH}^-]{\text{1. B}_2\text{H}_6} \text{R-CH}_2\text{-CH}_2\text{OH}}$$

Step 1: Hydroboration (B₂H₆ adds)

  • Syn addition
  • Boron adds to less substituted carbon

Step 2: Oxidation

  • Replaces B with OH
  • Retains stereochemistry

Net result: Anti-Markovnikov, syn addition of H-OH

Memory Trick: Hydration Methods

Markovnikov hydration:

  • “Marko Merits More” → H₂SO₄/H₂O or Hg(OAc)₂

Anti-Markovnikov:

  • “Boron Breaks the rule” → Hydroboration-oxidation

JEE Decision Tree:

Need alcohol from alkene?
│
├─ Markovnikov? → H₂SO₄/H₂O (watch for rearrangement!)
│                 or Hg(OAc)₂ (no rearrangement)
│
└─ Anti-Markovnikov? → B₂H₆ then H₂O₂/OH⁻

Reaction 7: Ozonolysis

Oxidative cleavage of C=C bond

$$\boxed{\text{R-CH=CH-R} \xrightarrow[\text{2. Zn/H}_2\text{O}]{\text{1. O}_3} \text{R-CHO} + \text{R-CHO}}$$

Mechanism:

Step 1: Formation of ozonide

R-CH=CH-R  +  O₃  →  Molozonide  →  Ozonide (5-membered ring)

Step 2: Reductive cleavage (Zn/H₂O)

Ozonide  +  Zn/H₂O  →  2 R-CHO (aldehydes)

If oxidative workup (H₂O₂):

Ozonide  +  H₂O₂  →  2 R-COOH (carboxylic acids)

JEE Application: Structure determination!

JEE Classic Question Type

Q: An alkene on ozonolysis gives acetone and butanone. What is the structure of the alkene?

Solution: Work backwards!

Products:

  • Acetone: (CH₃)₂C=O
  • Butanone: CH₃-CO-CH₂-CH₃

Reconnect at C=O:

    CH₃               CH₃
     |                 |
CH₃-C=O    +    O=C-CH₂-CH₃

Connect here!
     ↓

    CH₃        CH₃
     |          |
CH₃-C=C-CH₂-CH₃

Answer: 2-methylbut-2-ene

JEE Tip: Ozonolysis questions ALWAYS work backwards from products!

Reaction 8: Potassium Permanganate Oxidation

Cold, dilute KMnO₄ (Baeyer’s test)

$$\boxed{\text{R-CH=CH}_2 + \text{KMnO}_4 \xrightarrow{\text{cold, dilute}} \text{R-CH(OH)-CH}_2\text{OH}}$$

Observation: Purple color of KMnO₄ disappears (positive test for unsaturation!)

Hot, concentrated KMnO₄ (Oxidative cleavage)

$$\boxed{\text{R-CH=CH-R} \xrightarrow[\Delta]{\text{KMnO}_4} \text{R-COOH} + \text{R-COOH}}$$

Similar to ozonolysis but gives carboxylic acids directly.

Polymerization of Alkenes

Addition Polymerization

$$\boxed{n\text{CH}_2\text{=CH}_2 \xrightarrow{\text{catalyst}} (\text{-CH}_2\text{-CH}_2\text{-})_n}$$

Examples:

  1. Polyethylene (from ethylene)

    • Uses: Plastic bags, bottles

  2. Polypropylene (from propylene)

    • Uses: Containers, ropes

  3. Polyvinyl chloride (PVC) (from vinyl chloride)

    • Uses: Pipes, electrical insulation

  4. Polystyrene (from styrene)

    • Uses: Packaging, insulation

Related: Polymers

Common Mistakes to Avoid

Mistake #1: Confusing Markovnikov Orientations

Wrong: “H adds to more substituted carbon”

Correct: “H adds to less substituted carbon (the one with more H already)”

Easy way: Think about the carbocation intermediate!

  • More substituted carbocation = more stable
  • Halogen ends up there
Mistake #2: Forgetting Peroxide Effect Specificity

Wrong: “Peroxide effect works with all HX”

Correct: Only with HBr!

JEE trap question: “What happens when propene reacts with HCl in presence of peroxide?”

Answer: Normal Markovnikov addition (peroxide has no effect with HCl)

Mistake #3: Missing Carbocation Rearrangements

Always check if a hydride shift or methyl shift can form a more stable carbocation!

Example:

     CH₃
      |
CH₃-C-CH=CH₂  +  HBr
      |
     CH₃

Can give: CH₃-C(CH₃)₂-CHBr-CH₃  (after hydride shift!)

JEE Tip: Draw the carbocation and look for possible shifts before writing the product!

Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Nomenclature

Q: Write the IUPAC name of CH₃-CH=CH-CH₂-CH₃

Solution:

  1. Longest chain with C=C: 5 carbons (pentene)
  2. Number from the end closer to C=C
  3. Double bond between C-2 and C-3

Answer: Pent-2-ene

Common mistake: Don’t write “2-pentene” - use “pent-2-ene” format!

Problem 2: Reaction Type

Q: What type of reaction do alkenes typically undergo and why?

Solution:

Type: Electrophilic addition

Why?

  • π bond has high electron density
  • π electrons are loosely held (above/below plane)
  • Easily attacked by electrophiles (H⁺, Br⁺, etc.)

Key point: σ bond remains intact; only π bond breaks!

Level 2: JEE Main

Problem 3: Markovnikov's Rule Application

Q: What is the major product when 2-methylpropene reacts with HBr?

Solution:

Structure: (CH₃)₂C=CH₂

Step 1: Identify possible carbocations

Option 1: (CH₃)₂C⁺-CH₃  (3° carbocation) ← STABLE
Option 2: (CH₃)₂CH-CH₂⁺  (1° carbocation) ← UNSTABLE

Step 2: More stable carbocation forms → Br⁻ attacks there

Product: (CH₃)₂CBr-CH₃ (2-bromo-2-methylpropane)

Answer: (CH₃)₃CBr

This follows Markovnikov’s rule!

Problem 4: Anti-Markovnikov

Q: What is the product when propene reacts with HBr in presence of peroxide?

Solution:

Normal: CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (Markovnikov)

With peroxide: Free radical mechanism

Step 1: Br• adds to less substituted carbon

CH₃-CH=CH₂  +  Br•  →  CH₃-CH•-CH₂Br  (2° radical - more stable)

Step 2: Abstract H from HBr

CH₃-CH•-CH₂Br  +  HBr  →  CH₃-CH₂-CH₂Br

Answer: 1-bromopropane (anti-Markovnikov product)

Key: Br adds to terminal carbon!

Level 3: JEE Advanced

Problem 5: Structure Determination

Q: An alkene (C₅H₁₀) on ozonolysis gives two moles of the same product (CH₃-CO-CH₃). Identify the alkene.

Solution:

Product: Acetone appears twice

Work backwards:

CH₃-C(=O)-CH₃  +  CH₃-C(=O)-CH₃
        ↑               ↑
     Connect at C=O to form C=C

CH₃            CH₃
  \            /
   C==========C
  /            \
CH₃            CH₃

Answer: 2,3-dimethylbut-2-ene (tetramethylethylene)

Check: C₆H₁₂ → Wait, question says C₅H₁₀!

Let me recalculate…

Actually, if we get TWO MOLES of same product from ONE mole alkene:

Only possible if alkene is symmetrical:

(CH₃)₂C=C(CH₃)₂

This is C₆H₁₂, not C₅H₁₀!

Correct question interpretation: The alkene must be C₆H₁₂

Answer: 2,3-dimethylbut-2-ene

JEE Tip: Always verify molecular formula of your answer!

Problem 6: Mechanism-Based

Q: Arrange the following alkenes in order of stability:

  1. CH₂=CH₂
  2. CH₃-CH=CH₂
  3. CH₃-CH=CH-CH₃
  4. (CH₃)₂C=CH₂

Solution:

Use heat of hydrogenation data OR count substituents:

Substituent count:

  1. Ethylene: 0 substituents
  2. Propene: 1 substituent
  3. 2-butene: 2 substituents
  4. 2-methylpropene: 2 substituents

But wait! For (3) and (4):

  • 2-butene: Both R groups are on same side or different (cis vs trans matters!)
  • Assume trans-2-butene for comparison

General stability:

$$\text{More substituted} > \text{Less substituted}$$

For equal substitution:

$$\text{Trans} > \text{Cis}$$

Order of stability:

$$\text{trans-2-butene} > \text{2-methylpropene} > \text{propene} > \text{ethylene}$$

(3) > (4) > (2) > (1)

Explanation:

  • Hyperconjugation: More α-H atoms = more stable
  • trans has less steric strain than cis
  • 2-methylpropene has 2 × 6 = 12 α-H (from two methyls on same C)
  • trans-2-butene has 3 + 3 = 6 α-H (but trans geometry wins!)

JEE Advanced concept: Don’t just count alkyl groups; consider α-H atoms for hyperconjugation!

Quick Revision Table

ReactionReagentProduct TypeFollows Markovnikov?
HydrogenationH₂/Pt, Pd, NiAlkaneN/A
HalogenationX₂/CCl₄Vicinal dihalideN/A (anti addition)
HydrohalogenationHXAlkyl halideYes
Peroxide effectHBr/peroxideAlkyl halideNo (anti-Markovnikov)
HydrationH₂SO₄/H₂OAlcoholYes
OxymercurationHg(OAc)₂, NaBH₄AlcoholYes (no rearrangement)
HydroborationB₂H₆, H₂O₂/OH⁻AlcoholNo (anti-Markovnikov)
OzonolysisO₃, Zn/H₂OAldehydes/KetonesN/A (cleavage)
KMnO₄ (cold)KMnO₄/OH⁻ (cold)DiolN/A
KMnO₄ (hot)KMnO₄/OH⁻ (hot)Carboxylic acidsN/A (cleavage)

Quick Decision Tree

Adding to C=C double bond?
│
├─ Need alkane? → H₂/Pt (hydrogenation)
│
├─ Need dihalide? → Br₂/CCl₄ (halogenation)
│
├─ Need monohalide?
│  ├─ Markovnikov → HX
│  └─ Anti-Markovnikov → HBr + peroxide
│
├─ Need alcohol?
│  ├─ Markovnikov → H₂SO₄/H₂O (watch for rearrangement!)
│  │                or Hg(OAc)₂ then NaBH₄ (no rearrangement)
│  └─ Anti-Markovnikov → B₂H₆ then H₂O₂/OH⁻
│
└─ Want to break C=C?
   ├─ To aldehydes/ketones → O₃ then Zn/H₂O
   └─ To carboxylic acids → KMnO₄ (hot) or O₃ then H₂O₂

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:

Teacher’s Summary

Key Takeaways

1. Alkenes have C=C double bond (σ + π) - Formula: CₙH₂ₙ

2. Key Preparation Methods:

  • Dehydrohalogenation: R-CH₂-CH₂X + KOH(alc) → R-CH=CH₂ (Saytzeff rule)
  • Dehydration: R-CH₂-CH₂OH + H₂SO₄ → R-CH=CH₂ (watch carbocation rearrangement!)
  • From alkynes: Lindlar (cis) or Na/NH₃ (trans)

3. Characteristic Reaction: Electrophilic Addition

  • Mechanism: Electrophile attacks π bond → carbocation → nucleophile attacks
  • π bond breaks, σ bonds form

4. Markovnikov’s Rule: “Positive charge goes to more substituted carbon”

  • HX, H₂O/H⁺, Hg(OAc)₂ → Markovnikov products
  • HBr + peroxide, B₂H₆/H₂O₂ → Anti-Markovnikov products

5. Important Reactions for JEE:

ReagentProductRemember
H₂/PtAlkaneSimple addition
Br₂/CCl₄DihalideDecolorization test
HBrR-BrMarkovnikov
HBr/peroxideR-BrAnti-Markovnikov (only HBr!)
H₂O/H⁺AlcoholMarkovnikov + rearrangement possible
B₂H₆ then H₂O₂AlcoholAnti-Markovnikov, no rearrangement
O₃ then ZnAldehydes/KetonesStructure determination

6. Stability Order: More substituted > Less substituted

  • Tetra > Tri > Di > Mono > Ethylene
  • Trans > Cis (for same substitution)

7. JEE Strategy:

  • Addition reactions → Draw carbocation intermediate first
  • Stereochemistry → Track syn vs anti addition
  • Structure determination → Work backwards from ozonolysis products
  • Always check for carbocation rearrangements in acid-catalyzed reactions

“In alkenes, the π bond is your reaction site - everything adds across it!”

Master alkenes and you’ve mastered half of organic chemistry! Next, study alkynes to understand triple bonds and acidic character!

Interactive Demo: Visualize Reaction Mechanisms

Watch electrophilic addition reactions unfold step-by-step with animated electron movements.