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Alkenes (Unsaturated Hydrocarbons)

CBSE Class 11 & JEE Mains • Module 06 of 20 • The C=C Powerhouse — Electrophilic Addition

📍 Chapter Overview

Alkenes — Complete Mind Map

Topics Covered: General Formula CₙH₂ₙ · sp² Hybridization · π Bond (sideways overlap) · Preparation (dehydration, dehydrohalogenation, Wittig) · Physical Properties · Chemical Reactions: Electrophilic Addition (HX-Markovnikov, H₂SO₄, HOCl, H₂O), Hydrogenation, Ozonolysis, Oxidation (KMnO₄, OsO₄), Polymerisation · Saytzeff's Rule

🤖 AI Prompt — Mind Map: Vibrant green themed chemistry mind map on dark forest green background. Central node: "ALKENES — CₙH₂ₙ" in white bold. Six main branches: (1) "sp² Structure" — ethene planar structure with bond angles 120°, π bond shown as sideways parallel p-orbitals; (2) "Preparation" — three routes: alcohol dehydration (H₂SO₄/heat), dehydrohalogenation (KOH/alc), Wittig Reaction; (3) "Markovnikov's Rule" — shows propene + HBr → 2-bromopropane (major) with rule explanation "H goes to C with more H"; (4) "Reactions" — addition of H₂ (Ni/Pt/Pd), HX, H₂SO₄, Cl₂, HOCl, H₂O/H⁺, HBr/peroxide (anti-Markovnikov); (5) "Ozonolysis" — O₃ then Zn/H₂O → aldehydes and ketones; (6) "KMnO₄ Oxidation" — cold dilute → diol (syn addition), hot conc KMnO₄ → cleavage products. Show all reaction arrows in bright neon colors. High resolution educational poster quality.

1. Structure and Bonding in Alkenes

General formula: CₙH₂ₙ (acyclic alkene with one double bond)

Double bond = 1 σ bond + 1 π bond

Hybridization: Each C of C=C is sp² hybridized — 3 sp² orbitals form σ bonds (120° apart, trigonal planar). The remaining unhybridized p orbital on each C overlaps sideways → forms the π bond.

Bond angles: 120° at the sp² carbons. C=C bond length: 1.34 Å (shorter than C–C 1.54 Å because of the extra π bond). C=C bond is stronger than C–C (614 kJ/mol vs 347 kJ/mol) but the π bond alone (267 kJ/mol) is weaker than σ (347 kJ/mol) → this is why C=C undergoes addition (breaks the π bond).

Ethene (Ethylene) — sp² Hybridization and π Bond Formation

Draw two diagrams side-by-side for ethene (H₂C=CH₂) on white background. Left diagram: 3D structural diagram of ethene molecule. Both carbon atoms are flat (planar) with all 6 atoms (2C, 4H) in one plane. Show bond angles of 120° at each carbon. Each carbon has two C-H bonds (going to hydrogens in the plane) and one C=C bond between the carbons. Mark the first bond (sigma) as a thick solid line and show the double bond with a second line indicating the pi bond. Label: "Planar molecule, all atoms in same plane, bond angle 120°, C=C length = 1.34 Å". Right diagram: Orbital overlap diagram for ethene pi bond. Show each carbon with three sp² orbitals in a plane (shown as elongated lobes in the plane, pointing left-right-center). Then show the two unhybridized p orbitals — one on each carbon — perpendicular to the plane (shown as dumbbell orbitals above and below the plane). Show these two p orbitals overlapping sideways, above and below the plane, to form the pi bond. Label the pi bond with π symbol. Add: "sigma bond = head-on overlap (between sp² of C and sp² of C or sp³ of H)" and "pi bond = sideways p-p overlap". White background, educational chemistry orbital diagram quality.

2. Preparation of Alkenes

2.1 Dehydration of Alcohols

Conc. H₂SO₄ / heat: CH₃CH₂OH →(170°C, conc. H₂SO₄)→ CH₂=CH₂ + H₂O

Al₂O₃ / heat: ROH →(Al₂O₃, 350°C)→ alkene + H₂O

Saytzeff's Rule: When dehydration (or dehydrohalogenation) can give more than one alkene, the MORE SUBSTITUTED alkene (more stable, more alkyl groups on C=C) is the MAJOR product.

Example: butan-2-ol → mainly but-2-ene (CH₃CH=CHCH₃, di-substituted) not but-1-ene (CH₂=CHCH₂CH₃, mono-substituted).

2.2 Dehydrohalogenation (β-Elimination)

R–CH₂–CHX–R' + KOH/alcohol →(heat)→ R–CH=CH–R' + KX + H₂O

KOH in alcohol (alcoholic KOH) → favors E2 elimination → alkene. KOH in water → favors SN → alcohol.

Saytzeff's Rule applies: More substituted alkene is major product.

Example: 2-bromobutane + KOH/ethanol → but-2-ene (major) + but-1-ene (minor)

2.3 Cracking of Alkanes

Higher alkanes at high temperature → smaller alkanes + alkenes. (Also covered in Module 05.)

3. Physical Properties

First three (ethene, propene, butene) are gases. Higher are liquids/solids. Non-polar (slightly polar only if one end has substituents). Insoluble in water, soluble in organic solvents. Slightly higher BP than corresponding alkanes due to π electrons → slight polarizability. Geometric isomers (cis-trans) shown here.

4. Chemical Reactions — Electrophilic Addition

Why electrophilic addition? The π electrons of C=C are above and below the plane → electron-rich region → attracts electrophiles (E⁺). The electrophile attacks, breaking the π bond, and the electrons add across the C=C.

General pattern: C=C + E–Nu → E–C–C–Nu (anti-Markovnikov if radical; Markovnikov if ionic)

4.1 Addition of HX — Markovnikov's Rule

Markovnikov's Rule: In addition of HX to an unsymmetrical alkene, the H (electrophile) goes to the carbon with MORE hydrogens (already-hydrogen-rich C), and X goes to the carbon with FEWER hydrogens.

Scientific basis: H⁺ attacks the C that forms the MORE STABLE carbocation intermediate (more substituted = more stable 3° > 2° > 1°).

Example: CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane, major) NOT CH₃–CH₂–CH₂Br

Markovnikov Addition: HBr to Propene — Full Ionic Mechanism

Draw a detailed curved-arrow ionic mechanism for the addition of HBr to propene (CH₃CH=CH₂ + HBr) on a white background. Step 1 "Electrophilic attack (slow, rate-determining)": show propene structure (CH₃-CH=CH₂) drawn with double bond clearly visible. Draw a curved arrow from the π bond (above the C2=C3 bond) attacking the H of H-Br. Simultaneously show another curved arrow from the H-Br bond to Br, making Br become Br⁻. The result: a SECONDARY carbocation (CH₃-C⁺H-CH₃, positive charge at C2, the more substituted carbon). Label it "(2° carbocation — more stable, major pathway)". Below this, show the alternative: a primary carbocation (CH₃-CH₂-C⁺H₂, positive at C1, less stable). Label "(1° carbocation — less stable, minor)". Step 2 "Nucleophilic attack (fast)": show Br⁻ (with lone pairs) attacking the planar carbocation carbon from either face. Draw curved arrow from Br⁻ to the C⁺. Product: CH₃-CHBr-CH₃ (2-bromopropane, the Markovnikov product). Write the overall equation at the bottom. Include a stability comparison box: "2° C⁺ > 1° C⁺ explains Markovnikov selectivity". White background, bold curved arrows (full-headed for ionic mechanism), educational chemistry quality.

4.2 Anti-Markovnikov Addition of HBr — Kharasch Effect (Free Radical)

Condition: HBr + peroxide (ROOR) or UV light

Mechanism: Free radical chain. Br• (more stable radical) attacks C=C to give the MORE STABLE radical (at the more substituted C). Then H• is added from HBr at the less substituted C.

Result: Br goes to C with FEWER H atoms (anti-Markovnikov!)

Example: CH₃CH=CH₂ + HBr/peroxide → CH₃–CH₂–CH₂Br (1-bromopropane) — opposite to Markovnikov!

Important: This peroxide effect works ONLY with HBr, NOT with HCl or HI. HCl is too fast (Cl• adds immediately, no selectivity issue). HI is too slow (I• doesn't add easily to alkenes).

4.3 Addition of H₂SO₄ (Dilute)

CH₃CH=CH₂ + H₂SO₄ → CH₃–CH(OSO₃H)–CH₃ (Markovnikov addition of H and HSO₄⁻)

On hydrolysis (warm water): → CH₃–CHOH–CH₃ (2-propanol) + H₂SO₄ regenerated

This is how alcohols are made from alkenes industrially (acid-catalyzed hydration).

4.4 Addition of Halogens (Cl₂, Br₂) — Test for Unsaturation

CH₂=CH₂ + Br₂ (in CCl₄) → BrCH₂–CH₂Br (1,2-dibromoethane)

Bromine water (orange/brown) is decolorised — classic test for C=C double bond!

Mechanism: Br₂ is polarized by the π electrons → Br–Br becomes Brδ⁺–Brδ⁻ → the Brδ⁺ (electrophile) attacks the double bond → forms a bromonium ion (3-membered cyclic Br⁺ ring bridging the two C's) → Br⁻ attacks from the back (trans) → anti-addition product — the two Br atoms end up on OPPOSITE faces (trans / anti addition).

Bromonium Ion Mechanism — Anti-Addition of Br₂ to Ethene

Draw the bromonium ion mechanism for addition of Br₂ to ethene (CH₂=CH₂) on white background. Step 1: Show Br₂ approaching ethene. The π electrons of ethene polarize Br₂ (show δ+ on near Br, δ- on far Br with a curved arrow showing polarization). Step 2: Draw a curved arrow from the C=C π bond to the δ+ Br, simultaneously another curved arrow from Br-Br bond to the far Br (making it Br⁻). Show the product: a BROMONIUM ION — a 3-membered ring with Br⁺ bridging over both carbon atoms (top bridge), both carbons now with partial positive charge. Label: "Bromonium ion intermediate", circle the 3-membered ring with Br⁺ at top and two CH₂ groups at bottom. Step 3: Show Br⁻ attacking ONE of the carbons from the BACK side (bottom face, since bromonium blocks top face). Draw curved arrow from Br⁻ to C. Product: 1,2-dibromoethane with the two Br atoms on OPPOSITE sides (anti addition). Label: "Anti addition — Br⁻ attacks opposite face to bromonium". Final product: BrCH₂-CH₂Br with stereochemistry shown. Add note: "This proves anti addition via bromonium ion". White background, step-by-step, textbook quality.

4.5 Addition of HOCl (Hypochlorous Acid)

Cl₂ + H₂O → HOCl + HCl (in aqueous Cl₂ solution)

CH₃CH=CH₂ + HOCl → CH₃–CHCl–CH₂OH or CH₃–CH(OH)–CH₂Cl

Cl⁺ is the electrophile (Cl–OH → Cl is more +ve). Cl⁺ follows Markovnikov (goes to more substituted C). OH⁻ goes to the other C.

Product: chlorohydrin. Example: ethene + HOCl → 2-chloroethanol (HOCH₂CH₂Cl) — used to make ethylene oxide.

4.6 Catalytic Hydrogenation

C=C + H₂ →(Ni/Pt/Pd catalyst, heat)→ C–C (alkane)

Syn addition (both H atoms added to same face on catalyst surface).

Heat of hydrogenation measures alkene stability: greater the stability of alkene → smaller the ΔH of hydrogenation. More substituted alkene → lower ΔH → more stable.

Stability order: trisubstituted > disubstituted > monosubstituted > ethene.

4.7 Ozonolysis

Step 1: Alkene + O₃ → ozonide (unstable)

Step 2: Ozonide + Zn/H₂O (reductive workup) → aldehydes + ketones

Step 2b (oxidative workup, H₂O₂): → carboxylic acids (if CHO formed) + ketones

This reaction CLEAVES the C=C completely — each carbon of the original double bond becomes a carbonyl (C=O) group.

Ozonolysis of 2-Butene — Reductive and Oxidative Workup

Draw a detailed ozonolysis reaction diagram for (E)-but-2-ene (CH₃CH=CHCH₃) on white background. Start: draw skeletal structure of (E)-but-2-ene with the C=C clearly shown. Arrow 1 pointing right labeled "Step 1: O₃, CCl₄ (or CH₂Cl₂), −78°C". Show ozonide intermediate box (cyclic 5-membered ring with O-O-O across the former double bond — the 1,2,3-trioxolane ring structure). Then two arrows from the ozonide: Upper arrow: "Reductive workup: Zn/H₂O or Me₂S" → products: 2 molecules of acetaldehyde (CH₃CHO, ethanal). Draw structural formula of CH₃CHO twice. Label: "Both are aldehydes (terminal CH)". Lower arrow: "Oxidative workup: H₂O₂" → products: 2 molecules of acetic acid (CH₃COOH). Draw structural formula. Label: "Aldehydes oxidized to carboxylic acids". Add: "Rule: if the C of C=C has one H → gives aldehyde (reductive) or acid (oxidative). If C has no H (quaternary or disubstituted) → gives ketone regardless." Add example: "2-methylbut-2-ene: (CH₃)₂C=CHCH₃ → acetone (CH₃COCH₃) + acetaldehyde (CH₃CHO)". White background, clear reaction arrows, full structural formulas, educational chemistry textbook quality.

4.8 Oxidation with KMnO₄

Cold, dilute KMnO₄ (alkaline) — Baeyer's Reagent (violet/purple):

C=C + cold KMnO₄ → vic-diol (1,2-diol). Both OH groups added to SAME face (syn addition). KMnO₄ is DECOLORISED (purple to colourless) — test for C=C unsaturation!

Example: CH₂=CH₂ + cold KMnO₄ → HOCH₂–CH₂OH (ethylene glycol)

Hot, concentrated KMnO₄ (acidic):

Cleaves C=C completely. Each carbon becomes:

=CH₂ → CO₂ + H₂O (terminal carbon)
=CHR → RCOOH (gives carboxylic acid)
=CR₂ → RCOR' (ketone — not cleaved further)

4.9 Polymerisation

Alkenes undergo addition polymerisation under high pressure and catalyst → long-chain polymers.

n CH₂=CH₂ →(Ziegler-Natta catalyst)→ –(CH₂–CH₂)ₙ– (polyethylene/polythene)

n CH₂=CHCH₃ → polypropylene

Also: PVC from vinyl chloride (CH₂=CHCl), Teflon from tetrafluoroethylene (CF₂=CF₂).

Worked Examples

Ex 1 M: Predict the product of CH₂=CHCH₃ + HBr (a) without peroxide, (b) with peroxide.

Solution:
(a) Markovnikov → HBr + ionic mechanism → H to C1 (more H), Br to C2 → 2-bromopropane (CH₃CHBrCH₃)
(b) Anti-Markovnikov (Kharasch effect) → Br• adds to C1 (less substituted gives more stable radical at C2... wait — actually Br• adds to the TERMINAL carbon to give the more stable SECONDARY radical at C2... no). Let me clarify: Br• adds to C1 giving a secondary radical at C2, then H from HBr adds at C2? No — Br• adds to the less hindered C to give a secondary radical at the other C. Actually: Br• adds to C3 (terminal CH₂ = C1 in propene numbering) → •radical at C2 (secondary) → H from HBr removes H → gives 1-bromopropane (CH₃CH₂CH₂Br). Anti-Markovnikov product.

Ex 2 H: An alkene on ozonolysis (Zn/H₂O) gives CH₃CHO and (CH₃)₂CO. Identify the alkene.

Solution: Reconnect the carbonyls across the former C=C. CH₃CHO gives –CH(CH₃)– at one carbon. (CH₃)₂CO gives –C(CH₃)₂– at the other carbon. Alkene: CH₃CH=C(CH₃)₂ (2-methylbut-2-ene).

Practice Problems

E Q1. How do you distinguish between cyclohexane and cyclohexene using just one chemical test?

E Q2. Give Saytzeff product for dehydration of 2-methylbutan-2-ol with conc. H₂SO₄.

M Q3. An alkene (mol. formula C₅H₁₀) on ozonolysis gives methanal (HCHO) and 3-methylbutanal (CH₃CH(CH₃)CH₂CHO). Identify the alkene.

M Q4. Explain with mechanism why bromine addition to alkenes gives anti-addition product.

H Q5. 2-butene reacts with cold dilute KMnO₄. Draw the stereochemistry of the product (which face are the OH groups added to? What does this mean for the stereochemistry of the diol?)

H Q6. Why does the peroxide effect only work for HBr and NOT HCl or HI?

Applying Markovnikov to alkene + Cl₂ or Br₂ — Markovnikov applies ONLY to HX, H₂SO₄, HOX additions (reagents with H). Halogen addition (X₂ only) doesn't use Markovnikov — both X atoms are identical in Cl₂/Br₂.
Saying bromine addition gives syn product — WRONG! Bromine addition goes via bromonium ion → anti addition (trans product).
Forgetting reductive vs oxidative ozonolysis: Zn/H₂O → aldehydes; H₂O₂ → carboxylic acids from those aldehydes.
Applying peroxide (anti-Markovnikov) effect to HCl or HI — this ONLY works for HBr!
Saying Baeyer's reagent (cold KMnO₄) cleaves the double bond — cold/dilute KMnO₄ only gives diol (syn addition), NOT cleavage. Cleavage needs hot/concentrated acidic KMnO₄.

Reaction	Reagent	Product	Key Feature
Hydrogenation	H₂, Ni/Pt/Pd	Alkane	Syn addition
Halogenation	Br₂/CCl₄	Vicinal dihalide	Anti addition via bromonium
HX addition	HBr, HCl	Alkyl halide	Markovnikov
HBr + peroxide	HBr, ROOR	Alkyl halide	Anti-Markovnikov
Acid hydration	dil H₂SO₄, H₂O	Alcohol	Markovnikov
Ozonolysis (red.)	O₃, then Zn/H₂O	Aldehyde + ketone	Cleavage of C=C
Ozonolysis (ox.)	O₃, then H₂O₂	Acid + ketone	Aldehydes → acids
Cold KMnO₄	Cold dil. KMnO₄	Diol (glycol)	Syn addition, decolorises KMnO₄
Hot KMnO₄	Hot conc. KMnO₄/H⁺	Acids/ketones/CO₂	Cleavage of C=C