What is Markovnikov's Rule and what is its carbocation basis?

Markovnikov's rule states that during electrophilic addition of an unsymmetrical reagent (like HX) to an unsymmetrical alkene, the negative part of the addendum attaches to that double-bonded carbon which has the fewer number of hydrogen atoms. Mechanistically, electrophilic attack by H+ generates the more stable carbocation intermediate (3 deg > 2 deg > 1 deg), which is then captured by the halide nucleophile X-.

Why is the Peroxide Effect (Anti-Markovnikov addition) observed only with HBr and not with HCl or HI?

The peroxide effect operates via a free-radical chain mechanism involving two propagation steps: (1) addition of halogen radical to double bond, and (2) hydrogen abstraction from HX by the carbon radical. With HCl, the H-Cl bond is too strong (430.5 kJ/mol) making step 2 endothermic. With HI, the H-I bond is weak but I-I bond formation is preferred, causing iodine radicals to recombine into I2 rather than add to the double bond (step 1 endothermic). Only for HBr are both propagation steps exothermic.

How do Lindlar's catalyst and Na/liquid NH₃ reduction differ in reducing alkynes?

Partial hydrogenation of alkynes using Lindlar's catalyst (Palladised charcoal poisoned with quinoline or sulfur) yields cis-alkenes via syn-addition. In contrast, reduction of alkynes using sodium or lithium in liquid ammonia (Na/liquid NH₃ reduction) yields trans-alkenes via anti-addition.

What is Ozonolysis and how is it used to locate double bonds?

Ozonolysis involves the reaction of an alkene with ozone (O3) to form an ozonide, followed by reductive cleavage with Zn dust and H2O into aldehydes and/or ketones. By identifying the carbonyl fragments formed, the exact position of the C=C double bond in the original alkene can be unambiguously reconstructed.

§ 9.2NCERT Class 11 · Chemistry · Chapter 9

Alkenes: Structure, Geometrical Isomerism & Reactions – Masterclass

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 4 min read

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C), having the general molecular formula C_nH_2n. They are also known as olefins (oil-forming) because lower members like ethene react with halogens to form oily liquids.

1. Electronic Structure of the Double Bond

In ethene (CH₂=CH₂), both carbon atoms are sp² hybridized with planar trigonal geometry and bond angles of approximately 120°.

The double bond consists of one strong sigma (σ) bond (formed by head-on sp²−sp² overlap, bond enthalpy ~ 397 kJ/mol) and one weaker pi (π) bond (formed by lateral overlap of unhybridized 2p orbitals, bond enthalpy ~ 284 kJ/mol).
Total C=C bond energy is 681 kJ mol⁻¹, and the bond length is 134 pm (shorter than the C−C single bond of 154 pm).
The loosely held electron cloud of the π bond lies above and below the internuclear plane, making alkenes prime targets for electrophilic addition reactions.

2. Geometrical Isomerism (Cis – Trans Isomerism)

Rotation around a C=C double bond is restricted because rotation would require breaking the lateral overlap of the π bond (requiring ~ 284 kJ/mol). When each doubly bonded carbon is attached to two different atoms or groups (type: RCH=CHR or abC=Cab), geometrical isomerism arises:

Cis Isomer

Similar groups lie on the same side of the double bond:

Possesses a net dipole moment (μ > 0) due to reinforcing bond dipoles.
Higher Boiling Point: Stronger dipole-dipole attractions.
Lower Melting Point: Less symmetrical shape packs less efficiently into crystal lattices.
Higher solubility in polar solvents.

Trans Isomer

Similar groups lie on opposite sides of the double bond:

Zero or very low dipole moment (μ ≈ 0) due to opposing bond dipoles canceling out.
Lower Boiling Point: Weaker dipole attractions.
Higher Melting Point: Highly symmetrical molecular geometry packs tightly into crystalline lattices.
Lower solubility in polar solvents.

3. Methods of Preparation of Alkenes

A. Stereoselective Alkyne Reduction

Cis-Alkene: Alkynes hydrogenated with Lindlar's catalyst (Pd/CaCO₃ partially poisoned with quinoline or sulfur) yield exclusively cis-alkenes (syn-addition):
RC≡CR + H₂ → (Lindlar) → cis-RCH=CHR
Trans-Alkene: Alkynes reduced with Sodium / Lithium in liquid ammonia (Na/liquid NH₃ reduction) yield exclusively trans-alkenes (anti-addition):
RC≡CR + 2 [H] → (Na/liq. NH₃) → trans-RCH=CHR

B. Dehydrohalogenation (β-Elimination)

Alkyl halides heated with alcoholic KOH undergo dehydrohalogenation:

CH₃CH₂CH(Br)CH₃ → (alc. KOH, Δ) → CH₃CH=CHCH₃ (80%, Saytzeff major) + CH₃CH₂CH=CH₂ (20%)

Saytzeff (Zaitsev) Rule: In dehydrohalogenation, hydrogen is preferentially eliminated from that β-carbon which has fewer hydrogen atoms, yielding the more substituted, more stable alkene.

4. Chemical Reactions of Alkenes

A. Electrophilic Addition of Hydrogen Halides (HX)

Markovnikov's Rule: During the addition of an unsymmetrical reagent (HX) to an unsymmetrical alkene, the negative part of the addendum (X⁻) attaches to the carbon with fewer hydrogen atoms:

CH₃−CH=CH₂ + HBr → CH₃−CH(Br)−CH₃ (2-Bromopropane – Major)

Mechanism: Protonation of propene yields a 2° carbocation (CH₃−CH⁺−CH₃) which is more stable by hyperconjugation than the alternative 1° carbocation (CH₃−CH₂−CH₂⁺).

The Peroxide Effect (Kharasch Effect / Anti-Markovnikov Addition)

When addition of HBr (and only HBr) is carried out in the presence of organic peroxides (e.g., benzoyl peroxide (C₆H₅CO)₂O₂), the reaction proceeds via a free-radical mechanism giving the Anti-Markovnikov product:

CH₃−CH=CH₂ + HBr → (Peroxide) → CH₃−CH₂−CH₂Br (1-Bromopropane – Major)

Why only HBr? The two propagation steps are both exothermic only for HBr. For HCl, the H−Cl bond is too strong (step 2 endothermic). For HI, the I−I bond formation is favoured, causing I^• to recombine into I₂ rather than add to the double bond.

B. Oxidation & Baeyer's Test

Baeyer's Reagent: Cold dilute alkaline KMnO₄ (1%). Decolourization of purple KMnO₄ with formation of brown MnO₂ precipitate is a definitive laboratory test for unsaturation (syn-dihydroxylation to vicinal glycols):
CH₂=CH₂ + H₂O + [O] → (alk. KMnO₄) → HO−CH₂−CH₂−OH (Ethane-1,2-diol)
Acidic KMnO₄ Cleavage: Terminal =CH₂ oxidises to CO₂ + H₂O; =CH−R oxidises to carboxylic acid RCOOH; =CR₂ oxidises to ketone R₂C=O.

C. Ozonolysis

Alkenes add ozone to form cyclic ozonides, which upon cleavage with Zn dust and water yield carbonyl compounds:

R₂C=CHR' + O₃ → (Zn/H₂O) → R₂C=O (Ketone) + R'CHO (Aldehyde) + H₂O₂

Zinc dust is crucial to consume H₂O₂ and prevent secondary oxidation of aldehydes to carboxylic acids.

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