Alkenes: Preparation, Addition Reactions and Markovnikov

Alkenes are hydrocarbons with at least one carbon–carbon double bond (general formula CnH2n). The double bond's exposed pi bond donates electrons to electrophiles, which is why alkenes undergo addition reactions so readily.

Ethene is the most manufactured organic chemical in the world — the starting point for plastics, ethanol, and detergents. How alkenes are made and how they add reagents across the double bond unlocks a huge part of industrial chemistry.

Key Concepts

• Dehydration of alcohols: heating an alcohol with concentrated H2SO4 removes water and forms an alkene. Ethanol at 443 K (170°C) gives ethene. Secondary and tertiary alcohols dehydrate more easily, and Saytzeff's rule picks the more substituted alkene as the major product.
• Dehydrohalogenation: an alkyl halide heated with alcoholic KOH loses HX to give an alkene. Alcoholic KOH causes elimination; aqueous KOH would cause substitution instead.
• Markovnikov's rule: when HX adds to an unsymmetrical alkene, hydrogen attaches to the carbon that already has more hydrogens, via the more stable carbocation. Example: HBr + propene → 2-bromopropane.
• Peroxide effect (Kharasch effect): in the presence of peroxides, HBr adds to alkenes against Markovnikov's rule, giving the anti-Markovnikov product. It works only for HBr (not HCl or HI) and goes by a free-radical mechanism.

Formulas

• Dehydration: CH3CH2OH → CH2=CH2 + H2O (conc. H2SO4, 443 K). Temperatures in kelvin: 443 K = 170°C.
• Dehydrohalogenation: CH3CH2Br + KOH(alc) → CH2=CH2 + KBr + H2O.
• Markovnikov addition: CH3CH=CH2 + HBr → CH3CHBrCH3 (2-bromopropane). The H+ adds first to the terminal carbon, forming the more stable secondary carbocation CH3C+HCH3.
• Peroxide effect: CH3CH=CH2 + HBr →(peroxide) CH3CH2CH2Br (1-bromopropane), anti-Markovnikov, via Br· radical adding first to the terminal carbon.
• Bromine test: CH2=CH2 + Br2 → CH2BrCH2Br. One mole of ethene (28.05 g/mol) consumes exactly one mole of Br2 (159.81 g/mol).

Worked Example

A student bubbles ethene through bromine water until the colour just disappears, using 5.60 g of ethene. What mass of bromine reacted?

1. Molar mass of ethene C2H4 = (2 × 12.01) + (4 × 1.008) = 24.02 + 4.03 = 28.05 g/mol.
2. Moles of ethene = 5.60 ÷ 28.05 = 0.1996 mol, about 0.200 mol.
3. The addition is 1:1, so moles of Br2 consumed = 0.200 mol.
4. Molar mass of Br2 = 2 × 79.90 = 159.80 g/mol.
5. Mass of Br2 = 0.200 × 159.80 = 31.96 g, about 32.0 g. This stoichiometry is the basis of quantitative unsaturation tests.

Common Misconceptions

• Alcoholic and aqueous KOH do the same thing. Wrong. Alcoholic KOH causes elimination (alkene formation); aqueous KOH causes substitution (alcohol formation). The solvent decides the reaction family.
• The peroxide effect works for HCl and HI too. Wrong. Only HBr shows the peroxide effect, because only the H−Br bond is weak enough for the radical chain to propagate but strong enough to start it.
• Markovnikov's rule is about where bromine goes. It is about where the hydrogen goes first — H adds to the carbon with more hydrogens already, via the more stable carbocation; the halogen ends up on the more substituted carbon as a consequence.

Real-World Applications

• Plastics: ethene polymerizes to polythene and propene to polypropylene — the two most used plastics on Earth.
• Ethanol fuel: millions of tonnes of ethanol are made each year by acid-catalyzed hydration of ethene.

Alkenes are made by removing small molecules (water, HX, X2) and react by adding them back across the pi bond. Markovnikov's rule predicts the normal product via the most stable carbocation; the peroxide effect flips it for HBr via radicals.