14.2 Alkenes

Elimination removes atoms from neighbouring carbons to create C=C. Halogenoalkanes eliminate HX with hot ethanolic NaOH; alcohols dehydrate with heat and an acid or heated alumina; cracking also forms alkenes.

The reagent and conditions determine the route. Balance the equation and identify the small molecule removed—HX, H₂O or a fragment from a larger alkane.

CH₃CH₂Br + OH⁻(ethanol) → CH₂=CH₂ + H₂O + Br⁻. Ethanol can also dehydrate to ethene with concentrated sulfuric acid and heat.

Aqueous OH⁻ favours substitution more than hot ethanolic OH⁻, so do not ignore the solvent and temperature.

The electron-rich C=C attacks an electrophile, then a nucleophile bonds to the carbocation-like intermediate. Hydrogenation adds H₂ with Pt/Ni, hydration adds steam with phosphoric acid, and HX adds a hydrogen halide at room temperature.

The π bond is consumed and each alkene carbon gains a new atom or group. Write the reagent, catalyst and product rather than treating every addition as hydrogenation.

CH₂=CH₂ + H₂ → CH₃CH₃; CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH with H₃PO₄. Both use the same alkene but different reagents and products.

Electrophilic addition is not substitution, and steam hydration is reversible whereas catalytic hydrogenation is not described by the same equilibrium.

Bromine water is orange-brown. An alkene decolourises it because the electron-rich C=C reacts with Br₂ in an electrophilic addition reaction.

The test is evidence for a reactive C=C under the stated conditions, not a universal test for every unsaturated substance. Keep the reagent and observation together.

Ethene + Br₂ → 1,2-dibromoethane, so the orange colour disappears. An alkane does not rapidly decolourise bromine water at room temperature without radical conditions.

Decolourisation is not proof that a molecule contains only one double bond, and bromine water is not the same as bromide solution.

The π bond donates an electron pair to an electrophile. In Br₂ addition, a bromine electrophile is attacked and Br⁻ completes the addition; in HBr addition, H⁺ adds first and Br⁻ attacks the carbocation-like intermediate.

Curly arrows begin at the π bond or a lone pair and end at the electron-poor atom. The intermediate and its stability help determine which product predominates.

Ethene gives 1,2-dibromoethane. Propene gives mainly 2-bromopropane with HBr because the more stable secondary carbocation pathway is favoured.

Do not draw a radical mechanism for HBr addition, and do not start a curly arrow at a positively charged atom.

Alkyl groups have a positive inductive effect: they push electron density toward a positively charged carbon. More substituted carbocations are therefore generally more stable: tertiary > secondary > primary.

When HX adds to an unsymmetrical alkene, the pathway that forms the more stable carbocation is favoured. The hydrogen adds so that the halide ends up on the more substituted carbon: the Markovnikov product.

Propene + HBr gives mainly 2-bromopropane, because proton addition can form a secondary rather than primary carbocation.

This is a product-prediction rule, not a claim that the final product contains a free carbocation or that every addition gives one product only.