Revise Alkenes for Cambridge 9701 (14.2)

The carbon atoms in a C=C double bond are sp² hybridised. Each carbon atom uses its three sp² hybrid orbitals to form three σ bonds: one to the other carbon atom and two to other atoms (usually hydrogen). This results in a trigonal planar geometry around each carbon atom, with bond angles of approximately 120°. The remaining unhybridised p-orbital on each carbon atom lies perpendicular to this plane.

These two parallel p-orbitals overlap sideways, both above and below the plane of the σ bonds, to form a π bond. The π bond's electron density is concentrated in these two regions. This has two major consequences: firstly, rotation around the C=C bond is restricted, leading to the possibility of E/Z isomerism. Secondly, the π bond is weaker and more exposed than the σ bond, making it the site of chemical reactivity.

The electron-rich π bond makes alkenes nucleophilic, meaning they readily donate an electron pair. Consequently, they are attacked by electrophiles (electron-pair acceptors). In this reaction, known as electrophilic addition, the π bond breaks and a new molecule is added across the double bond. Let's examine the mechanism using the reaction between ethene and bromine.

The mechanism proceeds in two main steps. First, as a non-polar $Br_2$ molecule approaches the ethene, the electron-rich π bond repels the electrons in the Br-Br bond, inducing a dipole ($Br^{\delta+}-Br^{\delta-}$). The π electrons from the C=C bond attack the partially positive $Br^{\delta+}$ atom, and the Br-Br bond breaks heterolytically. This forms a bromide ion ($Br^−$) and a carbocation intermediate. In the second step, the lone pair on the bromide ion attacks the positively charged carbon atom of the carbocation, forming the final product, 1,2-dibromoethane.

When an unsymmetrical reagent like HBr adds to an unsymmetrical alkene like propene, two different products are possible. For propene ($CH_3CH=CH_2$) reacting with HBr, the products could be 1-bromopropane or 2-bromopropane. In practice, 2-bromopropane is the major product. This observation is summarised by Markovnikov's rule: the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms.

The underlying reason for this rule is the stability of the carbocation intermediate formed in the first step. Alkyl groups are electron-donating (a positive inductive effect), which helps to stabilise the positive charge on the carbocation. The more alkyl groups attached to the positively charged carbon, the more stable the carbocation. When H⁺ adds to C1 of propene, a secondary carbocation ($CH_3\overset{+}{C}HCH_3$) is formed. When H⁺ adds to C2, a primary carbocation ($CH_3CH_2\overset{+}{C}H_2$) is formed. The secondary carbocation is more stable, so it forms more readily, leading to the major product.

Carbocation Stability: Tertiary (3 R groups) > Secondary (2 R groups) > Primary (1 R group)