Electrophilic addition mechanismsEdexcel International A Level Chemistry: Revision notes

An electrophile is an electron pair acceptor. The C=C bond of an alkene is electron-rich, so it attacks electrophiles. This reaction type is electrophilic addition.

A curly arrow shows the movement of an electron pair. It starts from a bond or a lone pair (never from an atom or a positive charge) and points to where the pair ends up: an atom or the space between atoms to form a new bond.

When a bond breaks and both electrons go to the same atom, this is heterolytic fission, giving a positive and a negative ion.

Br₂ is non-polar, but as it approaches the C=C bond the electron-rich double bond induces a dipole: Brδ+–Brδ−.

1. A curly arrow from the C=C bond to the Brδ+ atom forms a C–Br bond. A second arrow from the Br–Br bond to the Brδ− atom breaks that bond, producing Br⁻.
2. This leaves the carbocation intermediate $BrCH_2CH_2^+$.
3. A curly arrow from a lone pair on Br⁻ to the positive carbon forms the second C–Br bond, giving 1,2-dibromoethane.

H–Br is polar, Hδ+–Brδ−, because bromine is more electronegative.

1. The arrow from the C=C bond goes to Hδ+. The arrow from the H–Br bond goes to Br, forming Br⁻ and the carbocation $CH_3CH_2^+$.
2. An arrow from a lone pair on Br⁻ to the C⁺ forms bromoethane, $CH_3CH_2Br$.

Carbocations are classified by the number of alkyl groups on the positive carbon: primary (one), secondary (two), tertiary (three). Alkyl groups release electron density towards the positive carbon (the positive inductive effect), spreading the charge and stabilising the ion.

Stability: tertiary > secondary > primary.

H⁺ can add to either carbon of propene.

• Adding to C1 gives the secondary carbocation $CH_3CH^+CH_3$, which is more stable and forms in greater amount. Br⁻ attacks it to give 2-bromopropane, the major product.
• Adding to C2 gives the primary carbocation $CH_3CH_2CH_2^+$, which gives 1-bromopropane, the minor product.

The more stable intermediate decides the major product.

If ethene is bubbled into bromine water, the products include 2-bromoethanol, $BrCH_2CH_2OH$, as well as 1,2-dibromoethane. If sodium chloride is also present, $BrCH_2CH_2Cl$ is formed too.

A one-step addition of both bromine atoms would give only 1,2-dibromoethane. The mixture is evidence for a two-step mechanism: $BrCH_2CH_2^+$ forms first and is then attacked by whichever nucleophile (Br⁻, H₂O or Cl⁻) is present.