As mentioned earlier, electrophilic addition is the characteristic reaction of alkenes and alkynes, in which the C=C π bond is replaced by two σ bonds. The π bond of an alkene results from the overlapping of the unhybridised p orbitals and provides regions of increased electron density above and below the plane of the molecule. These electrons are less tightly bounded than those in the σ bond, and hence are more polarisable and can interact with an electrophilic reagent. This forms the first part of an electrophilic addition in which the electrons are used to form a σ bond with the electrophile and leave the other carbon electron deficient, forming a carbocation. This carbocation then rapidly reacts with the nucleophile to form a second σ bond. This latter step is very much faster than the first, and thus the formation of the carbocation is the rate determining step.

Rate = k [Alkene] [E]

where E is the electrophilic reagent, and k is the rate constant.

Addition of hydrogen halides

The first type of electrophilic addition that we will be discussing is the addition of hydrogen halides to alkenes.

Alkenes react with HX (where X = Cl, Br, or I) in a two step reaction to produce halogenoalkanes.

The first step begins with the electrons in the C=C π bond being attracted towards the partial positive hydrogen atom in HX, causing H-X bond to break. A new C-H bond, while the two electrons in the H-X bond move to the halogen to form an anion.  In the second step, the halide ion rapidly reacts with the carbocation to form a halogenoalkane.

Both these steps are shown in the diagram below:

For a symmetrical alkene, the reaction is very straightforward. What happens if the electrophilic addition of a hydrogen halide occurs for an unsymmetrical alkene? This will lead to a regioselective reaction. This means that the H and X atoms are added to the carbon atoms to selectively form the more stable compound. This in turn leads us to the Markovnikov’s rule. The halide component of HX, bond preferentially at the more highly substituted carbon, whereas the hydrogen prefers the carbon which already contains more hydrogens.

Markovnikov’s rule

This selective formation of product is explained by the stabilities of the intermediate carbocations. We will consider the example of 2-methylbut-2-ene to explain this concept. Two different carbocations might be formed in this reaction. The first carbocation that is formed is a tertiary carbocation, whereby there are 3 electron donating alkyl groups to stabilise the cation by dispersing the charge. The alternative carbocation that is formed is a secondary carbocation, with 2 electron donating alkyl groups to stabilise the cation. The tertiary carbocation is hence more favourable and more likely to be formed.

 

This brings us to our next concept, the carbocation rearrangements.

Carbocation rearrangements

This concept is one that complements that of Markovnikov’s rule. Carbocation rearrangements is defined as the movement of a carbocation from a less stable state to a more stable one, through the use of various structural reorganizational “shifts” within the molecule. These rearrangements involve the movement of a hydrogen atom or an alkyl group. This give rise to the name of 1,2-hydride shift and 1,2-alkyl shift. An important thing to note is that the use of 1,2- indicates that the hydride or alkyl group moves to an adjacent carbon, and in fact not related to the numbering of the carbon chain.

In 1,2 hydride shift, the hydrogen atom moves with a pair of electrons to the adjacent carbon atom while in 1,2 alkyl shift, the alkyl group moves with a pair of electrons to the adjacent carbon atom. As a result, a more stable carbocation is formed. This is clearly illustrated with the diagram as shown below.

 

If a reaction involves an intermediate carbocation, it is a good practice to always check to see if the carbocation can rearrange to form a more stable carbocation. This is because carbocation rearrangements are extremely common in organic chemistry reactions.

Addition of halogens

Halogens such as bromine and chlorine can react readily with alkenes to produce 1,2-dihalogen derivatives. The halogen-halogen bond of Cl₂ and Br₂ is non polar, unlike the halogen halide bond in H-X which is polar. However, the halogen-halogen bond will become polarised as it approaches the π electrons of the double bond in the alkene. The electrons in the halogen-halogen σ bond hence become unequally shared and are distributed towards the halogen atom that is further away from the C=C bond. As a result, the dihalogen functions as an electrophile, in much the same way as a halogen halide.

The initial step is similar that to the addition of halogen halides, but what comes next is significantly different from the general concept that we have laid out previously. Instead of a carbocation, a cyclic halonium ion is formed instead. This is illustrated by the diagram as shown below, using the example of bromine.

This mechanism involves 2 steps.

In the first step, the C=C bond being the nucleophile, donates a pair of electrons to the partial positive bromine atom to form a C-Br bond and Br^–. At the same time, the partially positive bromine atom also donates a lone pair of electrons to one of the C atom to form a second C-Br bond. This occurs simultaneously and results in the formation of a bromonium ion instead of a carbocation as in the case of the addition of hydrogen halide.

In the second step, the Br^– will rapidly reacts with the bromonium ion to form a new C-Br bond. However, since the bridging halogen atom blocks
any further attack on the halogen-bonded face of the original double bond, so that when Br^– attacks it has to be from the opposite face. Of course, the Br₂ can react equally well with either face of the C=C bond in the initial attack, hence this means that the bromonium ion is formed as a racemate, which also means that the end product is similarly formed as a racemate.

Addition of halogens in the presence of water

When water is present in the reaction of halogens with a C=C bond, a different compound may be formed instead of the one shown above. This is because in an aqueous solution, even though water is not a very good nucleophile as compared to the halide, it is the most abundant available nucleophile and thus the predominant product formed will consist of an -OH group. H₂O

 

This occurs similarly to the mechanism as discussed in the addition of halogens, the only difference being water acts as the nucleophile that attacks the halonium ion.

This reaction will also have a further step, in which deprotonation occurs to remove one H atom from the -OH₂ to form -OH.