Electrophilic Addition to Alkenes Mechanism

Two characteristics of the double bond help us understand why these addition reactions occur:

  1. An addition reaction results in the conversion of one p bond and one s bond .

2. The electrons of the p bond are exposed. Because the pi bond results from overlapping of p orbitals, the p electrons lie above and below the plane of the double bond:

The first step of the reaction is a relatively slow addition of

the electrophilic proton to the nucleophilic alkene to form a carbocation

intermediate.

In the second step, the positively charged carbocation intermediate

(an electrophile) reacts rapidly with the negatively charged bromide

ion (a nucleophile).

Electrophiles include proton donors such as Brønsted-Lowry acids, neutral reagents such as bromine (because it can be polarized so that one end is positive), and Lewis acids such as BH₃, BF₃, and AlCl₃. Metal ions that contain vacant orbitals—the silver ion (Ag^_), the mercuric ion (Hg²⁺), and the platinum ion (Pt²⁺),  for example—also act as electrophiles .

Nucleophiles
Nucleophile means "nucleus loving" which describes the tendency of an electron rich species to be attracted to the positive nuclear charge of an electron poor species, the electrophile .
The nucleophilicity expresses the ability of the nucleophile to react in this fashion.
In general terms this can be appreciated by considering the availability of the electrons in the nucleophile. The more available the electrons, the more nucleophilic the system.
Hence the first step should be to locate the nucleophilic center.  At this point we will be considering Nu that contain lone pairs and may be anionic, however the high electron density of a  C=C is also a nucleophile. 

When H–Br reacts as an electrophile, it is attacked at H, losing Br–. 

So electrophilic addition of a proton (which is what this is) to an alkene gives aproduct best represented as a carbocation. 

This carbocation rapidly reacts with the bromide ion just

formed. Overall, H–Br adds across the alkene.

 This is a useful way of making simple alkyl

bromides.

The energetic s of this  reaction are shown below . Proton attack leads to formation of a transition state1. This is higher in energy.After a sigma bond is formed , a carbocation is formed which is lower in energy. Energetically it is higher then  either the reactants or the products. In step 2 nucleophillic attack forms transition state 2 silghtly higher in energy.Agter forming the final product,an alkyl halide its energy is lowered .

 an example to show electrophillic addition :

the bromine atom ends up on the more substituted carbon, why ?

Protonation at one end gives a stabilized, benzylic cation .

while protonation at the other would give a highly unstable primary cation, and therefore does not take place.

 The benzylic cation   gives the benzylic alkyl bromide.

Markovnikov’s rule

‘The hydrogen ends up attached to the carbon of the double bond that had more hydrogens to start with.’ 

I don’t suggest you learn this rule, though  you may hear it referred to. As with all ‘rules’ it is much more important to understand the reason behind it.

One way to state Markovnikov’s rule is to say that in the addition of HX to an  

alkene, the hydrogen atom adds to the carbon atom of the double bond that 

already has the greater number of hydrogen atoms.*

or

MARKOVNIKOV'S RULE

The hydrogen of the HX is added to the carbon which has the greater number of hydrogens and the halogen is added to the carbon which has the lesser number of hydrogens.


a example to show reaction of hydrogen iodide with an alkene.

Electrophilic addition to dienes

Conjugated dienes  , in the presence of proton (electrophile ) and nucleophile ,gives two  

carbocations .

Each gives a different addition product.

One is 1,2 addition product .if addition is taking place
between neighboring atoms .If addition takes place between
1st and 4th atoms then 1,4 addition product.