Electrophilic addition to a double bond is characterized by three main events; nucleophilic attack of the double bond on an electrophile, carbocation formation, and an additional nucleophilic attack on the carbocation. In these reactions, the pi bond serves as the nucleophile. Let's take a look at each of these steps.
Nucleophilic Attack and Carbocation formation
The first step initiates the process by breaking the pi bond and attacking the electrophile. However, this leads to a problem if the molecule is not symmetrical: WHERE THE HECK DOES THE CARBOCATION GO???? Well, the Markovnikov rule helps us out here. The rule basically says the more highly (hence the more stable) substituted carbocation is formed. So, the reaction would look something like this:
Here, you can see that the more stable, tertiary carbocation is formed over the secondary carbocation. This is also known as a regioselective process since "one region" is preferred over another. Ok, let's move on to the next step now.
Final Product Formation
The last step of this process is nucleophilic attack of a newly generated nucleophile on the carbocation. In the case of our example, this would be the bromide that was formed once the double bond attacked the proton.
More Than One Double Bond
Ok, so what happens when there is more than one double bond and they are conjugated? Wait, what's conjugated?
When there are two conjugated double bonds, we have the possibility of more than one product. In this case, we can get what is referred to as the 1,2-product or the 1,4-product. There is just one catch, sometimes we only get one or the other. So, this means we need to be able to predict which product will predominate.
Alright, let's start by describing how these guys get their name.
When looking at 1,3-butadiene above, we see that there are four numbered carbons. If the product is the result of the addition of H-Br across the first bond (either 1 and 2 or 3 and 4, there is no difference in this example), then you obtain the 1,2-product. Notice that the hydrogen and the bromide are on carbons 1 and 2. However, in the other product, notice that the hydrogen is on carbon 1 and the bromide is on carbon 4, hence the 1,4-product.
Ok, now that we have naming figured out, let's look at how we actually obtain the different products.
As you can see, we get the different products by drawing resonance structures. Now, let's look at the products that we do not see.
We do not see the above product because carbocation formed is so unstable. Not only is the carbocation primary, it is also not allylic or adjacent to a double bond. The double bond allows for resonance stabilization.
Next, we need to look at which product is more stable. In our example, the 1,2-product has a double bond that is substituted by two hydrogens on the end. However, the 1,4-product has alkyl groups on either side. The higher substitution about the double bond increases its stability. This is referred to as the Zaitsev rule. Therefore, the product with the most overall stability is the thermodynamic product and the other is the kinetic product.
Is there a way to predict which product will predominate? Of course! At higher temperatures and longer reaction times, the thermodynamic product will predominate. At lower temperatures and shorter reaction times, the kinetic product will predominate.
There you have it!