 Let's discuss a method of preparing haloalkanes from alkanes. So this method is called the electrophilic addition of Hx or hydrogen halides. So as you can see here, when hydrogen halide adds across the double bond of alkanes, we get a haloalkene. Now this is an electrophilic addition because you see the double bond which is electron rich will attack or will abstract an electrophile which is the proton here. So the proton of hydrogen halides adds across the double bond of alkanes to give an electrophilic addition product. So let's quickly look at the mechanism of this reaction. So the first step in this reaction is the protonation of double bond. That is when proton adds across the double bond of the alkene. Now the important thing to remember here is that when the pi electrons of the double bond attacks the electrophile here which is proton of hydrogen halides, the carbocation or the positive charge is formed on the most substituted carbon atom. In this case we have a symmetrical alkene so positive charge on either of the carbon atoms would give you the same product but when we have an unsymmetrical alkene it is important to remember that the positive charge forms on the most substituted carbon atom because that would generate the most stable carbocation. And in the next step, x-attacks the carbocation here or the carbon with the positive charge because this carbon is now electrophilic right, it is electron deficient. So nucleophile attacks this electron deficient carbon and gives us the final product which is nothing but a haloalkene. So let's take an example to understand this mechanism better. So here we have an unsymmetrical alkene and when this alkene reacts with hydrogen bromide HBr here, so the Hx that we have taken here is HBr. The first step is nothing but protonation of double bond right. So the pi electrons attack the electrophile which is proton here and this attack can produce two different types of carbocations. If the proton adds to the secondary carbon here then we get a tertiary carbocation right because the pi electrons shift in such a way that you get positive charge on the tertiary carbon. On the other hand if the proton adds to the tertiary carbon atom which is this carbon here, in that case we get a secondary carbocation. So here we have a tertiary carbocation and here we have a secondary carbocation. Now which among the two would be most stable? Obviously tertiary carbocations are more stable than secondary carbocations due to hyper conjugation and inductive effects. In the next step the halodion or bromide ion in this case attacks the carbocation to produce the corresponding haloalkene as shown here. Now you will notice that the product that is formed or the bromide formed from tertiary carbocation will be obtained in major amounts whereas those formed from the secondary carbocation would be formed in minor amounts. That is the major product is the one where the bromide ion attacks the most stable carbocation. Now this is also called Markovnikov product. Essentially electrophilic addition of Hx to alkenes follows the Markovnikov's rule. What does this rule say? It states that in an electrophilic addition to an alkenes the electrophile adds in such a way as to generate the most stable carbocation. So now that we have familiarised ourselves with the mechanism of this reaction let's quickly solve a couple of questions. So here we have two reactions. Now the question is to find out the major products formed in both of these reactions. So let's pause the video here and try figuring out the answer. Okay so the first step would be protonation of double bond or the attack of pi electrons on the proton and this would generate two different type of carbocations. So the two different carbocations formed would be a tertiary carbocation and a primary carbocation. Now which among the two is most stable? Obviously tertiary carbocation is much more stable than primary carbocation and that means the major product would be formed from this intermediate. So in the next step halide ion attacks this tertiary carbocation to give you the final product which is a tertiary alkyl iodide. So this would be the major product in this reaction. Now what about the second reaction? Here again the pi electrons can attack the electrophile to generate two different type of carbocations right? One is a tertiary carbocation and one is a secondary carbocation. So the carbocations would look like this tertiary. So this is a tertiary carbocation and this is a secondary carbocation. Here again a tertiary carbocation is much more stable than a secondary carbocation. That means the major product formed would be from the tertiary carbocation. So the final product here or the major product formed in this reaction would look like this. So this is the major product formed in the second reaction.