We will take a look at the next electrofilic addition which is called halah hydr formation. So here is an example of halah hydr formation. The double bond will react with a halogen Br2 or Cl2 as well as water. And the product has a Br attached to one double bonding carbon O to the other carbon. So we start with an alken and we ended up with a visinal halah hydrate. Okay, halah hydr refers to the molecule that have a h hallogen and alcohol at the same time. Haha hydr. And when the halogen and O group are on the adjacent carbon, it's the visinal halah hydr. The reg selectivity of this reaction, it follows the marovnikov's rule. We will talk about that. And the stereo selectivity of the reaction it's anti- addition. So this reaction is quite similar to the halogenation of alkenes. The addition of Br2. Actually the first step is exactly the same. It's just that we have another H2O that serves as a nucleophile in the second step. All right. So let's take a look at one more example here. Suppose the double bond is not symmetrical. All right. So when Br and H2 adds to the double bond, Br will go to the less substituted carbon and O will be attached to the more substituted carbon. This follows the marov mikov rule which I will explain in the mechanism. So the first step of the mechanism is the same as the mechanism for halogenation reaction. The double bond will donate the pi electron to one of the halogen molecule. So the halogen halogen bond breaks and as a result we will have that three membered ring bridged bremonium ion. Okay that's the our first intermediate bremonium ion. In the second step the bremonium ion will be attacked by water molecule. Okay. So here water will serve as the nucleophile to break the three member ring instead of Br2. All right. So in halogenation we have Br2 attack in this step. But when water is present water will attack. Okay. So as a result now we have Br on one carbon and a hydrronium ion attached to the other carbon. Now because the molecule is not symmetrical which carbon will O attack that is the key. Okay. The difference between these two carbons, one is a tertiary carbon, the left one. The right one is a primary carbon. The water is the nucleophile. Nucleophile will attack a more substituted carbon. In this case, it's the tertiary carbon. Why? Because in this three membered ring, we have a positive charge. Okay. So the two carbon atoms will help the bromine to share the positive charge. The left carbon, the tertiary carbon has the ability to take more partial positive charge because it's a tertiary carbon. So the left carbon, the tertiary carbon will be more positive than the primary carbon on the right. Okay. So for this case you can think of it this way. Suppose we are going to form a carboation on one of these two carbon atoms. Which one of the carboation will be more stable? Of course the tertiary will be more stable than primary. So that's the carbon the nucleophile will attack. Okay. So that's why we say the reaction follows the marovnikov rule. All right. So the last step, the third step of the mechanism is a quick deprotonation step. The alq hydrononeium ion will be deproonated by Br minus. So the proton is removed and the visinal ha halah hydrine form as the final product. Okay, three steps. Let's take a quick look at this example here. Can you get the product? Here we go. The stereo selectivity anti-addition. So the O and the Cl has to be anti- to each other. So they will be trans to each other. Okay. And the products is a pair of inantiomers.