Halohydrin - Online Tutor, Practice Problems & Exam Prep

Halohydrin formation involves adding a halogen to a double bond in the presence of water, resulting in a halohydrin with an alcohol and a halogen. The mechanism includes the formation of a halonium ion, where water acts as a nucleophile, typically attacking the more substituted carbon due to Markovnikov's rule. This reaction produces a racemic mixture of enantiomers, highlighting the importance of regioselectivity and stereochemistry in organic reactions. Understanding this process is crucial for mastering electrophilic additions and the behavior of nucleophiles in organic synthesis.

General properties of halohydrin formation.

Now I want to talk about a reaction that's so similar to halogenation that a lot of people get it confused with halogenation. It's just a little bit different because in this case, we're going to be adding a halogen in the presence of water to a double bond. And this reaction is going to be called a halohydrin formation. Okay? So let's go ahead and get started. Basically, the general reaction is that I still have my double bond. I still have my diatomic halogen. The only difference is that I'm running this reaction in the presence of water instead of an inert solvent. I actually have water present. What that means is that the water is going to wind up interacting and my end product is going to give me an alcohol on one side and a halogen on the other. This molecule right here is called, as a functional group, it's called a halohydrin. Okay? So let's go ahead and talk about the actual mechanism. The intermediate is once again going to be a bridged ion, just like before. The stereochemistry is going to be anti, just like before because anytime you're opening up a 3-membered ring, you're going to wind up getting anti products and my product is going to be a halohydrin. Okay? Are there going to be rearrangements? Nope. Okay. So no rearrangements because there's no carbocation. And then finally, this actually will have Markovnikov regiochemistry because now I am adding 2 different things. So now I do have to be aware of which one goes on the Markovnikov side and which one goes on the anti-Markovnikov side. But overall, we would say this is a Markovnikov reaction because it's driven by the stability of the intermediate. Okay?

Halohydrin Mechanism

So let's just go ahead and get started. The reaction is really straightforward. What I've got here is I've got the same situation where I've got a double bond and I've got a diatomic halogen and I've also got water, let's just say. So I've also got water. Which of these is going to react with my double bond? Well, water by itself doesn't really do anything to double bonds. Now if I had water and acid, that would be different. Okay? But this is just water by itself, so you can't really react with water in a double bond. But we know that we can react a diatomic halogen. So I'm going to draw my 3 arrows once again where the X is making a bridge to the double bond and it's also kicking out one of the X's as a leaving group. What this is going to give me is a bridge ion called a halonium ion once again and now we just have to figure out what's the nucleophile that's going to do the backside attack or the nucleophilic attack of this ring. Okay? So basically, I've got 2 nucleophiles. I've got X−, just like before when we were talking about halogenation, we've got the X− just like before. Nothing has changed. But now what I also have is I've got some water lying around. Now of these 2 different species, which of them do you think is going to be the stronger nucleophile? The X− or the X− is going to be much stronger. In fact, X− is one of the best nucleophiles around. Water is kind of like because X− has a negative charge. Water is neutral. So we would expect X− to be a lot stronger. So why wouldn't I just get the X− attacking and get a halogenation reaction like before? Well, the answer is that I will. Some of that actually will happen. Some of the X− will attack and I will get halogenation. But that's not going to be what happens the majority of the time. The majority of the time, this bridged ion is so unstable that it's going to react with the first nucleophile that it encounters, even if it's not the strongest. And what if I have, once again, how about if I have a billion times more of the water than I have of the X−? So what if there's water everywhere and there's only a few X−? Okay. What's going to wind up happening? What's going to wind up happening is that even though the X− is stronger, the water is just going to have an advantage because there's a lot more of it around. Because when I planned out my reaction, I used a little bit of diatomic halogen and I put a lot of water in there. Okay? So what that means is that in the second step, even though the X− is more stable and is more nucleophilic, my water is going to wind up attacking the most substituted side. Does that make sense? Because there's just a whole lot more of it around. Now for this water, is it going to attack the more substituted or the less substituted? It's still going to attack the more substituted side because that's the one with the most positive character. And remember the water has the electrons on it. So I go ahead and I make that bond, I break that bond and what that's going to give me is a major product that is a halohydrin. So if my water adds to the front, then that means that my X is going to add to the back. Okay? And that means that if my water was in the front, then my methyl group would also be towards the back. Okay? Now is this my final product? No. I still have one more step, unfortunately, because I added water, so I need to deprotonate. Right? I need to get rid of that H. What can I use to get rid of one of those H's? I could use the X−. So I'm going to use the X− in this last step to pull off an H and give the electrons to O. So now what I'm going to get is a product that looks like this. A Markovnikov alcohol and an anti Markovnikov halogen, that's anti. Do we have chiral centers here? Yep. So we have to draw both enantiomers. The other enantiomer would be that the water attacked from the bottom, x's now at the top, and that means that my x is now at the top. And these would be a racemic mixture because I really have no clue which side it came from. Does that make sense, guys? So I hope that you guys were able to see the similarities between halogenation and the halohydrin formation. The only difference is that I've got a lot of water around, so in that second nucleophilic attack, water is going to do the attack instead of the x−. Alright? So let me know if you have any questions on that, but if not, let's go ahead and move on.

Predict the product of the following reaciton. 

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Predict the product of the following reaction.

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Here’s what students ask on this topic:

What is halohydrin formation in organic chemistry?

Halohydrin formation is a reaction in organic chemistry where a halogen (such as Br₂ or Cl₂) is added to an alkene (a molecule with a carbon-carbon double bond) in the presence of water. This reaction results in the formation of a halohydrin, a molecule that contains both an alcohol (OH) group and a halogen (X) atom on adjacent carbon atoms. The mechanism involves the formation of a halonium ion intermediate, followed by nucleophilic attack by water, leading to Markovnikov regioselectivity and anti stereochemistry.

What is the mechanism of halohydrin formation?

The mechanism of halohydrin formation involves several steps: (1) The alkene reacts with a diatomic halogen (X₂) to form a halonium ion intermediate. (2) Water, present in excess, acts as a nucleophile and attacks the more substituted carbon of the halonium ion, following Markovnikov's rule. (3) This attack opens the three-membered ring, leading to the formation of a halohydrin with anti stereochemistry. (4) Finally, the water molecule is deprotonated to yield the final halohydrin product, which contains an alcohol and a halogen on adjacent carbons.

Why does water act as the nucleophile in halohydrin formation instead of the halide ion?

In halohydrin formation, water acts as the nucleophile instead of the halide ion (X^-) because it is present in much larger excess. Although the halide ion is a stronger nucleophile due to its negative charge, the sheer abundance of water molecules means that water is more likely to encounter and react with the halonium ion intermediate. This results in water attacking the more substituted carbon, leading to the formation of the halohydrin product.

What is the stereochemistry of the product in halohydrin formation?

The stereochemistry of the product in halohydrin formation is anti. This means that the alcohol (OH) group and the halogen (X) atom are added to opposite sides of the double bond. This anti addition occurs because the nucleophilic attack by water on the halonium ion intermediate opens the three-membered ring from the opposite side, leading to the formation of enantiomers. The reaction typically produces a racemic mixture of these enantiomers.

What is the role of Markovnikov's rule in halohydrin formation?

Markovnikov's rule plays a crucial role in halohydrin formation by determining the regioselectivity of the reaction. According to Markovnikov's rule, the nucleophile (water) will attack the more substituted carbon of the halonium ion intermediate. This is because the more substituted carbon has a greater positive character, making it more susceptible to nucleophilic attack. As a result, the alcohol (OH) group ends up on the more substituted carbon, while the halogen (X) is added to the less substituted carbon.