Halohydrin Formation

Indentical mechanism to halogention, except with water as the nucleophile in the second step. Why would water prefer to react as a nucleophile over a halogen anion? Let’s find out. 

Concept: General properties of halohydrin formation.      

Video Transcript

Now I'm going 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. 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.
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 three-membered ring, you're going to wind up getting anti products. And my product is going to be a halohydrin.
Are there going to be rearrangements? Nope. No rearrangements because there's no carbonyl, whoa, I said that completely wrong – no carbocation. And then finally, this actually will have Markovnikov regiochemistry because I am adding two 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. 

  • Opening of 3-membered intermediates/molecules always results in anti-addition.

General Reaction:

Concept: A worked-example of the halohydrin mechanism.     

Video Transcript

So let's just go ahead and get started. The reaction is really straight forward. 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. 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, but this is just water by itself, so we can't really react water and a double bond. But we know that we can react a diatomic halogen. So I'm going to draw my three 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 bridged ion called a halonium ion, once again.
And now I just have to figure out what's the nucleophile that's going to do the back side attack or the nucleophilic attack of this ring. Basically, I've got two 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 two different species, which of them do you think is going to be the stronger nucleophile? The X- or the water? What do you guys think? I've actually asked you guys a question like this before for another mechanism. The answer is that the X- is going to be much stronger. In fact, X- is one of the best nucleophiles around. Water is kind of like, eh 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 waters everywhere and there's only a few X-'s. What's going to wind up happening? Well, 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.
So what that means is that in this second step, even though the X- is more stable – 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. That means that if my water was in the front, then my methyl group would also be towards the back.
Now is this my final product? No. I still have one more step, unfortunately, because I added water, so I need to deprotonate. 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 the 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 you have chiral centers here? Yup, so we have to draw both enantiomers. The other enantiomer would be the water attacked from the bottom, the methyl is 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 are 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-.
So let me know if you have any questions on that, but if not, let's go ahead and move on. 

1. Electrophilic Addition

2. Nucleophilic Substitution (SN2) and Deprotonation

Problem: Predict the product of the following reaciton. 


Problem: Predict the products of the following reaction.


Halohydrin Formation Additional Practice Problems


Optically inactive ether from the reaction of (1R,2R)-2-bromocyclopentanol with aqueous NaOH

Watch Solution

Predict the products obtained from the reaction of 1-methylcyclohex-1-ene with diatomic chlorine in water. 

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For the transition state structure below, choose an organic reaction in which it is involved and give a chemical equation for your selected organic reaction that includes the following:

(i) structural formula(s) for the organic reactant(s); be sure to show stereochemistry appropriately when necessary

(ii) the experimental conditions (Give structural formulas for organic compounds. Give a chemical formula or inorganic reactants or catalysts . If heat and/or light is needed, be sure to indicate it appropriately.)

(iii) structural formula(s) for the major organic product(s); be sure to show stereochemistry appropriately when necessary. As we do for most organic reaction equations, the chemical equations that you give do not need to be balanced.

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Which is a bromohydrin?

a) 2-bromobutane

b) 3-bromobut-1-yne

c) 2-bromobut-1-ene

d) 4-bromopentan-3-ol

e) 2-bromobut-2-ene

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Read these directions carefully. Read these directions carefully. (It was worth repeating) For the reaction of an alkene with water in the presence of Cl2 shown below, fill in the details of the mechanism. Draw the appropriate chemical structures and use an arrow to show how pairs of electrons are moved to make and break bonds during the reaction. For this question, you must draw all molecules produced in each step (yes, these equations need to be balanced!). Finally, fill in the boxes adjacent to the arrows with the type of step involved, such as "Make a bond" or Take a proton away". MAKE SURE TO NOTICE THE QUESTIONS AT THE BOTTOM. If an intermediate or product is chiral, you only need to draw one enantiomer for this problem. For the product, you must draw both enantiomers and write "racemic" if appropriate.

During the reaction described by the above mechanism, say what happens to the pH of the solution _________________________________________________________________________________

Is this reaction catalytic in acid? ____________________________________________________

Watch Solution

Provide the major product for each of the following alkene reactions. 

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Which is the expected product of the reaction shown?

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Complete the following reaction and show the complete arrow-pushing mechanism required to produce the product. Show stereochemistry.


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Using arrows to show the flow of electrons, write a stepwise mechanism for the reaction given below. [If this reaction proceeds via a mechanism for free-radical chain reaction, give three termination steps and be sure to label each elementary step in your mechanism as either “initiation,” “propagation,” or “termination.”] Recall that HCl in H2O exists as H3O+ and Cl .

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Give all possible product/products and designate stereoselectivity &/or regioselectivity by using wedges and dashes. 


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Determine the mechanism and predict the product of the reaction:


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