Hydroboration: Videos & Practice Problems

Hydroboration-oxidation is a unique reaction that adds alcohol to alkenes via a four-membered transition state, resulting in syn addition. This process is anti-Markovnikov, favoring the addition of alcohol to the less substituted carbon. The first step involves the reaction of alkenes with boron sources like BH₃ or B₂H₆, followed by oxidation with hydrogen peroxide and a base to yield alcohol. Understanding this mechanism is crucial for organic synthesis, as it allows for selective functionalization of molecules.

General properties of hydroboration-oxidation.

Hey guys. So just to catch you up, now we've learned 2 different addition reactions that both add alcohol to double bonds. The first one was hydration and if you remember, that was a Markovnikov alcohol that could rearrange. Then we had oxymercuration. Remember that oxymerc was also a Markovnikov alcohol, but it couldn't rearrange because it didn't have a carbocation intermediate. Now what that's going to do is that's going to bring us to our 3rd way to add alcohol to a double bond, but this one is just going to be weird. It's just going to have a lot of difference from the other 2 and that's why it's going to be really useful to us because it's a very unique reaction. The name of this reaction is hydroboration oxidation. So let's go ahead and just learn the general features first. Okay? So first of all, the intermediate for hydroboration oxidation is not going to be a carbocation. It's actually going to be its own thing. It's going to be what we call a 4 membered concerted intermediate. Now I know that already sounds terrible. I'm going to explain to you guys how to draw it, but it is kind of weird. It's a very unique looking intermediate. Okay? By the way, this isn't considered one more thing, it's not considered an intermediate, it's actually considered a transition state. Okay? So that's another thing that you want to keep in mind. It's actually not an intermediate, it's a transition state. Okay? Now the stereochemistry for this is going to be very unique. It's actually going to be syn addition. Okay? Now remember that I told you guys for oxymerc that anti addition meant that you get trans products? Well, it's the same kind of thing for syn addition. All syn addition means is that you're going to get cis products. Okay? And once I show you the mechanism, that will make more sense. Okay? And then finally, what's the product going to be? Well, I already told you you're taking a double bond and you're making an alcohol out of it. So that part hasn't changed. The only thing is that the stereochemistry is changing and the intermediate or the transition state is changing. Okay? What else? Well, can this rearrange? Remember that what kind of intermediates like to rearrange? Carbocations. Do I have one? No. So we're not going to get any rearrangements here. Okay? Finally, this is probably the most interesting part of hydroboration. Hydroboration is going to be one of the only one of only 2 reactions we're going to learn that are anti-Markovnikov. Okay? So what that means is that it's going to prefer to add my alcohol to the least substituted position. And that's really only going to make sense once I explain the mechanism. Okay? But the reason this is important is because later on when we get into synthesis, synthesis is all about taking molecules and making them into what you want. So having a reaction that adds anti-mark is going to allow me to add to branches. It's going to allow me to add to things that are on the peripheral or on the outside of a molecule instead of the things that are towards the center. You're going to see what I mean by that later, but this is a very important reaction. Just keep that in mind. So let's go ahead and look at the general product of this reaction. I have that double bond that I've been using forever, the same double bond. And notice that I'm adding some interesting reagents. What I'm adding here is BH₃ or it says other boron source. Okay? Like oxymercuration, this is also going to be a 2 part reaction. Where in the first part, I'm going to add my boron and that's going to be the part that I call hydroboration. Hydroboration, the word bor comes from the word boron. So when I see that, I Okay? Now, the thing that is a little bit complicated about hydroboration is that different textbooks and different professors might have their own source of boron that they want to use. Okay? So most typically that source of boron is going to be BH₃ or another one that's very common is B₂H₆. Okay? Now what you'll notice is that the empirical formula of B₂H₆ is the same as BH₃. That's just a dimer. That just means you have 2 BH₃s together and it makes B₂H₆. Alright? If you see those things, they're the same thing. But some professors do get a little bit creative with this and they'll use some other reagents. So some other ones that I've seen that you should just be on the lookout for in case you have any online homework or in case you just want to like look this up with an online resource is one of them is catecholborane. This actually looks like this thing right here. You don't need to know how to draw it. Just recognize that catecholborane is one of the reagents that is used for this. Now I said you don't need to draw it unless this is the one your professor uses. If your professor is always using catecholborane, then obviously you should learn how to draw that. Okay? Another one is a molecule called 9-bbn, which honestly I'm not going to draw, but it is a molecule that has boron on it. And then finally, any molecule that has the molecular formula R₂BH. Okay? So that just means any source of boron that has a boron with 2 R groups sticking off of it. Okay? All of these could be used for the hydroboration step. The one that you're going to use is just go to class and just make sure that you know which one your professor talked about in class. Okay? So that's the hydroboration step. Now we're going to go to the second step. After we hydroborate, we're going to oxidize. Okay? And we're going to use hydrogen peroxide with a base as our oxidizing step. Okay? So we don't know a whole lot about oxidation reduction yet, so I'm just going to leave it right there for now. Now check out what my end products are. What I've got now is that even without knowing the mechanism, I can predict what my product is going to look like. Remember that we said that I'm going to get an alcohol? That's anti-Markovnikov and that's syn. So what that means is that my alcohol should go down here to the less substituted carbon and it should be cis to the hydrogen that it adds. Okay? So notice that cis, notice that right here I had a methyl group. Okay? If that methyl group the reason that that methyl group is being faced up towards the wedge here is because on the double bond I added 2 things. Remember that every addition reaction adds 2 single bonds to the double bond. I'm going to get an H on one side and an OH on the other. And those 2 need to be cis to each other because it's syn addition. So that's why I've drawn it like this. So does that make sense so far? Basically, we've got anti-Markovnikov alcohol that has syn addition. Alright?

Acid-catalyzed hydroboration-oxidation mechanism

Now we've got nothing left to do than to just go through the mechanism. And this is the part that's a little bit interesting. So let's go ahead and get started. Something that I told you guys that's very unique about Boron early in the semester, actually in the very first chapter, I talked about the way boron looks. Okay? And something that's unique about boron is that it has an empty p orbital. Since it has that empty p orbital, that's going to make it really good at doing what? Do you guys remember? It's going to be an amazing electron pair acceptor. In fact, BH3 is an extremely strong Lewis acid. Acid meant that I don't donate protons. Lewis means that I accept electrons. This is a really good electron pair acceptor. So when my double bond sees that empty p orbital, it's going to be like hey, I want to give my electrons to that p orbital. Okay? But my boron is going to have 2 different choices. Either it can go down here and basically make a bond to that carbon or it can go down here to the more substituted position and make a bond to that carbon. Okay? And it turns out that the one that it's going to prefer is going to be the one that is the least sterically hindered or the one that is the easiest to approach. So what that means is that my boron is actually going to choose to orient its p orbital right underneath the least substituted carbon.

What that means is that I'm going to get a transition state that looks like this, where basically I've got a partial bond to the b and my BH2 is there. Then I've got my H over here and I'm going to have a partial bond to my h. So what's basically going to happen, let me just show you the mechanism really, really quick, is going to be a cyclization reaction. So my BH3 goes like this. This is my BH3 and I have a double bond here. Right? And that double bond says okay, I'm going to give my electrons to that empty orbital and then this single bond says I'm going to give my electrons to this bond right here. So what that does is it's going to make a transition state that looks like this, where now I have my methyl group there and I'm going to have a partial bond to B, partial bond to H, partial bond. So all of these bonds are being broken and created at the same time. Okay? So that's what my transition state looks like.

Now does that make sense kind of how the double bond donates its electrons to the and then the donates its electrons to the bond that is breaking on the more substituted side. Okay? So that's why I get a transition state that looks like this. Now what you're going to notice is that the BH2 and the H are on the same side of the ring. They're cis. And the reason is because since it's making a ring, a ring can either be on the top or can be on the bottom. But it can't be trans. It can't be like one of them is at the top and one of them is at the bottom. That wouldn't make sense. Okay? So what that means is that that's why we get syn addition with hydroboration because of this 4 membered intermediate.

Now let's go to the oxidation step. So basically what happens after the transition state is that these bonds fully form. So what that means is that this bond fully forms and this bond fully forms, giving me just a single giving me just a single bond to BH2 on one side and a single bond to H on the other. Does that make sense? So basically the transition state just showed when all the bonds were being broken and made at the same time. Now my oxidation state just showed when all the were being broken and made at the same time. Now my oxidation step is going to have the final hydroboration done at the end. Okay? So now what we're going to do is we're going to do the oxidation step. And it turns out that for this step, just like oxymerc, you don't need to know the mechanism. The reason is because the mechanism is really, really long. It goes through what's called a tri triboroester and it's just a very, very long mechanism that professors don't require you to draw the whole thing. Okay? So all you're going to need to know is that you're going to use an oxidizing agent, H2O2 to turn this into an alcohol. Okay? And the base is going to help as well. Alright? So what that means is that at the end, I'm going to get something that looks like this. I'm going to get an alcohol in the least substituted position, and I'm going to get an H that assists to that. And then this is where my methyl group would go. Okay? And if I were to draw this out in an actual planar structure, what you would see is that it's going to look like the one that we had up above, where basically what we have is an o towards the back, an H towards the back, this has to do with the syn addition. Okay? And if those are in the back, that means that my methyl group must be going towards the front and that means that this H must be going towards the front as well. Okay? And if you look at this and if you look at this product up here that I drew, they're the same thing. Okay? So basically, what I was just drawing was the entire mechanism of the top general reaction. Does that make sense? Now keep in mind that this could have happened with any source of boron. It didn't just have to be BH3. The only difference would have been that I just have a different looking group in my intermediate or in my transition state. I'm going to have a slightly different looking group. But the boron is still going to have that p orbital that coordinates with the double bond. Alright? So I hope this mechanism wasn't too confusing, but it is supposed to be a little bit hard. This is one of the trickier mechanisms that we deal with in this chapter. So I hope that you guys didn't get too freaked out by it. Let's go ahead and move on.

Indentify the mechanism of reaction and predict the product.

• A. No Reaction
• B.
• C.
• D.

Answer: (C)

Predict the product of the following reaction.

• A.
• B.
• C.
• D.

Answer: (C)

Here’s what students ask on this topic:

What is the mechanism of hydroboration-oxidation?

The mechanism of hydroboration-oxidation involves two main steps. First, the alkene reacts with a boron source like BH₃ or B₂H₆, forming a four-membered transition state. This step is characterized by syn addition, where both the boron and hydrogen add to the same side of the double bond. The boron prefers the less substituted carbon due to steric hindrance. The second step is oxidation, where the organoborane intermediate is treated with hydrogen peroxide (H₂O₂) and a base, converting the boron group into an alcohol. This results in an anti-Markovnikov product, where the alcohol is added to the less substituted carbon.

Why is hydroboration-oxidation considered anti-Markovnikov?

Hydroboration-oxidation is considered anti-Markovnikov because the addition of the alcohol group occurs at the less substituted carbon of the alkene. This is due to the steric hindrance that influences the boron atom to attach to the less crowded carbon during the hydroboration step. The subsequent oxidation step retains this orientation, resulting in the alcohol being placed on the less substituted carbon, which is opposite to the Markovnikov rule where the more substituted carbon typically receives the substituent.

What are the stereochemical outcomes of hydroboration-oxidation?

The stereochemical outcome of hydroboration-oxidation is syn addition, meaning that both the boron and hydrogen add to the same side of the double bond. This results in the alcohol and hydrogen being cis to each other in the final product. This syn addition is a consequence of the four-membered transition state formed during the hydroboration step, which ensures that the substituents are added to the same face of the alkene.

What reagents are used in the hydroboration-oxidation reaction?

The hydroboration-oxidation reaction involves two sets of reagents. In the hydroboration step, common boron sources such as BH₃ (borane) or B₂H₆ (diborane) are used. Other boron reagents like catecholborane or 9-BBN (9-borabicyclo[3.3.1]nonane) can also be employed. In the oxidation step, hydrogen peroxide (H₂O₂) and a base, typically sodium hydroxide (NaOH), are used to convert the organoborane intermediate into the final alcohol product.

How does hydroboration-oxidation differ from oxymercuration-demercuration?

Hydroboration-oxidation and oxymercuration-demercuration both add alcohols to alkenes but differ in regioselectivity and stereochemistry. Hydroboration-oxidation is anti-Markovnikov and results in syn addition, placing the alcohol on the less substituted carbon. Oxymercuration-demercuration follows Markovnikov's rule, adding the alcohol to the more substituted carbon, and typically results in anti addition, where the substituents are trans to each other. Additionally, hydroboration-oxidation does not involve carbocation intermediates, thus avoiding rearrangements.