Hydroboration-Oxidation

This is the last of three ways to add alcohol to a double bond. This reaction creates alcohols that are much different from the first two methods, so pay attention to the differences!

Concept: General properties of hydroboration-oxidation.     

7m
Video Transcript

Hey, guys, so just to catch you up, now we've learned two 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 Markovnikov alcohol, but it couldn't rearrange because it didn't have a carbocation intermediate.
Now what that's going to do is it's going to bring us to our third way to add alcohol to a double bond, but this one's just going to be weird. It's just going to have a lot of differences from the other two 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.
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 four-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.
By the way, this isn't considered – one more thing, it's not considered an intermediate, it's actually considered a transition state. So that's another thing that you want to keep in mind. It's actually not an intermediate. It's a transition state.
Now the stereochemistry for this is going to be very unique. It's actually going to be syn addition. 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. And once I show you the mechanism that will make more sense.
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.
What else? Well, can this rearrange? Remember that what kind of intermediate likes to rearrange? Carbocations. Do I have one? No. So we're not going to get any rearrangements here.
Finally, this is probably the most interesting part of hydroboration, hydroboration is going to be one of only two reactions we're going to learn in orgo one that are anti-Markovnikov. What that means is that it's going to prefer to add my alcohol to the least substituted position. And that's going to only really make sense once I explain the mechanism.
But the reasons 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. So 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. 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.
Like oxymercuration, this is also going to be two-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 immediately know there should be a boron in this reaction. And that's a way that you can recognize it, kind of like we use for oxymerc that had a mercury. Hydroboration has a boron.
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. So most typically that source of boron is going to be BH3 or another one that's very common is B2H6. 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 two BH3's together and it makes B2H6. 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 one's 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 look this up with an online resource is one of them is catechol borane. This actually looks like this thing right here. You don't need to know how to draw it. Just recognize that catechol borane 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 catechol borane, then obviously you should learn how to draw that.
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. So that just means any source of boron that has a boron with two R groups sticking off of it. 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.
So that's the hydroboration step. Now we're going to go to the second step. After we hydroborate, we're going to oxidize and we're going to use hydrogen peroxide with a base as our oxidizing step. 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 products are going to look like. Remember that we said we're 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.
So notice that – cis – notice that here I had a methyl group. 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 two things. Remember that every addition reaction adds two sigma bonds to the double bond. I'm going to get an H on one side and an OH on the other. And those two 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, you've got anti-Markovnikov alcohol that has syn addition. 

General Reaction:

Concept: A worked-example of the acid-catalyzed hydroboration-oxidation mechanism.     

7m
Video Transcript

So now we've got nothing left to do than 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, OK? And something that's unique about Boron is that has an empty P orbital.... 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, OK? Remember that the definition of a Lewis acid is a good electron pair acceptor, OK? Remember that Bronsted Lowry acid means that I 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, OK? But my Boron is going to have two different choices, my Boron is going to say well could I go closer? Like the Boron basically my double bond is going to attack the empty orbital and my Boron is going to have two 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 positioned and make a bond to that carbon, OK? And it turns out that the one that is 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, alright? What that means is that I'm going to get a transition state that looks like this where basically I got a partial bond to the B and my BHQ is there then I've got my H over here and I'm going to have a partial bond to my H and this is going to have a partial bond there, OK? So it'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 here, right? And that double bond says OK I'm going to give my electrons to the 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 when I have my methyl group there and I'm going to have a partial bond to B, partial bond to H, partial bond to carbon and then a partial double bond so all of these bonds are being broken and created at the same time, OK? So that's what my transition state looks like now does that make sense kind of how the double bond kind of donates its electrons to the B and then the H donates its electrons to the bond that is breaking on the more substituted side, OK? 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 it 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, OK? So what that means is that that's why we get syn addition hydroboration because of this four-member intermediate, Ok? Or this four-membered transition state, OK? So is that making sense so far? Cool so that's what my intermediate looks like 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 bond to BH2on one side and a single bond to H on the other each does that make sense? So basically the transition state just showed when all of bonds are being broken and made at the same time now my oxidation step is going to have the final hydroboration done at the end, OK? 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, OK? The reason is because the mechanism is really really long it goes through what's called the triborel ester and it's just a very very long mechanism that professors don't require you to draw the whole thing, OK? So all you're going to need to know is that you're going to use an oxidation agent...Oxidizing agent H2O2 to turn this into an alcohol, OK? 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 stable position I'm sorry in the least substituted position and I'm going to get an H that is cis to that and then this is where my methyl group would go, OK? 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 OH towards the back, an H towards the back this has to do with the syn addition, OK? 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, OK? And if you look at this and if you look at this product up here that I drew they're the same thing, OK? So basically what I was just drawing was the entire mechanism of the top general reaction, does that makes sense? Now keep in mind that this could have happened with any source of Boron it didn't have to 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 in a slightly different looking group but the Boron is still going to have a 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, OK? This is one of the trickier mechanisms that we deal with in this chapter so I hope that you guys don't get too freaked out by it let's go ahead and move on.

1. Electrophilic Addition

2. Oxidation

Problem: Indentify the mechanism of reaction and predict the product.

5m

Problem: Predict the product of the following reaction.

10m

Hydroboration-Oxidation Additional Practice Problems

Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw the answer in skeletal form.

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Which of the following would be a reasonable synthesis of CH 3CH2CH2CH2OH?

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In a hydroboration-oxidation reaction, a hydrogen and an alcohol are added where there once was a double bond. Describe the regio- and stereochemical relationship between the two.

  1. Vicinal and trans 
  2. Geminal and cis
  3. Vicinal and cis
  4. Vicinal and racemic
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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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Provide the following when a product is given. If an organic reactant is missing, supply a structural formula; if an inorganic reactant (reagent) or catalyst is missing, simply give a formula. 

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Supply the missing inorganic reactant (reagent) or catalyst that is missing in the reaction given below.

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Draw the starting material that under the given reaction conditions, result in the following product.

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Fill in the box with the product of the following reaction of alkenes. Draw only the PREDOMINATE REGIOISOMER and indicate stereochemistry by drawing dashes and wedges where appropriate. When a racemic mixture is formed you must draw both enantiomers and write RACEMIC in the box. 

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Fill in the box with the product of the following reaction of alkene. Assume that the starting alkene is enantiomerically pure and optically active. Draw only the PREDOMINATE REGIOISOMER and indicate stereochemistry by drawing dashes and wedges where appropriate. In the box label the product as “optically active” or “optically inactive”.

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Show how you would carry out each of the following reactions. You do   NOT need to draw the mechanisms 

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Follows Anti-Markovnikov’s Rule

a)    Hydrogen goes to the double bonded carbon with _______ hydrogens. ( ______ substituted alkene carbon) 
b)    Hydroxide ion goes to the double bonded carbon with _______ hydrogens. ( ______ substituted alkene carbon). 

 

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Give the product, or products, including stereochemistry of the reaction of (Z)-3-methyl-2-pentene with the reagent below. If the products are a pair of enantiomers, you need to draw only one and state that the other enantiomer is formed. 

BH3, THF - H2O/H2O2, NaOH

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Predict the starting material of the following reactions.

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Predict the product(s) or reagents of the following reactions.

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Predict the product for the reaction below. 

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Complete the following reaction with the correct structure of the product. Don’t forget to specify the stereochemistry.

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What will be the major product of the following reaction? Pay careful attention to the stereochemistry of the product.

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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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