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

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
Watch Solution

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). 

 

Watch Solution

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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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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