A two-step reaction that converts an alkane double bond to a single bond, with regioselective and stereoselective addition of a hydroxyl group.
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- For the first example, Jay says that OH goes to the less substituted carbon. How come for Hydration (the prior video+last example), Jay says that OH goes to the more substituted carbon? Is it because this example of anti-Markov. while the the prior video was for Markov?(12 votes)
- That is correct. In the last video, the OH goes to the more substituted carbon because it is an example of Markovnikov. In this video, the OH goes to the less substituted carbon because it is an example of anti-Markovnikov.(11 votes)
- I wanted to clarify how you know that the reaction will be anti-Markovnikov vs Markovnikov. Do you simply memorize which types of reactions form which type of product? Or is there some other indicator of which product will be formed?(10 votes)
- If there is peroxide involved in the reaction, then it is anti-Markovnikov...
If not, then it is Markovnikov.
You can find out why this happens in a detailed reaction mechanism..
Basically it has to do with the fact that peroxide undergoes homolytic cleavage forming free radical, and thus making the reaction follow a free radical mechanism.(6 votes)
- At6:20, why are the methyl groups which are shown to have stereochemistry going into the page or coming out of the page on the left side not represented as such on the right side?(8 votes)
- Hello, I wanted to know why in a hydroboration oxidation reaction the reagent diisoamylborine,
[(CH3)2CH—CHCH3]2BH has a greater tendency for to give an Anti-Markonikov addition product over the 'traditional' BH3/THF reagents?(3 votes)
- It has to do with the mechanism itself. The reversal of the regiochemistry of addition (from markovnikov to anti) is the result of the reversal of the order in which the two components (OH & H) add to the alkene. In this example the OH adds first and then the H.(4 votes)
- at 1.39 min .. could you please explain how i decide which of the carbons more or less substituted, depends on what ?!(3 votes)
- The degree of substitution is the number of C atoms that are directly attached to an alkene carbon.
In 2-methylpropene, (CH₃)₂C=CH₂, C-1 has only H atoms directly attached to it. There are no C atoms, so C-1 is unsubstituted.
C-2 has two C atoms directly attached (the two CH₃ groups), so C-2 is disubstituted.(3 votes)
- At6:34you say that the products are enantiomers. How do you know the product isn't a racemic mixture like in hydration and hydrohalogenation? Does it have to do with the fact that it is a concerted reaction and doesn't utilize a carbocation?(2 votes)
- This reaction will produce a racemic mixture, which is defined as having equal amounts of each enantiomer. In other words, you can't have a racemic mixture without having enantiomers!(2 votes)
- Are the enatiomers racemic?(2 votes)
- Yes, the BH₃ has an equal probability of attacking from either face of the double bond.
So you get a 50 :50 mixture of the enantiomers.
That's a racemic mixture.(2 votes)
- Could someone briefly recap what a syn addition is?(1 vote)
- Syn addition involves both groups adding to the same side of a C=C double bond..
For example, in the addition of BH₃ to an alkene, both the H and the BH₂ groups add to the same side (both on the "top" or both on the "bottom") of the double bond.(3 votes)
- When it comes down to priority, which one has a higher priority over the other? Or does that not come into play here. Like regioselectivity.(2 votes)
- Is the OH and H always going to be added in syn format or could they be added in as trans?(1 vote)
- [Voiceover] Here's the general reaction for hydroboration-oxidation. We start with an alkene and in the first step this is our hydroboration-oxidation step we're adding borene which is BH3 and tetrahydrofuran which is THF. Our second step is the oxidation where we add hydrogen peroxide and a source of hydroxide ions, and we can see what our product would be. We're gonna add an OH and an H across our double bond. So that the double bond goes away and the OH adds to one of the carbons and an H adds to the other carbon. How do we know which carbon to add the OH and which carbon to add the H? We think about the regiochemistry for this reaction, and it turns out to be an anti-Markovnikov which means the OH adds to the less substituted carbon. To understand this we need to look at the mechanism for a hydroboration-oxidation which I put in the next video because it's way too long to fit into this video. I also in the next video will go into the details about the stereochemistry for this reaction. Turns out to be a syn addition which means the H and the OH add to the same sides. Alright! Let's look at an example of hydroboration-oxidation. So we'll look at this alkene here. Alright well we know, we know we're gonna add an OH and a H across our double bond, and we know the OH adds to the less substituted carbon. So I find my two carbons for my double bonds. So this carbon right here and then this one. Which one of those carbons is the less substituted carbon? Obviously the top carbon here is the less substituted carbon so that's the one where we're going to add the OH. So I can go ahead and draw my product. Alright we're adding the OH to the top carbon right there. Alright so that's this one in magenta, and the hydrogen would add to this carbon down here in red. So I'm just not gonna draw in the hydrogen, but that's where it would add. So there's our product. We don't really have to worry about stereochemistry for drawing our final product because this carbon in magenta.... Alright, this is not a chiral center and the carbon in red is also not a chiral center. Alright, let's look at an example where we do have to worry about the stereochemistry when we're drawing our products. Alright, so for this alkene right here. Alright, I look at my two carbons across my double bond, this carbon and this carbon. I know this is a hydroboration-oxidation. So I'm going to add my OH to the less substituted carbon, and this of course on the right would be the less substituted carbon. So just thinking about regiochemistry I'm gonna draw in my OH on that carbon. So I'm gonna draw in my product not thinking about stereochemistry yet. So the OH added to the carbon in magenta, which was our less substituted carbon, and when we look at our product-- Alright, so this is our carbon now. That carbon, that carbon I just circled is a chiral center. So, we do need to worry about stereochemistry. Alright this carbon over here on the left is not a chiral center. So we don't have to worry about the carbon in red. We do have to worry about the carbon in magenta. So let's think about the stereochemistry with a syn addition Adding the H and the OH to the same sides. Alright so we need to think about this alkene here and these carbons that I've marked in yellow, alright those are sp2 hybridized carbons. So the geometry around those carbons is plainer, and we add our H and OH, right. We're turning those into sp3 hybridized carbons. Alright so let's go ahead and sketch in one of our products here. So one of the possible products would be to add the H and the OH on the same side. I'm gonna draw them as wedges. So here's our hydrogen right as a wedge, and then here's the OH as a wedge. I already know which carbon to add which because we already figured that out, right. We know the OH adds to this carbon, the one on the right. We know the H added to the one over here in red. Alright, so I'm showing a syn addition of my hydrogen, of my H and of my OH here and so those are wedges, and so let's think about what's attached to those carbons. So let me go back up to here. This is the carbon in magenta. That's this carbon down here. What else is attached to that carbon? Well there's a hydrogen. Alright, and therefore if I think about the stereochemistry that hydrogen must be going away from me now for this product. We're going from an sp2 hybridized carbon in our alkene to a an sp3 hybridized carbon. So that hydrogen is going away from us. Let's think about, let's think about this other carbon. Let me make this other carbon over here red. So this carbon right here. Alright, this carbon. What else is attached to this carbon? Well there's a methyl group. Alright, and if this hydrogen is coming out at me this methyl group must be going away from me. So we have a methyl group going away from us in space, and so that's one of our possible products. Alright, I could draw this out the way we're used to seeing it where we would show the OH coming out at us in space. Alright, the OH is coming out at us. So that's one of our possible products. Alright our other products, I could show the H and the OH adding on the same side but I can show them adding as dashes. Alright so I could draw in our carbons here, and I can show this time, I can show the H adding as a dash, and the OH adding as a dash. Now that means that this hydrogen right here would have to be a wedge, right. So that hydrogen would have to be a wedge. So I draw in the hydrogen here as a wedge, and then for the other carbon, the one in red, so this one. This methyl group would have to be coming out at us now. Alright, so that methyl group is now coming out at us. I'm gonna draw that as a wedge. So our methyl group CH3 would be here. Alright so this molecule, just know the way to draw it would be like this, where your OH is going away from us in space. Right, because we already identified our chiral center. So this carbon up here, this carbon was our chiral center. So the OH could be a wedge or the OH could be a dash, and of course the relationship between our two products here, these are enantiomers of each other. Alright let's do one more example where we have to worry about the regiochemistry and the stereochemistry. Alright, so let's look at this one right here. So here's our alkene. We know this is a hydroboration-oxidation, and here's our double bond. So we have a carbon right here and we have a carbon right here. We know the OH adds to the less substituted carbon. Well, this carbon is of course the less substituted carbon. So just thinking about regiochemistry, we can draw in just thinking about regiochemistry right here. We know the OH adds to this carbon down here, and we have a methyl group up here, and let me go ahead and highlight our carbons once again. So the carbon in magenta is this one, and the carbon in red right here is this one. So the hydrogen would add to the carbon in red, and we could go ahead and draw in CH3 here just to clarify. Alright notice that we have two chiral centers. So the carbon in red is a chiral center, and the carbon in magenta is also a chiral center. Alright, so let's think about stereochemistry now. Okay so we're going to draw, let's sketch in one of our products here. We know that this is a syn addition, which means that the hydrogen and the OH, the H and the OH, add on the same side. I already know the OH adds to this bottom carbon here. So I'm gonna show the OH adding as a wedge. So I'm gonna draw a wedge in here, and that's my OH. My hydrogen adds to the other carbon, right. So this one right here and because it has to add on the same side as the OH I draw it as a wedge. Alright so let's think about this carbon in red. What else is attached to that carbon? Of course it's methyl group. Alright we're going from-- We're changing hybridization states. Alright, so now I can show that methyl group going away from us in space, right. So that's a CH3 right here. Alright, that's one of our possible products. Let's draw in the other one, alright. So the other possibility is the OH and the H add as dashes. Alright, so I can show them adding from the opposite side. So I can show the OH adding as a dash. I can show the H adding as a dash, and now when we think about this top carbon here, right, and the stereochemistry, this carbon in red, this methyl group must be coming out at us now in space. Alright, so for this the methyl group is coming out at us in space right here. So this would be a CH3, right. If you wanted to you could add in the hydrogen for the carbon with the OH on it, but I'm gonna stop right there because we have our products. Right, so what is the relationship between these molecules? They are enantiomers of each other. Alright, we have opposites, absolute configurations at both of our chiral centers.