There are two mechanisms by which each of the two enantiomers ...

Question

There are two mechanisms by which each of the two enantiomers can form in the reaction shown in Figure 9.37. Show them.

Accepted Answer

Welcome back. Everybody. Here's our next problem. It says provide the mechanisms that produce the two in Antium. And the following reaction, we're starting with a branch chain alkene. It has 123456 carbons in its main chain car between carbons, two and three, we have a double bond and carbon five has two metal grips attached to it. This is reacting with and we have this funny structure. We have a sort of a red dot attached to. So three H. This just means we have sulfonic acid. The so three H attached to a polymer, the polymer doesn't do anything, we can ignore it, but it's just telling us it's there. That's on top of the reaction arrow and below our reaction arrow, we have H2O 2 and H2O. So hydrogen peroxide and water. The products are two and anti immerse. The first one we see our double bond has been broken. It's now a single bond. And on each carbon, on carbon two, we have a solid wedge and then oh and on carbon three, a dashed wedge and oh the second man humor, carbon two has the dashed wedge and an oh carbon three has the solid wedge and an oh. So to determine the mechanism, we need to think about what type of reaction occurred. And we started with an alchem and we're ending with vicinal dial. So two alcohol groups added next to each other, next to each other, the double bond is gone and the alcohol groups are in the anti orientation towards each other. So what kind of reaction does that? Well, when you think of a vicinal diol from an alkene, you might think of the pinnacle rearrangement, but that adds alcohol groups in the sin uh orientation to each other. If we want an anti addition, we need to have an oxidation A RK. So that makes that oxide ring where we have the two carbons that were in the double bond linked to an oxygen and a little triangle ring. And then we have an acid catalyzed ring opening because when we open the ring like that, the oxygen, the oxygen groups will end up in an anti orientation to each other. And then finally, just a de protonation as our last step. So what do we need? And we have the reagents for an oxidation of Nalen Well, for that, we need a perox acid and we have this Sulfonic acid which is not a proxy acid, but we've handily got hydrogen peroxide there. So the sulfonic acid on its polymer, when combined with hydrogen peroxide, you will end up with a proxy acid So we'll just draw that got a red dot It will now become sulfur, still has its two double bonds to oxygen. But instead of an oh it now has an O and then oh so there's our paroxy acid there. So let's look at the mechanism that happens here. So we have our carbon chain. We've got some really complicated electron jumping around here. We have the formation of our ring. As you see, here's our perox acid. So the last oh on our perox acid, the electrons from that oxygen come down and form a bond with carbon two. Meanwhile, the electrons from the double bond between those carbons come up and make a second bond to that oxygen. There's the formation of the ring then that hydrogen that was there, its electrons jumped back onto the oxygen. The oh group was bonded to the second oxygen on our perox acid. The electrons that bonded those two oxygens together form a new double bond between that sulfur and oxygen. And then meanwhile, the top so double bond, the electrons from the double bond come over and snag a proton away. That was the final hydrogen on our perox acid. So a lot of arrows going on here. But the end result is we regenerate our original sulfonic acid. So we see it is a sulfonic acid catalyst. And I'm just going to draw that first even though that's the catalyst because I'm going to use it in the next step um to proton my ring. So I'm just putting it there for convenience. So I've regenerated. I have, I'm sort of reorienting the bonds here, but I still have two oh double bonds. And then I have an oh, it's different oxygens, but I've regenerated that sulfonic acid. And then I've produced two possible products here because the ring can be projecting forward out of the screen or it can be projecting backwards. So we'll just draw those products. So you can see these two possible projects products excuse the drawing. It's a little hard to tricky to draw those wedges when making a triangular ringing like this. But we say oxygen added either projecting forward or backwards with solid wedges or dash wedges and then each has two lone pairs. So oxygen has two lone pairs and two bonds as it likes to have. So the next step is we're going to weaken this ring a bit. We're in this acidic environment, we're going to proton the ring. So that's an easy step. The alone, one of the lone pairs on the oxygen can just snag a hydrogen from that sulfonic acid causing its electrons to jump back onto the oxygen and will end up with a proton ring. And it's the same obviously for both structures. So I'm just going to scroll down here to draw that. So you can see we now have a protein ring. So oxygen has a positive charge that it's not very happy about and we've generated a sulfonate there. So I, I forgot to put my little reaction arrow there. So now, don't forget we have water in our solution. So water can participate in a nucleic attack to break our ring here. Not that the water is going to attack the most substituted carbon. But looking at our molecule, we can see both of the carbons in the ring are secondary carbons, they're equally substituted. So we're going to end up with a reca mixture with the water attacking one half the time and the other half the time. So I'm going to show that with a red arrow bringing one of the lone pairs over to attack carbon two and then those electrons in the co bond going up to the oxygen, the ring breaking open and then I'll use a green arrow for the other reaction. I'll just take the other lone pair just to make it easier attacking carbon three at half the time. And then those electrons coming up to the oxygen, it's a little hard to see them there breaking the ring open on the other side again, because our carbons are equally substituted. It can happen either way we'll get a RM mixture of products. So let's look at that and I'm just going to change this really quickly, red and green is probably not the greatest contrasting color pair there. I will do red and light blue instead. And then looking at my red electron pushing if the water attacks at C two, then the ring is broken open there. So I end up still with a solid wedge on C three because that's still attached there. And now it's an an oh, so I, I have one alcohol group and then where my ring has broken open, I have my water molecule has attached and it's attached as a dashed wedge since it's coming in the anti fashion and it's O H two plus since that was a water molecule and then half the time. So that's my red product just going to circle it here. And then in my light blue product, I attacked the water attacked a carbon three. So carbon three is where the ring has broken open. So I have the solid wedge with an oh on carbon two, which was the oxygen that was in the ring. And then I have a dashed wedge with the 02 plus on carbon three. So we'll circle that in light blue. So you can see those are the products from those two different attacks there. These are in an timmers. I've been told that in my problem and you can see they are non superimposable mirror images. Now, do we need to go back and do this again? For back in the beginning when my oxide ring had the dash wedges was going behind. And the answer is no, because when you open that up in the anti fashion that will give the same exact two an anti ier as here. So I don't need to worry about that one. The end result will be these last two in anti ier. So let's look at the continued mechanism. I still got um a positive, I have still a proton on one of these oxygens. So we need to have a final last easy step of dep protonation. So let's just scroll on down. So we have a little more space. Well, what do we have? We have conveniently a sulfate ion that we generated earlier. So one of its lone pairs, it is a lone pair, negative charge can just come on up and grab a proton from that 02 plus group generating my vicinal diol and it will do the same for the other nant tumor draw that in my will. And I have regenerated my sulfonic acid. And I've ended up with my two and Ant Timor that I wanted as shown in the original question, I have the one product where carbon two has a dashed wedge attached to an oh, carbon three has a solid wedge attached to an oh and the other product where carbon two has a solid wedge to oh and carbon three has a dashed wedge to oh. So there's my mechanism again. Let me scroll up. So we can recall, I used the Perox Aid Acid method since I had sulfonic acid and hydrogen peroxide and generated that perox acid to form an apex ring from my alkene. That oxide ring was then had an acid catalyzed ring opening which causes the addition of a water molecule in an anti fashion from the oh, that's originally from my ring. And I end up with these vicinal dials in an anti arrangement two in anti imer. See you in the next video.

There are two mechanisms by which each of the two enantiomers can form in the reaction shown in Figure 9.37. Show them.
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Key Concepts

Enantiomers

Chirality

Reaction Mechanisms

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