Alkene cleavage using ozone. Created by Jay.
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- Where would something like this occur in nature? The upper atmosphere?(17 votes)
- Exactly. Ozone also forms during lightning strikes (3O2 --> 2O3), so theoretically it could also occur during lightning storms, albeit in small amounts.(30 votes)
- This reaction makes me want to cry(32 votes)
- It's okay, it gets a lot worse than this. I feel like you just have to think about how the electron pushing arrows are changing the mechanisms, the rest sort of follows. That's just what helped me.(1 vote)
- 2:07. Why does the lone pair of the oxygen attack the carbon?(10 votes)
- Because the carbon is electronically not rich as the double bond contribution interferes with the electron cloud density. while the oxygen is looking for an electron-deficit site to attack, which is in this case,the carbon centre(2 votes)
- At8:51, why does sulfur attack oxygen?(9 votes)
- Oxygen-oxygen bonds are weak and therefore unstable, which is why they can be attacked by the sulphur (from DMS).(2 votes)
- My professor is very into reagents. What are other common reagents for the second part of this addition reaction? Is there a way to only produce aldehydes?(3 votes)
- Probably the three most common reagents for reductive workup are Zn/acetic acid, dimethyl sulfide, and triphenylphosphine. To get only aldehydes, your starting alkene can have no more than one substituent on each alkene carbon,
You can use other reagents can reduce the carbonyl compounds directly to alcohols or to oxidize the aldehydes to carboxylic acids.(7 votes)
- at9:01, what keeps that negative oxygen from the already cleaved intermediate from earlier from just forming DMSO earlier on? It would seem counter intuitive to cleave, then reform, then cleave again. Not that nature always follows the easiest path given, but is there any specific reason that extra oxygen bonds to the carbon at6:50rather than just leaving to form DMSO?(3 votes)
- There is no DMS present during the reaction. The usual solvents is a mixture of methanol and dichloromethane. DMS is added for "reductive workup" only after the molozonide has formed.. It is easily oxidized to DMSO by the molozonide and any excess ozone.(6 votes)
- In the last part of this video where Jay expanded the product of the cyclohexene into a chain, why was the ketone/aldehyde components on either end still attached and not separate like in the general form of the Rx for the product? Thanks!(4 votes)
- because ozone only breaks the carbon-carbon double bond. the single bonds are left in tact so if you go the other way around the ring structure, you find they are still attached, but on opposite ends.(3 votes)
- what would the name of the molcule be at the end of the video? (13:40) Would the name be different for the straight chain and ring structure.(3 votes)
- I had the same question and I think the future video https://www.khanacademy.org/science/organic-chemistry/aldehydes-ketones may give us a clue since no one has answered you yet. We'll have to figure it out ourselves I guess. I'm also curious to know what it's relevance is.(1 vote)
- 13.41- "we add DMS in the second step"
can we add Zn, CH3COOH and H2O instead of DMS?(2 votes)
- At3:50, why would that happen?(2 votes)
- Any arrangement with three O atoms in a row is unstable. That is why ozone, :O-O=O:, is so reactive in the first place. The five-membered ring formed by the reaction of ozone with the ring, called a molozonide , is still highly unstable, because it still has 3 O atoms in a row. It could decompose back into ozone and the original alkene, but if it decomposes as in the video, we get more stable products — carbonyl groups and only two adjacent O atoms.(3 votes)
In this video, we're going to look at the cleavage of alkenes using a reaction called ozonolysis. So over here on the left I have my generic alkene, and to that alkene we're going to add O3 in the first step, which is ozone. And in the second step, we're going to add DMS, which is dimethyl sulfide. And be careful because there are different regions you could add in the second step, so make sure to learn the one that your professor wants you to use. If you use it DMS, you're going to get to a mixture of aldehydes and or ketones for your product, depending on what is attached to your initial alkene. So let's start by looking at a dot structure for ozone. So over here on the left is a possible structure for ozone. And we can draw a resonant structure by taking these electrons and moving them in here to form a double bond between those two oxygens. And now it pushed these electrons in here off onto the oxygen on the left. So let's go ahead and draw the other resonant structure. So now I would have a double bond between my two oxygens on the right. The oxygen on the far right has two lone pairs of electrons around it now. The oxygen in the center still has a lone pair of electrons on it, and the oxygen on the far left now has three lone pairs of electrons on it. So the oxygen on the far left now has a negative 1 formal charge, and the oxygen on the top here still has a plus 1 formal charge. And so those are our two resonant structures. Remember that the actual molecule is a hybrid of these two resonant structures. So let's go ahead and pick one of those resonant structures. I'm just going to take the one on the right. So let me just go ahead and redraw the resonant structure on the right. And so we're going to do the one that has the negative charge on the oxygen on the far left. And this top oxygen is a plus 1 formal charge. And the oxygen on the right has no charge in this resonant structure. So let's go ahead and draw in our alkene. And so here is our alkene with unknown substituents at the moment. We think about our mechanism, the negative charge on the oxygen-- this lone pair of electrons here-- is going to attack this carbon which would push these pi electrons off. And those pi electrons are actually going go to this oxygen right here which would push these pi electrons in here off onto this oxygen. So it's a concerted mechanism here. And so let's go ahead and draw the results of those electrons moving. So the oxygen on the left is now bonded to the carbon on the left. The carbon on the left now has a single bond to the carbon on the right. The carbon on the right is now bonded to this oxygen. And then these two oxygens are bonded to an oxygen in the center. For lone pairs of electrons, all of our oxygens are going to have two lone pairs of electrons. Like that. And so we had so many electrons moving, let's say if we can follow them. So let's color coordinate some electrons here. So I'm going to say that these electrons in blue, those are the ones that formed this bond between the oxygen and the carbon. And I'm going to say that these pi electrons here in red, those are the ones that formed this bond between this carbon and this oxygen. And then finally, these electrons-- I'm saying these are my pi electrons in here-- are going to move off onto this top oxygen. So you could say that those magenta electrons would be right there. And so now we have this structure. Now oxygen oxygen bonds are relatively weak. So they're unstable and so we have two oxygen oxygen bonds in this molecule. And so one of those oxygen oxygen bonds is going to break in the next step of the mechanism. And so I could pick either one since they're symmetrical. I'm just going to say that--- I'm just going to say that these electrons over here-- so I'm going to say that this oxygen oxygen bond is going to break in the next part of our mechanism. And so, if we think about these electrons and this oxygen moving in here, that would break this bond between the two carbons because there'd be too many bonds to the carbon on the right. So this bond is going to break, push those electrons into here, and then these electrons in green are going to come off onto the top oxygen [INAUDIBLE]. So our unstable oxygen oxygen bond breaks. And so let's go ahead and draw what we would get. Well, on the left side, we have a carbon. And now that carbon is going to have two bonds to the oxygen. And the oxygen is going to have two lone pairs of electrons. It would make a carbonyl compound. On the right, the carbon on the right is bonded to two other things and now has a double bond to this oxygen here. And now this oxygen has only one lone pair of electrons on it. And then this oxygen is bonded to the other oxygen, and that oxygen now has three lone pairs of electrons, which would give it a negative 1 formal charge. So we have a negative 1 formal charge here, this oxygen right here gets a positive 1 formal charge, and we form a carbonyl oxide on the right. So this charge compound on the right's called a carbonyl oxide. So in the next step of the mechanism, if we think about the carbonyl compound on the left-- so this carbonyl compound on the left here-- the double bond between oxygen and carbon, oxygen is more electronegative than carbon so oxygen's going to pull those electrons closer to it to give the oxygen a partial negative charge. Carbon's going to lose some electron density so this carbon's going to have a partial positive charge. And you can see that you have this partially negatively charged oxygen right next to a negatively charged oxygen on the carbonyl oxide. And so those, of course, are going to repel. So we're going to flip over the carbonyl compound on the left. So we're going to flip over this compound on the left and I'm going to draw the carbonyl oxide on the right the same way. So as we continue in our mechanism here, we're going to flip over the carbonyl compound on the left and we're going to keep the carbonyl oxide the same, and the same orientation like that. And so let's go ahead and put in all of our lone pairs of electrons and formal charges. So negative formal charge here, positive formal charge here, and two lone pairs of electrons on this oxygen. So let's get a little bit more room here. And once again, we can think about our carbonyl compound being polarized. Right? The oxygen is partially negative and this carbon right here is partially positive. And so in the next step of the mechanism, we're going to get a nucleophilic attack. We're going to get to this negatively charged oxygen is going to attack this partially positive carbon right here. And if the oxygen forms a bond with this carbon, that would be too many bonds to this carbon and so this lone pair of electrons, these pi electrons in here, are going to go after this carbon, which would push these electrons in here off on to that oxygen. So once again, we have a lot of electrons moving. So let's see if we can go ahead and draw the product and see if we can follow those electrons a little bit. So now we're going to have this carbon over here is bonded to an oxygen. This oxygen is now bonded to what used to be our carbonyl oxide carbon. This carbon is bonded to two other things, it's also bonded to an oxygen, and that oxygen is bonded to an oxygen which is now bonded to this carbon. So let me go ahead and put in our lone pairs of electrons. Each of our oxygens is going to have two lone pairs of electrons like that. And once again, let's see if we can follow those electrons because there was a lot that was happening. So let's go ahead and follow these electrons right here in blue. So the ones on the negatively charged oxygen, those the ones that are going to nucleophilic attack that carbonyl carbon right here and form of this bond. So this bond in blue represents those blue electrons. All right, if we follow these pi electrons here-- so these pi electrons and that carbonyl in red-- those are going to form a bond between this oxygen and this carbon like that. And then finally, these pi electrons in here on the carbonyl oxides are going to move-- those electrons are going to move off on to that oxygen like that. And so we have this compound now. So that's all for the first step, right? That's adding ozone the to our alkene. And now we can go ahead at our dimethyl sulfide. So dimethyl sulfide comes along-- let's go ahead and draw that in here-- so we'd have a sulfur. And then that sulfur is bonded to two methyl groups and the sulfur has two lone pairs of electrons like that. And so one of those pairs on the sulfur is going to attack this oxygen right here. So one lone pair is going to attack this oxygen. And once again, we have a weak oxygen oxygen bond. Right? So let me see if I can highlight that weak oxygen oxygen bond. So right here is our weak oxygen oxygen bond which is going to break. And those electrons are going to move in to here. And that would be too many bonds to this carbon, so these electrons in red are going to move into here. And then finally, the electrons in blue are going to come back off on to that top oxygen. So let's go ahead and draw-- let's go ahead and draw that. All right, so what happens now? Now we would have-- we would have a carbon over here on the left. It's now double bonded to this oxygen. And this oxygen gets two lone pairs of electrons. The carbon over here on the right is now doubly bonded to an oxygen like that. And then this carbon is also bonded to two other things. And the sulfur-- right-- the sulfur is now bonded to this oxygen here. The sulfur is bonded to this oxygen, this oxygen has three lone pairs of electrons, which gives it a negative 1 formal charge. The sulfur is bonded to two methyl groups and it still has a lone pair of electrons on it because it had two lone pairs to start with which gives the sulfur a plus 1 formal charge like that. And so if you just focused in on the sulfur compound up here, you could draw a resonant structure. You could move these electrons in here. And so if we drew a resonant structure for that, we would now have sulfur doubly bonded to an oxygen like that with two methyl groups and still having a lone pair of electrons on it. So we can now recognize that as DMSO, or dimethyl sulfoxide. And so we produce DMSO, but more importantly, we produce these two carbonyls down here. Once again, aldehydes or ketones depending on what is attached. Right? So we produce these aldehydes or ketones here. And so if we follow some of our electrons-- so these electrons in the red over here on the left, those moved in here on our carbonyl compound. And the electrons in green over here on the left, those moved in here to form this carbonyl compound. So after an extremely long mechanism, we finally get to our mixture of aldehydes hides ketones. Let's go ahead and do one example to make sure that you can apply this reaction. All right, let's look at an alkene. So here is our alkene. And to our alkene, in the first step, we're going to add ozone, so O3. And the second step, we're going to add dimethyl sulfide, so DMS. So the way to think about this reaction, or an easy way to think about it-- let me go ahead and redraw it here. We know that we're going to cleave the double bond. We're going to break the double bond here. We know that our double bond-- this is a methyl group, and over here is a hydrogen, right? We didn't draw it in over here on the left side but we know it's there. We're going to cleave this double bond. We're going to cleave it right here. And so one way to do this would be just to erase that part. Right? So if we think about erasing this part right here and adding in an oxygen to either one of those carbons here. So we're going to add in an oxygen right here. And add in an oxygen down here. And so it's a little bit hard to see so let me go ahead and redraw it. So down at the bottom we now have a carbonyl and that's a ketone. And up here at the top, we're going to have an aldehyde. And so this is your product. Now, this is probably not the easiest way to draw it. So let's go ahead and redraw it in a different way here. So if we're going to draw it as a straight chain, let's see how many carbons we would have to deal with. So we have one, two, three, four, five, six carbons. Let's go ahead and straighten out that molecule a little bit. So we would have a one and then we have a ketone here. So that's two carbons. Three, four, five, six carbons. And then we have our aldehyde right here. And so there's our aldehyde which would be this carbon right here. So a total of seven carbons. So one, two, three, four, five, six, seven. So this reaction produces an aldehyde and a ketone. So once again, be careful with the second step. And pay attention to what your professor wants to use.