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Current time:0:00Total duration:13:42

Video transcript

this video we're going to look at the cleavage of alkenes using a reaction called O's analysis so over here on the Left I have my generic alkene and to that alkene we're going to add oh three in the first step which is ozone and the second step we're going to add DMS which is dimethyl sulfide and be careful because there are different regions you could add the second steps and make sure to to learn the one that your professor wants you to use if you use DMS you're going to get 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 dot structure for ozone and we can draw a resonance structure by taking these electrons and moving them in here to form a double bond between those two oxygens and that would push these electrons in here off on to the oxygen on the left so let's go ahead and draw the other resonance structure alright 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 the negative one formal charge and the oxygen on the top here still has a plus one formal charge and so those those are our two resonance structures remember that the actual molecule is a hybrid of these two resonance structures so let's go ahead and and pick one of those resonance structures I'm just going to take the one on the right so let me just go ahead and redraw the resonance 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 one formal charge and the oxygen on the right has no charge in this resonance structure so let's go ahead and draw in our alkyne and so here is our alkyne with unknown substituents at the moment and we think about our mechanism the negative charge on the oxygen right this lone pair of electrons here is going to attack this carbon which would push these pi electrons off and those PI electrons reaction go to this oxygen right here which would push these PI electrons in here off onto this oxygen so it's a 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 right is now bonded to the carbon on the left the carbon on the left is 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 have so many electrons moving let's see if we can now let's see if we can follow them so let's color coordinate some electrons here and so I'm going to say that these electrons in blue all right 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 and red those are the ones that formed this bond between this carbon and this oxygen and then finally these electrons I'm setting 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 all right 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 right 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 on this oxygen moving in here all right 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 two so our unstable oxygen oxygen bond breaks and so let's go ahead and draw what we would get well on the left side right 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 so make a carbonyl compound on the right the carbon on the right is bonded to two other things and it is now it now has a double bond to this 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 a negative 1 formal charge here this oxygen right here gets a but gets a positive 1 formal charge we form a carbonyl oxide on the right so this is charged compound the writes called a carbonyl oxide so in the next step of the mechanism if we think about the carbonyl compound on the left right so this carbonyl compound on the left here the double bond routine oxygen and carbon oxygen is more electronegative than carbon so oxygen is going to pull those electrons closer to it to give the oxygen a partial negative charge carbon is going to lose some electron density so this carbon is going to have a partial positive charge and you can see that you have these this partial negative this partially negatively charged oxygen right next to a negatively charged oxygen on the carbonyl oxide and so those of course is going to repel so we're going to 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 the carbonyl compound on the left and we're going to keep the carbonyl oxide the same in 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 right we're going to get this negatively charged oxygen 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 this pi 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 if we can go ahead and draw the product and see if we can follow those 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 used to be our carbon dioxide carbon this carbon is bonded to 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 alright so let's go ahead and follow these electrons right here in blue so the ones on the negatively charged oxygen those are the ones that are going to nucleophilic attack that carbonyl carbon right here and form this bond so this bond in blue represents those blue electrons alright if we follow these PI electrons here so these PI electrons in that carbonyl in red right those are going to form a bond between this oxygen and this carbon like that and then finally and then finally these PI electrons in here on the carbonyl oxides are going to move those electrons are going to move off onto that oxygen like that and so we have this compound now so that's that's all for the first step right that's adding ozone to our alkyne and now we can go ahead and add our dimethyl sulfide all right so dimethyl sulfide comes along let's go ahead and draw that in here so we'd have a 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 the one lone pairs on the sulfur is going to attack this oxygen right here okay so one lone pair is going to attack this oxygen and once again we have a weak oxygen-oxygen bond alright so let me see if I'm going to highlight that weak oxygen-oxygen bond so right here is our weak oxygen-oxygen bond which is going to break right and those electrons are going to move in to here and that would be too many bonds to this carbon so these electrons in Reds are going to move into here and then finally they let the electrons in blue are going to come back off on to that top oxygen so let's go ahead and draw and let's go ahead and draw that alright so what happens now now we would have okay 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 has also bonded to two other things and the sulfur right the sulfur is now bonded to this oxygen here sulfur is bonded to this oxygen this oxygen has three lone pairs of electrons which gives it a negative one formal charge the sulfur is bonded to two methyl groups and it still has a lone pair of electrons on it alright because it had two lone pairs to start with which gives the sulfur a +1 formal charge like that and so if you just focus in on on the sulfur compound up here you could you could draw a resonance structure alright you could move these electrons in here and so if we drew a resonance structure for that we would now have sulfur double 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 it as DMSO or dimethyl sulfide 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 to follow some of our electrons all right so these electrons in red over here on the left right those moved in here on our carbonyl compound and the electrons in green over here on the left right those moved in here to form two on this carbonyl compound so after an extremely long mechanism we finally get our mixture of aldehydes or ketones let's go ahead and do one example to make sure that you can apply this reaction alright let's let's look at an alkene alright so here is our alkene and to our alkene in the first step we're going to add ozone so oh three and the second step we're going to add dimethyl sulfide so DM s so the way to think about this reaction or an easy way to think about it let me go ahead and redraw 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 right ah 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 one way to do this would be just to erase that part right so if we think about think about erasing this part right here and adding in an oxygen to either one of those carbons here so we're going to 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 let me go ahead and redraw it here so down at the bottom alright we now have a carbonyl and that's a ketone and up here at the top right we're going to have an aldehyde and so this is your product this is your product now this is probably not the easiest way to draw it alright 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 alright 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 our aldehyde right here and so there's our aldehyde alright which would be which would be the carbon right here so so a total of a total of seven carbons so one two three four five six seven so this reaction produced an aldehyde and a ketone so once again be careful with the second step make and pay attention to what your professor wants to use