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Current time:0:00Total duration:14:07

Video transcript

if oxides have a lot of ring or angle strain and this makes them very reactive towards ring opening so in this video we'll look at the ring opening reactions of Epoque sides using strong nucleophiles so in the first step we add a strong nucleophile to our epoxide and in the second step we add a proton source and the nucleophile is going to end up opening opening the ring and adding an anti-fashion to the o h that is created here and so in terms of regiochemistry that nucleophile adds to the least substituted carbon so it's going to add to this carbon right here if we go back and look at our epoxide on the left that's this carbon and this carbon is attached to two other carbons so it would be classified as a secondary carbon whereas the carbon on the right is attached to three other carbons and so that's a tertiary carbon so the mechanism for this reaction is is an sn2 type mechanism which means that our strong nucleophile is going to attack the least sterically hindered carbon which is the one on the left since a hydrogen takes up less space than some some bulky r group over here on the right and so that's the regio chemistry here the stereochemistry is an anti opening of the Ring for an anti addition of the nucleophile and this will affect possible chirality centers as we will see in later examples let's look at some examples of strong nucleophiles so for your strong nucleophile you could use an alkoxide anion so something like this over here you could use a grignard reagent which we know are sources of carbon ions right so we could have a negatively charged carbon like this and you didn't have to write your your your grignard reagent like this you could have written it like you know just showing it our MGP our hydride itself is not a strong nucleophile but if hydride is coming from something like lithium aluminum hydride it functions like a strong nucleophile so we could go ahead and and show that hydride is really coming from lithium aluminum hydride so if we go ahead and draw this in here aluminum is going to have a negative 1 formal charge and then lithium would have a plus 1 formal charge like that so lithium aluminum hydride is a source of a strong nucleophile it's just at the hydride and I on its own we've already talked about is not the best nucleophile and then of course you could have something like hydrogen sulfide anion or the hydro sulfide anion right here which also functions as a strong nucleophile let's take a look at the mechanism to form to form our alcohols here so we're going to once again start with our epoxide and I'll use the same the same Epoque side that we had before meaning there's going to be an R prime here and R double prime going away from us hydrogen coming out at us and our group going away from us like that so when I think about electronegativity differences I know that the oxygen is more electronegative than this carbon for example so the oxygen is going to pull the electrons in the bonds between the oxygen that carbon closer to the oxygen giving it a partial negative charge so pulling some electron density away from this carbon will give this carbon down here a partially positive charge and so that carbon is going to function as our electrophile so when the nucleophile comes along right to the nucleophile we'll have a lone pair of electrons at a negative 1 formal charge it has it has two options right it could attack the carbon on the left or it could attack the carbon on the right and since it's a strong nucleophile which precedes to be an sn2 type mechanism it's going to attack the carbon on the left so once again the carbon the left is the least sterically hindered right that hydrogen is is not very big compared to compared to some alkyl group over here on the right so the nucleophile is going to attack the electrophile like that and when that happens it's going to kick these two electrons in here off onto your oxygen so we can go ahead and draw the product of that nucleophilic attack all right so now we broke our we broke our epoxide ring there and we have an extra lone pair of electrons on that oxygen giving that oxygen a negative 1 formal charge these are groups on the carbon on the right haven't changed right so they are still there like that we've now added our nucleophile like that and that's going to push the hydrogen up relative to the plane mein and and then we have an R group over here like that so the nucleophile has to attack from below because the oxygen is on top and the nucleophile is going to attack the least sterically hindered carbon this ends up putting your nucleophile and your oxygen and tie to each other and in the second step we can add a proton source so something like water and we're going to get an acid-base reaction where our alkoxide is going to take a proton from water which kicks these electrons in here off onto the oxygen and that will form our final product with an alcohol so I'm going to have an alcohol right here and once again my R groups are still there so I have R Prime and R double Prime our nucleophile has added and our hydrogen is pushed up and our our group is still back here so that is the mechanism for the sn2 type reaction let's do let's do a problem which involves some stereochemistry so we can see we can see how the mechanism will determine the stereochemistry of the products here so I'm going to start out I'm going to start out I'm going to put my my my epoxide coming out at me in space like that and then at this top carbon here I'm going to have a methyl group going away from me in space so that's my that's my reactant and I'm going to use sodium ethoxide so go ahead and draw sodium ethoxide here and we're going to use ethanol as the solvent so we have ethanol here as our solvent like that so if we if we look at this at Bach side it's it's a little bit hard to see the stereochemistry so let's go ahead and redraw this molecule and we'll look at it from a different vantage point here so if I look at that Epoque side from a different vantage point I'm looking at it from above and down a little bit and that would put my POC side going up like that and then it was this carbon right here that's going to have the methyl group going down relative to the plane of the Ring and then this carbon over here on the left has a hydrogen so my first step would be to identify my nucleophile which of course is the ethoxide anion so I can go ahead and on my ethics ID and on in here and next I need to think about where that nucleophile will attack all right so I know I know that this top oxygen here being more electronegative than then the carbons is going to have a partial negative charge and so I have two options right I could be the carbon the left I could be the carbon on the right and if I think about the fact that this is a strong nucleophile sn2 type mechanism it's going to attack the the less sterically hindered carbon which of course is the carbon on the left which is going to have a partial positive charge and so we're going to get nucleophilic attack all right my nucleophile is going to attack this carbon right here which would kick these electrons in here off onto the oxygen so I can go ahead and draw the product of that right so I have my ring my ring like that and I opened up the epoxide ring so now now this this bond is to an oxygen it has three lone pairs of electrons like that which give it a negative 1 formal charge there's still a methyl group going down at this carbon and I've added my nucleophile to this carbon so I can go ahead and draw my oxygen and then an ethyl like that and then I pushed up the hydrogen alright so so when this reaction occurred the hydrogen that was down that relatives that playing the ring got pushed up so now there's a hydrogen up at that carbon and in my final step right I'm going to protonate the epoxide so lone pair of electrons will pick up a proton from ethanol which kicks these electrons off onto here and I form my alcohol as my product all right so when I'm drawing my product over here now I have an OHA and I still have a methyl group going down and over here I have an oxygen and then and then ethyl like that so if I wanted to draw my product as a two dimensional dot structure right I could I could look I could look this way I could look down on it like that and just go ahead and draw what I see when I'm looking down so I can draw my ring so here is my ring and if I look at this carbon right here there's an O H coming out at me so I can represent the O H coming out at me with a wedge right here and there's a methyl group going away from me at that carbon so this methyl group is going away from me so I can go ahead and put in a dash for that methyl group and when I look at the the other carbon here alright I have I have an O and then Ethel going away from me so I have this guy would be a dash going away from me like that so that of course is this portion of the molecule so that's my product when I look at what happened to these stereochemistry let's let's first start with this this top carbon right here this top carbon had a methyl group going away and an oxygen coming out coming out at you at that carbon and if we look over here that's that's the same thing that we have we had that we have a methyl group going away from us and then an oxygen coming out at us like that so it's the same absolute configuration at that carbon if we look at the other carbon if we look at this carbon down here on the left we had an oxygen coming out at us in space and on the right at this carbon now the oxygen is going away from us in space so we actually see inversion of configuration right so right here we have inversion of configuration in terms of the stereochemistry we went from a wedge to a dash and that's of course due to the mechanism okay we had a nucleophilic attack at a chirality Center and so we observe inversion of configuration at that chirality Center let's do one more example of a strong nucleophile attacking an epoxy ID and so this time let's use a grignard reagent so I'm going to start with a phenol magnesium bromide so we've already made phenol magnesium bromide as our as our grignard reagent and in the first step we're going to add ethylene oxide right the simplest Epoque side and our second step we're going to add hydronium so h3o plus and that'll be our source of proton but that has to be in a second step you can't add that at the same time because of there would be an acid-base reaction from your grignard reagent with your hydronium ions so if I'm if I'm looking at phenol magnesium bromide I could draw it like that or I could redraw it too to identify my carb anion all right so if I think about what's going to happen the electrons and the bonds between this carbon and magnesium and here carbon is more electronegative than magnesium so you could think about those two electrons as being on that carbon like this which would give that carbon a negative 1 formal charge and so now we have a we have a carb anion and then we'd have mg BR plus like that and that's going to be the nucleophile when we have our when we have our athlean oxide that comes along right so same idea oxygens more electronegative so partial negative on the oxygen and a partial positive on the carbons that are symmetrical so when I think about nucleophilic attack all right these electrons in blue are going to be my nucleophile they're going to attack my electrophile kicking these electrons off so when I draw the product of the nucleophilic attack I have my I have my ring here and and now I'm going to end up with an alkoxide that looks like this okay so let me go ahead and highlight the electrons in blue the electrons in blue formed a new bond a carbon-carbon bond right here so the electrons in blue formed a new carbon-carbon bond which is y which is y organometallics are are so useful right you can form these carbon-carbon bonds here and then I can see that the two carbons over here on the right this carbon and this carbon those are the ones that came from my ethylene oxide like that so in my in my last step right it's an acid-base reaction and this alkoxide is going to pick up a proton from h3o plus in the second step so a lone pair of electrons picks up a proton kicks these electrons and I can go ahead and draw the product all right so once again here's my ring and I now protonated my alkoxide to form and alcohol and the name of this alcohol all right so if I think about if I think about this as being carbon one and this is being carbon two i have a phenyl group coming off of ethanol so two phenyl ethanol would be the name of the product that's formed and two phenyl ethanol is famous for being a component of rose oil so this molecule smells a lot like roses and it's used it's used in the perfume industry so this is this is one way to make two phenyl ethanol there are actually several ways to do it but I think perfume perfume chemistry is extremely interesting