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Current time:0:00Total duration:13:54

Video transcript

this video we're going to look at how to determine if a reaction proceeds be an sn1 or an sn2 mechanism and also how to draw the product or product for those reactions and help us we're going to look at this sn1 versus sn2 summary and the first thing that we're going to look at is the structure of our substrates for example for this reaction down here we have a primary alkyl halide a primary substrate so we need to think about an sn2 reaction which requires decreased steric hindrance and that's what we have with a primary alkyl halide from an earlier video we know that an sn2 mechanism has our nucleophile attack at the same time that we get lost of a leaving group and for this reaction our nucleophile must be our solvent here which is ethanol so our nucleophile is going to attack and the oxygen is going to form a bond with this carbon which I'll go ahead and highlight in red so our nucleophile attacks at the same time we get loss of a leaving group these electrons come off to form the iodide anion which is an excellent leaving group let's draw what we would form so let's sketch in our carbon chain here and we know that a bond forms between the oxygen and the carbon in red so the carbon in red is this carbon and let's make these electrons magenta so those electrons form of bonds between the oxygen and the carbon in red the oxygen is still attached to this ethyl group here so let's draw in those two carbons and the oxygen is still bonded to this hydrogen so let's put in this hydrogen we still have a lone pair of electrons left on this oxygen so I'll put in that lone pair right here and that gives us a +1 formal charge on the oxygen next we need to make a neutral molecule for our product so we need to have another molecule of ethanol come along so let's draw that in here so ethanol is our solvent and this time the ethanol molecule is going to function as a base we need to take we need to take this proton here and these electrons are left behind on the oxygen so let's draw our fine product.we is sketch in our carbon chain we have our oxygen we have these two carbons and now we have two lone pairs of electrons on this oxygen so let's make these electrons blue so our second step is an acid-base reaction where we take a proton and these are these electrons in blue here to form our final product which is an ether notice we don't have to worry about any stereochemistry for our final products we don't have any chiral centers to worry about let's look at another reaction so for this reaction we're starting with a secondary alkyl halides if I look at my summary over here with a secondary substrate we could have either an sn2 mechanism or an sn1 mechanism so we need to look at a few more things first let's look at the nucleophile this is na plus and SH of - let me draw in the SH - here which that is going to be our nucleophile and that's a strong nucleophile a negative charge on a sulfur would make a strong nucleophile and for our solvent we saw in an earlier video DMSO is a polar a protic solvent which favors an sn2 reaction so with a strong nucleophile and a polar a protic solvent we need to think about an sn2 mechanism so we know our nucleophile attacks at the same time that we get loss of our leaving group so our nucleophile is going to attack this carbon so again I'll make this carbon red at the same time that we get lost of a leaving group so these electrons going to come off onto the bromine to form the bromide anion for this reaction we need to think about the stereochemistry of our sn2 reaction our nucleophile has to attack from the opposite side of our leaving group so we get inversion of configuration alright so if we have a chiral center we have to worry about our stereochemistry for this reaction so we have the bromine on a wedge so drawing the final product here we need to have the SH going away from us in space we put that on a - again we saw more details about this in an earlier video so we get inversion of configuration for this sn2 reaction first let's look at our alkyl halide the carbon that's bonded to our bromine is bonded to two other carbons so this is a secondary alkyl halide and so we know we could have either sn1 or sn2 we need to look at the nucleophile and the solvent next decide which mechanism it is our nucleophile will be formic acid which is a weak nucleophile and water is a polar protic solvent so we know that these two things favored an sn1 type mechanism the polar protic solvent water can stabilize the carbo cation that would result so the first step should be loss of our leaving group to form our carbo cations so these electrons come off onto our bromine to form the bromide ion and we're taking a bond away from this carbon in red so the carbon in red is going to have a +1 formal charge so let's draw our carbo cations let me put in this ring here with our PI electrons and then let me highlight our carbon in red the carbon in red is this one so that should get a +1 formal charge and let me highlight the other carbon over here so this carbon in magenta I moved it up to here to make it easier to see the mechanism so this is our secondary carbo cation but this is actually a benzylic carbo cation which makes it even more stable and we would normally expect the pi electrons in the ring can actually provide us with other resonance structures to stabilize this positive charge so I don't have the time or the space to show that here but we have our secondary benzylic carbo cation which will be our electrophile and our next step is to have our nucleophile attack our electrophile and our nucleophile is formic acid here we have two oxygens in formic acid and one of them is more nucleophilic than the other so it turns out that this carbonyl oxygen is more nucleophilic than this oxygen down here and we'll go into more detail in a couple of minutes but for right now let's show our nucleophile attacking our electrophile so a lone pair of Tron's on our oxygen are going to form a bond with this carbon in red so let's draw the results of our nucleophilic attack we have our benzene ring so I'll draw that in here and our carbon in red is this one and let's highlight the electrons on our oxygen so this lone pair of electrons in blue forms the bond with our carbon in red so now let's draw in our oxygen our oxygen still has a lone pair of electrons on it and our oxygen is double bonded to a carbon which is bonded to a hydrogen and on the right we have our oxygen and a hydrogen now we have a +1 formal charge on our oxygen so this oxygen has a +1 formal charge and this product is resonance stabilized so we can move in a lone pair of electrons here and push these electrons off onto our oxygen and let's go ahead and show the result of that so we'll put in our resonance brackets here and our resonance arrow so we have our benzene ring let's draw that in and we have our carbons we have a bond to this oxygen and I'll draw on that lone pair of electrons on that top oxygen here it should only be one bond now to this carbon I'll put in my hydrogen and now actually we have a double bond over here so let me draw everything in and let me show the movement of electrons so first let's start with these electrons which I will make magenta those electrons move into here to form a double bond and then let's highlight these electrons here in red so these electrons came off onto our oxygen right here and now we have a +1 formal charge on this oxygen so the product of nucleophilic attack by the carbonyl oxygen is resonance stabilized and then to go to our product just think about a base coming along something like water and we could take this proton and then these electrons would be left behind these electrons will be left behind on that oxygen so let's draw in our product we would have a benzene ring so here's our benzene ring we would have an oxygen here and then we would have our carbonyl and then our hydrogen so this oxygen has two lone pairs of electrons and so does this one so we could highlight some of those electrons when we make them green so these electrons in green here came off onto this oxygen and that gives us our product notice that in our product we have a chiral Center so this carbon right here is a chiral Center we have four different groups attached to it so we would expect to get a racemic mixture we should get both enantiomers here because remember going back to our carbo cation this carbo cation is planar and our nucleophile can attack from either side if you're wondering why this oxygen is not as nucleophilic as the carbonyl oxygen let's show the result of what would happen if this oxygen attacked our positively charged carbon so let me draw in our benzene ring so let's sketch that in and we would form a bond between the oxygen and let me highlight our carbon which I made red below so this is our carbon in red and I'm going to show this lone pair of electrons forming a bond with that carbon so that's bonded to this oxygen and the oxygen is bonded to our carbon eel which is bonded to our hydrogen and our oxygen still has a hydrogen on it and a lone pair of electrons so there's still a hydrogen on it and there's still a lone pair of electrons which gives this oxygen a plus 1 formal charge notice this positively charged oxygen is right next to this carbonyl carbon and we know that this carbonyl carbon is partially positive because this carbonyl oxygen withdraws electron density from it so you have a positively charged oxygen next to a partially positively charged carbon and we know that like charges repel so having these two positive charges next to each other would destabilize this structure and so that's the reason why by this oxygen is not the nucleophile the carbonyl oxygen is let's look at one final example so for this alkyl halide this is a tertiary alkyl halide and a tertiary substrate means think about an sn1 mechanism so our first step here would be loss of a leaving group these electrons come off to form the iodide anion and we're taking a bond away from this carbon in red to form a carbo cation let's get some more room let's go down here and let's draw our carbo cations so we have a six membered ring so let me draw in our six membered ring here and our carbon in red is this carbon so that carbon gets a plus 1 formal charge so we have a tertiary carbo cation which is relatively stable for a carbo cation and in our next step our nucleophile attacks our electrophile and our nucleophile is our solvent which is methanol so our nucleophile is going to attack our electrophile our oxygen is going to form a bond with this carbon in red so let's draw draw the results of our nucleophilic attack so put in this ring here move the methyl group over to make room for the oxygen so here's that oxygen let's highlight the carbon in red and that's also highlight some electrons so these electrons in magenta form the bond between the oxygen and the carbon in red the oxygen is still bonded to a hydrogen and a methyl group so let's put those in so here is our hydrogen and here is our methyl group we still have a lone pair of electrons on the oxygen so let's put on that lone pair of electrons and that gives the oxygen a plus 1 formal charge to get to our product we need to use we need a neutral products so another molecule of methanol comes along but this time instead of acting like a nucleophile it's going to act as a base and take off that proton so here is our molecule of methanol and we're going to take this proton and leave these electrons behind on the oxygen so let's draw in our final product up here so here is our ring and let's move up a little bit so let's move up and our ring we have this methyl group here and we have our oxygen and another methyl group two lone pairs of electrons on the oxygen let's show those electrons so these electrons in here in blue come off on to the oxygen to form our final product notice that for our final product we don't have any chiral centers to worry about so we don't need to worry about specifying any stereochemistry