Video transcript
- [Instructor] In thisvideo, we're going to look at how to determine if a reaction proceeds via an S N 1 or an S N 2 mechanism and also how to drawthe product or products for those reactions. To help us, we're gonna look at this S N 1 versus S N 2 summary and the first thing thatwe're going to look at is the structure of our substrate. For example, for this reaction down here, we have a primary alkylhalides, a primary substrate, so we need to thinkabout an S N 2 reaction which requires decreased steric hindrance and that's what we havewith a primary alkyl halide. From an earlier video, weknow that an S N 2 mechanism has our nucleophileattack at the same time that we get loss of a leaving group. And for this reaction, our nucleophile must be oursolvent here, which is ethanol. So, our nucleophile is going to attack and the oxygen is gonnaform a bond with this carbon which I'll go ahead and highlight in red. So our nucleophile attacks, at the same time, we getloss of leaving group. These electrons come offto form the iodide and ion which is an excellent leaving group. Let's draw what we would form. 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 a bond between the oxygen and the carbon in red. The oxygen is still attachedto this ethyl group here, so let's draw in those two carbons. And the oxygen is stillbonded to this hydrogen. So let's put in this hydrogen. We still have a lone pair ofelectrons left on this oxygen, so I'll put in that lone pair right here. And that gives us a plus oneformal charge on the oxygen. Next, we need to make a neutralmolecule for our product, so we need to have anothermolecule 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 this proton here and these electrons areleft behind on the oxygen. Let's draw our final product. We sketch in our carbon chain. We have our oxygen. We have these two carbons. And now we have two lone pairsof electrons on this oxygen. Let's make these electrons blue. So, our second step isan acid-base reaction where we take a proton and these are, theseelectrons in blue here, to form our final productwhich is an ether. Notice we don't have to worryabout any stereochemistry for our final product. We don't have any chiralcenters to worry about. Let's look at another reaction. For this reaction, we're starting with asecondary alkyl halides. If I look at my summary over here with a secondary substrate, we could have either an S N 2 mechanism or an S N 1 mechanism, so we need to look at a few more things. First, let's look at the nucleophile. This is N a plus and SH minus, so let me draw in the SH minus 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, wesaw in an earlier video, DMSO is a polar A product solvent which favors an S N 2 reaction. So with a strong nucleophileand a polar A product solvent, we need to think about in S N 2 mechanism. So we know our nucleophile attacks at the same time that we getloss of our leaving group, so our nucleophile is goingto attack this carbon. So again, I'll make this carbon red. At the same time that weget loss of leaving groups, so these electrons are gonnacome off onto the bromine to form the bromide and ion. For this reaction, we need to think about the stereochemistry of our S N 2 reaction. Our nucleophile has toattack from the opposite side of our leaving group, so we get inversion of configuration. So if we have a chiralcenter, we have to worry about our stereochemistryfor this reaction. So we have the bromine on a wedge, so drawing the final product here, we need to have the SHgoing away from us in space, so we put that on a dash. Again, we saw details aboutthis in an earlier video. So we get inversion of configuration for this S N 2 reaction. First, let's look at our alkyl halides. The carbon that's bonded to our bromine is bonded to two other carbon. So, this is a secondary alkyl halide. And so we know we couldhave either S N 1 or S N 2. We need to look at thenucleophile and the solvent next to decide which mechanism it is. Our nucleophile will be formic acid which is a weak nucleophile and water is a polar product solvent. So we know that these two things favor an S N 1 type mechanism. The polar product solvent water can stabilize the carbocationthat would result. So the first step should beloss of our leaving group to form our carbocations. These electrons come off onto our bromine to form the bromide ion and we're taking a bond awayfrom this carbon in red. So the carbon in red is gonna have a plus one formal charge. Let's draw our carbocations. Let me put in this ringhere with our pi electrons. And then let me highlightour carbon in red. The carbon in red is this one. So that should get aplus one formal charge. And let me highlight theother 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 carbocation, but this is actuallya benzylic carbocation which makes it even more stable than we would normally expect. The pi electrons in the ring can actually provide us withother resonant structures to stabilize this positive charge. So I don't have the time orthe space to show that here, but we have our secondarybenzylic carbocation which will be our electrophile and our next step isto 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 morenucleophilic than the other, so it turns out that this carbonyl oxygen is more nucleophilic thanthis oxygen down here and we'll go into more detailin a couple of minutes, but for right now, let'sshow our nucleophile attacking our electrophile. So, a lone pair of electrons on our oxygen are gonna form a bondwith this carbon in red. So let's draw the resultof 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 theelectrons on our oxygen. So, this lone pair of electrons in blue forms a bonds with our carbon in red. So now, let's draw in our oxygen. Our oxygen still has a lonepair of electrons on it and our oxygen is doublebonded to a carbon which is bonded to a hydrogen. And on the right, we haveour oxygen and a hydrogen. Now we have a plus oneformal charge on our oxygen. So this oxygen has aplus one formal charge. And this product is resonance stabilized, so we can move in a lonepair of electrons here and push these electronsoff onto our oxygen. And let's go ahead andshow the results of that. So we'll put in ourresonance brackets here and our resonance arrow. So we have our benzene ring, so let's draw that in. And we have our carbons. We have a bond to this oxygen and I'll drawn on thatlone pair of electrons on the top oxygen here. Should only be one bondnow to this carbon. I'll put in my hydrogen, and now actually, we havea double bond over here. 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 moved intohere to form a double bond. And then let's highlightthese electrons here in red. So these electrons came offonto our oxygen right here. And now, we have a plus oneformal charge on this oxygen. So the product of nucleophilic attack by the carbonyl oxygenis resonance stabilized. And then to go to our product, just think about a base comingalong, something like water, and we could take this proton and then these electronswould be left behind. These electrons would beleft 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 twolone pairs of electrons and so does this one. So we could highlightsome of those electrons. Let me make them green. So these electrons in greenhere came off onto this oxygen and that gives us our product. Notice that in our products, we have a chiral center. So this carbon righthere is a chiral center. We have four differentgroups attached to it. So we would expect toget a racemic mixture. We should get both in the enatiomers here because remember, goingback to our carbocation, this carbocation is planer and our nucleophile canattack from either sides. If you're wondering why this oxygen is not as nucleophilicas the carbonyl oxygen, let's show the result of what would happen if this oxygen attacked ourpositively charged carbon. Let me draw in our benzene ring. Let's sketch that in. And we would form a bondbetween the oxygen and, let me highlight our carbonwhich I made red below, so this is our carbon in red. And I'm gonna show thislone pair of electrons forming a bond with that carbon. So, that's bonded to this oxygen and the oxygen is bonded to our carbonyl 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 oxygena plus one formal charge. Notice this positively charged oxygen is right next to this carbonyl carbon and we know that this carbonylcarbon is partially positive because this carbonyl oxygen withdraws electron density from it. So you have a positively charged oxygen next to a partiallypositively charged carbon and we know that like charges repel. So having these two positivecharges next to each other would destabilized this structure. And so that's the reason why 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 S N 1 mechanism. So our first step here wouldbe loss of a leaving group. These electrons come offto form the iodide and ion and we're taking a bondaway from this carbon in red to form a carbocation. Let's get some more room. Let's go down here and let's draw our carbocation. So we have a six-membered ring, so let me draw in oursix-membered ring here. And our carbon in red is this carbon, so that carbon gets aplus one formal charge. We have a tertiary carbocation which is relativelystable for a carbocation. And in our next step, our nucleophile attacks our electrophile and our nucleophile is oursolvent, which is methanol. So our nucleophile is goingto attack our electrophile. Our oxygen is gonna form abond with this carbon in red. So let's draw the resultsof our nucleophilic attack. I'll put in this ring here, move the methyl group overto make room for the oxygen. So here's that oxygen. Let's highlight the carbon in red. And let's also highlight some electrons. So, these electrons inmagenta form the bond between the oxygen and the carbon in red. The oxygen is still bonded toa 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 pairof electrons on the oxygen, so let's put in thatlone pair of electrons and that gives the oxygena plus one formal charge. To get to our product, we need a neutral product, so another another moleculeof methanol comes along but this time, instead ofacting like a nucleophile, it's gonna act as a baseand take off that proton. So here is our molecule of methanol and we're gonna take this proton and leave these electronsbehind on the oxygen. So let's draw in ourfinal product up here. So, here is our ring and let's move up a little bit. So, let's move up. In our ring, we havethis 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. These electrons in here in blue come off onto the oxygento form our final product. Notice that for our final product, we don't have any chiralcenters to worry about, so we don't need to worry about specifying any stereochemistry.
