 Okay, we are going to continue with lecture seven today and finish it up and we're also including in that lecture eight which is really just a couple of things for me to say about salvation. So when you look online for the notes for these lectures, I combine them together. There is no longer a link to separate notes for lecture eight. Okay, so we were talking about neighboring group participation in ionization reactions and we had mentioned things like mustards where nitrogen lone pairs can assist in the ionization of leaving groups like chloride. We had talked about acetate groups bending over with lone pairs on carbonyls making five membered rings when they participate in the ionization and so instead of talking about lone pairs I want to shift over and talk about double bonds and when they can help groups to ionize away and I want to urge you to look for homo allyl groups when you look at ionization reactions and here's an example of that. Let's take a look at the rates for SN1 ionization, the loss of this to give a carbocation for these two types of norbornal systems. They look very similar. The only difference is that this one has a double bond here at this position and when you look at the relative rates for SN1 ionization to make a carbocation, when you have that double bond over there at the two, three position of the norbornal system, the rate is the relative rate, it looks a lot faster. Why is there this huge difference in rate for the ionization of this tosyloxie group over here? Obviously it has to do with the double bond and this is purely an example of something akin to neighboring group participation. It's sort of related to what we had over there with that nitrogen mustard except now this involves a double bond. Let me go ahead and draw out some explicit participation. I'll draw a little bit of arrow pushing to see what would happen if I did use that double bond to help push out the leaving group. What would we see happening here? I'll try to draw some of those structures that you would get and I hope you'll immediately see what's going on here to be something you're already familiar with. Now you can't tell from my arrow pushing which bond I'm forming which carbon atom I'm forming a bond to. You can't tell whether I'm forming a bond to here or here. So arbitrarily I'm going to make the carbocation end up at this position and why is that a special carbocation? It's got that amazing cyclopropyl carbinyl setup right here. These bonds in this three-membered ring are strained. They're donating in to that carbocation. So it's a great carbocation that should form that's stabilized in low on energy or conversely this pi bond is indeed helping to push out that leaving group and it buys you 100 billion fold in rate acceleration. And of course there's another resonance structure I could draw that would put the three-membered ring bond to here and leave the carbocation on top. So this is nothing more than forming a cyclopropyl carbinyl cation. It's just that we didn't have a cyclopropane in our starting material so maybe you didn't see that. When you look at the problem set that I just assigned you all over the place I have these tricks embedded in here because I'm counting on the fact that you're not good at seeing this. And so your job is to try to see that stuff. This is like Sudoku except instead of paying money to buy a Sudoku puzzle I'm giving you free puzzles to work on and you should be paying me money. For why don't I use it? Well this is a reaction we're breaking the tosyloxie group so that's no longer why don't I? Oh there's another resonance structure. Why didn't I just don't have the time. You can draw another resonance structure where this is bonded over here. This one? Oh no because this oxygen is no longer bonded to that carbon that's this is the cation that results from the ionization. Okay so what's another secret way that I've hidden this stuff in your problems? I'm going to ask you to compare the fates of two different and related substrates. This is not benzyl we would refer to this as homo-benzyl. Homo means you've got one extra carbon atom. So instead of having allyl if you have homo-allyl that means you have one extra carbon. This is not benzyl this is homo-benzyl there's an extra carbon atom in there. So let's try to imagine an SN1 ionization in the presence of formic acid is your solvent that's your nucleophile and your solvent we call that solvolicis and so let's imagine if we do a displacement reaction and what we're going to do is we're going to imagine what would happen if I put a C13 label right at that position? Right there is the benzylic position if I had a C13 label carbon 13. What I'll show you is that the end products here depend on the substituent on this aromatic ring. Really you get the same kind of phenethyl substituted product in both cases but you don't know that anything interesting has happened here until you do that C13 labeling experiment. If I label these with a C13 at the benzylic position the nitro group shows you that it's still there at the benzylic position didn't move anywhere nothing special but when I look at the substrate that has the methoxy group there what I find is that that label is now scrambled it's 50% C13 label here and 50% C13 labeling up there and so what's going on with this? Let me just be clear that that's a C13 label at that position. How can you get 50% scrambling at that position? Let's try to draw this out I'm going to draw the substrate in this I'm going to try to imagine what we would see if I looked at that edge on so I'll try to draw this sort of three dimensionally and so here's a picture of that aromatic benzene ring if we were to stand on the side of that and view that edge on and I'll draw that side chain substituent sticking off of here with the tosylate leaving group. So there's that benzene ring side on and if I look at the methoxy substituted system I should be reminded that there's a resonance structure that I can draw in which I engage the lone pairs on the oxygen atom so the lone pairs are nucleophilic we know that so let's draw a resonance structure of this I'll take these electrons right here and I'll donate them into the aromatic ring and I think you've probably done a lot of these types of resonance structures in order to explain the rates of electrophilic aromatic substitution why you favor ortho parasubstitution those are common concepts from sophomore organic chemistry. So when I draw the resonance structure for this hopefully it'll be totally clear why we're seeing this scrambling effect so now when I draw that resonance structure what I can see is that there's a in that resonance structure there's a carbanion located here at the position para to the methoxy group so here's the resonance structure and so you can see why for the methoxy substituted substrate but not for the nitro substituted substrate you can expect this kind of participation like that in other words the aromatic ring the electrons in the pi system of the aromatic ring can help to push out this leaving group and so the intermediate that you generate has this structure that at first looks insane and we call this particular ion not an arinium ion we call this a phenonium ion it's a well-known kind of intermediate this is called a phenonium ion intermediate there's 26 to 27 kilocalories per mole of ring strain in a three-membered ring but this is really just another cyclopropyl carbonyl it's that same kind of cyclopropyl carbonyl I can take that positive charge and roll it over here with a resonance structure and show that it's right next to those very nucleophilic cyclopropane bonds and there's another resonance structure where it's located right here on the other side next to the cyclopropane bond so that's a super stabilized kind of structure and so that's why it's in particular when you have this methoxy group here you really can easily form that type of phenonium ion which is really just a cyclopropyl carbonyl cation and so now when your nucleophile comes in of course you have no way to control whether the nucleophile is attacking the labeled carbon or the unlabeled carbon because these look pretty similar those two carbons there might be a slight isotope effect but it would also work for ortho that wasn't on the problem set was it okay so let's take a look at sigma bonds I don't even know what through space most molecules are mostly empty space electrons are very small so maybe everything is through space what I want to pay attention to here is the effects of conformation and medium sized rings and this is the idea we want to consider is well-aimed carbon hydrogen bonds ones that are poised I want to think about the ionization of tosylates in acetic acid so acetic acid is the solvent and it's the nucleophile in this reaction of course it's a boring transformation but these kinds of experiments really inform us about chemical structure and what I want to look at is how the rates vary as we vary the ring size if you change this ring size from 3 to 4 to 5 to 6 to 7 every kind of ring size in between what you find is that for some ring sizes this becomes much faster and so the question is why should that be faster for those particular sizes of rings so 11 membered rings aren't quite as fast 12 membered rings actually seem pretty normal like it's more closer to a 6 membered ring what's special about these types of ring sizes so in order to understand why you get this rate acceleration for these particular ring sizes we need to draw the confirmations of these kinds of rings and it's closely related the reasoning for this kind of rate of ionization is closely related for why it is so hard to cycle eyes to make 8, 9 and 10 membered rings so I'm going to start off by drawing what you would get if you had 2 chair rings used to each other we call this a decalin ring system it's a 10 carbon bicyclic ring system when you have 2 chairs fused that's called decalin and the first thing that I know here with decalin I'll draw the hydrogens at the bridgehead because it helps me to better see the confirmation of this molecule the first thing that I see is every and this is why chairs are so great every single carbon-carbon bond has a staggered orientation so if I draw a Newman projection for any one of the bonds looking down the axis of any one of these bonds what I would see is perfect staggering between these bonds that minimizes the amount of torsional strain in any system so this type of bicyclic system is perfect but what happens if I don't have this carbon-carbon bond in the middle what happens if I have 10 carbon atoms and no bond in the middle it will not be a comfortable situation so if you want you can just well I don't want to draw that bond in the middle I'm simply going to draw again this this decalin ring system but if I don't have a bond in the middle and I still have these hydrogens sticking straight up and straight down and everything else is perfectly staggered the question is where am I going to put these hydrogen atoms which now have to exist in the middle of that ring the problem is they're trying to exist in the same locations in space this is the confirmation that minimizes torsional strain and so now if I want to replace this carbon-carbon bond with two H's that's a problem and it's exactly this that makes it very hard to synthesize 10-membered rings through cyclization we call those bumping interactions between the two hydrogens we call those trans-annular interactions trans-annular means across the ring so they're bumping into each other and so if you take one of these groups down here and you replace this with a tassel group and you look at it at the ionization of that group the carbocation that exists in this ring actually ends up being stabilized by the electrons in the opposite CH bond I mean these are right in each other's faces and the electrons in the CH bond are reaching across the ring and donating into the antibonding orbitals they're in the cation and they're there in the starting materials and they're there in the transition state lowering the energy for ionization of the tassel group so having that extra electron density and my drawing doesn't do it justice you have to imagine this CH bond poised above the antibonding orbital ready to overlap so you can see the effects of even sigma bonds helping to assist in pushing out leaving groups any kind of donation of filled orbitals into unfilled orbitals is a stabilizing interaction okay so that's all we have to say about neighboring group participation with lone pairs with pi bonds and ultimately if the positioning is just right even sigma bonds can have that kind of effect yes no I mean my confirmation doesn't do justice to this I mean maybe if I put this maybe if I put this other H going downward I mean I again my drawing doesn't do my drawing doesn't allow you to see which of these CH bonds is poised best above sigma star okay so I had originally another lecture that had just a couple of things to it on salvation and it's not really a full lecture so you can consider this to be lecture eight but we just have a few things to say about salvation so I want you to think about the SN1 transition state for ionization of bromide from t butyl bromide in the transition state the bond distance the distance between bromide and carbon is getting longer you're starting to build up more partial positive charge on this carbon more partial negative charge on this carbon and in particular I want to look at the rates for this reaction as we change the solvent the relative rates actually the rate constants so we'll look initially at chloroform and then we'll compare that to formic acid and the rates of reaction in formic acid which is a reasonably polar solvent so if I assign a relative rate of 1 in chloroform for this ionization reaction it turns out that the rate in formic acid is substantially bigger and I want you to remember this difference in rates what if you could accelerate your Ph.D. by a factor of 200,000 right or what if you had to wait over the weekend for your reaction to finish I'm basically giving you a recipe for how to make your reactions faster there's something in here your reactions you don't have to suffer with the reaction conditions you find in the literature you can take basic concepts and apply them to how to make things faster why is this so much faster it has to do with this remember this Coulomb's law or this variant of the Coulomb's law where you say something about charge 1 times charge 2 divided by the distance between the charge but there's also another factor in there either permittivity or dielectric constant or some version of that that has to do with the polarity of the solvent the dielectric constant for the solvent how good is the solvent at screening charges so the charges can't see each other it takes energy to pull a minus charge away from a plus charge that's very very energetically costly to take a plus and a minus and pull them apart and if you could somehow decrease it so those charges couldn't see each other then you'd make it easier to pull the charges apart let's go ahead and take a look at some dielectric constants for some various solvents I'm going to give you a table here we're going to compare everything to the gas phase what's the dielectric constant for the gas phase it's not zero it can't be zero because if it's zero you'd be dividing by zero it's one it's nothing can be lower than that so if you think about nonpolar solvents that are similar to the gas phase hexane is a super nonpolar solvent most things don't dissolve in hexanes but if more things dissolve in hexanes and you wanted a nonpolar solvent you might be inclined to use that there's other solvents that aren't quite as nonpolar but are more commonly used ether diethyl ether has a dielectric constant of only four THF is more polar than ether tetrahedral furan but still it's sort of in this range where it's I wouldn't call that a nonpolar solvent but it's not very polar I would refer to hexanes and ether as nonpolar when somebody tells me nonpolar solvent those are the common types of solvents that use for chromatography or for extractions that you would think is something as nonpolar THF is just slightly polar enough where I wouldn't call it nonpolar dichloromethane also has a dipole moment and it's a polar solvent the two chlorine carbon bonds have dipoles associated with them and that's got about the same dielectric constant bulk dielectric constant as THF and now I'm going to show you a big jump to some common solvents I would refer to these as polar solvents methanol dimethyl formamide acetonitrile dimethyl sulfoxide and they all have dielectric constants are in this 30 to 40 range so I'll give you the numbers here 33 for methanol 37 for DMF acetonitrile and nitrile is 38 I never remember the exact numbers for these but I remember that the DMSO is more polar than the rest of those as a group I would consider all of these polar solvents and DMSO the most polar among them if I try to think about polar solvents to run my organic reactions in DMSO is probably about the most polar solvent that I know of to run reactions in except for one there's one other solvent that I don't have on this list that is the most polar of all water of course and it's totally in a different class here it's 78 so why don't you run more reactions like SN1 reactions in water well most things aren't soluble in water and of course water is reactive as well it's a nucleophile and an electrophile so that's why you don't use water more often okay so in other words if you change from the gas phase to DMSO you could expect things to be how much faster for an ionization reaction not 10 of just 50 if you go from 1 to 50 you can expect changes on the order of a factor of 50 what's up with that? That's a lot bigger than 30 or a factor of 40 or even a factor of 78 there must be something else involved here that's related to solvents it's not just this is an important part of why formic acid is faster but this dielectric constant thing can't be the whole story so why else might that ionization be so much faster in formic acid I'm going to redraw that transition state where this is where this T-butyl group is starting to become flat and that bromide is walking away and starting to pick up its negative charge if you're in formic acid formic acid is very powerful at hydrogen bonding so let me draw one of the and there's lone pairs on here I'll draw some of them let's just imagine some of those hydrogen bonds let me draw it in a different color so it really gets emphasized here's a hydrogen bond and if this is surrounded by formic acid molecules that can form very powerful hydrogen bonds to that bromide in a way it makes it a better leaving group importantly those hydrogen bonds are stabilizing the transition state as this bromine is starting to walk away as it's starting to walk away with its negative charge those hydrogen bonds are already there assisting they were there in the starting material they're there in the transition state they're definitely there when the Br minus is floating all around on its own helping to stabilize the Br minus so H bonds can stabilize the transition state and what you have to imagine is not just one hydrogen bond I won't draw the whole formic acid molecule here but you can imagine this thing surrounded by hydrogen bonds helping that to leave so the effects of polar solvents are partially due to dielectric constant but they're also due to very explicit effects that involve interactions between filled orbitals and unfilled orbitals like hydrogen bonds okay so I want you to be cautious well not overly cautious there's exceptions to this this idea that polar solvents lead to rate accelerations of course an SN2 reaction doesn't have this degree of reliance on dielectric this is what I'm really talking about here is SN1 reactions to make carbocations maybe I should just make this B of 3 Eth rate so of course you can distill that's a just it always baffles me it's a distillable liquid but it's a common source of boron trifluoride you put it into your reaction and it dissociates to give you a very powerful Lewis acid so there's a reversible dissociation of boron trifluoride Eth rate what's the effect of going to a what's the effect of polarity on this reaction so up above what we said is that if you have a polar solvent that it screens these charges as they're starting to pull away but in this case you're not making ions you start with an ion and you go to something that's neutral and so here polar solvents will decrease the rate of this ionization polar solvents will decrease the rate of that ionization because the polar solvents will stabilize the starting materials and destabilize the more and more you go in the transition state towards those two products so in most cases ionization leads to ions but here's a case where you start with ions or something that's ionic and you go to something that's neutral. Okay so let's look at some other explicit types of solvent effects like these formation of these formation of those hydrogen bonds so THF and dichloromethane have very similar dielectric constants just as a judge of solvent polarity but if you look at THF it's totally different from dichloromethane. Usually when you run reactions in dichloromethane and you add salts, potassium carbonate, sodium phosphate, they're totally insoluble in dichloromethane. They just sort of spin around at the bottom of your reaction flask as this white solid that doesn't do anything. If you put lithium bromide in THF it's totally soluble. It dissolves like it's dissolving in water and you can't justify that based on dielectric constant. The reason why that's true is because lithium is a second row atom and it wants eight electrons just like every other second row atom. So as many solvent molecules as you can fit around there, you will find them attached to THF. I'm just drawing two molecules here attached to THF. If I want to match the charges, if I leave the charges off those are dative bonds and if I want to have regular Lewis structure bonds I have to put a minus 2 on there. Maybe that'll make you uncomfortable but if you want you can just leave the bonds as data bonds and don't put a charge. So lithium will happily take four ligands on it in order to reach that octet. When you look up crystal structures of lithium-containing ions you generally see four bonds to lithium at least or the important thing is you see eight electrons around lithium. So lithium wants those electrons and we'll talk more about lithium and organolithium compounds and this desire to have extra bonds to lithium later. Let me do this over on this other board here because I think I'm going to need a little more room. I want to look at two different SN2 reactions in order to take alcohols and convert them into alkyl chloride. So let's imagine taking this asymmetric alcohol, there's a stereogenic center in here that allows us to keep track of what's going on and if you treat this with thynial chloride with SOCl2 you can replace the hydroxy group with chloride. Ultimately this kind of a reaction goes through a chlorosulfite ester. I'm not going to draw all the arrow pushing for that. I assume you've covered this at some point in your sophomore organic chemistry course. It's a lone pair on sulfur. It's not important for this reaction. Okay so you go through this chlorosulfite ester and then there's a chloride floating around that we've just displaced and I'm going to draw it right here and so when you use thynial chloride on secondary alcohols you generally get inversion of configuration. So what you get is the chloride that's got the opposite configuration because it displaces the leaving group with inversion. So this is just a simple SN2 reaction here. So you get inversion of configuration to get that final alkyl chloride. And now you can use that to make a grignard reagent or do whatever you're going to do with that. So now let me contrast this. Same starting material, same reagent. For whatever reason you decided to use dioxane for this reaction you get a different result. It turns out that in this case the dioxane actually acts as a nucleophile that displaces the chlorosulfite intermediate to give you an intermediate like this. Before the chloride can push out the chlorosulfite from the back there's already a dioxane molecule waiting right there and it displaces the chlorosulfite to give you this oxonium ion intermediate. So now when the chloride comes in finally the chloride makes its way to the other side of the molecule it's like oh I got beat out by the dioxane solvent. Now the chlorosulfite comes in and when it displaces still in SN2 reaction now you end up with an overall retention of configuration because this process involved a double displacement. You displace once in an SN2 reaction with a dioxane. You displace a second time with the chloride so the overall result is retention of configuration. Okay so these, this is an effective solvent. It's not dielectric constant. The solvent is actually participating in the reaction and you have to look out for that in a lot of different reactions. Benzene if your stuff is soluble then benzene. It depends on what your alcohol soluble is. It would be a common one. Well or nowadays toluene because benzene is toxic. Okay so why are solvent effects great? Solvent effects are the reason why chemists will not suddenly all be out of jobs, right? If it weren't for solvent effects just about every transition state and every intermediate we could draw into a computer and hit the button and do ab initio calculations on. Except for solvent effects. For solvents you have to know how the solvents are going to participate and draw them specifically in in every conceivable configuration. And nobody can do that. So electronic structure calculations. Every time you see somebody say oh I calculated this with 8 hard tree, Fox 631G star, B3 lip, LAN, LDZ. You know something like that. When you see electronic structure calculations they are virtually never taking solvent into account. And so these huge differences in reactivity that you can see with solvent factors of hundreds of thousands in rate those will all be absent. When you look at electronic structure calculations. Okay so that's the end of I guess lecture 7 slash 8. And that is the end of the material for the first exam. So when we come back on Monday we're going to start talking about displacements in addition to sigma star orbitals and that will not be on the exam that's on Wednesday. That's this is where we end for exam one material. Okay that's it. I'll see you guys. Enjoy your Friday.