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Current time:0:00Total duration:12:26

Video transcript

what I want to do in this video is to try to figure out what type of reaction or reactions might occur if we have what is this is one two three four five it's in a cycle this is bromo cyclopentane if we have some bromo cyclo cyclopentane dissolved in our solvent is dimethyl formamide sometimes you'll see that just written as DMF and I've actually drawn the formula for it here so we can think about what type of a solvent it is and also in our solution we have the methoxide ion so we also have the methoxide ion right here so let's think about what type of reaction might occur and just to narrow things down we'll think about it in the context of the last four types of reactions we've looked at so this might be an sn2 reaction an sn1 reaction and e2 reaction or an e1 reaction we're going to look at all the clues and figure out what's likely to occur and then actually draw the mechanism for it occurring now the first thing that since they gave us the solvent and other things that are in the solvent let's think about how those might affect the reaction so if we look at this solvent right here and whenever you look at any of these reactions when you look at the solvent you just want to think about is it protic or not and protic means it has hydrogen's that can kind of be released or that have the electrons could be nabbed off and these protons could just float around and if we look over here we do have hydrogen's but all of the hydrogen's are bonded to carbon and carbon is unlikely to just steal a hydrogen's electrons and let the hydrogen float around carbon is not that electronegative if you had hydrogen's bonded to an oxygen that'd be a different question then you would be you would have a protic solvent but in this case all the hydrogen's bonded to carbons not likely to get their electrons nabbed off and float around as free protons so this is an a protic solvent this is an a a protic solvent now we've gone over this a little bit with sn2 and sn1 but the same idea applies in order to have an sn2 or an e2 in order to have an sn2 or an e2 reaction you have to have either a strong nucleophile or a strong base and the same thing could actually be both although not they're not always correlated we've seen that before now if you had a protic solvent it would stabilize the strong base or the strong nucleophile the protons would react with them they would take the electrons from that strong base or that strong nucleophile so in order to have an sn2 and e2 you have to have no protons flying around so you need an a protic solvent so this a protic solvent will favour sn2 or or an e2 reaction now so our mind is already thinking an sn2 or e2 let's think about what it what what let's think about the reactants themselves so over here we have the methoxide ion we have the methoxide ion and let's think about whether it's a strong or weak well think about first is a strong or weak nucleophile what's actually a pretty strong nucleophile it is a strong nucleophile strong nucleophile so that would go put us in the direction of an sn1 so we have two data points I'm sorry for an sn2 we have two data points for sn2 because remember it has to just kind of go in there and be active it's not too big of a molecule so it's not going to be hindered but it's also an extremely strong base even stronger than hydroxide so it's also an extremely extremely extremely strong base which might lead us or that does imply that we're going to have an e2 reaction now the last thing we need to think about is the carbon where the leaving group might leave from and immediately when you look at the bromo cyclopentane there's only one functional group attached to the chain and that is the bromo group right here right there it is attached to this carbon we could call that the alpha carbon and it is a secondary carbon this carbon right here is bonded to one two other carbons so that actually so this is this alpha carbon let me write it this way this alpha carbon carbon is a secondary secondary carbon and that kind of makes it neutral in this mix if it was a methyl or primary carbon it would favor it would favor sn2 actually I mean methyl don't think you could have what is an sn2 and if it was a tertiary carbon it would favor sn1 or e1 because it would it would favor a stable carbo cation the leaving group could just leave and if this guy was bond to another carbon v-very it would be very stable but in this situation it's a secondary carbon bonded to two carbons it's a little bit neutral it doesn't it doesn't any of these reactions might occur but when we look at all of the other data points they're pointing at both sn2 or e2 if strong nucleophile / base we have an a protic solvent it's going to be sn2 or an e2 reaction so let's actually draw the reaction so let me do the the sn2 first so let me do it in orange so if we were to have an sn2 reaction sn2 let me redraw the molecule let me draw the cyclopentane part I want to make sure let me draw it the same way I had it drawn up there so the Pentagon is facing upwards and then we have our bromo group right there so we have our methoxide ion we have our methoxide ion right over here so ch3 OH - or another way we could view it is that this oxygen has one two three four five six seven valence electrons with a negative charge one of these electrons right over here this can attack it can attack the substrate right over there that carbon right when that happens simultaneously simultaneously this bromine is going to be able to nab is going to be able to nab an electron from that same carbon and then we are going to be left with the bromine now becomes the bromide anion it now has you know it had 1 2 3 4 5 6 7 valence electrons 1 2 3 4 5 6 7 now it nabbed one more electron making it bromide now it has a negative charge and if we were to draw if we were to draw the Chane it would look like this if we were to draw the chain well we could draw it on this side I might as well draw it on this side just so it's attacking from the other side this isn't the chiral substrate so we don't have to be too particular about how we draw the connections to the carbon we're not actually even showing anything popping in or out but we would have the methoxide ion or now it's weed now it's bonded so it's no longer an ion so it's o ch3 ch3 just like that has bonded to this carbon obviously implicitly this carbon had another hydrogen that we are not showing so that just that quickly that was the sn2 reaction that is the mechanism now let's think about what the e2 reaction to do the e2 properly to give it justice we're gonna have to draw some of the hydrogen's so on the e2 reaction let me draw that in blue the e2 reaction let me draw the cyclopentane part let me draw it big actually over here it's less important to draw it too big so let me draw the Pentagon the Pentagon just like that that is the bromine three four five six and it has a seventh valence electron right over here this is the alpha carbon that right there is the alpha carbon and then there are two beta carbons there are two beta carbons right over there and there they each have two hydrogen's on them they each have two hydrogen's I know it's becoming a little hard to read they each have two hydrogen's on them and in an e2 reaction the strong base will react let me make it a little cleaner than that let me get rid of the beta the beta makes it a little dirty okay so they each have two hydrogen's on them they each have two hydrogen's on them now in an e2 reaction the strong base over here the methoxide ion was acting as a strong nucleophile and e2 it's going to act as a strong base it's going to nab off a hydrogen off of one of the beta carbons and you might wanna say okay which one let's look at zaitsev's rule it doesn't matter these are symmetric they they're both bonded to two other carbons they both are bonded to the same number of hydrogen's it doesn't matter it's actually going to be random which one and you actually won't be able to tell the difference because it's symmetric so let's just draw it like this let me draw the methoxide ion one two three four or anionic maybe I should say five six and then it has one bond to the ch3 it has a negative charge very very very strong base it can go over here and nab the hydrogen and leave hydrogen's electron behind so it can maybe I'll take a color this electron can be given to the hydrogen so that it forms a bond with it hydrogen's electron let me do this in a suitably different color hydrogen's electron that is sitting right over there can now be given to the alpha carbon it can now be given to the alpha carbon to form a double bond and now that the Alpha carbon is getting that electron now the bromo group can leave it's a decent leaving group and that was another thing that we should think about in our equation but a good leaving group actually favors all of the reactions sn2 e2 sn1 e1 and so they're the carbons getting the electron and then the bromine the bromine can then take this carbons electron and just in one step that's what's distinctive about the e2 and the sn2 reactions all of the reactions are involved in the rate determining step and there's really is only one step just like that after that happens what we're left with is the methoxide anion takes the hydrogen so it becomes methanol let me draw that so it becomes methanol so it had one two three four and then five that's this one right there but then this guy goes in bonds with the hydrogen this guy goes and bonds with the hydrogen just like that and hydrogen leaves its electron behind though and let me let me draw the the cyclopentane part now and so the cycle of pentane looked like this before if I just focus on the ring now this guy was bonded to a hydrogen he was bonded to he was bonded to this hydrogen over here but now that electron is gone to be used to form a bond with this alpha carbon right over here let me draw the alpha carbon so the alpha carbon is right over there obviously implicitly at every one of these edges we have a carbon but now a double bond is going to form with that alpha carbon we could just draw it like that a double bond obviously there's another carbon here I could write up another carbon over there and now this double bond will form and now the bromide has left it's taken an electron with it from that carbon now that the carbon doesn't need it it was already starting to hog it because it's so electronegative so that's bromine takes that orange electron takes that orange electron now it is bromide and we're done and so just to go back to the original question here which reaction is likely to occur which mechanism it's actually both sn2 and e2 you would see a mix of both of these occurring because you have all of the environmental factors that would enable both and so you would have both of these mechanisms here's the let me separate them out here is the sn2 reaction you would have the sn2 reaction occurring in your whatever your vial or your pot or whatever your you're making all this stuff occur in and you would also have you would also have your e2 reaction so you would see some of all of these some of all of those products and these products right over there