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Current time:0:00Total duration:13:18

Video transcript

let's think about what type of reaction might occur if we have this molecule right over here I won't go through the trouble of naming it would take up too much time in this video but it's dissolved in methanol it's dissolved in methanol and we talked about what type of reactions we're going to pick between sn2 sn1 e 2 and E 1 reactions now that may be the first place to start or the place I like to start so just look at the solvent itself and when we're trying to decide what type of reaction will occur the important thing to think about is is the solvent protic or a protic and if you look at this solvent right here this is methanol it is protic it is parodic in case you don't remember what protic means it means that there are protons flying around in the solvent that they can kind of go loose and jump around from one molecule to another and the reason why I know that methanol is protic is because you have a ha you have hydrogen bonded to a very electronegative atom in oxygen so every now and then in one of the methanol molecules the oxygen can steal the oxygen can steal hydrogen's can steal hydrogen's electron and then the hydrogen itself without the electron the hydrogen proton will be flying around because it doesn't have a neutron so this is a protic solvent now you might say well does anything with a hydrogen would that be protic and the answer is no if you have a bunch of hydrogen's bonded to just carbons that is not protic carbon is not so electronegative that it could steal a hydrogen's electron and have the hydrogen's floating around so a big giveaway is hydrogen bonded to a very electronegative atom like oxygen so this is protic and when we think about product out of all the reactions we studied that favors that favors so this this makes us well even a better way to think about it is it dis favors so it tells us that we it's it's unlikely unlikely to have an sn2 or an e2 reaction and the logic there is an sn2 reaction needs a strong nucleophile an e2 reaction needs a strong base now if you have protons flying around the nucleophile or the base is likely to react with the proton so it wouldn't it would not be likely to react with the substrate itself so product solution you're unlikely to have an sn2 or e2 but you are likely to have what you are likely to have is an sn1 or an e1 reaction both of these need the leaving group to leave on its own and actually having protons around might help to stabilize the leaving group to some degree so it makes sn2 e2 unlikely sn1 e1 a little more likely so so far these are good candidates now the next thing to think about is to just look at the leaving group itself or see if there is and there you see if there is even a delete a leaving group and over here everything we see on this molecule is either a carbon or hydrogen except for except for this iodine right here except for this iodine right here and we know that iodide is a good leaving group so this so far well a good leaving group does not it does not make it any less likely that you'd have sn2 or e2 both of those can do well with a good leaving group but then it's a necessary requirement for an sn1 or an e1 reaction remember an sn1 and e1 and both of them the first step is that the leaving group leaves on its own that is the rate determining step so that's a requirement for sn1 or e1 so it still looks likely we haven't seen anything that would make us think that we wouldn't have an sn1 or e1 this is a good leaving group let me write here good good leaving good leaving group now the last thing we can think about right here is the carbon that we might be leaving from so far everything is pointing in the sn1 and e1 direction the kind of the final thing is when this leaving group leaves it's going to form a carbo cation from the carbon that it's bonded to right now and in order for that carbo cation to be reasonably stable at minimum it should be a secondary carbo cation bonded to at least two carbons but ideally would be bonded to three carbons via tertiary carbon now the carbon that the leaving group is bonded to the carbon that the leaving group is bonded to is a tertiary carbon it's bonded to one two three carbons so it is a tertiary carbon so it would actually it can actually be a stable carbon cations it's a tertiary carbon which it once again favors sn1 and e1 so all of the clues here tell us that sn1 and e1 are going to happen and actually they'll both happen so let's think about the mechanism so the very first step in the very first step the leaving group leaves in either one of these reactions so if we look at the iodine it already has 7 valence electrons 1 2 3 4 5 6 7 and in the first step the rate determining step of the sn1 or e1 reaction the iodine is going to nab an electron off the carbon I'll do that electron in in green right over there that's going to get nabbed onto the iodine to make iodide and then that carbon is going to lose an electron and become a carbo cation so after that very first step we have something that looks like this so that's our molecule so you can just so clear what we did this arrow I'll do all the way over here so it's clear that this is our this is our next step and what's happened here the our car our tertiary carbon has lost its electron so now this carbon right here has a positive charge it's a tertiary carbo cation and now the iodine has become iodide it has left it has left the molecule so it had its original 7 valence electrons 1 2 3 4 5 6 7 it nabbed one more electron from the carbon and now it is I want to do that in green and nab one more electron from the carbon now it is the iodine and if the iodide anion and the next so this is this step right here is common to both sn1 and e1 reaction the leaving group has to leave now after this they start to diverge in an sn1 the leaving group essentially gets substituted with a weak nucleophile in an e1 a weak base trips off one of the beta hydrogen's and forms an alkene so let's do them separately let's do them separately so over here I'm going to do the sn1 I'm going to the sn1 and on the right hand side I will do I will do be one reaction so let me start over here so the sn1 is starting over here at this step I'll just redo this step over here so this has a positive charge that has a positive charge here the iodide has left so the iodide has left I don't have to draw up its valence electrons anymore and what's going to happen next we're going to get substituted with the weak base and the weak base here is actually the methanol the weak base here is the methanol so let me draw some methanol here so let me draw some methanol it's got two unbonded pairs of electrons and one of them it's a weak base it was willing to give an electron but you know it does it has a partial negative charge over here because oxygen is electronegative but doesn't have a full negative charge so it's not a strong nucleophile but it can donate an electron to this carbon cation and that's what is going to happen it will donate an electron to this carbon cation and then after that happens after that happens it will look like this so that's our original molecule now this now this magenta electron has been donated to the carbo cation the other end of it is this blue electron right here on the oxygen it is now bonded that is our oxygen here's that other pair of electrons on that oxygen and it is bonded to a hydrogen and a methyl group and then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there oh I want to be very clear here the oxygen was neutral the methanol here is neutral it is giving away an electron to the carbo cation the carbo cation had a positive charge because it had lost it originally now it gets an electron back it becomes neutral the methanol on the other hand was neutral gives away an electron so now it becomes it now is positive so now you might have another methanol you might have another methanol molecule sitting out here someplace sitting out here someplace that might also nab the proton off of this pause Aion so this one right here it would NAB it or would bond with it would give the electron to the hydrogen proton really the hydrogen's electron it gets nabbed by the oxygen and so then that it then that becomes neutral so then the final step in the final step it'll all look like this we have that over here the methanol that it originally bonded has lost its hydrogen so it looks like this we just have the oxygen and the ch3 there ch3 it is now neutral because it gained an electron when it's when that hydrogen proton was nabbed so if you wanted to draw it it has actually those two extra electrons just like that and then if you want to draw this last methanol it's now a positive cation so it looks like this so it's o h CH 3 h and then his unbonded pair right there and now this has a positive charge so that was the sn1 reaction now the other reaction that's going to occur is the e1 reaction so II 1 reaction once again our first step open I didn't want to do that our first step looks like that so that is our first step let me get everything straight so the leaving group had left so they both in each situation the leaving group had left our iodide is up here our iodide is up here and in an e1 reaction you don't get substituted what happens is one of the beta carbons gets a hydrogen swiped off of it by a weak base now let's think about what a beta carbon is the alpha carbon is the carbo cation carbon that's right over there that's the alpha carbon beta carbons are one carbon away so this is a beta carbon this is a beta carbon and then that is a beta carbon now this carbon over here is not bonded any hydrogens it's only bonded to other carbons so that one cannot lose any hydrogen's and then we have to pick between this carbon that's bonded to three hydrogen's and this carbon that's actually bonded to two hydrogen's I didn't draw it before but it's implicitly there it's bonded to two hydrogen's now side selves rule tells us that the dominating a product is going to be produced when the carbon that has less hydrogen's loses a hydrogen so lot of this this carbon has two hydrogen's this one has three so the one that's going to lose the hydrogen to the base or more likely to lose the hydrogen to the base is the one that has two hydrogen's not three the fewer hydrogen's so our base in this case is once again the methanol acted as a nucleophile on the F and one X is a weak acted as an weak nucleophile in sn1 now it'll act as a weak base so we have methanol right over here that's a hydrogen ch3 has some electrons right over here has a partial negative charge it will give it will give one of the electrons to the hydrogen just to the hydrogen proton the hydrogen is not going to take its electron with it that electron is then going to be given to the carbo cation to the carbo cation to make it neutral and it will form a double bond so after that happens we are left with something that looks like this this was our original molecule now this carbon right here this carbon right over here in yellow it has lost this hydrogen this hydrogen back here is still there I could draw it if I like I don't need to it just makes the thing messy but it has now formed a double bond with the primary carbon it has now formed a double bond that electron that it had bonded with what the hydrogen is now being was now given to the carbo cation it has a double bond with that and then you have your methanol you have your methanol has gained has now turned into a positive ion it has now turned into a positive ion ch3 and now it has this bond it has this bond with the hydrogen I don't even make that electron that it gave away in magenta and then it has that extra lone pair that extra lone pair of electrons so in this circumstance we looked at all the clues all of the clues were against sn2 and e2 they favored sn1 e1 so if you would actually make this if you were actually try to see what happens in this reaction you would get products of both sn1 and e1 reactions these products and these products over here