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Current time:0:00Total duration:8:53

Video transcript

let's try to come up with a reaction of when we have this molecule right here reacting with sodium sodium methoxide in a methanol solution or with methanol as the solvent with methanol as the solvent just so we get a little practice of naming let's see this is 1 2 3 4 carbons so it has Butte as a prefix no double bonds or triple bonds so it is Butte Ain and we have a choral group here so if we start numbering at the side closest to it 1 2 so it's 2 chloro this is 2-chloro 2-chloro butane so let's think about let's think about what might happen here so the first thing let me just let me just redraw the molecule right there the first thing you need to realize is this sodium this sodium methoxide it is a salt it is in when it's not dissolved it's made up of a positive sodium cation let me draw them right here a positive sodium cation a positive sodium cation and a negative methoxide anion so let me draw the methoxide part right here so a negative methoxide anion and then it has so normally we'd have just two pairs but now we have three pairs here the oxygen has an Electra electron the oxygen has an extra electron actually let me draw another electron as the extra electron this will be useful for our mechanism so it could be any of them but the oxygen has an extra electron it has a negative charge in solid form when they're not dissolved they form an ionic bond and they form a crystal like structure it's a salt but when you dissolve it in something in a solution in this case we have methanol as the solution they will dissociate from each other and what you have right here is this methoxide right here this is a very very very strong base so this right here this right here is a strong this is a strong base and if you use the Lewis definition of a base that means it really really really wants to give away this nucleus to something I'm sorry this electron to something else if you use the bronsted-lowry definition this means that it really really really wants to take a proton off of something else and in this situation that is exactly what it will do and I'll just give you one I'll actually give you the the the most likely reaction to occur here and we'll talk about other reactions and why this is the most likely reaction in future videos so it wants to nab a proton it is a strong base it wants to give away this electron so let's say that it gives away this electron it gives away to this electron to this hydrogen right over here now this hydrogen already had an electron if it's getting an electron from the methoxide from the methoxide anion the methoxide base then it can give away its electron to the rest of the molecule so now it can give away this electron to the rest of the molecule now carbon won't need the electron carbon doesn't want to have a negative charge maybe simultaneously that electron goes to that carbon right over there but once again this carbon doesn't want it it only has four bonds but what we see is we had this chloro here this is a highly electronegative I guess you can call it a group right now the chlorine is very electronegative so the whole time the chlorine was already tugging on this electron right here so now all of a sudden this is all happening simultaneously when an electron becomes available to this carbon then now this chlorine this carbon doesn't need this electron anymore the chlorine already wanted it so now the chlorine the chlorine can take the electron and just like that in exactly one step let's think about what happened if all of this this whole chain reaction and I shouldn't call it a chain reaction the simultaneous reaction occurred what are we left with what are we left with so let me let me redraw my 2-chloro butane but lamb we have to change it now so now the chlorine this has disappeared so let me clear it that's disappeared the chlorine has now left it is now left this chlorine is now up here it is left it had this electron right over there which is right over there and then the other electron it was paired with that it was forming a bond is now also on the chlorine is now also on the chlorine so now it becomes a chloride anion so now it is a chloride anion if I want to draw the rest of let me draw the rest of the valence electrons 1 2 3 4 5 6 and it had the seventh one right here 1 2 3 4 5 6 7 8 now so it now has a negative charge now this electron up here let me be now let me clear let me clear this part out as well so let me clear that out as well now this electron right here this magenta electron this is now given to this carbon so let me draw it here so that magenta electron is now given to that carbon and if we look at the other end the one that was paired with those bond with it still is bonded with it so that is that green electron is now still on that carbon and now they are bonded and now they're forming a double bond so now they are forming a double bond this all of a sudden has become an alkene and now the methoxide took the hydrogen the methoxide took the hydrogen so let me redraw the methoxide so Oh ch3 the oxygen had 1 2 3 4 5 it had 6 ll actually see it had 8 valence electrons electrons one with the carbon and then all of these six unpaired ones neutral oxygen has six but now it gave one of them away to a hydrogen it gave that green electron over there to the hydrogen it grieved that green electron over there did this hydrogen let me do it I'll do this I'll make this hydrogen the same color to this hydrogen right over here so now this hydrogen is now bonded with it this bond hydrogen is now bonded with it so what are we left with we have a chloride anion we have a chloride anion so this is chloride chloride anion we now have methanol this was a strong base now it has become it's its conjugate acid so now we have methanol methanol which is the same as the solvent same solution so it kind of just or actually same as a solvent so it's now mixed in and now we're left all off with one two three four still four still we still have four we still have four carbons but now it's a it's a it's an alkene we have a double bond so we could call this butte 2e or sometimes called 2-butene so let's think about what happened here it all happened simultaneously both reactants were present there was actually only one step so this is the rate determining step so if we were to try to name it would probably have a two in at some place and something got eliminated the chloride the chloride or I guess you could call it the Coulomb the chloro group got eliminated it left it was a leaving group in this situation so this this was eliminated this was eliminated it eliminated and this type of reaction where something is eliminated and both of the reactants are participating in the rate determining step and we only had one step here so that was the rate determining step is called an e2 reaction e2 it is called an e2 reaction and once again he stands for elimination elimination and the two stands for that both both reactants involved in the rate determining step we had one reactant and two reactants they were both involved in the rate determining step II - just like sn2 I mean it's obviously not the same reaction but sn2 had substitution we were leaving group left nucleophile came it was a substitution using a nucleophile so that's the s and the N and then the two was we had both reactants it was all happening simultaneously I'll leave you there we're going to go into more detail on the different types of e - reactions which ones are favored and then we'll talk a little bit about e1 and you could maybe guess what that's going to be based on our experience with sn2 and than one