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Current time:0:00Total duration:9:22

Video transcript

let's think about what will happen if we have this molecule ELISA name it we have one two three four five carbons no double bond so five tells us pent it's pentane and it has two groups on the number three carbon one two three doesn't matter which side we start counting from we have a bromo group and we have an ethyl group two carbons right there so on the three carbon so we have three bromo three bromo three ethyl three ethyl pentane right here so we have three bromo three ethyl pentane dissolved in a solvent of in a solvent and this right here it's an alcohol and it has two carbons right there two carbons so meth so it is ethanol so this right there is ethanol so let's think about what might happen if we have three bromo three ethyl pentane dissolved in some ethanol now ethanol already has a hydrogen it's not super eager to get another proton although it does have a partial negative charge it is polar oxygen is very electronegative so it has a partial negative charge so maybe it might be willing to take on another proton but it doesn't want to do so very badly so it's actually a weak base so ethanol right here is a weak base so it's not strong enough to just go nabbing hydrogen's off of carbons like we saw in an e2 reaction so it's just going to sit passively here and maybe wait for something to happen now what might happen well we have this bromo group right here we have this bromine and the bromide anion is actually a pretty good leaving group it's a fairly large molecule it's able to keep that charge because it's spread out over a large electron cloud and it's connected to a tertiary carbon right this carbon right here this carbon right here is connected to one two three carbon so if it were to lose its electron if it were to lose that electron right there it would be might not like to do it but it would be reasonably stable it's within the realm of possibilities it could occur so maybe in this first step since bromine is a good leaving group and this carbon can be stable as a carbo cation maybe and bro the bromine with all is already more electronegative so it was already hogging this electron maybe it takes it all together so let me draw so neutral bromine has one two three four five six seven valence electrons maybe it swipes this electron from the carbon and now it'll have eight valence electrons and become bromide so what happens now and of course the ethanol did nothing it's a weak base wasn't strong enough to react with this just yet so what is happening now so this is going to be the slow reaction so right here what I said you know this isn't going to happen super fast but it could happen this is actually the rate determining step rate determining rate determining not determining determining this is the rate determining step so what happens after that so let me just paste everything again so this is our set up to begin with but now that this little reaction occurred what will it look like well the bromine has left so let me clear that out let me clear out the bromine edit clear and actually took an electron with it so it's bromide so let me draw it like this so I'll do it in blue so this is the bromine so the bromine is right over here it had one two three four five six seven valence electrons it swiped this magenta electron from the carbon now it has eight valence electrons it has a negative charge the carbon lost an electron so it has a it has a positive charge and it's you know somewhat stable because it's a tertiary carbo cation so now let's think about what's happening and I want to point out one thing in this first step of a reaction only one of the reactants was involved this rate determining the slow step of reaction this doesn't occur nothing else will but now that this does occur everything else will happen quickly but in our rate determining step we only had one of the reactants involved so it's analogous to the sn1 reaction but we're going to see here is that we're actually eliminating and we're going to call this an e1 reaction we're going to see that in a second actually the elimination has already occurred the bromide has already left so hopefully you see why this is called an e1 reaction it's elimination efore elimination and the rate determining step only involves one of the reactants right here it didn't involve it didn't involve in this case the weak base but now that the broom let's now that the bromide has left let's think about whether this weak base this ethanol can actually do anything it does have a partial negative charge over here it does have a partial negative charge and on these ends has partial partial positive charges so it is somewhat attracted to hydrogen or to protons I should say two positive charges but you know not so much that it can swipe it off of things that don't that aren't reasonably acidic now that this guy is a carbo-cation this entire molecule actually now becomes pretty acidic which means it wants to give away protons or I guess another way you could view it is it wants to tog it wants to take electrons depending on what you want whether you want to use the bronsted-lowry definition of acid or the Lewis definition either way it wants to give away this well it wants to give away a proton it could be this one it could be that one it has excess positive charge it wants to get rid of its excess positive charge so it's very it's reasonably acidic enough so that it can react with this weak base and what you have now is the situation what you could have now is the situation where on this partial negative charge of this oxygen let's say let me pick a nice color here let's say this purple electron right here it is it can be donated or it will it will swipe the hydrogen proton then hydrogen's electron hydrogen's electron will be taken by the larger molecule in fact it'll be attracted to the carbo cation so it will go to the carbo cation just like that now in that situation what occurs well what's our final product so let me draw it here so this this part of the reaction is going to happen fast the rate determining step happens slow the leaving group had to leave the carbo cation had to form that's not going to happen superfast once that forms it's not that stable and then this thing will happen this is fast so let me just paste everything actually let me just read let me just paste everything again aid the drawing so now so now we already had the bromide it had left the bromide had left and now the hydrogen is gone let me clear this the hydrogen from this carbon from that carbon right there is gone now this this electron that electron is still on is still on this carbon but the electron that was with this hydrogen is now on the carb Oh what was the carbo cation so that electron right here is now over here and now this bond this bond right over here is this bond we formed an alkene and now the what was an ethanol now a hydrogen proton and now becomes a positive cation so let me draw that so this right here this electron this electron ends up being given so it's no longer with the ethanol it's no longer with the ethanol so it's not no longer with the ethanol it gets given to this hydrogen it gets given to this hydrogen right here that hydrogen right there and now they have formed a new bond and since it since this oxygen gave away an electron it now has a positive charge it now has a positive charge and all along the bromide anion had left in the previous step so the bromide anion is floating around with its eight valence electrons one two three four five six seven and then it has this one right over here that makes it negative and then our reaction is done we had a weak base we had a weak base and a good leaving group a tertiary carbon and the leaving group left we only had one of the reactants involved it was eliminated and then once it was eliminated then the weak base was able to take was then able to take a hydrogen off of this molecule and that allowed this molecule to become an alkene formed a double bond this is called an E and I already told you this is an e1 reaction efore elimination in this case of the halide one because the rate determining step only in involved one of the molecules it did not involve the weak base we'll talk more about this and especially different circumstances where you might have well one the different types of you one reactions you could you could see what you know which hydrogen is going to be picked off and all of the things like that and we're going to see when II 1 e2 sn1 or sn2 what what's kind of the environment or the reactants that need to be there for each one of those to occur in different circumstances