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Current time:0:00Total duration:12:39

Video transcript

let's look at the mechanism for an e1 elimination reaction and we'll start with our substrates so on the Left let's say we're dealing with an alkyl halide so the carbon that's bonded to our halogen would be the alpha carbon and the carbon next to that carbon would be the beta carbon so we need a beta hydrogen for this reaction the first step of an e1 elimination mechanism is loss of our leaving group so loss of leaving group let me just write that in here really quickly and in this case the electrons would come off onto our leaving group in the first step of the mechanism so we're taking a bond away from this carbon the one that I've circled in red here so that carbon is going from being sp3 hybridized to being sp2 hybridized so now we have a carbo cation and we know that carbo cations sp2 hybridized carbons have planar geometry around them so I've attempted to show the planar geometry around this carbo cation so that's the first step loss of the leaving group to form a carbo cation in the second step our base comes along and takes this proton which leaves these electrons behind and those electrons move in to form our alkene so this is the second step of the mechanism which is the base takes or abstract the proton so base takes a proton to form our alkene and let me go ahead and highlight those electrons so these electrons here and magenta moved in to form our double bond and we form our product we form our alkene so the first step of the mechanism the loss of the leaving group this turns out to be the rate determining step so this is the slow step of the mechanism so if you're writing a rate law the rate of this reaction would be equal to the rate constant K times the concentration of your substrate so that's what that's what studies have shown that these mechanisms depend on the concentration of only your substrate this over here on the left so the first order with respect to the substrate and that's because of this determining step at the loss of the leaving groove is the rate determining step and so the concentration of your substrate your starting material that's what matters your base can't do anything until you lose your leaving group and so since the base does not participate in the rate determining step it participates in the second step the concentration of the base has no effect on the rate of the reaction so it's the concentration of the substrate only and since it's only dependent on the concentration of the substrate that's where the one comes from an e-1 I'm go ahead and write this out here so in e1 mechanism the one it comes to the fact this is aut molecular a uni molecular rate law here and E comes from the fact that this is an elimination reaction so when you see II 1 that's what you're thinking about it's an it's an elimination reaction as uni molecular the overall rate of reaction only depends on the concentration of your substrate so if you increase let's say you have to say this was your substrate right here and you increase the concentration of your substrate let me just write this down if you increase the concentration of your substrate by a factor of 2 you would also increase the rate of reaction by a factor of 2 so it's first order with respect to the substrate so this is some general chemistry here if you increase the concentration of your base by a factor of 2 you would have no effect on the overall rate of the reaction so let's talk about one more point here in the mechanism and that is the formation of this carbo cation since we have a carbo cation in this mechanism you need to think about the possibility of rearrangements in the mechanism and you need to think what would form what substrate would form a stable carbo cation so something like a tertiary substrate forming a tertiary carbo cation would be favorable for an e1 mechanism here we have a tertiary alkyl halide and let's say this tertiary alkyl halide undergoes an e1 elimination reaction so the carbonate spawns the iodine must be our alpha carbon and then we would have 3 beta carbons so that's the beta carbon that's the beta carbon and that's a beta carbon so the first step in an e1 mechanism is loss of our leaving group so if I draw on the lone pairs of electrons in here on iodine I know that these electrons and this bond would come off onto ion to form the iodide anion so let me draw that in here so we would make the iodide anion and let me highlight our electrons so the electrons electrons in this bond come off on to the iodine to form the iodide anion and this is an excellent leaving group iodide is an excellent leaving group and you know that by looking at PKA values the iodide anion is the conjugate base of a very strong acid H I with a peak approximate PKA value of negative 11 so H is very good at donating a proton which must mean the conjugate base is very stable so the i'side anion is an excellent leaving group so if we lose the iodide anion that means we're going to have a carbo cation so we lost a bond to this carbon in red so we're going to form a carbo cation let me go ahead and draw that in so this is a planar carbo cation and so the carbons let me go ahead and highlight it here the carbon in red has a plus 1 formal charge it lost a bond so that's the first step of an e1 elimination mechanism the second step of an e1 elimination mechanism is the base comes along and it takes a proton from a beta carbon so our base in this case would be ethanol let me go ahead and draw in lone pairs of electrons on the oxygen so notice we're also heating this reaction so the ethanol is going to function as a base so ethanol is not a strong base but it can take a proton so let me go ahead and draw in approach on right here and a lone pair of electrons on the oxygen is going to take this proton and the electrons would move into here to form our alkene so let me go ahead and draw our product let me put that in here and let me highlight some electrons so the electrons in bluh moved in here to form our double bond so a couple of points about this reaction one point is when you're looking at sn1 mechanisms the first step is loss of a leaving group to form your carbo cation so when you get to this carbo cation you might think well why is that all acting as a base here why couldn't it act as a nucleophile and the answer is the ethanol certainly can act as a nucleophile and it would attack the positively charged carbon and you would definitely get a substitution product for this reaction as well so if ethanol acts as a nucleophile you're going to get you're going to get a substitution reaction an sn1 mechanism if the ethanol acts as a base you're going to get an e1 elimination mechanism so here we're just going to focus on the elimination products we won't worry about the substitution products but we'll talk about this stuff in a later video because that would definitely happen all right something else I want to talk about is we had three beta carbons over here if I looked at these 3 beta carbons and I just I just picked one of them right I just said that this carbon right here let me highlight it I just took a proton from this carbon but it doesn't matter which which of those carbons that we take a proton off off because of symmetry and then go ahead and draw this in over here so this is my carbo cation let's say let's say we took a proton from this carbon so our weak base comes along and takes a proton from here and these electrons would move into here that would give us that would give us the same product alright so this would be let me go and highlight those electrons so these electrons in dark blue would move in to form our double bond but this is the same as that product alcohols can also react via an e1 mechanism the carbon that's bonded to the O H would be the alpha carbon and the carbon next to that would be the beta carbon so reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene and sometimes phosphoric acid is used instead of sulfuric acid so we saw the first step of an e1 mechanism was loss of a leaving group but if that happens here if these electrons come off onto the oxygen that would form hydroxides as your leaving group and the hydroxide anion is a poor leaving group and we know that by looking at PKA values down here is the hydroxide anion it is the conjugate base to water the water is not a great asset and we know that from the pKa value here so water is not great at donating a proton which means that the hydroxide anion is not that stable and since the hydroxide anion is not that stable it's not a great leaving group so let's go ahead and take off this arrow here because the first step is not loss of a leaving group the first step is a proton transfer we have a strong acid here so if uric acid and the alcohol will act as a base and take a proton from sulfuric acid now its form water as your leaving group and water is a much better leaving group than hydrogen the hydroxide anion and again we know that by pKa values water is the conjugate base to the hydronium ion h3o plus which is much better at donating a proton the pKa value is much much lower and that means that water is stable so the first step the first step when you are doing an e1 mechanism with an alcohol is to protonate the o H group so here's our alcohol and the carbon bonded to the O H is our alpha carbon and then these carbons next to the alpha carbon would all be beta carbons we just saw the first step is a proton transfer a lone pair of electrons on the oxygen take a proton from sulfuric acid so we transfer a proton and let's go ahead and draw in what we would have now so there'd be a plus 1 formal charge on the oxygen so let's highlight our electrons in magenta these electrons took this proton to form this bond and now we have water as a leaving group let me just fix this hydrogen here really fast and these electrons can come off onto our oxygen so that gives us water as our leaving group and let me go ahead and draw in the water molecule here and let me highlight electrons the electrons in light blue and this bond came off on to the oxygen which armed water and we know water is a good leaving group we took a bond away from this carbon in red so that carbon would now be a carbo-cation so let me draw in our carbo cation here so the carbon in red is now positively charged so let me draw in a +1 formal charge on that carbon the next step of our mechanism we know a weak base comes along and takes a proton one of the protons on one of the beta carbons over here so let's just say it's this one and I'm just going to draw in a generic base so now a generic base right here which is going to take this proton and then these electrons are going to move in to form our product so let me draw our product in here and let me highlight those electrons so the electrons in let's use green this time the electrons in green moved in here to form our double bonds to form our out our alkene so I just put a generic base let me go ahead and talk about the base for a second here I just put in a generic base sometimes you might see water acting as a base sometimes you might see hso4 - right the conjugate base - sulfuric acid acting as a base different textbooks give you different things I don't think it really matters but one of those acting is a weak base it's probably water takes this proton to form your alkene sometimes this reaction is called a dehydration reaction since we lost water in the process