the dehydration of alcohols to create alkenes is an elimination reaction the mechanism is going to depend on whether or not you start with a primary alcohol or a secondary or tertiary alcohol so before you decide which one you're going to use let's make sure we're all on the same page about what those words mean this molecule the oh is connected to this carbon the question is how many other carbons is it connected to in this case it's only one other carbon so it's a primary alcohol here the oh is connected to that carbon it's connected to one two other carbons that's secondary o h is connected to this carbon it's connected to one two three other carbons that's tertiary and here the oh is connected to that carbon which is only connected to one other carbon and so it's primary these two that one and that one will react via an E2 mechanism the other two will react via E1 and we'll get to that now an E2 mechanism the two represents bimolecular that means there's two things colliding in the rate limiting step the rate limiting step I believe is the second one here so let's just talk about what's happening here first the very first step of this is for um some acid I'm going to use h2so4 here and I'm going to draw it in a weird way I just want to emphasize that the H isn't connected to anything other than the other oxygen here and the lone pair on the oh will take that hydrogen away and leave a minus charge on the anion the products we get out of that are the same long carbon chain we start with but now instead of it being an oh group we have what's called an alkoxonium alkyl oxonium ion it's basically a carbon chain where the oh has been converted to an H2O and there's still a covalent bond between the oxygen and the carbon but because water is such a great leaving group it's it's a severely weakened Bond and it's very open to being broken though if one lone pair on the oxygen and a formal charge of plus one that's countered by the extra minus charge left on whatever the conjugate base of the acid you used was now the question is if we're going to eliminate this part of the molecule we need to eliminate a hydrogen off of the next carbon in order to leave behind the double bond I'm going to draw in those hydrogens here that next carbon here has one two extra hydrogens on it now technically this one does as well but I didn't need to show those what's going to happen here is the conjugate base of the acid which again the acid gave away its H to make the o h into oh2 is going to steal one of the hydrogens on the carbon next to the carbon that has the oh on it it's stealing the H and those electrons have they need to go somewhere they're attracted to this Bond because the electronegative oxygen is sucking electron density in this direction so those electrons end up going into or between those two carbon atoms but each carbon can't have more than four bonds so this covalent bond breaks and those electrons are not absorbed but they're taken by the oxygen because oxygen is electronegative you have the flow of electrons heading towards the electronegative atom you are breaking that covalent bond between carbon and hydrogen you're breaking this covalent bond between oxygen and carbon but then you're forming a double bond between the two carbons that are there in the end our products here I still have the long carbon chain oh am I missing one more there we go except now I have a double bond at the end the hso4 minus took its extra H so it's now h2so4 which is actually how it started so it's a catalyst and you have H2O as a product ah look dehydration because you're taking a water out of the molecule total now granted this hydrogen came from the acid and uh the other the new hydrogen that's attached to this hso4 it came from the carbon chain but hydrogens are not like you can't tell the difference between those unless you make one some kind of weird isotope and that's probably how we figured out what the mechanism was but the point is that you're protonating the oh to make it an oh2 group super easy to leave as long as you pluck a hydrogen off of the next carbon in the chain electrons flow around to create that double bond it's beautiful please note that in what I believe is the rate determining step which is this one to that one you have two molecules colliding that makes it a bi-molecular reaction and that's why it's called E2 I want to talk about E1 I bet you oh that's not it's not very sheet you want to talk about E1 I bet you do let's go so for E1 the one here represents unimolecular it is a single thing that is doing its magic in the rate limiting step okay so what actually ends up happening here we're still going to have to protonate the oh I'm going to give you h2so4 again I don't know I'm just picking an acid the lone pair on the water still steals the H you still end up with the alkyl oxonium ion here I've got oh2 I'm just going to be a little uh there we go we got our plus charge on the oxygen there and of course we have the hso4 minus that's left over okay but that's easy it's a strong acid gives away its H piece of cake right now for secondary or tertiary alcohols what happens is that this oh2 actually falls off of the molecule entirely the only electron flow Happening Here is going to be electrons flowing out of the covalent bond between carbon and oxygen straight to the water this is a una molecular decomposition because it's one molecule breaking apart into two what you end up with is that same carbon chain but now you've broken the water molecule off and you're left with a positive charge on this carbon the carbon that the oh fell off of this is why or rather this carbocation intermediate is the reason this mechanism only occurs for secondary interchury alcohols a primary carbocation a plus charge on a carbon that isn't connected to more than two other carbons is not very stable at all you need that carbon that has the positive charge to be connected to at least two other carbons in order for it to be stable enough to operate as this intermediate great now I want to point out that you have two hydrogens connected to this carbon and two hydrogens connected to this carbon which hydrogen gets plucked off of the molecule will depend on a few different factors like if there's blocking groups or steric hindrance etc etc but if you're uh for a molecule this simple where it's just two H is over here and two H is over here you can probably get away with grabbing either one in fact let's show that happening to both now the H2O or maybe we'll use the hso4 here something needs to steal this hydrogen away I think I'm going to choose the water to make that happen um the water molecule is going to steal let's say that H away where are the electrons from this covalent bond going to go well they're going to flow towards the positive charge because electrons are negatively charged and negatives and positives subtract each other right if you pluck a hydrogen off of this carbon here then your resulting molecule will have a double bond between those two carbons Now is it going to end up being CIS or trans as far as I recall trans is preferred as the product it's not uh specific in that it's a hundred percent trans but trans is the major product as opposed to CIS okay you could just as easily have had the water pluck off one of these hydrogens in that case you would have ended up with your double bond between these two carbons the carbon that had the positive charge and the carbon you pluck the hydrogen off of and again I'm going to draw that as trans now this because you have hydrogens on both of the carbons on either side of the carbocation I could imagine both of these products getting made and realistically I could also Imagine I just want to make sure I draw this correctly I could also Imagine a CIS form also being there but as a minor product now I got to draw this one so I'll draw the double bond there one two something like that that's the same number of carbons and this is also a minor product um the real summary here though is that secondary and tertiary alcohols dehydrate via a carbocation intermediate it's that the formation of the carbocation that makes this an E1 mechanism and the fact that that isn't stable for primary theoretical carbocations is the reason that this is E1 and this one's E2 I don't know if that last sentence made sense but you've already watched 11 minutes of this video so I'm sure you're on board thanks for being with me and best of luck