Dehydration reactions eliminate alcohols, yielding double bonds.
Concept: General features of acid-catalyzed dehydration.7m
Now we're going to talk about a way that we can make double bonds out of alcohols and this is through a mechanism called acid catalyzed dehydration. So how does this work? Remember back to when I talked about leaving groups in the past and leaving group is even a concept from acid/base chemistry. All it means is that it's something that really doesn't want to leave, something that really doesn't want to get a negative charge. Another word for leaving group is just conjugate base. So if you maybe don't really remember exactly what a leaving group is, just think conjugate base, back to acids and bases.
Basically, alcohols are really bad leaving groups or they're really bad conjugate bases. They hate to become OH-. Why? Because OH- is actually a really strong base. Remember that you always want to go from stronger to weaker? If we're making OH- hydroxide, that's a really strong base. So this is not going to be very favored to just leave a molecule.
But it turns out that there is one thing we can do to make alcohol a better leaving group and that is to use acid. If we can use some kind of acid, we can actually convert alcohol into an awesome leaving group. That awesome leaving group would just be that we add an H to it, we protonate it, so it turns into water. And water is an awesome leaving group because it's neutral. It loves to leave and it loves to just be by itself in solution.
So here's the general formula. I'm not going to show you the mechanism just yet. But basically what we have is we have some kind of alcohol and we have some kind of acid over water. Now, in this case, I just put the general HA to mean really any acid, but your common acids are going to be – sorry abut that, any of the strong acids, so H2SO4 is seen all the time Any of the hydrogen halide strong acids. I've also seen phosphoric acid. This is not really a strong acid, but it's still strong enough to make the reaction go. These are all very frequently used acids for acid catalyzed dehydration.
What we're basically doing is we're taking an alcohol. We're going to be removing two sigma bonds. We're going to be removing the alcohol and a beta-hydrogen so this is my alpha, this is my beta-carbon, alpha and beta relative to the alcohol. We're going to be taking away two different sigma bonds and we're going to be making one pi bond instead. Now that general reaction of taking away two sigmas and making one pi is an elimination. So this is actually going to be an elimination reaction. Hope that's making sense so far.
Now there's another reaction that you may have already learned or that you will learn soon and that's actually the opposite. It's called acid catalyzed hydration. Acid catalyzed hydration is a reaction where we go from a double bond and we go back to the alcohol. This is actually what's called an addition reaction. If you remember back to the general types of reactions we talked about, that when you take one bond and you make it into two, that's actually addition. When you take two bonds and you make it into one, that's elimination. So hopefully, that makes sense.
So basically, the addition part is going to be the hydration. The elimination part is the dehydration. How do you know which one it is? How do you know if it's going to be a hydration or a dehydration? You just look at what you're starting with. In this case, since I have my alcohol, I know that I'm starting with an alcohol and I'm going to try to eliminate it with an acid to become a double bond. However, if I was starting with a double bond and I used acid, then I could add water to it and that would become a hydration.
In this step, in this video, we're just going to learn about dehydration, but I'm just asking you to keep this in the back of your mind later on for when you have to do practice problems with hydration that this is the way the way that we tell the two reactions apart.
So let's just talk more in depth about dehydration. There's a few facts I want you to know. First of all, the more R groups on that alcohol, the easier it's going to be to dehydrate. This is just a fact that might come up on maybe a conceptual part of your exam or your professor might even ask you – give you four different alcohols, which one's the easiest to dehydrate. Tertiary would be the easiest, secondary, primary and then – is the worst and then – oops, sorry. I don't know what I'm trying to draw here. But basically, that methyl can't even happen because if it's a methyl alcohol or methanol that can't even eliminate because it's only got one carbon. So basically, the easiest one is tertiary, the worst one is primary. That's the first thing.
The second thing is that the specific elimination mechanism that we use is going to depend on how easily the molecule's going to form a carbocation. And the understanding of carbocations is kind of essential to these two mechanisms.
All that means is that if you guys remember back to the trend of carbocation stability that tertiary carbocations are the most stable. And primary carbocations are the worst. And then obviously methyl is even worse than that, but like I said, methyl doesn't even get counted because we can't use it. So it's really the same trend as the one about ease. Basically, what we're saying is that the easier it is to make a carbocation, a positive charge on a tertiary carbon, the easier – I'm sorry, not only the easier it's going to be to make it, but also it's going to change the mechanism as well.
So what I want you to know for this section is that secondary and tertiaries make pretty good carbocations. And then primaries make pretty bad carbocations. So I would say these two are good and that primary is bad and then obviously methyl is out of here.
Recall that for elimination to take place, you need a good leaving group. Alcohols are terrible leaving groups, but in the presence of acid, they can be converted into water, which is an amazing leaving group.
If an alcohol can form a stable carbocation, the E1 mechanism will be favored. If it can’t, then the mechanism will follow an E2 pathway. Let’s start off 1° ROH, which usually follow E2.
Concept: Dehydration of 1° alcohols: The E2 Mechanism6m
So let's go ahead and start off with my primary mechanism. So what if I have a primary alcohol. We know it's not that easy to dehydrate. It doesn't make good carbocations. So what mechanism are we going to use? It turns out that for this mechanism, because we don't want to use a carbocation, we're going to use the E2 mechanism. The E2 mechanism, if you guys remember back to elimination is a concerted mechanism. That means everything happens at once.
So where are we going to start with this/ Well, if you guys remember, anytime that we have an acid catalyzed reaction, anytime it says acid catalyzed – this might be the first time you've heard me say this because maybe you haven't watched the other videos or it's cool. That means that you're always going to get a protonation step at the very beginning. Protonation just means I'm adding a proton to something. Whenever you hear me say acid catalyzed, even this applies for orgo two, think there's going to be a proton added to something.
In this case, I'm starting off with an acid. Notice that the acid I'm using here is H3O+. Now that's actually a conversation in and of itself because you're probably thinking, “Hey, Johnny,what happened to the whole HA? Why isn't it sulfuric acid?” Well, because if you think about it, let's say I was using sulfuric acid, H2SO4 plus water. What's going to happen is that any time you have any acid in water that is always going to be in equilibrium with the deprotonated form. What that means is that I'm going to get H3O+ plus the anion so that would be basically, HSO4-. Does that make sense?
Basically, what I'm trying to show you is that no matter what acid you use, it could be HI, it could be sulfuric acid, you're always going to have H3O+ as part of the solution because you're going to get the water grabbing a proton from the strong acid and turning into H3O+. So here when I'm using H3O+, this is just going to be our general acid that we use for these reactions. But if your professors asking you to draw the mechanisms, then obviously use the specific acid that he gave you.
Now we're going to go ahead and start. The first step is protonation. Do you guys see anything that's nucleophilic or that has electrons that could grab this electrophilic H?Now keep in mind there's a strong dipole pulling away from the H so that I have a partial positive here and a partial negative on the O.
And the answer is that yes, I have this oxygen with two lone pairs. It's very nucleophilic. It can go ahead and grab this H. But if it makes a bond to the H, we have to break a bond to the rest of the water so what we're going to get at the end of this first step is we're just going to get a water molecule like that. And then plus we would get plus another molecule of water actually because we just removed a proton.
So now here's the next step. In the next step, I've got this water molecule. Notice that I've got this down here. And now I'm just drawing one of the H's. That H was always there. I'm just going to draw it right now.
It turns out that in this next step, I'm going to do beta-elimination. Now beta-elimination remember is just based off of your alpha is wherever your leaving group is. In this case, alcohol was a terrible leaving group, but now I just protonated it, so it became a great leaving group. So I'm going to count this as my alpha-carbon. My beta-carbon is the one right next to it. Any carbon right next to the alpha is beta. That has a hydrogen on it. Can I eliminate? Yes.
So now I'm going to do a concerted three-step mechanism where I do my nucleophile to the H. Make a double bond. Kick out the leaving group. And what I'm going to get at the end is just a double bond plus, check it out, an acid that I originally had and now a molecule of water. The molecule of water came from the leaving group. The acid came from the nucleophile that attacked that H.
Now, what's important about the acid is that notice that the name of this reaction was acid catalyzed hydration. That means – remember the definition of a catalyst is that it's something that can neither be consumed nor destroyed in the reaction. All it does is it lowers the activation energy of the actual reaction.
So it's really important that at the end of this reaction, you're ending up with the same acid that you started with. So if I started with H3O+, I have to make sure to get one molecule of H3O+ at the end. Why? Well, because if you just didn't have H3O+ present at the end, it wouldn't be a catalyst anymore. It would be something else. It would be a reagent. But since I'm regenerating it, not you know it's a catalyst.
I hope that makes sense. That was the E2 mechanism. That only applies, just so you guys know, that only applies to primary alcohols. So if it's a primary, you'd use this. But what about if it's a secondary or a tertiary, well, you're going to have to find out. That's what our next topic is all about. So let's go ahead and see what happens when you have a secondary or a tertiary alcohol.
E2 Concerted β-Elimination:
Concept: Dehydration of 2° and 3° alcohols: The E1 Mechanism.6m
The biggest difference between primary alcohols and secondary and tertiary alcohols is that secondary and tertiary, I told you guys, can make more stable carbocations. So you already might be figuring that the mechanism is going to include a carbocation for these structures and that's actually right. So let's go ahead and get started.
This mechanism that I'm going to show you right now only applies to secondary and tertiary alcohols. And these are going to follow the E1 dehydration mechanism, meaning that if you remember back to E1, E1 is actually going to be a two-step reaction and it's going to involve a carbocation intermediate. So let's go ahead and get started.
In my first step, I'm starting off with an acid, again, notice that once again I'm using H3O+ which is just my generic acid in water. That could have been any strong acid to begin with. So now I've got alcohol. I've got my acid. Your first step of any acid catalyzed mechanism is to protonate. So do you see anything that I could protonate easily? Yes, you do. There is an alcohol that has very nucleophilic lone pairs.
So I can take those lone pairs, one of them, and I can go and use it to make a new bond to that very electrophilic H or very acidic H. What that's going to do is it's not going to give me a leaving group that instead of being terrible, alcohol sucks, it's actually going to be really awesome because it's just going to be water. And remember that water is a good leaving group because it's neutral after it leaves and what's better than to be neutral. That means it really wants to get off there.
Now I've got my water. So here's the second step, the second step, I just said this was E1, by the way, that was the first step is protonation, but now we're actually on to the E1 stage. At this point, we need to do the first step and because this is E1, maybe you guys can predict what that first step will be. And if you don't know, that's fine. Maybe you didn't watch that video or you haven't learned it yet.
But it's just going to be that the water leaves all by itself because it's so stable it doesn't really need anything to kick it out. For the E2 reaction, I need to have a concerted three arrows. But in this case, I have such a great leaving group, it's just going to take off by itself and what I'm going to wind up getting is a carbocation. Notice that, in this case, that's a secondary carbocation, which is pretty stable.
Now that I have a carbocation, there's one thing you guys should be watchful for. Do you guys remember anything tricky about carbocations? Maybe you don't remember and that's fine. But just so you guys know, carbocation like to do one thing, they like to shift. Carbocations love to shift to more stable positions. That means that at the end they're actually going to generate constitutional isomers. So that's something really important to know.
In this case, would this carbocation shift? No, because it's secondary. It doesn't matter, but – and there's no where else that it can shift that it would become more stable. But if I had a methyl group around or something like that, this would be able to shift and become more stable carbocation.
So now that we are mindful of that, we're onto the last step. The last step is beta-elimination just like normal E1. So what can I do to eliminate it? I can use that conjugate base of my acid, which is just water, pretty good nucleophile. It's okay nucleophile. And I'm going to go ahead and say that my carbocation is the alpha-carbon, meaning that any carbon that's attached to it would be beta. So this is beta and this is beta.
My question is do any of these beta-carbons have at least one hydrogen on them? And actually they do, they both have at least one hydrogen. Is there a difference between which one I use? Could I use the top one or the bottom one and get the same thing? Yeah, it doesn't matter because both of them would lead to the same product.
So let's just go ahead and use the bottom one. So what that means is I would take my nucleophile, grab an H and make a double bond. So what that's going to make is a new double bond, so this is the pi bond. This is my elimination product. Also what we call dehydration product because dehydration is just the name for an elimination with water or with alcohol. And we would get the leaving group originally which was water, remember that water left in that first step. And then we would also get H3O+.
What's the significance of H3O+? That that is my catalyst once again. Remember that we said that it was important to get the same catalyst that you started with because that's the whole definition of a catalyst.
So I hope that makes sense guys. It's pretty related to the other elimination reactions that we have to learn in orgo one. If you're struggling with these mechanisms, that means you have to go back or go forward or whatever direction you need to go in your curriculum to learn more about elimination. But hopefully, this wasn't so bad.
So let's go ahead and do some practice problems based on everything we just learned.
Formation of a Carbocation (Slow Step):
E1 β-Elimination (Fast Step):
Concept: An extra note of caution with 1° alcohols.2m
Let's do some cumulative practice based on everything we've learned from acid catalyzed dehydration and make sure to be mindful of all the different details I taught you about which mechanism you would use and what would happen in terms of how many steps they would have and stuff like that.
Now there is one instruction that I want to give you and that's remember that I told you that primaries would do an E2 reaction or mechanism and then secondaries and tertiaries would perform an E1. Now that is almost always true, but there is one exception to that. That is going to be if you have a primary alcohol that can rearrange to a tertiary carbocation through a shift, then it's actually going to do a carbocation mediated E1 mechanism instead.
So you're going to have to use that information to maybe determine this first one. Notice that it is primary. Figure out this would usually be E2, if it could shift to a tertiary position, then you should actually use E1. But I'm going to let you guys figure that out.
Second one, same thing. You'll have to figure out what mechanism and everything that would happen in between. So go ahead and get started on this first one and then I'll show you guys how to do it.
Remember how I mentioned that 1° alcohols usually follow E2?
This isn’t the case of 1° alcohols that can rearrange to 3° alcohols. Since the 1,2-rearrangement creates a super stable carbocation, the reaction will follow the E1 pathway.
Problem: Predict the major product of the reaction6m
Problem: Predict the major product of the reaction4m
DRAW THE MOLECULAR STRUCTURE OF THE FOLLOWING REACTION PRODUCT(S).
Two products, both of molecular formula C10H18, from reaction of 1-(1,1-dimethylethyl)cyclohexanol (1-tert-butylcyclohexanol) with concentrated H2SO4.
Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw all answers in skeletal form.
Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw your answer in skeletal form. You will be graded on the product your draw from the reaction no other information is needed for this question.
What would be the optimal conditions to effect the following transformation?
A) Dilute H2SO4
B) Concentrated H2SO4
C) Dilute HBr
D) Concentrated HBr
Acid-catalyzed dehydration of 3-methyl-2-butanol gives three alkenes: 2-methyl-2-butene, 3-methyl-1- butene, and 2-methyl-1-butene. Propose a mechanism to account for the formation of each product.
For each of the following, supply a structural formula for the major organic product(s) when the product(s) is/are not given; if no reaction occurs, write N.R. Give the best possible answers. Be sure to show stereoisomers properly when necessary.
Complete the mechanism for the following acid-catalyzed dehydration of a alcohol reaction. Draw all the arrows to indicate movement of electrons, write all lone pairs, all formal charges, and all the products for each step. In the dotted box write which mechanistic element is involved in each step. If a racemic mixture is formed you must draw both enantiomers by drawing dashes and wedges and write RACEMIC. Draw the expected reaction coordinate diagram for the above reaction. Assume the above reaction is overall exothermic.
Provide a detailed mechanism for the following transformation using a curved arrow notation.
For the following dehydration:
A. Predict the product of the following reaction.
B. Construct a reaction mechanism that accounts for the product proposed.
C. Construct a reaction coordinate diagram for the reaction. Label all axes. Note that the reaction has three steps: The 1st is fast and endothermic. The 2nd is slow and endothermic. The 3rd is fast and exothermic. The overall reaction is exothermic.
D. Using complete sentences state how you have shown that the 1st and 3rd steps are fast and that the 2nd reaction is slow.
What is/are the major product(s) of the following reaction? What mechanism does it follow (E1, E2, SN1, or SN2)? (Hint: Note the temperature and reaction time).
Predict the organic product(s) of the following reaction. When appropriate, be sure to indicate stereochemistry. Be sure to indicate the major product if more than one product is formed. Draw all answers in skeletal form.
Select the product(s) from the following reaction.
Propose a mechanism for the following reaction
Draw the mechanism for the following elimination reaction, without a carbocation rearrangement, and in the presence of sulfuric acid, H2SO4, as reagent. Indicate if it follows an E1 or an E2 mechanism.