Practice: Predict the product of the following reaction.
|Ch. 1 - A Review of General Chemistry||4hrs & 48mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 19mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 25mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 32mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
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|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|Addition Reaction||7 mins||0 completed|
|Markovnikov||5 mins||0 completed|
|Hydrohalogenation||7 mins||0 completed|
|Acid-Catalyzed Hydration||17 mins||0 completed|
|Oxymercuration||20 mins||0 completed|
|Hydroboration||27 mins||0 completed|
|Hydrogenation||7 mins||0 completed|
|Halogenation||6 mins||0 completed|
|Halohydrin||17 mins||0 completed|
|Carbene||13 mins||0 completed|
|Epoxidation||8 mins||0 completed|
|Epoxide Reactions||9 mins||0 completed|
|Dihydroxylation||9 mins||0 completed|
|Ozonolysis||7 mins||0 completed|
|Ozonolysis Full Mechanism||25 mins||0 completed|
|Oxidative Cleavage||8 mins||0 completed|
|Alkyne Oxidative Cleavage||6 mins||0 completed|
|Alkyne Hydrohalogenation||3 mins||0 completed|
|Alkyne Halogenation||2 mins||0 completed|
|Alkyne Hydration||6 mins||0 completed|
|Alkyne Hydroboration||3 mins||0 completed|
This is the first of three ways to add alcohol to a double bond. It is similar to hydrohalogenation in terms of mechanism, however it will require a protonation and deprotonation step since it is acid-catalyzed.
Concept #1: General properties of acid-catalyzed hydration.
Guys, now we're going to discuss what might be the most important addition reaction of this entire section and that's called acid catalyzed hydration. Keep in mind that of the three methods to make alcohol that we're going to learn in this section, acid catalyzed hydration is considered the first of three ways to make alcohols by addition. So let's go ahead and get started.
So, guys, acid catalyzed hydration is more similar than it is different to the addition reactions that we learned so far. The intermediate is the same as the ones that we've seen so far, which is that it is a carbocation. Now if the intermediate is a carbocation, what do you think that says about its ability to do shifts or rearrangements? Totally. This thing is going to rearrange all the time.
Now that could actually be a bad thing because remember that rearrangements lead to unpredictable outcomes for our reactions. It could lead to a reaction or a product that we weren't expecting. So we may actually see that the ability to rearrange is a drawback that we have to fix later on, but we'll get there later.
Now, one thing about carbocations and their stereochemistry is that they're very unpredictable. Carbocations, remember that they're trigonal planar, so they can be attacked either from the front or from the back. What that means is that the stereochemistry of our products is really still going to be unknown. Unknown because we don't know where it's hitting. There may be several chiral centers produced. So I don't want to say anything more than just unknown. There's usually a big mixture of chiral products in these reactions.
As we said earlier, acid catalyzed hydration or simply known as hydration, is the first of three methods to make what? Alcohol. So we know that my product is going to be an alcohol. Now is this going to be a Markovnikov alcohol or an anti-Markovnikov alcohol? Well, it turns out that in this section we're going to learn how to do both. We're going to learn how to do Mark and anti-Mark alcohols. But hydration is definitely a Markovnikov reaction. It's a Markovnikov alcohol because your carbocation is always going to want to form in its most stable location.
So before we go to the general reaction, let's just read off a few bullet points because they will be helpful for us. Now, notice that our reagents are H2O and H2SO4. Now, put more generally, that's the most likely way to see it, but it's simply going to be H2O with some form of acid or HA. Recall that when I talked about dehydration. I said dehydration was H2O and HA, so actually the reagents haven't changed at all since dehydration. They're the same.
So you might be wondering, “Johnny, how can I tell, if the reagents are exactly the same, how can I tell if it's going to be a hydration or a dehydration? How do you know?” Well, you look at what you're starting with. Since you're starting with a double bond that means that a double bond is going to hydrate to an alcohol. So it's the same reagents as acid catalyzed dehydration except starting with a double bond instead of starting with an alcohol. Cool.
So you look not at the reagents, but you look at the starting molecule to know if you're going to go to hydration or dehydration. Just think about hydration means you're adding water and look at what we just did. We're adding a water. We're going to add an H and we're going to add an OH so that means we're hydrating the double bond. We're adding a water to it.
Now in terms of the general mechanism, it's also going to be very predictable. It's going to be the same as hydrohalogenation, which you guys might remember is just the simple reaction of a double bond with HX, which is really our example reaction for addition. It's the same general mechanism as that except that we're going to use water as the nucleophile. Instead of using X- and attacking X-, we're going to attack water instead.
Then guys, remember, I stated this when we talked about dehydration, but remember that every acid catalyzed mechanism always is very predictable. It always starts with something and it always ends with something. If the name is acid catalyzed, that means it always begins with protonation and it ends with deprotonation. That's because if it's acid catalyzed, that means you're always going to put a proton on something and at the end, since it's a catalyst, you have to take that proton off so you can regenerate your acid.
So let's just look at the general reaction here. The general reaction states that a double bond with water and acid is going to do what? Well, notice that this is my Markovnikov site and this is what I would call my anti-Markovnikov site. So this is my Mark site. So even without knowing the mechanism, because I told you guys all the things about the properties of the reaction, I can predict that my product is going to look like what? I'm going to have a Markovnikov alcohol. An alcohol in the most substituted place and then the H of the water will be in the least substituted place. So I'm adding water. I'm adding H2O, but the OH forms Mark and the H forms on the other position.
Now the last thing you might be wondering – not wondering about me, is wondering what's up with the squiggly line because it looks super weird. You may have not seen that before. You may have forgotten what that was. Just remember guys, if you see a squiggly line like that, it means that it's unknown stereochemistry. It means that it could be towards the front. It could be towards the back.
You're not expected to draw all the stereoisomers because that could be a huge burden. All you need to know is that that H went somewhere. And that's why I put the stereochemistry unknown because you're not – there could be up to four different stereoisomers on these things, so I don't think you're expected to draw all of them, just to know that several could form.
Awesome guys. So that's it for this. So then let's move on to the next problem where we're going to draw the mechanism at length.
This reaction uses the same reagents as acid-catalyzed dehydration, so how do you know which reaction to use? Just look at what you are starting with:
Note: The squiggly line on the product just means “indeterminate stereochemistry”. We aren’t sure where that –H will add, so we’ll just draw it on a squiggly line.
Example #1: A worked-example of the acid-catalyzed hydration mechanism.
So what I want to do for this question is I just want to show you guys the full mechanism, OK? I'm going to draw the full mechanism step by step and you guys are just going to follow me as we go along, alright? So in my first step I've got my double bond and my double bond is a nucleophile, I'm looking for a electrophile something that I can grab an H from, OK? And that's going to be my Sulphuric acid so I go over here, OK? And now notice that I drew the sulfuric acid a little bit weird, OK? This actually just has to do with the fact that that the way that sulfuric acid is drawn right here is very difficult to take a proton off of it, well it still makes sense chemically it's better if you draw in this notation so you can actually see where is the H attached to, oh it's attached to an O it's not attached to an S, OK? So I know this is a little bit tricky but I am going to ask you guys to memorize how to draw sulphuric acid when you're trying to deprotonate it, OK? I know this sucks but this is going to make your life easier when it comes to actually drawing the product, OK? So here we go we've got our double bond and I'm looking for my electrophile that's going to be this H, OK? Because my H is acidic so I take my double bond, I grab an H give the electrons to the O, what I wind up getting is a Carbocation right here, Cool? Alright so I've got my Carbocation what's my next step, I've got that plus I've also got OSO3H negative, OK? So is that going to attack? Well not yet, OK? Because on top of that this Carbocation is not very stable, it's secondary and if it moves over it could become tertiary so what we're going to do here is a shift, what kind of shift would it be? Well it was because this is a methyl group I only have methyl groups surrounding this carbon so what that means is that I can't do my Hydride shift, remember that Hydride shifts are always going to be preferred but I don't have any hydrogens on that carbon so that means I'm forced to do a methyl shift so I'm going to grab any of these methylÕs it doesn't matter which one and I'm going to move it over to this carbon, what this is going to give me is a constitutional isomer at the end that looks like this 1, 2 CH3 shift and now what I'm going to get is a new carbocation that looks like this, where now my positive charge is down there, OK? Why is that? Because I used to have 4 bonds right there and I only have 3, OK? So now I've got my carbocation and now we're faced with the dilemma what attacks the carbocation, OK? And I've basically got two different nucleophiles, I've got the anion from my acid and I've also got water, OK? One of these nucleophiles is negatively charged and then one of them is natural so which of these do you think is the stronger nucleophile? It's usually going to be the negative charge, so I would say that actually the conjugate base of my acid, OK? The OSO3H negative would be the stronger nucleophile and I would expect it to attack but we've got another variable in the mix, OK? And this is actually going to be a repeating theme throughout this section, it turns out that even though the OSO3H negative is the stronger nucleophile because it's negatively charged there's just going to be a whole lot more water so imagine that I've got one of these guys because I just donated the acid whatever and then I've got like a billion of these guys, OK? Imagine that I've got a billion water molecules and I've got only one of the OSO3H negative even though the sulfuric acid anion is going to be a stronger nucleophile there's not a whole lot of it to go around so what that means is that this Carbocation is very easy to attack, it's very easy for a water to attack as well so it's just going to attack with the first thing that collides with it and because of the way that I have this solution mixed up I'm always going to have way more water than I am going to have anion so what that means is that even my water is weaker it's going to be the one that attacks, OK? And that all has to do with the ratio of water to acid and just you know from hydration that usually going to be a lot water and only a little bit of acid just to catalyze the reaction, OK? So I hope that makes sense why the water attacks and why the sulfate doesn't, OK? So now what we're going to do is we're going to draw the final or not the final but the new product which is that I have an OHH positive, OK? Because the O still has those 2Hs on it, now that O has a formal charge because it doesn't want to have 3 bonds so can you guys tell me what the last step is going to be in this reaction? The last step always going to be in any acid catalyzed mechanism it's always going to be deprotonation, OK? So that means that the last step I need to take away a proton from something and regenerate the catalytic acid, OK? And what's that going to be? Well the catalytic acid in this case was the OSO3H negative, OK? so in this final step what I can do is I can take that negative charge and I can grab that H and deprotonate it and finally get my product, my product is going to be an alcohol, OK? Plus it's going to be my sulfuric acid, alright? So notice that what did we do? we went ahead we took a double bond at the beginning, we reacted with aqueous acid, what we wind up getting was an alcohol at the end a markovnikov alcohol that shifted so can rearrange and we got our acid back at the end knows that we started off with H2SO4 and we ended up with H2SO4 that means that it was a catalyst, OK? Because we got the same amount of acid at the end as we had at the beginning, alright? This is a very common reaction in this section I hope you guys understood this mechanism it's really important that you guys practice it though, OK? So let's go ahead and move on to the next topic.
Practice: Predict the product of the following reaction.
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