Addition of Water (Acid-Catalyzed) Mechanism. Created by Sal Khan.
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- Hello all. So what makes the pi bond between the two carbons less stable/prone to elicit these reactions? Is it to do with these carbons being less saturated and more exposed to other molecules/atoms? Is it to do with the increased electron density in a sp2 hybridised carbon (making it more attractive to an electrophile, e.g. hydrogen proton)? Is a pi bond easier to break than a sigma bond? Thanks!(2 votes)
- The electrons in a σ bond are directly between the two nuclei.
The electrons in a π bond are "off-to-the-side" and further away from the two nuclei, so their attraction for the two nuclei is less.(8 votes)
- Why doesn't attack OH- the carbocation at07:05? it is much simpler, isn't ? I mean in the end of the theory Sal tells us there is a free proton left and this will bind to the OH- anyway, right? Would be the same result...(2 votes)
- Because there are so many more water molecules than hydroxide ions. In a 1 mol/L of HCl, for example, the concentration of water molecules is about 55 mol/L, while the concentration of hydroxide ions is about
10⁻¹⁴ mol/L. The chances of the cation colliding with a water molecule are about 6 .x 10¹⁴ times more likely than of its colliding with a hydroxide ion.(5 votes)
- just to clarify, the first molecule could've been calleded Z-2-methylpent-2-ene as well right?(4 votes)
- Z/E notation is usually used when you have 4 different substituents of of a double bond. In this case, the two groups bonded to the "left" side of the double bond are both methyl groups so you can't say that one is heavier than the other. This makes the Z/E notation redundant. Another way to look at it is if Z-2-methylpent-2-ene were correct, then E-2-methylpent-2-ene would also be correct.(16 votes)
- at8:41, why do we only want a hydroxyl group? Why can't the water be left attached to the Carbon and leave the molecule as is with a + charge on the Oxygen. As in why is the Hydrogen moved to another water molecule to form Hydronium?(2 votes)
- Good question!
One way to think about this is as an acid-base equilibrium with competition between two acids –
H₃O⁺and the protonated 2-methylpentan-2-ol. If you consult a pKa table§, then you will find that protonated organic alcohols typically have pKa values around -2, while
H₃O⁺has a pKa of -1.7. This means that the protonated alcohol is a slightly stronger acid and thus the equilibrium will be shifted towards water. In addition, there are typically many more water molecules than alcohol molecules, which also makes it more likely for the proton to end up in an
The protonated 2-methylpentan-2-ol is a stronger acid because a hydronium molecule can more easily share its excess positive charge with other water molecules through hydrogen bonding.
See this explanation for details:
§Example pKa table:
- Aren't hydronium ions naturally in water or how do we get them?(2 votes)
- Two water molecules collide at the right angle and speed and form one hydronium ion and one hydroxide ion. This happens in the given aqueous solution.
- At7:28Sal says that one electron is moving over ( the orange one) yet he draws a double barbed arrow. I am confused because I can't understand if he is moving just one electron ( only the orange one) or a pair of electrons( orange+ green). Could some one clarify?(2 votes)
- It should be a pair of electrons (for example, orange and green) that forms the bond between the oxygen and carbocation, not just the single orange electron.(3 votes)
- At7:40, if two oxygen electrons are shared, then why isn't a double bond formed??(1 vote)
- so the carbocation will always be the most "substituted" carbon?(1 vote)
- When you say always, bite your lip.
There is an exception to every rule in organic chemistry. Often there are dozens of them.
When we predict the product, we naturally look toward the more highly substituted carbocation possibly formed. That is a very good assessment Alfonso.
You will learn about the other possibilities as well. The opposite situation you describe leads to a situation known as "Anti-Markovnikov" regiochemistry in the product(s)....e.g. the alcohol or halide addition product formed is on a lesser substituted carbon than originally expected.
You'll eventually learn 3 major reasons how this is possible:
1) Radicals 2) Boron 3) Carbocation Rearrangements (hydride and alkide shifts)
I think you're going to do very well. keep it up :)(3 votes)
- At2:12, how did the hydronium molecule become positive ?(1 vote)
- The oxygen in water used one of its lone pairs to form a bond to H+ (so just a proton without an electron), so overall hydronium has one more proton than it has electrons, hence the +1 charge.(3 votes)
- [Voiceover] Anytime you're trying to come up with a mechanism for a reaction, it's worthwhile to study a little bit of what you are starting with and then thinking about what you finish with and think about what is different. So what we're starting with, we could call this one, two, three, four, five, so this is, let's see, we have methyl group on the number two carbon, it is a pentene, and that is double-bond between the number two and number three carbons, so this is two-methyl-pent-2-ene. So that's what we start with, we're in the presence, we're in an acidic environment, we've got what's gonna be catalyzed by our hydronium here, and we end up with this, and how is our product different from what we started with? Well the double bond is now gone, the number three carbon gains this hydrogen, and now the number two carbon gains a hydroxyl group. So one way to think about this is, in the presence of an acid, it's acid-catalyzed, we have gained two hydrogens and an oxygen, which is what we've gained, what could be used to make a water. And this is actually called an acid-catalyzed addition of water. The water isn't sitting on one part of the molecule, but if you take the hydrogen we added, and the hydroxyl we added, if you combine them, that's what you need to make a water. So let's think about how we can, how this actually happens in the presence of our hydronium. So let me redraw this molecule right over here. So we copy and paste it, so that's not exactly it yet, that is just with the single bond. So let me draw, woops, wrong tool. Let me draw the double bond there. And now let me put it in the presence of some hydronium. Alright, so we have an oxygen bonded, two. So this would just be water, as oxygen has two lone pairs, but hydronium is a situation where oxygen is sharing one of those lone pairs with a hydrogen proton, thus making the entire molecule positive, because the hydrogen proton is positive. So there you go, this now has a positive charge. And this can be pretty reactive, 'cause we know that oxygen is quite electronegative, it lives to keep its electrons. So what is there was a way, what if there's a way for the oxygen to take back the electrons in this bond right over here, the two electrons in this bond. Well what if one of these carbons, especially the ones that have the double bonds, what if some of the electrons from this double bond could be used to snab, to take that hydrogen proton, and then oxygen can hog its electrons again. And you might say, "oh that's reasonable, but which of these carbons would actually do it?" And to think about which of those carbons would do it, we have to turn to Markovnikov's rule. Markovnikov's rule tells us that, look, if you have a reaction like this, and alkene reaction, the carbon that already has, that's already attached to more hydrogens is more likely to gain more hydrogens, the carbon that's attached to more functional groups is more likely to gain more functional groups. Another way to think about it is think about, well that is the order of the carbons? Because the higher order of carbon, the more stable it will be if it forms some type of cation. So if you look at this carbon right over here, our number two carbon, let me circle it, our number two carbon is bonded to one, two, three carbons. So this is a tertiary carbon. This one right over here, this carbon on the other side of the double bond is only bonded to one, two carbons. So this is a secondary carbon. So the tertiary carbon is going to be more stable as a carbocation, you can think of it as it can spread the charge a little bit. So it would be more likely to lose the electrons in one of these, in one of these bonds, and so the way that we can think about this mechanism, and it might be a little bit clearer when we form the carbocation, is let's have, going to do this in blue. Let's have these two electrons that form this bond, well now they form a bond with that hydrogen, and now the oxygen can take back these two electrons, and what are we going, what is going to result? And I'm drawing it in equilibrium, remember all of these things are going back and forth depending on how things bump into each other, but what are we left with? So let me copy and paste this again and I'm copying and pasting in a way that, just so I, this is the backbone, and I'll add what I need to add. So once this happens, we have this carbon, the number three carbon, now, woops I keep using the wrong tool. The number three carbon now forms a bond with this hydrogen just like that. This carbon, our number two carbon, has lost an electron, it's no longer sharing this bond. And so now it is going to have a positive charge, it is a carbocation and once again is a tertiary carbocation, it is bonded to one, two, three carbons. That is stable, more stable than if we did it the other way around, if this one grabbed the hydrogen somehow, then this would be a secondary carbocation, it'd be harder for it to spread that positive charge around. And what about our, what about our molecule up here? Let's see what it looks like now. We have our oxygen bonded to the two hydrogens, it had one of those lone pairs, and now the electrons in this bond are now going to form another lone pair. So it took back an electron, or you can think of it, it gave away a hydrogen proton. And so this is now just neutral water, and we see that we have a conservation of charge here, this was positive in charge, now our original molecule is positively charged. And what feels good about this is we're getting, we're getting close to our end product, at least on our number three carbon, we now have, we now have this hydrogen. Now we need to think about, well how do we get a hydroxyl group added right over here? Well we have all this water, we have all this water floating around, let me, I could use this water molecule but the odds of it being the exact same water molecule, we don't know. But there's all sorts of water molecules, we're in an aqueous solution, so let me draw another water molecule here. So the water molecules are all equivalent, but let me draw another water molecule here. And you can imagine, if they just bump into each other in just the right way, this is, water is a polar molecule, it has a partially negative end near the oxygen because the oxygen likes to hog the electrons, and then you have a partial positive end near the hydrogens 'cause they get their electrons hogged, so you can imagine the oxygen end might be attracted to this tertiary carbocation, and so just bumping it in just the right way, it might form a bond. So let me say these two electrons right over here, let's say they form a bond with this, with that number two carbon, and then what is going to result? So let me draw, so what is, what is going to result, let me scroll down a little bit, and let me paste, whoops, let me copy and paste our original molecule again. So, here we go. So what could happen? This is the one we constructed actually, so we have the hydrogen there. We have the hydrogen, now this character, so we have the water molecule, so oxygen bonded to two hydrogens, you have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting. And so it now forms a bond. Woops, let me do it in that orange color. It now, it now forms an actual bond. And we're really close to our final product, we have our hydrogen on the number three carbon, we have more than we want on our number two carbon, we just want a hydroxyl group, now we have a whole water bonded to the carbon. So somehow we have to get one of these other hydrogens swiped off of it, well that could happen with just another water molecule. So let's draw that. So another water molecule someplace, I'll do the different color just to differentiate, although as we know water, well it's hard to see what color is water if you're looking at the molecular scale. So here we go, and we're really in the home stretch at this point. You have another water molecule, let's say, let me pick a color. So let's say these electrons right over here, maybe they form a bond with that hydrogen proton, and then these, the electrons in that bond can go back to form a lone pair on that oxygen, and then what are we left with? And this really is the home stretch. So we are in equilibrium with, so let me draw my five carbons, so let's see, I have... H3C, carbon, carbon, CH2, CH3, I have a CH3, I say H3C instead of CH3, I wrote it that way just so it's clear that the carbons are bonded to the carbons, you have the original hydrogen right over there, you have the one that we just added as part of this mechanism, you have this orange bond to now, this hydroxyl group, the hydroxyl group, and it had one lone pair before, it had one lone pair before, but bound to both of the electrons from this bond to form another lone pair. To form another lone pair, which I am depicting in pink, and then this water is now, this water molecule is now a hydronium molecule. So let me draw that, so this is now, oxygen, hydrogen, had one lone pair that didn't react, and it had one lone pair that I put in blue that is reacting with this hydrogen proton. With that hydrogen, just like this, and so since it got the hydrogen proton it's giving its, sharing its electrons now, now this has a positive, this has a positive charge, just like that. I have to be very careful, in the last step I forgot to draw the positive charge! We always wanna make sure that your charge is being conserved, we started off with the positive charge on the hydronium, then we have the positive charge on the tertiary carbocation right over here on our number two carbon, and now we have the positive charge, would be right over here, because that oxygen, what we saw before that oxygen, which this water molecule was neutral, but you could say, you could view it is "well now it's going to be, it's now sharing these two electrons instead of keeping them," so you can view it is maybe it's giving away an electron, and so now it becomes positive. And then, and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium. But just like that we are done. We have added a hydroxyl group and a hydrogen combined, that's a water, so that's why we call it addition of water, and it was catalyzed by acid, so it's an acid-catalyzed addition of water.