Addition of water (acid-catalyzed) mechanism | Organic chemistry | Khan Academy

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 think 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. Let's see, we have a methyl group on the number two carbon. It is a penene, and that's a double bond between the number two and number three carbons. So, this is two methyl-2-pentene. That's what we start with.

We're in the presence of an acidic environment; we've got it going to be catalyzed by our hydronium here. 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 hydroxy group.

One way to think about this is that in the presence of acid, it's acid-catalyzed. We have gained two hydrogens and an oxygen, which is, well, we've gained the build, what could be used to make water. 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 hydroxy we added, if you combine them, that's what you need to make water.

Let's think about how this actually happens in the presence of our hydronium. So, let me redraw this molecule right over here. So, let me copy and paste it. That is not exactly it yet; that is just with the single bond.

Let me draw—oops, wrong tool—let me draw the double bond there and now let me put it in the presence of some hydronium. All right, so we have an oxygen bonded to—so this would just be water. Oxygen has two lone pairs, but hydronium is a situation where oxygen is sharing one of those lone pairs with the hydrogen proton, thus making the entire molecule positive because the hydrogen proton is positive.

There you go; this now has a positive charge, and this can be pretty reactive because we know that oxygen is quite electronegative. It likes to keep its electrons. So, what if there was a way for the oxygen to take back the electrons in this bond right over here, the two electrons in this bond?

What if one of these carbons, especially the ones that have the double bonds, could be used to snatch that hydrogen proton, and then oxygen can hog its electrons again? You might say, well, that's reasonable, but which of these carbons would actually do it?

To think about which of those carbons would do it, we have to turn to Markovnikov's rule. Markovnikov's rule tells us that if you have a reaction like this, an alkene reaction, the carbon that already has 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 to consider the order of the carbons because the higher order of the carbon, the more stable it will be if it forms some type of cation. 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. This is a tertiary carbon.

This carbon on the other side of the double bond is only bonded to one, two carbons. So, this is a secondary carbon. The tertiary carbon is going to be more stable as a carbocation. You can think of it as being able to spread the charge a little bit, so it would be more likely to lose the electrons in one of these bonds.

The way that we could think about this mechanism might be a little bit clearer when we form the carbocation. Let's have—I'll do this in blue—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.

What is going to result? I'm drawing an equilibrium; remember, all these things are going back and forth depending on how things bump into each other. But what are we left with?

Let me copy and paste this again, and I'm copying and pasting it in a way that I’ve just—this is kind of—you can view this as a backbone, and I'll add what I need to add. So, once this happens, we have this carbon, the number three carbon. Now, whoops, 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, it is a tertiary carbocation; it is bonded to one, two, three carbons. That is 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 would be harder for it to spread that positive charge around.

What about our molecule up here? Well, 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 not going to form another lone pair, so it took back an electron, or you could think of it as it gave away a hydrogen proton.

So, this is now just neutral water. We see that we have a conservation of charge here. This was positively charged; now our original molecule is positively charged, and what feels good about this is we're getting close to our end product.

At least on our number three carbon, we now have a hydrogen. Now, we need to think about how do we get a hydroxy 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 are all sorts of water molecules; we're in an aqueous solution. So, let me draw another water molecule here. There; the water molecules are all equivalent.

Let me draw another water molecule here, and you can imagine it's just that if they just bump into each other in just the right way, this 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 because they get their electrons hogged.

You can imagine the oxygen end might be attracted to this tertiary carbocation. So, just bumping in in just the right way, it might form a bond. So, let me—let's say these two electrons right over here—let's say they form a bond with this number two carbon.

What is going to result? Let me draw—what is going to result? Let me scroll down a little bit, and let me copy and paste—oops, let me copy and paste our original molecule again. So, there we go.

What could happen? Let me construct it, actually. So, we have the hydrogen there; we have the hydrogen now, this character. 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 the reacting and so it now forms a bond.

Oops, let me do it in that orange color. 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 hydroxy group.

Now, we have a whole water bonded to the number two 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 it in a different color just to differentiate—although, as we know, it's hard to say even 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—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 electrons in that bond can go back to form a lone pair on that oxygen.

What are we left with? This really is the home stretch. So, we are in equilibrium with—so let me draw my five carbon.

Let’s see, I have H3C-carbon-carbon car C H2-CH H3. I have a C H3. 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 the hydroxy group. The hydroxy group had one lone pair before, but now it took both of the electrons from this bond to form another lone pair—another lone pair, which I am depicting in pink.

Then, this water is now... or this water molecule is now a hydronium molecule. So, let's draw that. This is now oxygen-hydrogen-hydrogen; it had one lone pair that didn't react, but it had one lone pair that I put in blue that is reacting with this hydrogen proton—just like this.

Since it got the hydrogen proton, it's giving its electrons away or sharing its electrons. Now this has a positive charge, just like that. Oh, I have to be very careful in the last step. I forgot to draw the positive charge.

You always want to make sure that your charge is being conserved. We started off with a positive charge on the hydronium. Then we have the positive charge on the tertiary carbocation right over here on the number two carbon.

Now we have the positive charge—it would be right over here because that oxygen, which we saw before, was neutral. But you could say you could view it as, well, now it's going to be giving or sharing these two electrons instead of keeping them.

So, you could view it as maybe it's giving away an electron, and so now it becomes positive. 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 hydroxy group and a hydrogen combined—that's water. So, that's why we call it the acid-catalyzed addition of water.