Reaction of an alkene with a diatomic halogen and water, converting the double bond to a single bond with halogen and hydroxyl substituents. Created by Jay.
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- 3:00Why wont Br- attack the partially positive C instead of H-OH?(20 votes)
- It can. Indeed, Br- is a stronger nucleophile than water, because of its negative charge. But there are so many more water molecules present than bromide ions that the chances of attack by a water molecule are much greater than by a bromide ion.
For example, the concentration of water may be about 55 mol/L The concentration of Br2 may be about 0.1 mol/L. and only a very small fraction of these exist as Br- ions at any given instant.(36 votes)
- I don't understand the selectivity of the nucleophilic attack at6:44.
1. Shouldn't the secondary carbocation (left) be more positive because it is less neutralized by hyperconjugation?
2. Shouldn't the secondary carbocation (left) be more accessible because there is no methyl group in the way?
Yet water nucleophillically attacks the tertiary carbocation (right). Why?(13 votes)
- Consider the cyclic bromonium as a resonance hybrid. One contributor has the + charge on the Br. A second contributor has no bond between Br and C-2, but has a formal + charge on the 2° carbon atom C-2.
The third contributor has no bond between Br and C-1, but has a formal + charge on the 3° carbon atom C-1.
Since the 3° carbocation (C-1) is more stable, it is a major contributor to the resonance hybrid.
So when the water molecule is looking for a positive site to attack, it will then go to C-1 where the positive charge is greatest.(16 votes)
- If both C's are "equally" substituted but one has an ethyl and the other a methyl, which will the H2O attack?(3 votes)
- You can consider the Br⁺ in the cyclic bromonium ion to be an "enhanced" leaving group.
So the reaction will be similar to an SN1 reaction, and the carbon that forms the more stable carbocation will be the one that is preferentially attacked.
An ethyl group provides a little more electron density and stabilization than a methyl group (for example, propionic acid is weaker than acetic acid).
You will get attack on both carbons, but the major product will have the OH on the carbon with the ethyl group.(5 votes)
- 4:02Why the halide ion formed won't act as a nucleophile and attack the hydrogen from the H2O added to the compound?(2 votes)
- The halide ion formed can act as a base and attack the hydrogen from the H2O added to the compound. But H2O is more likely to attack because:
1. The H2O molecules vastly outnumber the X- ions.
2. H2O is a much stronger base than X-.(6 votes)
- At06:34, you mentioned that the the carbon on the right is a partial carbocation. I'm a little confused on how that carbon, as well as the one on the left, have partial positive charges.(2 votes)
- The Br atom has a formal positive charge. Also, it is a fairly electronegative atom. The Br wants electrons! Is gets some of this electron density by drawing the electrons in the C-Br bonds closer to itself. This decreases the electron density around the two carbon atoms and gives them a partial positive charge.(3 votes)
- At8:29, would it be correct to say that we did not have to draw it upside down like that? We could have done it in the standard way and it should have also produced the same enantiomer, right? Thanks in advance.(2 votes)
- Can someone explain me this experiment?
I used both Cl2(g) and Cl2(aq) as an oxidant, but I never used Halogens on an alkene.
When I bubble Cl2 into the Alkene, both Cl Atoms will bond with the alkene, but what will happen if I use Cl2(aq)?
Obviously water is a stronger nucleophile than Cl- because Cl- has got an electron oktett,but as long as there are other Cl2-Molecules in the solution, the Cl2 is will be a stronger nucleophile than H20, right?
So what will be our main product if we react an Alkene with Chlorine water/Cl2(aq)? (in a solution satured with Cl2)(1 vote)
- Chlorine is an electrophile. It adds first to the double bond to form a chloride ion and an intermediate cyclic chloronium ion. As you say, water is a stronger nucleophile than chloride ion, so it is water that attacks one of the carbon atoms of the chloronium ion. You end up with a Cl on one C atom and an OH on the other C atom ( a chlorohydrin ). If your alkene is ethene, the product is 2-chloroethanol.(3 votes)
- Why is the addition of water in the first reaction not a second step? My organic teacher stresses that the addition of water must come after the first reaction is complete, but I'm not sure why or if it is even in this type reaction?(1 vote)
- I'm pretty sure I understand this, but I just want to make sure.2:12, Jay says that the halogen attracts the magenta electrons closer to it. Wouldn't it attract both sets of electrons closer to it and you would have a choice of which carbon the water would attach. If I'm wrong and for some reason the magenta electrons are more attracted to the halogen than the other set, then why?(1 vote)
- You are correct. In this case the substrate is symmetrical, so the halogen attracts both electrons equally.(2 votes)
- Also, I can't seem to find what the rule on regioselectivity is: I do know that the product Jay drew on the left is the major organic product and the right is the minor product. But why? I looked in my textbook, dummies book and online & can't seem to find it. Thanks!(1 vote)
- The Br⁺ in the cyclic bromonium ion is a much better leaving group than a Br is, so the reaction behaves as if it were an SN1 reaction.
You can think of it as if the Br is leaving first to form the more stable carbocation (tertiary), so the water attaches to the tertiary carbon.(2 votes)
Here's the general reaction to make halohydrins from alkenes. So if I start with my alkene on the left, and I add a halogen to it and some water, you can see that an OH and a halogen are added anti to each other. So anti, or on opposite sides of where the double bond used to be. The mechanism for this reaction starts off the exact same way the halogenation reaction did. And so we have our halogen approaching our alkene. And we saw in that video that the halogen is usually, of course, nonpolar, because those two atoms have the exact same electronegativity. So these blue electrons in here are pulled with equal force to either halogen, so it's a nonpolar molecule. However, if the pi electrons in my alkene here get too close to the electrons in blue, we saw how that could induce a temporary dipole on the halogen molecule. So those electrons in blue are repelled by the electrons in magenta and push closer to the top halogen, which gives the top halogen a partial negative charge and leaves the bottom halogen with a partial positive charge. The bottom halogen is now an electrophile, so it wants electrons. It's going to get electrons from those pi electrons here, which are going to move out and nucleophilic attack that partially positively charged halogen atom. And then this lone pair of electrons is going to form a bond with this carbon at the same time these blue electrons move out onto the halogen. So when we draw the result of all those electrons moving around, we're going to form a bond between the carbon on the right and the halogen, and we use the magenta electrons to show that. So there's now a bond there. And so we used red electrons before to show these electrons in here forming a bond with the carbon on the left. That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge. We called this our cyclic halonium ion in an earlier video. And if I think about that cyclic halonium ion, I think about the halogen being very electronegative. It's going to attract, I'll say the electrons in magenta again just to be consistent, closer towards it. So it's going to take away a little bit of electron density from this carbon right down here. So I'm going to say this carbon is partially positive. It's going to have some partial carbocationic character. So in the next step of the mechanism, water's going to come along. And water's going to function as a nucleophile. So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here. And so when that lone pair of electrons on oxygen attacks this carbon, that's going to kick the electrons in magenta off onto your halogen. And so let's go ahead and draw the product. We're going to have, on the left carbon, this halogen now used to have two lone pairs of electrons. It picked up the ones in magenta, so now it looks like that. On the right, we still have the carbon on the right bonded to other things, except now it's bonded to what used to be our water molecule. So the oxygen is now bonded to the carbon. And there's still one lone pair of electrons on that oxygen, giving it a plus 1 formal charge. So let's go ahead and highlight these electrons here in blue. Those electrons in blue are the ones that formed this bond between the carbon and the oxygen. So we're almost done. The last step of the mechanism would just be an acid-base reaction. So another water molecule comes along, and one of the lone pairs of electrons on the water molecule is going to function as a base and take this proton, leaving these two electrons behind on the oxygen. And we are finally done. We have formed our halohydrin, right? So I have my halogen on one side. And then I now have my OH on the opposite side, like that. Let's go ahead and do an example so we can examine the stereochemistry a little bit more here. So if I start with an alkene, and to this alkene we are going to add bromine and water. And we're going to think about doing this two different ways here. So we'll start with the way on the left. So Br2 and H2O. And then we'll come back and we'll go ahead and do this on the right. So BR2 and H2O. So on the left side, I'm going to think about the formation of that bromonium ion here. So I'm going to once again look at this molecule a little bit from above, so looking down. And I'm going to say the bromonium ion is going to form this way. So the bromine's going to form on top here. And so there's going to be two lone pairs of electrons on that bromine. It's going to have a plus 1 formal charge. And if I look at this carbon right here, that's this carbon. So I'm going to say that my methyl group is now going down in space. So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here is H2O, and I think about which carbon will the oxygen attack? So I have two options, right? This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the option is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation, or a partial carbocation in character. So you could think about it as being like a partial secondary carbocation on the left here, if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red. And if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character, and that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons here we kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. OK, so what would we have here? Let's go ahead and draw our ring. And the bromine is going to swing over to the carbon on the left. It's now going to have three lone pairs of electrons around it, like that. And the methyl group that was down relative to the plane is going to be pushed up when that water nucleophilic attacks. And now the methyl group is up, and this oxygen is now going to be bonded to this carbon. And so we still have our two hydrogens attached to it, like that. And there's a lone pair of electrons on this oxygen, giving it a plus 1 formal charge. So once again, let's go ahead and highlight those electrons. I'll draw them in blue here. These electrons right here, those are the ones that formed this bond. So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here. So on the left I showed the bromine adding from the top. If I think about the alkene portion of my starting material, well, there's a chance that the bromonium ion could form from below that plane as well. So let's go ahead and draw the result of that over here on the right. So I'm showing another bromonium ion that is possible. And this time the bromine is going to add from below the plane, like that. It's going to get two lone pairs of electrons and have a plus 1 formal charge just like usual. And I will say that this is the carbon that has the methyl on it, OK? So let's go ahead and draw that in as well. So let's see, I'll put it like that. So that's my CH3. And this time, when I think about where will water attack-- so let's go ahead and think about water as my nucleophile-- it's the same idea if I compare the carbons on either side. So I compare this carbon with this carbon. It's the carbon on the right that's going to be the most stable partial carbocation, right? So I'll draw a partial carbocation here. It would be the most stable one. So that's where my nucleophile is going to attack. So I can think about this lone pair of electrons on oxygen that are going to attack right here, which would kick these electrons off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. So let's see what that would look like. So I have my ring like that. And now I'm going to have my oxygen bonded here. It still has two hydrogens attached to it. It has a lone pair of electrons on it, which give it a plus 1 formal charge. And when the oxygen attacked, that is going to push down this methyl group. So this methyl group is going to be pushed down relative to the plane of the ring. So now we have a methyl group down at this carbon. And the bromine is going to swing over to the carbon on the left. And so that's the position of my bromine now. And so, in the last step-- now I have these two guys right here-- they're both going to lose a proton in the next step, right? So it's an acid-base reaction. And we could show water coming along. So for the molecule on the left, water comes along. And water's going to take this proton. These electrons are going to kick off on to that oxygen. And we can draw that product. So if we were to draw that product, we would look down this way. And we would treat this as being the top carbon here. So there's a methyl group going up at that carbon. So I can say that this is going to, after it loses a proton, so that carbon in blue is going to have a methyl group up relative to the plane of the ring. And it's going to have an OH group down. So this OH is going to be down relative to the plane. So I can go ahead and put OH going away from me in space. And then this bromine over here. This bromine is going to be coming out at me. So I can go ahead and show a wedge for that bromine. So that's one of our possible products. Over here on the right, if this is what happens on the right, I do the same thing. I put my eye right here and I stare down, with this carbon being the top carbon. So I can go ahead and draw my cyclohexane ring. And I can see that this time my OH will be coming out at me after it loses a proton. So if I wanted to, I could go ahead and draw water in here and show the last step of the mechanism. Lone pair of electrons taking this proton, leaving these electrons behind, giving me an OH coming out at me in space. And then this methyl group would therefore be going away from me in space. So I can go ahead and show that methyl group as a dash. And then finally, this bromine over here would be going away from me. So that would be a dash on my ring, like that. When I finally get to my two products, I can analyze them. And I can see that they are enantiomers to each other. So they're different molecules. And we can look at the absolute configurations really fast. I can see that I have a carbon coming out at me on a wedge and an oxygen going away from me. And it's been reversed over here on the right. This time the oxygen is coming out at me on a wedge, and the carbon is going away from me on a dash. When I look at the bromine, it's coming out at me on the left, and the bromine's going away from me on the right. So I can see that I have different absolute configurations at both chirality centers. And so these two will be enantiomers to each other.