Two-step reaction in which an alkene is converted to an alkene with a hydroxyl substituent (which may tautomerize to a carbonyl). Created by Jay.
Want to join the conversation?
- Why is this reaction anti-Markovnikov?(5 votes)
- Whenever you see the following reagents, it will be an anti-markovnikov addition.
2. H2O2/NaoH/H2O(19 votes)
- What is the purpose of THF and hydrogen peroxide in this reaction? Someone else answered that the boron in BH3 is oxidized to an OH group to make the overall reaction anti Markovinikov and the OH- group is the base in base catalyzed tautomerization but I don't see where the other molecules come in.(3 votes)
- BH3 is violently reactive on its own, so THF is used to stabilize the molecule. The BH3 essentially accepts an electron pair from the THF to complete its octet and form a stable BH3-THF complex. Think of THF as a calming agent. As far as the others are concerned: NaOH serves to remove a proton from H2O2, making it HO2, a better nucleophile (conjugate bases are always stronger nucleophiles). HO2 then attacks the boron to complete the rearrangement reaction, and the O is ultimately protonated with H2O to form a neutral alcohol. Keto-enol tautomerism kicks in here and the (C=O) bond is highly favored.(4 votes)
- In the resonance structures, the first one should be more stable than second [as structure 1 has -ve charge on more electronegative element ie.oxygen & hence conc. of enol >conc. of aldehyde [in products] but it is not so.why?(2 votes)
- We must consider not just the stability of the O atom, but the stability of the system as a whole. A C=O bond is more stable than a C=C bond, so the position of equilibrium favours the aldehyde.(4 votes)
- Are you able to use hydroboration on a symmetrical internal alkyne? If so would it also be a mixture of both ketones, or would BH3/H2O2 force the ketone to create a more stable molecule?(3 votes)
- Why does he bring up resonance structures, and how are you supposed to know when to do a resonance structure?(3 votes)
- The molecule wishes to become more stable, which means to shift the double-bond to the oxygen; from that point, it's much easier to stabilize it.
If you can't see the clues (example: an atom with lone pairs of electrons nearby a double bound) which tell you that there might be a resonance structure, you should go back and re-watch the video about Resonance Structures.(1 vote)
- In my textbook instead of using BH3 it uses Sia2BH, which I believe is a dialkylborane. I want to know if I can use this reagent instead of what was shown in the video?(2 votes)
- Sia₂BH or disiamylborane is bis(1,2-dimethylpropyl)borane. It is a hindered borane, so it is selective for the hydration of terminal alkenes and alkynes. If you have a terminal alkene or alkyne, then you can use Sia₂BH. If not, then you have to use BH₃/THF.(3 votes)
- Would it not be a more efficient way of teaching regiochemistry to have the alkyne have a more subsitued carbon and a less subsitiued carbon? That way you could see the regiochemistry taking place?(1 vote)
- With an alkyne, all you can have is a terminal or an internal triple bond.
The regiochemistry is important only for a terminal alkyne.
You can't make reliable predictions for an internal alkyne, because there is an alkyl group on each end of the triple bond.
Steric factors then become more important than electronic factors.(4 votes)
- can you get a geminal diol, by hydroboration of the enol?
the geminal diol would not be stable right? and could throw out water which would also lead to a keton/aldhyde. or is this reaction unlikely?(2 votes)
Here's the hydroboration-oxidation of alkyne reaction. So we start with our alkyne. And, once again, usually it's a terminal alkyne. So here's your hydrogen here. And on the other side of our triple bond, let's say there's some R group attached to this carbon. So two are alkyne. We're going to add, in the first step, borane, BH3, in THF, as our solvent. And in the second step, we're going to add hydrogen peroxide and hydroxide anion OH minus. And we're going to get addition of water across our double bond. So it's very similar to the last reaction. The difference here is the regiochemistry. This regiochemistry is actually anti-Markovnikov addition. So this is antimark's rule. So anti-Markovnikov, meaning the OH is going to add to the least substituted carbon, which, of course, must be the carbon over here on the left. So we end up adding H2O, water, across our triple bond. So now there's a double bond present, and then there's an OH in our molecule. So just like in the last video, we form an enol here like that. And the enol, of course, is not the most stable form. We're going to get some rearrangement, and we're going to end up with an aldehyde tier. So we can see we get an aldehyde functional group from a terminal alkaline undergoing this reaction. Now, sometimes if you use borane in this first step over here, sometimes borane is so reactive that it's actually going to add two molecules across your triple bond, as your triple bond consists of two pi bonds. So sometimes you'll see a different molecule used. Sometimes you'll see a dialkyl borane. So put two alkyl groups. I'll put R2BH and if you make them really bulky the steric hindrance prevents the second addition of the borane. Because in the mechanism, you want only one addition of the borane. And the borane is going to add onto the same carbon that the OH does. So check out the video on hydroboration-oxidation of alkenes for much more detail of that first mechanism that we discussed. So the net result of this reaction is to form an aldehyde from your terminal alkyne. So let's look at a reaction here. So let's make our triple bond right here like that, and we'll make it a terminal alkyne, and we'll make this side a benzene ring. So we'll make a rather large R group here. So let me put in some electrons on my benzene ring. So let's just go like that. All right, so if I react this alkyne with either a dialkyne borane or I'll just put borane here, and you could use a dialkyne borane if you want to-- whatever your professor wants to use-- and THF. And second step, hydrogen peroxide and OH minus, like that. So the first thing you have to think about is, what's going to happen? I'm going to be adding water across my triple bond. And I need to think about what side gets the hydrogen and what side gets the OH. So, once again, I have two choices. I could add the OH onto the carbon on the right, or I could add the OH to the carbon on the left. And the regiochemistry here is anti-Markovnikov, so I'm going to add that OH to the least substituted carbon, which, of course, is the carbon on the right side. So I'm going to end up adding the OH to the carbon on the right side. So let's go ahead and draw our double bond in here. First, let's go ahead and draw our benzene rings. So I'll go ahead and make it like this. And that benzene ring is connected to a carbon. This carbon right here. There's now a double bond presence like that. And I added the OH to the least substituted carbon. I added the OH to the carbon on the right like that. And I add a hydrogen to the carbon on the left. So that's adding my H20 across my triple bond. And then there was still a hydrogen on this carbon on the right like that. So that's my enol intermediate like that. So let's take a look at what's going to happen to that enol intermediate. It's going to rearrange. And on the last video, we saw an acid-catalyzed tautomerizaton. In this video, base is present. So we have our hydroxide anions in the solution. So let's go ahead and redraw our enol, and show the enol reacting with base. So let's go ahead and draw the enol here again. So I'll draw that enol on the right side. So there's my benzene ring. And I have my double bond in here like that. And I know I have an OH like that. And put in my lone pairs of electrons. And I'm going to react this with hydroxide anions. So hydroxide anions are floating around. Hydroxide is negatively charged. So hydroxide will function as a base. So let me get in my lone pairs of electrons here on my oxygen. And I'm going to pick up that proton and kick these electrons off onto this oxygen. So let's go ahead and draw the results of that acid-base reaction. So it's an equilibrium. So let me go ahead and draw my equilibrium arrows like that. So what's going to happen? I have my benzene ring still. And I have my double bond. And let's see, now I have my oxygen with three lone pairs of electrons around it like that, so negatively charged. So now I can draw a resonance structure for this molecule. So let's go ahead and put in our resonance brackets here like that. So what can I do to spread that formal charge out throughout the molecule? Well, I could take a lone pair of electrons right here, and I could move it into here to form a double bond. And now I give too many bonds to this carbon. So the pi bond here is going to break and the two electrons here are going to move out onto this carbon like that. So let's go ahead and draw the results of the movement of those electrons. I still have my benzene ring like that. And there's my carbon skeleton here. Now I have a carbon double bonded to an oxygen. There used to be three lone pairs of electrons around this top oxygen. One moved in to form the double bonds, and now there's only two lone pairs like that. And the electrons that used to be in this pi bond have moved onto this carbon right here. So this carbon right here is still bonded to another hydrogen. So the two electrons moving out onto it are actually going to form a carbon anion here. So this is actually a negatively charged carbanion like that. And water is now present. OH minus fix of a proton become H2O. So now water is floating around like that. So now water is actually going to function as an acid. It's going to donate a proton. And this lone pair of electrons are going to take that proton. So they take that proton and kick these electrons in here off onto the oxygen atom like that. This is an acid-base reaction. So this is going to be at equilibrium. I'm going to go ahead and put in our other resonance bracket like that. So what are we going to get? And we still have our benzene ring untouched over here. And let's see, now we have our carbonyl like that. And we just added on a proton to this carbon right here like that. So now we are done. And you can see this is an aldehyde. I didn't draw on the hydrogen over here, so you can leave that out if you want to, but if you want to draw on your hydrogen, there's a hydrogen there. There's also a hydrogen on this carbon right here. I just didn't draw it in to show some clarity in the mechanism. So the end result is we form a an aldehyde from a terminal alkyne. And we showed this tautomerization using base. So let me just go ahead and write one more thing here. This is an example of a base-catalyzed tautomerization, where we go from the enol form to, in this case, an aldehyde as your product, which is more stable because of the carbon-oxygen double bond.