Electron-pushing mechanism of a hydroboration-oxidation reaction.
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- This is one of the more complicated mechanisms that I've seen here. Can someone tell me the significance of this particular reaction in real life? For example, does it occur or affect certain biological processes? Knowing these things really help me appreciate and remember the information better.(14 votes)
- not in biological processes but it is useful in generation of alcoholwhich tends to form keto enol form which is industrially very important for the synthesis of big molecule(6 votes)
- I noticed that NaOH is included in the list of reagents. What happens with the sodium throughout this process? What is the purpose of the NaOH if there is an OH- is made during the alkyl shift?(3 votes)
- The Na⁺ ions are simply spectator ions during the whole process. Their function is to balance the charge of the negative ions.
The NaOH serves two main functions:
1. It converts the hydrogen peroxide into a more nucleophilic hydroperoxide ion.
HO⁻ + HOOH → HOH + ⁻OOH
2. It converts the borate ester to the alcohol.
3HO⁻ + 3(RO)₃B → [3RO⁻ + H₃BO₃] → 3ROH + BO₃³⁻
You definitely need NaOH for Step 1.
Although OH⁻ is formed during the alkyl shift, the extra OH⁻ is useful in driving the reaction to the right.(15 votes)
- Is this always anti-markovnikov reaction ?(4 votes)
- BH3 cannot exist on its own, it always exists as diborane, then how can we isolate BH3 for the mechanism ?(5 votes)
- Also, the THF here supports boron to be present as BH3. In normal situations, boron would dimerize to form B2H6. but in the presence of THF, it is stabilized and present as BH3. So that's why we also need the BH3.(4 votes)
- why must alkyl group shift to OOH which oxygen already have complete octet state ? Could carbon just leave and be protonated by H2O producing alkane instead ?? It seems more reasonable....(4 votes)
- The O-O bond is the weakest bond present.
When the OH⁻ group departs, it leaves a positive charge on the blue O atom.
The alkyl migration satisfies the oxygen and also gets rid of the formal charge on the B atom.(5 votes)
- Do we eventually get the 3 alcohols that were attached to the Boron (as alkoxides), or do we only get 1?(4 votes)
- I am lost. What do you mean that the process happens two more times? How can those H's in BH2 be replaced by those alkane groups?(3 votes)
- If 1 H can be replaced, then 3 H can be replaced.
Just as RCH=CH₂ + BH₃ → RCH₂-CH₂BH₃, we can get
3RCH=CH₂ + BH₃ → 3(RCH₂-CH₂)₃B.(1 vote)
- Is there anywhere y'all recommend finding practice problems for alkene reactions(2 votes)
- Here are some links that you might find useful.
- after formation of trialkylborane, why is OOH- added to Boron? because most of OH- became OOH-? then I want to ask same question about trialkoxybornae getting OH-. is it about small amount of OH- that's formed in process of migration of alkyl group?
so What I'm guessing is that when both OH- and OOH- are present, stronger nucleophilic OH- is added to boron. is it correct?(2 votes)
- [Voiceover] We've already seen the general reaction for a Hydroboration-oxidation and in the previous video we did this as one of our practice problems. We started with this alkene and we got this alcohol with the OH added on to the less substituted carbon. Let's take a look at the mechanism for this reaction. So in step one we add our borane BH3, and if you look down here at the dot structure boron is sp2 hybridize which means trigonal plane or geometry around the boron and also an MTP orbital, which is capable of accepting a pair of electrons because as you can see boron has only six electrons around it and the fact that it lacks and octet makes it very reactive. Actually, borane can react with itself and that 's why we have the THF here to stabilize it. So when borane approaches our alkene... Alright, so here's our alkene. So we have the two methyl groups right here, and then we have the two hydrogens on this carbon. Alright so here are the two hydrogens. The borane approaches the carbon on the left side of our double bond. So the boron gets closer to the carbon on the left side, and one reason for that is because these methyl groups are relatively bulky, right. So there's some steric hindrance that prevents the boron from getting to close to the carbon on the right side of the double bonds, and the pi electrons are going to attack the boron. Right, so the pi electrons are going to attack the empty orbital of the boron and we're gonna form a bond between the carbon on the left side of the double bond and the boron. So these pi electrons are going to move into here, and as that bond between this carbon and this boron is forming we are withdrawing electron density from the carbon on the right. So I'm only going to put a partial positive charge here, because this is pretty much a concerted mechanism. This step is concerted but as this bond forms right, we're increasing the partial positive charge on this carbon and that triggers a hydride shift. Remember what hydride is. It's a hydrogen with two electrons, so a negative one formal charge. So we can think about a hyrdride being right here. So let me mark these electrons in blue. Right, so these electrons down here in blue are attracted to the developing partial positive charge. Alright so we're going to form a bond right here. Between this carbon, right, between this carbon and the hydrogen as the electrons in blue move in here. So the formation of this partial positive charge triggers the hydride shift and we move those electrons in blue there to form a bond. So let's draw the product. So on the left side of where the double bond used to be, we would have BH2 and on the right side we would have H. We would have a methyl group coming out at us in space and a methyl group going away from us in space. On the left side we would have a hydrogen coming out at us and a hydrogen going away from us. So let's follow those electrons along. The electrons in magenta right here form this bond between the carbon and the boron and the formation of the carbon-boron bond triggers the hydride shift. So the electrons in blue move into here and we get that. Alright later in this mechanism we'll see that the OH is gonna go where the boron is, right. So the OH is going to go there and so now we can talk about the regiochemistry and the stereochemistry for this reaction. So let's first think about regiochemistry. Why does the OH add on to the less substituted carbon? Well that's because of two reasons, right. One reason would be the steric hindrance, right. So these methyl groups prevent the borane from getting on or getting close to the carbon on the right side of the double bond. So steric hindrance is one reason, and also the formation of the, the developing partial charge on this carbon. It's better stabilized by the presence of these methyl groups, right. So, we have two reasons steric hindrance and the stabilizing factor of these methyls. Right, stabilizing the developing partially positive charge that explains why the boron adds on to the carbon on the left side of the double bonds, and then eventually where of course the OH goes. So that explains the regiochemistry, and the stereochemistry we see the OH and the H add on to the same side because this is pretty much concerted. Alright, this pretty much happens at the same time and so your stereochemistry is locked in at this portion of your mechanism. Let's continue on with the mechanism. Let me redraw this guy right here. So this is a little bit too complicated. Let me make it a little bit easier to look at. So we would have our methyl groups right, and then we would have our BH2 right here. This process occurs two more times. Alright, repeat times two because there are two more hydrogens on the boron, and so we actually form a trialkylborane. So let me draw in the boron with three alkyl groups on it. So it should look like this to form a trialkylborane. Some textbooks just leave it as a monoalkyl borane and proceed to the oxidation part of this reaction, but most of the time it's a trialkylborane. So that's what I'm gonna show you here. When you move on to the oxidation step, remember this is the second step here. You have hydrogen peroxide and you have hydroxide ions. So let's draw those out. Alright, we have some hydrogen peroxides. Let me draw the dot structure for hydrogen peroxide. Put in low end pairs of electrons on the oxygens, and we have hydroxide so OH minus. Hydroxide's gonna function as a base. It's gonna take a proton from hydrogen peroxide leaving these electrons behind on the oxygen. So we would make the hydro peroxide ion. So I'm sketching that in over here. Three lone pairs of electrons on the oxygen on the right therefore a negative one formal charge. So we can show these electrons in magenta coming off on to the oxygen. So we have the hydro peroxide ion which is gonna function as a nucleophile and our nucleophile is going to attack the empty orbital of our borons. The nucleophile attacks the trialkylborane, and let's show the result of that. So, now there's a bond between the boron and the oxygen. Right, then there's an oxygen oxygen bond and then a hydrogen here and let me draw alone pairs of electrons on the oxygens the lone pairs of electrons here and then boron still has our three alkyl groups. So, I'm sketching in the alkyl groups here like that. The next step of the mechanism... Oh, we have a negative one formal charge on the boron here. The next step is the migration of an alkyl group. So this is the weird part. Alright so these electrons right here are actually going to form a bond with the oxygen. At the same time these electrons are gonna come off on to this oxygen. So, we form hydroxide. Let me go ahead and show that first. That's maybe the easier part to understand of what's going on here. So we have three long pairs of electrons on the oxygen, negative one formal charge, we're gonna make these electrons green. These electrons in green here come off on to the oxygen. So the oxygen oxygen bond is weak. So, it's relatively easy to break this bond, and let me highlight the other electrons here. Well, these electrons in magenta formed this bond between the oxygen and the boron, and now we get our alkyl group migration and I'll show these electrons in the blue. So the electrons in blue are forming a bond between this carbon and this oxygen. So let me draw that out. It's pretty hard to see. Let me go ahead and draw what we have and then we'll highlight some electrons here. So now we have our oxygen bonded to this group and our boron has two alkyl groups still bonded to it like that. So let's follow those electrons. The electrons in magenta represent the bond between the oxygen and the boron, and the electrons in blue, right, would be these electrons right here. So it's this carbon is now bonded to this oxygen. So the migration of the alkyl group removes the formal charge on the boron and it breaks the weak oxygen oxygen bond and kicks off hydroxide. Alright so now this process happens two more times. Alright cause we have these two other alkyl groups, and so when that happens two more times we get, let me go ahead and draw it in, a trialkylborane. So we'd have an oxygen here and then we have our group coming off of that and then over here we would have our oxygen and then we have our group coming off of that and then also down here. Alright, our next step is the hydroxide anion functions as a nucleophile. So the hydroxide attacks the boron. Once again our boron has an empty orbital. So nucleophilic attack let's draw the results of our nucleophilic attack. Now we have OH bonded to the boron and the boron still has all of these groups around it. So let me sketch all of those in, and you can see why this mechanism is getting rather cumbersome at this point. Right, drawing in all these groups is a little bit annoying here but we still have a negative one. I should say we have a negative one formal charge and a boron and we still have some lone pairs of electrons on this oxygen and I'm putting those in because in the next step to get rid of the negative one formal charge these electrons come off on to the oxygen and that would form alkoxide anion. Now we have this oxygen would have a... Let me do a better job of drawing in our alkoxide anions. So we'd have three lone pairs of electrons on our oxygen. That's a negative one formal charge, and let me highlight those electrons. Alright, so these electrons in here in magenta come off on to the oxygen and reform our alkoxide anion. We are almost done, right. So in the last step we have water present. So water can donate a proton to our alkoxide anion. So the alkoxide anion picks the proton up from water. We protonate the alkoxide anion. We form our alcohol and so we're finally done. Very long mechanism a lot of drawing but we formed the alcohol that we know cause we know the OH added on to, let me highlight it here, this carbon. So again you could probably do this mechanism just doing the monoalkyl borane and I'm sure some professors will let you do that. Most of the time it's the triakylborane. So I drew all this out even though it was... It definitely got a little tedious here, but there's your mechanism for hydoboration-oxidation.