so we've been looking at electrophilic addition reactions and one of the last reactions we looked at was how to convert an alken to an alcohol and to do that we used the acid catalyzed addition of water so we had an electrophile which was our hydrogen ion that we added to our alken and then water came in as our nucleophile and added to the carbocation and produced the alcohol so when we saw this happen we saw that the hydrogen that electrophile was going to add to the carbon that already had the most hydrogen's bound so let's say we wanted to make something like one butol well with the reaction we've learned we wouldn't be able to because that's not how the mechanism of that reaction worked so what we're going to do today is we're going to look at another way to convert an alken to an alcohol and the reaction we're going to look at is two successive reactions those two successive reaction are known as hydroboration oxidation so there's two parts basically the first part that we're going to see is a hydroboration and the second part is an oxidation now when we look at this reaction keep in mind that you that an atom or a molecule doesn't actually need a positive charge to be a nucleophile or to be an electric file excuse me so the compound we're going to be looking at Boran which is bh3 is an electrophile because because boron has an incomplete octet Boron and this Boran as such is going to be SP2 hybridized so it's going to have three SP2 hybridized bonds to hydrogen to the sigma of hydrogen Sigma bonds to hydrogen to the S orbital and then there's going to be an empty p orbital and that empty p orbital is going to be able to accept a share in an electron pair so just like the methyl cation is a is an electrophile because it has a positive charge it also has this unoccupied p orbital that can accept a share in an electron pair and we see the same thing for Boran even though it's not positively charged so what we're going to look at in this lecture is we're going to see that a reaction that we have involves Boran and that Boran is going to be our our electrophile that adds to our that adds to our alken so before our electrophile was a hydrogen ion now it's boring because it has the same um incomplete octet that can share that can accept a share in an electron pair and so that's going to add to an our alken and what we're going to see in this reaction is that our nucleophile is not water here but a hydride ion that's bound to the Boron okay so we're going to look at this hydroboration oxidation reactions that happens and figure out exactly how we can make something like one butenol so this Hydro boration oxidation reaction happens by two steps this is indicated by the one and the two so what this means is that one step has to completely be over before we add the reagents for the next step it's basically a series of sequential reactions we do this reaction first labeled one followed by the reaction that's labeled to so there's two parts to this reaction um and the first step is the addition of Boran in thf tetrahydrofuran um to the alken the second step is the addition of an aquous solution of sodium hydroxide indicated by this o minus and H2O and hydrogen peroxide so the first step is that hydroboration the second step is going to be the oxidation and ultimately what's going to happen here is we're going to have an alcohol that's formed from these two sequential steps now something that's unique about this alcohol is that notice that it's kind of the opposite addition of what we would expect so if we had an acid catalyzed addition of water we would expect to form two propenol we would expect our electrophile our hydrogen to addh here and then our nucleophile our water would adhere to that carbocation so we couldn't ever form one propenol if it from the acid catalyzed addition of water but if we use Boran tetrahydrofuran followed by that oxidation with an aquous solution of sodium hydroxide and peroxide we can make one propenol we can add the alcohol to the first carbon rather than the second carbon and we're going to see why that's the case now something else that's important is boring and tetrahydrofuran the purpose of tetrahydrofuran thf is to stabilize the molecule to stabilize Boran if that wasn't there then we would actually form something called dorine and dorine is a very dangerous very explosive flammable and toxic gas so this tetrahydrofuran just makes the Boran a little bit safer to work with because it is so reactive it tends to form or try to form bonds and instead of of forming an unstable bond with it with another molecule of itself will give it something to form bonds with okay so borine comes in the form of the solubilized tetrahydrofuran borine mixture so as I mentioned already we actually get two separate two different products whether we undergo the acid catalyzed addition of water with an alken or whether we undergo hydroboration and oxidation so notice we start with the same alen in both cases but under the acid catalyzed addition of water conditions we produce two propenol under the hydroboration oxidation conditions we produce one propenol and this allows us to put the alcohol on a different place within that molecule so if we wanted to form one propenol instead of two propanol we couldn't use the reaction conditions for the acid catalyzed addition of water we would have to use the hydroboration oxidation reaction conditions so you can see here that the alcohol that's formed from this reaction is going to kind of as you can kind of see have the hydrogen and O group switched compared to the alcohol that's formed from the acid catalyzed addition of water you might think well Dr Wise how in the world are these reactions happening then if nucleophiles attack electrophiles right shouldn't we produce the same thing and the answer is yes and no so yes nucleophiles attack electrophiles but here the electrophile is going to be something different and here the electrophile is going to um add to the SP2 carbon that's bound to the most hydrogens so our electrophile in the case of the acid catalyzed addition of water was hydrogen ion and it added to the carbon that was bound to the most hydrogens it added here now the same thing is going to be true for the hydroboration oxidation the electrophile is going to add to the SP2 carbon bound to the most hydrogens which is still this one but our electrophile is not a hydr a hydrogen ion here our electrophile is this Boran okay so our Boran our electrophile adds here and that's the reason why that the ox the O group is on the end because the o group replaces that electrophile that was added that we're going to see in this mechanism so to summarize again in the acid catalyzed addition of water we have the hydrogen ion that's the electrophile and water that's the nucleophile but in the hydroboration oxidation we're going to see that our Boran is the electrophile and the O group is going to subsequently take its place and our nucleophile is going to be a hydride ion okay so we're going to understand how this hydroboration oxidation reaction happens by looking at the mechanism for these two sequential steps in the next few slides of this lecture so our first step of the reaction is where we add Boran in tetrahydrofuran to an our to our alkane and this first step of the reaction is something called a concerted reaction and in a concerted reaction all of the bond making and bond breaking processes occur in the same step so we're going to see multiple things happening here because it all happens in one step of this reaction and so ultimately what's happening in this step one is that as your Boron right this is the specific atom that's the electrophile accepts those Pi electrons so we still have our nucleophile attacking our electrophile right so we've started with our nucleophile here attacking our electrophile that accepts those P electrons from the alken and it forms a bond with one of those SP2 hybridized carbons and as that happens it releases a hydride ion to the other SP2 hybridized carbon so this is all happening at once a bond is forming here between the pi electrons and Boron and a bond is breaking here and reforming a bond between the carbon and this hydride ion so remember a hydrogen with its two electrons it's going to be negatively charged and it's a hydride ion okay so this is all happening at once and from this reaction what we see is that this um Boron dihydride here is going to be adding to our SP2 carbon that's bound to the most hydrogens right our electrophile is adding and our hydride ion which is our nucleophile kind of two nucleophiles in this step is going to to be adding to this carbon so the next thing we're going to look at is why the Boron is going to add to this carbon and the hydride ion is going to add to this carbon right because theoretically you could have it the opposite way right what if the electrons here the pi electrons were going to form a bond between this carbon and the Boron right and the hydride would add here but it turns out it doesn't happen like that and there's a reason for that that we're going to look at so that addition of Boran and I hope I'm not boring you with this lecture um and the addition of hydrogen bromide follow that same rule the electrophile is going to add to the SP2 carbon bound to the most hydrogens so when we're dealing with an alken and those Pi electrons you're always going to have that electrophile add to the SP2 carbon bound to the most hydrogens and let's look at why that is the case so let's look at first the addition of hbr one of those familiar mechanisms that we've looked at already now when we look at this mechanism what we see here is that during the transition state when we add the electrophile to the carbon that's already bound to the most hydrogens we still have a partial positive charge on this other carbon right we have a carbocation like transition state here so as we form a bond between this um one of these carbons the SP2 hybridized carbons of the alken and our electrophile we get a positive charge on the other carbon it's carbocation like okay and we see that here as well right so we have this propan and you can see that if we form a bond with our electrophile in our first carbon here we're going to have somewhat of a positive charge in this transition state on a secondary carbon right there's two carbons attached to it but if we were to have our hydrogen that was bound to the carbon that is going to um so so if we were to have the um the hydrogen binding to our least substituted carbon then we'd have the partial positive charge forming on the more substituted so and this would be a primary carbocation is like thing that would form and remember the more stable carbocation or carbocation like substance is going to be the one that has it more substituted okay so because we're forming something that looks similar to a carbocation because we're forming somewhat of a positive charge on that carbon the transition state that's going to be the most stable is the one that's going to add the electrophile to the most substituted carbon because that means that that positive charge that's happening is going to be on the carbon that's bound to the most alkal groups okay so this is a less stable transition state and this is a more stable transition state and what we um see Happening Here is the same thing with the addition of Boran our electrophile is boron here and so we're adding our electrophile to the more or sorry to the less substituted SP2 hybridized carbon because adding our electrophile to the less least substituted SP2 hybridized carbon means that we're putting that positive charge somewhat of a positive charge even though we're not ever fully forming a carbocation here but that transition state is carbocation like and it's going to be more stable to put that somewhat of a positive charge as that occurs on the more substituted of those two carbons so just to kind of summarize What's Happening Here the acid catalyzed addition of water and the hydroboration follow the same r rules the electrophile is binding to the carbon that's the least substituted the carbon that has the most hydrogen detached and that's because the transition state that forms is going to be more stable when that occurs when the electrophile adds to the SP2 hybridized carbon it's already bound to the most hydrogens because the transition state is carbocation like it has somewhat of a positive charge on a carbon and it's going to be the most favorable the most stable to put that positive charge on a more substituted carbon so what we have when this first step of the reaction is over is we formed an alkal borine that alkal substituted bh2 and that alkal substituted bh2 can still react there's still two hydrogens that are attached to the Boron so it can react with another molecule of the alken to form a dialy Boran right so this is what we form in our first step it can react again and form another molecule with an alal group that's been already attached and another hydrogen here and then there's still another hydrogen that can be removed and so it can react again a dialy borine can react with another molecule of the alkane to form a trial Boran and at this point the trial Boran can't react anymore because all of the hydrogens are gone because remember we needed a hydride ion to leave to be a nucleophile for this second carbon here so in all of those reactions we have the Boron that's adding to the SP2 carbon that's bound to the most hydrogens and the hydride ion right the Boron with the hydride the hydrogen is going to add to the other SP2 hybridized carbon so when we think about the alkal Boran and the dialy Boran those are bulkier than just Boran right the alkal Boran and the dialy Boran are because they have a carbon group off of the side they're much bulkier than our original starting point which was boring so ultimately what I'm trying to say here is that if we start with Boran if we start with Boran we have a few unwanted products that you can that we're actually going to produce since it has three potential hydride ions you can form more substituted carbon groups and sometimes you might want that but other times you might not only one hydride ion is needed for this process of hydroboration so often times what you see is this compound being used for a hydroboration reaction this comp compound is called 9 BBN or nine borao 3.3.1 nonin we'll go with its nickname here right 9bb n but understand what it is we have Boran essentially but it's going to be disubstituted there's two alkal groups that are attached to this Boran and one hydride ion and so there's a couple reasons why this is advantageous first of all there's only one hydride ion so you're only going to get your Al alkan that is going to have Boron on only one of those carbons right you're not going to get like dialy borine trial borines right you're not going to get borons with three alkal groups attached which is a good thing depending on what you want right if you just want to add water to that double bond or not water if you just want to add an O group to that double bond you don't want the disubstituted or the tri substituted Bor um alkal borines you just want one thing put on that and so if you just want one thing put on that you need to use something that just has one hydride ion so the advantage of this is that it has only one hydride ion it's going to only going to add one time to your alken the other reason why this is advantageous is because it has very bulky R groups and because of its very bulky R groups it's going to have a much stronger preference for the less substituted SP2 carbon so ultimately why we would use this is to only get one of our Pro one product not to have a non-specific reaction where we would have more alkal groups added to our compound than what we want and we have more of a preference for that less substituted SP2 carbon because it's bulkier it's more likely to add to the end of the molecule rather than in the middle which is again what we want because we already have a way to form an alcohol with the least or sorry with the more substituted carbon right we don't have a mechanism yet well we're learning a mechanism to add to the least substituted carbon and form an alcohol from that so the mechanism from this dialy Boran is going to be the same you have your Boron which is still your electrophile so the nucleophile your P electrons are going to attack your Boron your electrophile and then in this concerted reaction again you have at the same time this is happening at the same time a bond is forming between the less sorry yeah the less substituted carbon and the Boron another bond is forming between this new nucleophile and this forming electrophile okay remember this is going to be the one that has somewhat of a positive charge on it because it's going to there's a bond that's going to be forming here so those electrons are breaking here adding here so we de somewhat of a positive charge and so we've got now a new nucleophile that's forming and a new electrophile that's forming which attacks here so again the products that we just looked at for our previous mechanism for hydroboration are the same as what we would get from the hydroboration of a dialy borine and then what's formed then is our trial borine but this this trial borine these two alkal groups is what we want here right we want the um 9 BBN where we have those two alkal groups in the Boron so the mechanism with BB9 9 BBN sorry is the same mechanism as what we saw earlier so we've talked about the hydroboration we now know that we can add that Boran to our Al and we're going to form this trial borine but we haven't produced our alcohol yet so what else do we need to do right we need to replace our Boran with our oh group with our alcohol sorry to form our alcohol and so the next uh step the next part once we've completely reacted and formed our trial borine we need to take that compound that we formed and put it in an aquous solution of sodium hydroxide and hydrogen peroxide and what that does is it replaces that br2 group with an O group and so essentially what we're doing in this next step is actually forming our alcohol now this is called an oxidation reaction and the reason why it's called an oxidation reaction is because the number of carbon oxygen bonds are going to increase so when we say oxidation we're simply talking about a reaction that increases the number of carbon oxygen bonds or decreases the number of carbon hydrogen bonds so again ultimately what we see here is we've taken our product from our first step of our sequential reaction our trial borine and now we're replacing this borin with an O group to form our alcohol and this is done in the presence of aquous sodium hydroxide and peroxide and in that case we will form our alcohol and so we're going to look at the mechanism on the next step slide so it looks like there's a lot happening right but we've still got the same concepts of our nucleophile reacting with our electrophile so ultimately here what you need to realize is where your nucleophile is and where your electrophile is we've already seen that Boron is going to be our electrophile and our nucleophile here is going to be the hydrogen peroxide ion that forms in a in a in a basic solution here and so our hydrogen peroxide ion is going to basically be hydrogen peroxide but without one of its hydrogens hydrogen peroxide is H2O2 and so if it's lost one of its hydrogens because it's a basic solution it's going to just have its um electrons remaining have the one of the oxygen negatively charged and then the O attached here now this is a very unstable species so it's very highly reactive and so you've got a very strong nucleophile reacting with our electrophile here so the pair of electrons this lone pair of electrons on this uh first oxygen atom is going to form a bond with the Boron because remember that boron has an empty p orbital that means it can act as an electrophile and so we formed now a bond between the oxygen of the hydrogen peroxide ion and this Boron atom now the next thing that's going to happen in this reaction is a one two alkal shift and this one two alkal shift is going to displace a hydroxide ion and what did does this does here is it allows the Boron to no longer be negatively charged so Boron because it's um towards the left side of the periodic table it's actually very unstable with a negative charge on it it doesn't handle negative charge well so it's actually not very happy in this state this is a high energy State and so what's going to happen here is this one two alkal shift where the alkal group is going to shift over to this oxygen so the alkal group takes its electrons with it and shifts to the oxygen and then there's a pair of electrons here with the between the two oxygens because this is a highly unstable Bond that's then going to go onto this particular oxygen so let's look at the compound we formed now right we've shifted the alkal group over and we've gotten rid of that hydroxide ion okay so our Boron is back to three bonds where it's happier than having a negative charge but still can be an electrophile um because it only has three bonds and now we have a new nucleophile we've put in solution we have our nucleophile here and so we have our hydroxide ion that is now our nucle nucleophilic sub species that's going to attack our electrophile so after that shift the hydroxide actually adds to the Boron and reestablishes that negative charge but as we as we've mentioned this isn't very stable this isn't a stable situation so what's going to happen is that the alkal group with its oxygen attached the Oxygen's going to take its electrons take its alkal group and leave so we've eliminated now an Al coxide ion and so we have NOW Boron with its two alkal groups and now the O group attached and the AL coxide ion and then what happens next is that after that Al coxide ion is eliminated and Boron is no longer negatively charged it can gain a proton from the solution from water potentially to form the alcohol okay and so what you can see here is that a lone pair on the alkoxide ion is going to attack a hydrogen on the water and then the electrons that were existing between the hydrogen and the oxygen are now going to belong on the oxygen so we've produced hydroxide and the alcohol so the hydroboration that first step we looked at was adding the electrophile to the alken and now the Second Step we've looked at the oxidation part is replacing that Boran with an O group and that o group is going to end up attached to the SP2 carbon that's already bound to the most hydrogens again because it replaces Boron which is the electrophile in that first reaction the hydroboration reaction so the second step is the tricky step right there's a lot happening here to oxidize the um compound that we formed from our first step to our alcohol but if you follow the nucleophiles and electrophiles and how they react with each other then you can follow the mechanism of the reaction to producing the alcohol right so starting with the hydrogen peroxide ion right that's where we um have our nucleophile we know that Boron is our electrophile right it's attacking that and then we're going to follow that uh form a bond here between these two realize that after we form this Bon's really unstable with the negative charge so it's going to shift around so Boron doesn't have a negative charge anymore right and one way to do it is that this could come off it could be reversible and as you can see these arrows are reversible all of these reactions are going to be reversible so we could start with an alcohol and go backwards and form this um compound with Boron attached to it right but ultimately again all that's happening is that we are um arranging this so that we're forming more stable compounds we're taking our alil group shifting it over getting rid of the hydroxide so that we form NOW Boron with three bonds rather than four but we've got a new nucleophile right our electrophile is still Boron and then we're just attacking our nucleophile electrophile reforming Boron with a negative charge again not stable so now instead of the O group getting kicked off the alkal the alkoxide ion gets kicked off and then once the AL oxide ion gets kicked off and we've separated it just protonating it's going to form our alcohol now this reaction was very advantageous to us because there was no carbocation rearrangement because we're not actually forming a carbocation there is no rearrangement of carbocations right there's no carbocation rearrangement because no carbocations are actually formed from that reaction so what we what this means is that when we have these compounds we actually get what we expect we get the alcohol ultimately on the carbon that's going to be bound to the most hydrogens rather than having to worry about the carbo rearrangements that are present when we have um the um acid catalyzed addition of water so this section only has two homework problems if you're looking for a few more you can always look in the end of chapter problems I'd be happy to help you with any of those as well that you're working through so I know those reactions were a lot especially the second one so do let me know if there's any questions or clarifications that you guys need