[Music] hey everybody welcome to the hydroboration oxidation lab here's a quick outline of what i'll be going over in this video i'm going to go over the reaction itself talk about the procedure and what we'll be doing in this lab with that i'll be going over markovnikov versus anti-markovnikov reactions and taking a quick look quick review about lewis acids and lewis bases and then i'll show you the actual experiment and at the end of the video i'm going to take a quick look at the nmr just because the carbon nmr on these molecules is a little bit tricky so i'll review that a little bit this is the reaction that we will be doing we're taking one octane and converting it into one octanol the first step would be the hydroboration step and then the second step will be the oxidation step notice that in the balanced equation we have these threes here and that's just because the boring compound has three hydrogens and so it can interact with three different alkenes simultaneously and that's shown here so if we look at the table initially we might assume that the boring thf complex is our limiting reagent because it has a smaller amount of millimoles but we have to remember that the one octane is actually used up three times as quickly and so that'll be our limiting reagent in this reaction the hydroboration oxidation reaction is an anti-markovnikov reaction meaning that the group that we are adding is going to be added to the less substituted side of the alkene usually in reactions we have a markovnikov reaction where the group we are adding is being placed onto the more substituted carbon and that just has to do with the carbocation intermediate it's more stable for that positive charge to be placed on a more substituted carbon so the group is added to that more substituted carbon but in our reaction we are placing the alcohol on the terminal carbon and that just has to do with how the boring interacts with the alkene and then gets substituted with the hydrogen peroxide we'll be manipulating some lewis acid lewis base chemistry in this reaction so for a quick review lewis acids are electron acceptors and lewis bases are electron donors so for example if we look at the boring molecule boron can make three bonds so bonds with three hydrogens meaning that it is sp2 hybridized and there's an empty p orbital so this will act as a lewis acid because electrons will be donated into that mtp orbital okay here's a quick look at the hydroboration step we're going to be using a bh3 thf complex boring as itself comes as a toxic colorless gas actually called diborine and when exposed to a lewis base like tetrahydrofuran or thf it forms a stable complex and that's more easily and safely handled than just a diaborium gas this complex is violently reacted with water so we want to make sure that our reaction vessel is dry so we're going to put our glassware in an oven to make sure no water is on it and we'll set up our reaction apparatus like so we'll have our conical vial where our reagents will be attached to a claisen head adapter which has two parts the first part will be attached to a drying tube so that it can absorb water from the atmosphere the second part will be attached to a cap and a septum so that we can add our reagents through a needle and a syringe therefore not exposing them to any moisture in the atmosphere once the reagents have been allowed to spin for about 45 minutes we'll add two drops of water to stop the reaction and move on to the next step which will be the oxidation step for this second reaction we'll be using hydrogen peroxide and sodium hydroxide the hydroxide will activate the hydrogen peroxide by deprotonating one of those oxygens so that it connects as a good nucleophile and attack the boron and eventually switch it out to become the desired alcohol and we'll allow this to reflux in this second apparatus for about an hour once the entire reaction is completed we will extract our product using ethyl ether or diethyl ether which is less dense than water so when we do our extractions the organic ether layer containing our product will be on top of the aqueous layer we'll rinse those extracts with some hcl just to remove any leftover base from the oxidation step then we'll dry that organic layer with sodium sulfate to remove any water and evaporate the ether solvent hopefully leaving our octanol with possibly just some unreacted one octane from the initial reaction to purify our product and to separate it from any leftover octene we are going to set up a plug which is similar to a chromatography column in the sense that we were going to manipulate you know something like polarity or in this case lewis acid lewis base properties to separate our product from any contamination and for this plug we're going to use silver nitrate treated activated silica gel wow that's a mouthful so this is what the plug will look like we'll plug our pasture pipette using some cotton then add a small layer of ottawa sand and add our silver nitrate treated silica gel after which we'll add another little layer of sand after this we can go ahead and add our our product on top and we'll loot that through the plug using an ether pentane solution the silver nitrate is crucial for this because if we use just normal silica gel which is a polar compound we would just be separating our two compounds based off of their polarity in which case octane being less polar would come out of the plug first would enter the conical vial and just be sitting there when the the actual product octanel came through and would contaminate it all over again but silver is a transition metal with a non-filled d orbital so when we elute the column the more loosely held pi electrons on the one octane will actually complex with the silver allowing the one octanol to loot out of the column by itself and once we've gathered about five milliliters of solution we'll evaporate the ether pentane solvent leaving our purified product and then we'll characterize that product just to make sure that we did produce the desired one octanol by using infrared spectroscopy i'm going to put all of the glassware in the oven this includes the conical vial with a spin vane the clayson adapter and the drying tube and we'll leave them in there for about 10 minutes just to make sure everything is dry here i'm just going to set up the apparatus real quick making sure after adding the claisen adapter to have the cap and septum so i can add the reagents i'll add calcium chloride to the drying tube to absorb water from the atmosphere just to make sure that water doesn't react with our boring complex i'll be using a needle and a syringe to get the reagents and just to equalize the pressure within the reagent bottles i'll be adding about the same amount of air as i will be removing reagent since the boring complex is violently reacted with water both of the reagents for the hydroboration step will be in a ventilated hood and each will have a cap and septum so that they can be removed with a needle therefore minimizing exposure to the atmosphere before adding the boring i'm going to be putting the conical vial into an ice bath because the bourne complex is very reactive we want to control the rate of reaction as much as possible and i'll also be adding the the reagent dropways over a period of five minutes just so that the energy released from this reaction doesn't cause runaway reactions that we don't want once everything has been added i can now take the conical vial out of the ice bath and let the reagents react together for about 45 minutes at room temperature once that is done i'll add two drops of water to hydrolyze any unreacted borane and stop the reaction now we can move on to the oxidation step of this reaction so i'll be adding the sodium hydroxide and the hydrogen peroxide this is also very reactive so i'll be adding the hydrogen peroxide dropwise over a period of about 10 minutes [Music] and we'll let that reflex for an hour now once the reaction is completed i'll take it off the heat so everything can cool down for our extractions and we'll be using ethyl ether for set extractions i'll remove the spin vane and then add a little bit of ether just so we can see the organic layer a little bit better while doing these extractions then i can remove the organic layer on top containing our organic product and then repeat this process two more times it was really difficult to see the separation of the two layers so it's very possible i got some water in there but that's what drying with sodium sulfate is for right [Music] this time i added some distilled water so we could see the layers a little bit better i'll rinse the combined ether extracts with some hcl to remove any base that might have gotten moved over along with the organic phase i'll go ahead and remove that aqueous acid layer and then do multiple washes with some distilled water until the remaining aqueous layer after those washes is neutral to a ph strip this means that all the acid has been removed and the aqueous layer is no longer acidic we're looking for a yellowish green color so that looks great i can remove the remaining water now and then i'll go ahead and move the organic layer to an erlenmeyer flask where we can dry it with sodium sulfate and i'll add that until it's free flowing i'll go ahead and remove the ether now and after doing so i'll actually add a little bit more to the sodium sulfate just to reclaim any product any one octanol that might have been left behind now that we have our dry organic layer i can go ahead and evaporate the solvent using a stream of air and some gentle warming which you can see the evaporation happening here a little bit i'll be setting up the plug using a pasture pipette so i'm going to try and get this cotton to the bottom which apparently is a struggle after that i'll add some ottawa sand and then the crucial silver nitrate treated silica gel and then at the top i'll add a little bit more ottawa sand to finish it off to the remaining product i'll add 500 microliters of pentane and then take that solution and add it to the plug so we can then elute our product for that i'll be using a 1 to 4 ratio of diethyl ether pentane solution and i'll attach a bulb to the top of the pipette so i can push that through i'll gather about five milliliters of solution just to make sure that i got enough product out and then we can evaporate the solvent again hopefully leaving behind our purified product and it looks like i barely got enough to take an ir which is totally fine i'll weigh the full conical vial or the conical vial with the product in it which looks to be 26.321 grams now i can run a quick ir to see how well i isolated the product so i definitely was able to produce the one octanol product but it looks like i still have some one octane contamination left over which you can see in two spots the peak at 1641 would be for the carbon carbon double bond and the peak at 30 78 would be for the hydrogens coming off of the alkene i'll weigh out the empty conical vial and that'll be it for this experiment okay i want to go over a few things real quick on the nmr and i'll just be focusing on this one so let's take a look at the hmr to figure out which molecule we're looking at we don't see any peaks above five so that means we don't have any alkene hydrogens on this molecule so this is going to be our octanol product i'm not going to spend much time going over the hmr but i will point out that there are eight unique hydrogens and only five peaks meaning that there have to be overlapping peaks somewhere which makes sense because if we look at this peak here it has a very big integration meaning that there are multiple hydrogens attributed to that peak so what i would do is save this peak for the end and after labeling all the other peaks whatever hydrogens are left over will all be jumbled up into this one big peak i want to focus more on the carbon nmr though just because there are some tricky things in here first off if you look to the right you are given the actual positions of the peaks and the first three around 77 are going to be for the solvent used so that would be cdcl3 meaning that the remaining eight peaks will be for the eight actual carbons on our molecule you are also given a reference molecule above and this is of octane so obviously it's a little bit different than our molecule because it doesn't have the oh group on it but if we cut the molecule in half and look at the four left carbons here five through eight they should match up fairly similarly to the four carbons on the left side of octane because they're far enough from that oh group that they're not going to be affected but if we look at the four carbons on the right side of the molecule they're going to be different because of that electronegative oxygen on that side of the molecule so let's start off by looking at carbons 1 and 8. normally without the oxygen on the molecule we would expect to see both of them around 14.2 but one being directly attached to an oxygen is going to be the most deshielded carbon on this molecule around 63. so we can go ahead and label that peak as carbon 1 whereas carbon 8 we would still expect to see around 14.2 which we see an actual value here so we'll label the most shielded peak as carbon 8. now if we look at carbon 7 we would expect it to have a chemical shift of 22.9 which we look at the actual positions we see one right around there so we'll go ahead and label that peak as carbon 7. as for carbon 6 we would expect it to be around 32.2 but we actually see two peaks around that range so it's hard to tell which one is which um so we'll come back to that but if we look at carbons four and five both of them are far enough from the oxygen where they wouldn't really be affected and we expect to see both of them around 29.5 and just our luck we see two peaks right on top of each other around there so we can label those now four is on the side of the oxygen so we can assume that it would be slightly more detailed than five so we'll label it as such now we have carbons 2 3 and 6 left we know that 6 has to be one of the peaks around 32 but other than that it's going to be hard to argue which one goes where to do so i'm actually going to look at the carbon nmr for a molecule that we've already labeled before from the fischer esterification lab and that molecule is isopentyl alcohol i'm looking at this molecule because we can assume that carbon 3 would be somewhat similar to carbon 2 on one octanol because it is also two bonds away from the alcohol group so if we look at the peak for carbon 3 it's the second most deshielded peak on this nmr and it's around 40 ppms so if we look back at carbon 2 on one octanol we can assume that it's going to be the second most deshielded and whereas is peak doesn't quite reach 40 it is the most deshielded out of the rest of the other carbons so we'll go ahead and label that as carbon 2 meaning that the other peak around 32 would be our carbon 6 and then the last remaining peak would be carbon 3. now you may be wondering why carbon 3 shows up around 25.9 when without the alcohol group we would expect to see it around 32.2 you're not alone in this i'm not really sure what's going on but again we argued the peaks using molecules we've seen before and we knew their peaks for sure and now you can use the same strategy that i showed you in this video to assign the peaks on the carbon nmr for one octane you