in this lecture we'll be reviewing the theory and concepts surrounding liquid liquid extraction in which we utilize the differences in solubility of compounds in various solvents that are immiscible with one another liquid liquid extraction is a common simple and effective method of separating complex mixtures of components for example if we have three different types of molecules dissolved in one particular solvent and we add in a separate solvent in which one particular type of molecule is more soluble we can effectively achieve a separation let's take a look at an example that you may have experienced in the kitchen let's say you have some vegetable oil and sugar and you're planning on making a quick bread however because of some mistake you accidentally pour the sugar into the vegetable oil at the wrong time now you have to go about the business of trying to separate that sugar from the oil however as you try to scoop out the sugar that's remaining in the bottom of the solution you notice that no matter how much of the granular sugar solids you get out the vegetable oil always tastes sweet so then the question is what do we do we can taste the sugar remaining in the vegetable oil but the particles are just too small to get out this is where a liquid liquid extraction can help us so we have this sugary oil mixture and we remember from our general chemistry days that sugars are carbohydrates and carbohydrates are decorated with lots of hydroxy groups or alcohols and we know that sugar is very soluble in water in fact sugar is much more soluble in water than it is oil this is where that liquid liquid extraction can help us now it's important to note that water and oil are immiscible or not soluble with one another so when we pour in our water to our sugary oil mixture we notice that the water is on the bottom and the oil is on top this is a result of the density difference between the two different types of solvents now if we were to simply pour off the oil from the water what we would notice is that the water has taken on a slightly sugary flavor we also noticed that the oil still is sweet to the taste so we know that there's more sugar still dissolved in the oil so what we'll need to do is shake or agitate the two layers together to increase the contact area between our two phases essentially what we're making sure is that the sugar which is more soluble in the water is able to come into contact with the sugar molecules and hence the sugar will go into the aqueous or water layer more easily now we can separate the two we taste the oil phase and note that sure enough it still tastes like vegetable oil but there's no sugar whereas the water phase is now very sweet what we've essentially done is extracted the sugar from the oil using a liquid liquid extraction by separating out the two layers we accomplish the separation of the sugar from the vegetable oil and if we wanted to we could boil off the water and retrieve our sugar and use our oil as necessary liquid liquid extraction is the transfer of a solute substance from one liquid phase into another liquid phase according to solubility we're talking about solutes it's the substance that's dissolved in one particular solvent so for example when we looked at the sugary oil the solute substance would be the sugar extraction is a very useful tool if you choose suitable solvents extraction can be used to separate a substance selectively from a mixture or remove unwanted impurities from a solution when we get into the laboratory we'll find that salts and acids are very common side products for a variety of different reactions we can simply use an aqueous workup in order to remove those salts or acids from the solution essentially removing the side products away from the product that we want in practice one phase is invariably going to be a water or aqueous phase and the other is going to be an organic solvent these need to be mutually exclusive or immiscible with one another if our two solvents dissolve each other that's the problem and we're not going to be able to use those two particular solvents as our liquid liquid extraction solvents so again water and some sort of a greasy organic solvent are going to be used in conjunction to achieve our separation whether or not this method succeeds depends upon the difference in the solubility we're going to have essentially a ratio of how much of the compound is in one solvent as compared to another solvent and we quantify this as the distribution coefficient this is a very important term that you'll want to take note of when it comes exam time when a dissolved compound is shaken in a separatory funnel between two immiscible solvents the solvent will distribute itself between the two solvents according to what we call the partition coefficient or distribution coefficient again normally one of these solvents is going to be water and the other is going to be a water immiscible organic solvent now when we talk about water in terms of solvents we're also talking about all aqueous solutions for example in the laboratory you may perform a separation using a sodium bicarbonate solution again it's assumed that the chemist would know that that's an aqueous or water based solvent or you may be performing an acid wash of your reaction mixture that would be a weak hydrochloric or acetic acid solution but it's still considered the water portion of our solvent system most organic compounds are going to be soluble in organic solvents however there are some organic compounds that are soluble in water sugar is a very common example of an organic compound that is soluble in water again if our compound is more soluble in the organic portion it will be in that particular layer notice here that we're using ether i'll put that in quotes ether is short for diethyl ether one of the most common organic solvents diethyl ether has a lower density than water so the organic layer is on the top and has correspondingly more of those organic soluble molecules present in it however at any one particular time there will be some of our organic molecules that are dissolved in the aqueous layer if however our compound is more soluble in the aqueous layer or water layer we would see a correspondingly larger number of those molecules in that lower aqueous layer as compared to the upper organic layer the universal rule for liquid liquid extractions is that at a certain temperature usually room temperature the ratio of concentrations of a solute in each solvent is always constant meaning it's unchanging and this is what we call the distribution coefficient it's the concentration of the solvent in one particular solvent compared to the concentration in the other immiscible solvent and it's also important to know that these solvents have to be immiscible if we have one solvent that is soluble or miscible with the other solvent we cannot perform a liquid liquid extraction let's say we have a compound that has a distribution coefficient equal to 2 between solvent 1 and solvent two now it's important to note that by convention the organic solvent is always going to be solvent two and the aqueous solvent or water is going to be solvent one what that's going to do is allow us to typically have a number that is greater than one you might remember that organic compounds are going to be more soluble in the organic solvent so that means our numerator is always going to be larger often times much larger than the denominator that we have here in this particular example there's not a big solubility difference between our example compound in that we only have twice as much in the organic layer as we have in the aqueous layer the distribution coefficient is considered to be a constant again we're looking at the concentration in one layer compared to a concentration in another layer and it all depends on the intermolecular attractive forces between the molecules of the solvent and our solute so using our example of a kd ratio of 2 if there are 30 particles of a compound distributed between equal volumes of solvent 1 and solvent 2 for a distribution coefficient of 2 we would necessarily have to have twice as many particles in the upper organic layer as we have in the lower aqueous layer so these 30 particles would divide themselves 20 to the organic and 10 to the aqueous and we would observe that distribution coefficient of two and this is going to be true no matter how many particles we have in our two solvents if we have 300 particles we're going to see exactly the same distribution coefficient we're always going to have twice as much of that particular compound in the organic layer as compared to the aqueous layer so if we have 300 particles 200 particles are expected to be in the organic layer 100 particles in the aqueous layer and we see that we have the distribution coefficient of two now things get kind of interesting when we change up the volumes of the different solvents so remember that we're looking at a concentration so what we've assumed or eliminated here is the fact that for concentration we're dividing both of these by 100 but since the volumes of the solvent are the same those essentially cancel out we're just looking at the number of particles but let's look and see what happens when we double the volume of our organic solvent so let's say we still have those same 300 particles but we've increased the volume of the organic solvent to 200 well the distribution coefficient is in fact going to remain the same because there is more organic solvent now we'll have a correspondingly larger number of particles but you'll note that since we're dealing with a ratio of concentrations we're going to divide this 240 by 200. the 60 particles in the aqueous layer is divided by 100 here our solvent volumes do not cancel and yet we still have the exact same distribution coefficient in the previous example with our 300 particles we saw that we were able to remove or extract more particles than we were able to with that original 100 milliliters of solvent so again it's more efficient to use more of the organic solvent in order to get out more of the stuff that we want however let's see if we can't increase the efficiency by extracting twice with 100 milliliters of solvent so what's going to happen well if we do it once with 100 milliliters of solvent we're going to remove 200 of our molecules that leaves 100 molecules remaining in the aqueous solvent now we add in our second 100 milliliters of solvent and we're going to see the exact same distribution coefficient being observed we're going to be able to extract 67 more particles from that remaining solution so now we have 200 particles in one 100 milliliter volume of our diethyl ether and now we have an additional 67. 267 is more than 240. it looks like doing the extraction more times with the same overall volume of solvent is more efficient so let's go ahead and review our example with the 100 particles say these 300 particles are dissolved in that 100 milliliters of aqueous solvent but then we add in our 200 milliliters of organic solvent the diethyl ether what we'll observe is that there are 240 particles in the 200 milliliters of organic solvent and 60 particles in the lower aqueous solvent but now let's go ahead and split up our 200 milliliters of solvent and perform what looks to be the more efficient type of extraction using the same overall amount of solvent so again rather than putting in 200 milliliters we would put in only 100 milliliters we would expect to see 200 particles in that organic layer again we would have to open our stopcock and drain the two different layers into different containers the organic layer would have our 200 particles in it then we take our second amount of diethyl ether another 100 mils and we treat the organic phase with this new portion of 100 milliliters of organic solvent again we're going to be partitioning those 100 particles between our two different solvents with the kd ratio of two so what we'll observe is that there are now 67 particles in the organic layer and only 33 particles remaining in the aqueous layer so 200 particles plus 67 particles gives us more particles of the original 300 the general rule is that it's more efficient to carry out two or more extractions with half or a third the volume than one large volume of solvent if you extract twice with half the volume the extraction is more efficient than if you extract once with a full volume if we extend this out extraction three times with a third the solvent would be more efficient extracting five times with a fifth the volume would be even more efficient essentially it comes down to what's more important the solvent or your time liquid liquid extraction really doesn't take that much time to perform and so most chemists will find that it is more efficient in terms of time and money to perform several smaller extractions than just one big one so the take home is the greater the number of small extractions the greater the quantity of the solute that gets removed typically an organic chemist is going to perform an extraction three times with a third the volume so again bigger or more volume is not always better more extractions with smaller amounts of solvent is going to be more efficient and get you more of the compound that you want so it was established previously that most organic compounds are going to be more soluble in the organic solvent typically this distribution coefficient is going to be greater than 4 but many times it can be measured in the thousands or could even be immeasurable so organic compounds soluble in organic solvents however there are certain classes of organic compounds notably organic acids and bases that can be reversibly altered to make them more water soluble this introduces the concept of a chemically active extraction or what we typically refer to as just an acid-base extraction this is a very powerful technique that allows you to separate very complex mixtures of organic compounds simply by removing or adding a proton to an acid or a base which completely switches their solubility so then the question becomes how do you go about changing the solubility of a particular compound using your knowledge of its acidity or basicity the two classes of organic compounds that can be reversibly converted between organic soluble and water-soluble are the organic acids and the organic bases organic acids include carboxylic acids and phenols you'll note that both of these functional groups include protons with relatively low pkas the carboxylic acids being around 4 and phenols being around 10. we can remove these protons relatively easily using bases like sodium hydroxide sodium bicarbonate or sodium carbonate the other classic compounds are the organic bases typically amines you may remember aniline in which we have an amino group that features the nitrogen with a lone pair of electrons we can add acid to these compounds and convert them into something that is water soluble for both organic acids and organic bases we have to convert the neutral organic material into an ionic form again the ionic form is going to be the form of our molecule that is water soluble so for organic acids we're going to be adding bases we need to remove the proton in order to form the anion of the organic acid this can be accomplished by adding in sodium hydroxide now notice we're not using pure sodium hydroxide this is an aqueous solution of sodium hydroxide so again five percent of relatively weak solution sodium hydroxide however it's a fairly strong inorganic base and does a very efficient job of removing the protons from our carboxylic acids now again the carboxylic acid in its neutral form is soluble in the organic phase and insoluble in the aqueous phase however once our base does its job of removing the proton it converts it into the carboxylate anion the carboxylate anion is in fact insoluble in the organic phase and is now soluble in the water so again we have converted something from organic soluble into aqueous soluble by making the corresponding ion carboxylic acids are also strong enough organic acids that we can in fact use even weaker bases than sodium hydroxide to perform this ionization so we can use sodium carbonate or sodium bicarbonate to achieve this effect adding in our sodium bicarbonate removes the proton the original compound that was insoluble in an aqueous phase is now made soluble in an aqueous phase so again we're going from neutral to ion neutral to ion organic soluble to aqueous soluble in each case let's look at an example that you might find in your typical organic laboratory let's say that you have a sample of naphthalene and benzoic acid that is dissolved in dichloromethane or methylene chloride methylene chloride is a very common organic solvent it's very nonpolar in nature our two organic compounds will both be soluble in the dichloromethane what if we want to separate these two compounds what we need to do is again take our solution and add in one of our bases we can use either sodium hydroxide or sodium bicarbonate and extract the benzoic acid away by converting it into its salt form so again the two compounds originally are both neutral and both soluble in our organic phase when we add in our base we remove the proton selectively from our carboxylic acid that benzoic acid that converts it into the water soluble carboxylate and so now we have the naphthalene in the lower organic phase and our sodium benzoate that conjugate base of our benzoic acid present in the aqueous phase now it's important to note that methylene chloride and most all haloalkanes are going to be more dense than water that's why this example is a little bit different the organic phase is on the bottom because it's more dense than water in order to affect our separation of course we would have to open up our stopcock and selectively drain the two layers into different containers but again we've achieved a separation of the two different organic compounds the second class of organic acids we looked at were the phenols these are weak organic acids their pkas are typically between about 9 and 11 and dependent upon the types of other functional groups we have on the ring structure in addition to the alcohol interestingly phenol the parent compound is partially water soluble you can get about 1 gram of phenol to dissolve in 15 milliliters of water this is due to the fact that the alcohol group can hydrogen bond in that water solvent however if we have any groups attached to the aromatic ring along with the alcohol it renders it water insoluble now because phenol is a weak organic acid we do have to use stronger bases sodium hydroxide is the most common one if we try to use a weak acid such as sodium bicarbonate or sodium carbonate then in fact we will not remove the proton there will be no acid-base reaction and the phenol will remain in the organic layer however when we use sodium hydroxide it'll efficiently remove the proton making the phenylate ion which is in fact water soluble so again we go from the neutral compound phenol which is soluble in the organic phase and we convert it into the ion which is aqueous soluble let's take a look at another example in which we attempt to separate a carboxylic acid and a phenol from one another that are dissolved in dichloromethane the benzoic acid has a pka right around 4 where this methoxyphenol that we have has a pka of around 10. now remember both of these in their neutral form are soluble in the organic layer the dichloromethane so what if we want to separate these two compounds again initially they will both be dissolved in the organic layer if we add in sodium hydroxide we can remove protons we can convert neutral into ionic however in this case because we used a stronger base we're going to deprotonate both the carboxylic acid as well as the phenol and we have not achieved a separation yes we have converted both compounds into their ionic forms and they're aqueous soluble but they're both still in one layer so no amount of playing with the stopcock or draining one layer off or another will allow us to separate these two compounds so we'll have to essentially use a different tact in order to separate a carboxylic acid from a phenol so let's try that separation again we have benzoic acid and methoxyphenol dissolved in dichloromethane we know that the sodium hydroxide was too strong a base there was no selectivity both the carboxylic acid and the phenol were converted into their ions and became aqueous soluble so instead let's use a weaker base such as sodium bicarbonate again we start out with both of the organic compounds dissolved in the organic layer by using the sodium bicarbonate we convert only the carboxylic acid into its ionic form so again stronger acid will react with the weak base whereas the less acidic phenol will not be deprotonated so here we've achieved a separation we've converted our carboxylic acid into the ion and made it aqueous soluble whereas the phenol is still dissolved in the organic layer because it was not deprotonated by this weaker base so we've been looking at organic acids and using bases to convert those acids into conjugate bases or anions which are aqueous soluble now let's take a look at organic bases for an organic base we're going to be converting the neutral base into its ionic form or what we call the anilinium so again these amines can be converted into water-soluble compounds by adding an acid to them when they accept a proton they become the ammonium or anilinium ions they go from being organic soluble in their neutral form to aqueous soluble in their ionic form same concept we just have to use a different material in order to convert it from the neutral to the ionic form let's look at a separation problem that features both an organic acid as well as an organic base dissolved in dichloromethane when we add the two solid compounds into our dichloromethane our organic solvent they'll both be dissolved neatly how do we go about separating them again we look at our two different possible groups we've got an acidic proton here and a basic nitrogen up at the top using our knowledge of the acid-base characteristics we could add in a five percent solution of hydrochloric acid to convert the base into its corresponding anilinium ion which would be aqueous soluble when we do so what we observe is just exactly what we expect we have made the ion which is aqueous soluble it will migrate into the aqueous layer leaving the carboxylic acid in the lower organic layer again we can open up the stopcock and using two different containers separate one from the other again we've performed an extraction in this case we have extracted out our organic base that chloroanalyn i briefly mentioned at the beginning of the video that we can separate out some pretty complex mixtures of compounds just by using their acid and base characteristics let's say we have a methylene chloride solution in which we have all four different types of compounds that we've looked at so far the four classes are the amines the carboxylic acids the phenols and neutral compounds so again we have organic bases organic acids and neutral compounds by using our knowledge of pkas and the acid base characteristics we can in fact separate out all four of these different components by selectively using an acid or a base extraction solvent so in this particular case we would start off by adding in dilute hydrochloric acid this would convert the organic base our amine into the corresponding ammonium or lineam ion so again in our separatory funnel we would have the aqueous layer sitting on top and all three of the other components are going to remain in the lower aqueous layer we separate the two liquids and we have isolated our amine the next thing we would do is use a weak base in order to remove the more acidic carboxylic acid from the solution so we add in a dilute solution of sodium bicarbonate this is going to selectively deprotonate only the strongest acid and so we would convert the carboxylic acid into the corresponding carboxylate that would be dissolved in the aqueous phase and our other two components are going to remain in the organic phase in order to remove the phenol from the neutral naphthalene we would need to add in sodium hydroxide in order to convert that phenol into the phenolite which is aqueous soluble the only thing remaining in the organic layer now is our neutral compound the naphthalene so again we would have one say erlenmeyer flask containing our protonated organic base we would have another flask with our deprotonated carboxylic acid a third flask with our deprotonated phenol and finally the last flask with our naphthalene so again we can separate very complex mixtures using our knowledge of the acid or base characteristics of our organic compounds so here we have our different ions and of course the neutral component sitting in either the aqueous phase or in the organic phase for the neutral compound now it's important to remember that these are not the original forms of our compounds they have been chemically altered and are in their salt form so what we're going to need to do is convert these first three back into their neutral forms the salt forms can be very useful for example many pharmaceutical compounds will exist and be administered as their salt form so for example if you go and look at a bottle of medication you'll oftentimes see that the pharmaceutical compound is actually the hydrochloride salt that's quite literally this guy right here kind of an interesting side note so we're going to need to convert these back now the neutral compound of course is not uh going to need to be converted back because it's still of course neutral but we will need to boil off our organic solvent in order to recover the solid material but to recover the others if we were just to boil off the aqueous phase we'd be left with a problem because it's not the original neutral compound so the next thing that we have to do to our compounds that have been chemically altered is to isolate or recover them by converting them back so let's see how we go about doing that the organic base was the first compound that was removed from our four component mixture and we did that by adding in hydrochloric acid to convert it into its anilinium salt so again we have an aqueous acidic solution so if we want to obtain the neutral form we need to quite literally neutralize the acid and we'll do that by adding in a base typically we're going to do this using the minimum volume possible so for example in this case we would add in say a five percent sodium hydroxide solution that will remove the proton from our anilinium giving the nitrogen back its lone pair of electrons and the neutral compound will precipitate uh oftentimes uh in in dramatic form from our solution remember it's uh in an aqueous phase and we are quite literally converting it into something that is aqueous insoluble and so we will recover a precipitate now here we show the precipitate at the bottom more often than not what we will observe is the organic compound forming on the top layer because most of these are going to be less dense than water and are going to float on the surface as a solid so again we started out with the acid it was aqueous soluble and we converted into something that is neutral and aqueous insoluble and so all we have to do is perform say a vacuum filtration and we can recover our neutral amine so let's turn our attention to some of the practical aspects of recovering our organic base from this acidic solution remember when the solution's acidic our organic compound is going to be protonated and in the ionic form and it's going to be dissolved in the aqueous phase so we're going to be adding in a base in order to neutralize the acidity of our solution and this will this will give us our neutral organic compound back so we add in sodium hydroxide but it's not just any concentration of sodium hydroxide we want to make sure that we use a more concentrated form with more base present and this is due to the fact that we have a volume limitation there's only a certain amount of liquid that we can fit into our erlenmeyer flask so if we use a weak basic solution say a one molar or a one normal sodium hydroxide solution our final volume may be too large we we may risk not neutralizing all the acid leaving some of our compounds still dissolved in the aqueous phase so we're going to use a more concentrated basic solution in order to get this job done again we want to minimize the amount of solvent we have to add in the other couple of things we need to worry about are making sure that we have efficient mixing of the base and the acid so we're going to be swirling this a lot and then the last thing we need to worry about is the heat that's being generated during this acid-base reaction so we need to maybe have an ice bath or some sort of a water bath in order to maintain reasonable temperatures last but not least we're also going to be checking the ph of the solution we need to add in enough of our sodium hydroxide solution to get the ph up to above 10 in order to make sure that we've removed every single proton that we can from our anilinium ions to generate the neutral organic base all right so we've now recovered our organic base from the solution and we have it in the neutral form possibly sitting on the filter paper in the buchner funnel the next thing that we need to do is address the two organic acids that we extracted from that mixture of four different compounds so those carboxylic acids and the phenols you might remember we used bases in order to deprotonate them and convert them into either the carboxylate or the phenylate and in order to get them back in their original neutral form we need to replace those protons so we're going to be neutralizing our basic aqueous solution with an acid so typically we'll use just a solution of hydrochloric acid in order to do this and what we're going to see is that the for example the benzoate when we add that proton back into our molecule it's going to convert it into something that's aqueous insoluble and it will quite literally just generate itself on the surface of that aqueous solution so again aqueous soluble into something that is aqueous insoluble and we'll see it precipitate the same thing is true for the phenolates so the phenylalate is going to be aqueous soluble it's just going to be in solution and then when we protonate it we will generate the neutral phenol form which is of course aqueous insoluble so it will also precipitate so again we go from the basic media in which those organic compounds are soluble into an aqueous media in which the neutralized or neutral organic compound is going to be aqueous insoluble so again we can recover both our carboxylic acids and our phenols by just giving them back their protons all right so here we have one of our basic solutions that has let's say that this has the benzoate dissolved in it there are some practical things that we need to concern ourselves with so we're going to be giving this benzoate back its proton essentially just like we looked at before we have to pay attention to the volume of our flask we're going to be using relatively concentrated hydrochloric acid so we'll want to use say 6 molar or 6 normal hydrochloric acid so that we do not run the risk of over topping our erlenmeyer flask now just like we talked about previously acid-base reactions are typically very exothermic so we're going to have to worry about heat and we're also going to have to concern ourselves with the swirling and mixing we want to make sure that we distribute all of our hydrochloric acid throughout the solution in order to achieve efficient protonation of all of the sodium benzoate present in solution the last thing we'll need to make sure of is that we have a ph down at around two or three so that we can ensure all of the benzoate has been protonated one of the most useful and important pieces of glassware the organic chemist will have in the laboratory is the separatory funnel so here you see four different examples you'll see everything from 50 ml separatory funnels all the way up to two liter separatory funnels it all depends on the volume of solvents and dissolved compounds that you're working with the one that most of you will see in the laboratory is this one right in the middle right here and this has a stop going through the center of it so this is the valve at the bottom of our separatory funnel that allows us to drop or remove layers very accurately and efficiently so again if we have two layers sitting in our separatory funnel we can carefully open our stopcock and drain off one layer and be able to see uh exactly the partition between the organic phase and the aqueous phase so that's that's why it has this taper right here we're not just pouring one solvent off of the other where we're using this funnel in order to identify the partition between the aqueous and the organic phase so with regard to the stop you'll notice that we have a nice white teflon stopcock right here it has a nut at the end and this is threaded right here the the stopcock itself the body is threaded through the center right here and we also have a black o-ring and a white washer that's made of teflon right there it's important that you check your separatory funnels and check all these components to make sure that they are present and in good condition otherwise things can go the wrong way very quickly when you're performing a liquid liquid extraction i want you to take note of the center right here when the hole in the stopcock is lined up with the body of your separatory funnel it's going to release liquid so of course if you're pouring something into the separatory funnel you're going to want to make sure that the stopcock is closed with the the handle that we have right here typically being perpendicular to the body of the stopcock when we're setting up our station to perform a liquid liquid extraction we need to have a couple of things available to make this a little bit more time efficient so for our separatory funnel we're going to want to have some sort of a support device you can use a funnel support clamp or you can use a ring on a ring stand that has some little pieces of rubber that will help support the glass of your separatory funnel so you'll notice we have our solvents available we have our stopper that's very important and you'll also notice that we have a funnel up here that will allow us to without making a mess deposit our liquids into the separatory funnel now again when we're getting ready to pour our liquids into the separatory funnel we want to have the stopcock in the closed position now if you are steady of hand you can sometimes uh get away with emitting the funnel that we have right here most of my students have very steady hands and they'll simply pour it directly into the separatory funnel so if you trust yourself that is just fine you can do that so we pour in the liquid to be extracted this might be the organic solvent that contains several different components and then we're going to add in our extraction solvent this could be an acid or a base depending on what sort of functional group we're either trying to protonate or deprotonate then we cover this over with a ground glass stopper to seal it up and we can take it off and start agitating or shaking the immiscible liquids in order to get efficient transfer of organic materials from one phase to another it's very important to stopper and shake the separatory funnel this ensures that there will be good surface area contact between our two immiscible solvents if we do not or fail to shake the two liquids in our separatory funnel there's a high likelihood that we will not extract all of the material that we want to extract so we'll put the stopper in and keep our finger on that stopper that's very important there's nothing magical about the ground glass you you need to hold it there otherwise if you invert it as you're shaking it it could quite literally fly off or pop off and break so if you're familiar at all with the james bond movies one of his famous lines is shaken not stirred well that applies right here we want to shake these two materials together to get good transfer of the organic molecules from one phase to the other now the other thing that we need to take note of is that when we are mixing there is quite literally heat generated the heat of mixing and so because of this our organic solvents will oftentimes be volatilized they'll be converted into gases and that could potentially pressurize our separatory funnel because of course we have it sealed up in order to relieve this pressure what we're going to do is carefully invert the stopcock as you see right here always holding on to that stopper and we're going to open up the stopcock in order to release any gas that gets generated during the course of mixing if you don't do this you'll oftentimes see leakage out of your ground glass stopper due to over pressurization so make sure that any time you agitate or shake the separatory funnels that you invert it and open the stopcock and kind of burp the separatory funnel so to speak the transfer of solutes from one phase to another is not instantaneous we need to make sure that we shake the funnel vigorously for a few seconds release pressure shake it again release pressure we want to do this for approximately half a minute or so this allows the solute to equilibrate between the two different solvents we also want to make sure that we're venting after each shaking this is especially true when we are removing protons from carboxylic acids because we frequently use sodium carbonate and sodium bicarbonate and these will generate a lot of carbon dioxide gas so if you are using sodium bicarbonate or carbonate along with that carboxylic acid then you're going to have extra pressure and you're going to need to vent an extra few times so now that we have thoroughly shaken and mixed our two immiscible solvents together we need to allow the mixture to separate into its two distinct layers now one of the things that you will probably notice during the course of the layer separating is the emulsion that we typically observe as the two layers are separating so this will appear as bubbles or kind of a hazy cloudy layer you need to make sure that you wait for that hazy cloudy bubbly layer to separate into the distinct layer so you should have only a lower layer and an upper layer you should not observe any haziness if there is haziness or bubbles you need to just wait a few minutes until the emulsion dissipates then we're going to remove our stopper up at the top we if we don't remove the stopper will create a vacuum and you'll open the stopcock and essentially nothing will happen you'll be like oh my goodness gravity just broke but it's in fact just the vacuum that's been pulled popping off this stopper will relieve the vacuum and allow the stopcock to release the two different materials so it's very faint in this particular picture but there is the partition between the lower layer and the upper layer so we would very carefully open up our stopcock turn it you don't have it's not completely on or completely off you can allow a trickle or drops of material to go through the stopcock as you will so you can do it fast you can do it slow but you need to accurately close the stopcock when you reach the partition between the two layers because again if we're going to do a separation we need to have one layer in one container and the other layer in another container and again once you do this you can isolate the materials by adding an acid or a base or perhaps by boiling off the organic solvent now one other thing to note is that if you get a little carried away and say you you let the top layer go through into into the bottom layer easy fix you just pour all the material back in the separatory funnel allow the two layers to form and then try again