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Describe with equations how you might separate a mixture of aniline, $p$ -cresol, benzoic acid, and toluene using ordinary laboratory reagents....

Question

Describe with equations how you might separate a mixture of aniline, $p$ -cresol, benzoic acid, and toluene using ordinary laboratory reagents.

Describe with equations how you might separate a mixture of aniline, $p$ -cresol, benzoic acid, and toluene using ordinary laboratory reagents.



Answers

Describe with equations how you might separate a mixture of aniline, $p$ -cresol, benzoic acid, and toluene using ordinary laboratory reagents.

So first thing we want to separate the I'm and sand we want you is magnetic Magnum. Wait. Why is it only I'm can be tracked by magnetic Bad sand One Sorry. Attracted by magnet. So which is common sense? Um and B is about separate, assigned and sound. So I will choose Use water to dissolve south where Santana dissolved Now sand and then you can recover the south by evaporate, you know, now evaporate the water every water so you can gather sound again and then say is about separate the ah, I separated the okay. Separate the components having so I'll choose is boning massive mattered thio separate. Since when you ball the the ink the dye will precipitate And, uh, Walter Well, we well evaporated and you can cover the water by the circulation. I mean condensation, which is ah, part of process of the distillations. Our father come confusion so you can use content used condensed cover. The water evaporated during the separation processes and the fennel long we want separate a Hey, Liam and oxygen. So the possible way is you know, cold. The temperature called the temperature to liquefies to liquefied. They could find a helium and oxygen and then use distillation, says the this to gas. It is a different ball in point. So you well, you know, hey, that different. It's a different temperature it during different temperatures, you can, you know, clacked each of the different temperature separately to family, separate the helium and oxygen.

So describe the properties of bending that makes that made chemist think it was not. And Al Keen with several double one. So who caught at the structure opposed in 18 sixties had a molecular Formula 686 which is unstable and highly reactive because of the many double bond.

All right. In this question, we are told that we have a solid mixture of benzo, kassid and charcoal and were asked to separate them. So we're also told that Ben's OIC acid is soluble in warm water, whereas charcoal is not soluble. So how do we separate these two components since we know that took acid insoluble if we add if we add water? Hot water is particular hot water to this solid mixture of benzo kassid of charcoal. We know that the Ben's OIC acid with dissolve and the charcoal won't. So, um, if we have, let's say this cube represents have been so casted a charcoal group. If we splash some water on it and we put it through a filter paper soon that, uh, this little black figure over here hiss, filter paper. What we would expect is that the charcoal we'll stay. This part of this is just charcoal chocolate stand solid form. And that been said, Kassid will seep through the same filter paper and I'll be and dissolved in the water, and that's it. So just add hot water and that filter. Thank you

Okay. This problem is asking us to separate these three compounds from a mixture case. When are very left. We have our form, Ethel Fennel. In the middle, we have Arbenz oak acid and on the right. We have our Halloween. Okay, so obviously, these are all three different compounds. They're gonna have a little bit of different activities in different solvents. However, they are all organic compounds. So that means that they're gonna be soluble in an organic solvents such as a neither case we can use th f um dia feel if there it doesn't really matter, we just need to make sure that all of these three are gonna be soluble in one particular solvent. So first step is we're gonna dissolve all these in our solvent. So let's just say it's gonna be di ethyl ether. Okay? So that means that we're gonna combine all of them, and they're all gonna be within that mixture combined with my docile either. Okay, so that's the first step. And then now we have to separate them. So we're gonna start off with the one that is most easy to separate, and that is are particularly acidic. Ben's OIC acid. So benzo CASS It is a little bit unique because, as I said before, it's the most acidic. So that means that if we have a weak enough base, our weak enough base is going to only deep protein. Eight. That Ben's OIC acid So the week base that we're gonna be using is sodium bicarbonate and a H C 03 So the reason that we're using sodium bicarbonate is because if this were appropriated, it would turn into Htoo three, which would eventually turned into a 20 and Co two, otherwise known his water and carbon dioxide. But the PKK of the hypothetical H two C 03 is approximately 6.35 Okay, so that means that we're going from a PK of approximately 4.74 point 82 a PK of 6.35 So obviously we know that it is most favorable to go from a stronger acid into a weaker acid in terms of our equilibrium. Whereas if we had our federal our four methyl phenyl, that hasn't PK of roughly 10 ish, which would be going in the reverse direction, right? I would be going from 10 to 6.35 which is not the favorable direction. And then this is just not acidic. Okay, so that's the second step in which we use our sodium bicarbonate to deep protein. It are benzo kassid. Okay, so that would turn into this compound. Okay, So I would have my benzene over here, and then I would have money. Part box late. Okay, So the question is, what's gonna happen next case? Obviously we have and lets you say this is all performed in a test tube, for example. So this is my terrible drawing of a test tube, But bear with me. Okay, So let's just say a very first step was this right. We had all three compounds 123 with our diethylene there. It was just all within this test tube. And then after that first step after report in our n h n a h c 03 that means that is going to react with my benzo. CASS said that means it's going to deep protein it that bones. Okay, so this is gonna be the second step after that second step of adding that mixture. Right? So we're gonna have a little bit higher solution. It's going to be composed of my organic solvent composed of fo. Either might Halloween like this as well as my for fennel, therefore in a metal fennel. So this is a little bit small, but this Theo age And then over here we had the metal group. So these three are still within the solution, Kate. But after we have two d protein ate the benzo kassid, that's gonna be on the bottom layer, okay? And I'll explain why, Okay, so again we have our deep protein ated benzo acid, which is just the oh minus case of the reason that that goes on the bottom or could even go in the top. It just depends on what's gonna occur, but it's most likely going to go on the bottom due to the densities of the different compounds. But basically this deponent benzo CASS, it is going to be separated from these three compounds simply because it's now considered a quiz, right? So we have our organically here, and then this is going to be our quiz later. And the reason that it is it goes into the acquis layer is because after we use our and H c 03 to deep protein eight hour benzo casted. As I said before, this pro nated is going to turn into water and Co two case of water, obviously, and we know that's the major compound needed for in a quiz layer. And then the deponent version of car box of carb oxalic acid in my car box like that is an ion. We know that ions specifically molecules with negative and positive charges. Those are soluble in water, right? So we have salt. We know that salt is very soluble in water, just anything with a negative or positive charge. So that means that it's going to be within that same layer of water. So then we can just use attest to or starting a test to a pipe at, for example, for example. So here's my drawn of pipette that is going to squeeze out this acquis layer, right? We can put it all up into the steps test tube, and then we can put it into another test, too. Are bigger whatever we want. So this is my new layer of my former. A quiz later. Okay, so this is composed in my car box late I on my deponent benzo kassid. And then we can just use HDL in order to protein. Ate that to get it back into my Benz oak acid. Okay, so that would go back to my benzo kassid. Okay, so now we're stuck with this layer. Okay? So that the organic layer after the step to So in order to separate my foreign four method fennel from my Halloween, we have to look closely at that oxygen, right? The difference between this compound and this compound is the presence of this oxygen a k a. The fennel component. So fennel is much more acidic than might Halloween. Right? Because we have the oxygen connected to the hydrogen. The hydrogen is prone to getting deep protein ated. Right. So all we need to dio is used something like sorting hydroxide in order to D protein ate that fennel. So the next step after technically after we separate the acquis later from this one is going to be to use sodium hydroxide case. Well, that's gonna do it is going to create this in which we have that organically up here. Okay, so the organic layer is composed with diarrhea either as well as my tolerance. Okay. And then on this bottom layer, the quiz layer. So again, this is organic. And then this is a quist. In the acquis layer, we have water. Can We got water from the protein ation of my sodium hydroxide right here, as well as the deep protein ated fennel. So here we have my fennel depreciated as well. Is that method group in that fourth position? So again, the reason that this is in the Equus layer is because we have that Oh, minus. And, um, tends on the minus charge in water. Right. A quiz. So that is gonna be separated from my organic layer. Okay, so once again, we can use a pipette to separate those two layers my quiz layer from my organic, and then we would fill that pipe up, and we get this. We just put that chris late into a beaker. So now this is composed of my water as well as my the protein ated for metal fennel. And then we just add something like H CEO. That's going to protein eight, my deep rooted for metal fennel. Okay. And then we can separate that and get back to our starting material, but just separated from the other 32 okay. And last but not least, we have this. So this layer, we need to separate that di ethyl ether from my Ptolemy. Okay, so in order to do that, all we have to do is use the difference in boiling points. Okay, so we have that new are the same test he was before, except it's composed of my tolerance and my diet for either. So these are gonna have different boiling points. Okay, so it's gonna be a little bit difficult. Teoh use different saw liability rules because they're both considered to be relatively non acidic, are relatively non basic. So it just be difficult to separate those based on those terms. So what we're gonna do is we're going to boil this. So here's my example. Fire and we're gonna have we could use fractional distillation, for example. Er could use a variety different things. All were our basic gonna be doing. It's separating them based on their differences in boiling points, and then we can get our desired product, and that would be the whole process


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