Mastering Organic Chemistry with Dr. Frank

Acid-Base Reactions Problem Set – The planning of a reactive extraction flowchart

Estimated read time: 1:20

Summary

Dr. Frank presents an engaging and structured acid-base reaction problem set video, utilizing the technique of liquid-liquid extraction to separate organic and aqueous substances. The video begins by outlining the process, inviting learners to tackle practice problems, and providing solutions to compare. Dr. Frank then delves into the technique of liquid-liquid extraction, explaining the importance of solubility and electrical charge in organic chemistry. He emphasizes the role of acid-base reactions in changing a molecule’s solubility, highlighting the importance of pH levels in deciding a molecule’s behavior. As the video progresses, Dr. Frank guides viewers through separating multiple molecules using acid-base chemistry, creating distinct layers of organic and aqueous solutions via fun and interactive flowcharts.

Highlights

Dr. Frank's charisma makes complex chemistry topics accessible and enjoyable. 🧑‍🔬
The video utilizes visual flowcharts to effectively explain liquid-liquid extraction. 📊
Dr. Frank guides viewers step-by-step in solving acid-base problems, ensuring understanding. 🧭
The content explores the reversible nature of acid-base reactions exceptionally. 🔄
Molecules are cleverly separated into distinct solutions using pH manipulations. 🌊

Key Takeaways

Dr. Frank's video offers a comprehensive look into acid-base reactions, focusing on liquid-liquid extraction techniques. 📚
The video emphasizes the interplay between solubility and electrical charge in determining reactions. ⚖️
Understanding the pH impact on organic molecules' behavior is essential in separating substances. 🧪
The flowchart method provides a clear and practical approach to solving acid-base reaction problems. 🗺️
The video is structured to facilitate learning through problem-solving and explanation. 💡

Overview

In Dr. Frank's acid-base reactions video, the intricacies of liquid-liquid extraction are unraveled with his characteristic teaching style. Beginning with an introduction to problem sets, he challenges viewers to pause and engage with the material actively, offering solutions shortly after. This hands-on approach immerses students in the subject, encouraging them to learn through doing and immediate application.

The heart of the video elucidates the process of liquid-liquid extraction, a crucial technique used in organic chemistry to separate compounds based on their solubility preferences. As Dr. Frank explains, the key lies in understanding how molecules behave differently in aqueous and organic layers, influenced by their ionization states. Visual aids and flowcharts enhance comprehension, demystifying a sophisticated concept.

Finally, Dr. Frank applies the theoretical knowledge to practical examples, detailing how specific reaction conditions can manipulate molecule solubility. By showcasing the reversible nature of acid-base reactions, he presents an efficient strategy for solving complex chemistry problems, ultimately empowering students with the tools to separate organic from aqueous layers effortlessly.

Chapters

00:00 - 00:30: Introduction and format explanation The introduction chapter features Dr. Frank who outlines the structure of a video lesson focused on acid-base reactions in organic chemistry. The format involves presenting practice problem sets on the screen, followed by encouraging viewers to pause the video to solve them before the solutions are revealed. This interactive approach allows learners to engage actively with the material.
00:30 - 01:00: Display and answer problem sets This chapter focuses on the display and resolution of problem sets. Initially, answers are shown on the screen for comparison against expected results. In the latter segment, the instructor returns to methodically solve and explain each problem, providing insights into their reasoning.
01:00 - 02:00: Liquid-liquid extraction basics The chapter introduces the concept of liquid-liquid extraction, which is a purification technique frequently used in organic chemistry. It explains how this technique helps in separating chemical entities by leveraging their differing solubilities in various solvents.
02:00 - 02:30: Acid-base reaction explanations This chapter covers the basic concepts of acid-base reactions, particularly focusing on the process of liquid extraction. The explanation begins with the typical setup of a liquid extraction which involves mixing two different organic solvents in a piece of glassware, generally a separatory funnel. Often, one of the solvents used is water due to its poor miscibility with many organic solvents. The chapter likely elaborates more on this fundamental process and its significance in acid-base reactions.
02:30 - 03:00: Separation of molecules based on acidity and basicity The chapter discusses the separation of molecules based on their acidity and basicity within a system involving an aqueous phase and an organic solvent. It highlights how different chemical entities prefer to dissolve in either of the two solvents, indicated as yellow and pink, depending on their properties. This preference is utilized to separate compounds effectively.
03:00 - 04:00: Acid-base chemistry and solubility changes This chapter discusses the principles of acid-base chemistry and its influence on solubility changes during liquid extraction processes. It explains how neutral organic molecules tend to dissolve in the organic phase, whereas ionized molecules, regardless of being positively or negatively charged, favor dissolving in the aqueous phase, exemplified by green dots.
04:00 - 05:00: Incident of acidic hydrogens and basic atoms The chapter discusses a technique where molecules are made to preferentially dissolve in either an organic or aqueous layer, allowing physical separation since these layers do not mix. This process ideally yields purified molecules in both layers. The chapter then ties this concept into acid-base reactions, explaining how molecules undergo changes when they donate or remove a hydrogen atom.
05:00 - 06:00: Determining behavior as a function of pH This chapter delves into how the behavior of chemical species changes as a function of pH, focusing on the electrical charge transitions. It explains how molecules can transition between neutral and ionized forms, affecting their solubility in organic versus aqueous environments. The key concept is the manipulation of molecular solubility for separation processes.
06:00 - 07:00: Flowchart for reactive extraction The chapter discusses the concept of using reactive extraction through acid-base reactions to manipulate the solubility of molecules. By understanding the properties of molecules, you can strategically alter their solubility in either organic or aqueous layers. This concept is visually represented through a graph, which has been covered in previous material, that aids in comprehending the effects of changing the solution's pH relative to the pKa of the acid or its conjugate acid.
07:00 - 08:00: Acid-base reaction reversibility The chapter on 'Acid-base reaction reversibility' explains the conditions under which the acid or base form of a substance will dominate. If the pH is lower than the pKa value of the substance, the acid form, depicted in pink, will predominate. Conversely, if the pH is higher than the pKa value, the base form, shown in blue, will be more prevalent. The text briefly touches on the electrical neutrality of the acid or base form, suggesting that this property might be relevant to its dominance in a given pH environment.
08:00 - 10:00: Separation strategy for different molecule sets This chapter discusses strategies for separating different molecule sets based on their acid-base chemistry. By exploiting the differences in solubility between organic and aqueous solutions when molecules are protonated or ionized, it is possible to selectively separate molecules into distinct layers. The chapter also highlights the reversible nature of acid-base reactions, using a neutral molecule in the example provided.
10:00 - 11:00: Conclusion and practical tips The chapter discusses the process of deprotonating molecules in basic conditions to make them ionic and water-soluble, and how reprotonating them in acidic conditions returns them to their electrically neutral form, making them soluble in organic solvents. A practical application is provided through a question that involves separating three molecules into distinct organic layers based on their behavior in water at different pH levels.

Acid-Base Reactions Problem Set – The planning of a reactive extraction flowchart Transcription

00:00 - 00:30 "Hi everybody, Dr. Frank here, and today we're going to do some practice problem sets on the field of acid-base reactions in organic chemistry. So, the format of this video will go as follows: once I'm done talking in about 20 or 30 seconds, I'm going to display on the screen some practice problem sets for you to work on. Once on the screen, these problem sets are going to be there for approximately 10 seconds. I then invite you to pause the video and answer each question to the best of your abilities. Following just 10 seconds, I will then display the final answer for each part
00:30 - 01:00 of the problem sets on the screen so that you may compare your answer to the expected answer. Finally, in the second part of the video, I will be back on the board to solve each part of the problem set one by one and explain the reasoning behind each question.
01:00 - 01:30 So, before we get started, I want to go over a few general considerations regarding the question at hand. Ultimately, this question has us consider liquid-liquid extraction, which is a pretty common purification technique used in organic chemistry lab that allows us to separate chemical entities based on their relative solubilities for different solvents.
01:30 - 02:00 So, ultimately, when we do a liquid extraction, we start by mixing two different organic solvents in an appropriate piece of glassware, usually a separatory funnel. And so, this little drawing here symbolizes my two solvents in my separatory funnel. Almost 100% of the time, one of our solvents in a liquid-liquid extraction is going to be water because water doesn't really mix well with a lot of organic solvents. That's what we call
02:00 - 02:30 our aqueous phase. Otherwise, well, we're going to have an organic solvent symbolized here in yellow. The idea here is that different chemical entities are going to preferentially dissolve in either of these solvents. So, it's certainly not impossible to have, for instance, an entity that would preferentially dissolve in the organic layer, which I'm representing here with pink dots, while you might have an entity that will preferentially dissolve
02:30 - 03:00 in the aqueous phase, which I'm representing here using green dots. As a general rule of thumb, when you're doing a liquid extraction, if you have a neutral organic molecule, neutral organic molecules tend to preferentially dissolve in the organic phase. If you have an ionized molecule, whether it be positively or negatively charged, it doesn't really matter, they tend to preferentially dissolve in water.
03:00 - 03:30 And so, the neat idea here is that, well, if you can force the molecules to preferentially dissolve in one layer over the other, you can then physically separate the organic layer from the aqueous layer because they don't tend to mix. And then, you can obtain ideally purified molecules in both layers. So, how does that all relate to acid-base reactions? Well, you have to realize, based on the generic equations that I've written here on the board for acid-base reaction, when you have a molecule and you either donate a hydrogen atom or remove a hydrogen atom,
03:30 - 04:00 usually something happens to your electrical charge. You usually go from either a neutral species to an ionized species or an ionized species to a neutral species. And so, here, you have to understand, well, if I'm changing the electrical charge of my molecule, I am changing its relative solubility for either organic or water. I can start from a neutral organic molecule that is preferentially soluble in the organic layer and make it preferentially soluble in the water phase, which I can then separate from the rest of my molecules.
04:00 - 04:30 And so, that's the key idea here. Knowing your molecules, you can exploit acid-base reactions to selectively and reversibly change their respective solubilities for either the organic or the aqueous layer. Everything you need to know here can be rationalized using this little graph here, which we've seen in previous videos. Ultimately, what this tells me is that if I have a molecule in an aqueous solution, if the pH of my aqueous solution is inferior to either the pKa of my acid or the pKa of my conjugate acid if I'm
04:30 - 05:00 dealing with a base, then the acid form shown here in pink in all four equations is going to dominate. On the other hand, if the pH is superior to the value of the pKa, which is this region here, then the base form shown here in blue, again in all four cases, is going to dominate. And so, here, depending on whether our acid or basic form is either electrically neutral
05:00 - 05:30 or ionized, then ultimately our solubility for organic or water is going to change. And so, here, knowing all of that, you can rationalize a plan that will allow you to selectively separate molecules into distinct organic and aqueous layers simply by exploiting their acid-base chemistry. Finally, note that an acid-base reaction is a fully reversible process. If I take this first reaction as an example, if I start with a neutral molecule,
05:30 - 06:00 yes, I can deprotonate it in basic conditions to make an ionic, water-soluble species. But I can also use acidic conditions to reprotonate my conjugate base and go back to my initial electrically neutral molecule that is soluble in an organic solvent. Question one wants us to separate the three following molecules into three distinct organic layers. And so, ultimately, what we need to know here is we need to determine how each molecule behaves in water as a function of pH.
06:00 - 06:30 And for that, well, we need to figure out for each molecule which is the most acidic hydrogen, which is the most basic atom. So, let's start by doing that. Let's start with molecule 1. So here, I have some bad news and I have some good news. The bad news is I'm willing to bet that you actually picked the wrong acidic hydrogen, and I can't blame you for that. And I've mentioned that in a few videos before. I've often said, you know, more often than not, 80-90% of the time, the most
06:30 - 07:00 acidic hydrogen is going to be the hydrogen that is bound to the heteroatom. In occurrence, that'd be the aniline here. In this specific scenario, you'd be wrong. There's a more acidic hydrogen, and that's going to be the hydrogen that is bound to the carbon alpha to the carbonyl. Carbonyl does a really good job at stabilizing electrons by resonance, meaning that this hydrogen is more acidic. We're looking at a pKa of approximately 20. Compare that to aniline that has a pKa of about 30-31. The good news is, it doesn't matter if you do everything else
07:00 - 07:30 correctly. It will not change your final answer. So, we'll come back to that in a few minutes. Now, in terms of basicity, you have two potential candidates: you have the nitrogen and the oxygen. Both have lone pairs available for sharing. More often than not, everything else being equal, your nitrogen is going to be more basic than oxygen, especially here, we're dealing with a carbonyl. Carbonyl oxygens are seldom basic. And so, here, indeed, the most basic position is this nitrogen.
07:30 - 08:00 If you protonate it, you create an anilinium. The pKa of this anilinium is around four. Now, let's look at molecule 2. So here, we have an obvious hydrogen that does stick out, and here, there's no trap. The most acidic hydrogen is indeed the phenol, the hydrogen that is bound to the oxygen. Phenols usually have a pKa of approximately 10.
08:00 - 08:30 In terms of basicity, we have three potential candidates: the oxygen and the two bromine atoms. You'll remember we saw that in previous videos, halides or halogen atoms in organic molecules have effectively zero basicity. They are not willing or inclined to share their electrons. You'll never see a halogen atom getting protonated in an organic molecule. And so here, that essentially leaves us with the oxygen. If you protonate that oxygen, you're going to make an oxonium. Oxoniums we're looking at a pKa of approximately -4.
08:30 - 09:00 Finally, let's look at molecule number 3. Here, there are no hydrogens bound to heteroatoms, so we have to look at whatever is bound to carbon. And so here, we have carbons that are essentially sp3, sp2, and sp in nature. You'll remember the heavier the s character of your orbital, the more acidic the hydrogen is going to be. So here, your most acidic hydrogen is going to be the alkyne hydrogen, which has a pKa of about 25. In terms of basicity,
09:00 - 09:30 we have the choice between the nitrogen and the oxygen. Remember that nitrogen is pretty much always more basic than oxygen, especially when you're dealing with a carbonyl. Now comes the tricky part. A lot of students make this mistake. Here, very often students see that functional group and just see, 'Oh, that's an amine.' Amines, once protonated, have a pKa of 10. This is completely wrong here. This is not an amine, this is an amide! That carbonyl is pulling on these electrons quite strongly,
09:30 - 10:00 which means they are less available to be shared. This nitrogen is much less basic. If you protonate the amide, the conjugate acid is gonna have a pKa around -1, -2. Okay, now that we've identified all of the acidic and basic positions in each of our molecules, we can determine how they're going to behave as a function of pH. So starting with molecule number 1, molecule 1 is not really acidic.
10:00 - 10:30 The most acidic hydrogen has a pKa of 20, which tells me that if I want to ionize, if I want to remove a hydrogen, I need to increase the pH to at least 21 or 22, which I'll never be able to do. And so here, that's why it doesn't really matter if you identify this hydrogen or this hydrogen as the most acidic. This is even less acidic, it's even less inclined to be deprotonated. So here, you can't negatively charge this molecule in water. However, this nitrogen is quite basic. The conjugate acid of this has a pKa of 4. What that tells me is that if I lower the pH to 3 and below, the solution is going to
10:30 - 11:00 be acidic enough to force this aniline to be protonated and thus water-soluble. In the case of the phenol, the second molecule here, this hydrogen is moderately acidic. It has a pKa of 10, which tells me that if I take the pH of my solution and I raise it to at least 11,
11:00 - 11:30 then the water is going to be basic enough to deprotonate, create a negatively charged phenoxide, which is going to be water-soluble. As for the basic position, this is not a good base. The conjugate acid of this basic atom has a pKa of -4, which means that if I want to force this molecule to be protonated and positively charged, I would have to lower the pH to -5 or -6, which I'll never be able to do. So the basic reaction is nothing to worry about. Finally, we have molecule 3. If I want to ionize, I need to either have
11:30 - 12:00 very, very basic conditions. I need to have a pH of 26 to remove this hydrogen, or I need to have a pH of at least -2 or -3 to force a hydrogen on this nitrogen and make it positively charged, which you'll never be able to achieve. So here, this molecule will not participate in acid-base reactions in water, not to any meaningful extent anyway. Now we have all of our data, we can proceed with the flow chart. So here, the idea is pretty simple. We're going to have an organic solvent. Ethyl acetate is a good
12:00 - 12:30 universal choice. You can use dichloromethane (DCM), or diethyl ether. These are all great organic solvents that usually do a good job at dissolving organic molecules and that are immiscible with water. And so here, we're going to use an aqueous phase to try to separate these molecules. So in this case, for instance, I'm going to start with 1 mol/L HCl. 1M HCl is quite acidic. It's going to have a pH of roughly 0 or 1.
12:30 - 13:00 And so I'm going to have two phases, aqueous and organic. So these two molecules don't care about acidic conditions. They're inert. They are effectively going to stay in their original neutral form, which means that they are going to be preferentially soluble in the organic solvent, represented here. On the other hand, this molecule is basic. So under acidic conditions, it will acquire a proton
13:00 - 13:30 on that nitrogen atom and form an anilinium species that is positively charged. Now, remember, the question asks you to have an organic solution of each molecule. Ultimately, you need to have each molecule back into its original neutral state. So here, we're not in organic solvents, we're in water. We have a charged molecule. This is where acid-base reactions are great. They're reversible. I can easily take that hydrogen away and regenerate
13:30 - 14:00 my neutral molecule. All I need to use here is an excess of a strong base, such as sodium hydroxide. Again, have an organic solvent such as ethyl acetate or whatever, and then I'm going to have again an aqueous and an organic layer. My aqueous layer is going to have excess sodium hydroxide. I'm going to have sodium chloride too that results from the HCl and the NaOH reacting together. And then ultimately, this molecule is going to be deprotonated back to its original form, as such.
14:00 - 14:30 Now we still have a mixture of these two organic molecules. Now you have to remember, the phenol is mildly acidic, so it will react under basic conditions. So here, I can use, for instance, 1M sodium hydroxide and ethyl acetate. 1M sodium hydroxide is going to have a pH of around 13 or 14. And so here that's more than basic enough to deprotonate my phenol and
14:30 - 15:00 form the corresponding phenoxide, as shown here. In this case, the amide, like I said earlier, is just not reactive to either acids or bases. It remains under its neutral form and it remains in the organic solvent. And so ideally, I have an organic solution that has the pure amide in there, so I am done with the amide.
15:00 - 15:30 Finally, I just need to finish with the phenoxide. The phenoxide is negatively charged and dissolved in water. I want to bring it back into an organic solvent. So here, I need to use excess acid. If I use excess acid again in the presence of an organic solvent, such as ethyl acetate, then I'm going to have an aqueous layer and an organic layer. My aqueous layer will have excess sodium chloride, excess acid, and whatever. I'm going to have my neutral phenol
15:30 - 16:00 in the organic layer. Hiding a little bit at the back but this is the neutral phenol that I had here. And so, that's it! I'm done. I have successfully separated the three molecules here into three distinct organic solutions simply by exploiting acid-base reactions. Note that the exact order of the steps is not that important. Here, I started with acid to extract the anilinium. I could have very well started with sodium hydroxide here to extract the phenol and then use acid to extract the anilinium.
16:00 - 16:30 Ultimately, I would still end up with my three compounds in three different layers. Question two has us separate the following four molecules into four distinct solutions. Now, note that the question doesn't specify that it has to be four organic solutions. So, ultimately, if you finish with the molecule in the neutral form in an organic solvent or in the ionic form in water, good enough for the question. Here, we just want to separate each molecule into a distinct,
16:30 - 17:00 separate one. And so here, as always, we need to figure out what is going to be the behavior of our molecules in water as a function of pH. And for that, we need to figure out what's the most acidic hydrogen and what's the most basic atom in each of our molecules. So, let's start by doing that. So, molecule number 1, we have a very obvious candidate, which is going to be the hydrogen that is bound to oxygen. Oxygen is quite electronegative, pulls on the electrons,
17:00 - 17:30 acidifies the hydrogen, especially if you have the carbonyl here that further pulls on electron density. And so, here we're looking at a carboxylic acid. Carboxylic acids have pKa values of approximately 4 to 5, so let's call it 4.5. Now, in terms of basicity, we have potentially a couple of atoms. We have nitrogen and two oxygens here in our nitro group, and we have two oxygens in the carboxylic acid group. What you have to realize here is that nitro groups are very much like halides or halogen
17:30 - 18:00 atoms in organic molecules. They have essentially zero basicity. Right off the bat the nitrogen atom already has four bonds, it has no lone pair available to share. So, the nitrogen cannot play the role of a base. Now, you do have a negative charge. That negative charge is delocalized between the two oxygens by resonance, meaning that they're not really available for sharing. So, the nitro group has zero basicity. That leaves us with the two oxygens here. Realistically, the carbonyl
18:00 - 18:30 is going to be the most basic atom. That being said if you protonate the carbonyl and make an oxonium, that protonated carbonyl has a pKa of approximately -5. So, it's not a particularly basic atom. In terms of molecule 2, here we have the choice for acidic hydrogen in terms of either the hydrogens in the amine group or the alcohol group here. Oxygen is more electronegative than nitrogen, pulls on the electrons more strongly, and has a stronger acidifying effect. Now,
18:30 - 19:00 you have to realize this here is not a carboxylic acid like what we had here. There's an extra carbon that separates the alcohol from the carbonyl. And so, ultimately here, our alcohol or the hydroxyl group pretty much behaves like an alcohol, which has a pKa of approximately 16. In terms of basicity, we have the choice between nitrogen and two oxygen atoms. Nitrogen tends to be more basic because it's less electronegative, more willing to share the electrons. Especially
19:00 - 19:30 here, we're looking at a carbonyl. Carbonyls are not basic. And so, in the case of our amine, if we protonate it, we're going to make an ammonium. Ammoniums have a pKa of approximately 10. Now, looking at molecule number 3, there are no hydrogen atoms bound to heteroatoms. So, we have to look at hydrogens that are bound to carbon. We have the choice between sp3 and sp2 carbons. The heavier the s character, the more acidic your hydrogen is going to be.
19:30 - 20:00 Realistically, your most acidic hydrogen is this hydrogen here since it's in close proximity to the electron-withdrawing carbonyl group. Either way, we're looking at a pKa of approximately 40. In terms of basicity, we have the choice between three different oxygen atoms here. Realistically, the most basic oxygen is going to be the methoxy group. Carbonyls have very low basicity. This oxygen here, the lone pairs of electrons, are delocalized quite strongly
20:00 - 20:30 with the carbonyl. Yes, you do have delocalization of your electrons with the aromatic ring, but usually the carbonyl does a better job at delocalizing electrons than the aromatic ring would. And so, realistically, this oxygen is the more basic one. If you do protonate it, you make an oxonium. Oxoniums have pKa values of usually around -4. Finally, molecule 4. Molecule 4 has one hydrogen that is going to stick out from the others in terms of acidity, and that is going to be the hydrogen bound to oxygen. So,
20:30 - 21:00 that's our phenol group. Phenols usually have a pKa of 10. In terms of basicity, we have two potential atoms, the oxygen and the nitrogen. As always, nitrogen tends to be more basic. And so, here, that is still going to be the case. That nitrogen is quite basic. It's an aniline group. If you protonate your aniline, you're going to make an anilinium. Aniliniums have a pKa of approximately 4, unlike, say, your oxonium, which would have a pKa of like -3, -4, -5.
21:00 - 21:30 So now that we have all of this information, now we can figure out how each of the molecules is going to behave as a function of pH. And so, molecule number 1 has a fairly acidic hydrogen, the carboxylic acid. And so, here, as long as the pH is over 5.5–6, the solution is going to be basic enough that your hydrogen is gone, it's deprotonated, it's going to be negatively charged, and it's going to be water-soluble. In
21:30 - 22:00 terms of basicity, it's not something we have to worry about. The most basic atom is realistically this oxygen here, which has a pKa of -5. Which means I would need to have a pH of at least -6, -7 to force this to be protonated and positively charged. Not something that's going to happen. For molecule number 2, in terms of an acid reaction, we don't really have to worry about it. The alcohol has a pKa of 16, which means that if I need to deprotonate this alcohol
22:00 - 22:30 to a significant level, I need a pH of 17, 18, 19. Not going to happen. I don't have to worry about this. On the other hand, the amine here is moderately basic. And so, here, as long as my pH is inferior to 9, so 9, 8, 7, acidic enough that my amine is going to be protonated as an ammonium and thus water-soluble. Molecule 3, we don't have to worry too much about it from an acid-base reaction point of view. The most acidic hydrogen has a pKa of 40. The most basic position has a corresponding pKa of -4.
22:30 - 23:00 It means that if I want to deprotonate or protonate my molecule, I need to have respective pH values that are respectively 41, 42 or -5, -6, neither of which is going to happen in water. And so, finally, molecule 4 is a little bit interesting. It has two reactive positions, acidic and basic, both of which are susceptible to react in water. So, on one hand, you have the phenol. The phenol is moderately acidic. So, if you put it in a solution with a pH of at least 11,
23:00 - 23:30 it's going to be basic enough that you're going to deprotonate your phenol and make it negatively charged. On the other hand, you have the aniline. If your solution is sufficiently acidic, so 3 or less, then your solution is going to be acidic enough to force a hydrogen here and make it positively charged and thus water-soluble. So now that we have all this information, now we can start doing our flow chart. Let's start with that. Let's say that I'm starting with my organic solvent again. I'm going to use ethyl acetate,
23:30 - 24:00 although there are a few options here that would be valid choices of solvents. And here, I'm going to use, to start, one molar HCl, which again would have a pH of approximately 0 to 1, depending on how accurate you are in your preparation. Again, I'm going to have an aqueous and an organic layer. So, if I have a pH of 1, let's say 1, what's going to happen? This molecule doesn't care. This molecule doesn't care. These two molecules, however, molecule 2 and 4, will react under acidic
24:00 - 24:30 conditions. And so, here, you end up forming the ammonium version of molecule number 2 like this, and then you're going to form the anilinium version or conjugate version of molecule number 4 as such. And then here, while molecules 1 and 3 are essentially going to be unaffected
24:30 - 25:00 by the acid treatment, they're going to remain electrically neutral and thus soluble in an organic solvent as such. So now we have two separate solutions that each contain two different molecules. So here, I'm going to start with the aqueous solution. It doesn't really matter which one I start with at this point. But now, in my aqueous solution, I have a mixture of two molecules that are both positively charged, and I want to separate each of them. Well, again, I can
25:00 - 25:30 use acid-base reactions to separate them. So here, I'm going to go the other way. I'm going to use very strong basic conditions. I'm going to use, let's say, 1M sodium hydroxide, pH of 13, 14. And so, here, what's going to happen? Under basic conditions this hydrogen, this extra hydrogen, is going to go away, and I'm going to reform the neutral amine here. Otherwise, under basic conditions, not a lot more is going to
25:30 - 26:00 happen. At a pH of even 14, there's only a very, very, very, very small fraction of my alcohol here that's going to be deprotonated. We're talking less than one percent. So, for the most part, my molecule is going to be electrically neutral and thus soluble in an organic solvent as such here. On the other hand, what about this molecule? Well, under basic conditions, sure, you're going to deprotonate the anilinium and reform the neutral aniline. But then,
26:00 - 26:30 you're going to have a negatively charged phenoxide and thus water-soluble. This is my aqueous, this is my organic. And so here I'm already halfway done. Remember, the question only wants you to separate each molecule into a different solution, whether it be aqueous or organic. So I have an aqueous solution of my phenolaniline here, I have an organic solution of my amine here. Now we have to take care of the first organic extract. We have a mixture of this acid here and this ester.
26:30 - 27:00 Well, similarly, we can further separate our molecules on the basis of acid-base reactions. Here again we are going to use basic conditions, such as 1M sodium hydroxide. I'm going to have an aqueous solution and an organic solution. Under basic conditions, you'll remember, this carboxylic acid is going to get deprotonated to form the corresponding carboxylate, as shown here, which is negatively-charged.
27:00 - 27:30 As for the ester, it doesn't react. It's not acidic, it's not basic enough, there's nothing that HCl or sodium hydroxide will have it do, and it will remain in the organic layer. So now, we are effectively done. We've separated each of our four molecules into four different layers, four different solutions. Note that I could have flipped the order of my flowchart. Here I started with acid. I could have very well started with sodium hydroxide. In that
27:30 - 28:00 case, I would have extracted the carboxylate and the phenoxide into that aqueous layer, and then, by using strongly acidic conditions, I would have obtained the anilinium in the aqueous layer and the neutral carboxylic acid in the organic layer. In that case, I would have had the amine and the ester in this organic layer, and by using a little bit of HCl, I would have extracted the ammonium
28:00 - 28:30 salt in the aqueous layer, along with the neutral ester in the organic phase like we have here. Finally, note that all of these reactions are still reversible. So, if I was really interested in obtaining the neutral version of each of my molecules, all I would have to do is reverse the acid-base reaction. So, in the case here, I have a carboxylate. If I want to have the carboxylic acid, all I need to do is put a lot of HCl in there. I'm going to force the pH to be very low. I'm going to force this molecule to go back to the neutral carboxylic acid
28:30 - 29:00 form. The only tricky part here would be the phenol here. Your pH can't be too acidic, or else you protonate the nitrogen, but it can't be too basic either, or else you deprotonate the phenol. And so, practically speaking, it just means that you would have to be very careful in controlling the pH of your aqueous solution to force the pH to be roughly around 7. If you're at a pH of 7, none of these positions are going to be ionized. You're
29:00 - 29:30 going to have a neutral molecule, and then you'll be able to extract it into an organic layer.