organic one. So there's a lot of people in here. If you guys have questions, feel free to type them into chat.
And then we'll, we have this semester review broken down by section. So we'll pause at the end of each section, sort of take time questions there. And we'll go through the questions that you guys leave in chat.
We'll go through the questions that were put into the Google form that we sent out. And if you want, during this time, we even encourage you to unmute yourself and ask questions if you have them. So if you don't have anything to add, we can get started with the first experiment.
Oh, wait, we have, we'll just fix. Yep. Yep.
Okay. So as far as the exam goes, you will have 60 minutes. So the exam starts at 7 30 PM on Monday, December 2nd. Please note. There was information sent out like a few weeks ago saying it was 6 p.m.
This has changed. So please know it is 7.30 p.m. It will be 60 minutes long. So it will go from 7.30 to 8.30 p.m.
on Monday. There will be 35 questions and it will be an all multiple choice exam. You do not need to bring a Scantron.
We will have Scantrons provided. All you need to bring are writing utensil and ID. And then if you want, you can bring a calculator. The calculators.
The calculations should be pretty simple, but we do recommend still that you bring a calculator. It may help you speed things up and make sure that you're sure of the answers that you're putting down. It doesn't matter what kind. Graphing calculator is fine.
Scientific calculator, basic for function is fine. More details on this can also be found on your Canvas page. So whenever you guys have like the regular experiments, if you just scroll all the way to the bottom, so below the last experiment, which was the elimination experiment, There was a week that was like no labs and then Thanksgiving week. And then at the bottom, there's one that says final exam.
It has all of this information. It has the first page of the exam, which is basically just instructions and a place to write your name. And then the final page is a periodic table.
So you know what you'll have access to in terms of like reference information for the periodic table. What's the best way to prepare? So.
we recommend that you study the discussion questions. These 35 questions that are on the exam are all structured very similarly to the discussion questions, except that they're multiple choice. So if you recall throughout the semester, whenever you were doing your discussion question, they're very much focused on the concepts behind the chemistry that we did. So if you think about some of the mechanisms that we looked at, if you think about like how polarity impacts some of our experiments. like how polarity impacts the reaction or the TLC plate.
Those are the things that we're going to be testing on. So very much concept based. Right.
Analysis section. So whenever you get data from your experiment, for example, if something moved a lot on a TLC plate, what does that tell you about your compound? Is it polar?
Is it nonpolar? Things like that. So now we'll get into the first experiment, which was thin layer chromatography.
Cool. So way back in the day, you did your TLC experiment. So we'll go through like the kind of slides that you guys saw and part of our like post-lab review and concepts, etc.
So just a review of polarity, you guys have been working with it kind of all semester at this point. But so nonpolar compounds have only hydrocarbons, and it can have elements with low electronegativity. So some examples are like hexane, it's all So carbon-hydrogen bonds, same as benzene, but then iodobenzene is considered pretty nonpolar because iodine doesn't have a very high electronegativity. If you think about the electronegativity trend, the highest is fluorine, and then it gets less so as you go down the halogens or less so as you go left on the periodic table. So iodine, since it's not very electronegative, you can still call this entire compound pretty nonpolar.
You also have polar compounds, and those have carbon atoms bonded to electronegative heteroatoms. And so these are things like ethyl acetate, dichloromethane, and triethylamine, because you have like those CO bonds or those C-chlorine bonds or like the nitrogen-carbon bonds. So that would give you like that electronegative difference to give you a polar bond. And then you also want to think about is this molecule symmetrical where all those dipole moments cancel out. You can think about when we did the WebMO exercises for chromatography, we looked at dipole moments.
That also tells you about if the molecule overall is polar. And then you have very polar compounds, and those usually have a like carbon heteroatom bond and a heteroatom to like a hydrogen bond. So you can think about that.
Like acetic acid has an ethanol and aniline are all in this very polar category. Acetic acid has those like carbon oxygen bonds and an oxygen hydrogen bond, making it more and more polar. And you can see that those polar bonds are all concentrated in one part of the molecule.
So it's not canceling out with anything. And so here's just you guys can practice this. You could like. after this and we send it out, you could pause the video and try to sort these into nonpolar or polar.
So just so you know, like this first one, it's all carbon-hydrogen. You have some carbon-oxygen bonds and oxygen-hydrogen bonds, so it's very polar. Here, think about what kind of bonds you have.
It's going to end up being polar because you don't have those heteroatom hydrogen bonds. Again, same thing here. all carbon hydrogen for the second to last one, and then the last one you have a lot of different polar bonds putting it in the very polar category. So just some more practice for you guys with polarity. It's a good thing to review just for all of your organic stuff.
Cool. So now we're going to talk about chromatography. So this is a general separation technique and so your stationary phase for normal phase chromatography, your stationary phase is polar. It's an immobilized material, and it causes the compounds to resist movement. And then your mobile phase is relatively non-polar to your stationary phase, and that helps the compounds travel further.
So if you start with your compound mixture here on the left, and you start adding your mobile phase, you're going to move things throughout the column. column or like a TLC plate or something like that. So compounds will travel a different distance based on their intermolecular forces with the stationary and mobile phases.
And this all depends on polarity. So as things travel through the column, they'll stick, quote unquote, to either the mobile phase, they'll stick to the stationary phase a different amount. And so the thing that traveled the least.
has the greatest affinity for the stationary phase and the lowest affinity for the mobile phase, and that's going to be your most polar compound because it's sticking to your polar stationary phase the most. And then at the end, the thing that traveled the furthest has the greatest affinity for the mobile phase and the lowest affinity for the stationary phase, and that's going to be your least polar compound because it's like going with your nonpolar mobile phase. So this is like the theory behind like any of the chromatography things that we used. So for TLC specifically, your stationary phase was your silica TLC plate, and then you spotted your compound mixture on the bottom, and then you had a relatively nonpolar mobile phase. Again, this is nonpolar in comparison to the TLC plate.
And so Everything depends, how long it moved depends on the polarity. And so it's kind of like, like dissolves like. So if something's very polar, it's going to stick to your polar stationary phase. If it's less polar, it will go with your less polar mobile phase. So this, again, this is for normal phase, TLC, and chromatography.
You guys also had questions about reverse phase chromatography, where things are just the opposite. As you run your plate, the spots will spread out if you have a good solvent system, so they'll stick to the similar polarity as the compound. Okay, so you can have multiple situations.
So for this most left plate, your mobile phase is too nonpolar. Your compounds are not moving enough with your nonpolar mobile phase. And so you can think about like when you did your experiment, this is like when you did 100% hexanes.
Your compounds. were more polar like the plate than they were non-polar like your hexane solvent. The furthest right, everything, your mobile phase is too polar. So your compounds are traveling the most with your like polar solvent because the polarity of your solvent is really similar to the polarity of your compounds and your plate is still like too polar.
polar to like make a difference there. And then in the middle you have a mobile phase that is just right so you balance the polarity of the solvent with the polarity of your compounds to get a good separation of spots. So your goal with like a TLC plate, if you're just trying to identify things, is a separate like the most distance between spots that you can get.
This changes when you do a TLC plate to try to determine a column chromatography solvent though. And so for our experiment, we thought about fluorine, fluorenone, and benzoic acid and where we would expect them on our TLC plate. And so you guys ran a mixture and standards of each of these compounds.
And the goal was to identify which spot was which on the plate. and you also wanted to optimize the solvent system. So you should kind of know concepts that go with how to optimize the solvent system like we just talked about and how to identify spots on a TLC plate given standards and also thinking just about their polarity.
So here you can see that benzoic acid is going to be the most polar of these compounds, so it's going to be the furthest down on the plate because it's going to stick to the plate. And then fluorine is the least polar. It has no carbon heteroatom bonds.
So it's going to travel the furthest with the nonpolar solvent. So your solvent options were hexane and ethyl acetate. You should be able to think about solvent polarity and solvent systems with things other than just hexane and ethyl acetate. And that was probed in your guys'discussion questions as well.
And so if you think about which of these will carry the spots further on the plate, it's going to be ethyl acetate because it's the most polared. And then we have questions that you guys submitted about TLC. So we kind of talked about whether or not polar or non-polar compounds travel further on a TLC plate.
Again, for normal phase, polar compounds stick to the polar plate, which they won't travel very far. So non-polar things travel the furthest. A spot on a normal TLC plate, the lower spot is going to be more polar.
It's sticking to the plate more. And then talk about the location of spots based on a polarity of the TLC plate. This is kind of like all going back to the idea of like dissolves like.
You can adjust your solvent system to change the RF values of your spots. So if you use, if you think about like, you run something in 50-50 hexane ethyl acetate and you have a spot of 0.2 and 0.9, if you use a less polar solvent system, so if you go maybe just 100% hexanes, the compounds are going to travel less on the plate because Basically, everything is more polar than hexane. The compounds are not going to travel with the solvent very far at all.
They just don't really like the solvent, like dissolves like, so they're going to stick to that plate. But if you change your solvent system to like 100% ethyl acetate instead, the solvent, it's still less polar than the plate and it has a pretty similar polarity to polar compounds. So it's going to...
push, it's going to bring those spots further up the plate altogether. So it all goes back to polarity. So if there's any chat questions about it.
Okay, so if we want to think about reverse phase TLC plates, if we go... Okay, so if this were a reverse TLC plate, we would have a nonpolar stationary phase and a polar mobile phase. So that means that nonpolar compounds are going to stick the most to the plate because they're the most similar. But then polar compounds are going to travel the furthest. So if we have this plate as a normal phase TLC and then we did the same mixture of compounds in a reverse plate, we would have the red spot on the bottom, the blue spot in the middle, and the yellow spot on the top.
because the least polar thing would stick to the plate the most and not travel up with the solvent. And then one of your discussion questions for TLC also talked about like reverse phase. So you should like review that with the comments from your TA. Cool. All right.
So let's talk about the main concepts. surrounding the extraction experiment. So as you guys know, well throughout the semester, we did a lot of extractions. And so the main driving force behind that was our exploit of having two layers. So we have an organic layer and a water layer or aqueous layer.
Whenever we have different compounds, they will migrate to one of the two layers. And physically separating the layers is how we do the extraction to separate them. So how do we know where things get dissolved? And so this, I think, was the one place we had the most questions on. So the organic phase will tend to have organic compounds.
For our purposes, we'll simplify that to compounds that have no charge, so neutral charge. In the aqueous phase, we'll tend to have inorganic salts and salts of organic compounds. So for example, and we'll get to this here in a couple of slides, we had, we use benzoic acid this semester. And so benzoic acid is an organic compound. So I'll just skip ahead and come back.
So benzoic acid, when it has this proton is a neutral organic molecule. But when you take a base, right? So a base likes to take protons away from other things. If we take this proton away, and leave the lone pair of electrons on the oxygen, benzoic acid becomes charged.
It gains that negative charge. And so then it would switch over from being an organic compound to a salts of an organic compound. So now that it's charged, it would migrate over to the aqueous phase. And so that was one of the things that we did during this experiment. One thing to consider.
as well, remember, is that when you have your two phases, you want to consider which solvent is on top and which solvent is on bottom. And the way to tell that is the density of the solvent. So if we give you a solvent during the discussion questions, you had the luxury of being able to Google, like, what is the density of ethyl acetate? If we were to provide you with a solvent that you haven't used on the exam, will give you the density of that. But as a general rule of thumb, organic solvents are less dense than water.
So organic solvents are going to be on top of water. Mainly throughout the semester, we use ethyl acetate. So that's what we observed.
The ethyl acetate layer was on top and the water layer was on bottom. There's one exception, which are chlorinated solvents, such as dichloromethane, the formula which is written here. Chlorinated solvents tend to be heavier than water.
So then... those two layers would flip and water would be on top, dichloromethane would be on bottom. That was one of the questions from the discussion questions where we said, if you switch out ethyl acetate for a different solvent such as dichloromethane, where would the organic layer be?
The organic layer would actually be on bottom in this case. So there was some confusion about that, and that was where the inspiration for that question came in. Okay.
So what happens when you treat benzoic acid with a strong base? You turn it into a benzoate, which is a salt. So remember, salts dissociate in water. And if we disregard this sodium ion, we have benzoate, which is just benzoic acid. But instead of the OH here, we have O minus.
So we have a negatively charged compound. This is a salt of an organic compound. So it goes to the aqueous layer.
Then what we did was we neutralize it again. We put that proton back with hydrochloric acid to make benzoic acid. Benzoic acid is soluble in organic layer, right?
Again, because it is a now, once again, a neutral organic compound. So it goes back into that ethyl acetate organic layer. And then that's how we did the purification there and the separation there.
So just to recap real quick, is sodium benzoate soluble in an aqueous environment? Yes, because it is charged. It is a salt of an organic compound. Is benzoic acid soluble in an aqueous environment? The answer would be no, because it is, once you put the proton back, it is neutral.
So it will stay in the organic phase. It will not be in the aqueous environment. That was really the main question that we got.
So can you go over the properties of organic and aqueous layer? What is present in each layer? That's sort of what I talked about.
And that's the main concept behind extraction that you want to review in terms of that. Are there any questions about the extraction process concepts relating to extraction? I'll give it just a few seconds.
But if there are no questions to that, we can move on to the next, the concepts from the next section. Cool, so after you guys did your extraction, you did column chromatography. So most of the concepts are the same as TLC.
So you have like the separation of compounds for purification. So analytical TLC will kind of tell you and help you identify compounds preemptive, like presumptively, not for sure. And then column chromatography is a purification method. So we can think about if we have our compound mixture, what, and the TLC plate looks like this, what is the column going to look like slash what compounds are going to come off the column first. So again, we have our polar stationary phase.
So it's silica and loaded in a glass column and a nonpolar mobile phase. So it's like an upside down TLC. And so we load our compound mixture at the top. And then we use gravity to help push those compounds through the column.
And then you collected the fractions underneath the column. And so that'll be a combination of solvent and your compounds that you're trying to purify. So as you're running the column, think about what is going to elute from the column first.
So come off of the column first. So if our Red spot traveled the furthest in our TLC. That means that it's going to have the greatest affinity for the mobile phase. So it's going to travel the fastest in our column as well.
So we can run our column. We're adding solvent. We're collecting spots. We're collecting compound. And then you saw that the compounds came off the column in the same order that they were in the TLC.
So whatever traveled the furthest in TLC will come off first. But it was the red spot. The red spot came off first. So if we're trying to determine a good solvent for our good solvent system for our column, we can use TLC to try to like test different solvent systems. So for a column you want like the target RF you want on your.
plate is around 0.35. This just lets the compound separate the best in like a larger scale than our TLC. And so if we have our current RF value as 0.85 but we want 0.35, do we think the eluent should be more polar or less polar? So we want them to travel less on the plate, so we want to lower the polarity.
So if we're using hexane and ethyl acetate, how should we adjust the ratio? So you want more of your nonpolar solvent. So that means we want to increase the like hexane in that ratio.
And so that's how we would move those spots down on the plate and make it a better solvent system for our column. And so you used a couple different TLC techniques during column chromatography as well. So after you ran your column, you used a like quick and dirty TLC plate to tell whether or not the fractions had anything in them. So you spotted them individually on the plate and then you didn't put this plate in a chamber to run.
And then after you determined which fractions had any compound at all, you spotted those on like your... TLC plates that you were used to like running and solving. So here you had In this example, there were compounds in fractions three through eight, so these were each spotted on TLC plates, and that can tell you whether or not things are pure, whether those fractions are pure.
So in this case, you can think about three and four only have one spot in the lanes of the higher compound, so those are going to be pure. And then six through eight each only have one spot as well, so those will be pure. But five, since it has two spots in its lane, shows that there are two different compounds in fraction five.
So you would not combine that with anything to like collect your pure compound at the end. So after you determine which fractions were pure, you combined your pure fractions to see to make sure they were still pure after you combine them. So like B1 could be three and four combined. And then B2 is six through eight combined. And then you compare them to a co-spot to make sure that the identity is the same that you predicted.
So like your data might have looked something like this. You had your quick and dirty plates. So you might have had spots in 11 through 20. And then you would try to identify which. fractions were pure.
And so in this case, you can see that this first fraction, in this case, it's labeled five is pure, and then 16 is pure. But the ones in between are all impure because they have multiple spots in their lanes. So overall, we only had like... One question about this, and it was asking the relationship between a elution speed and its R compared to that compound's RF value. So if something has a high RF value, it's going to elute out the column the fastest, and so it's going to have the lowest elution time because it spent the least amount of time on the column.
And so that's going to act in the same way as TLC would. So if something elutes off the column really fast. It has a high affinity for the mobile phase, and so that would be a least polar compound.
So if there's any questions on column chromatography, we can wait a second for any of those, but it's very similar to TLC. There's also reverse phase columns, and they act in the same way as reverse phase TLC. Cool. All right, let's talk about major concepts behind recrystallization.
So in this experiment, we took benzoic acid and we did a recrystallization. So recrystallization is a purification technique. There are four main principles to a good recrystallization solvent. So number one, obviously the solvent should not chemically react with your desired compound.
We don't want to change the compound that we want to isolate. The solvent should dissolve the compound when hot, and this will become important later when we talk about hot filtration and cold filtration. But it should dissolve your compound when hot. It should not dissolve your compound when cold. And then in relation to the impurities present, it should either dissolve all the impurities or not dissolve the impurities at all.
So we'll get into this when we talk about actually going through the benzoyl gas. So if. we represent our insoluble impurities and you guys have seen this graphic before so i'll just uh briefly go through it but insoluble impurities are purple benzoic acid is here in blue ins soluble impurities are here in yellow so the yellow compounds i've depicted as sort of mobile flowing around because they're dissolved the benzoic acid and insoluble impurities i've depicted as uh rigid sort of more rigid structures as crystals because they are not dissolved.
There's still solid crystals present in solution. So when you heat it up, the insoluble impurities remain insoluble. Think the sawdust or wood chips, if you threw that in the hot water and then soluble impurities like sugar, for example, will remain dissolved, whether it's hot or cold, but benzoic acid, because we, we chose the property that it should dissolve when hot, it will go from that crystal form to, um, the dissolved form.
Okay, so what happens whenever we do our hot filtration? Whenever we do our hot filtration, anything that's not soluble, aka anything that is a solid, will be trapped on our filter paper. So whenever we do the hot filtration, the only thing that is insoluble is the purple insoluble impurities. So that remains trapped on our filter paper. And then we remove that from the equation.
And then we're left with the dissolved benzoic acid and the dissolved soluble impurities. Then we cool it down until it's cold, until it's ice cold. And then we have...
Here, our soluble impurities. So remember, our sugar is dissolved, even though it's cold. And then our benzoic acid, which has very low solubility in water at this temperature, is solid. So again, I've depicted that as a crystalline structure.
Then, since we know that benzoic acid is a solid at this temperature, we can filter it out on the filter paper, physically remove it from the soluble impurities that are dissolved in water. And then we are left with a filter paper with our solid benzoic acid crystals remaining in the funnel. So that's what we took to get our percent recovery.
And that's what we did during this experiment. So again, recrystallization is a purification technique. It's one of four purification techniques that we talked about this semester. And that's how we purified it away from our impurities. One of the things that we did with our purification was we wanted to see how pure it is.
The way we did that was with melting point. So with melting point, we really want to talk about it as a range. I know we call it a melting point, but it really should be a melting point range.
On the low end of that range, that's when the first crystal begins to melt. And then on the upper end of the range, that's when the last crystal is melted. And so if your compound is pure, This range typically is very narrow, and that's why a lot of people may refer to it as a melting point. Because if the compound is pure, oftentimes that range is one or two degrees Celsius.
So I don't remember the numbers off the top of my head, but if your compound is, like if the low end of the range is 133 degrees Celsius and the upper end is 134 or 135 degrees Celsius. Typically, we'll refer to that compound as having a very narrow melting point, and that is indicative that your compound is pure. This is just sort of how to visualize that. You won't be expected to know this during the exam, but it's a good way of seeing whenever the first crystal begins to melt.
So that will not happen until sort of this third panel right here, crystals begin to melt, and then once it's completely melted, it's a liquid. So what happens if we introduce an impurity into our sample, right? Because we talked about what the melting point range looks like when it's pure. What happens if it's impure?
Two things typically will happen. You might observe that the melting point range will broaden. So earlier we talked about the melting point of a pure compound being narrow, one or two degrees Celsius in difference.
If it's impure, if the sample's impure, the melting point range will broaden. So 133 to 134 might begin to look like 130 to one. 40. So that would be a range of 10 degrees Celsius. And we would say that compound is probably impure if it has such a broad melting point range. So that's one thing, melting point broadening.
The second thing is melting point depression. So the observed melting point will begin to decrease if you introduce impurities. So If our expected melting point for a pure compound is 133 degrees to 134, if it has melting point depression, we might see a shift where our sample has a melting point range of 125 to 135. So it begins melting at a much lower temperature than we would expect it to if it were pure. So why is that the case? I don't remember if I have slides.
I do. Okay, that's fine. So if you have an impurity present, you're getting in the way of the compound trying to form crystals. So I guess over here, if our compound wants to form these nice crystal lattices, introducing an impurity might get in the way and that sort of disrupts the structure of the crystal lattice.
And so it'll be easier to break that up. So you don't have to go as high of a temperature in order to break up that crystal lattice structure. So that's why when you... introduce impurities, think it's easier to break them apart, you need to use less energy to do so, the melting point range will be lower.
So melting point depression and melting point broadening are the two things. that happen with an impure sample. We also did percent recovery.
So we get this question throughout the semester all the time. Make sure you know how to calculate percent recovery. So if you have your pure sample, think however much you ended with at the end of your purification. So pure benzoic acid crystals. Divide that by your initial sample, what you started with at the beginning.
That makes sense, right? Because you're trying to calculate how much you recovered. So if you ended up with one gram and you started with two grams, it's one over two, which is 50% after multiplying by 100%. So I think what you ended up with divided by what you started with.
That is your percent recovery. Those are the main concepts for recrystallization. Does anybody have any questions regarding... this experiment or any concepts related to this experiment.
Okay. Cool. So now we kind of finished most of our like technique labs and experiments. So then you started like doing things that we might do in lab. So first you guys did a natural products isolation.
So you had dried orange peel and you wanted... The goal was to do a microwave facilitated extraction of limonene from these orange peels. And then you did a workup with liquid-liquid extraction to try to obtain limonene and identify it via TLC analysis.
So overall, natural products in organic chemistry are utilized in countless commercial processes like essential oils and things like peppermint oil. That's extracted from peppermint leaves mint leaves um and then like rubber entire like rubber in general originally it was isolated from rubber trees so things like that and then historically and currently an efficient way to isolate these products is a major area of research in like chemistry and other industrial like jobs and then so you In addition to efficiently isolating these, some natural products are converted to molecules that have a similar function, but they can improve on other characteristics. So an example of this would be salicylic acid is found naturally in willow bark and had been used for thousands of years to soothe pain as a pain reliever.
And then eventually biturine co. It synthesized acetylsalicylic acid. So you can see like the base molecule of that salicylic acid still here. And they functionalized it to reduce like any previous side effects. So it's a lot safer to use and not, it still does the job of like pain relief.
So that's another reason that like natural products are important in chemistry. And then so for the the Limonene isolation, we did a solid liquid extraction in the microwave. So it's kind of like when you make tea. So you make tea and you like have all these like leaves and stuff in a teabag.
And then the compounds will seep out of the teabag into your like mug of tea, basically. And so that's kind of like what we were doing with limonene is you had basically. you were making like orange peel tea and that limonene in the orange peels was being extracted out of the orange peels into solution. And that's how you were able to like separate out limonene after doing the like extraction workup afterwards.
So you can see like our orange peels at the bottom of our microwave tube. And then eventually that limonene was separated out. And so overall, you did like a microwave-assisted extraction.
So again, using the microwave, it's a more modern piece of technology and it is a different way to add energy into our sample that might be more precise or more efficient than maybe just heating things up like you would like boiling water instead. And so you filtered everything, did a workup with extraction, and then you used TLC and compared this to a standard. So overall, you want to think about what information did you gain from your TLC plate.
Imagine that it worked great. And then how you determine that. So...
Overall, you could, when you compared your extracted sample to your limiting standard, you could try to have a presumptive identification of limiting by comparing RF values. So if E is our extracted sample and then S is our limiting standard, you can compare the RF values and think about, do we think that limiting is present in our sample? You can't say for sure. because TLC doesn't give you 100% certainty, but you can get a good idea of it. So in this sample, we'd say, yeah, limonene's probably there.
The RF values are really similar. So that's all we have for natural products isolation. If we have any questions about it, that's cool.
Basically just extraction. It's more extraction. Yeah. Okay.
So Let's talk about the nucleophilic substitution experiment. What was the main thing, what was the main concept behind this lab? Well, it's a nucleophilic substitution. So we typically think of two main ones, especially an organic one. So there are SN1 substitutions and SN2 substitutions.
So in this lab, we'll go over the SN1 substitution, which was a reaction that we did. And then later, we'll talk about SN2 later. So this was the reaction that we did. We did...
a substitution about here where the water was the leaving group, and then we had either a chloride ion or a bromide ion attack as the nucleophile. So looking at an SN1 substitution, it happens through two steps. So think unimolecular, two steps. In this case, in this example that I've chosen, just we'll treat... the chloride ion as the leaving group and the water will be the nucleophile in this case.
So the key point of an SN1 reaction is that in the first step the leaving group leaves. So this chloride ion will take this pair of electrons with it. Can I draw?
Or do I pull up the pen? On the top maybe? I know how to do it on the mom. Okay, whatever. We won't worry about it.
Okay. So this chloride ion will take this pair of electrons with it when it leaves. And so now this carbon is missing out on a bond.
So it likes having four bonds, one to hydrogen, one to methyl, one to benzene. And then in this case, it's missing a bond. So this is a carbocation and the chloride would be a negatively charged ion.
So this carbocation intermediate is a flat molecule. So once it becomes a flat molecule, think water can attack from either side of the flat molecule. It can attack from in front, in front, or from behind.
So the stereochemistry here is scrambled. You have a 50-50 mix of water that has attacked from the front and water that has attacked from the back. So these are the hallmarks of an SN1 reaction.
So remember, unimolecular, two steps. Two steps meaning you create this intermediate. And then later we'll talk about the SN2 reaction where it happens with a bimolecular, one step. And then we talk about transition state between that one step.
So yeah, that's the main concept behind the nucleophilic substitution. we'll talk about a little bit of the analysis that we did. So we went through gas chromatography and mass spec. So I know a lot of you might not have talked a lot about mass spec, especially in the context, like during lecture, you might not have gone through it. So I will basically only talk about what you need to know in terms of this lab.
So grass chromatography works a lot like thin layer chromatography or column chromatography, where things will separate on its properties. And so whenever things move faster or move slower, that's how we separate things across the column. The detector in this case is sensitive enough to actually know how much we have of a compound.
So the larger the peak, the more of that compound that we had. So that is gas chromatography in a nutshell. Then we'll talk about mass spectrometry, which will basically...
help us identify our compounds based on their mass or mass to charge ratio, but essentially mass. Each compound has a distinct molecular weight, and that will be the readout on the mass spectrometer. And we can see the molecular weight and correspond that to the compounds that we had. And that's what we did during the semester. One thing we talked about was compounds can fragment.
I won't spend too much time on this, but sometimes when... a little bit of a compound falls off or gets chopped off based on the ionization process, that can lead to different molecular weights that you see on the mass spectrum. But the molecular ion peak, which is the whole molecule intact, usually can be seen here at the far right end. Okay, so one of the main concepts we talked about in terms of mass spectrometry was isotopic distribution of bromine and chlorine. The main point here was that bromine has two major isotopes that are stable, and these two major isotopes are present in similar amounts.
So whenever you see the molecular ion peaks for a brominated compound, they will exist as a one-to-one set of peaks. This is different when you have chlorine. So in chlorinated compounds, chlorine has two isotopes, but they are not in equal amounts. One of them is much more common than the other.
And so that's why when you see a set of peaks with chlorine in it, the ratio of peaks will be one to three. And that can be observed here. So on the bottom, you can see here that this set of peaks are in a one to one ratio.
And so peak two, which I'll go to in a second, but peak two represents the brominated compound because the... the compounds are present in one-to-one, the weights are present in one-to-one ratio. On the top, which is peak one, we can see that these two peaks are present in one-to-three ratio.
So that's how we know that this compound is chlorinated and the bottom compound is brominated. So we just said peak one was chlorinated, peak two was brominated, and we can see that peak one is much larger than, or peak one is much larger than peak two. So we know peak one is much more.
uh, present in the solution based on what we talked about gas chromatography. And that can also be observed here. So peak one was much larger. Like I said, it is, uh, two times the size. So 66% of the solution and then peak two, which was the brominated compound is 33% of the solution.
So that's how we can look at gas chromatography and mass spectrometry in tandem and sort of analyze our data. So this is basically what we did during the nucleophilic substitution experiment. And then we sort of did that later as well during the alkene addition experiment, too, which we will talk about.
Oh, sorry. Sorry. There was a question in the chat.
So it says smaller peaks are BR and taller peaks are CL. No. So that might be a miscommunication.
I hope this was clear. It doesn't necessarily mean that smaller peaks are BR and taller peaks are CL. What matters is the ratio of the peaks. So the ratio of these two peaks is the same. There's about the same height of this peak and the same height of this peak.
That's how you know it's a brominated compound because peaks are in one-to-one ratio. This peak seems taller, but it's really just because the ratio is different. This is three times the size of this peak right here.
And so it matters what that is. Okay, the other graph that I said. So this graph, right? So in this graph specifically, this peak is larger because they're...
Because there was just more amount of the chlorinated compound than the brominated compound. If we ran another sample where this peak is smaller and this peak ends up being larger, all that means is we had more of the brominated compound than we did of the chlorinated compound. It doesn't necessarily mean that the chlorinated compound peak will always be taller. It's representative of how much we've added or how much is in solution.
So, right, if I take water and ethanol and I mix them together. It entirely depends on how much I've added, right? If I can choose to add, you know, 10% water, 90% ethanol, or I can choose to add 10% ethanol, 90% water, the graphs will just reflect what's in the solution, but it doesn't necessarily mean that the chlorinated peak will always be taller.
I hope that's clear. Does that make sense? Yeah. It's just a reflection of what we have in solution. What are the peaks that you are supposed to look at specifically?
Like, how do we know which ones to use? Good question. So like I mentioned earlier, whenever you look at a compound, it has a particular molecular weight.
So carbons have a molecular weight of 12, hydrogens have a molecular weight of one, and then bromines have either a molecular weight of 79 or 81. So if you add them all up, so carbon one, two, three, four, five, six, six carbons. However many hydrogens there are plus 79 will equal 164. So that's the molecular weight of the compound. It weighs 164. So if you see along this X axis, this is mass. If you add them up and you see along this X axis, what the mass is, you would see one six, what was it? 164. Now, this compound is not what we used during class.
So this graph is what we did use during class. That's why you don't see 164. 164 is just an example. But along this x-axis, you would see here, oh, hey, 164, we know we're looking for this peak, whereas with this isotope for bromine, it would be 166. So that's how you know what peaks to look for is.
It is the molecular weight of the compound that you're looking at. That's the peak that you're looking for. So the difference between these two are one-to-one.
So if we add them up, 164 and 166, one-to-one ratio. Awesome. Good questions. Let's talk about alkene addition. Yeah.
And we'll have more mass spec for alkene addition too that we'll talk about. So cool. So the second reaction that you guys did this semester was alkene addition. So we took like... isolated limonene that we had from earlier in the semester, and we added, used NBS to make this bromohydrin.
So this was a two-step synthesis. So our first step was our alkene addition, and then the next step we did an epoxide formation. So the first week we did our alkene addition, which is what we're going to talk about right now.
So... Overall, our reaction, we started with limonene, added NBS in water and acetone at room temperature to give this bispromohydrin product. And you want to think about, is this the only product that you like made or expect to make? So that was part of your analysis for your ILN Pro was to think about what other side products could have been made.
So what you could have done is you could have noticed how we added NBS. across both of our alkenes in this scheme shown. But some side reactions that could happen, what if you only added NBS to one of the alkenes? So if you only add it to this alkene in the ring, then you would only have the bromine and hydroxide added on this portion and the alkene would have stayed. over on the right.
Or you could have the opposite where you only add the out only add NBS across this double bond and so the one in the ring would be unreacted. So you could have had three different products but our goal was and with the equivalents that we used was to get this bisbromohydrin. So bis is just that there are two bromohydrins on there.
So After our reaction, we did our workup where we extracted our mixture of compounds with organic solvent, and then we removed that solvent via rotovap. So that was another thing that you guys learned how to use this semester was the rotovap. Then after our workup, we did our analysis using TLC and GCMS.
And so for TLC, the idea is, do you have a pure compound at the end of your workup? And do you have any leftover starting material? So if NBS is our pure NBS, S is our pure limonene, and then M is our...
our mixture of compounds at the end of our workup. So you can see that there are multiple spots in that lane so it's not pure. And you can think about do any of these spots line up with any of our starting materials. So it seems like maybe NBS might be lining up with either this top spot or this bottom spot, probably more the top spot. So you might have leftover NBS which makes sense because you added a lot of extra NBS than you needed.
So you can think about purity from that standpoint. And then we also had GC analysis for this. So overall for mass spectrometry, you have, like Brian was saying earlier, you look for like the molecular weight of that compound.
So if we have our molecular weight of 165 as our average molecular weight, you can see this kind of doublet. These two peaks down at 164 and 166 because that bromine has that 50-50 ratio. But sometimes in mass spec, parts of the molecule will break off and give different masses, which is why we see all these other peaks.
So like this 135 and 137, you just lost a couple of methylene groups, which is why you have your average molecular weight of 136, but you still have that. one-to-one ratio telling you you have a bromine. So that's why even if you didn't see your molecular weight, you could still say, oh, I have a brominated product there. And so since we have a dibrominated product, we have to think about for like a mass, mass plus two and mass plus four peak, that'll all show up together.
So if we kind of go through this example, we have our, this molecule, it has two bromines. The average molecular weight is 258. So you can think about if I have both of these as bromine 79s, you can have one in one as bromine 79 and bromine 81, or you can have both bromine 81s. So Bromine, the compound with both bromine 79s is going to be the lowest mass number of any of these.
So that's going to be our M peak. And so that molecular weight would be 256. And then here you have one and one on both of these molecules. And each of those molecular weights is 258. And notice how there's two possibilities of having this compound.
and only one of the normal m. And then if we get to both with bromine 81, we have our molecular weight of 260. And so you can kind of think about the ratio of the number of times you could get this product as 1 to 2 to 1. So you should have a group of peaks with these mass numbers that are this. m plus 2 should be twice as tall as our m or our m plus 4. So that's kind of how you would identify a dibrominated product in your mass spec. So if we think about just our chromatogram, you can think about it almost like a sideways TLC plate.
So each of these peaks is going to be a different compound. So more peaks indicate a less pure sample. So that's how you can think about purity from your GC.
And then looking at our mass spec, we can help identify if your target compound is there. So since we're looking for a dibrominated product, you're going to look for that 1 to 2 to 1 peak. And you can see down here at 297, you kind of see that. You see this little tick mark is about half as tall as this middle one, which again is the same height as this last one.
So even if that mass number doesn't line up exactly, you can see that pattern of the dibromination and identify you at least have something with a dibrominated that has two bromines on it. And so do we have any questions about this stuff? You should be able to identify which part of a reaction is the reaction itself, the workup or the purification and the analysis.
So a lot of the times during our post-lab reviews, We have this panel on the left side of the screen telling you like, oh, this is the part where you're like actually doing the reaction. And then the workup is when you quench any reagents. So you stop the reaction and you separate out the desired compounds from everything else. But then sometimes we didn't do a purification.
And so here we didn't do a purification because we ended up with a mixture of compounds. We ended up with our three possible products. And then during other experiments, we did do a purification.
Like our elimination lab, we did a distillation as a purification. So you should be able to identify those parts during in your ILN pros. Hopefully you like labeled those and then your TA can tell you whether or not like that's correct or not correct too. Cool.
Glad that helped. I guess I just want to add to that last point real quick. So again, the focus of this exam is going to be conceptual. And so while you might need to know like analytical techniques that we did, for example, I talked earlier about melting point range. What information does that tell us conceptually, right?
It tells us about the purity of the compound. So that's an analytical technique we were analyzing. the compound that we had. And then things like distillation and recrystallization, those were things that we did to purify the compound away from other impurities, right?
So think about in that sense, but we will focus mainly on conceptual things rather than like experimental techniques. But yeah, all good points. Okay, let's talk about the epoxide formation.
So the main thing in this experiment was a an SN2 reaction. So earlier I talked about SN1 reactions, and I'll sort of compare that to the SN2 reaction here in this section. So the epoxide formation reaction took our product from last week. We'll call that our starting material for this week.
So that's this bisbromohydrin. And so we take these two functional groups and form epoxides out of them. So how do we do that? I'll get to that in a second.
Okay. So let's just take, for example, a molecule that has one set, one bromohydrin. So in the compound that we use during class, we had two, but in the example, I'll just take one. So what does the SN2 reaction look like?
Earlier, I mentioned that it is bimolecular, right? SN2 means bimolecular one step. It goes from this product to, or this starting material to this product in one step. And it will pass through a transition state when that happens. So looking at the mechanism for this, we had introduced a strong base of sodium hydroxide.
So again, what do bases like sodium hydroxide do? They like to take protons away. So we take away this proton, and the proton leaves with sodium hydroxide.
There's a lone pair of electrons. that will remain on the oxygen, right? Because before it was shared between the oxygen and hydrogen, that lone pair of electrons will now go to the oxygen, which has a negative charge. So now this lone pair of electrons will come down and be shared across this double bond at the same time that this bromine leaves.
And so that's why you end up with this epoxide formed where the location of the oxygen is the same side as where the hydroxy was behind the page. That's why it's represented as dashes. So why is this important in an SN2 reaction?
Whenever you have your geometry of a compound, so tetrahedral geometry, you have the R group in plane. We'll say this methyl group's in front, this hydrogen's behind. Water, in this case, the water is already attached to something, but the OH can come in and attack at this center.
It has to attack at the same time that the leaving group, which is the halogen, leaves. So down here, the leaving group was bromine. Here, I've just written it as a chlorine. So if the OH comes in and attacks here at the same time that this leaves, we can represent it with a transition state sort of when it's in the middle.
So partway while this bond is being formed, This bond is being broken between the carbon and the chlorine. And you can see that... It has to happen from opposite sides because if the bonds are being formed and broken at the same time, there's not really space on the right side of the molecule for water to come in and attack the carbon. So it has to happen from the same side.
And that's why we see this flip in stereochemistry, this quote unquote flip, where the water is substituted from the opposite side of the chlorine is substituted. So here in the same way, water comes in. and bromine has to leave from in front of the page because water attacked from behind the page over here.
This is in stark contrast to the SN1 reaction, whereas earlier we formed an intermediate that was flat and water could attack from in front or behind the page and that's why we had scrambling in that stereochemistry here, but in the SN2 reaction it has to happen from... it has to result in flipping of the stereochemistry. So that was the SN2 reaction. What did we talk about in terms of analysis for this lab? So one of the ways that we did was we looked at it by TLC.
So this will call back to a lot of things that Mary talked about, where we look at the functional groups of this compound, right? So we have a lot of hydroxies, so O carbon-oxygen bonds and oxygen-to-hydrogen bonds, and bromines, carbon-bromine bonds. This is a pretty polar molecule. There's a lot of polar bonds, so it's a polar molecule. And that's why...
you might have seen it stayed at the bottom of the plate. Polar molecules stuck to the polar TLC plate, and it didn't move. Looking at the product, there are carbon-oxygen bonds, but for the most part, this is much more nonpolar. And that's why we saw that shift.
Whenever we went from the starting material to the product, we got less polar. And when it became less polar, this compound tended to move more along the TLC plate. So if we wanted to increase the RF value of this compound, how would we do that? Well, we want to push it further up the TLC plate, right? So if we want to push it further, we want it to interact more with the mobile phase, we would increase the polarity of the solvent and that would help push things further up the TLC plate.
So that's just sort of a callback to the neurochromatography. It all ties together, I promise. Okay, and again, we look at gas chromatography once more.
Think of gas chromatography just as another TLC plate. Every peak is a spot on your TLC plate, and this happens to be one of them. This one is nice.
So if you add up all of the weights of all these compounds, it adds up to 168.2, where we can see this is the peak that we're looking for. And this peak is just one peak. There's no bromine on this compound. So we don't see any isotopic distribution.
It comes out nicely. It's one peak. It's 168.2 right where we expect it. So that's so we can look at this spectra and say we have our compound here. Now on our TLC plate, our gas chromatogram, we do see some other peaks.
So we have some impurities present, right? But it looks a lot cleaner than our other one for alkene addition. So I would call this more pure of a reaction solution.
than the one for alkene addition, for example, because that one, I'll just go back to it quick. This one had a lot of spots on our TLC plate. There were a lot of different compounds and it was fairly impure. So that's sort of some analysis that you can do with that. This set of discussion questions was where we did calculation of a percent yield of multi-step reaction, right?
Because it was a multi-step reaction. We had starting material one, this was the product of the first reaction. And then the second reaction happened. This is the product of our second reaction. So just going over that again to calculate the percent yield of a, excuse me, of a multi-step reaction, you just multiply the yields together.
So if you had a 50% yield in your first reaction, so let's say we started with four grams of our starting material. If you had a 50% or four moles, we'll say four moles, because if you want to use grams, you have to do conversion. If we started with four moles of our starting material and we had 50% yield, we'd end up with two moles. And then if you have another 50% reaction of the two moles of our intermediate here, then we would end up with one mole of our compound.
So we ended up with a fourth of what we started with. And so that checks out. If you take 50% times 50%, you'll end up with 25%, which is a fourth of what you started with.
And again, you may have to do conversions from moles to grams, but that should be simple math that you can do with your calculator. And then that brings us to the last one, last section, elimination. Cool.
Last experiment was elimination. You guys just did this a week or two ago. So again, we have our reaction, workup, purification, and analysis. So our reaction, we had the acid-catalyzed dehydration of cyclohexanol.
So notice that since it's acid-catalyzed, acid-catalyzed? We don't have to think about whether or not this acid is a limiting reagent or not. Because the acid is regenerated, that's the definition of like a catalyst.
So we have cyclohexanol. You can see where that water comes off dehydration to make our alkene at the end. And so I thought I would change this.
Anyway, and so you... We did this reaction in the microwave. And so you wanted to think about what was all present and how that led you to the next steps of your like workup. So after our reaction, we had liquids and solids present. So that's why we had to filter things afterwards.
So we filtered and we did a micro extraction. And then during our micro extraction, if we go back to our original extraction experiment, you want to think about. which layer is the aqueous layer and which layer is the organic layer. And so you can think about the densities of your solvents. So since cyclohexane is a liquid, we can think of it as a solvent in our extraction.
So water, since it's more dense, will be on the bottom and our cyclohexane would be on the top. And then you obtained a crude yield. So this is before you finish your purification. you have a crude yield.
And then you did a new purification technique, which was distillation. So it's the purification of different liquids, of liquids based on their differing boiling points. And so for distillation to be effective, you should have different boiling points, different boiling points of the liquids you want to separate. So if we think about cyclohexanol, cyclohexene, and water. Overall, there's a difference of about 30 degrees between those boiling points, with cyclohexene as the lowest, around 82, and cyclohexanol as the highest, around 116, I believe.
And then water is at 100. So that is a big enough difference between all those boiling points to separate out that cyclohexene. And because cyclohexene is the lowest boiling point, it's going to distill off first because it's the easiest to boil. So it's going to boil, the vapors are going to go up into the distillation head and collect in the cooler air and condense on the sides, which you then saw the cyclohexene dripping down the sides and pooling in the collection well.
So biggest thing is that the lowest boiling point liquid will distill off first. So, and then after we did our purification, we did some qualitative tests. So this is a different type of analysis than we used before.
So first we did our bromine test. And so for a positive result, the red, like brownish red bromine color disappeared quickly because it was reacting with that alkene. And then once it had reacted, that bromine was all used up, so you didn't have any leftover like red or brown color. So that's how you could tell that you had either like an alkene or an alkyne.
You can think about the reaction, does it make sense to get an alkyne for this? Not really. So you have an alkene. And then if that like red-brown color persisted in the test tube, then you can see that that bromine did not react. So you probably don't have an alkene or don't have very much of an alkene to get rid of that bromine.
So that would be a negative test and you don't have your predicted product. The thing with qualitative tests is this bromine will go and react with other functional groups as well. So it's not 100% you would only have an alkene. So for this bromine test, you If your alkene is like really sterically hindered, it might be unreactive because that bromine just doesn't have enough room to go in and react. And so that's the reason that you might have still have an alkene, but see a negative test.
In our case, since we had cyclohexene, there's. No real steric hindrance. So that's not really a worry that we would have for this reaction, for this like experiment. But the color also may react with other func-the bromine may react with other functional groups, such as phenols, enols, amines, aldehydes, and ketones.
So for example, if we had acetone, which is a ketone, in our test tube when you're doing this reaction, the bromine might go react with that. So even if you didn't have an alkene present, you could still get a positive test if you had acetone in your test tube. So that's kind of things you need to watch out for with like a qualitative test. So in addition to the bromine test, we did a permanganate test.
And so we have our alkene will go be oxidized. by our potassium permanganate, which is a bright purple color. So if this reaction does take place, then the purple color will disappear because that potassium permanganate is being used up and a brown precipitate will form. And that brown precipitate is this manganese oxide.
And so that manganese oxide would be an indication that your permanganate did react to give with your like alkene. So if your purple color persisted and didn't disappear or turn brown, then you don't have like that functional group that's reacting with the permanganate. So that would mean assuming there's nothing else in your test tube that would react, that means you don't have an alkene present.
So there are like inconclusive or false results with the permanganate test as well. And the biggest thing that we wanted to think about for our experiment was it reacts with primary and secondary alcohols. So our starting material is cyclohexanol.
That's a secondary alcohol. And so that permanganate could go react with cyclohexanol and give a positive test. And so even if you didn't have cyclohexene present, if you still had starting material present, then... Right.
you would get a positive test and give you the false idea that you have an alkene present. So that's something you would want to watch out for with a permanganate test. And so the question we had submitted was going over the chemistry behind the bromine tests and the permanganate tests.
And so I just summarized them in a different way. I'm not going to read through it again since I just covered it. But if you want a different way of presenting that information, here it is for you guys.
And then... So, so overall, for knowing the different steps of like, a reaction, workup, purification, and analysis, you would just need to be able to like, categorize and say that, oh, like, recrystallization is a purification technique, or something like that. Or like, maybe they like, give an outline of like, some student did this set of things and it might ask what is like what is a viable purification technique yeah right so for example if we gave you uh a mixture of ethanol and water how would you how would you perform that purification well you wouldn't say recrystallization because you can't crystallize ethanol or water um you would do a distillation right You would use the difference in boiling points between ethanol and water and do a distillation. Similarly, if we give you like a solid compound, you wouldn't do a distillation with that.
You would probably want to either run a column to separate it on the column or do a recrystallization to purify that. So in a way, it's sort of both. So you don't need to know the exact compounds that we worked with, but you should know like the the i guess the benefits of each purification technique and what you would do with that yeah and when they would be useful in an experiment yeah good questions that's all that we had for experiments and then we have other questions that people submitted so do you want to start with sure i'll start yeah so uh what are the main topics we should focus on for the review we've sort of talked about this throughout but the only like we will only be covering what was present during the lab portion of this class.
So some of this material might overlap with what you talked about during lecture, but. If we didn't talk about it during lab, we won't ask you about it on the exam. Right.
I'll just do one more. OK, how should I approach questions about reaction mechanisms or products? Should I focus on drawing them or explaining them verbally? So in a multiple choice format, there's only so many ways we can ask you to draw, quote unquote, draw a mechanism.
So I would focus more on. understanding the mechanisms and why they happen the way they do. So sort of like you're saying, explain them verbally because right in a multiple choice format, there's just only so many ways we could go about doing that. Yeah.
And then be able to identify whether or not that mechanism is correct or like how it would change to make it correct. So we will not be providing a practice exam or practice problems. You guys have discussion questions. The exam is written by the same people who write the discussion questions, and it's multiple choice, so it's a different format.
The exam is not designed to trick you guys. We are not out here to get you. We want to set you guys up for success in taking the exam.
And so someone asked if it's more experiment-based or concept-based, so we've kind of been talking about that. So it's not, you're not going to be asked for the specific amounts of reagents. that we used in an experiment.
They're not going to ask you how much sodium sulfate did we use in our dehydration or elimination experiment. That's not what we're going to ask, but you should talk about like, oh why did you use the solvent system for the separation of these two compounds or something like that, but in a multiple choice format or like why a certain technique was used. And then So you are response. Someone asked, like, how much like reaction types, mechanisms and principles like that, that they might be asked about.
And so you're responsible for the types of reactions that we have done this semester in lab. Outside of that, we aren't going to ask you. So we did like additions, substitutions, eliminations.
We did epoxide formation. So if we did that type of reaction in lab, you should know, like. how that works, like the mechanism and like be able to predict the products of that.
You want to do the last? Yeah. So is theoretical and actual yield going to be tested? Yes. So just to briefly go over it again, theoretical yield, think about how much you're expecting to get, assuming everything, assuming the reaction goes perfectly, how much you're expecting to get.
Actual yield is how much you measured out during class. And then percent yield is just actual yield divided by theoretical times 100. Will we be required to do calculations? Yes, I would say there definitely will be stuff about like RF values, like yields. So recommend you to bring your calculators, although I think the math isn't too complicated, but go ahead and bring your calculator.
Identify solvents and products. So, yeah. If we give you a scheme, you should know what your solvents are, what your products are, what your reactants are.
If we give you a set of compounds and if so, if we give you an alkene and we say we're reacting with NBS, right, you should be able to predict the products because that's a reaction that we did during class. So it won't be exactly the same, but it'll be a very similar. Should I be prepared to explain what might go wrong during an experiment and how to troubleshoot some of the issues?
Yeah. So remember, these types of questions showed up during the discussion questions where it was like, based on this result, like if I were to cut out the hot vacuum filtration step from an experiment, what might go wrong? So sort of things like that. If you were to do XYZ to an experiment, modify it in some way.
Where might that change the experiment? There's one more question in the chat as well. Can you list out the names of the different techniques in the lab?
So I think we spent the past hour and a half doing that. So if you want to just scroll through all the sections, those are all the techniques that we did during lab. And they're also the names of the experiments and all the analysis you did.
Yeah, but one more page of questions. And so you should know the general safety considerations for the experiments. So like we used hexanes and ethyl acetate a lot this semester. So you should probably know that they're flammable. things like that.
So you should know the safety considerations for things that we use or maybe things that come up in the discussion questions. You should understand the goal of each experimental technique like recrystallization helps to purify solids, but you don't need to memorize that we were doing forming a bispromohydrin from limonene. You don't need to memorize that. And then So focus on studying the discussion question assignments the most, since those also address concepts that were in the analysis of the ILN pros.
And also now the discussion question assignments for all days through the epoxide formation have been published. So the only thing you don't have access to is the elimination experiment, which you turned in this week. So your TAs have not had time to grade it yet because it just got turned in.
So other than that, you have. everything available to you. We're trying to get that out by Monday, though. That's the goal. That's the goal.
We'll see if hopefully that happens. But yeah. Other resources to use to study for the exam. Your lab manual, the post lab review videos, this video, the introduction videos, discussion questions, Island Pros.
You're allowed to like look things up online to study. I like Ken Libre a lot. Sometimes AI is not good at chemistry, so I'd be a little hesitant about that, to be honest. But if there's like a reputable website that you've been using to help study for like lecture, for example, that would still be helpful for lab. Yeah.
Okay. Are there any specific steps from the procedure that we expected to describe or explain? So you don't need to know. So.
I'll say it like this. So if you remember during the recrystallization experiment, we go through hot vacuum filtration, then we do a cold vacuum filtration. So you need to know those steps in general and why they're beneficial to your experiment. But you don't need to know things like, oh, we at this point, we should be adding two grams of sodium sulfate to do the extraction.
I think. You should know the steps, but it's a very broad picture of the steps. So yeah, general procedure for running a successful column, right? You should aim for RF of about 0.35. You should start by adding your compound at the top.
Obviously, you want your top compound to be on the column, and then you add solvent to move it through. But you don't need to know exactly how much at each point, how much solvent you're adding. Just generally speaking.
You have your compound at the top, then you add solvent and you collect at the bottom, general things like that. Will the exam require me to make predictions based on the experimental setup? Right.
So yeah, I think I just mentioned this. So if I have a bromohydrin and I treat it with sodium hydroxide, what should I expect the product to be? That should be an epoxide.
That's an experiment that we did during lab. So we should be aware that treating this type of... compound with sodium hydroxide will result in formation of an epoxide SN2 reaction. Awesome. If there's any more questions, we'll give it maybe just a few seconds in case people are still typing, but we appreciate that you stuck with us for an hour and a half to listen to us yap about these experiments and these concepts.
So hopefully you learned something. If you have questions, feel free to ask your TAs. ask us email us with questions and we'll ask um and this video will be posted to canvas so please like we did talk at you for a very long time so if you want to just scroll through the parts that you're worried about studying and just revisit the parts revisit those um sections and there will be office hours next week on monday and tuesday and on december 2nd on monday um so if you want like further help there's also office hours that you can still go to.
And again, the exam is at 7.30 p.m. on December 2nd. It will be posted under the lab exam section. Yeah, it'll be at the bottom of that page on Canvas.
Yeah. And we'll also email it out to you as soon as YouTube finishes processing it. So like always, we'll email it to you and it'll be available on Canvas.
Awesome. thank you thanks good luck on the exam and have a good thanksgiving holiday