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0:00
- Hi, everyone. My name is Jolie Ren and I am an MCAT tutor for Shemmassian Academic Consulting. Today I will be going over an organic chemistry MCAT passage. And so what we're going to do is that I'll walk through my thought process as if I got this passage on the exam, and I will include any important points we should know. So with that being said, let's get started.
0:24
(keyboard clicking)
0:30
So from a first glance at this passage, it looks pretty short. It has three paragraphs and a figure. Sometimes you'll get passages this short.
0:42
And this is also an organic chemistry passage with a mechanism in the figure which tends to make people uncomfortable. But that's okay, we're gonna go through this together.
0:55
So, starting from the first paragraph. "In a new study, researchers aim to synthesize diindolylmethane, or DIM, derivatives based thiazole as a new class of urease inhibitors. Diindolylmethane is a natural product alkaloid used extensively in medicinal chemistry." So before we highlight or make notes or anything, when going through a passage, the goal is to stay focused and efficient. And so for that, we want to highlight terms that we're familiar with.
1:27
So the idea is that clinging onto familiarity in a passage we encounter for the first time helps us stay calm and discern what kinds of questions we'll get, especially because you'll never get the same passage twice. Also, we want to highlight terms with abbreviations, as well as numbers. So if the question asks for them, we wouldn't need to waste time hunting for these in the passage. So for example, in paragraph one, it introduces the research topic, which is DIM. So we could highlight that.
2:01
And it also seems like there's a lot of background info here, like urease inhibitors, diindolylmethane again. But here we're just looking for the main topic. And they only gave us two sentences, so there's really not much else we could extract from this paragraph, other than they're focusing on DIM as possible urease inhibitors.
2:27
So, moving on to the second paragraph. "Researchers first synthesized several diindolylmethane based thiazole analogs, one through 18, and characterized them with various spectroscopic techniques such as 1HNMR, 13CNMR, and EIMS.
2:46
Then, the analogs were evaluated for urease, or jack bean urease, inhibitory potential. All compounds showed excellent to moderate inhibitory potential with an IC50 value within the range of 0.5, plus or minus 0.01, to 33.20, plus or minus 1.20 micromolar,
3:09
compared with the standard thiourea, 21.60, plus or minus 0.70 micromolar. The synthetic pathway for these analogs is shown in figure one." So that was quite a lot of information. We're given numbers, the analogs. So whenever it feels like a lot, I like to go back and pick out parts that are important for the research study. So that would be the independent variable and the dependent variable. So, the independent variable here would be the analogs.
3:43
So the several diindolylmethane based thiadiazole analogs, number 1 through 18.
3:51
And they're testing for urease inhibitory potential, so that'd be our DV.
3:59
So evaluated for urease inhibitory potential, that's our dependent variable.
4:07
So another side note is that the MCAT will give you a passage topic that most people will not be familiar with. Like, urease inhibitors or diindolylmethane derivatives is not on the outline for an exam. And that is the gist of the exam, knowing what to do with unfamiliar information and picking out the big picture within it. So here we figure out the independent variable, the dependent variable, and how they tested it. So with these techniques that I'm outlining right now.
4:40
And then how do they measure inhibitory potential? They found the IC50 for these analogs
4:50
compared with the control. So the standard thiourea, which was this number. And then as we covered before, it'd also be good to highlight numbers. So in case, if the question asks for them later, we don't need to waste time going back to find these.
5:08
All right. So overall, this paragraph shows us more information about what the researchers are doing, measuring urease inhibitory potential with IC50 values.
5:24
All right, so now we're gonna analyze the figure.
5:28
Before we analyze any figure, we would want to read the figure caption, and this is "DIM-bearing thiadiazole analog synthesis." So we're just getting a synthetic scheme of this reaction.
5:44
So before we talk about the figure, I want you to pause the video and see what we're doing in the first step.
6:00
All right. So, welcome back. So what are we doing in the first step? So we have two of these molecules
6:11
with bromine sticking out of it, plus one of these.
6:18
And then we look at the product in the end. So we have like, a bigger molecule, basically. And so we're essentially combining these three starred molecules. And so another question you might be asking is what is missing? Or how do they even combine those in the first place? So looking at the molecule, we have the aldehyde here, but not the aldehyde here, or in this molecule.
6:53
So basically, these molecules were able to connect because of this aldehyde. So the carbon in the middle looked like it came from the aldehyde, since the aldehyde is missing.
7:08
And so, moving onto the second step.
7:15
Basically, when looking at a mechanism or synthesis, we wanna ask ourselves like, what is missing? What is being added? What is changing between the reactant and the product? And so for here, we're adding this sulfur variety.
7:31
And then looking at the product,
7:34
the overall structure appears to be the same, but then we had this new thing right here. So because of this new thing right here, was there anything that was replaced or missing?
7:50
So if we go back to the reactant before that,
7:54
yes, so it looks like this COOH mo-ID was missing.
8:01
And so it looks like it was a substitution reaction that they did. Not entirely sure what the mechanism might be, but now we know that COOH is out and this new reactant is in, and it turned into a five-member read.
8:21
So that's the second step.
8:24
And in the third step,
8:27
again, the difference between the reactant and the product is now that this molecule is added.
8:39
And so that came from the reactant here.
8:43
So basically, all these molecules, this synthesis reaction was a series of substitution reactions. In the first step, they combined three molecules, and they kept on replacing that group in the corner with another group.
9:01
So I just have a bird's eye view of the synthesis. Again, not entirely sure what the mechanism might be, but that's okay. And sometimes the questions don't even ask about the figure. But if they do, you can always come back to it.
9:16
So now, onto the third paragraph.
9:21
"Compound eight was the most potent inhibitor amongst all derivatives, with an IC50 of 0.50, plus or minus 0.01 micromolar.
9:32
In the next step. researchers used x-ray crystallography to determine structure activity relationships for the compounds. Researchers discovered that compound eight forms a pi-stacking interaction with a conserved Y87 residue within urease. Mutation of Y87 to alanine decreases compound eight potency by 3.7 fold.
9:56
The key binding interactions of most active compounds with urease were confirmed through molecular docking studies." Okay. So when reading between paragraphs in all science sections, and even CARS, we wanna see, like, what is different? What is the same? Are they being more specific, less specific? Are they switching topics? So for here, they're being more specific. They're still analyzing these compounds as potential urease inhibitors. So now we're given that compound eight was the most potent,
10:33
and they gave us the IC50 value for that.
10:38
And so now they're exploring compound eight a little more deeply here. So we know that it forms a pi-stacking interaction with a conserved tyrosine 87 residue. Side note, you should know all your amino acids, their structures, abbreviations, both three-letter and one-letter abbreviation.
11:01
And then now we also get more information about a mutation. So if it's mutated to alanine, it decreases compound eight's potential.
11:15
Yeah. So basically, noting compound eight, its interaction with the conserved tyrosine residue, and the mutation to alanine. So all of those are super important in this paragraph.
11:28
All right, so now we can go onto our questions. So when approaching these, note one, keywords. What is the question really asking? The MCAT likes to put a lot of fluff into the questions, and so your job is to either reword it or pick out keywords to make it make sense.
11:55
The second thing you want to ask is can I find information in the passage to help me out? Or is this something that I should know? And then third, when going to the answer choices, what can we eliminate?
12:12
All right.
12:15
So, question one. "Which of the following is not a strong acid?" Okay. So keywords, not. So anytime they have the word not, except, always, note that because it is a very common mistake.
12:34
And then the other thing is we're looking for something that is not a strong acid. So can we find this in the passage or is this a knowledge question? So there isn't really anything in the passage that hints towards information that helps us know whether or not something is a strong acid or not. And so this is something that we should know.
12:59
So strong acids include HCl, HBr, HI,
13:07
so the halides except for hydrofluoric acid.
13:14
Sulfuric acid, H2SO4.
13:24
And then HNO3.
13:29
So all of these are strong acids.
13:34
And so the one that is not on that list is ascetic acid, which is CH3COOH.
13:45
So, this question is fairly straightforward. So when you're reviewing, let's say you missed this question and you want to review it. So you should not only just review the correct answer, you should also review the incorrect answers as well. So for a content question like this one, I review all the common strong acids and all the common weak acids.
14:15
And another strong acid is perchloric acid, so HClO4.
14:24
So generally, these six acids are the strong acids that I should know for the exam.
14:31
All right. So, question two. "During the first step of the reaction in figure one, which of the following functional groups present on the reactants is not present in the product?" Okay, so again, there's a not.
14:48
So we're gonna look for one of the functional groups that is not in a product,
14:57
and then in the first step.
15:01
So again, looking at the keywords in the question is super important because it guides you to the portion of the passage that you need to look at.
15:13
And so, because this does ask us about the passage/figure, we need to go back in the passage to find this answer. So in figure one, the first step is the one that we went through in blue. And so now they're asking us, okay, so comparing the reactant and the product, which of the functional groups is not there? So let's go through the answer choices one-by-one and eliminate everything from there.
15:44
Okay, so A, aromatic ring. So do the reactants and the products have an aromatic ring? And yes, they do. So this is an aromatic ring. This is also an aromatic ring. So yes, they both do have an aromatic ring. And because the question says not, we can cross that out.
16:12
So what about an aldehyde?
16:15
So in the reactant, the aldehyde is here. The CHO, as aldehydes are often abbreviated.
16:24
But in the product, we do not have an aldehyde.
16:30
And earlier, we did talk about how potentially, how the aldehyde allowed the three molecules to combine with one another, and how that carbon in the middle was from the aldehyde.
16:44
So B is a potential answer, but because we want to err on the side of safety, let's make sure that answer choices C and D are completely wrong.
16:59
So choice C, carboxylic acid. So they're often abbreviated as COOH,
17:07
which is here in the reactant and here in the product. So there is carboxylic acid in there, so we can cross that out.
17:17
What about an amine? So amine groups can be RNH2, R2N.
17:28
As long as nitrogen is attached to three different things,
17:34
that's an amine group. Or R3N, NH3.
17:42
So these are the amine groups. In the reactant, we have an amine group here.
17:48
And then in the product, we have an amine group here. And then these were from the two molecules that were there. So we do have an amine group,
18:02
so we can cross that out. And so we confirm that aldehyde is not there, and is therefore the correct answer.
18:11
All right. So in case you miss this question, if it were a content review, content question, go back and review the functional groups that are relevant to the MCAT.
18:25
If it's because you didn't know how to interpret the figure, I would really highlight keywords and figure out a pathway that would take you to efficiently interpret a figure,
18:40
and then go back to the answer choices.
18:45
All right. So, question three. "Which of the following urease mutations at position 87 is most likely to maintain pi-stacking interactions with compound eight?" Okay. So looking at this question, it seems like they pulled it out of thin air. Where are they getting pi-stacking interactions from? How do mutations have to do with interactions with compound eight? So this one will cause a little more thinking, but not to worry. If we go over that strategy with picking out keywords,
19:16
figuring out if it's passage-related or knowledge-related or both and eliminating answers, we can figure this out. So the keywords would be urease mutations at position 87,
19:34
pi-stacking interactions with compound eight.
19:39
So it's most likely...
19:42
And so usually passage-based questions
19:48
are confined to one paragraph or figure in the passage. And so we know that in paragraph three, they talked about compound eight and position 87.
20:01
So what exactly is a pi-stacking interaction?
20:08
So these interactions are formed between aromatic rings. So let's say we have an aromatic ring on the bottom and then one on top of it.
20:21
And so these pi bonds form interactions with each other,
20:27
like so.
20:33
And so that's what they mean by a pi-stacking interaction. And so now they're asking us,
20:44
if the amino acid at position 87 were mutated to something else. So Y87 to what?
20:56
Which one would maintain a pi-stacking interaction with compound eight?
21:02
So compound eight, why would it do pi-stacking interactions? But looking back at the figure, the ending product here is for compounds one through 18. And so we know that compound eight has to have aromaticity in there.
21:22
So now we, basically we just reinterpreted the question so that it makes more sense to us.
21:33
So which amino acid in question three can do a pi-stacking interaction? That is essentially what the question is asking.
21:43
So this requires knowledge of amino acid structures.
21:48
So alanine has one carbon sticking out from, as its R group.
21:57
Proline looks like this.
22:02
Its R group looks like that, with three carbons. And then threonine looks like this.
22:13
And then phenylalanine is just a phenol ring.
22:22
So because pi-stacking interactions are formed between aromatic rings, phenylalanine is the only one that can do that, because it's the only one out of that cohort of amino acids that has an aromatic ring. So we would go with D.
22:40
All right. So lastly, question four. "Researchers want to decrease the rate of product formation from the second step of the reaction in figure one."
22:51
So keywords, second step, figure one. "Which of the following would not decrease the rate of compound two formation, according to Le Chatalier's Principle?" So again, we're looking for something that would not,
23:06
and then specifically we're looking at the rate of compound two formation in terms of Le Chatalier's Principle. Okay. So remember that Le Chatalier's Principle says that if equilibrium is disturbed
23:26
by a change of conditions somehow,
23:29
the equilibrium shifts to counteract that change in order to reestablish equilibrium.
23:35
So kind of like homeostasis, but in chemistry.
23:39
So looking at answer choice A, it says to decrease the concentration of the bromide-containing initial reactant.
23:51
So before that, let's go to figure one, step two.
23:57
So the bromine-containing initial reactant is actually in step one.
24:04
So if you decrease it here, that's where the initial stress is. So we're decreasing it here. And because of that, the equilibrium will shift so that it'll counteract that stress. So it'll make more of that initial reactant to maintain those equilibrium concentrations. Because it will shift to the left towards the initial reactants, you'll have less of compound two formation. Yeah, but we're looking for something that will not decrease it. So A would be out.
24:44
All right, now onto B. What if we decrease the concentration of the aldehyde and the carboxylic acid-containing initial reactant? Okay, so now they're asking about this product. Because you need both of those reactants to eventually form compound two, decreasing this will also shift the reaction to the left
25:11
and will decrease the rate of compound two formation. So B is wrong.
25:19
So what about C? What if we decrease the concentration of compound two? So here, if we decrease the concentration of that,
25:30
this is the stress that we're putting on our system. And so the system will make more of this and shift to the right to reestablish those concentrations. So it looks like C is a contender for a right answer,
25:46
because it will not decrease the rate of compound two formation. It will increase it.
25:54
So let's double check D. What if we decrease the concentration of compound one? So let's find compound one, which is here. So if we decrease that, the reaction will shift to the left to make more of that compound one. And because the equilibrium is shifting backwards
26:21
to reestablish those concentrations, you're making less of compound two. So we can cross out D
26:29
and choose C as our final answer. So as we saw, there was a mix of questions. Some only required knowledge of strong acids, but some required a little more thought, like in the pi-stacking question or the Le Chatalier's Principle. So if you get stuck on a hard question, not to worry. One strategy I like to share is, so for example in this passage, let's say if I find question three hard, I would just do questions one, two, and four first, and then come back to three later
27:02
when I feel more confident or have better context to these questions. If you enjoyed this walkthrough, be sure to like this video and subscribe to our channel so that you don't miss out on our future MCAT videos. And be sure to also check out our high-deal strategies in our MCAT 528 series, and click the link in the description box to sign up for our practice questions, so that you can get every last bit of practice in before testing. Happy studying.
27:30
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