For each structure,
1. star (*) any asymmetric carbon at...

Question

For each structure,

1. star (*) any asymmetric carbon atoms.

2. label each asymmetric carbon as (R) or (S).

3. draw any internal mirror planes of symmetry.

4. label the structure as chiral or achiral.

5. label any meso structures.

(a)

(b)

Accepted Answer

Hey everyone for each of the following structures assigned the configuration of each asymmetric carbon determine the structure as Cairo or a Cairo, show any plane of symmetry and label any measles structure. And now there are many things we want to do here. Let's just break it down step by step. First of all, thinking about configuration, we have to understand that if there's an asymmetric carbon, it implies that 60 carbon as Cairo and Cairo implies that there must be four unique or different substations bonded to that specific carbon atom. Looking at structure A we noticed that there is just one carbon atom which has four different substitutions is here in the middle because we have four different groups that that should we have a chloral group, a methyl group, an alcohol group and an ethyl group. And that makes this carbon atom a Cairo carbon atom or a stereo center. Now having identified our carbon, our carol carbon atom, we now wish to assign the configuration set and still that we have to recall the con single pre log rules and take care of the priorities of these substations. In other words, we wish to look at the bonded atoms and the first atoms bond. It's that carl carbon atom are carbon oxygen, another carbon atom and chlorine. And I will think about it this way which of them has the highest atomic number and that's chlorine. So it gets priority number one. Following the following substitution would be priority number two. That's oxygen because it's higher in atomic number than any of the two carbon atoms and now between the two remaining carbon atoms because there's a tie they have exactly the same atomic number of six. The longer straight alcohol chain would get a higher priority. So priority number three goes to ethel and priority number four goes to muscle because muscle is a shorter L kill chain. Now let's think about it this way I have my priorities. I have to make sure that priority number four or the metal group that I have here is pointing in the back or away from the viewer. In this case it's just sufficient to notice that it's on a dash. So that's perfect for us because the rules say that if priority number four is pointing away, we're ready to trace the path for substitutions from 1 to 2 to three and understand what kind of rotation we have. So I have a rotation, clockwise and if it's a clockwise rotation, I can immediately say that the configuration of this compound is our. So this is the first answer for party. We know that the configuration of this compound is our And I now need to think about the structure and identify whether or not it's Cairo to do that. I actually have multiple ways of doing this. And one of them is to actually try and draw its mirror image and think of something which correspond to the definition of finance humors. So what I'm going to do now I'm going to redraw that structure and I'm going to draw the mirror plane and what I'm going to do now, I'm going to draw its mirror image. So chlorine is in plain that I have an ethyl group in plain. A wage is pointing up and then I have a metal group on a dash. Now the question is, are these super imposible mirror images of each other? The answer is actually no. And to make it very clear you want to think about it this way? What if you imagine there's a vertical line or a horizontal line sometimes along which you're rotating your compound 1 80 degrees and you're trying to superimpose them. If you transform it, if you essentially rotated 180 degrees with respect to the vertical line, you will obtain uh this looking structure. First of all, you will swap the methyl group a DSL group and you will still have chlorine in plane. But now, because the wage group was pointing up now it will be pointing down so it will be on a dash. And the methyl group that was pointing down specifically this group will now be pointing up. And now, if you try to take this structure and superimpose it with the original structure, notice that they're non super imposible because the older age group alone is pointing in different directions. So what we can conclude from here is that this structure must also be Cairo because the two mirror images that we have drawn our non super imposible mirror images of each other. So it's a pair of an ant humors. Now, if it's Cairo, there's no internal plane of symmetry. So we can also add a comment that there's no plane of symmetry. And because there's no internal plane or line of symmetry, we can essentially say that such a compound is not miso. A miso compound implies that a structure is a Cairo. So if we have previously determined that the structure is actually Cairo, it's it's sufficient to understand for us that Cairo implies not miso. So we can also say that this structure is definitely not miso because it's Cairo. So these two together imply exactly the same thing there. They act like synonyms. Wonderful. So we are done with party. This was our party and now we are ready to look at part B for part B. The structure that we have is actually given on a Fischer projection. And what I'm going to do, I'm going to redraw it real quick on its fisher projection form and I'm going to explain how to read it so that you could easily assign configuration set. Now let's look at it this way. This is a official projection and if we have official projection, you have to recall that in particular the vertical lines that you're seeing these vertical bonds are your dashes and the horizontal ones are your wedges. So instead of drawing solid lines, you can visualize it this way and to make it a cleaner drawing, I'm just going to enjoy it. I have hydrogen on a dash. I have a metal group on another dash. I have a car boxing group on a wedge and I have an amino group on another wedge. I need to assign priorities to my compound. Of course I first of all have to understand that there is just one carol center here because there are four unique submissions in particular. Those are hydrogen amino group, car boxes group and the methyl group over here. So these four unique submissions make the middle carbon a Cairo carbon atom and now according to the con Ingle prelaw groups that once assigned priorities. Now we can essentially start from um the easiest one and the lowest priority of course goes to hygiene that be priority for because hygiene is the lowest in atomic number. And now looking at the other ones, If I have a tie between two carbon atoms, I can immediately tell that Nigerians priority one because it's greater an atomic number than any of the two carbon atoms. And between these two, priority number two will actually go to curb box cell group for by priority number three for the metal group. And we can easily see why because if you draw the structures for the car boxes group over here and the metal group, it's sufficient to notice that the main difference is that it doesn't really matter which bond you take. But let's suppose you're looking at the C. H bond and you have a competition between ch and ceo whether you take this ceo or under part of ceo, oxygen versus hydrogen is higher in priority. So that oxygen will be higher in priority and therefore the car boxes group itself will take a higher priority. And now that we have our priorities, we have to think about the rotation. And we're actually really happy to see that priority number four is on a dash. And because it's on a dash, I can assign my rotation directly. I will trace the path from position number one towards position number two and then towards position number three. And I will notice that this rotation is again our it's a clockwise rotation. So I understand that for part B. My answer is that this compound is also our. It has an R configuration. So now let's think about it this way. We want to understand whether it's Kyrill or not and we again can draw its mirror image and we can just try to superimpose its mirror image on itself. If I again draw that structure in its official projection, we can just draw, it's for fisher projection and we have hydrogen amino group, car box cell group and the muslim group. Now I'm going to draw airplane and I'm going to see the car box group reflected that way amino group. And then hydrogen and muscle. Now what if I flip it, What if I flip the second structure that I've drawn. You might say that I will see exactly the same thing as I see in the first structure. But recall that you will also flip your bonds Now if these are your dashes and vertical lines and these are your wedges which you of course have exactly the same thing over here. When you flip your compound with respect to a vertical line, you will change the way your bond, your bonds look like the two bonds that were pointing up. Will actually start pointing down. And you will not see the exact same structure that you're seeing on the left. You will actually start seeing this. You will start seeing age pointing up instead the methyl group pointing up and the two substations on the left and on the right pointing down now. So that'd be your C. 00. H. After you flip it. And that would be your and age two. And notice because these bonds in particular here and here, they're pointing in opposite directions. That means these two structures are not super imposible mirror images of each other. And as a result, this structure is also Cairo. So we can label it as Cairo. And now we're ready for an easy part if it's Cairo, we can immediately state what it's not me. So right and simultaneously, if it's not miso, then it has no plane of symmetry. So that's our final answer. Now let's just summarize our findings for part A. We said several things. We said that the structure itself has an R configuration, it's Cairo it has no plan of symmetry, and it's not a meso compound. Now, for part B, we said that this structure is also are in configuration. It's Cairo it's not missile, and again, it has no plan of symmetry. The correct choice for this multiple choice question is B. And in case you have any questions, feel free to drop your message down below. Thank you for watching this video.

For each structure,
1. star (*) any asymmetric carbon atoms.
2. label each asymmetric carbon as (R) or (S).
3. draw any internal mirror planes of symmetry.
4. label the structure as chiral or achiral.
5. label any meso structures.
(a)
(b)

Key Concepts

Asymmetric Carbon Atom

Cahn-Ingold-Prelog Priority Rules

Chirality and Meso Compounds

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