- Drawing enantiomers
- Cahn-Ingold-Prelog system for naming enantiomers
- R,S system
- R,S (Cahn-Ingold-Prelog) naming system example 2
- R,S system practice
- More R,S practice
- Fischer projection introduction
- Fischer projection practice
- Optical activity
- Optical activity calculations
Enantiomers are stereoisomers that are non-superimposable mirror images, meaning that one enantiomer will be the mirror image of the other enantiomer. In order to draw an enantiomer, you can determine the stereocenter, then swap the two groups attached to the stereocenter.
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- Hi, I had a doubt in dextrorotatry and laevorotatory molecules.
How would you differentiate between (+) and (-) molecules? For eg, at5:23, which one is dextrorotatory, and which one is laevorotatory?
(Note: I understand the definitions of each, just having difficulty identifying them from their structural formulae)
- So the (R,S) system for naming the configuration of stereoisomers is different from the (+,-) terms. Dextrorotary and levorotary corresponds to the angle an enantiomer rotates polarized light. Dextrorotary meaning a positive angle and levorotary meaning a negative angle. A molecule being R or S is independent of it being + or -. An R enantiomer of one molecule could be + while the R enantiomer of another molecule could be -. The only way to tell if a molecule is + or - is by experimental measurements. So we can determine whether the molecules at5:23are R or S, but not + or - just by looking at their structures. Hope this helps.(5 votes)
- I'm trying to look at the 1,2-dibromocyclohexane example from the point of chirality. (You can find it at4:00). Since both carbons attached to the bromine atoms are chirality centers, by the formula #(isomers)=2^(#centers) given in an earlier video, wouldn't there be four isomers for the 1,2-dibromohexane ring?
Number one: "top one" sticking back, "lower one" protruding out (in video)
Number two: both sticking back
Number three: "top one" protruding out, "lower one" sticking back (in video)
Number four: both protruding out
Perhaps the ones with both in the same direction are not sterically favorable, but aren't they isomers nonetheless?(3 votes)
- The Formula 2^(#centres) here, take the number of chiral centers i.e., differently asymmetric. In this Case as It's a cyclo-hexane ring, Both the 1 and 2 Bromo position are identical and thus there are only one pair of enantiomers.
Actually, for this structure there are 3 Unique Optical Isomers. 2 of them are Enantiomers, and one is mesformic structure, which is optically inactive. The structures two and four defined by you are essentially the same if you look at them from different perspective, and also one and three, The structures one and three, are the same, and their mirror images are same too. It's all about which Bromine you choose first. I would suggest drawing out each structure by naming one Br alpha and the other one Beta.(4 votes)
- why is the pentabromin ring an enantiomer ? it schould have a chiral center right ? but isnt there just 3 different groups attached to it ?(2 votes)
- So if you rotate a molecule, do you form another enantiomer or is it the same molecule?(2 votes)
- At6:37you drew two enantiomers of the original compound.So ,in exam do I have to draw the both the structures or anyone of them?
Which one will be the best to draw?(2 votes)
- There’s only two enantiomers full stop. The ones he draw in on the right are the same thing from a different perspective.
It depends what your exam is asking you to do as to what you need to do, we can’t answer that.(2 votes)
- In the second example he did, aren't those two carbons connected to the same groups though?(2 votes)
- Do you mean the one with the ring and Br?
Look at the chiral carbons individually. Each are bonded to 4 different groups: a Br, an H, a C bonded to Br,C,H and a C bonded to C,C,H.
They are definitely chiral.(2 votes)
- Is there no shortcut using Cahn-Ingold nomenclature to determine if a meso compound is present?
I thought there was some trick, like, "if EVERY stereocenter switches (goes from R to S)" then its meso, or like "every EXCEPT one stereocenter switches"-- or is that just a trick to determine enantiomer vs. diastereomer?
Thanks for any help provided! :)(2 votes)
- Why does it matter so much about the direction that they are pointing? Does my question make sense? I’m kind of bad with words.(2 votes)
- It doesn't, really. He's just explaining that no matter how we change the orientation of the molecule (the direction the molecule or its groups are pointing in), the enantiomers cannot superimpose each other. Like our hands, for example. No matter how we rotate each hand, neither of them will align perfectly. They're mirror images of each other. Also, yes it makes sense and I don't think you're bad with words.(1 vote)
- [Instructor] We've already seen in an earlier video, that enantiomers are stereoisomers that are nonsuperimposable mirror images of each other. Let's say we want to draw the enantiomer of this compound on the left. One way to do it would be reflect this compound in the mirror and if you look at this carbon skeleton, here we have our carbon skeleton with our OH group coming out at us in space. That's this model on the left. There's our carbon skeleton with our OH coming out at us in space. If we reflect this compound in the mirror, we'll see the enantiomer in the mirror. Our mirror image on the right is nonsuperimposable upon our model on the left. Let's just draw what we see, we see our carbon skeleton, like this, so let's draw that. So there is our carbon skeleton, and our OH group is coming out at us in space. So we could represent that with a wedge, let's fill in our wedge here. Now let's draw our OH, and so this drawing on the right, is the enantiomer to the drawing on the left. There's another way to represent the enantiomer on the right and to do that, let's check out the video. So, in the video I imagine an axis going through this carbon and then I rotate about this axis to give us another viewpoint of our other enantiomer. So here's a model of our enantiomer, and you can see our carbon skeleton with our OH coming out at us, attached to this carbon and then a hydrogen going away from us in space. If we imagine rotating about an axis, through this carbon. Let's go ahead and do that. We'll see another way to look at the enantiomer. So now we have, for our carbon skeleton, you can see our carbon skeleton looks like this now, and then as this carbon has the OH going away from us in space, and the hydrogen coming out at us. So here we have some pictures from the video to help us with our drawings. We can see that this picture is this compound, if you look at that carbon skeleton and you can see the OH coming out at us in space. So in the video, we took this compound and we rotated it to give us this image on the right. So these are just two images of the same compound. And this gives us another way to draw our enantiomer. This time our carbon skeleton's going like that. Let me go ahead and draw in our carbon skeleton, and our OH group is going away from us in space. So we have to put in the OH group with a dash, like that. So these two drawings, represent the same compound, the enantiomer that we were trying to draw. There are two main ways to draw an enantiomer, at least two ways that I like to use. The first way is to reflect the compound in a mirror> That's what we did first, we took this compound and we reflected it in the mirror and we drew what we saw and that gave us this drawing of the enantiomer. We've seen that this drawing on the right, is the same thing as this drawing on the left here and notice the difference between this drawing and our original compound. The carbon skeletons are the same if you look at these two, the only difference is that we changed the wedge to a dash. So that's another very convenient way to draw an enantiomer. If your starting with a wedge, change it to a dash. If your starting with a dash, change it to a wedge. Now let's draw the enantiomer of this compound. The first method that we'll use is the mirror method. So here's a simplified representation of our compounds so ignoring things like confirmations of our ring. On this carbon, our Bromine is going down in space so that's this carbon, so you can see that Bromine's going down. And on this carbon Bromine's going up in space, that's this carbon with our Bromine going up. In the mirror we can see the mirror image or the enantiomer. So let's go ahead and draw our enantiomer here. So we draw our cyclohexane ring, and at this carbon, we have our Bromine going down in space, so let's go ahead and put our Bromine going down. Then at this carbon, we have our Bromine going up in space, so that's a wedge. We draw in our Bromine here, let me fill in this wedge, and let's put in the Bromine. So this on the right, is the enantiomer. Sometimes you don't have model sets or a mirror, but you can draw the mirror image by just using this drawing on the left. Imagine a mirror right here, and just as you see up here, this Bromine is reflected in the mirror, this Bromine is reflected in the mirror, right. So these two, these Bromine's are reflecting each other. This carbon is opposite of this carbon, so that's this carbon and this one, and this carbon is opposite of this one, so this carbon is opposite of that one. Just a few tricks to help you draw the mirror image. There's another way to represent our enantiomer on the right, and let's go to the video to see the other way. So here we have our two enantiomer's, you can see they're mirror images of each other. If I rotate the enantiomer on the right, you can see it from a different view point. Notice that both chiral center's have been inverted and just to prove that these are enantiomer's of each other, let's try to superimpose one on top of the other. Notice how you can't do it. These are nonsuperimposable mirror images of each other. This picture on the right in the video, shows you the relationship between our two enantiomer's. So thinking about reflecting the molecule on the left in a mirror, we can see the enantiomer on the right. But we took this molecule on the right, this enantiomer and rotated it into this position. So now let's draw the enantiomer from this perspective. We start out with our cyclohexane ring and you can see at this carbon our Bromine is going down in space so that must be a dash, so there's my Bromine. And then at this carbon, our Bromine is coming up in space, coming out at us, so that's a wedge. Let me go ahead and draw in our wedge, and we'll put in our Bromine. So this is just another way to represent our enantiomer's on the right. So this drawing and this drawing are two different ways to represent the same molecule. So this is the enantiomer to the compound on the left. Let's look at our original compound and compare this drawing on the right. Notice that this carbon your Bromine is coming out at you in space. Where as with this carbon your Bromine is going away from you in space. At this carbon your Bromine is going away from you in space, at this carbon your Bromine is coming out at you in space. So to draw our enantiomer, we just invert all chirality center's. So if you have a wedge, change it to a dash. If you have a dash, change it to a wedge. So this way is often easier, so just make sure to invert all your chiral center's to draw the enantiomer. For bicyclic compounds it's easiest to use the mirror method. So if our goal is to draw the enantiomer of this compound, we can imagine a mirror right here. And we can use this picture above as a guide. You can see that the Hydrogen is reflected in the mirror, and the Chlorine is reflected in the mirror, let's go ahead and draw those in. So we draw our Hydrogen here and then our bond going down to this carbon and then the Chlorine going straight down from here. Notice how this Hydrogen is reflected and this Chlorine is reflected. Next let's think about reflecting this carbon, so we need to draw a line in this way right here. So now this carbon is reflected with that one. Let me highlight those carbons right here. I'm going to extend this line out a little bit so we can see where the horizontal is approximately, like that. So that's this carbon and this carbon, so that's carbon's reflected in our mirror. Next, let's draw a line up in space relative to that line like that so that takes us to this carbon which is reflecting this one in our mirror. Now let's worry about this one, so this should go up in space, let me draw a horizontal line down here just to help us with our drawing. So approximately horizontal at this point. We know we want to go up in space from this point, so let's do that, so that's this line. This carbon is reflecting this carbon. Next, let's go ahead and reflect our top carbon here. Let's draw this line over to here, so you can think about this top carbon reflecting that one. Let's draw our lines over here, so from here and then back down to here. It's more of a drawing exercise really, we're going to go down a little bit, we'll leave that line broken so we can see that one's going behind. And then we'll draw this to the carbon in the back and then we know what draw down from this line to the carbon in the back and then we can connect those line. Finally, we've drawn our enantiomer. So on the right is the enantiomer to the compound on the left. And just to prove that this on the right is the enantiomer, let's look at a video where I tried to superimpose the mirror image on original compound. Here we have our two enantiomer's, so mirror images of each other that are nonsuperimposable, just to prove they're nonsuperimposable, I'll rotate the enantiomer on the right and try to superimpose it onto the one on the left. You can see that they don't match up. So these are enantiomer's