Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: deutdeut on January 19, 2007, 10:52:19 PM
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I want to ask if a molecule has 2 chiral centers, like 1,2-dichloro-1-methylpropane, how I determine whether it is chiral by considering its three-dimensional structure?
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Basically the same way you would check another molecule for chirality: take its mirror image and see if you can superpose the two. Usually molecules with multiple chiral centers will be chiral; however, it is not uncommon for these types of compounds to be meso (http://en.wikipedia.org/wiki/Meso_compound) (i.e. achiral). So, you definitely should test these molecules for chirality.
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But how can I draw the 3-D structure accurately, like the bond is pointing in or out, for the molecule and the mirror image to see whether they can superimpose?
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There's two ways. One way is preferentially easier than the first.
The first way is to draw the compound, then next to it draw the mirror image. If you can mentally rotate one of them 180° clockwise, counterclockwise, or however, and cannot superimpose it on the other, then you know it is chiral.
The second, easier, way, especially for more complex molecules, is to use Cahn-Ingold-Prelog (priority) rules for naming enantiomers. An enantiomer is basically a chiral compound.
First by assigning what group is the highest priority, then what group is the second highest priority, and from there you follow the directions on how naming the enantiomer.
An overview of this method can be found here (http://members.aol.com/logan20/configur.html#intro).
If one compound is (R,S) and the other (S,R), you have a chiral compound. Likewise, if one is (S,S) and the other (R,R), you have a chiral compound. See a pattern here?
This takes some time to understand, because Cahn-Ingold-Prelog (priority) rules are paramount.
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But how can I know which bond is in or out of paper?
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But how can I know which bond is in or out of paper?
Whichever you choose. ;D
If you take the time to go through the rules from that link I gave, you would know to put the lowest priority group in plane and the highest priority group out of plane.
Out of plane is represented by a solid wedge, whereas in plane is represented by a dashed wedge. As long as you reflect the mirror image in accordance to what you gave in-plane and what you gave out-of-plane, the compound name by (R,S) configuration does not change.
As long as you stay consistent, it does not matter which group you say is out of plane and which you say is in plane.
But beware! Enantiomer naming rules follow a specific format, where the highest priority group is out of plane and the lowest priority group is placed in plane.
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Then take the example of 2,3 dichlorobutane. The 2 chiral carbons are attached by the same substituents (-CH3 , H and Cl), then is it a chiral molecule? If the molecule has 2 chiral centers and attach the same substituents, they are chiral or not?
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Then take the example of 2,3 dichlorobutane. The 2 chiral carbons are attached by the same substituents (-CH3 , H and Cl), then is it a chiral molecule? If the molecule has 2 chiral centers and attach the same substituents, they are chiral or not?
No you're missing the point. :-\
You have to know how to name an enantiomer to understand. Just follow the link I gave.
here (http://members.aol.com/logan20/configur.html#intro)
I cannot go on about this because it would take me two pages to explain the method. If you take the time to read the information in the link, it would take about the same amount of time as it would if you were to read all of it by me explaining it in detail here.
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So, 2,3 dichlorobutane is a symmetric molecule, hence it's achiral. Is it right?
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Because it has two stereogenic carbons, there will be 4 diastereomers of 2,3-dichlorobutane: (R,R), (R,S), (S,R), and (S,S). You should check whether each diastereomer is chiral (hint: two are chiral and two are not).
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Because it has two stereogenic carbons, there will be 4 diastereomers of 2,3-dichlorobutane: (R,R), (R,S), (S,R), and (S,S). You should check whether each diastereomer is chiral (hint: two are chiral and two are not).
Four stereoisomers, not four diastereomers. Diastereomers are defined as stereoisomers that are not enantiomers.
I guess in this case there are only 3 though.
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Because it has two stereogenic carbons, there will be 4 diastereomers of 2,3-dichlorobutane: (R,R), (R,S), (S,R), and (S,S). You should check whether each diastereomer is chiral (hint: two are chiral and two are not).
Four stereoisomers, not four diastereomers. Diastereomers are defined as stereoisomers that are not enantiomers.
I guess in this case there are only 3 though.
Yeah this cmpnd is meso.
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That is, it is achiral?
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That is, it is achiral?
Yes (http://en.wikipedia.org/wiki/Meso_compound).
Meso compounds meet the requirement of having stereogenic centers, but because their mirror image is superimposable, it's optically inactive.
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Then if I take the example of 2-chloro-3-methylbutane, it's not a symmetric molecule. Hence despite it has 2 chiral centers, it's still chiral. Is it right? Because the mirror image cannot superimpose on the molecule itself?
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Then if I take the example of 2-chloro-3-methylbutane, it's not a symmetric molecule. Hence despite it has 2 chiral centers, it's still chiral. Is it right? Because the mirror image cannot superimpose on the molecule itself?
Right.
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No, then you would only have one chiral center! The 3-position would have 2 methyl groups on it.
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Right. I didn't draw that. My mind's eye sucked that time.
:)
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I want to ask if a molecule has 2 chiral centers, like 1,2-dichloro-1-methylpropane, how I determine whether it is chiral by considering its three-dimensional structure?
The fact that a molecule is chiral or not doesn't depend on the presence of stereocenters, because:
1. there are lots of chiral molecules that have no stereocenters at all;
2. there are lots of achiral molecules that have stereocenters.
(By the way, this shows how incorrect is the use of the term 'chiral center': one calls 'chiral center' a point of an achiral molecule).
The only certain rule to follow (as given by k.V.) is to check if the molecule is superimposable to its mirror image. If it is, then it is achiral. If it is not, then it is chiral. More technically, one could say that molecules are achiral when they have at least one symmetry element of the second order (a rotoreflexion axis, i.e. a plane, a center... of symmetry).
I disagree with the use of the CIP rules in this case, because they'll cause confusion when you have no stereocenters or when you consider meso compounds.
I hope I didn't scramble your certainties...
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I disagree with the use of the CIP rules in this case, because they'll cause confusion when you have no stereocenters or when you consider meso compounds.
CIP rules still apply to achiral molecules. There shouldn't be a problem. It's the best way to define precisely which stereoisomer you are dealing with.
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I disagree with the use of the CIP rules in this case, because they'll cause confusion when you have no stereocenters or when you consider meso compounds.
CIP rules still apply to achiral molecules. There shouldn't be a problem. It's the best way to define precisely which stereoisomer you are dealing with.
I didn't write that CIP rules don't apply to achiral molecules, and it's true that absolute configuration of stereocenters is precisely defined by the CIP descriptors. This is just stating the obvious.
I wrote that I disagree with their use in this case, i.e. for determining if a molecule is chiral or not.
Not only because the symmetry considerations are the primary reference for chirality, but also because there are at least two examples where CIP rules seem to complicate the problem.
One is atropoisomers, like many binaphthyl systems. There you have stereoisomers without stereocenters. Stereogenic axes were defined to get around the issue, but I think it's still much simpler to use the old mirror image method. Besides, with the very good freeware applications (like ACD ChemSketch), it's easy for everyone to visualise them.
The other is meso compounds. Take for instance meso-tartaric acid, which is (2R,3S)-2,3-dihydroxysuccinic acid. Someone inexperienced could erroneously think that (2S,3R)-2,3-dihydroxysuccinic acid does exist and is the enantiomer of the first! And in fact, unless one is a genius, how does he know that the two molecules are actually the same without drawing them?
But still, this just my view, someone may like CIP rules so much that all these problems don't bother him.
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I didn't write that CIP rules don't apply to achiral molecules, and it's true that absolute configuration of stereocenters is precisely defined by the CIP descriptors. This is just stating the obvious.
I wrote that I disagree with their use in this case, i.e. for determining if a molecule is chiral or not.
Ah, I see. Yes, I agree that it is not the way to determine whether or not a molecule is chiral. I thought you meant that it wasn't applicable at all to meso compounds.
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But I have an example: Consider a compound consists of a C=C double bond. There are four groups of atoms attchong the carbons: On the top left, there's -COOH group, top right, -CH2CH3group, bottom left, a H atom and bottom right, -CH2CH3 group. This molecule should not have any element of symmetry and yet I can still superimpose the mirror image with the molecule and still it is achiral!
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That molecule has a mirror plane that contains the C=C bond and the carbon atoms of your substituents.
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But I have an example: Consider a compound consists of a C=C double bond. There are four groups of atoms attchong the carbons: On the top left, there's -COOH group, top right, -CH2CH3group, bottom left, a H atom and bottom right, -CH2CH3 group. This molecule should not have any element of symmetry and yet I can still superimpose the mirror image with the molecule and still it is achiral!
Why do you say this molecule 'should not have any element of symmetry' ?
If you find it difficult to visualise it just on paper, try ACD ChemSketch: it's freeware and it gives you a very nice 3D image of your molecules. You can even draw enantiomers, rotate them in space and verify that they don't overlap.
If instead you're talking about asymmetry in conformers of your molecule, then it's a different problem, and normally in stereochemistry you consider an averaged conformer which has the highest possible symmetry, without being completely unreal (e.g., you don't draw planar cyclohexanes...).