stereochemical interpretation in bridged bicyclics

I have an interesting conundrum for you. To the human eye, the three structures below look identical. JChem, however, interprets them differently. Structure 2 is interpreted as having no stereochemistry at all, and structure 3 is interpreted as having opposite stereochemistry from structure 1. In fact, if you delete the H atoms from structure 3, Marvin will make the bonds to the bridging CH2 group into hashed bonds.





I believe the root of the misinterpretation has to do with the fact that the two nonstereo bonds bisect the two stereo bonds.





Unfortunately, telling my students just to show the stereochemistry of the bridging group, not the H atoms, doesn't work. Can you come up with a solution that solves this problem?

You are right. The first and third molecule has opposite stereochemistry. The second one has also stereochemical meaning but Marvin interprets stereochemistry if the center atom has one and only one wedge. (Centers defined with more wedges could lead to misleading interpretation.)


Let's see the first and the third molecules.


Here is two examples which shows why Marvin interprets stereochemical representation of this molecules as it is.


At the first molecule the hydrogen is not shown. So it is automatically at the opposite side of the wedge. The Cl, Br and center atoms define a plane.


At the second example the Cl and Br is closer to us and the I and H is further.


In this case no plane is defined.


I tried to show this on the 3D depicted images...

Quote:

Unfortunately, telling my students just to show the stereochemistry of the bridging group, not the H atoms, doesn't work.

Why?


Don't let them draw the H. If you have one center and 3 ligand the stereochemistry is determined without doubt.

Your method works fine for acyclic compounds, but it does not work well for bridged bicyclic ones. Here's why. In the structures shown, the 120° angles about the six membered ring and the very wide angles around the bridging CH₂ group let the reader know that the CH₂ group is out of plane and the six-membered ring is in plane. If the CH₂ group is out of plane, it must be up or down, and the H's must be opposite.





Another way of describing the problem is as follows. If the compound were not bicyclic, then the two groups attached to the top and bottom C's of the ring would both be pointing outside of the hexagon. One would be bold, the other hashed, and there would be no stereochemical ambiguity. Because of the bridge, however, the CH₂ group ends up pointing into the hexagon, and that creates a stereochemical ambiguity. If you pulled the CH₂ group so far to the left or right that it was outside the hexagon, again there would be no stereochemical ambiguity. It is only because it is inside the hexagon that the interpretation goes awry. Again, the human eye can recognize this feature from the bond angles, but Marvin is not doing so.





Perhaps you can write a separate subroutine for dealing with the stereochemistry of bridged structures?

The ambiguity happens only if all the 4 ligand is shown. Do you agree?


I understand that the ring defines a plane and in this case if you show only 3 ligand from the 4 (not showing the H) the chirality should be clear.


What dou you think?


I have some doubt separating ring system and non ring systems. The ring systems are usually deformed which would lead to many cases.


By the way do you mean the C at the top of the ring shoul have different chirality in the attached picture?

The configuration is assigned correctly when the bond to the outside and the bond to the inside are the same stereo.





The configuration is assigned correctly if only the bond to the inside is stereo.





When the bond to outside the ring is one stereo, and the bond to inside the ring is the other, Marvin refuses to assign R or S. That's not so good, because our teaching program doesn't show R/S labels, and if the student isn't aware they may be making a mistake, they wouldn't think to turn them on. Any person looking at such a structure would know that it is the same as if the bond to the outside were not stereo. This is where if you move the bonds from inside the ring to outside the ring, the configurations are suddenly assigned. (See picture.)





The most serious problem comes when the bond to the outside is stereo and the bond to the inside is not. Most people would interpret the bond to the inside as having the opposite stereo from the bond to the outside. Marvin and JChem interpret the bond to the inside as having the same stereo as the bond to the outside, regardless of whether the inside bonds are moved to the outside.





Again, I don't know that there's anything practical to do about it. I just wanted to bring your attention to it.