[marzike]

http://i118.photobucket.com/albums/o88/Shrewstur/chiral.png

In E, there's 4 stereoisomers, whereas for F, there's only 2.  This must mean that there's 2^2=4, so 2 asymmetric centers for E... and 2^1=2, so 1 asymmetric center for F.  I'm not understanding how to tell if they're asymetric in a cyclic, though.  Where does the substitutent end and start?

For instance, in E... I would argue that there are zero asymmetric centers.  For the Carbon that's attached to Br, it's other three substituents are a Hydrogen, the entire ring going clockwise, and the entire ring going counterclockwise.  The ring is the same thing.  So it's impossible for an cyclic carbon to have an asymmetric center.

I know how to find asym. centers for acyclic structures. I exaggerated the futility of understanding this in the above paragraph, but I really do need more of an explanation than "pretend it's not a cyclic and do it like normal."

~~
I remember my professor using the example of a spy "circling around" the cyclic structure.  I can't remember if she meant to use that for asymmetric centers, though.  What might I be mistaking this with?

[jdpaul88]

First step is to think about what the asymmetrical carbon is attached to

For The OH it is connected to CH(Br)CH(OH)CH2

[Dan]

http://i118.photobucket.com/albums/o88/Shrewstur/chiral.png

For instance, in E... I would argue that there are zero asymmetric centers.

Do you mean E or F?

For E (2-bromocyclohexanol) both centres are clearly asymmetric. Let's take C1 as an example: substituents are OH, H, CH₂R and CHBrR - you can see simply by writing them down that there are 4 different groups attached to C1. R here refers to "the rest of of the ring".

This must mean that there's 2^2=4, so 2 asymmetric centers for E... and 2^1=2, so 1 asymmetric center for F.

You need to be very careful making assumptions. If a compound has two stereoisomers, there are a number of explanations for this - one explanation is that there is one asymmetric centre, but that is not the only possible reason. This is a sloppy short cut that is going to hinder your understanding of stereochemistry, so I'd advise forgetting it now.

F does not have any asymmetric centres (it has two pseudoasymmetric centres).

I really do need more of an explanation than "pretend it's not a cyclic and do it like normal."

I don't really see how this is possible - the methods for identifying stereoisomers of molecules do not change depending on whether or not the molecule is cyclic.

[marzike]

Hm, okay so "R" being the rest of the ring.  For E, now with C2, the substituents are Br, H, CH2(OH)R, and CH2(R).  These are 4 different substituents, so its an asymmetric center.  Neither of the two will have a meso compound because they have different substituents than the other.


And F's pseudoasymmetric centers...one carbon has substituents OH, H, CH2R, and CH2R.  The other carbon has substituents CH3, H, CH2R, and CH2R.  Is this the reason why it's not technically considered to be asymmetric?  How much must you dwelve into the "R" before just considering it to be the same thing?  I imagine there must be scenarios, possibly with longer rings or maybe cyclohexanes, in which you couldn't just stop at examining as "one member of the ring to the left... and the rest" but rather "one member of the ring to the left, and the next member... and then the rest."

So, why does F has stereoisomers?  Because it has pseudoasymmetric centers I'm guessing (there I go with a rushed assumption  ), but what makes it a pseudo. to  begin with?  Why not just let this be one of the many problems where a person could say "no stereoisomers" as their final answer?

~~~
I notice that jdpaul made it simple by only looking at one thing next to the asymmetric center, as opposed to the entire ring.  And right now, you're saying that whether or not a compound is cyclic or not doesn't affect how you identify stereoisomers for it.

.,.,.,.,.CH2 - CH3
..,..,..,.|
CH3 - C - CH2 - CH3
..,...,.,.|
.,...,...H

This is a structure of hexane, which I found online, that does not have an asymmetric center.  Based on the position of the carbon, it's clear that the center is attached to a methyl, hydrogen, and two identical ethyl groups.

But then looking at a carbon in a cyclic, two of the substitutents on the sides will eventually touch and loop around the other.  This is why I believe identifying substituents for cyclics must be different.

[sjb]

Hm, okay so "R" being the rest of the ring.  For E, now with C2, the substituents are Br, H, CH2(OH)R, and CH2(R).  These are 4 different substituents, so its an asymmetric center.  Neither of the two will have a meso compound because they have different substituents than the other.


And F's pseudoasymmetric centers...one carbon has substituents OH, H, CH2R, and CH2R.  The other carbon has substituents CH3, H, CH2R, and CH2R.  Is this the reason why it's not technically considered to be asymmetric?  How much must you dwelve into the "R" before just considering it to be the same thing?  I imagine there must be scenarios, possibly with longer rings or maybe cyclohexanes, in which you couldn't just stop at examining as "one member of the ring to the left... and the rest" but rather "one member of the ring to the left, and the next member... and then the rest."

Technically, you would have to reach the end of the molecule to spot the difference, so something like the middle carbon of 1,n-dideuterio-(2n+1)-ane is still an asymmetric centre, even for very large n

So, why does F has stereoisomers?  Because it has pseudoasymmetric centers I'm guessing (there I go with a rushed assumption  ), but what makes it a pseudo. to  begin with?  Why not just let this be one of the many problems where a person could say "no stereoisomers" as their final answer?

The ring is not planar, look into things like chair and boat conformations, and cis and trans configurations

~~~
I notice that jdpaul made it simple by only looking at one thing next to the asymmetric center, as opposed to the entire ring.  And right now, you're saying that whether or not a compound is cyclic or not doesn't affect how you identify stereoisomers for it.

.,.,.,.,.CH2 - CH3
..,..,..,.|
CH3 - C - CH2 - CH3
..,...,.,.|
.,...,...H

This is a structure of hexane, which I found online, that does not have an asymmetric center.  Based on the position of the carbon, it's clear that the center is attached to a methyl, hydrogen, and two identical ethyl groups.

But then looking at a carbon in a cyclic, two of the substitutents on the sides will eventually touch and loop around the other.  This is why I believe identifying substituents for cyclics must be different.

This is not hexane per se, but 3-methylpentane. But you are right, this does not have any (pseudo-)asymmetric centres. For cyclics, you might want to think about torque or something - at each carbon (in this case) always go clockwise, then you can see there is a slight difference

[Dan]

Hm, okay so "R" being the rest of the ring.  For E, now with C2, the substituents are Br, H, CH2(OH)R, and CH2(R).  These are 4 different substituents, so its an asymmetric center.  Neither of the two will have a meso compound because they have different substituents than the other.

OK

And F's pseudoasymmetric centers...one carbon has substituents OH, H, CH2R, and CH2R.  The other carbon has substituents CH3, H, CH2R, and CH2R.  Is this the reason why it's not technically considered to be asymmetric?

The C is not attached to 4 different groups.

  How much must you dwelve into the "R" before just considering it to be the same thing?

All the way round. E.g. Cycohexanol does not have an asymmetric centre, but cyclohex-3-enol does. If you trace a path from the oxygenated centre around the ring, the path is different depending on which way you go.

So, why does F has stereoisomers?  Because it has pseudoasymmetric centers I'm guessing (there I go with a rushed assumption  ), but what makes it a pseudo. to  begin with?

The centres are pseudoasymmetric because the difference in the clockwise and anticlockwise paths around the ring differ only in the sterochemical orientation of the substituent at the 4-position. This is perhaps easier to see for a linear molecule such as D-ribitol - the central C atom of which is pseudoasymmetric.

Why not just let this be one of the many problems where a person could say "no stereoisomers" as their final answer?

Because it is not true! The two isomers, traditionally referred to as cis and trans, are different molecules.

I've drawn some pictures that might help you:

[marzike]

Thanks guys, I think I understand well enough.