Are these cyclohexanes truly chiral or is it a trick question?

[physickkksss]

Homework Statement

Looking at questions 18-20 on this webpage:
http://www.utdallas.edu/~scortes/ochem/OChem1_Lecture/exercises/ch5_stereo1.pdf

(Answers given on end of handout)

Homework Equations

Meso compounds have an internal plane of symmetry

The Attempt at a Solution

Compounds #18 and #20 seem to have an internal plane of symmetry, but the answers say that they are not meso

Am I missing something here or is that a typo on the sheet?

 

[nobahar]

Hello!

From http://books.google.co.uk/books?id=...CEAQ6AEwAw#v=onepage&q=meso compound&f=false:

"To be a meso compound, a molecule must also have chiral isomers."

Hope that helps!

 

[physickkksss]

Not really sure what that means..."chiral isomers"

For example, wouldn't the molecule in Q#20have chiral isomers, if you change one of the methyl groups to another configuration?

 

[nobahar]

This is my understanding of meso compound, enantiomers, etc. Maybe chemistree or Borek could weigh-in if need be.

There is steroisomerism and structural isomerism: structural isomerism is when you have the same number of atoms and the same number of each type of atom in the molecules but the way they are connected up is different (see http://en.wikipedia.org/wiki/Structural_isomer). Stereoisomerism is when you have the same number of atoms and the same number of each type of atom in the molecules AND the molecules are connected up the same way, but the positions in space of the atoms relative to one another is different (see http://en.wikipedia.org/wiki/Stereoisomerism).

Chirality is to do with mirror-images: if you move one of the functional groups - such as a methyl group - to a different carbon you have structural isomerism, but they aren't mirror images. If you take a molecule and draw its mirror image, and you can't rotate it such that you end up with the original (i.e they are not superposable) then you have a chiral (as oppose to an achiral) molecule - the molecules are enantiomers.

This can be extended: you can get molecules that are connected in the same way and that differ in the position of the atoms relative to one another in space (stereoisomerism), but are not mirror-images of one another: these are diastereoisomers.
When a compound has enantiomers, and these enantiomers have a diastereoisomer that is superposable on its mirror-image, then the 'version' of the molecule that is superposable on its mirror-image is the meso compound. You can see this in figure 3.4 in the link above: for tartaric acid, there are enentiomers, and these enantiomers are diastereoimsomers to the meso compound, the relative arrangement of the atoms in space is different between all three, but the mirror image of the meso compound can be rotated such that is is superposable on the original.

Sorry if that explanation is confusing. I recommend the wikipedia links and the link to the textbook above.

 

[nobahar]

I figured it might help to do two examples - 19 and 20 - from your questions.
If you draw 19, and then draw its mirror image, you'll find you can rotate the mirror-image to match the original, these are superposable, it meets one requirement for being a meso comppound, but it also needs diastereoisomers that are enantiomers. The way the atoms in the molecules are connected needs to be kept the same, but the positions in space can be changed: it wouldn't bemuch use to move both the Cls behind the plane of the screen, because you can rotate the molecule to get the same compound. Instead, move one Cl behind the plane of the screen and keep the other Cl pointing out of the plane; then draw the mirror-image of this. You'll find they are not superposable, they are eneantiomers and both are diastereoisomers of the compound drawn in 19 (where both Cls are coming out of the plane of the screen). 19 also meets the second requirement: it has diastereoisomers that are enantiomers. Therefore, 19 is a meso compound.

For 20, if you draw its mirror-image, you'll find you can rotate the molecule to get the original. It meets the first requirement for a meso compound. To get a diastereoisomer of 20, do the same thing as for 19: keep one Cl coming out of the plane of the screen and draw one Cl going behind the plane of the screen; then draw its mirror-image. You'll find you can rotate it to overlap. These are therefore not enantiomers, and therefore 20 is not a meso compound because there are not diastereoisomers of 20 that are enantiomers.

Hope that helps! (and I hope its right!)

 

[chemisttree]

Homework Statement

Looking at questions 18-20 on this webpage:
http://www.utdallas.edu/~scortes/ochem/OChem1_Lecture/exercises/ch5_stereo1.pdf

(Answers given on end of handout)

Homework Equations

Meso compounds have an internal plane of symmetry

The Attempt at a Solution

Compounds #18 and #20 seem to have an internal plane of symmetry, but the answers say that they are not meso

Am I missing something here or is that a typo on the sheet?

1,4-disubstituted cyclohexanes are not chiral because they have identical substitution around the putative 'asymetric' carbons... which aren't asymetric at all. Bicyclo[2.2.2]octane similarly has no chiral centers.

This is a cruel and twisted version of the 'meso trap'.