Meso Compound and Enantiomer

[WOAHHI]

I'm kind of confused with this problem - here's the image

http://img257.imageshack.us/img257/2872/ts100.jpg

These look like meso compounds so I chose that as the answer but the answer is that these are enantiomers. I can see that they are mirror images and not superimposable but I thought the internal plane of symmetry which makes these meso compounds means that there are no enantiomer?

Is it possible for meso compounds to still have enantiomers? Or do they only have diastereomers? Is this compound not meso since the two methyl groups don't point in the same direction? ..

I think maybe I am not looking at the pictures properly so if anyone could give some clarification that would be great =] thanks!

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[Valigator00]

I'm kind of confused with this problem - here's the image

http://img257.imageshack.us/img257/2872/ts100.jpg

These look like meso compounds so I chose that as the answer but the answer is that these are enantiomers. I can see that they are mirror images and not superimposable but I thought the internal plane of symmetry which makes these meso compounds means that there are no enantiomer?

Is it possible for meso compounds to still have enantiomers? Or do they only have diastereomers? Is this compound not meso since the two methyl groups don't point in the same direction? ..

I think maybe I am not looking at the pictures properly so if anyone could give some clarification that would be great =] thanks!

There is no internal plane of symmetry in the above picture. They are not mirror images of each other. So they are enantiomers. Meso compounds are optically inactive, enantiomers are optically active. It is impossible for meso compound to be an enantiomer.

 

[kkmaths90]

The compound looks like trans b/c one CH3 is going up and other one is going down. Trans gives you enantiomers.

 

[The Anhedonia]

in addition to the good tips listed, you should right out the S or R configurations of each chiral carbon to cleary show that they are enantiomers (Then you'll see that they can't be superimposed on top of each other)

 

[wodehouse]

There is no internal plane of symmetry in the above picture. They are not mirror images of each other. So they are enantiomers. Meso compounds are optically inactive, enantiomers are optically active. It is impossible for meso compound to be an enantiomer.

First of all, these compounds are drawn in the worst possible perspective, so the question is bs. They ARE mirror images, and therefore enationers. A meso compound has no enantiomer because its mirror image is itself. A meso compound can have a diastereomer, however.

 

[Valigator00]

First of all, these compounds are drawn in the worst possible perspective, so the question is bs. They ARE mirror images, and therefore enationers. A meso compound has no enantiomer because its mirror image is itself. A meso compound can have a diastereomer, however.

What i meant to say is they are "non-superimposible" mirror images of each other.

 

[nze82]

Here's something I posted a while ago. Hope it helps:
http://forums.studentdoctor.net/showthread.php?t=628475&highlight=meso+compounds

 

[WOAHHI]

If both of the Methyl groups of say the first compound on the left were pointing up would it be cis instead of trans and therefore have a mirror plane of symmetry?

 

[WOAHHI]

in addition to the good tips listed, you should right out the S or R configurations of each chiral carbon to cleary show that they are enantiomers (Then you'll see that they can't be superimposed on top of each other)

How would i assign R and S to this compound if I do not know which parts are coming out of the plane and which is going in? How would I assign the methyl groups as going into the board or out of the board (solid or dashed wedges) based on the way its written now? Thanks!

 

[kkmaths90]

Well like i said one methyl group is going up and other one down so, if you top put top one on wedge then put bottom one on dashed line . and Draw enantiomer of that. Then compare.

 

[UCB05]

How would i assign R and S to this compound if I do not know which parts are coming out of the plane and which is going in? How would I assign the methyl groups as going into the board or out of the board (solid or dashed wedges) based on the way its written now? Thanks!

Given the way the molecules are drawn, you could start by assuming that since the methyl group is drawn down, the -H should be pointing up, and vice versa with the methyl pointing up. With that in mind, orient yourself to the chiral carbon and notice that the two internal cyclohexane ring bonds point away from you.