Organic Chemistry: Stereoisomers, Enantiomers + Diastereomers help!

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thenextchemist
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Is this correct for 1,2,3-trichlorocyclohexane
And are all of these meso ?

Thank you

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charco
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(Original post by Bloom77)
Is this correct for 1,2,3-trichlorocyclohexane
And are all of these meso ?

Thank you

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thenextchemist
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thenextchemist
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(Original post by charco)
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It is now, thanks!
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_NMcC_
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(Original post by Bloom77)
Is this correct for 1,2,3-trichlorocyclohexane
And are all of these meso ?

Thank you

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I worked out however that not all of them are Meso. Some of them are Meso, others are enantiomers, and some enantiomers turn out to be the same as others, so be careful. Draw out each possible configuration (there are 8 possible stereoisomers (2^3), there are 3 stereogenic/chiral centers. Where U = Up, D = down; UUU, DUU, UDU, UUD and each ''mirror image'' = 8.

The ones that you can't superimpose one mirror image compound on an other are enantiomers, check each set you find by rotation to see if they are different or the same. The ones that you can superimpose on each other are Meso compounds. The ones that are different in configuration and not mirror images are Diastereomers (i.e you will have drawn them above and beneath each set automatically).

So go through each one and see which ones are superimposable on each other etc
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thenextchemist
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(Original post by _NMcC_)
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I worked out however that not all of them are Meso. Some of them are Meso, others are enantiomers, and some enantiomers turn out to be the same as others, so be careful. Draw out each possible configuration (there are 8 possible stereoisomers (2^3), there are 3 stereogenic/chiral centers. Where U = Up, D = down; UUU, DUU, UDU, UUD and each ''mirror image'' = 8.

The ones that you can't superimpose one mirror image compound on an other are enantiomers, check each set you find by rotation to see if they are different or the same. The ones that you can superimpose on each other are Meso compounds. The ones that are different in configuration and not mirror images are Diastereomers (i.e you will have drawn them above and beneath each set automatically).
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Meso compounds should have a symmetry in between (regardless of the wedges and hashes) right? So, wouldn't all of them be meso?
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_NMcC_
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(Original post by Bloom77)
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Meso compounds should have a symmetry in between (regardless of the wedges and hashes) right? So, wouldn't all of them be meso?
"(regardless of the wedges and hashes)"

Nearly. Look at the 1st set of compounds under enantiomers, you can draw a plane of symmetry through the compound (vertically through the middle Chlorine) and therefore you can superimpose one of them on top of the other, so those two are actually the same Meso compound, just rotated.

The plane of symmetry has to be able to fold the compound with the bonds. You do have to look at the bonds when you are drawing planes of symmetry. Don't just look at the cyclohexane ring when considering planes of symmetry.

There's another meso compound under 'diastereomers', try flipping and superimposing the right on the left. So each hashed bond becomes a 'thick' bond and the 'thick' bond becomes a 'hashed'.

An enantiomer is a non-superimposable mirror image compound.

The second set down under enantiomers are enantiomers. There's no mirror image plane in the molecule (you can't fold it like a piece of paper) and if you try rotating the right 180 degrees and place it on the left, you will find they won't fit very well. So they are enantiomers. Remember: You do have to consider the hashed and wedged bonds when considering symmetry.

It's like trying to place one hand on top of the other. You can but they don't really fit very well. They would if we had 5 fingers of the same length and no thumbs, we would have meso hands.

****Note the edits, I read through your reply again and noticed your mistake.
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thenextchemist
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(Original post by _NMcC_)
"(regardless of the wedges and hashes)"

Nearly. Look at the 1st set of compounds under enantiomers, you can draw a plane of symmetry through the compound (vertically through the middle Chlorine) and therefore you can superimpose one of them on top of the other, so those two are actually the same Meso compound, just rotated.

The plane of symmetry has to be able to fold the compound with the bonds. You do have to look at the bonds when you are drawing planes of symmetry. Don't just look at the cyclohexane ring when considering planes of symmetry.

There's another meso compound under 'diastereomers', try flipping and superimposing the right on the left. So each hashed bond becomes a 'thick' bond and the 'thick' bond becomes a 'hashed'.

An enantiomer is a non-superimposable mirror image compound.

The third set down under enantiomers are enantiomers. There's no mirror image plane in the molecule (you can't fold it like a piece of paper) and if you try rotating the right 180 degrees and place it on the left, you will find they won't fit very well. So they are enantiomers. Remember: You do have to consider the hashed and wedged bonds when considering symmetry.

It's like trying to place one hand on top of the other. You can but they don't really fit very well. They would if we had 5 fingers of the same length and no thumbs, we would have meso hands.

****Note the edits, I read through your reply again and noticed your mistake.


Hi, sorry about the late reply!
I forgot about this question.

You said: An enantiomer is a non-superimposable mirror image compound.

And then: The third set down under enantiomers are enantiomers.

But, the third set are meso compounds which are superimposable so they can't be enantiomers, right?

Thanks for your assistance!
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_NMcC_
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(Original post by Bloom77)
Hi, sorry about the late reply!
I forgot about this question.

You said: An enantiomer is a non-superimposable mirror image compound.

And then: The third set down under enantiomers are enantiomers.

But, the third set are meso compounds which are superimposable so they can't be enantiomers, right?

Thanks for your assistance!
Sorry, badly worded. I meant the second set under enantiomers. (3rd from 'diastereomers')
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