Enantiomer drawings - Confused by my textbook

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F1's Finest
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I understand why the mirror plane is drawn, as the other side is the enantiomer. But what do the last two molecules prove?

Also, why have they been drawn? Also, why is the book saying that the enantiomer can be obtained by 'reversing two substituents' and 'by reversing the wedge and dotted lines'? Surely we can just draw the mirror plan to obtain the enantiomer???

Any help would be appreciated.!

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(Original post by James A)
I understand why the mirror plane is drawn, as the other side is the enantiomer. But what do the last two molecules prove?

Also, why have they been drawn? Also, why is the book saying that the enantiomer can be obtained by 'reversing two substituents' and 'by reversing the wedge and dotted lines'? Surely we can just draw the mirror plan to obtain the enantiomer???

Any help would be appreciated.!

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They are just trying to get you familiar with how to spot enantiomers. The three on the right are all the same enantiomer, they have merely been rotated so they look different, but the spatial relationships between the groups is the same - hence they are the same enantiomer.

If they were two different enatiomers, they could not simply be rotated to form each other.

The other thing this diagram shows is that to obtain one entantiomer from the other, all you need to do is exchange the positions of two of the groups. Yes, you can do this by just visualising the mirror image, but swapping two groups is an easier way of doing it.
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(Original post by Plato's Trousers)
They are just trying to get you familiar with how to spot enantiomers. The three on the right are all the same enantiomer, they have merely been rotated so they look different, but the spatial relationships between the groups is the same - hence they are the same enantiomer.

The other thing this diagram shows is that to obtain one entantiomer from the other, all you need to do is exchange the positions of two of the groups. Yes, you can do this by just visualising the mirror image, but swapping two groups is an easier way of doing it.
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(Original post by James A)
Life saver. Cheers buddy.

Have some rep!

EDIT: PRSOM , dammit
You're welcome! Any time
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(Original post by Plato's Trousers)
You're welcome! Any time
Hey again, I got a question (it's an easy concept but I can't get my head round it).

http://science.uvu.edu/ochem/wp-cont...lmolecule1.png

As this has two groups which are the same, this molecule has to be a superimposable mirror image, right?

However I'm confused when I read up on the topic and it says that if you place the mirror image over the original image and it should be identical. So surely this isn't a superimposable mirror image? I'm kinda confused
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(Original post by Plato's Trousers)
You're welcome! Any time
I just found this!

http://2012books.lardbucket.org/book...260048d343.jpg

I think this pretty much explains my question above!!! I guess it was my lecture slides that drew the molecule wrong!
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(Original post by James A)
Hey again, I got a question (it's an easy concept but I can't get my head round it).

http://science.uvu.edu/ochem/wp-cont...lmolecule1.png

As this has two groups which are the same, this molecule has to be a superimposable mirror image, right?

However I'm confused when I read up on the topic and it says that if you place the mirror image over the original image and it should be identical. So surely this isn't a superimposable mirror image? I'm kinda confused
Yes, in this case, the mirror images are superimposable. So the molecule is not chiral. In order to have non-superimposable mirror images, you need to have four different groups attached to the carbon atom. In this case, you only have three different groups.

The best analogy for a chiral molecule is your two hands. They are kind of the same, but they are not superimposable.
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(Original post by Plato's Trousers)
Yes, in this case, the mirror images are superimposable. So the molecule is not chiral. In order to have non-superimposable mirror images, you need to have four different groups attached to the carbon atom. In this case, you only have three different groups.

The best analogy for a chiral molecule is your two hands. They are kind of the same, but they are not superimposable.
Ahh yes, I see Thanks for clarifying!
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