Organic Chemistry Isomer Question help

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betterfuture

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There is a particular question I am working on in my Kaplan review and was stumped on it.

It states, for example, (2R)(3S) dihydroxybutanedioic acid AND (2S)(3R) dihydroxybutanedioic acid are meso compounds AND the same molecule. I do not understand how it can be meso and the same molecule.

This is how I drew it: http://classconnection.s3.amazonaws.com/106/flashcards/563106/jpg/tartaric_acid_21342996404630.jpg

Also, when I tried doing the absolute configurations of the chiral carbons I got (2S) and (3S) AND NOT (2R) and (3R). Am I missing something?

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Okay, first things first. You don't understand what a meso compound is. Compounds that are meso to each other are, in fact, the same molecule. By definition. Basically all you're doing is looking at it from a different perspective.

Now, draw your molecule using dashes and wedges. You should have two carboxylic acids on the ends with two methylene groups in between. Each methylene group will have a hydroxyl on it. Now, let's say the left hydroxyl points into the page (dash) and the right one points out of the page (wedge). That means when you flip the stereochemistries, the left one will point out of the page and the right one will point in. So let's focus on the first one for now. Left hydroxyl goes into page and the right one comes out. Okay, now imagine that you are looking at the molecule from the other side of the page - that is, from under the plane of the page. If you did that, then the "left" hydroxyl is now on your right and the "right" hydroxyl is now on your left. Moreover, the one on your left is now pointing out at you while the other one is going back into the page. In other words, it's the same as the other stereoisomer you drew above!

Do the drawings and imagine looking at it from the other side of the page. You can also use your organic chemistry kit if that helps.
 
Okay, first things first. You don't understand what a meso compound is. Compounds that are meso to each other are, in fact, the same molecule. By definition. Basically all you're doing is looking at it from a different perspective.

Now, draw your molecule using dashes and wedges. You should have two carboxylic acids on the ends with two methylene groups in between. Each methylene group will have a hydroxyl on it. Now, let's say the left hydroxyl points into the page (dash) and the right one points out of the page (wedge). That means when you flip the stereochemistries, the left one will point out of the page and the right one will point in. So let's focus on the first one for now. Left hydroxyl goes into page and the right one comes out. Okay, now imagine that you are looking at the molecule from the other side of the page - that is, from under the plane of the page. If you did that, then the "left" hydroxyl is now on your right and the "right" hydroxyl is now on your left. Moreover, the one on your left is now pointing out at you while the other one is going back into the page. In other words, it's the same as the other stereoisomer you drew above!

Do the drawings and imagine looking at it from the other side of the page. You can also use your organic chemistry kit if that helps.

Oh I see now. But what about the absolute configurations?
 
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As in assigning R and S. Plus, how do we know which hydroxl group is wedge and which is dash from a simple line structure drawing?
 
By assigning R and S... Draw the butanedioic acid first. Then you have to decide how to orient C2 and C3. You're attaching a hydroxyl group, so that's top priority. Then second is the carboxylic acid. And finally is the methylene group. Assign R and S using the same rules you would any other molecule.
 
I was confused because I forgot what meso was. Orgo was so, so long ago ...

I'm 100% good now. I think the "behind the page" technique could be helpful but for me it's easier just to work from one side of the page and manipulate the molecule as such. To each his/her own.
 
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