Chiral vs. Meso Question..

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Avery07

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From Chad's Video's:

Screenshot2010-07-17at33606PM.png


The molecule on the right, as pictured, is R,R; however, if you rotate C3, wouldn't that flip the Br behind the plane and make it R,S?

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Rotation about a single bond or atom won't affect absolute configuration because all three substituents that get rotated get rotated together. When you assign configuration you still have to picture the lowest priority substituent pointing away from you and you still get the same answer: R or S.

In this case, (R,S), which is identical to (S,R), would be meso and optically inactive. (R,R) and (S,S) are enantiomers, both optically active.
 
Rotation about a single bond or atom won't affect absolute configuration because all three substituents that get rotated get rotated together. When you assign configuration you still have to picture the lowest priority substituent pointing away from you and you still get the same answer: R or S.

In this case, (R,S), which is identical to (S,R), would be meso and optically inactive. (R,R) and (S,S) are enantiomers, both optically active.

Haha I'm not thinking today.. thanks. :thumbup:
 
The molecule on the right, as pictured, is R,R; however, if you rotate C3, wouldn't that flip the Br behind the plane and make it R,S?

The molecule on the right (the Fischer projection) is actually S,R (the top stereocenter is S, the bottom stereocenter is R.
The molecule on the left is R,R.
 
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