Destroyer Ochem 72-Help!

[sweetdaises2000]

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Ok guys, so I'm working through the destroyer ochem problems and I'm not satisfied with the answer to this question. I understand that if there is more than one chiral center, then diastereomers are possible because of the 2^n rule for stereoisomers. So I see how that applies to choice c. But what about choice D? Its not a meso compound since Br2 adds anti. So the two carbons that Br2 adds to should be chiral-2 chiral centers-possible diastereomers right??
Anyone who has destroyer, please take a look and explain if you can
Thanks!

 

[lex1489]

Tricky one! I stared at it for a while trying to figure it out.

In choice D the two chiral carbons have to be added anti, but there are no other groups so the 'diastereomer' of the 1S,2R isomer is just 1S,2R named in the opposite direction-- it's the enantomer.

Choice C actually has 3 chiral carbons. The ethyl group can be above or below the ring, thus R or S, and the adjacent Br can also be R or S. So you can have ethyl up, adjacent Br up, or ethyl up, adjacent Br down. Everything enantomer is just a form of one of these choices, so you do in fact have 2 stereoisomers.

 

[sweetdaises2000]

Thanks lex. That was really helpful. So you're basically saying that since Br2 must be added anti and there are no other groups on the ring (for choice d), in one view. Br could be coming towards you on the first carbon but going away from you on the second carbon; or in the second view(like if you turned the molecule around, or read it from right to left), br is going away from you, or coming towards you, so there are only 2 possible ways to manipulate the molecule?
Basically upon turning the molecule around, you get the same molecule. So there can only be that molecule and its enantiomer? Am I understanding this correctly? Stereochem always trips me up...sorry 🙂

 

[sweetdaises2000]

And for choice c, since there are 3 chiral centers. 2^3=8 so there are 8 possible stereoisomers? Which you could get from having various combinations of ethyl up, br down and ethyl up br up..and so forth?

 

[lex1489]

I think you got the first one. To help, try drawing it with a Br and a Cl anti to each other. You should then be able to draw all 4 stereoisomers (2 diastereomers and their enantomers).

For choice C, there *should* be 8 possible stereoisomers, but because of the molecule's symmetry there are only 2 real possibilities and the rest turn out to be superimposable on one of the first 2.