Br2 + alkene

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

joonkimdds

Senior Member
10+ Year Member
15+ Year Member
Advertisement - Members don't see this ad
which one do u guys think is the right answer?
I know that it is anti-addition but I can't tell which is anti when they put it in this fischer projections.
 

Attachments

  • w12121.JPG
    w12121.JPG
    11.6 KB · Views: 239
the correct answer would be E. With anti addition, you will always get enantiomers. A and D are enantiomers of each other. B is clearly wrong because Br2 attaches to 2 different Cs. Choice C is wrong because both of the chiral carbons are 'S' whereas with Choice A and D, they are R and S. At a first glance I would pick C just because the Br's "appear to be" anti, but if you compare the R and S configurations, they are both S, which would not happen with anti addition.
 
Advertisement - Members don't see this ad
the correct answer would be E. With anti addition, you will always get enantiomers. A and D are enantiomers of each other. B is clearly wrong because Br2 attaches to 2 different Cs. Choice C is wrong because both of the chiral carbons are 'S' whereas with Choice A and D, they are R and S. At a first glance I would pick C just because the Br's "appear to be" anti, but if you compare the R and S configurations, they are both S, which would not happen with anti addition.

U r right, but if I change this fischer projection into newman projection, both Br are in eclipsed position so it looks like there was a syn addition. Maybe addition happen only when they are in staggered position?
 
the way i thought about it was....Br2 adds across the alkene so that the product has one of the Br coming out the plane (wedge) and the other Br going into the plane (dashed)....so the carbons must have different absolute configurations. One should be R and one should be S and it doesn't matter which carbon has which configuration just as long as the other carbon has the opposite configuration.

So going through answer choices, the only ones that meet that criteria are A & D, so E is your answer.
 
of course. because before the reaction, it was an alkene and both of the carbons were sp2 hybridized and achiral. the only new addition is the Br, so as long as they are different (R and S), you know its anti. If it helps you, do this problem using the wedge and dotted line method, and you will see that if its anti, both of the chiral carbons have to have a different conformation.