[kriggy]

Hi guys, I just need to check something:
If I have alkene such as this one:
R¹HC=CHR² then when I do addition of Br₂ I get those products:

if R¹=R² and they are Z (cis) then I get single product
if R¹=R² and they are E (trans) then I get mixture of enantiomers (asuming there is no other chiral center)
if R¹≠R² then I get some mixture of products since the olefin is not symetrical and the reaction centers arent the same and it doesn matter whether the olefin is E or Z. I could in theory get single product given the right substituents.

I am not 100% sure and because Im making study material that will be used by few of my friends i want to be 100% sure

thanks

[discodermolide]

I think you will scramble the stereochemistry.
The reaction goes via an epibromonium ion. Now if this opens before attack of the Br^- then it will scramble. This depends on the nature of the R groups which may or may not stabilise the ion!
But if the R groups are not the same then it may well depend on steric factors? Giving some selectivity but not 100%.
I'm dredging my memory (only 1 active brain cell left) here perhaps look in March?

[kriggy]

Oh, I din´t know that the epibromonium ion can open before the attack of Br^-. I was under impression that the cyclic intermediate is opened by attack of Br^-, if that is the case than was my reasoning correct?
If its not the case then it doesnt matter because i get secondary carbocation and that allows for attack from both sides and get same mixture of products no matter if the alkene is E or Z (if R¹=R²?

I dont have March, we were using rather old book for the mechanism class that dont describe it in much detail.

[discodermolide]

Well opening before attack etc probably bring us into the discussion of classical Vs non-classical systems.
Getting a carbocation will in any case give a mixture.

I  have March so if you pm me arrangements could be made!