[icyhead]

Thanks for the link

[icyhead]

Id just like to check, for the bromination of alkenes, lets say of a symmetrical Z alkene, do the 2 stereoisomeric products result from
A. the reaction mostly happening in an Anti method, with a minority Syn method.
or
B. are there 2 anti products?

Im asking this because although I think i was taught there are 2 anti products, but id just like to check is this due to the first bromine being able to bond with either of the carbons (previously attached to both)... does this result in the mix of products?

[Fridushka]

i think the reason for OMs to be a good leaving group is that it has a lot resonance structures although i dont know the exact number

[Dan]

Id just like to check, for the bromination of alkenes, lets say of a symmetrical Z alkene, do the 2 stereoisomeric products result from
A. the reaction mostly happening in an Anti method, with a minority Syn method.
or
B. are there 2 anti products?

Im asking this because although I think i was taught there are 2 anti products, but id just like to check is this due to the first bromine being able to bond with either of the carbons (previously attached to both)... does this result in the mix of products?

There are two anti products. If you consider the bromonium ion intermediate it can be opened by S_N2 attack of bromide from either side of the ring - this leads to two enantiomeric products in equal proportion (for a symmetrical Z alkene) - ie. a racemic mixture.

If you're happy with Cahn-Ingold-Prelog rules - taking Z-but-2-ene and bromine as an example - the bromonium ion will have R,S configuration. S_N2 displacement (proceeding with inversion) will happen at either the R centre or the S centre giving S,S-2,3-dibromobutane or R,R-2,3-dibromobutane, respectively.