[dorratz]

Hi

I got a question about this halohydrin formation:


Can someone please explain me the mechanism of the red products?

This is my solution - as you can see only one of my products is shown in the original solution.

I think that the MeOH can only attack from bottom of the partially positive carbon, because if it will come from the upper side it will connect to the positive bormide ion, and this is less stable the first option.

Thanks
Dor

[kriggy]

Actualy, the first one and the last one are those you made. Your assuption about MeOH attack is correct, however, the bromide can attack from both sides as shown in picture. I think you can figure out the rest

[dorratz]

Ok, so how do I know if I should the second way of the bromide to attack as you showed? is there any guideline except of doing and then find out?

can you give me another exercise like that? I couldn't find any other examples with 4 products like that.

Thanks
Dor

[kriggy]

Basicaly, every time this will happen except in cases when you use some kind of chiral catalyst or there is another center of chirality in the alkene. In those cases, one of those isomers might be formed preferentialy over the other one

[dorratz]

Hey,

I couldn't find any other examples like that where I can find 4 products. All the time I see just one side attack of the bromine ion.

Can you please give me some more examples?

Thanks
Dor

[kriggy]

I suggest looking into some stereochemistry book.
I will try to look into one I have at school but can promise anything

edit: ok I did some googling for you and you might read into this:
http://www.masterorganicchemistry.com/2013/03/06/bromination-of-alkenes-how-does-it-work/
http://www.masterorganicchemistry.com/2013/03/15/bromination-of-alkenes-the-mechanism/