[adeelonks]

According to my synthesis lab class, when brominating p-methoxycinnamic acid it can undergo both syn- and anti- addition...  However I don't understand how the syn addition is working.

Could anybody be able to explain this to me?
Thanks for any help, and this is a nice looking forum you guys got here. I plan on sticking around for sure.

Adrian

[azmanam]

premature bromonium opening to give resonance stabilized carbocation?  then unselective attack by Br-?

[adeelonks]

And this would be aided by the lone pairs on the methoxy perhaps allowing for additional resonance stabilization once the bromonium opens up?

[azmanam]

right, that's my thinking.  but I'm not 100% sure.

[sjb]

Something like the attached? Formation of the bromonium ion occurs as per regular olefins, but it is further stabilised by the OMe. Attack of the Br- can occur from either side of the new exocyclic C=C bond, but I don't think you'll get a racemic mixture here.

[azmanam]

but I don't think you'll get a racemic mixture here.

Why not?  You drew bromonium up - it could just as well be back.  You should get a statistical mixture of all 4 possible compounds: 2 syn and 2 anti.

[sjb]

but I don't think you'll get a racemic mixture here.

Why not?  You drew bromonium up - it could just as well be back.  You should get a statistical mixture of all 4 possible compounds: 2 syn and 2 anti.

Sorry, perhaps I need to rephrase - I don't think you'll get a racemic mixture at the alpha-position relative to a fixed beta, as if you look at the newman projection there is probably a more favoured trajectory of at attack - a bit like Burgi-Dunitz on carbonyls, but not necessarily that angle.

[Squirmy]

You will get a racemic mixture since you start with achiral materials.

You may not get a 1:1 mixture of syn/anti, but the syn will form as 1:1 mix of enantiomers and the anti will form as a 1:1 mix of enantiomers.

[adeelonks]

Thanks for the responses everyone!

Been a lot of help in aiding my understanding.