[nj_bartel]

Erythro MP: 204 ^oC
Threo MP: 95 ^oC

In the most stable conformation, it would seem to me like threo- (R,R/S,S) would have the higher melting point because the bromo groups are on the same general side of the molecule, creating a larger dipole and polarization effect.  However, this is pretty obviously far from right, so what's really going on?

Thanks.

[macman104]

No, the erythro type has them on the same side.  Read here:

http://wwwchem.csustan.edu/chem3012/Dibromocinnamic%20Acid.htm, scroll down almost to the end.

I also imagine that with the erythro type you can imagine a structure where the bromines are up, and another molecule can line up nicely "below" the other one in a stacking like pattern.  With the threo, that isn't as possible.  I don't have a solid referenceable reason why the large melting point difference though.

[nj_bartel]

I have a hard time interpreting those diagrams for some reason.  The threo- is (R,R/S,S), correct?

 http://img185.imageshack.us/my.php?image=dibromocinnamicacidyp1.jpg

And that is the R,R stereoisomer?  They look like they're on the same side to me, when everything is in staggered conformation - both coming outward from the plane of the page.

Where is my reasoning going wrong? =/

[macman104]

Correct, that's the threo.  And I know see how you were interpreting it.  I was picturing it more eclipsed up, when you have the R and S conformation, then both bromines are up, and the H's are down.  I tried playing around with it in Chem3D, to see if I could come up with a good conformer explanantion, but I'm failing at it.  Sorry, lol.

[nj_bartel]

Np, thanks for the effort