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Stereochemistry of Bromine Addition to trans-Cinnamic Acid

  1. Nov 30, 2003 #1
    We determined our melting point to be 199 C. Which would correspond to (2R,2S) and (2S,3R) 2,3-dibromo-3-phenylpropanoic acid (mp 202-204 C) but we are trying to determine whether this is a syn or an anti addition.

    Any suggestions???

  2. jcsd
  3. Nov 30, 2003 #2
    Which is the mp of the former? If it's not apparent which mp the measured is closer to than you have to say that it was inconclusive. You know it is anti based on the mechanism but you can tell that from this experiment.
  4. Nov 30, 2003 #3
    The other has a mp of 98 C. The melting points from 5 different experiments all came up between 197 and 198.

  5. Nov 30, 2003 #4
    Well if I had to pick a difference of 100 degrees or 4 degrees I'd go with the latter. That's acceptable experimental error.
  6. Dec 1, 2003 #5
    Yes, but from looking at the models I do not understand, why it is not a 50/50 attack from both sides of the double bond. And, was also thinking that it might be a 80/20 split between the 2.

  7. Dec 1, 2003 #6
    I'm not sure I understand the question. You do get a 1:1 mix of the enantiomers, they have the same melting point. There's no way you would get a 80:20 mix because the cinnamic acid is achiral, the bromine can react initially to either face just fine.

    The alternative to the anti (2S, 3R) and (2R, 3S) product would be the cis product (2S, 3S + ent.) and you would see any of that from this reaction.
  8. Dec 2, 2003 #7
    We got melting of the 200 C. So we know that it is either the syn or the anti addition.

    The problem I am having is that after the double bond is broken the molecule is free to rotate, which should allow for a 50/50 split. This did not appear to happen, So maybe there is something in the solution that keeps the bond from rotating. I know that when the Br joines, it joins between the 2 carbons which would cause a trigonal planar situation, but when that is broke the bond should be free to rotate???

  9. Dec 2, 2003 #8
    You stated earlier that the other option has a mp of 98. Or was that a typo? If so than you can't say anything based on mp.

    When the double bond is broken the single bond is NOT free to rotate. It's stuck in the three membered ring bromonium ion, which is quite stable. The ring does not open until the bromide attacks on the opposite face. Bromine addition only gives the anti product.
  10. Dec 2, 2003 #9
    I agree, our melting poing is the 200 and that would be the product.

    Based on this mp it would be the erythro pairs, which I thought was the syn version of this product.

    Does the anti addition not result in the anti product?
  11. Dec 2, 2003 #10
    No, the anti product is the erythro product. Remember that the starting material is trans, make the anti product model and then rotate the bond, you'll get the erythro product.

    Yes, anti product = anti addition.
  12. Dec 3, 2003 #11
    That clears it up. I was confused by looking at the fisher projection with the Br's on one side.

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