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quick o-chem question...

Discussion in 'DAT Discussions' started by prydA, May 10, 2008.

  1. prydA

    prydA 2+ Year Member

    Feb 7, 2008
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  3. I'm not sure why they switched 2 of the substituents because the molecule on the left and the Fischer projection are actually enantiomers.
  4. mddang

    mddang 5+ Year Member

    Dec 29, 2007
    Houston, TX
    hmm.. that first Fischer projection makes sense... I don't get why they did that whole 'one exchange' thing... The Fischer on the far right is its enantiomer (I think...).

    Just check for R/S. The two on the left are S, the ones on the right are R.
  5. prydA

    prydA 2+ Year Member

    Feb 7, 2008
    oh sorry, the molecule on the right is another molecule the problem was comparing the left molecule to =).. so i guess whichever molecule is facing back (with the dotted line) will be drawn downwards in the Fischer projection?

    and i have a couple more questions...

    2. problem..
    the text says they are both meso compounds, and diastereomers of each other. why?? i cant find the internal line of symmetry nor can i determine the R/S configurations to determine the stereochemistry.

    not too different from problem #2... how are they diastereomers?

    btw, is a very good site for review of stereochemistry, with explanations and practice probs.

    thanks for the help so far guys!
  6. 2 is meso because the symmetry is from the top down. Think of cutting the methyl and hydrogen down and through the middle of the double bond. I actually don't think the compounds in 3 are diastereomers because they do not have a chiral center (stereocenter, w/e). Since they are 1,4 substituted, each way around the ring is the same substituent. These examples are not the best way to learn stereochemistry.
  7. prydA

    prydA 2+ Year Member

    Feb 7, 2008
    yeah i agree the questions are kind of unusual..

    i have YET another question if you guys don't mind..

    after looking thru achiever and destroyer, i observe that..
    HBr, hv adds anti-Markovnikov to a double bond (achiever ochem, test #1, #82)
    Br2, hv adds to the most substituted carbon (destro road map)
    Br2, hv adds Markovnikov (destro road map)

    so... my question is does HBr add anti-Markovnikov with hv while Br2 adds Markovnikov? or is one of these sources wrong?

    sometimes i hear that hv adds anti or sometimes not. and how about Cl/hv? thanks again so much!
  8. Alkane halogenation is a very bad way to prepare alkyl halides because it is so unpredictable. Chlorination with hv is tough because primary hydrogens are the least reactive but there are usually many of them. Tertiary are usually most reactive but there aren't many of them, so you get mixtures of products. Alkane bromination is much better because it gives the more substituted halide. Now for alkenes: HX will add Markovnikov and X2 will also add normally (most substituted and X's are trans). When you have an alkene and Br2 or NBS and hv (I've never seen HBr for radical alkene bromination) it adds to the allylic position and the double bond is retained. If the allylic position is quaternary, there is no hydrogen to be substituted so it must go to the other allylic position. When you have this reaction though, you also have to think of resonance at the radical intermediate stage. A secondary radical has resonance with a primary radical which is more easily brominated (steric reasons). For example, 1-butene reacted with NBS and hv will give (Z)-4-bromo-2-butene as the major product (rearranged) and 3-bromo-1-butene as the minor product. Another twist on this halogenation is that an alkene reacted with HBr in the presence of a peroxide (R-O-O-R) will be anti-Markovnikov addition through a radical mechanism.

    Hope this helps a little.

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