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Organic Chemistry question!!!

Discussion in 'Pre-Medical - MD' started by yalla22, Apr 25, 2007.

  1. yalla22

    yalla22 Senior Member 7+ Year Member

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    Jan 6, 2003
    For Orgo lab buffs out there, I have my final exam tomorrow and have a few urgent questions about hydroborations:( I understand the mechanism but dont understand how the lab procedure fits into the mechanism...
    For example, why do we add NaOH along with the hydrogen peroxide? Why is diethyl ether added? And what are we extracting from what during the last phase of the experiment?

    Any help would be appreciated!!!!
     
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  3. alicias108

    alicias108 2+ Year Member

    19
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    Apr 18, 2007
    Indiana State University
    A hydroxide ion, in your NaOH, is generally always there to take a proton. I believe the hydrogen peroxide would be an oxidizing agent. Hard to tell without knowing what your starting product was and what you were supposed to end up with. Really hard to tell without reading the lab manual too. I just looked in mine..that reaction sounds similar..what is it entitled in your book?

    Alicia
     
  4. yalla22

    yalla22 Senior Member 7+ Year Member

    911
    0
    Jan 6, 2003
    Basically you take 1-octene and convert it to 1-octanol...we add borane THF solution with 1-octene and stir for 45 minutes. then is says we have to add NaOH and hydrogen peroxide to the solution ( i understand why we add h2o2 but not the naoh)....then we have to stir for another hour, cool to room temp and then add diethyl ether to extract it..then it says to do another extraction with hcl (why??)
     
  5. ADeadLois

    ADeadLois Senior Member 10+ Year Member

    3,162
    5
    Dec 18, 2005
    The ether extracts the product and then gets boiled off. The HCl removes remaining OH. Two years removed from Organic Synthesis, that's the best I can do.
     
  6. Nevadanteater

    Nevadanteater biochemical engine 7+ Year Member

    306
    2
    Jun 18, 2006
    MDApps:
    Et2O = solvent (easily removed at the end of the reaction)

    ------------------------

    The intermediate of this reaction is B(OR)3

    The B is HUGELY electro positive (or...non-electronegative). It likes being surrounded by e- donating oxygen atoms. We can think of BO3 as a good leaving group. OH- attacks the R and OB(OR)3 leaves. This happens two more times resulting in Na3BO3.

    In otherwords, the OH is what actually forms the alcohol.

    Because this is done in base the newly formed alcohol gets deprotonated (R-O-) and must be extracted using HCl (forming R-OH)

    N/A
     
  7. alicias108

    alicias108 2+ Year Member

    19
    0
    Apr 18, 2007
    Indiana State University
    Whoops..I think Nevada is right. I'm wrong.
     
  8. Janny

    Janny 2+ Year Member

    52
    0
    Feb 26, 2007
    http://www.organic-chemistry.org/namedreactions/brown-hydroboration.shtm

    See the above webpage. Basically you need to understand the whole mechanism of the rxn to understand the NaOH. Adding H2O2 oxidizes the intermediate boron species (remember 3 alkenes per BH3) to boronic ethers then NaOH hydrolyzes that species and liberates boronic acid and the free alcohol(s). In the last stage of your experiment you are extracting the alcohol (ether soluble) from the boronic acid (water soluble).:luck:
     

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