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quick question~ help! sos!

Discussion in 'DAT Discussions' started by ippie, Jul 25, 2006.

  1. ippie

    ippie ippie
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    (CH3)2C=CH2 + hot KMnO4 --> (CH3)2C=O + CO2
    right?

    then,
    (CH3)2C=CH2 + O3 / Zn ---> (CH3)2C=O + ???

    A. CO2
    B. formaldehyde

    What's the answer? Please, help!!!
     
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  3. dat_student

    dat_student Junior Member
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    Formaldehyde (i.e. O=CH2) :) [ozonolysis]
     
  4. djeffreyt

    djeffreyt Senior Member
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    I thought ozonolysis of a terminal alkene made a CO2...but I'm only guessing at what I remember
     
  5. dat_student

    dat_student Junior Member
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    No CO2 :) I am 200% sure. "B. formaldehyde" is the correct answer ;)
     
  6. RozhonDDS

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    you have the substitutues or R-groups, and 2 H's which yields H2-c=0 or formeldhyde
     
  7. Clapton

    Clapton Member
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    With a KMnO4, don't you get a cis-diol? O3 always results in an oxidative cleavage.
     
  8. RozhonDDS

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    no, if you used kmno4 you would get a ketone and co2
     
  9. Banana Berry

    Banana Berry Junior Member
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    I thought the first one was ketone + CO2.
    The second one is formaldehyde.

    KMnO4 is a strong oxidizing agent and heat promotes it.
     
  10. yorkiepoo

    yorkiepoo Member
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    I thought that KMnO4 would give carboxylic acid group, no?
     
  11. grapeflavorsoda

    grapeflavorsoda Senior Member
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    KMnO4 is a strong oxidizing agent. so when formaldehyde is formed initially, it is first oxidized to formic acid. subsequently formic acid will be then oxidized into bicarbonate.

    as you should know, bicarbonate equilibrates with CO2 and aqueous content of bicarbonate is very low due to its low solubility in water. thus CO2 will bubble out of the solution.

    Although ozone itself is strong oxidizing agent, using Zn (Zn 2+ ion has one of the lowest reduction potential, thus Zn is a very good reducing agent) along with ozone makes the reaction very mild, meaning not as oxidizing. Thus one of the initial product of the both reaction, which is formaldehyde, will not be oxidized further.

    on the side note, in order for a typical transitional metal complex oxdizing agent to oxidize the carbonyl carbon, the carbonyl compound should have the alpha hydrogen(s).

    The reason for this is that, usually it is the hemiacetal form the carbonyl compound that reacts with highly oxidized metal in the oxidizing compound. The metal-oxygen ligand bonding is quite strong(plz refer to hard-soft acid base principle)so its bond will be stable. To compensate for the shift of electron density in the metal-oxygen bond, oxygen shifts the electron density from adjacent carbon, which is the carbonyl carbon in this case.

    The carbonyl carbon in turn withdraws from its adjacent bonds. Usually alpha hydrogen and carbonyl carbon bond is weak enough for this process to occur. On the contrary, C-C bond is not as weak as C-H bond.
     
  12. grapeflavorsoda

    grapeflavorsoda Senior Member
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    only if the temp. is very low.
     
  13. Clapton

    Clapton Member
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    I did not know that, but I guess in every other case you'll have a diol depending on what the co-solute is.
     
  14. grapeflavorsoda

    grapeflavorsoda Senior Member
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    for both oxidative cleavage and dihydroxylation rxns with potassium permanganate, the solvent system involved is water. but for dihydroxylation rxn pH should be greater than 8, usually done by adding KOH etc.

    free hydroxide is better ligand than alcohol group, thus will bind to manganate atom faster, depleting the oxidizing agent before it can act on the dihydroxylated diol product. If there aren't enough ligand(which occurs at low pH) to deplete the permanganate, permanganate will simply oxidize the dihydroxylated alkene product, diol.

    rxn pathways obviously depend on the temperature.

    so it is combination of both temp. and pH that control the rxn pathway taken by KMnO4 when reacted with alkene; high temp low pH leads to oxidative cleavage of alkene, whereas low temp with high pH leads to dihydroxylation of alkene when reacted with KMnO4.
     
  15. dental#1

    dental#1 Fla DDS
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    Grape you explained that perfectly! I am sitting here studying Ochem now and your comments are much appreciated!
     

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