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Structure of 2,4-Pentanedione

  1. Feb 7, 2005 #1
    Is 2,4-Pentanedione a ring structure or just a hydrocarbon chain with double-bonded O's on C 2 and 4? When showing the reaction with the Iodoform test what would the chemical equation be? I know it is

    2,4-Pentanedione

    NaOH
    --> I know how this would work with a single double-bonded O but how
    I2 would it work with two? Would the result be...

    CH3CIONaCIOCH3 with 2 CHI3 (iodoform) derivatives in the solution? Please help. Thanks. :bugeye:
     
  2. jcsd
  3. Feb 8, 2005 #2

    Gokul43201

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    I believe there may be two products possible, but I'm only guessing here :

    [tex]H_3C-C(=O)-CH_2-C(=O)-CH_3 \xrightarrow {NaOH, ~I_2} NaO-C(=O)-CH_2-C(=O)-ONa + H_3C-C(=O)-CH_2-C(=O)-ONa [/tex]
     
  4. Feb 8, 2005 #3

    chem_tr

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    Yes, Gokul is right; the products are carboxylates and iodoform, aka [tex]CHI_3[/tex].
     
  5. Feb 8, 2005 #4
    Thank you! :)
     
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