Is 2,4-Pentanedione a Ring or Hydrocarbon Chain in the Iodoform Test Reaction?

[MusicMonkey]

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.

 

[Gokul43201]

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

$$H_3C-C(=O)-CH_2-C(=O)-CH_3 \xrightarrow {NaOH, ~I_2} NaO-C(=O)-CH_2-C(=O)-ONa + H_3C-C(=O)-CH_2-C(=O)-ONa$$

 

[chem_tr]

Yes, Gokul is right; the products are carboxylates and iodoform, aka $$CHI_3$$.

 

[MusicMonkey]

Thank you! :)