Organic Chemistry: Haloform Reaction

AI Thread Summary
The discussion centers on the conditions required for a compound to yield a positive Haloform Test, specifically the Iodoform Test. A methyl ketone, characterized by a C=O group adjacent to a methyl group, is essential for a positive result. The presence of a CH3CO- group is necessary, as it enhances the acidity of the methyl protons through tautomerization to an enol form. Additionally, the presence of a strong electron-withdrawing group (X) is important for the test's success. The mechanism involves the iodination of the methyl group to form a triiodomethyl intermediate, followed by nucleophilic attack and base removal of H+ ions, ultimately leading to the formation of iodoform. It is also noted that using an excess of hypoiodite is advisable to avoid false positives due to reactants that can reduce hypoiodite.
maverick280857
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Hi

What are the necessary and sufficient conditions for a compound to give a positive Haloform Test (Iodoform test for instance)?

Is it necessary to have a CH_{3}CO- group or is it necessary to have a XCH_{2}- group where X is some group? Is the occurence of a CH_{3}CO- group a sufficient condition?

I would be very grateful if someone could help me with this...

Thanks and cheers
Vivek
 
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Haloform test is ideally positive for methyl ketones; the C=O group attached to a methyl group increases the overall acidity of methyl protons due to tautomerization to enol form. So the X you wrote should be a strong electron withdrawing substituent, or at least a group with a highly electronegative atom.

However, you should take into account that the reactant is alkali iodine, or hypoiodite; so any reactant capable of reducing hypoiodite would consume it, causing a positive error. It would be wise to include a bit excess of hypoiodite.

Try to find the mechanism responsible for iodoform formation, the mechanism probably goes from an enol intermediate.

Good luck.
 
Thanks chem_tr...

I am referring to the reaction of a given compound with KOH/I2 which we call the Iodoform Test. The mechanism as I know it involves first the iodination of the Me group attached to the carbonyl carbon to a CI3 group followed by "base removal of H+ ions" with a I- nucleophile successively attacking the intermediate to finally form iodoform (precipitate).

So it now does appear that having a MeCO- group is a necessary condition...

Thanks and cheers
vivek
 
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