Organic Chemistry: Haloform Reaction

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SUMMARY

The Haloform reaction, specifically the Iodoform Test, requires the presence of a methyl ketone (CH3CO-) group as a necessary condition for a positive result. The reaction mechanism involves the iodination of the methyl group adjacent to the carbonyl, followed by base-induced removal of protons and nucleophilic attack by iodide ions. A strong electron-withdrawing group (X) is also essential to enhance the acidity of the methyl protons. Alkali iodine or hypoiodite must be present, and excess hypoiodite is recommended to avoid false negatives.

PREREQUISITES
  • Understanding of the Haloform reaction and Iodoform Test
  • Knowledge of organic reaction mechanisms, particularly nucleophilic substitutions
  • Familiarity with the properties of methyl ketones and their reactivity
  • Basic principles of acidity and tautomerization in organic compounds
NEXT STEPS
  • Research the detailed mechanism of the Iodoform Test, focusing on enol intermediates
  • Study the effects of different electron-withdrawing groups on the reactivity of methyl ketones
  • Explore the role of alkali iodine and hypoiodite in organic reactions
  • Investigate other tests for ketones and their mechanisms
USEFUL FOR

Chemistry students, organic chemists, and laboratory technicians involved in organic synthesis and testing of functional groups.

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 occurrence 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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