ah ha. How about the reaction between Halides and CH3COOH.
It isn't mentioned in Step by step but in your link so I want to
know it.
No iodoform generated. Under alkaline conditions, carboxylic acids
are deprotonated, generating carboxylate ions, which are thus
anionic, rendering the alpha C's protons non-acidic.
The first step of the iodoform mechanism (for methyl ketones;
for methyl alcohols there is an additonal prior step which oxidizes
the methyl alcohol to methyl ketone) involves alpha deprotonation
of the methyl ketone's alpha protons (by the OH- base) to generate
the enolate ion nucleophile, which subsequently attacks the
diiodine electrophile.
This substitution of the alpha protons with iodine atoms repeats
itself until the CH3 group becomes the CI3 group, which is a far
better leaving group because CI3- anion has a much lower
charge density compared to the CH3- anion.
The penultimate step involves the OH- nucleophile attacking the
methyl ketone's acyl/carbonyl C atom, eliminating CI3 as CI3-. The
final step involves a proton transfer from the (newly generated)
carboxylic acid, to the CI3-, generating the yellow precipitate of
triiodomethane (aka iodoform).
However (considering the first step of deprotonation), if you
were to use a methyl carboxylic acid (instead of a methyl alcohol
or a methyl ketone), a dinegative anionic charge on the conjugate
base is simply too destabilizing. Hence methyl carboxylate ions
cannot be further deprotonated, and therefore the iodoform reaction
cannot proceed with methyl carboxylic acids.
Methyl amides and esters, also do not undergo the iodoform
reaction, because the carbonyl O atom already bears additional
electron density by resonance (as the O atom of the ester, or N
atom of the amide, already donates its lone pair by resonance to
the carbonyl group). Hence methyl esters and amides also (like
carboxylate ions), are unable to be further deprotonated, and
therefore unable to undergo the iodoform reaction.
