Question

Which molecules will undergo haloform reaction?

a) CH3COCH3
b) acetophenone
c) CH3CH2CHO
d) CH3COOH
e) CH3Ctriple bondN

Answer 1

Only those capable of forming an enolate, as that is what acts as a nucleophile in this reaction. Typically, that would be if they are methyl ketones.

So, acetone and acetophenone (`A,B`).

Also:

• propionaldehyde would not work because it is more likely to be an electrophile than a Bronsted acid to the `"OH"^(-)`.
• acetic acid would rather be a Bronsted acid at the carboxyl `"OH"`, but you want it to do so at the `alpha` proton instead...
• methyl nitrile is probably not reactive enough to do this; nitrogen is not as electronegative as oxygen, so the nitrile carbon is not as electrophilic as in ketones.

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The haloform reaction is an enolate reaction.

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(click to zoom in.)

1. A strong base (usually just `"NaOH"`) comes in to deprotonate the `alpha`-carbon, forming the enolate.
2. This is followed by nucleophilic attack upon a diatomic halogen molecule (repeat steps 1 and 2 two more times until the `alpha`-carbon is saturated with halogen atoms).

3. Strong base comes in (e.g. `"NaOH"`) as a nucleophile this time (due to the loss of all `alpha`-protons).

   However, this step is thermodynamically difficult in base, since the three halides make the carbonyl carbon more electronegative via inductive electron withdrawal, lessening its susceptibility to nucleophilic attack.

4. Once the tetrahedral intermediate does form, the resultant `""^(-)"CX"_3` is a good (enough) leaving group since there are three electron-withdrawing groups on the carbon. One may need to push the equilibrium with additional `"OH"^(-)`.

5. The mechanism finishes with a deprotonation of the resultant carboxylic acid to form the haloform and the carboxylate.