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The haloform reaction is usually done in a solution of NaOH and $\ce{I2}$ (or indeed, any halogen). Now the first part of the reaction mechanism in the wikipedia page and in the most organic books describes the reaction between the enolate and a halogen molecule ($\ce{X2}$).

haloform reaction mechanism

[Picture source: Wikipedia, URL:https://upload.wikimedia.org/wikipedia/commons/8/8d/Haloform_Schritt_1.svg ]

Now, one of the things that I read in the wikipedia page is that the haloform reaction can also be done with sodium hypochlorite($\ce{NaOCl}$). We also know that hypohalite ions are produced in a solution of base and halogen, which is what we use in case of haloform reaction. So, I was wondering, is there any way to draw the mechanism with hypohalite only, without bringing in the halogen molecule? What would that mechanism look like?

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    $\begingroup$ I'd expect same thing, just with OH- leaving instead of halide anion. $\endgroup$ – Mithoron Feb 26 at 22:42
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The mechanism of haloform reaction will also be almost similar in the case if we use hypohalites (here, it is $\ce{NaOCl}$).

In the first step, instead of $\ce{OH-}$, the acidic $\alpha$ proton will be taken up by $\ce{OCl-}$ (Remember, $\ce{HOCl}$ is a weak acid (but not as weak as water), so it's conjugate base i.e $\ce{OCl-}$ is a moderately strong base to take up a proton), and $\ce{HOCl }$ will be formed, which will be attacked by the enolate formed in the previous step, and $\ce{OH-}$ will leave to form the $\alpha$-haloketone. The reaction will proceed further two steps to form the tri-halo substituted ketone, and correspondingly $\ce{OH-}$ will be generated. These generated $\ce{OH-}$ can then attack the electrophilic carbonyl carbon give the haloform and the salt of the organic acid .

The mechanism is shown below,

enter image description here

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  • $\begingroup$ This is quite interseting but is there any way to experimentally prove that this is the mechanism? Do you know any research paper on this? $\endgroup$ – Shoubhik Raj Maiti Mar 16 at 21:40
  • $\begingroup$ @ShoubhikRajMaiti This might be helpful pubs.acs.org/doi/pdf/10.1021/cr60052a001 . Although you might need to search through the references a lot. $\endgroup$ – Soumik Das Mar 17 at 5:03

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