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?

  • 1
    $\begingroup$ I'd expect same thing, just with OH- leaving instead of halide anion. $\endgroup$ – Mithoron Feb 26 at 22:42

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.