# Mechanism with hypohalite in haloform reaction

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 ($$\ce{X2}$$).

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?

• I'd expect same thing, just with OH- leaving instead of halide anion. – 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 .