# 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 '19 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,

• 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? – Shoubhik R Maiti Mar 16 '19 at 21:40
• @ShoubhikRajMaiti This might be helpful pubs.acs.org/doi/pdf/10.1021/cr60052a001 . Although you might need to search through the references a lot. – Soumik Das Mar 17 '19 at 5:03
• @SoumikDas in the second step of the mechanism, its OH group to be attacked and not the halide atom, because halide groups have a more electron withdrawing power than hydroxyl groups, and so, in ClOH, OH is more electrophillic and hence its attacked. – Hon Bisaso Ismail Dec 15 '19 at 5:28
• @HonBisasoIsmail Seriously?? Think about it once again what you are saying – Soumik Das Dec 16 '19 at 6:55
• yes we all know that oxygen is more electronegative than chlorine, but my argument is that despite that fact, also when it come to functional groups, halides are more electron withdrawing than hydroxyl group – Hon Bisaso Ismail Dec 16 '19 at 18:15

From another vantage, departing from the classic organic chemistry depiction, I start with 2012 article in Atmospheric Chemistry and Physics 12, pages 6237–6271 (full paper ) regarding the haloform reaction in atmospheric conditions (and a path variation below for natural waters), to quote:

“Halocarbons can then be formed through the reaction of HOCl, HOBr and HOI with the dissolved organic matter through the haloform Reaction (R6, written for X = Cl), or less likely, through addition Reaction (R7).

RCOCH3 + 3 HOCl → RCOOH + CHCl3 + 2 H2O (R6) “

Based on the following comment on the reactivity of HOCl contained in this 2017 compilation by ScienceDirect at to quote:

“The reactivity of HO-Cl mimics those of the two-electron oxidation reaction exhibited by peroxynitrite (HO–NO2) or hydrogen peroxide (HO–OH) whereby the homolytic cleavage of O–X (X=N, O, or Cl) to form the HO and Cl pair could impart a diversity of chemical reactions with a variety of biomolecules.”

Based on the above and other mentioned radical pathways, I suggest the following largely radical pathway, letting G = RCO for convenience, assuming the presence of sunlight (or lab light) containing UV, as part of a hypochlorous acid activation path:

HOCl + hv → •OH + •Cl

Reference: See Scheme I and also a link to a free pdf download of the full paper, which I recommend, as this advanced work is replete with references to rRHS (radical Reactive Halogen Species) attacking organics.

Or, an alternate route per other sources including to quote “and HOCl can react in chloride containing aerosols to form Cl2. Cl2 can then escape to the gas phase and photolyze, releasing Cl atoms which may then reform HOCl” , and a similar observation in this full paper at, which also cites a role for natural organic matter (NOM) and HCO3- (contained also in some chlorine bleaches with added Na2CO3) acting as a catalyst.

Then, the proposed action of the hydroxyl radical (as frequently employed in environmental remediation, see as an example here ) together with the chlorine radical could continue as follows:

GCH3 + •OH + •Cl → •G + •CH2Cl + H2O

HOCl + hv → •OH + •Cl

•CH2Cl + •OH + •Cl → •CHCl2 + H2O

HOCl + hv → •OH + •Cl

•G + •OH → GOH

•CHCl2 + •Cl → CHCl3

Resulting in the cited net reaction for the haloform reaction in an assisted UV reaction (via possibly free liberated chlorine, NOM, HCO3-,..., per above references):

$$GCH3 + 3 HOCl + hv → GOH + CHCl3 + 2 H2O$$

Hypochlorous acid can also apparently be radicalized in the dark in the presence of transition metals occurring in natural waters in a Fenton-type reaction also per this reference:

$$Fe(2+)/Cu(+) + HOCl → Fe(3+) + •ClOH-$$

As such more generally, in the presence of a source of electrons (e.g. , Fe(2+) → Fe(3+) + e- ) and a medium that supports an electron’s half-life (like organics including humic acid and ethanol):

$$HOCl + e- (aq) → •ClOH-$$

With the created chlorine hydroxyl radical anion (above) displaying a pH dependence, with some authors assuming, contextually, an appropriate pH condition, citing different Fenton-type reaction products via hypochlorous, more explicitly:

At pH > 5 (Source: Scheme 1 reference above to quote: "Above pH >5 ketones and aldehydes are halogenated by electrophilic substitution at the α carbon"):

•ClOH- → •OH + Cl- (k = 6.1×10^9)

Source: See Reaction 6 at Supplement Table S1 at and click on Supplement pdf to download)

For completeness, at pH < 5: $$•ClOH- + H+ → H2O + •Cl (k = 21×10^9)$$ Interestingly, the presence of chloride ion, can also be impactful, albeit slow, per the reaction: $$•ClOH- + Cl- → •Cl2- + OH- (k = 1.0 ×10^4)$$ Where the dichloride radical anion (•Cl2-), while being less chemically reactive is, however, more stable and provides a source of •Cl (which slowing reacts with water liberating the hydroxyl radical at pH >5): $$•Cl + H2O → •ClOH- + H+ (k = 2.5×10^5)$$ The Science Direct compilation also relatedly postulates a radical argument pathway with HOCl, to quote: “Reaction of HOCl with thiols yields RSOH via the radical–radical coupling reaction of the intermediates, RS• and HO•” displayed in Equations (2.77) to (2.79).

More simply, starting with Cl2 + NaOH implies: $$Cl2 + NaOH = NaCl + NaOCl$$ with the above reaction being the basis for the commercial production of aqueous chlorine bleach.

So, equivalently, we should be able to start with commercial bleach (which is actually not pure NaOCl, but sodium hypochlorite plus NaCl and even possibly Na2CO3).

Interestingly, if there is an electro/photochemical redox aspect to the haloform reaction (leading to radical and non-radical associated reactive halogen species), then the presence of sodium chloride, which is an excellent electrolyte, may be significant as could be carbonate radicals.

Working with freshly prepared AgOCl (stable in the presence of Ag2O) in place of NaOCl/NaCl would be one way to test, as the formation of insoluble AgCl and Ag2CO3 removes chloride and carbonate ions.