It is a common fact that $\ce{Cl2}$ reacts to form $\ce{HCl}$ and $\ce{HOCl}$ when dissolved in water. However, how exactly does the reaction proceed? The first step could either be that the oxygen atom in $\ce{H2O}$ attacks $\ce{Cl2}$, forming $\ce{Cl-}$, and $\ce{H2OCl+}$, which then loses a proton to become $\ce{HOCl}$.

But it could also be that chlorine attacks the $\ce{H}$ in $\ce{H2O}$, forming $\ce{Cl+}$, $\ce{HCl}$, and $\ce{HO-}$. The first reaction seems more probable (based on the products), but I am not sure.


1 Answer 1


According to this paper, the attacking species is $\ce{OH-}$ rather than $\ce{H2O}$ on $\ce{Cl2}$

The paper also says that:

The proposed mechanism also provides a good explanation for the high initial rate of reaction. At the instant the strong chlorine solution is mixed with the distilled water the hydroxyl ion concentration is considerably higher than that calculated from $\ce{kw}$ and the hydrogen ion concentration, for it takes a finite time for the hydrogen and hydroxyl ions present to react to reduce the ion product to the equilibrium value. During this period chlorine molecules also react with hydroxyl ions at approximately the same rate as hydrogen ions are reacting. So for this initial period a higher rate of hydrolysis of chlorine is expected than that found after the hydroxyl ion has reached its equilibrium concentration.

It also says that:

During the course of the reaction the equilibrium concentration of hydroxyl ion is only about 10^(-12) M. Since about 10^-(3) mole per liter of chlorine must react before final equilibrium is reached, new hydroxyl ions must be supplied by the dissociation of water molecules. This could easily complicate the reaction kinetics, if the dissociation rate were not rapid enough. However, calculations assuming that the dissociation of water is a normal bimolecular reaction with an activation energy equal to the endothermicity of the reaction show that the total hydroxyl ion required may be formed in 10^(-5) sec. This is so rapid that the steady state concentration of hydroxyl ion should not be measurably different from the equilibrium value.

What this stuff basically means is that using $\ce{OH-}$ as the nucleophile, as it can explain the initially high rate of reaction(at the back of our minds, we know that $\ce{OH-}$ is a better nucleophile than $\ce{H2O}$). When $\ce{Cl2}$ is added to the distilled water, we find that $\ce{[OH-]}$ is more than that what would be expected at equilibrium value, so $\ce{OH-}$ and $\ce{H+}$ keep combining to reduce the ion product, while at the same time, $\ce{OH-}$ also reacts with $\ce{Cl2}$. So at this time, rate of reaction will be quite high. Also, the ionization of water to provide the $\ce{OH-}$ and $\ce{H+}$ will be keeping up with the consumption. Also, the reaction rate tells us that collisions in this reaction are very effective, and lead to reaction almost all the time

So, essentially, to explain the initial high reactivity, we take the reaction mechanism to be a coupled equilibria of $\ce{Cl2 + OH- <--> HOCl + Cl-}$ and $\ce{ H2O <--> H+ + OH-}$

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    $\begingroup$ I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please note that in English, there is a space after the punctuation mark. $\endgroup$ Commented Nov 25, 2019 at 15:38
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    $\begingroup$ On the matter at hand, I doubt there is enough evidence for your proposed mechanism. It is one path in a complex coupled-equilibrium reaction. Neglecting the existence of $\ce{H+(aq)}$ or $\ce{OH-(aq)}$ is probably not doing the system justice. $\endgroup$ Commented Nov 25, 2019 at 15:42
  • $\begingroup$ @Martin - マーチン♦ I appreciate the edit,and the reason I neglected H+(aq) or OH-(aq) is because the maximum amount of these ions produced by neutral water would be 10^(-7) M,which is not a lot of amount unless the solution is quite dilute. $\endgroup$ Commented Nov 25, 2019 at 15:48
  • $\begingroup$ Dear Yusuf, Organic chemists do make up lot of plausible stories with their arrow notations but that does not mean that the reaction proceeds that way. $\endgroup$
    – AChem
    Commented Nov 25, 2019 at 20:27
  • $\begingroup$ @M.Farooq So, is my suggestion correct or incorrect, objectively speaking, and if wrong, why? $\endgroup$ Commented Nov 25, 2019 at 21:00

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