I'm in trouble with the oxidation of NaI by $\ce{NaClO}$, the reaction should give $\ce{I+}$ which then reacts by electrophilic substitution with an aromatic complex. As of now I've come to:

$$\ce{ ClO- (aq) + 2I- (aq) + 2H+(aq) -> I2 (s) + Cl- (aq) + H2O (l)}$$

But I can't then find a way to obtain then the electrophilic $\ce{I+}$ that should react.


1 Answer 1


I think the reaction should be done in basic solution, not acidic solution. "One equivalent each of sodium iodide (2.77 g) and sodium hydroxide (0.74 g) was added, and the solution was cooled to 0 C. Aqueous sodium hypochlorite (34.50 g, 4.0% NaOCl) was added dropwise over 75 min at 0-3 C" J. Org. Chem. 1990,55, 5287-5291.

So perhaps:

$$\ce{ ClO- (aq) + I- (aq) + H2O (l) -> I+ (aq) + Cl- (aq) + 2OH- (aq)}$$

Edit: after looking at the cited article more closely, the hydroxide was only being added because of the acidity of the iodophenol product, so you could write

$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq) -> I+ (aq) + Cl- (aq) + H2O (l)}$$


  • $\begingroup$ You'd think the active species was iodine monochloride. I+ on its own looks improbable, it's too Lewis-acidic to survive on its own, especially in water. $\endgroup$ Dec 19, 2014 at 15:48
  • 1
    $\begingroup$ @AbelFriedman In section 16.2.10 of Inorganic Chemistry by Gopalan, it says $\ce{I+}$ can stablized by solvent molecules and there are species such as $\ce{I_2+}$, $\ce{I_3+}$, etc. $\endgroup$
    – DavePhD
    Dec 19, 2014 at 16:10

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