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How does the oxidation of NaI by sodium hypochlorite to give I+ work?
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+ Redox
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Lance Chambers
How does the oxidation of NaI by sodium hypochlorite to give I+ work?
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.
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
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) wasadded dropwise over 75 min at 0-3 C" J. Org. Chem. 1990,55, 5287-5291.
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
Youd think the active species was iodine monochloride. I+ on its own looks improbable, its too Lewis-acidic to survive on its own, especially in water.More
@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.More
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
$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq) -> I+ (aq) + Cl- (aq) + H2O (l)}$$
also.
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) wasadded 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
$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq) -> I+ (aq) + Cl- (aq) + H2O (l)}$$
also.
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