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In the laboratory, I performed an eletrophilic aromatic subsitution by reacting $\ce{I+}$ with salicylamide. The mechanism for this reaction is pretty straightforward, but it's the mechanisms for the supplementary reactions used in this lab that I can't seem work out.

  1. In order to prepare the electrophile $\ce{I+}$, sodium iodide was reacted with sodium hypochlorite:

    $$\ce{ ClO- (aq) + I- (aq) + 2H3O+(aq) -> I+ (s) + Cl- (aq) + 3H2O (l)}$$

    The equation (I think) is right, but can anyone provide some mechanistic insight or literature on how the hypochlorite manages to oxidize the Iodide?

    Does it first have to go into an iodine ($\ce{I2}$) form? The reason I propose this is because as the hypochlorite was added to the salicylamide/sodium iodide absolute ethanol solution, color change was observed from dark red-brown to increasingly lighter shades of yellow.

    It seems a similar question was asked about a year ago, but I was hoping to take it a step further.

  2. Once the reaction had finished, a 10% (w/v) sodium thiosulfate solution was added. I assume this was because thiosulfate is a well known bleach (hypochlorite) neutralization agent: $$\ce{HOCl + 2 S2O3^2- -> Cl- + S4O6^2- + OH-}$$

    Assuming the above equation is correct, how exactly (mechanistically) does thiosulfate neutralize hypochlorite/hypochlorous acid? The $\ce{S-S}$ bond of the tetrathionate is whats tripping me up.

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The order of events is probably:

Protonation of the hypochlorite ion

$$\ce{ClO- + H3O+ <=> ClOH + H2O}$$

Nucleophilic displacement of chloride by iodide

$$\ce{I- + ClOH \phantom{0}<=> Cl- + IOH}$$

Protonation of hypoiodous acid

$$\phantom{000}\ce{IOH + H3O+ <=> IOH2+ + H2O}$$

Fragmentation of the $\ce{IOH2+}$ cation

$$\phantom{00000}\ce{IOH2+ <=> I+ + H2O}$$

Alternatively, the second step could have been protonation of hyochlorous acid, followed by displacement of water by iodide, and then fragmentation of iodine monochloride:

$$ \begin{align} \ce{ClOH + H3O+ &<=> ClOH2+ + H2O}\\ \ce{I- + ClOH2+ &<=> ICl + H2O}\\ \ce{ICl &<=> I+ + Cl-} \end{align}$$

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  • $\begingroup$ Great answer, but I'm not sure if it successfully accounts for the brown and purple color observed which would mean there was triiodide ion or elemental iodide present and not just I- (which is colorless). $\endgroup$ – Nova Aug 12 '16 at 5:06
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Reaction of $\ce{I-}$ with $\ce{ClO-}$ is a redox reaction.

$$ \begin{align} \ce{ClO-(aq) + H2O(l) + 2e- &<=> Cl-(aq) + 2OH-(aq)} &&E^\circ = \pu{0.90 V}\\ \ce{2I-(aq) &<=> I2 (s) + 2e-} &&E^\circ = \pu{-0.54 V}\\ \ce{ClO-(aq) + H2O(l) + 2I- (aq) &-> Cl-(aq) + 2OH-(aq) + I2(s)}&&E^\circ_{\text{rxn}} = \pu{0.36 V} \end{align}$$ The reaction is spontaneous because $E^\circ_{\text{rxn}}$ is positive value. This produces $\ce{I2}$ in situ in amounts large enough to show its brown color in solutions ($\ce{I2}$ is purplish in solid state). In low concentrations it is yellowish. It is important to note that, this reaction did not use acid, thus acid is not used here for the mechanism.

The alicylamide nucleus is already activated by phenolic $\ce{-OH}$ group so that it does not need catalyst to iodonize in the presence of $\ce{I2}$. This reaction leaves $\ce{HI}$ as a byproduct, which react with $\ce{NaOH}$ in the solution. Once iodination reaction is completed the solution is still basic.

The final solution should contain both $\ce{I2}$ and $\ce{ClO-}$, if alicylamide is the limiting reagent. Thus, following redox reactions would happen when aqueous $\ce{S2O3^2-}$ is added to the reaction mixture.

$$ \begin{align} \ce{ClO-(aq) + H2O(l) + 2e- &<=> Cl-(aq) + 2OH-(aq)} && E^\circ = \pu{0.90 V}\\ \ce{2S2O3^2-(aq) &<=> S4O6^2-(aq) + 2e-} && E^\circ = \pu{-0.09 V}\\ \ce{ClO-(aq) + H2O(l) + 2S2O3^2-(aq) &-> Cl-(aq) + 2OH-(aq) + S4O6^2-(aq)} &&E^\circ_{\text{rxn}} = \pu{0.81 V} \end{align} $$ and, $$ \begin{align} \ce{I2 (s) + 2e- &<=> 2I-(aq)} &&E^\circ = \pu{0.54 V}\\ \ce{2S2O3^2-(aq) &<=> S4O6^2-(aq) + 2e-}&&E^\circ = \pu{-0.09 V}\\ \ce{I2(s) + 2S2O3^2-(aq) &-> 2I-(aq) + S4O6^2-(aq)}&&E^\circ_{\text{rxn}} = \pu{0.45 V} \end{align}$$

Both reactions are spontaneous because $E^\circ_{\text{rxn}}$ values for both are positive. Thus, medium is still basic and need acid ($10\%$ $\ce {HCl}$) to neutralize.

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