My textbook says that alcohols react with sodium bromide and sulphuric acid to form the corresponding alkyl bromides. At the same time they also mention that sulphuric acid mustn't be used with potassium iodide ($\ce{KI}$), because $\ce{HI}$ formed would be oxidized to $\ce{I2}$. Won't $\ce{HBr}$ formed from $\ce{NaBr}$/$\ce{H2SO4}$ oxidise to $\ce{Br2}$?

  • $\begingroup$ The answer might be lying in the bond enthalpy of H-I and H-Br and the less stability of HI as compared to HBr. $\endgroup$ – Abhishek Pallippara gopakumar Apr 4 '16 at 15:57
  • 2
    $\begingroup$ The original question has already been answered, but one way to work around the problem is to use phosphoric acid to make the hydrogen iodide. See en.wikipedia.org/wiki/Hydrogen_iodide. $\endgroup$ – Oscar Lanzi Sep 23 '16 at 12:58

Iodide ions are much more powerful reducing agents than bromide ions:

$$\ce{2I- ->I2 + 2e-} \hspace{1cm}E^{o}=-0.54\ \mathrm{V}$$

$$\ce{2Br- ->Br2 + 2e-} \hspace{1cm}E^{o}=-1.0873\ \mathrm{V}$$

Iodide ions are so powerful, in fact, that they are capable of reducing sulfur from a +6 oxidation state in $\ce{H2SO4}$ all the way to -2 in $\ce{H2S}$ via the following half reaction:

$$\ce{HSO4- +9H+ +8e- -> H2S + 4H2O} \hspace{1cm}E^{o}=+0.80\ \mathrm{V}$$

The full reaction being being:

$$\ce{HSO4- +9H+ +8I- -> H2S + 4I2 +4H2O}\hspace{1cm}E^{o}=+0.26\ \mathrm{V}$$

A quick glance at the reduction potentials tells you that the analogous bromide reaction has a negative reduction potential and is therefore non-spontaneous.



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