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 $\ce{Br2}$?
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$\begingroup$ 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 P GApr 4, 2016 at 15:57
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3$\begingroup$ 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 LanziSep 23, 2016 at 12:58
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$\begingroup$ What book is this? Would you share its title? $\endgroup$– ChengJul 26 at 11:42
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1 Answer
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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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$\begingroup$ What about this video? Solid Halides with Concentrated Sulfuric Acid youtu.be/0vT0pkKMTDI Isn't bromine produced, even without heating? Perhaps the OP should have mentioned cooling the reactants to prevent oxidation of bromide from happening? $\endgroup$– ChengJul 26 at 11:59