organic chemistry - Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

i) $\ce{R-CH_2-OH + HBr ->[H_2SO_4] R-CH_2-Br + H_2O}$

ii) $\ce{R-CH_2-OH + HI -> R-CH_2-I + H_2O}$

What role does sulfuric acid play in the first reaction? Why are we not using it in the second one?

Answer

The synthesis of alkyl halides from the corresponding aliphatic alcohols using concentrated hydrohalogen acids was investigated by Klein, Zhang and Jiang.^[1] They note:

[W]e found that the reflux of 1-butanol ($\pu{2.34 g}, \pu{31.5 mmol}$) with $48~\%$ hydrogen bromide ($\pu{7 mL}$) for $\pu{4 h}$ on a $\pu{120 °C}$ oil bath only gave low yield of 1-bromobutane ($54~\%$) and moderate purity ($93~\%$). The yield of 1-bromobutane was improved to $82~\%$ with $90~\%$ purity by adding additional sulfuric acid ($98~\%, \pu{1 mL}$). Interestingly, we found that 1-iodobutane could be directly synthesized using 1-butanol and hydriodic acid ($\ce{HI}, 57~\%$) by reflux without adding an additional acid in $80~\%$ yield with $98~\%$ purity.

They do not offer any reasoning why switching from $\ce{HBr}$ to $\ce{HI}$ gives better yields. However, we may use our chemical reasoning to deduce the reason. As has been pointed out and explained multiple times on this site, the acidity of hydrohalogen acids increases from fluorine to iodine: $\ce{HI}$ is a stronger acid than $\ce{HBr}$ is. This means that hydrogen iodide should protonate a greater percentage of alcohol molecules than hydrogen bromide — and it requires the collision of a protonated alcohol and the halide anion for the reaction to proceed ($\mathrm{S_N2}$ mechanism).

Remember that all hydrohalogen acids are gases at room temperature and standard pressure and thus need to be dissolved in water to give the actual acidic solution. While sulphuric acid itself may be less acidic than both pure $\ce{HBr}$ and $\ce{HI}$, adding it to the solution will increase the solution’s overall acidity since it comes with no added water. Hydrogen iodide in itself is acidic enough to promote the reaction as the experimental evidence shows so only in the case of hydrogen bromide is additional acidity needed for the reaction to occur.

Interestingly, the authors also state that alcohols with more than five carbon atoms give worse yields due to their low solubility in $\ce{HI}$. This can be overcome by adding phosphoric acid to the mixture. Again, phosphoric acid may not be strong but it features a low water content. The added acidity may protonate more alcohol molecules allowing the $\mathrm{S_N2}$ reaction to proceed. I point to the other answers as to why using sulphuric acid instead of phosphoric acid is a bad idea in the case of hydrogen iodide.

Thus, to sum up:

• hydrogen iodide is sufficiently acidic to protonate the corresponding alcohol and drive the reaction. More acids are not needed.


• hydrogen bromide is not acidic enough to drive the reaction. Adding sulphuric acid protonates the alcohol partially and allows the reaction to proceed.


• in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidise iodide (while sulphuric acid does).

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[1]: S. M. Klein, C. Zhang, Y. L. Jiang, Tetrahedron Lett. 2008, 49, 2638–2641. DOI: 10.1016/j.tetlet.2008.02.106.