About 5 months ago, I answered a very interesting question albeit a very simple one.
The question was why sulfuric acid is used to convert any alcohol to bromoalkane but not iodoalkane. Jan's answer quite pointed out the explanation:
- 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 oxidize iodide (while sulfuric acid does).
Though I am satisfied with Jan's excellent answer, I further pointed out that sulfuric acid may react with hydrobromic acid to form bromine(aq.) and sulfur dioxide. This bromine may oxidize alcohol to form aldehyde and carboxylic acid and they two will form ester. This was my proposed explanation but I couldn't find any experimental result to back my explanation and heavily relied on theory.
Now recently Nurdrage, a renowned chemist famous for his synthesis of pyrimethamine has uploaded a video on Youtube about the synthesis of bromoethane and bromohexane from ethanol and hexanol using sulfuric acid. He also explained the concern of forming bromine as side product from the reaction of hydrobromic acid and sulfuric acid. He showed a purification technique by adding sodium bicarbonate to react with bromine to form sodium hypobromide.
Now my question is:
Is my explanation of forming ester instead of bromoalkane correct enough? Is there any possibility of forming minute amounts of ester? The reaction may be slow but it can happen before any purification is done (reacting bromine with sodium bicarbonate). If so, what can be done to remove any ester or prevent the reaction of ester formation?
Also related: Why is alcohol boiled with HBr for preparing alkyl bromide?
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