Ethers can be cleaved (broken down) by HBr and HI reagents to give alkyl bromides and alkyl iodides.
In the first step of the mechanism oxygen gets protonated by the acid to become a good leaving group. Halogen can then attack the carbon with ROH pushing the ROH off and creating the first alkyl halide. The remaining alcohol gets protonated again and the same reaction happens to it, creating the second alkyl halide. The mechanism shown is for two Sn2 reactions, but Sn1 can happen as well if the alcohol is tertiary.
Reagent: HX (X=Br or I)
Product: Break the bonds between oxygen and the two carbons it is attached to. Attach X to each carbon broken off.
In the mechanism for this reaction, first oxygen gets protonated to become a better leaving group. X- can then attack carbon that has OHR+ group and the bond between oxygen and carbon breaks giving us the first product. This process repeats again with OH getting protonated to become a better leaving group and then X attacking the carbon that the OH2+ is attached to, and OH2+ leaving.
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