In the reaction of cyclic ethers with hydrogen bromide (HBr), the process begins with the protonation of the ether oxygen, resulting in a protonated ether and the formation of a bromide ion (Br^-) as a nucleophile. This sets the stage for a backside attack, where the bromide ion attacks one side of the ether, leading to the cleavage of the ring and the generation of a bromine atom on one end and a hydroxyl group (OH) on the other. The structure can be represented as a straight-chain molecule with four carbon atoms, one bromine, and one alcohol group.

However, the reaction does not conclude here. The presence of the alcohol allows for further reaction with another equivalent of HBr, converting the alcohol into a better leaving group, water (H₂O). This results in the formation of another bromide ion, which can then perform a second nucleophilic substitution (SN2) reaction. The final product of this sequence is a terminal dihalide, characterized by bromine atoms at both ends of the carbon chain.

It is important to note that when HBr is indicated in a reaction, it is typically assumed to be in excess unless specified otherwise. This means that the reaction is expected to proceed to completion, yielding the terminal dihalide product. If only one equivalent of HBr were used, the product would differ, as the second equivalent would not be available for the reaction.

In summary, the reaction of cyclic ethers with HBr leads to the formation of terminal dihalides through a series of nucleophilic attacks and the conversion of alcohols into leaving groups. Ethers are generally more challenging to synthesize than to react with, highlighting their unique reactivity profile in organic chemistry.