Final answer:
Ethanol (EtOH) is the compound that reacts with bromine in a polar protic solvent, leading to substitution and formation of ethyl bromide. Water and methanol do not participate in a similar reaction under the same conditions.
Step-by-step explanation:
Among the compounds listed (H₂O, CH₃OH, and EtOH), ethanol (EtOH, which is CH₃CH₂OH) is known to react with bromine (Br₂) in a polar protic solvent. This reaction typically involves the formation of an alkyl bromide through the substitution of the hydroxyl (-OH) group in ethanol. Bromine, as a halogen, will react in the presence of polar protic solvents like water or methanol because these solvents can stabilize the intermediates that form during the reaction. However, water (H₂O) and methanol (CH₃OH) do not significantly react with bromine in such a manner because their -OH groups are not suitably positioned for the same type of substitution reaction as with EtOH.Moreover, as alkyl halides formation is facilitated by having a good leaving group, ethanol's -OH can become a better leaving group when treated with a strong acid, such as H₂SO₄, to form a protonated alcohol, which can then undergo nucleophilic substitution with Br₂ to form ethyl bromide (CH₃CH₂Br).