Final answer:
When CH3-Br reacts with a polar protic solvent, a typical SN1 reaction could lead to the formation of CH3OH, whereas the formation of CH3O- may result from SN2 mechanism in the presence of strong base. Formation of CH3O+ is highly unlikely, and CH3OCH3 would not form directly from just the solvent reaction.
Step-by-step explanation:
When CH₃-Br reacts with a polar protic solvent, such as water, one plausible reaction is a nucleophilic substitution (SN1 or SN2 mechanism), depending on the conditions and the nature of the solvent. Since the question does not specify the type of solvent beyond it being polar and protic, nor the conditions (temperature, concentration), we can consider general possibilities. In a typical SN1 reaction, the bromide ion would leave, forming a carbocation which could then be attacked by a nucleophile like water forming CH₃OH or methanol; this corresponds to option 1. However, in a strong basic or highly nucleophilic environment, CH₃O- (methoxide ion) could form if the solvent also acts as a base, corresponding to option 2. Formation of CH₃O⁺, a carbocation, is highly unlikely because methyl cations are extremely unstable, so option 3 can be discounted. Option 4, the formation of CH₃OCH₃ or dimethyl ether, would require a series of steps including a substitution followed by a dehydration step, and is not a direct result of reaction with only a polar protic solvent.