This reaction is an example of a nucleophilic substitution reaction, specifically an SN2 reaction. In this case, the nucleophile is the hydroxide ion OH- provided by the potassium hydroxide KOH , and the electrophile is the alkyl bromide CH3CH2CH2-Br .Mechanism of the reaction:1. The hydroxide ion OH- attacks the electrophilic carbon atom the one bonded to the bromine from the backside, opposite to the leaving group Br- . This results in the formation of a transition state where the carbon is partially bonded to both the hydroxide and the bromine.2. As the bond between the carbon and the hydroxide strengthens, the bond between the carbon and the bromine weakens, and the bromide ion Br- leaves as a leaving group.3. The final product is formed when the bond between the carbon and the hydroxide is fully formed, and the bromide ion has completely left.Based on this mechanism, the major product of this reaction is the substitution of the bromine atom with the hydroxide group, resulting in the formation of an alcohol: CH3CH2CH2-OH 1-propanol There is no rearrangement in this reaction, as the SN2 mechanism involves a direct substitution without the formation of any carbocation intermediates that could undergo rearrangement.