The reaction described is a dehydrohalogenation, a process involving the elimination of a halogen (Br) facilitated by an excess base, KOH in ethanol. This type of reaction characteristically produces alkenes through the simultaneous removal of a hydrogen atom and the halogen atom. The excess reagent drives the formation of a conjugated diene product. The elimination proceeds to yield a product where two double bonds are conjugated with the phenyl group. The resultant product is identified as 6-Phenylhepta-2,4-diene, due to the conjugation occurring at positions 2 and 4.
