The major product of the following reaction is:
The given reaction involves a dehydrohalogenation process, where a halogen (Br) is eliminated in the presence of an excess base, KOH in ethanol. This reaction typically leads to the formation of alkenes by the elimination of H and Br atoms.
Since the reagent is in excess, it leads to a conjugated diene product.
The elimination occurs in such a way that the resulting product has two double bonds conjugated with the phenyl group.
The correct product formed in this reaction is 6-Phenylhepta-2,4-diene, as it has the conjugation in positions 2 and 4.