The reaction involves dehydrohalogenation of Ph-CHBr-CH₂-CH₃ with alcoholic KOH. This is an E2 elimination where the base abstracts a β-hydrogen. The molecule has two β-carbons: one adjacent to the bromine in the ethyl group (CH₂-CH₃) and the phenyl ring. The major product follows Zaitsev's rule, forming the more substituted alkene. The phenyl group stabilizes the double bond via resonance, making Ph-CH=CH-CH₃ the major product.

Final Answer: Ph–CH=CH–CH₃