The major product of the following reaction is:

We need to determine the major product formed when the given dibromo compound undergoes dehydrohalogenation with alcoholic $$\text{KOH}$$ under heating ($$\Delta$$).

Reaction Mechanism:

1. Double $$\text{E2}$$ Elimination ($$-2\text{HBr}$$):

   Alcoholic $$\text{KOH}$$ is a strong base used to induce elimination reactions. Since the substrate contains two bromine leaving groups on adjacent carbons along with accessible $$\beta$$-hydrogens, it undergoes a double dehydrohalogenation process, losing two molecules of $$\text{HBr}$$.




2. Thermodynamic Driving Force (Conjugation):

   The elimination is highly regioselective and follows Saytzeff's rule to produce the most thermodynamically stable alkene system possible. By eliminating both bromine atoms systematically, a highly extended conjugated diene system is formed.

   $$\text{Dibromo Substrate} \xrightarrow{\text{Alc. KOH, }\Delta} \text{Highly Conjugated Diene Product} + 2\text{HBr}$$



3. Resonance Stabilization with the Benzene Ring:

   The newly formed pair of carbon-carbon double bonds ($$\text{C=C}$$) aligns perfectly with the adjacent phenyl ($$\text{Ph}$$) ring. This creates an uninterrupted, delocalized $$\pi$$-electron cloud spanning across the entire molecular framework, providing immense resonance stabilization energy.

Conclusion:

The reaction yields a stable, extended conjugated diene system directly attached to the benzene ring due to the thermodynamic advantage of resonance delocalization.

Answer: Option C