The major product of the following reaction is:

Step 1: Identify the starting material

From the image: A dibrominated compound with structure showing:

• Benzene ring
• Two Br atoms on adjacent carbons (vicinal dibromide)
• Alkyl chain

Likely structure: $\ce{C6H5-CHBr-CHBr-CH2-CH2-CH2-CH3}$ or similar

Step 2: Reaction with excess alcoholic KOH

Vicinal dibromide + excess alcoholic KOH → double elimination → alkyne

But the products listed are dienes, not alkynes. This suggests the starting material might have additional double bonds or the elimination creates a conjugated diene.

Actually, with excess base and heat, we can get:

1. First elimination: dibromide → alkene
2. Second elimination: if there are more β-hydrogens → diene

Step 3: Determine the product

If the starting material is a dibromide on a longer chain near the phenyl group:

$\ce{C6H5-CHBr-CHBr-... + 2 KOH(alc) ->[heat] C6H5-CH=CH-...}$

With excess base, further elimination can occur if there are more β-H atoms, forming a conjugated diene.

The product 2-phenylhepta-2,4-diene suggests:

• 7 carbons total (hepta)
• Phenyl group at position 2
• Double bonds at positions 2-3 and 4-5
• Conjugated diene system

Structure: $\ce{C6H5-C(=CH-CH=CH-CH2-CH3)-CH3}$

Wait, that doesn't match. Let me reconsider.

2-Phenylhepta-2,4-diene:

     Ph
      |
  1   2   3   4   5   6   7
  CH3-C=CH-CH=CH-CH2-CH3

This has phenyl at C-2, double bonds at C2-C3 and C4-C5.

Answer: (a)

Key Points:

• Vicinal dibromides undergo double elimination with strong base
• Excess alcoholic KOH + heat → maximum elimination
• Conjugated dienes are more stable (thermodynamic product)
• Zaitsev's rule: more substituted alkene is favored