The major product of the following reaction is:

Step 1: Identify the starting material

From the image: Dibromide with phenyl group

Likely structure: $\ce{Ph-CHBr-CHBr-CH2-CH3}$ (vicinal dibromide)

Step 2: Reaction with excess alcoholic KOH

Alcoholic KOH (excess) + heat → Double elimination

Vicinal dibromides undergo two E2 eliminations to form alkynes

Step 3: Mechanism

First elimination: $\ce{Ph-CHBr-CHBr-R ->[KOH] Ph-CH=CBr-R + KBr + H2O}$

Forms vinyl bromide intermediate

Second elimination: $\ce{Ph-CH=CBr-R ->[KOH/Δ] Ph-C≡C-R + KBr + H2O}$

Forms alkyne

Step 4: Final product

Product: Phenylacetylene derivative (alkyne)

Structure: $\ce{Ph-C≡C-CH2-CH3}$

Answer: (d) Alkyne structure

Key Concepts:

Vicinal dihalides → Alkynes:

General reaction: $\ce{R-CHX-CHX-R' + 2 KOH(alc) ->[excess/Δ] R-C≡C-R' + 2 KX + 2 H2O}$

Requirements:

• Excess strong base
• Heat
• Two equivalents of base needed

Mechanism:

• First E2: Forms vinyl halide
• Second E2: Forms alkyne
• Vinyl halides are less reactive but react under forcing conditions

Alternative with geminal dihalides: $\ce{R-CX2-CH2-R' + 2 KOH(alc) -> R-C≡C-R'}$

Also forms alkynes via similar double elimination.