Identify the major product in the following reaction.

Step 1: Identify the reaction

Cyclopentyl bromide with methyl substituent + $\ce{OH-}$ in ethanol

Ethanol solvent with strong base → E2 elimination favored

Step 2: Determine structure

From the image: Cyclopentane ring with:

• Br at one position
• $\ce{CH3}$ at adjacent or nearby position

Likely structure: 1-bromo-2-methylcyclopentane or similar

Step 3: Apply E2 elimination

E2 elimination requires:

• Anti-periplanar geometry (H and Br on opposite sides)
• Removal of β-hydrogen

Zaitsev's rule: More substituted alkene is favored

For 1-bromo-2-methylcyclopentane:

Elimination can give:

1. 1-Methylcyclopentene (internal double bond, more substituted)
2. Methylenecyclopentane (exocyclic double bond, less substituted)

By Zaitsev's rule, the more substituted alkene is major product:

1-Methylcyclopentene (double bond in the ring, with methyl substituent)

Answer: (d)

Key Points:

• E2 elimination: Strong base, good leaving group, β-hydrogen
• Zaitsev's rule: More substituted alkene is thermodynamically favored
• Hofmann product: Less substituted alkene (with bulky base or poor leaving group)
• Cyclic systems: Internal double bond usually more stable than exocyclic