For elimination reactions with strong, non-bulky bases like \(OH^-\) or \(EtO^-\), always prioritize the Zaitsev product (the most stable/substituted alkene). Bulky bases like \(t-BuO^-\) would favor the Hofmann product (least substituted).
Step 1: Understanding the Concept:
The reaction involves a secondary alkyl halide (1-chloro-2-methylcyclopentane) reacting with a strong base (sodium hydroxide, NaOH) in a protic solvent (ethanol, \(C_2H_5OH\)). Under these conditions, the \(E2\) (bimolecular elimination) pathway is favored over substitution. Step 2: Key Formula or Approach:
1. Reaction Type: \(E2\) Elimination.
2. Regioselectivity Rule: Zaitsev's Rule (the most substituted alkene is the major product). Step 3: Detailed Explanation:
In the given cyclic structure, the chlorine atom is at the \(C1\) position and a methyl group is at the \(C2\) position.
During the \(E2\) mechanism, the base (hydroxide ion) abstracts a proton from a \(\beta\)-carbon relative to the chlorine. There are two types of \(\beta\)-hydrogens available:
1. At \(C2\): Abstraction of the tertiary hydrogen at \(C2\) leads to the formation of a double bond between \(C1\) and \(C2\). This results in 1-methylcyclopentene, which is a trisubstituted alkene.
2. At \(C5\): Abstraction of a secondary hydrogen at \(C5\) leads to the formation of a double bond between \(C1\) and \(C5\). This results in 3-methylcyclopentene (or 2-methylcyclopentene depending on numbering), which is a disubstituted alkene.
According to Zaitsev's rule, the trisubstituted alkene is thermodynamically more stable and therefore forms as the major product. Step 4: Final Answer:
The major product is 1-methylcyclopentene, which corresponds to option (B).
