See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When a secondary alkyl halide undergoes elimination with a strong base (KOH in ethanol, heated), an E2 elimination occurs. The major product is determined by Zaitsev's rule, which states that the more substituted (more stable) alkene is the major product. Step 1 – Identify the substrate: The starting material is 1-phenyl-2-fluoropropane: Ph-CH2-CH(F)-CH3. The fluorine is on C2 (secondary carbon). Step 2 – Identify possible elimination pathways: - Beta-hydrogens available on C1 (PhCH2-) and on C3 (CH3). - Elimination toward C1 gives: Ph-CH=CH-CH3 (internal, conjugated with phenyl ring — more substituted and conjugated alkene). - Elimination toward C3 gives: Ph-CH2-CH=CH2 (terminal, less substituted alkene). Step 3 – Apply Zaitsev's rule and conjugation stabilization: The internal alkene Ph-CH=CH-CH3 is both more substituted AND conjugated with the phenyl ring, making it significantly more stable than the terminal alkene Ph-CH2-CH=CH2. Therefore, elimination toward C1 (giving the conjugated, internal alkene) is strongly favored. Step 4 – Geometry: Under E2 conditions with KOH/EtOH/heat, the trans (E) isomer of Ph-CH=CH-CH3 is the predominant stereoisomer due to lower steric strain, consistent with option (b) which shows the trans internal alkene. Step 5 – Why other options fail: - Option (a): Terminal alkene — less substituted, not conjugated with Ph; minor product by Zaitsev's rule. - Option (c): SN2 substitution with EtO- giving ether; F is a poor leaving group for SN2 under these strongly basic/heated conditions, and elimination dominates anyway; not the major product. - Option (d): Hydrolysis/substitution to give alcohol; not favored under anhydrous KOH/EtOH conditions, and elimination dominates. Therefore, the correct answer is B.