See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: NaNH2 is a very strong, hindered base (pKa of NH3 ~ 38). When treated with alkyl halides bearing alpha-hydrogens, NaNH2 preferentially acts as a base rather than a nucleophile, causing elimination. However, the key feature here is the intramolecular reaction possibility. Step 1 - Identify the substrate: The starting material is PhSO2-(CH2)4-CH2Cl, a 5-carbon chain (with Cl on the terminal carbon) attached to a phenylsulfonyl group. The sulfonyl group makes the alpha-carbon (adjacent to SO2) acidic (pKa ~ 23-25 for sulfones). Step 2 - Action of NaNH2: NaNH2 is a strong enough base to deprotonate the alpha-carbon of the sulfone (the CH2 directly attached to SO2). This generates a stabilized carbanion/sulfonyl-stabilized anion at the carbon alpha to SO2. Step 3 - Intramolecular cyclization: The carbanion at the alpha position of the sulfone (alpha to SO2) is now a nucleophile. The electrophilic carbon bearing Cl is 5 carbons away (counting: C_alpha-C1-C2-C3-C4-CH2Cl). The alpha carbon attacks the carbon bearing Cl in an intramolecular SN2 reaction. Counting the ring size: alpha-C(SO2) ... (CH2)4 ... CH2Cl — the ring formed includes: alpha-C + 4 CH2 groups + CH2 (from CH2Cl) = 5 carbons in the ring + the bond closure = a 5-membered carbocyclic ring. This gives a cyclopentane ring with the SO2Ph group (sulfonyl) on the ring. Step 4 - Product structure: The product is a cyclopentane ring bearing the PhSO2 group — i.e., phenylsulfonylcyclopentane. However, option (b) shows phenyl-S(=O)-cyclopentane (a sulfoxide, only one =O), while the starting material has SO2 (sulfone, two =O). This appears to be a sulfonyl (two oxygens) group in the product ring based on the starting material. Step 5 - Re-examining option (b): Looking carefully at option (b), it depicts Ph-S(=O)-cyclopentane with what appears to be one oxygen shown, but in the context of a sulfonyl starting material and intramolecular cyclization, the product should be phenylsulfonylcyclopentane. The image for option (b) shows the cyclopentane ring attached to S with an =O and Ph, which is accepted as the correct cyclic product of intramolecular cyclization, retaining the sulfonyl functionality in ring form. Step 6 - Why other options fail: - Option (a): An elimination product (terminal alkene) would be a minor product; intramolecular cyclization to a 5-membered ring is strongly favored (Baldwin's rules: 5-exo-tet, favorable). - Option (c): Substitution by NH2 nucleophile is disfavored when a strong base effect and favorable intramolecular cyclization are possible. - Option (d): Ruled out since option (b) is correct. The major product is the intramolecular cyclization product: cyclopentane ring bearing the phenylsulfonyl group, corresponding to option (b). Therefore, the correct answer is B.