See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Intramolecular SN2 cyclization (Williamson ether synthesis intramolecular version). When a molecule contains both a nucleophilic group (OH) and a leaving group (Cl) in the same molecule, base (NaOH) can deprotonate the hydroxyl to form an alkoxide, which then attacks the carbon bearing the leaving group intramolecularly to form a cyclic ether. Step 1: Identify the structure of 4-chloro-1-butanol. The compound is: HO-CH2-CH2-CH2-CH2-Cl (numbering: C1 bears OH, C4 bears Cl). Step 2: Action of NaOH. NaOH deprotonates the hydroxyl group at C1 to generate the alkoxide: -O-CH2-CH2-CH2-CH2-Cl. Step 3: Intramolecular SN2 reaction. The alkoxide oxygen (nucleophile) attacks C4 (the carbon bearing Cl, the leaving group) in an intramolecular backside attack. The chain between O and C4 has four carbons (O-C1-C2-C3-C4), forming a 5-membered transition state and yielding a 5-membered cyclic ether (oxolane/tetrahydrofuran). Step 4: Count ring atoms. The ring contains: O, C1, C2, C3, C4 = 5 atoms total -> 5-membered ring = tetrahydrofuran (THF). Why other options fail: - Option (a) tetrahydropyran is a 6-membered cyclic ether, which would require 5-chloro-1-pentanol (a 5-carbon chain between OH and Cl). - Option (c) 3-methyloxetane is a 4-membered ring requiring a different substrate. - Option (d) 2-ethyloxirane is a 3-membered ring requiring a different substrate. Therefore, the correct answer is B.