See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When a beta-halo enone (alpha,beta-unsaturated ketone bearing a leaving group at the beta-carbon) is treated with a strong, bulky base such as tert-butoxide, the reaction pathway depends on the substrate geometry and base properties. Step 1 - Identify the substrate: The starting material is 3-chlorocyclohex-2-en-1-one. This is a cyclohexenone where the C2=C3 double bond is conjugated with the C1 ketone, and there is a chlorine leaving group at C3 (the beta-carbon relative to the carbonyl). Step 2 - Identify the reagent: t-BuO(-) is a strong, hindered base in t-BuOH. It favors elimination (E2 or E1cb) over substitution due to steric bulk. Step 3 - Reaction pathway: With a beta-halo enone, the base can promote elimination of HCl. In 3-chlorocyclohex-2-en-1-one, there is already one degree of unsaturation (the C2=C3 double bond). Elimination of the C3-Cl and an adjacent C-H (specifically at C4, the allylic/vinylogous position) would extend the conjugated system. Alternatively, a vinylogous E2 or an E1cb-type elimination occurs: the base removes a proton from C4 (alpha to the existing double bond), and the chloride departs from C3, generating a cross-conjugated or fully conjugated triene/dienone system. This gives 2,4-cyclohexadien-1-one, which is the dienone tautomer readily tautomerizing to phenol, but under these conditions the immediate product is the phenoxide anion (option b) because t-BuO(-) is basic enough to deprotonate the phenol product (pKa ~10) immediately upon formation. Step 4 - Why phenoxide (option b)? The elimination produces 2,4-cyclohexadienone (phenol tautomer). Since t-BuO(-) is present as base (pKa of t-BuOH ~19), it will immediately deprotonate phenol (pKa ~10) to give phenoxide anion (PhO(-)). Step 5 - Why other options fail: - Option (a): Would require SN2 substitution of Cl by t-BuO(-), which is disfavored due to the bulky base and vinyl/allylic position; also t-BuO(-) strongly prefers elimination. - Option (c): Phenol (neutral) would be the product only if no base were present to deprotonate it; since excess t-BuO(-) is present, phenol is immediately converted to phenoxide. - Option (d): Requires SN2 at a saturated carbon with no conjugation change, which contradicts the elimination pathway favored by bulky base. Therefore, the correct answer is B.