See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is a cyclohexane ring with a fused epoxide (at C1-C2) and a bromine substituent (on a wedge, axial or equatorial) at another carbon on the ring. Step 2 - First mole of NaOCH3: NaOCH3 is a strong base/nucleophile. The first equivalent performs an E2 elimination or SN2 on the carbon bearing bromine. Since bromine is on the ring with a specific stereochemistry, the base (OCH3-) abstracts a proton beta to Br in an E2 fashion, generating a double bond (alkene) adjacent to the epoxide, or alternatively opens the epoxide. More precisely, the first mole of NaOCH3 acts as a base to eliminate HBr from the bromocyclohexane portion, generating a cyclohexene ring that still contains the epoxide. Step 3 - Second mole of NaOCH3: The second equivalent of NaOCH3 opens the epoxide via SN2 (or the newly formed allylic system undergoes further elimination). Given that 2 moles of base are used and the product after H3O+ workup is phenol (a fully aromatic compound), both steps together produce two eliminations leading to a fully conjugated system (benzene ring) with a hydroxide equivalent. The sequence is: (1) first NaOCH3 eliminates HBr to give an epoxycyclohexene, (2) second NaOCH3 opens the epoxide or promotes another elimination, and the intermediate (A) upon acid workup (H3O+) gives a phenol via aromatization. Step 4 - Formation of phenol: With two equivalents of base causing two dehydrohalogenation/elimination steps involving the epoxide and the bromide on the cyclohexane scaffold, and subsequent acid workup of the resulting enolate or allylic alkoxide intermediate (A), the product is phenol (monohydroxybenzene). The driving force is aromatization of the cyclohexadienone-type or cyclohexadienol intermediate to give the stable benzene ring with one OH = phenol. Step 5 - Why other options fail: (a) 1,2-cyclohexanediol would result from simple hydrolysis of the epoxide alone, without using 2 moles of base or eliminating Br. (c) 2- or 4-methoxyphenol would require incorporation of OCH3 into the ring, which does not occur here since the methoxy group acts only as a base (as NaOCH3) and is not incorporated into the product. (d) Resorcinol (1,3-dihydroxybenzene) would require two OH groups, but only one oxygen source (the epoxide oxygen) is present in the starting material. Therefore, the correct answer is B.