See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The Williamson Ether Synthesis is the standard method for preparing unsymmetrical ethers. It involves the reaction of an alkoxide ion (nucleophile) with an alkyl halide (electrophile) via an SN2 mechanism. The key requirement is that the alkyl halide must be primary (or methyl) to avoid elimination side reactions, and the alkoxide should be the one derived from the more hindered or less reactive alcohol. Step 1 - Identify the target molecule: Phenyl benzyl ether is C6H5-O-CH2C6H5, which contains a phenoxy group (C6H5O-) connected to a benzyl group (-CH2C6H5). Step 2 - Analyze option (b): C6H5O(-)Na(+) is sodium phenoxide (phenoxide ion as nucleophile) and C6H5CH2Cl is benzyl chloride (primary alkyl halide as electrophile). The phenoxide ion attacks the primary carbon of benzyl chloride in an SN2 reaction: C6H5O(-) + ClCH2C6H5 → C6H5OCH2C6H5 + Cl(-). This is ideal because benzyl chloride is a primary halide, making SN2 favorable, and sodium phenoxide is a good nucleophile. Step 3 - Analyze option (a): C6H5Cl is chlorobenzene (aryl halide) with C6H5CH2O(-)Na(+) (benzyl alkoxide). Aryl halides are very unreactive toward SN2 substitution under normal conditions due to the sp2 carbon and resonance stabilization of the C-Cl bond. This reaction would not proceed well. Step 4 - Analyze option (c): 2C6H5Cl + Na2O would involve two aryl halides, which are resistant to nucleophilic substitution. This is not a viable route. Step 5 - Analyze option (d): 2C6H5MgBr + CH2O would give a Grignard addition to formaldehyde, yielding a primary alcohol (C6H5CH2OH) after workup from one equivalent, not an ether. This does not produce phenyl benzyl ether. Conclusion: Option (b) uses sodium phenoxide (a good nucleophile) reacting with benzyl chloride (a reactive primary alkyl halide) via SN2, which is the correct and efficient Williamson synthesis of phenyl benzyl ether. Therefore, the correct answer is B.