Benzyne mechanism for nucleophilic substitution in o-bromochlorobenzene

Steps:

• Compound J (C₆H₄BrCl) is o-bromochlorobenzene, reactive under strong basic conditions.
• Hot concentrated NaOH in ethanol acts as a base, abstracting ortho-H to Br and eliminating Br⁻ to form 3-chlorobenzyne.
• OH⁻ adds to the benzyne triple bond at either carbon, followed by H⁺ protonation.
• Products are 2-chlorophenol and 3-chlorophenol; X is one such hydroxyhalobenzene.

Why B is correct:

• B depicts the skeletal formula of 3-chlorophenol, formed by OH⁻ addition to the benzyne carbon distant from Cl (standard addition in unsymmetrical benzyne).

Why the others are wrong:

• A shows 4-chlorophenol, impossible from o-isomer benzyne.
• C shows 1,2-dihydroxybenzene, requiring double substitution not favored here.
• D shows bromo-substituted phenol, incorrect as Br is eliminated preferentially.

Final answer: B