See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: HBr reacts with alcohols to convert -OH groups to -Br groups. Phenols (ArOH) are generally resistant to substitution by HBr under mild/moderate conditions because the C-O bond is strengthened by resonance with the aromatic ring. However, benzylic alcohols (ArCH2OH) readily react with HBr because the benzylic C-O bond is easily cleaved to form a stabilized benzylic carbocation, giving ArCH2Br. Step 2 - Reaction A: The starting material is 3-(hydroxymethyl)phenol, i.e., a benzene ring bearing -OH (phenolic) at C1 and -CH2OH (benzylic alcohol) at C3. When treated with HBr/heat, the benzylic alcohol (-CH2OH) is selectively converted to -CH2Br, while the phenolic -OH remains intact (phenols do not react with HBr under these conditions to give ArBr). Product A = benzene ring with -OH at C1 and -CH2Br at C3. Step 3 - Reaction B: The starting material is 3-methoxyphenol, i.e., a benzene ring bearing -OH (phenolic) at C1 and -OCH3 (aryl methyl ether) at C3. When treated with HBr/heat, aryl methyl ethers (ArOCH3) are cleaved by HBr to give ArOH and CH3Br. The phenolic -OH remains intact. Product B = benzene ring with -OH at C1 and -OH at C3, i.e., resorcinol (benzene-1,3-diol). Step 4 - Matching to options: Product A = 3-(bromomethyl)phenol [ring with OH and CH2Br]; Product B = benzene-1,3-diol [ring with two OH groups]. This matches option (b). Step 5 - Why other options fail: - Option (a): Product B shown as 3-bromophenol, but HBr cleaves the methyl ether to give -OH, not direct ring bromination. - Option (c): Product B shown with Br on ring and OH, implying direct electrophilic substitution replacing OCH3 with Br, which is not how HBr/Delta works. - Option (d): Product A retains CH2OH and gains ring Br, and Product B gains ring Br — neither is correct under these conditions. Therefore, the correct answer is B.