See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Fischer esterification under acidic conditions (dry HCl gas) with ethanol. The starting material is RO-CH2-CO2H, which is an alpha-alkoxy acetic acid containing an ether linkage (R-O-) and a carboxylic acid group (-CO2H). Step 1: Identify the reactive functional group. The carboxylic acid (-CO2H) is the group that undergoes esterification with ethanol under acid catalysis (dry HCl). The ether linkage (R-O-CH2-) is generally not reactive under simple Fischer esterification conditions. Step 2: Reaction with EtOH/dry HCl. The carboxylic acid reacts with ethanol via Fischer esterification: -CO2H + EtOH -> -CO2Et + H2O. The R-O- ether group on the other side of the molecule remains intact because simple ethers are not cleaved under these mild acid conditions (dry HCl gas in the presence of EtOH is not strongly nucleophilic enough to cleave a simple alkyl ether under these conditions; ether cleavage typically requires concentrated HI or HBr at high temperatures). Step 3: Evaluate the product. The ester formed is RO-CH2-CO2Et, where the ether oxygen still bears the R group and the carboxylate is now an ethyl ester. This corresponds to option (b). Step 4: Why other options fail: - Option (a) RO-CH2-CO2R: This would require the carboxylic acid to esterify with R-OH (not ethanol), which is not present. - Option (c) EtO-CH2-CO2Et: This would require both the ether R-O- to be replaced by EtO- AND esterification. Ether cleavage and re-etherification does not occur under simple dry HCl/EtOH conditions. - Option (d) R-O-C(=O)-O-Et: This is a carbonate ester, which is not the product of simple Fischer esterification of an alpha-alkoxy acid. Therefore, the correct answer is B.