See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify aspirin's structure: Aspirin (acetylsalicylic acid) is 2-(acetyloxy)benzoic acid. It contains two functional groups: an ester group (acetate, -OC(=O)CH3) and a carboxylic acid group (-CO2H) on an ortho-substituted benzene ring. This corresponds to structure (a). Step 2 - Determine behavior in a basic environment (intestine): In a basic (alkaline) environment, acidic protons are removed by base. The relevant acidic functional groups in aspirin are the carboxylic acid (-CO2H, pKa ~ 3.5) and the ester oxygen (not acidic under physiological conditions). The phenol ester (acetate) is more resistant to simple deprotonation under mild basic conditions compared to the carboxylic acid. Step 3 - Which group gets deprotonated first? The carboxylic acid (-CO2H) has a pKa of approximately 3.5, making it the most acidic group. In the basic environment of the intestine (pH ~ 6-8), the carboxylic acid will be deprotonated to form the carboxylate anion (-CO2^-). The acetate ester group (-OC(=O)CH3) is not a simple acid and would not be deprotonated under these conditions; it remains intact. Step 4 - Evaluate each option: (a) Shows -OC(=O)CH3 and -CO2H: This is the neutral (protonated) form, present in acidic stomach, not the intestine. (b) Shows -OH and -CO2CH3: This represents a different compound (methyl salicylate derivative), not aspirin's intestinal form. (c) Shows -OC(=O)CH3 and -CO2^-: The acetate ester is retained and the carboxylic acid is deprotonated to carboxylate. This is correct for the basic intestinal environment. (d) Shows -O^- and -C(=O)OH: This would require deprotonation of the ester oxygen, which is not chemically reasonable under intestinal pH conditions. Step 5 - Conclusion: In the basic intestinal environment, the carboxylic acid of aspirin loses its proton to become a carboxylate anion (-CO2^-), while the acetate ester group remains unchanged, giving structure (c). Therefore, the correct answer is C.