See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: When a proton (H+) attacks a molecule, it attacks the site with the highest electron density (most basic/nucleophilic site). For N-methyl-4-pyridinone, we must consider the resonance structures and the distribution of electron density. Step 1: Identify the molecule. The compound is 1-methyl-4(1H)-pyridinone (N-methyl-4-pyridinone). It has a pyridine ring with a ketone-like carbonyl at C-4 and a methyl group on nitrogen. Step 2: Consider the resonance structures of 4-pyridinone. The molecule can be represented as a hybrid between the carbonyl form and a zwitterionic aromatic form where the oxygen carries negative charge and nitrogen carries positive charge. In the aromatic resonance structure, oxygen bears a full negative charge (O-), making it highly electron-rich. Step 3: Compare electron density at each labeled site: - Site a (oxygen): The oxygen has lone pairs and participates in resonance; it bears significant negative charge through the zwitterionic resonance structure (O- form). This makes it the most electron-dense site. - Site b (C-3): A carbon on the ring, less electron-dense than oxygen. - Site c (C-2): A carbon alpha to nitrogen, somewhat electron-dense but less so than oxygen. - Site d (nitrogen): The nitrogen is already carrying a methyl group and in the zwitterionic resonance structure bears a positive charge, making it less favorable for protonation. Step 4: The 4-pyridinone system is known to protonate preferentially on oxygen (site a) because: (i) Oxygen is more electronegative and holds lone pairs readily available for proton acceptance. (ii) O-protonation gives a fully aromatic, stabilized cation (the protonated form becomes analogous to a pyridinium-like aromatic system), providing extra resonance stabilization. (iii) N-protonation (site d) would disrupt the aromaticity and is disfavored because N already bears a methyl group and the positive charge on N in the resonance form makes it less basic. Step 5: Why other options fail: - b and c are carbon sites; carbon is not a typical site for protonation in aromatic/conjugated systems compared to heteroatoms. - d (N-CH3): Nitrogen is already substituted (tertiary-like), and in the dominant resonance form of 4-pyridinone the nitrogen bears positive character, making it a poor site for H+ attack. Conclusion: H+ attack is most favorable at site a (the oxygen), because O-protonation leads to an aromatic, stabilized oxypyridinium cation, and oxygen has the highest electron density in this molecule. Therefore, the correct answer is A.