See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The preferred site of protonation is the site that gives the most stable conjugate acid (most stabilized cation after protonation). Compound (i) - N-methylacetamide: The two basic sites are: - Position 1 (carbonyl oxygen, O) - Position 2 (amide nitrogen, N) In an amide, the nitrogen lone pair is delocalized into the carbonyl group via resonance (N lone pair -> C=O), making nitrogen a poor base. Protonation at oxygen (position 1) gives a resonance-stabilized oxocarbenium/iminium cation: CH3-C(OH)=N(CH3) <-> CH3-C(+)(OH)-N(CH3), where positive charge is delocalized. Protonation at nitrogen (position 2) gives a simple ammonium-type species with no special stabilization and disrupts the amide resonance. Therefore, oxygen (position 1) is the preferred site of protonation for amides. Compound (ii) - 2-aminopyridine: The two basic sites are: - Position 3 (ring nitrogen, pyridine-type N) - Position 4 (exocyclic amino group, NH2) Protonation at the ring nitrogen (position 3) gives a conjugate acid where the positive charge on N is stabilized by resonance donation from the exocyclic NH2 group at position 2 (para-like relationship), forming an aminopyridinium ion that is strongly stabilized by the push of the NH2 lone pair into the ring toward the protonated N. Protonation at the exocyclic NH2 (position 4) would give an ammonium-type species but would simultaneously remove the NH2 lone pair from resonance with the ring, and the resulting species is less stable because there is no comparable stabilization. The ring nitrogen of 2-aminopyridine is significantly more basic (pKa ~6.7) than a simple aniline nitrogen, precisely because the ring nitrogen protonation is stabilized by the amino group resonance. Therefore, position 3 (ring N) is the preferred site of protonation. Combined answer: Position 1 (oxygen of amide) and Position 3 (ring nitrogen of 2-aminopyridine) → answer (a) 1 and 3. Why other options fail: - (b) 2 and 4: Both wrong - amide N is not preferred over O, and exocyclic NH2 is not preferred over ring N. - (c) 1 and 4: Position 1 correct but position 4 wrong. - (d) 2 and 3: Position 3 correct but position 2 wrong. Therefore, the correct answer is A.