See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Basic strength of amines depends on availability of the lone pair on nitrogen. Greater electron density on N → stronger base. Aromatic amines are weaker than aliphatic amines due to resonance delocalization. Among cyclic nitrogen compounds: pyrrole N lone pair is part of aromatic pi system (very weak base, pKa ~-3.8), pyrrolidine is a saturated amine (strong base, pKa ~11.3), and imidazole has two nitrogens – one pyrrole-type (lone pair in pi system, not basic) and one pyridine-type (lone pair available for protonation, pKa ~7). Analysis of option (c): The order given is pyrrole > pyrrolidine > imidazole. - Pyrrole: pKa(conjugate acid) ≈ -3.8 (extremely weak base; lone pair delocalized into aromatic ring) - Pyrrolidine: pKa(conjugate acid) ≈ 11.3 (strong aliphatic amine) - Imidazole: pKa(conjugate acid) ≈ 7.0 (moderately basic; pyridine-type N available) The correct order of basic strength should be: pyrrolidine > imidazole > pyrrole. The given order in (c) states pyrrole > pyrrolidine > imidazole, which is completely wrong. Pyrrole is the weakest base (lone pair in aromatic system), pyrrolidine is the strongest (sp3 N, no resonance withdrawal), and imidazole is intermediate. Verification of other options: (a) Diphenylamine < aniline < cyclohexylamine: Correct. More phenyl groups = more delocalization = weaker base. Cyclohexylamine is aliphatic = strongest. (b) Pyrrole < pyridine < piperidine: Correct. Pyrrole lone pair is in pi system (weakest), pyridine lone pair is in sp2 orbital perpendicular to pi system (intermediate), piperidine is saturated (strongest). (d) 4-nitroaniline < aniline < 4-methylaniline: Correct. NO2 is electron-withdrawing (decreases basicity), CH3 is electron-donating (increases basicity). Therefore, the correct answer is C.