See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Bronsted base strength of nitrogen-containing compounds depends on the availability of the lone pair on nitrogen for accepting a proton (H+). A more available (less delocalized) lone pair means a stronger base. Step 1: Analyze each compound. - (a) Aniline (PhNH2): The lone pair on nitrogen is delocalized into the aromatic benzene ring via resonance. This significantly reduces its availability for protonation. pKa of conjugate acid ~ 4.6, making it a weak base. - (b) Ammonia (NH3): The lone pair is localized on nitrogen with no resonance delocalization. pKa of conjugate acid ~ 9.25. Moderate base. - (c) Pyrrole: The lone pair on nitrogen is part of the aromatic pi system (contributes 2 electrons to the 6 pi-electron aromatic system of the five-membered ring). This lone pair is fully delocalized into the ring and is NOT available for protonation. Pyrrole is actually a very weak base (pKa of conjugate acid ~ -3.8). - (d) Pyrrolidine: A saturated five-membered ring containing NH. There is NO aromatic ring or pi system to delocalize the lone pair. Additionally, the alkyl groups (the four CH2 groups forming the ring) donate electron density inductively to nitrogen, increasing the availability of the lone pair. pKa of conjugate acid ~ 11.27, making it a strong base among these options. Step 2: Rank the bases. Pyrrolidine > Ammonia > Aniline > Pyrrole Step 3: Why others fail. - Aniline loses basicity due to resonance with the benzene ring. - Pyrrole's lone pair is integral to aromaticity and essentially unavailable. - Ammonia is a decent base but lacks the inductive electron-donating alkyl groups present in pyrrolidine. - Pyrrolidine benefits from both localization of the lone pair (no resonance) and inductive donation from four surrounding CH2 groups, making it the strongest base. Therefore, the correct answer is D.