See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The extraordinarily high basicity of 1,8-bis(dimethylamino)naphthalene (proton sponge) compared to 1-(dimethylamino)naphthalene arises from two cooperative factors that are both rooted in the unique peri geometry of the 1,8 positions on naphthalene. Step 1 - Steric inhibition of resonance in the free base: In 1-(dimethylamino)naphthalene, the lone pair on nitrogen can delocalize into the aromatic ring via resonance, reducing the electron density on nitrogen and therefore reducing basicity. In 1,8-bis(dimethylamino)naphthalene, the two bulky NMe2 groups at the peri positions sterically force each other out of the plane of the naphthalene ring. This twisting prevents effective overlap of the nitrogen lone pairs with the aromatic pi system, i.e., resonance donation into the ring is inhibited. As a result, both lone pairs are localized on the nitrogen atoms and are fully available for protonation, greatly increasing basicity. Step 2 - Relief of steric strain upon protonation: In the free base, the two NMe2 groups crowd each other at the peri positions, creating steric strain. Upon protonation of one nitrogen, an N-H...N hydrogen bond forms between the protonated and unprotonated nitrogen. This intramolecular hydrogen bond stabilizes the conjugate acid (protonated form) and relieves some steric strain, further driving the equilibrium toward proton uptake and thus increasing the apparent basicity enormously. Step 3 - Why other options fail: (a) Resonance would decrease basicity by delocalizing the lone pair away from nitrogen, so resonance cannot explain increased basicity. (c) The ortho effect applies to substituted anilines on a benzene ring and involves inductive/field effects of ortho substituents; it does not account for the 10^10 magnitude increase seen here. (d) Hyperconjugation involves C-H bonds adjacent to a pi system and is not relevant to the nitrogen lone pair availability in this context. Therefore, the correct answer is B.