See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Analysis of Nitrogen Basicity in LSD **Identify the three nitrogens:** - **N(1)**: Secondary amine in indole ring (aromatic $\pi$ system) - **N(2)**: Tertiary amine with $N$-methyl group (saturated) - **N(3)**: Tertiary amide nitrogen (bonded to carbonyl) **Basicity ranking:** 1. **N(3) is NOT basic** — Amide nitrogen has its lone pair delocalized into the $C=O$ carbonyl through resonance ($C=N^+$ ↔ $C-N^-$). This stabilizes the lone pair and makes it unavailable for protonation. **Amides are extremely weak bases.** 2. **N(1) is weakly basic** — Indole $N-H$ is part of an aromatic system where the lone pair contributes to $\pi$ aromaticity. Protonation would disrupt this aromatic stabilization, making it less basic than saturated amines. 3. **N(2) is most basic** — Tertiary aliphatic amine with no resonance stabilization or aromaticity constraints. Its lone pair is freely available for protonation with no competing electronic effects. **Answer: (B) 2** The secondary/tertiary saturated amine N(2) is most basic because its lone pair is not delocalized (unlike N(1) in aromatic indole or N(3) in amide), making it most available for accepting a proton.