See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Acidic Strength Order Analysis **Identify the conjugate acids:** - **(a)** $-CH_2NH_3^+$ (primary alkylammonium ion) - **(b)** Pyridinium ion ($N^+$ in pyridine ring) - **(c)** Anilinium ion ($NH_3^+$ attached to benzene ring) **Apply conjugate acid-base principles:** Stronger acid = weaker conjugate base. The order depends on how well the positive charge is stabilized. **Analysis:** 1. **Base (c) vs (a):** Aniline's lone pair is delocalized into the benzene ring's π-system, making it a weak base. Therefore, anilinium (c) is a **strong acid**. 2. **Base (a) vs (b):** Pyridine is much weaker than alkylamines because nitrogen in pyridine is $sp^2$-hybridized with its lone pair in a p-orbital already part of aromaticity. Alkylamines are $sp^3$-hybridized with very basic lone pairs. So pyridinium (b) is more acidic than alkylammonium (a). 3. **Base (b) vs (c):** Aniline is moderately basic; its lone pair has some aromatic character but less than pyridine. So anilinium (c) is the strongest acid overall. **Correct acidic strength order:** $$c > b > a$$ **Answer: (b)** ✓ The resonance stabilization of the conjugate base (aromaticity restoration in aniline and pyridine) makes anilinium and pyridinium much more acidic than simple alkylammonium ions.