See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Solution **Understanding protonation states at different pH values:** At **pH < 1.82** (very acidic): - The $pK_a$ of the carboxyl group ($-COOH$) is ~1.82 - Since pH < $pK_a$, the carboxylic acid remains protonated: $-COOH$ (neutral) - The amino group ($-NH_2$) is protonated: $-NH_3^+$ (positively charged) - The imidazole ring nitrogen is also protonated: $N-H^+$ (positively charged) This matches the **given structure** with all three ionizable groups protonated. **At pH > 1.82** (less acidic): - pH > $pK_a$ of carboxylic acid means the carboxyl group **loses its proton** - The carboxyl group deprotonates: $-COOH \rightarrow -COO^-$ (negatively charged) - The amino group remains protonated: $-NH_3^+$ (still positively charged, since $pK_a \approx 9$) - The imidazole ring remains protonated: $N-H^+$ (still positively charged, since $pK_a \approx 6$) **Why option B is correct:** Option B shows $-COO^-$ (deprotonated carboxyl), $-NH_3^+$ (protonated amino), and $-N-H^+$ (protonated imidazole), which is the correct ionization state when pH > 1.82.