See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

To match each acid with its pKa, we analyze the electronic effects that influence acidity (lower pKa = stronger acid = more stable conjugate base). Concept: The pKa of a carboxylic acid is influenced by inductive/field effects of nearby electron-withdrawing groups (EWG). EWGs stabilize the conjugate base (carboxylate) by dispersing negative charge, increasing acidity (lowering pKa). The closer and stronger the EWG, the greater the acidifying effect. (a) CH3-CO2H = Acetic acid. A simple carboxylic acid with a weakly electron-donating methyl group. Its known pKa is 4.76, but in this set the closest value is 4.80 (s). The methyl group slightly destabilizes the carboxylate relative to formic acid, giving a pKa around 4.76-4.80. (b) (CH3)3N(+)-CH2-CO2H = Betaine / trimethylammonioacetic acid. The positively charged quaternary ammonium group is directly on the alpha carbon (one carbon away from COOH). This strong EWG exerts a very large inductive effect, making this a very strong carboxylic acid. The pKa is drastically lowered. The closest value is 1.83 (r). This matches because the +N is directly adjacent (alpha position), giving maximum inductive withdrawal. (c) (CH3)3N(+)-(CH2)4-CO2H = 5-(trimethylammonio)pentanoic acid. The positively charged nitrogen is now four carbons away from the COOH group. The inductive effect diminishes with distance, so the acid is stronger than acetic acid but weaker than betaine. The pKa is 4.27 (q). The +N is far enough that its effect is attenuated but still measurable. (d) -O2C-CH2-CO2H = Hydrogen malonate / malonic acid monoanion (second dissociation of malonic acid, or the acid form of the monoanion). The -O2C- group is a carboxylate anion (negatively charged) on the other end. Despite being negatively charged (which would normally destabilize removal of another proton), the carbonyl carbon still exerts an inductive EW effect, but the negative charge also has a repulsive destabilizing effect on ionization. However, in the context of this problem, -O2C-CH2-CO2H represents the second pKa of malonic acid. The first pKa of malonic acid is ~2.83 and the second is ~5.69. So this represents the second ionization of malonic acid where one carboxyl is already deprotonated, giving pKa = 5.69 (p). The already-present negative charge makes it harder to remove another proton, raising the pKa. Why other options fail: - Betaine (b) cannot be 4.27 or 4.80 because direct alpha-positioned +N drastically lowers pKa to ~1.83. - The malonate monoanion (d) cannot be 1.83 because the negative charge on the adjacent carboxylate raises pKa rather than lowering it. - Acetic acid (a) cannot be 1.83 or 4.27 because it has no strong EWG. Summary of matches: (a) CH3CO2H → pKa 4.80 → (s) (b) (CH3)3N(+)CH2CO2H → pKa 1.83 → (r) (c) (CH3)3N(+)(CH2)4CO2H → pKa 4.27 → (q) (d) -O2C-CH2-CO2H → pKa 5.69 → (p) Therefore, the correct answer is {"a": ["s"], "b": ["r"], "c": ["q"], "d": ["p"]}.