See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The first ionization constant (Ka1) of a dicarboxylic acid depends on the distance between the two carboxylic acid groups. When the two -COOH groups are closer together, the inductive/field effect of one group on the other is stronger, making the first proton easier to remove (higher Ka1, lower pKa1). As the chain length increases and the two -COOH groups move farther apart, the stabilizing inductive effect decreases, Ka1 decreases, and pKa1 increases. Reasoning: - Oxalic acid (ethanedioic acid, 2 carbons): The two -COOH groups are directly bonded to each other. The strong electron-withdrawing inductive effect of one -COOH on the other makes the first ionization very easy. Ka1 is very large, so pK1 is the smallest. - Malonic acid (propanedioic acid, 3 carbons): One -CH2- separates the two -COOH groups. The inductive effect is weaker than in oxalic acid. Ka1 is intermediate, so pK2 is intermediate. - Heptanedioic acid (pimelic acid, 7 carbons): Five -CH2- groups separate the two -COOH groups. The two carboxyl groups are far apart; the inductive effect of one on the other is negligible. Ka1 is smallest (most similar to a monocarboxylic acid), so pK3 is the largest. Thus: pK1 < pK2 < pK3, which is option (b). Why other options fail: - (a) pK1 > pK2 > pK3: This is the reverse of the correct trend. - (c) pK3 > pK2 = pK1: Malonic and oxalic acids do not have equal pKa1 values; oxalic acid is significantly stronger. - (d) pK3 > pK1 > pK2: This incorrectly places malonic acid as having the smallest pKa1, which is wrong. Therefore, the correct answer is B.