Animated Solution for Chemistry - Organic Chemistry: Which dicarboxylic acid in presence of a dehydrating agent is least reactive to give an anhydride?

When dealing with dicarboxylic acids and dehydrating agents, the outcome heavily depends on the relative positions of the two carboxyl groups. The question asks us to identify which of the given dicarboxylic acids is the least reactive towards forming an anhydride. To answer this, we need to understand the stability of the rings that would be formed.

Blanc's Rule

In organic chemistry, Blanc's rule is a fantastic predictive tool. It states that when dicarboxylic acids are heated (often with a dehydrating agent like acetic anhydride):

- $1,4$ and $1,5$-dicarboxylic acids readily lose a molecule of water (dehydration) to form highly stable $5$- or $6$-membered cyclic anhydrides. - $1,6$ and $1,7$-dicarboxylic acids, on the other hand, prefer to undergo both dehydration and decarboxylation (loss of ${\text{CO}}_2$). Instead of forming unstable $7$- or $8$-membered anhydrides, they form highly stable $5$- or $6$-membered cyclic ketones.

Let's look at the given options:

- Succinic acid is a $1,4$-dicarboxylic acid. It readily forms succinic anhydride, a stable $5$-membered ring. - Phthalic acid has two carboxyl groups on adjacent carbons of a benzene ring. It easily forms phthalic anhydride, another stable $5$-membered ring. - Adipic acid (Hexanedioic acid) is a $1,6$-dicarboxylic acid. Its structure is $\text{HOOC}-({\text{CH}}_2{)}_4-\text{COOH}$.

According to Blanc's rule, when adipic acid is heated with a dehydrating agent, it does not form a $7$-membered anhydride. Instead, it loses both ${\text{H}}_2\text{O}$ and ${\text{CO}}_2$ to form cyclopentanone, a stable $5$-membered cyclic ketone.

Because adipic acid forms a ketone rather than an anhydride, it is the least reactive towards anhydride formation among the given choices. Therefore, adipic acid is the correct answer.